From mboxrd@z Thu Jan 1 00:00:00 1970 X-Msuck: nntp://news.gmane.io/gmane.comp.tex.context/52804 Path: news.gmane.org!not-for-mail From: Xan Newsgroups: gmane.comp.tex.context Subject: Urgent: Strange index problems Date: Tue, 08 Sep 2009 19:58:20 +0200 Message-ID: <4AA69B3C.9040609@telefonica.net> Reply-To: mailing list for ConTeXt users NNTP-Posting-Host: lo.gmane.org Mime-Version: 1.0 Content-Type: multipart/mixed; boundary="------------030701000305020104030602" X-Trace: ger.gmane.org 1252480501 32513 80.91.229.12 (9 Sep 2009 07:15:01 GMT) X-Complaints-To: usenet@ger.gmane.org NNTP-Posting-Date: Wed, 9 Sep 2009 07:15:01 +0000 (UTC) To: mailing list for ConTeXt users Original-X-From: ntg-context-bounces@ntg.nl Wed Sep 09 09:14:52 2009 Return-path: Envelope-to: gctc-ntg-context-518@m.gmane.org Original-Received: from balder.ntg.nl ([195.12.62.10]) by lo.gmane.org with esmtp (Exim 4.50) id 1MlHOF-0001BG-Vl for gctc-ntg-context-518@m.gmane.org; Wed, 09 Sep 2009 09:14:52 +0200 Original-Received: from localhost (localhost [127.0.0.1]) by balder.ntg.nl (Postfix) with ESMTP id 0FBD9C9ADD; Wed, 9 Sep 2009 09:14:49 +0200 (CEST) X-Virus-Scanned: Debian amavisd-new at balder.ntg.nl Original-Received: from balder.ntg.nl ([127.0.0.1]) by localhost (balder.ntg.nl [127.0.0.1]) (amavisd-new, port 10024) with LMTP id i2t4xM1u9soj; Wed, 9 Sep 2009 09:14:31 +0200 (CEST) Original-Received: from balder.ntg.nl (localhost [127.0.0.1]) by balder.ntg.nl (Postfix) with ESMTP id C44FBC9A8A; Wed, 9 Sep 2009 09:14:30 +0200 (CEST) Original-Received: from localhost (localhost [127.0.0.1]) by balder.ntg.nl (Postfix) with ESMTP id E5A35C9A63 for ; Tue, 8 Sep 2009 19:58:33 +0200 (CEST) X-Virus-Scanned: Debian amavisd-new at balder.ntg.nl Original-Received: from balder.ntg.nl ([127.0.0.1]) by localhost (balder.ntg.nl [127.0.0.1]) (amavisd-new, port 10024) with LMTP id PMT32+P+Z3Cz for ; Tue, 8 Sep 2009 19:58:25 +0200 (CEST) Original-Received: from ctsmtpout2.frontal.correo (outmailhost.telefonica.net [213.4.149.242]) by balder.ntg.nl (Postfix) with ESMTP id 054E9C9A47 for ; Tue, 8 Sep 2009 19:58:24 +0200 (CEST) Original-Received: from [172.26.0.7] (83.55.61.50) by ctsmtpout2.frontal.correo (7.2.056.6) (authenticated as dxpublica) id 4A1E4E5F0173ED3B for ntg-context@ntg.nl; Tue, 8 Sep 2009 19:58:24 +0200 User-Agent: Thunderbird 2.0.0.23 (X11/20090817) X-Mailman-Approved-At: Wed, 09 Sep 2009 09:14:25 +0200 X-BeenThere: ntg-context@ntg.nl X-Mailman-Version: 2.1.12 Precedence: list List-Id: mailing list for ConTeXt users List-Unsubscribe: , List-Archive: List-Post: List-Help: List-Subscribe: , Original-Sender: ntg-context-bounces@ntg.nl Errors-To: ntg-context-bounces@ntg.nl Xref: news.gmane.org gmane.comp.tex.context:52804 Archived-At: This is a multi-part message in MIME format. --------------030701000305020104030602 Content-Type: text/plain; charset=ISO-8859-15; format=flowed Content-Transfer-Encoding: quoted-printable Hi, I have a document and I get this error: publications : warning: cite argument riley-tesi unknown on 139 references : unknown reference [][0] section : 1.4 M=E9s d'una secci=F3 section : 1.5 $p_w$ section : 1.6 altres {vertical mode: \tracingstats} {\tracingpages} {\tracingoutput} {\tracinglostchars} {\tracingmacros} {\tracingparagraphs} {\tracingrestores} {\showboxbreadth} {\showboxdepth} {\tracinggroups} {\tracingifs} {\tracingscantokens} {\tracingnesting} {\tracingassigns} {into \tracingassigns=3D2} {\errorstopmode} {\tracingonline} {changing \tracingonline=3D1} Completed box being shipped out [33.33] Memory usage before: 5614&778894; after: 919&777521; still untouched: 930= 938 title : - Refer\dochar {232}ncies [34.34] title : - =CDndex alfab=E8tic (./memoria.tuo ! Extra }, or forgotten $. \doregistertexthowto ...?id #1\c!textcommand }{#2} \dostopattributes \dohandleregisterentry ...exthowto {\v!index }{#1} \fi \egroup \!!doneafal= se \c!entryb ...ry {\v!index }{mitjan respecte de $k} \global=20 \firstregisterentr... \c!entrya \c!entryb \c!entryc \c!entryd \secondoftwoarguments #1#2->#2 =20 \dodosetpageregisterpageA ...\c!entryc \c!entryd } \global \let \c!entrya=20 \re... ... l.167 ...e{index}{,}{24}{2--0-1-3-0-0-0-0--32}{32} % ? x {/usr/share/texmf/fonts/enc/dvips/lm/lm-ec.enc}{/usr/share/texmf/fonts/en= c/dvip s/lm/lm-mathsy.enc}{/usr/share/texmf/fonts/enc/dvips/lm/lm-mathit.enc}{/u= sr/sha re/texmf/fonts/enc/dvips/lm/lm-rm.enc}{/usr/share/texmf/fonts/enc/dvips/l= m/lm-m athex.enc}< /usr/share/texmf/fonts/type1/public/lm/lmsy10.pfb> Output written on memoria.pdf (34 pages, 424021 bytes). Transcript written on memoria.log. TeXExec | runtime: 18.095553 xan@rulot:~/Desktop/TFM/TFM/nou$ The problem appears after section \section{$\varphi_k$}, because if I=20 put \stoptext after, all is ok. Anyone could see the problems? I attach=20 the file. Please, help me. Xan. --------------030701000305020104030602 Content-Type: text/plain; name="memoria.context" Content-Transfer-Encoding: quoted-printable Content-Disposition: inline; filename="memoria.context" % interface=3Den output=3Dpdftex %\environment capcalera.context % Cap=C3=A7alera % Regime \enableregime[utf] % Choose a font \setupbodyfont [cmr,11pt] % cmr, 11pt % Be tolerant with paragraph building \setuptolerance [horizontal,verytolerant,stretch] % Choose a language, and associated hyphenation rules. %\language [ca] \mainlanguage[ca] % Page number \setuppagenumbering [location=3D{footer}] % White space between paragraphs %\setupwhitespace [big] % Paper size \setuppapersize [a4] % Margins %\setuplayout [grid=3Dyes, footer=3D0.5\footerheight, header=3D0.5\heade= rheight] %\setuplayout[footer=3D2cm, header=3D2cm] %\showlayout %\showframe %\showsetups % Format de marges %\setuplayout[topspace=3D1.5cm, % marge d'adalt %margin=3D1.5cm, %marges dels costats %header=3D1.0cm,%separaci=C3=B3 entre adalt i primera l=C3=ADnia %footer=3D1.0cm,%separaci=C3=B3 entre abaix i darrera l=C3=ADnia %width=3Dfit,height=3Dfit,backspace=3D2cm] % Enable colors and activate hyperlinks \setupcolors [state=3Dstart] \definecolor[lightblue][r=3D0.5, g=3D0.5, b=3D1.0] %\setupinteraction [state=3Dstart, color=3DlightBlue] %\setupurl[style=3Dsmall, space=3Dyes] \setupurl[space=3Dyes] % Enumerate the URLs \useURL[bib:crm-link][http://www.crm.cat/Conferences/0405/WordProblem/pub= lications.htm][][http://www.crm.cat/Conferences/0405/WordProblem/publicat= ions.htm] \useURL[bib:bridson-tutorial-link][http://people.maths.ox.ac.uk/\~{}brids= on/papers/bfs/][][http://people.maths.ox.ac.uk/\~{}bridson/papers/bfs/] \useURL[bib:bernasconi-tesi-link][http://www.math.utah.edu/\~{}sg/Papers/= bernasconi-thesis.pdf][][http://www.math.utah.edu/\~{}sg/Papers/bernascon= i-thesis.pdf] \useURL[bib:quasi-isometries-link][www.math.utah.edu/\~{}malone/QI/notes.= pdf][][www.math.utah.edu/\~{} malone/QI/notes.pdf] \useURL[bib:rees-link][http://msp.warwick.ac.uk/gtm/][][http://msp.warwic= k.ac.uk/gtm/] %\useURL[bib:open-problems-link][http://www.math.mcgill.ca/\~{}alexeim/Pu= blications/All_files_new/Openproblem_final_40.pdf][][http://www.math.mcgi= ll.ca/\~{}alexeim/Publications/All_files_new/Openproblem_final_40.pdf] % Fonts %% Chapters... \setupheads[3=3Dflushleft] \setuphead[chapter][style=3D{\bfd}] \setuphead[section][style=3D{\bfc}, header=3Dnomarking] \setuphead[subsection][style=3D{\bfb}] \setuphead[subsubsection][style=3D{\bfa}] %\setuphead[section][textstyle=3Dbold] %% Italic in emph (by default is slanted) \setupbodyfontenvironment[default][em=3Ditalic] % Bibliography options % BIBTEX \usemodule[bib] \setupbibtex[database=3Dmemoria,sort=3Dauthor] \setuppublications [alternative=3Dams,sorttype=3Dbbl, criterium=3Dcite]% \setupheadtext[ca][pubs=3DRefer=C3=A8ncies] \setuppublicationlist[authoretallimit=3D3] \setuppublicationlist[authoretaltext=3D{\it\ et al.}] \setuppublicationlist[authoretaldisplay=3D1] %Indentation \setupheads[indentnext=3Dyes]=20 \setupindenting[yes,small,first] %\setupformulae[indentnext=3Dyes] % Vertical spaces between paragraphs \setupwhitespace[small] %Itemize \setupitemize[each][indentnext=3Dno,margin=3D2em] % [identnext=3Dyes,marg= in=3D2em] \setupitemize[each][headstyle=3Dbold] %\setupitemize[a][left=3D(,right=3D),stopper=3D] % Mathematical packets %\usemodule[newmat] %\usemodule[math-ams] % Heads and footers %\setupfootertexts[][{\tfxx \currentdate}] %\setupfootertexts[\pagenumber/\lastpage] %\setupfooter[text][before=3D\hrule] %\setupheader[text][after=3D\hrule] %\setupheadertexts[{\tfx M=C3=A0ster de Matem=C3=A0tiques}][{\tfx \jobnam= e.\ConTeXt{}.\currentdate}] %\setupheadertexts[][{\tfx \currentdate}] % hyphenating \hyphenation{do-cu-ment} \hyphenation{pro-ble-ma} \hyphenation{es-crip-tu-ra} \hyphenation{ge-ne-ra-lit-za-ci=C3=B3} \hyphenation{cor-res-po-nents} \hyphenation{cor-res-po-nent} \hyphenation{pa-rells} \hyphenation{ge-ne-rat} \hyphenation{re-so-lu-ble} \hyphenation{ge-ne-ra-dors} \hyphenation{re-pre-sen-ta-rem} \hyphenation{cons-ta} \hyphenation{e-xis-tei-xen} \hyphenation{e-qui-va-lent} \hyphenation{res-pec-ti-va-ment} \hyphenation{res-pec-te} \hyphenation{a-sin-cr=C3=B2-ni-ca} \hyphenation{sin-cr=C3=B2-ni-ca} \hyphenation{par-ti-cu-lar} \hyphenation{ge-ne-ra-lit-z=C3=A0} \hyphenation{co-mo-di-tat} \hyphenation{lo-ga-rit-me} \hyphenation{lo-ga-rit-mes} \hyphenation{am-pla-da} \hyphenation{su-fi-cient-ment} % Modules \usemodule[tikz] \usemodule[pgfmath] \usetikzlibrary[arrows,calc,decorations.pathmorphing,decorations.markings= ] %\usetikzlibrary[trees] %\usetikzlibrary[mindmap] % AMSTHM equivalent %% Exercici \defineenumeration [exercici] [text=3D{Problema},headstyle=3Dbold,between=3D\blank,titledistance=3D0e= m,textdistance=3D1em, stopper=3D{.\space},location=3Dserried,left=3D{\bgroup\bf},right=3D{\egro= up},width=3Dfit,before=3D{\bgroup\startframedtext[background=3Dscreen,fra= me=3Doff,width=3Dbroad]},after=3D{\stopframedtext\egroup}] %% Lema=20 \defineenumeration [mylema] [text=3D{Lema}, % Qu=C3=A8 es mostra before=3D{\blank[big]}, % abans de lema, un bigskip after=3D{\blank[big]}, % despr=C3=A9s de lema, un bigskip headstyle=3Dbold, % Negreta per la cap=C3=A7aleras %between=3D\blank, % Entre Lemmes una l=C3=ADnia en blanc titledistance=3D.5em, % espai entre n=C3=BAmero i par=C3=A8ntesis. textdistance=3D.5em, % espai entre ) i text stopper=3D{.\space}, % Com acaba. Despr=C3=A9s de par=C3=A8ntesis un '= =2E' location=3Dserried, width=3Dfit, % que ocupi tot l'espai style=3Ditalic, % estil del text title=3Dyes, % si puc posar o no arguments opcionals titlestyle=3Dbf, % estil del t=C3=ADtol way=3Dbytext, % enumerar en tot el document conversion=3Dnumbers,indenting=3Dyes] % enumera amb arabic %% Proposici=C3=B3, corol=C2=B7laris, teoremes. %% Comparteix els nombres amb lema %% Si volem que vagin a part, hem de posar 'number=3Dproposition' \defineenumeration [myproposition] [mylema] [text=3D{Proposici=C3=B3}] \defineenumeration [mycorollary] [mylema] [text=3D{Corol=C2=B7lari}] \defineenumeration [mytheorem] [mylema] [text=3D{Teorema}] \defineenumeration [myconjecture] [mylema] [text=3D{Conjectura}] %% Definici=C3=B3 \defineenumeration [mydefinition] [mylema] [text=3D{Definici=C3=B3},style=3Dtf,titlestyle=3Dbf,indenting=3Dyes] \defineenumeration [mynotation] [mydefinition] [text=3D{Notaci=C3=B3},style=3Dtf,titlestyle=3Dbf,indenting=3Dyes] \defineenumeration [mynota] [mydefinition] [text=3D{Nota},style=3Dtf,titlestyle=3Dbf,indenting=3Dyes] %% Demostraci=C3=B3 \defineenumeration[mydemo][text=3D{Demostraci=C3=B3.\space},number=3Dno,l= ocation=3Dserried,width=3Dfit,headstyle=3Ditalic,indentnext=3Dyes,between= =3D\blank,textdistance=3D.5em,closesymbol=3D{\mathematics{\Box}},style=3D= normal,indenting=3Dyes] % Table of contents %% chapter =3D bold. \setuplist[chapter][style=3Dbold,width=3D10mm]=20 \setuplist[chapter][before=3D\blank] %% dots between... and subsubsubsection are not listed \setupcombinedlist[content][level=3D4,alternative=3Dc]=20 %% section =3D bold. % width=3D 10mm --> less space between num-letter %% line break after section. \setuplist[section][width=3D10mm]=20 %\setuplist[section][before=3D\blank] %% margin =3D 10 mm. Put the subsection just bottom section. \setuplist[subsection][margin=3D10mm,width=3D10mm] \setuplist[subsubsection][margin=3D20mm,width=3D13mm] %\setuplist[subsection] %[distance=3D1em] % section =3D bold. %=20 % Aix=C3=B2 ho trec d'un manual: %\setuplist[subsection] % [margin=3D1em, % numbercommand=3D\NumCom] %\def\NumCom#1{\hbox to 2em{\hfill #1}} % Set "=C3=8Dndex" like "=C3=8Dndex de continguts" \setupheadtext [ca] [content=3D=C3=8Dndex] % Definitions/abbreviations \define[1]\dist{d(\sigma_g(#1), \sigma_h(#1))} \define[1]\imp{{\bgroup\startframedtext[background=3Dscreen,frame=3Don,wi= dth=3Dbroad]#1\stopframedtext\egroup}} %\define[1]\imp{{\bgroup\startframedtext[background=3Dcolor,backgroundcol= or=3Dlightblue,frame=3Don,width=3Dbroad]#1\stopframedtext\egroup}} % SPLIT \def\startsplit {\startalign} % no number by default \def\stopsplit {&\doalignNR[+][]\crcr % for a number on last line \stopalign} % GROUP FOR FORMULAS WITH CASES. \definemathmatrix[GROUP][left=3D{\left\{\,}, right=3D{\right.},align=3D{l= eft},style=3D\displaystyle,distance=3D0.2em] % GATHER \definemathalignment[gather][n=3D1,align=3Dmiddle] % Other %\setupunderbar[alternative=3Db] % Fix underline style % For putting underline with spaces: \underbar{\dorecurse{40}~} % Define new register for the Index of Symbols \defineregister[mysymbol][mysymbols] % Setup figures: %\setupfloat[figure][spacebefore=3D5*big,spaceafter=3Dbig] \setupheadtext [ca] [figures=3DLlista de figures] % Start the text \starttext \version[concept] %\input memoria-preliminars.context \chapter{La meva contribuci=C3=B3} En aquest cap=C3=ADtol es detallar=C3=A0 la meva contribuci=C3=B3 al camp= de la Teoria Geom=C3=A8trica de Grups i, en concret, al problema de la p= araula. De forma general, aquesta contribuci=C3=B3 consisteix en la millo= ra de l'ordre de la funci=C3=B3 de Dehn per als grups que admetin una sec= ci=C3=B3 geod=C3=A8sica $\sigma$ tal que $\varphi_{\sigma}(n) < n-1$, per= a $n$ suficientment gran, i en diverses generalitzacions de l'amplada d'= una secci=C3=B3. \section{Grups amb seccions geod=C3=A8siques d'am\-plada no molt gran} Si un grup $G$ admet una secci=C3=B3 geod=C3=A8sica $\sigma$ tal que $\Ph= i_{\sigma}(n) < n-1$ per a $n$ suficientment gran, aleshores $G$ t=C3=A9 = el problema de la paraula resoluble i, a m=C3=A9s, la seva funci=C3=B3 de= Dehn $\delta_G$ =C3=A9s tal que $\delta_G (n) \preceq n!$ (Teorema~\in[t= hme:teorema-riley]). En aquesta secci=C3=B3, millorarem aquesta fita supe= rior de la funci=C3=B3 de Dehn per als grups que admetin una secci=C3=B3 = geod=C3=A8sica $\sigma$ tal que $\varphi_{\sigma} (n) < n-1$, per a $n$ s= uficientment gran. En primer lloc, demostrarem una s=C3=A8rie de lemes que ens conduiran a v= eure que, per a qualsevol presentaci=C3=B3 ${\cal P}$, la funci=C3=B3 $\t= ext{area}_{\cal P}$ =C3=A9s subadditiva per a certa operaci=C3=B3 entre p= araules nul-homot=C3=B2piques, el que implicar=C3=A0 una desigualtat de l= a funci=C3=B3 de Dehn (Proposici=C3=B3~\in[thmi:desigualtat-dehn]). \startmylema[thmi:area-concatenacio] Siguin $G$ un grup, ${\cal P} =3D \l= angle X \mid R \rangle$ una presentaci=C3=B3 finita de $G$ i $u, v, w \in= {(X \cup X^{-1})}^*$ paraules nul-homot=C3=B2piques per ${\cal P}$. Si $= w =3D u v$ dins el grup lliure $F(X)$,\footnote{Recordem que dues paraule= s $w_1$, $w_2$ sobre ${(X \cup X^{-1})}^*$ s=C3=B3n iguals dins el grup l= liure si $[w_1]_{\sim} =3D [w_2]_{\sim} \in F(X)$.} aleshores=20 \startformula \text{area}_{\cal P} (w) \leq \text{area}_{\cal P}(u) + \text{area}_{\cal= P}(v). \stopformula \stopmylema \startmydemo Si $\text{area}_{\cal P}(u) =3D N$ i $\text{area}_{\cal P}(v= ) =3D M$, aleshores \startformula \startmathalignment \NC u \NC =3D \prod_{i=3D1}^N x_i^{-1} r_i x_i, \NR \NC v \NC =3D \prod_{j=3D1}^M y_j^{-1} s_j y_j, \NR \stopmathalignment \stopformula% per a alguns $x_i, y_j \in F(X)$, $r_i, s_j \in R_*$, on aquestes igualta= ts s=C3=B3n dins el grup lliure $F(X)$. Com que $w =3D u v$ tamb=C3=A9 di= ns el grup lliure, aleshores \placeformula[-] \startformula \startsplit \NC w =3D u v \NC =3D \bigl( \prod_{i =3D 1}^N x_i^{-1} r_i x_i \bigr) = \cdot \bigl( \prod_{j=3D1}^M y_j^{-1} s_j y_j \bigr) \NR \NC \NC =3D (x_1^{-1} r_1 x_1)\cdots (x_N^{-1} r_N x_N) \cdot (y_1^{-1}= s_1 y_1) \cdots (y_M^{-1} s_M y_M) \NR \NC \NC =3D \prod_{k=3D1}^{M+N} z_k^{-1} t_k z_k \stopsplit \stopformula on \startformula z_k =3D \startcases \NC x_k \MC 1 \leq k \leq N \NR \NC y_{k-N} \MC N+1 \leq k \leq N+M, \NR \stopcases t_k =3D \startcases \NC r_k \MC 1 \leq k \leq N \NR \NC s_{k-N} \MC N+1 \leq k \leq N+M. \NR \stopcases \stopformula Llavors, per definici=C3=B3, $\text{area}_{\cal P}(w) \leq N+M =3D \text{= area}_{\cal P}(u) + \text{area}_{\cal P}(v)$. \stopmydemo \startmylema[thmi:area-conjugats] Siguin $G$ un grup i ${\cal P} =3D \lan= gle X \mid R \rangle$ una presentaci=C3=B3 finita de $G$. Si $w \in {(X \= cup X^{-1})}^*$ =C3=A9s una paraula nul-homot=C3=B2pica per ${\cal P}$ i = $x \in X \cup X^{-1}$, llavors \startformula \text{area}_{\cal P}(x^{-1}wx) \leq \text{area}_{\cal P}(w). \stopformula \stopmylema \startmydemo Suposem que $\text{area}_{\cal P}(w) =3D N$. Aleshores exist= eixen $x_i \in F(X)$ i $r_i \in R_*$, amb $i \in \{1, \ldots, N\}$, tals = que \placeformula[-] \startformula w =3D \prod_{i=3D1}^N x_i^{-1} r_i x_i, \stopformula on aquesta igualtat =C3=A9s dins el grup lliure $F(X)$. Aleshores, dins e= l grup lliure, tenim que \placeformula[-] \startformula \startsplit \NC x^{-1} w x \NC =3D x^{-1} \bigl( \prod_{i =3D 1}^N x_i^{-1} r_i x_i= \bigr) x \NR \NC \NC =3D x^{-1} (x_1^{-1} r_1 x_1)\cdots (x_N^{-1} r_N x_N) x \NR \NC \NC =3D (x^{-1} x_1^{-1} r_1 x_1 x) (x^{-1} x_2^{-1} r_2 x_2 x)\cdo= ts (x^{-1} x_N^{-1} r_N x_N x) \NR \NC \NC =3D \prod_{i=3D1}^N x^{-1} x_i^{-1} r_i x_i x \NR \NC \NC =3D \prod_{i=3D1}^N (x_i x )^{-1} r_i (x_i x), \stopsplit \stopformula per la qual cosa tenim que $\text{area}_{\cal P} (x^{-1}wx) \leq N =3D \t= ext{area}_{\cal P}(w)$, que =C3=A9s el que vol=C3=ADem veure. \stopmydemo \startmylema[thmi:lema-tecnic] Siguin $G$ un grup, $X$ un conjunt finit d= e generadors de $G$ i ${\cal P} =3D \langle X \mid R \rangle$ una present= aci=C3=B3 finita de $G$, $g_i \in G$, amb $i \in \{1, \ldots, 6\}$, i els= camins $\gamma_{1, 2}$, $\gamma_{2, 3}$, $\gamma_{3, 4}$, $\gamma_{4, 5}= $, $\gamma_{5, 6}$, $\gamma_{6, 1}$ i $\gamma_{2, 5}$ els camins dins el = graf de Cayley $\Gamma_{G, X}$ que uneixen, en aquest ordre, els parells = de punts $(g_1, g_2)$, $(g_2, g_3)$, $(g_3, g_4)$, $(g_4, g_5)$, $(g_5, g= _6)$, $(g_6, g_1)$ i $(g_2, g_5)$, respectivament (tal com es representa = a la figura). \placefigure [none,here] [fig:figura-de-6] {Esquema dels 6 punts} {\startcombination[1*1] { \starttikzpicture[scale=3D1.2] % Els punts \filldraw (0,0) circle (2pt); \filldraw (2,0) circle (2pt); \filldraw (4,0) circle (2pt); \filldraw (4,2) circle (2pt); \filldraw (2,2) circle (2pt); \filldraw (0,2) circle (2pt); % Les l=C3=ADnies aleat=C3=B2ries entre punts \draw [decorate,decoration=3D{random steps,segment length=3D2mm, amplitud= e=3D2pt}] (0,0) -- (2,0); \draw [decorate,decoration=3D{random steps,segment length=3D2mm, amplitud= e=3D2pt}] (2,0) -- (4,0); \draw [decorate,decoration=3D{random steps,segment length=3D2mm, amplitud= e=3D2pt}] (4,0) -- (4,2); \draw [decorate,decoration=3D{random steps,segment length=3D2mm, amplitud= e=3D2pt}] (4,2) -- (2,2); \draw [decorate,decoration=3D{random steps,segment length=3D2mm, amplitud= e=3D2pt}] (2,2) -- (0,2); \draw [decorate,decoration=3D{random steps,segment length=3D2mm, amplitud= e=3D2pt}] (0,2) -- (0,0); \draw [decorate,decoration=3D{random steps,segment length=3D2mm, amplitud= e=3D2pt}] (2,0) -- (2,2); % el sentit \draw [decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow= [blue,line width=3D.7mm]{>}}}] (0,0) -- (2,0); \draw [decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow= [blue,line width=3D.7mm]{>}}}] (2,0) -- (4,0); \draw [decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow= [blue,line width=3D.7mm]{>}}}] (4,0) -- (4,2); \draw [decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow= [blue,line width=3D.7mm]{>}}}] (4,2) -- (2,2); \draw [decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow= [blue,line width=3D.7mm]{>}}}] (2,2) -- (0,2); \draw [decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow= [blue,line width=3D.7mm]{>}}}] (0,2) -- (0,0); \draw [decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow= [blue,line width=3D.7mm]{>}}}] (2,0) -- (2,2); % Els noms \draw (0, -0.3) node {$g_1$}; \draw (2, -0.3) node {$g_2$}; \draw (4, -0.3) node {$g_3$}; \draw (4, 2.3) node {$g_4$}; \draw (2, 2.3) node {$g_5$}; \draw (0, 2.3) node {$g_6$}; % Els noms dels camins \draw (1, 0) node[above] {$\gamma_{1,2}$}; \draw (3, 0) node[above] {$\gamma_{2,3}$}; \draw (4, 1) node[left] {$\gamma_{3,4}$}; \draw (3, 2) node[below] {$\gamma_{4,5}$}; \draw (1, 2) node[below] {$\gamma_{5,6}$}; \draw (0, 1) node[right] {$\gamma_{6,1}$}; \draw (2, 1) node[right] {$\gamma_{2,5}$}; % \draw[very thin,color=3Dgray] (-5.1,-5.1) grid [step=3D1] (5.9,5.9);= % \draw[->] (-5.2,0) -- (6.2,0) node[right] {$x$}; % \draw[->] (0,-5.2) -- (0,5.2) node[above] {$y$}; % r =3D \frac{-1}{3} x + 3 %\filldraw (3,2) circle (2pt); %\filldraw (-3,4) circle (2pt); %\draw (-6,5) -- (6,1); %\draw (1, 3.5) node {$r$}; \stoptikzpicture} { } \stopcombination} Per a cadascun d'aquest camins $\gamma_{i, j}$, sigui $w_{i, j}$ la parau= la corresponent ($\gamma(w_{i, j}) =3D \gamma_{i, j}$ per a cada $(i, j) = \in \{(1,2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 1), (2,5)\}$). I siguin = $u, v$ i $w \in {(X \cup X^{-1})}^*$ les paraules definides com: \startformula \startmathalignment \NC u \NC =3D w_{6,1}^{-1}w_{5,6}^{-1}w_{2,5}^{-1}w_{1,2}^{-1}, \NR[+] \NC v \NC =3D w_{4,5}^{-1} w_{3,4}^{-1} w_{2,3}^{-1} w_{2,5},\NR[+] \NC w \NC =3D w_{6,1}^{-1} w_{5,6}^{-1} w_{4,5}^{-1} w_{3,4}^{-1} w_{2,= 3}^{-1} w_{1,2}^{-1}.\NR[+] \stopmathalignment \stopformula Aleshores $u, v, w$ s=C3=B3n nul-homot=C3=B2piques per ${\cal P}$ i a m=C3= =A9s, \startformula \text{area}_{\cal P}(w) \leq \text{area}_{\cal P}(u) + \text{area}_{\cal = P}(v). \stopformula \stopmylema \startmydemo De forma =C3=B2bvia tenim que $u$, $v$ i $w$ s=C3=B3n nul-ho= mot=C3=B2piques per ${\cal P}$, ja el seus camins dins el graf de Cayley = formen cicles (per exemple el cam=C3=AD corresponent a $u$ forma un cicle= amb punt inicial i punt final $g_1$, perqu=C3=A8 =C3=A9s composici=C3=B3= de camins de $g_1$ a $g_6$, de $g_6$ a $g_5$, de $g_5$ a $g_2$ i, finalm= ent, de $g_2$ a $g_1$). D'altra banda, dins el grup lliure $F(X)$ tenim que \startformula w =3D (w_{6,1}^{-1}w_{5,6}^{-1}w_{2,5}^{-1}w_{1,2}^{-1}) w_{1,2} w_{2,5} = (w_{4,5}^{-1}w_{3,4}^{-1}w_{2,3}^{-1}w_{2,5})w_{2,5}^{-1}w_{1,2}^{-1} =3D= u w_{1,2}w_{2,5} v (w_{1,2}w_{2,3})^{-1}. \stopformula Per tant, dins $F(X)$, $w =3D u x^{-1} v x$ amb $x \in F(X)$. Llavors, ap= licant el lemes~\in[thmi:area-concatenacio] i \in[thmi:area-conjugats], t= enim que \startformula \text{area}_{\cal P}(w) =3D \text{area}_{\cal P}(u x^{-1}v x) \leq \text{= area}_{\cal P}(u) + \text{area}_{\cal P}(x^{-1} v x) \leq \text{area}_{\c= al P}(u) + \text{area}_{\cal P}(v). \stopformula \stopmydemo \startmydefinition Siguin $G$ un grup, $X$ un conjunt finit de generadors= de $G$ i ${\cal P} =3D \langle X \mid R\rangle$ una presentaci=C3=B3 fin= ita de $G$. Dues paraules $w_1, w_2 \in {(X \cup X^{-1})}^*$ nul-homot=C3= =B2piques per ${\cal P}$ s=C3=B3n {\em congruents}\index{paraules+nul-hom= ot=C3=B2piques+congruents} si, i nom=C3=A9s si, dins el graf de Cayley $\= Gamma_{G, X}$ existeixen punts $g_1, g_2, g_3, g_4, g_5, g_6 \in G$ i cam= ins $\gamma_{1,2}, \gamma_{2,3}, \gamma_{3,4}, \gamma_{4,5}, \gamma_{5,6}= , \gamma_{6,1}, \gamma_{2,5}$ que uneixen els parells de punts $(g_1, g_2= )$, $(g_2, g_3)$, $(g_3, g_4)$, $(g_4, g_5)$, $(g_5, g_6)$, $(g_6, g_1)$ = i $(g_2, g_5)$, respectivament, tals que les seves paraules corresponents= , que indicarem amb $w_{i, j}$, on $(i, j) \in \{(1,2), (2, 3), (3, 4), (= 4, 5), (5, 6), (6, 1), (2,5)\}$, satisfan \startformula \startmathalignment \NC w_1 \NC =3D w_{6,1}^{-1}w_{5,6}^{-1}w_{2,5}^{-1}w_{1,2}^{-1}, \NR[+]= \NC w_2 \NC =3D w_{4,5}^{-1} w_{3,4}^{-1} w_{2,3}^{-1} w_{2,5}.\NR[+] \stopmathalignment \stopformula En aquest cas, indicarem amb $w_1 \sharp w_2$\mysymbol{$u \sharp v$} a la= paraula definida com \startformula w_1 \sharp w_2 =3D w_{6,1}^{-1} w_{5,6}^{-1} w_{4,5}^{-1} w_{3,4}^{-1} w= _{2,3}^{-1} w_{1,2}^{-1}. \stopformula \stopmydefinition Del lema previ i d'aquesta definici=C3=B3 tenim que si $u$, $v$ s=C3=B3n = paraules congruents, aleshores $u \sharp v$ =C3=A9s nul-homot=C3=B2pica p= er ${\cal P}$ i $\text{area}_{\cal P}(u\sharp v) \leq \text{area}_{\cal P= }(u) + \text{area}_{\cal P}(v)$, o sigui, tenim que la funci=C3=B3 $\text= {area}_{\cal P} \colon \{w \in {(X \cup X^{-1})}^* \mid \text{nul-homot=C3= =B2pica per } {\cal P} \} \to \naturalnumbers$ =C3=A9s subadditiva per a = paraules congruents, o sigui, subadditiva per l'operaci=C3=B3 $\sharp$. \startmylema[thmi:subadditivitat-area] Siguin $G$ un grup, $X$ un conjunt= finit de generadors de $G$, ${\cal P} =3D \langle X \mid R \rangle$ una = presentaci=C3=B3 finita de $G$, $\sigma \colon G \rightarrow {(X \cup X^{= -1})}^*$ una secci=C3=B3 geod=C3=A8sica de $G$ respecte de $X$. Aleshores= , per a tota paraula $w \in {(X \cup X^{-1})}^*$ nul-homot=C3=B2pica per = ${\cal P}$, existeixen $u_k$ paraules nul-homot=C3=B2piques per ${\cal P}= $, amb $k \in \{1, \ldots, {l(w)}^2/2\}$, de longitud $l(u_k) \leq 2\varp= hi_{\sigma}({l(w)}/2)+2$ tals que \startformula \text{area}_{\cal P} (w) \leq \sum_{k=3D1}^{{{l(w)}^2}/2} \text{area}_{\c= al P}(u_k). \stopformula \stopmylema \startmydemo Sigui $w$ una paraula nul-homot=C3=B2pica per ${\cal P}$. Si= $w =3D \varepsilon$, aleshores el resultat =C3=A9s obvi, ja que $\text{a= rea}_{\cal P} (\varepsilon) =3D 0$ i la suma de la dreta =C3=A9s zero (el= conjunt d'=C3=ADndexos =C3=A9s buit). Per tant, podem suposar que $l(w) = \geq 1$. Per tant, $w =3D x_1 \ldots x_r$, amb $r \geq 1$, $x_i \in X \cu= p X^{-1}$ i $i \in \{0, \ldots, r\}$. Per a tot $i \in \{0, \ldots, l(w) -1\}$, considerem la paraula $v_i$ def= inida per la concatenaci=C3=B3 seg=C3=BCent: \startformula v_i =3D \sigma_{\pi(w(i))} \cdot x_{i+1} \cdot \sigma_{\pi(w(i+1))}^{-1}.= \stopformula A la Figura~\in[fig:figura-area-u] es mostra el seu corresponent cam=C3=AD= $\gamma(v_i)$ dins el graf de Cayley $\Gamma_{G, X}$. Per a cada $i \in = \{0, \ldots, l(w) -1\}$, $v_i$ =C3=A9s nul-homot=C3=B2pica per ${\cal P}$= , ja que forma un cicle dins el graf de Cayley ($\gamma(v_i)$ passa per $= 1$, $\pi(w(i))$ i $\pi(w(i+1))$). I, a m=C3=A9s, per construcci=C3=B3 \startformula w =3D v_0 \sharp (v_1 \sharp (\ldots, \sharp(v_{l(w) -1 })\ldots ). \stopformula Per tant, per aplicaci=C3=B3 reiterada del Lema~\in[thmi:lema-tecnic], \placeformula[form:desigualtat-w-vi] \startformula \text{area}_{\cal P} (w) \leq \sum_{i=3D0}^{l(w)-1} \text{area}_{\cal P}= (v_i). \stopformula \placefigure [here] [fig:figura-area-u] {El cam=C3=AD $\gamma(v_i)$, el qual passa per $1$, $\pi(w(i))$ i $\pi(= w(i+1))$.} {\startcombination[1*1] {\starttikzpicture[scale=3D1.1] % Els punts \filldraw[color=3Dblue!50] (0,-4) circle (2pt); \filldraw[color=3Dblue!50] (0.4216,3.9603) circle (2pt); % primer punt: a= valuo ({3*sin(\t r)},{4*cos(\t r)}); a t =3D 0.141 \filldraw[color=3Dblue!50] (-0.4216,3.9603) circle (2pt); % primer punt: = avaluo ({3*sin(\t r)},{4*cos(\t r)}); a t =3D -0.141 % Les l=C3=ADnies entre els punts \draw (-0.4216,3.9603) -- (0.4216,3.9603); \draw plot[domain=3D-3.141:-0.141,smooth,variable=3D\t] ({3*sin(\t r)},{4= *cos(\t r)}); \draw plot[domain=3D0.141:3.141,smooth,variable=3D\t] ({3*sin(\t r)},{4*c= os(\t r)}); \filldraw[color=3Dblue!50] (0,-4) circle (2pt); % perqu=C3=A8 me quedi el= punt damunt. % Els combings % Dibuixo: % amb y la l=C3=ADnia recta que uneix els dos punts, directament % per x faig un funci=C3=B3 del sinus (sin nx + ax =3D k) \draw plot[domain=3D0:0.4216,smooth,variable=3D\t] ({-0.857727*\t -sin (7= =2E31228*\t r) },{18.8812*\t -4 }); \draw plot[domain=3D0:0.4216,smooth,variable=3D\t] ({+0.857727*\t +sin (7= =2E31228*\t r) },{18.8812*\t -4 }); % El sentit del cam=C3=AD entre a i b=20 \draw[decorate,decoration=3D{markings,mark=3Dat position .55 with {\arrow= [green,line width=3D1mm]{>}}}] (0.4216,3.9603) -- (-0.4216,3.9603); % el sentit d'omega \draw[decorate,decoration=3D{markings,mark=3Dat position .9 with {\arrow[= red!50,line width=3D1mm]{<}}}] plot[domain=3D-3.141:3.141,smooth,variable= =3D\t] ({3*sin(\t r)},{4*cos(\t r)}); % Els punts de les cel=C2=B7les % Calcul els combings per a y=3D 0 i y=3D1 %\filldraw (-1.181475, 0) circle (2pt); %\filldraw (1.181475, 0) circle (2pt); % Els noms \draw (0, -4.3) node {$1 \in G$}; \draw (2.7, -3) node {$\gamma(w)$}; \draw (-1.18, 0)[left] node {$\sigma_{\pi(w(i+1))}$}; \draw (1.18, 0) node[right] {$\sigma_{\pi(w(i))}$}; \draw (-1,4.3) node {$\pi(w(i+1))$}; \draw (1,4.3) node {$\pi(w(i))$}; \draw (0, 0) node {$\gamma(v_i)$}; % El sentit dels combings \draw[decorate,decoration=3D{markings,mark=3Dat position .4 with {\arrow[= green,line width=3D1mm]{>}}}] plot[domain=3D0:0.4216,smooth,variable=3D\t= ] ({-0.857727*\t -sin (7.31228*\t r) },{18.8812*\t -4 }); \draw[decorate,decoration=3D{markings,mark=3Dat position .7 with {\arrow[= green,line width=3D1mm]{>}}}] plot[domain=3D0:0.4216,smooth,variable=3D\t= ] ({+0.857727*\t +sin (7.31228*\t r) },{18.8812*\t -4 }); \stoptikzpicture} { } \stopcombination} \indentation Pel Lema~\in[thmi:lema-distancia-menor-w-2], \placeformula[form:distancia-menor-que-lw-2] \startformula d_{G, X} (\pi(w(i)), 1) \leq l(w)/2, \stopformula per a tot $i \in \{0, \ldots, l(w)\}$. Com que $\sigma$ =C3=A9s geod=C3=A8sica, aleshores la longitud de $\sigma= _i$ =C3=A9s menor o igual que $l(w)/2$, per a tot $i \in \{0, \ldots, l(w= )\}$. Indiquem amb $x_i^{j)}$ la paraula (de com a m=C3=A0xim una lletra)= corresponent al cam=C3=AD geod=C3=A8sic que va des de $\sigma_i(j)$ a $\= sigma_i(j+1)$, amb $i \in \{0, \ldots, l(w)\}$ i $0 \leq j \leq l(w)/2-1$= ($x_i^{j)}$ coincideix amb la lletra $(j+1)$-=C3=A8ssima de $\sigma_i$ q= uan $j \leq l(\sigma_i)-1$ i amb $\varepsilon$ altrament). Per a tots $i \in \{0, \ldots, l(w) -1\}$ i $0 \leq j \leq {l(w)}/2$, sig= uin $\overline{v}_{i, j}$ la paraula corresponent a un cam=C3=AD geod=C3=A8= sic que va des de $\sigma_i (j)$ fins a $\sigma_{i+1} (j)$, i $u_{i, j}$ = la paraula nul-homot=C3=B2pica per ${\cal P}$ formada per la concatenaci=C3= =B3 \startformula x_i^{j+1)} \cdot \overline{v}_{i, j+1} \cdot {\left(x_{i+1}^{j+1)}\right)= }^{-1} \cdot \overline{v}_{i, j}^{-1}, \stopformula els camins de les quals es poden observar a la Figura~\in[fig:figura-area= -dos]. Per contrucci=C3=B3, per a tot $i \in \{0, \ldots, l(w)-1\}$, \startformula v_i =3D u_{i,0} \sharp (u_{i,1} \sharp (\ldots, \sharp(u_{i, l(w)/2-1})\l= dots ), \stopformula per la qual cosa, pel Lema~\in[thmi:lema-tecnic], \placeformula[form:desigualtat-vi-uij] \startformula \text{area}_{\cal P} (v_i) \leq \sum_{j=3D0}^{l(w)/2-1} \text{area}_{\ca= l P} (u_{i, j}). \stopformula \noindentation Aleshores, combinant (\in [form:desigualtat-w-vi]) i (\in[= form:desigualtat-vi-uij]), tenim que \startformula \text{area}_{\cal P} (w) \leq \sum_{i =3D 0}^{l(w) -1} \sum_{j =3D 0}^{l= (w)/2 -1} \text{area}_{\cal P} (u_{i,j}).=20 \stopformula \placefigure [here] [fig:figura-area-dos] {El cam=C3=AD $\gamma(u_{i, j})$ corresponent a la paraula $u_{i, j}$.}= {\startcombination[1*1] {\starttikzpicture[scale=3D1.1] % Els punts \filldraw[color=3Dblue!50] (0,-4) circle (2pt); \filldraw[color=3Dblue!50] (0.4216,3.9603) circle (2pt); % primer punt: a= valuo ({3*sin(\t r)},{4*cos(\t r)}); a t =3D 0.141 \filldraw[color=3Dblue!50] (-0.4216,3.9603) circle (2pt); % primer punt: = avaluo ({3*sin(\t r)},{4*cos(\t r)}); a t =3D -0.141 % Les l=C3=ADnies entre els punts \draw (-0.4216,3.9603) -- (0.4216,3.9603); \draw plot[domain=3D-3.141:-0.141,smooth,variable=3D\t] ({3*sin(\t r)},{4= *cos(\t r)}); \draw plot[domain=3D0.141:3.141,smooth,variable=3D\t] ({3*sin(\t r)},{4*c= os(\t r)}); \filldraw[color=3Dblue!50] (0,-4) circle (2pt); % perqu=C3=A8 me quedi el= punt damunt. % Els combings % Dibuixo: % amb y la l=C3=ADnia recta que uneix els dos punts, directament % per x faig un funci=C3=B3 del sinus (sin nx + ax =3D k) \draw plot[domain=3D0:0.4216,smooth,variable=3D\t] ({-0.857727*\t -sin (7= =2E31228*\t r) },{18.8812*\t -4 }); \draw plot[domain=3D0:0.4216,smooth,variable=3D\t] ({+0.857727*\t +sin (7= =2E31228*\t r) },{18.8812*\t -4 }); % Theta_ij \draw[decorate,decoration=3D{random steps,segment length=3D2mm,amplitude=3D= 2pt}] (-1.181475, 0) -- (1.181475, 0); \draw[decorate,decoration=3D{random steps,segment length=3D2mm,amplitude=3D= 2pt}] (1.161048, 1) -- (-1.161048, 1); % el sentit d'omega \draw[decorate,decoration=3D{markings,mark=3Dat position .9 with {\arrow[= red!50,line width=3D1mm]{<}}}] plot[domain=3D-3.141:3.141,smooth,variable= =3D\t] ({3*sin(\t r)},{4*cos(\t r)}); % el sentit de \theta_ij \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green,line width=3D1mm]{<}}}] (-1.181475, 0) -- (1.181475, 0); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green,line width=3D1mm]{>}}}] (1.181475, 0) -- (1.161048, 1); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green,line width=3D1mm]{>}}}] (1.161048, 1) -- (-1.161048, 1); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green,line width=3D1mm]{<}}}] (-1.161048, 1) -- (-1.181475, 0); % Els punts de les cel=C2=B7les % Calcul els combings per a y=3D 0 i y=3D1 \filldraw[color=3Dblue!50] (-1.181475, 0) circle (2pt); \filldraw[color=3Dblue!50] (1.181475, 0) circle (2pt); \filldraw[color=3Dblue!50] (1.161048, 1) circle (2pt); \filldraw[color=3Dblue!50] (-1.161048, 1) circle (2pt); % Els noms \draw (0, -4.3) node {$1 \in G$}; \draw (2.7, -3) node {$\gamma(w)$}; %\draw (-1.18, 0) node[left] {$\sigma_{\pi(w(i+1))}(j)$}; %\draw (1.18, 0) node[right] {$\sigma_{\pi(w(i))}(j)$}; %\draw (-1.18, 1) node[left] {$\sigma_{\pi(w(i+1))}(j+1)$}; %\draw (1.18, 1) node[right] {$\sigma_{\pi(w(i))}(j+1)$}; \draw (0,0.5) node {$\gamma(u_{i,j})$}; \draw (-1,4.3) node {$\pi(w(i+1))$}; \draw (1,4.3) node {$\pi(w(i))$}; \draw (0,1.5) node {$\gamma(v_{i, j+1})$}; \draw (0,-0.5) node {$\gamma(v_{i, j})$}; \draw (1.18, 0.5) node[right] {$\gamma(x_i^{j+1)})$}; \draw (-1.18, 0.5) node[left] {$\gamma(x_{i+1}^{j+1)})$}; \stoptikzpicture} { } \stopcombination} \indentation Per a tots $i \in \{0, \ldots, l(w)-1\}$ i $j \in \{0, \ldot= s, l(w)/2 -1\}$, per definici=C3=B3 de $\varphi_{\sigma}$ i per~{(\in[for= m:distancia-menor-que-lw-2])}, $l(\overline{v}_{i, j}) \leq \varphi_{\sig= ma}(l(w)/2)$, per la qual cosa $l(u_{i, j}) \leq 2 + 2 \varphi_{\sigma}(l= (w)/2)$. Per tant, reindexant aquest sumatori amb la bijecci=C3=B3 \startformula \{u_{i, j} \mid 0 \leq i \leq l(w) -1, 0 \leq j \leq l(w)/2 -1\} \longle= ftrightarrow \{u_k \mid 1 \leq k \leq {l(w)}^2/2\}, \stopformula existeixen com a m=C3=A0xim ${l(w)}^2/2$ paraules nul-homot=C3=B2piques p= er ${\cal P}$, $u_k$, tals que $l(u_k) \leq 2 + 2 \varphi_{\sigma}(l(w)/2= )$ i \startformula \text{area}_{\cal P}(w) \leq \sum_{k=3D1}^{{l(w)}^2/2} \text{area}_{\cal = P}(u_k). \stopformula \stopmydemo \startmyproposition[thmi:desigualtat-dehn] Siguin $G$ un grup, $X$ un con= junt finit de generadors de $G$, ${\cal P} =3D \langle X \mid R \rangle$ = una presentaci=C3=B3 finita de $G$ i una secci=C3=B3 geod=C3=A8sica $\sig= ma \colon G \to {(X \cup X^{-1})}^*$. Aleshores \startformula \delta_{{\cal P}} (n) \leq \frac{1}{2} \delta_{{\cal P}} (2 \varphi_{\sig= ma}(n/2)+2) \cdot n^2. \stopformula \stopmyproposition \startmydemo Sigui $w$ una paraula nul-homot=C3=B2pica per ${\cal P}$. Pe= l Lema~\in[thmi:subadditivitat-area], tenim que existeixen $u_k$ paraules= nul-homot=C3=B2piques per ${\cal P}$, amb $k \in \{1, \ldots, {\lvert w = \rvert}/2\}$, de longitud $l(u_k) \leq 2 \varphi_{\sigma}({l(w)}/2) +2$ t= als que \startformula \text{area}_{\cal P}(w) \leq \sum_{i=3D1}^{{{l(w)}^2}/2} \text{area}_{\ca= l P}(u_k). \stopformula Com que $l(u_k) \leq 2 \varphi_{\sigma}({l(w)}/2) +2$, aleshores $\text{a= rea}_{\cal P}(u_k) \leq \delta_{{\cal P}}(2 \varphi_{\sigma}({l(w)}/2) +2= )$. Per tant, \placeformula[-] \startformula \startsplit \NC \text{area}_{\cal P}(w) \NC \leq \sum_{k=3D1}^{{{l(w)}^2}/2} \text{= area}_{\cal P}(u_k) \NR \NC \NC \leq \sum_{k=3D1}^{{{l(w)}^2}/2} \delta_{{\cal P}} (2 \varphi_= {\sigma}({l(w)}/2) +2) \NR \NC \NC \leq \frac{1}{2} {l(w)}^2 \cdot \delta_{{\cal P}} (2 \varphi_{= \sigma}({l(w)}/2) +2). \stopsplit \stopformula \indentation Llavors \placeformula[-] \startformula \startsplit \NC \delta_{{\cal P}} (n) \NC =3D \max \{ \text{area}_{\cal P}(w) \mid w= \text{ nul-homot=C3=B2pica per }{\cal P}, l(w) \leq n\}\NR \NC \NC \leq \max \{ \frac{1}{2} {l(w)}^2 \cdot \delta_{{\cal P}} (2 \va= rphi_{\sigma}({l(w)}/2) +2) \mid w \text{ nul-homot=C3=B2pica per } {\cal= P}, l(w) \leq n\} \NR \NC \NC \leq \frac{1}{2} n^2 \cdot \delta_{{\cal P}} (2 \varphi_{\sigma}= (n/2) +2). \stopsplit \stopformula \stopmydemo Aquesta recursi=C3=B3 d=C3=B3na lloc a una fita superior de la funci=C3=B3= de Dehn per als grups tals que admetin una secci=C3=B3 geod=C3=A8sica $\= sigma$ tal que $\varphi_{\sigma} (n) < n-1$ per a $n$ suficientment gran.= \definemathcases[displaycases][style=3D\displaystyle] \startmylema[thmi:equacio-funcional] Sigui $F\colon \naturalnumbers \to \= reals $ una funci=C3=B3 que cumpleix la recursi=C3=B3 \startformula F(n) =3D F(n-2) + 2 \ln n + \ln \frac{1}{2}. \stopformula Aleshores \placeformula[-] \startformula F(n) =3D \startdisplaycases \NC F(0) + 2 \ln n!! - \frac{n}{2} \ln 2 \MC \text{si } n \text{ pa= rell} \NR \NC F(1) + 2 \ln n!! - \frac{n+1}{2} \ln 2 \MC \text{si } n \text{ se= nar} \NR \stopdisplaycases \stopformula on $n!!$ denota el doble factorial, definit recursivament per $0!! =3D 1$= , $1!! =3D 1$, $n!! =3D n \cdot (n-2)!!$. \stopmylema \startmydemo Com que la recursi=C3=B3 $F(n) =3D F(n-2) + 2 \ln n + \ln \f= rac{1}{2}$ =C3=A9s d'ordre $2$, per la Teoria d'Equacions en Difer=C3=A8n= cies, la soluci=C3=B3 d'aquesta recursi=C3=B3 =C3=A9s =C3=BAnica si es co= neixen les condicions inicials $F(1)$ i $F(0)$. Per tant, basta comprovar= que si $F$ t=C3=A9 aquesta forma, aleshores $F$ compleix la recursi=C3=B3= , el que es pot veure amb un simple c=C3=A0lcul. \stopmydemo \startmytheorem[thmi:Theorema-n!!-Presentacions] Siguin $G$ un grup, $X$ = un conjunt finit de generadors de $G$, ${\cal P} =3D \langle X \mid R \ra= ngle$ una presentaci=C3=B3 finita de $G$ i $\sigma \colon G \to {(X \cup = X^{-1})}^*$ una secci=C3=B3 geod=C3=A8sica tal que existeix un $n_0 \in \= naturalnumbers$ tal que $\varphi_{\sigma}(n) < n-1$ per a tot $n \geq n_0= $. Aleshores existeix $C$ constant, que nom=C3=A9s dep=C3=A8n de $n_0$ (i= de ${\cal P}$), tal que \startformula \delta_{{\cal P}} (n) \leq C \cdot \frac{(n!!)^2}{2^{n/2}}, \stopformula per a tot $n \geq n_0$. A m=C3=A9s, $C$ satisf=C3=A0 que \startformula C \geq \frac{(\delta_{{\cal P}}(n_0)+1)\cdot 2^{\frac{n_0+1}{2}}}{(n_0!!)= ^2}. \stopformula \stopmytheorem \startmydemo Per la Proposici=C3=B3~\in[thmi:desigualtat-dehn], tenim que \startformula \delta_{{\cal P}} (n) \leq \frac{1}{2} \delta_{{\cal P}} (2 \varphi_{\sig= ma}(n/2)+2) \cdot n^2. \stopformula Com que $\varphi_{\sigma}(n) < n-1$ per a tot $n \geq n_0$, llavors $\var= phi_{\sigma}(n) \leq n-2$, ja que la funci=C3=B3 $\varphi$ nom=C3=A9s pre= n valors naturals. Per aix=C3=B2, per a tot $n \geq n_0$, tenim que $2 \v= arphi_{\sigma}(n/2) +2 =3D 2 \varphi_{\sigma} (\lfloor n/2 \rfloor) + 2 \= leq 2 \lfloor n/2 \rfloor -2 \leq n-2$. Per tant, $\delta_{{\cal P}}$ sat= isf=C3=A0 la desigualtat \placeformula[desigualtat-delta-p] \startformula \delta_{{\cal P}} (n) \leq \frac{1}{2} \delta_{{\cal P}} (n-2) \cdot n^2,= \stopformula per a tot $n \geq n_0$. Sigui $f\colon \naturalnumbers \to \naturalnumbers\setminus \{0\}$ una fu= nci=C3=B3 tal que compleix que \placeformula[desigualtat-f] \startformula \startGROUP \NC f(n) =3D \frac{1}{2} f(n-2) \cdot n^2, \NR \NC f(n_0) \geq \delta_{{\cal P}}(n_0) .\NR \stopGROUP \stopformula La desigualtat~(\in[desigualtat-delta-p]) implica que $\delta_{{\cal P}}(= n) \leq f(n)$ per a tot $n \geq n_0$. Vegem-ho per inducci=C3=B3 sobre $n= $: \startitemize[1] \item Si $n=3D n_0$, aleshores $\delta_{{\cal P}}(n_0) \leq f(n_0)$ per c= onstrucci=C3=B3 de $f$. \item Suposem-ho cert fins a $n$ i provem-ho per a $n+1$. Aplicant hip=C3= =B2tesi d'inducci=C3=B3 i (\in[desigualtat-delta-p]), tenim que \startformula f(n+1) =3D \frac{1}{2} f(n-1) \cdot (n+1)^2 \geq \frac{1}{2} \delta_{{\ca= l P}} (n-1) \cdot (n+1)^2 \geq \delta_{{\cal P}} (n+1). \stopformula \stopitemize \indentation Considerem la funci=C3=B3 $F \colon \naturalnumbers \to \rea= ls$ definida per $F(n) =3D \ln f(n)$. $F$ est=C3=A0 ben definida, ja que = $\text{Im } f =3D \naturalnumbers \setminus \{0\}$. Per~(\in[desigualtat-= f]) prenent logaritmes i operant, tenim que $F$ compleix que \placeformula[-] \startformula \startGROUP \NC F(n) =3D F(n-2) + 2 \ln n + \ln \frac{1}{2}, \NR \NC F(n_0) =3D \ln f(n_0). \NR \stopGROUP \stopformula Pel Lema~\in[thmi:equacio-funcional], $F$ =C3=A9s de la forma \placeformula[-] \startformula F(n) =3D \startdisplaycases \NC F(0) + 2 \ln n!! - \frac{n}{2} \ln 2 \MC \text{si } n \text{ pa= rell} \NR \NC F(1) + 2 \ln n!! - \frac{n+1}{2} \ln 2 \MC \text{si } n \text{ se= nar} \NR \stopdisplaycases \stopformula Prenent $f(n_0) \geq \delta_{{\cal P}}(n_0) + 1 > 0$, $F(0) =3D \ln C_1$ = i $F(1) =3D \ln C_2$ amb $C_1$ i $C_2$ constants que nom=C3=A9s depenen d= e $n_0$ i de $\delta_{{\cal P}}$ i que satisfan \startformula \startmathalignment \NC C_1 \NC =3D \frac{f(n_0) \cdot 2^{n_0/2}}{(n_0!!)^2}, \NR[+] \NC C_2 \NC =3D \frac{f(n_0) \cdot 2^{\frac{n_0+1}{2}}}{(n_0!!)^2},\NR \stopmathalignment \stopformula aleshores tenim que $F(n_0) =3D \ln f(n_0)$. Notem que =C3=A9s necessari = prendre $f(n_0) > 0$ per assegurar l'exist=C3=A8ncia de $\ln C_1$ i $\ln = C_2$ i que sempre podem fer aquesta elecci=C3=B3 perqu=C3=A8 $f(n_0) \geq= \delta_{{\cal P}}(n_0)$. Per tot aix=C3=B2, $F$ t=C3=A9 la forma \placeformula[-] \startformula F(n) =3D \startdisplaycases \NC \ln C_1 + 2 \ln n!! - \frac{n}{2} \ln 2 \MC \text{si } n \text{= parell} \NR \NC \ln C_2 + 2 \ln n!! - \frac{n+1}{2} \ln 2 \MC \text{si } n \text{= senar} \NR \stopdisplaycases \stopformula \indentation De forma clara, $F(n) \leq \ln C_2 + 2 \ln n!! - \frac{n}{2}= \ln 2$, per la qual cosa tenim que \startformula \delta_{{\cal P}} (n) \leq f(n) =3D e^{F(n)} \leq C_2 \cdot \frac{(n!!)^2= }{2^{n/2}}, \stopformula per a tot $n \geq n_0$. Llavors si diem $C=3D C_2$, tenim el que vol=C3=AD= em. \stopmydemo Hem de notar que la Proposici=C3=B3~\in[thmi:desigualtat-dehn] tamb=C3=A9= implica fites inferiors sobre $\delta_{\cal P}$: tot grup $G$ admet una = secci=C3=B3 $\sigma\colon G \to {(X \cup X^{-1})}^*$ tal que $\varphi_{\s= igma}(n) \leq n$ per a tot $n \in \naturalnumbers$ (Teorema~\in[thme:brid= son-finitament-presentat]). Per tant, per a tota presentaci=C3=B3 ${\cal = P} =3D \langle X \mid R \rangle$ de $G$ i per a tot $n \in \naturalnumber= s$, tenim que \startformula \delta_{\cal P} (n) \leq \frac{1}{2} \delta_{\cal P} (n+2) \cdot n^2, \stopformula ja que $2 \varphi_{\sigma}(n/2) + 2 \leq 2 (n/2) + 2 =3D n+2$. Per tant, \startformula \delta_{\cal P} (n) \geq \frac{ 2 \delta_{\cal P} (n-2)}{(n-2)^2}. \stopformula Aquesta recursi=C3=B3, per=C3=B2, d=C3=B3na lloc a una fita inferior molt= grollera, la qual, tot d'una, es converteix en una fita trivial. \startmytheorem[thmi:fita-funcio-Dehn] Sigui $G$ un grup finitament gener= at. Si existeix una secci=C3=B3 geod=C3=A8sica $\sigma$, respecte d'algun= conjunt de generadors finit de $G$, tal que $\varphi_{\sigma}(n) < n-1$ = per a $n$ suficientment gran, aleshores $G$ =C3=A9s finitament presentat = i la funci=C3=B3 de Dehn de $G$, $\delta_G$, safisf=C3=A0 que \startformula \delta_{G} (n) \preceq \frac{(n!!)^2}{2^{n/2}}. \stopformula \stopmytheorem \startmydemo Pel Teorema~\in[thme:bridson-finitament-presentat], $G$ =C3=A9= s finitament presentat. A m=C3=A9s, per aquest mateix teorema, si $\sigma= $ =C3=A9s una secci=C3=B3 respecte d'un conjunt de generadors $X$, alesho= res existeix una presentaci=C3=B3 finita de la forma ${\cal P} =3D \langl= e X \mid R \rangle$. Sigui $n_0$ tal que $\varphi_{\sigma}(n) < n-1$ per a tot $n \geq n_0$. P= el Teorema~\in[thmi:Theorema-n!!-Presentacions] existeix una constant $C_= {{\cal P}, n_0}$, que dep=C3=A8n de ${\cal P}$ i de $n_0$, tal que=20 \placeformula[fita-delta-p] \startformula \delta_{{\cal P}} (n) \leq C_{{\cal P},n_0} \cdot \frac{(n!!)^2}{2^{n/2}}= =2E \stopformula per a tot $n \geq n_0$. De forma clara, podem prendre $C_{{\cal P}, n_0}$= prou gran per a qu=C3=A8 aquesta desigualtat es compleixi per a tot $n \= in \naturalnumbers$. Com que $\delta_G$ =C3=A9s la classe d'equival=C3=A8ncia de les funcions = de Dehn de les presentacions finites de $G$ m=C3=B2dul $\simeq$, tenim qu= e existeix $C' > 1$ tal que \startformula \delta_{G} (n) \preceq C' \frac{(n!!)^2}{2^{n/2}}. \stopformula Com que per a qualssevol funcions $f, g \colon \naturalnumbers \to \natur= alnumbers$ creixents i $C > 1$ constant, $f \preceq g$ implica que $f \pr= eceq C g$ (ja que $f(x) \leq k g(kx+k) + kx + k \leq k Cg(kx+k) + kx + k$= , per alguna constant $k > 0$), aleshores aix=C3=B2 implica que \startformula \delta_{G} (n) \preceq \frac{(n!!)^2}{2^{n/2}}. \stopformula \stopmydemo Pel Teorema~\in[thme:teorema-riley], si $G$ admet una secci=C3=B3 geod=C3= =A8sica $\sigma$ tal que $\varphi_{\sigma} (n) < n-1$ per a $n$ suficient= ment gran, aleshores $\delta_{G} \preceq n!$. Vegem que aquesta fita =C3=A9= s m=C3=A9s grollera que la que hem obtingut en el resultat anterior. \startmylema[thmi:lema-funcions-no-preceq] Siguin $f, g \colon \naturalnu= mbers \to \naturalnumbers$ tals que $g(n) \geq n$, per a tot $n \in \natu= ralnumbers$ i $f$ i $g$ s=C3=B3n creixents. Si $f(n)/g(n) \geq h(x)$ amb = $h(x)$ una funci=C3=B3 tal que \startformula \lim_{x \to \infty} f^2(x)/h(kx) =3D \infty, \stopformula per a tota constant $k > 2$, aleshores $f \not \preceq g$. \stopmylema \startmydemo Demostrem-ho per reducci=C3=B3 a l'absurd. Si $f \preceq g$= , aleshores existeix $k > 0$ tal que \startformula f(x) \leq k g(kx + k) + kx + k, \stopformula per a tot $x \in \naturalnumbers$. Podem suposar $k > 2$. Com que $g(n) \= geq n$, aleshores \startformula \startsplit \NC f(x) \NC \leq k g(kx+k) + kx + k\NR \NC \NC \leq k g(kx+k) + g(kx+k)\NR \NC \NC \leq (k+1) g(kx+k)\NR \NC \NC \leq 2k g(kx+k). \stopsplit \stopformula Per tant, $f(x)/g(kx+k) \leq 2k$, per a tot $x \in \naturalnumbers$. Com = que $k > 2$, aleshores $kx + k \leq k^2x$ i, com que $g$ =C3=A9s creixent= , aleshores $g(kx+k) \leq g(k^2 x)$. Per tant, \startformula f(x)/g(k^2 x) \leq f(x)/g(kx+k) \leq 2k. \stopformula Per veure que aix=C3=B2 =C3=A9s impossible, basta veure que per a tot $k = > 2$, el l=C3=ADmit del quocient $f(x)/g(kx)$ tendeix a infinit quan $x$ = tendeix a infinit. Com que $f(x)/g(x) \geq h(x)$, aleshores $g(x) \leq f(= x)/h(x)$. Per tant, com que $f$ =C3=A9s creixent, \startformula \frac{f(x)}{g(kx)} \geq \frac{f(x)}{h(kx)/f(kx)} =3D \frac{f(x) f(kx)}{h(= kx)} \geq \frac{f^2(x)}{h(kx)}. \stopformula \stopmydemo \startmyproposition Sigui la funci=C3=B3 $F\colon \naturalnumbers \to \na= turalnumbers$ definida com \startformula F(n) =3D \frac{{(n!!)}^2}{2^{n/2}}. \stopformula Aleshores, per a $n$ suficientment gran, tenim que $F(n) < n!$, ${(n!!)}^= 2 \not \leq n!$ i $n! < e^{kn^3}$, i, a m=C3=A9s, $n! \not \not \preceq F= (n)$. \stopmyproposition \startmydemo En primer lloc, vegem que existeix una funci=C3=B3 $G \colon \naturalnumb= ers \to \naturalnumbers$ tal que \startformula \frac{F(n)}{n!} =3D \frac{(n!!)^2}{n! \cdot 2^{n/2}} \leq G(n), \stopformula i tal que \startformula G(n) \sim C \cdot \frac{n^{5/2}}{2^{n/2}}, \stopformula per a qualque $C > 0$ constant. Diferenciem els casos parell i senar. Com que $(2n)!! =3D 2^n \cdot n!$, = llavors, si $n$ =C3=A9s parell, $n!! =3D 2^{n/2} \cdot (n/2)!$. Si $n$ =C3= =A9s parell, usant la f=C3=B3rmula d'Stirling, $n! \sim \sqrt{2 \pi n} \c= dot {(n/e)}^n$, tenim que \placeformula[-] \startformula \startsplit \NC \frac{(n!!)^2}{n! \cdot 2^{n/2}} \NC \sim \frac{2^n \pi n \cdot {(= \frac{n/2}{e})}^n}{\sqrt{2 \pi n} \cdot {(n/e)}^n \cdot 2^{n/2}}\NR \NC \NC \sim \frac{\sqrt{\pi} \cdot \sqrt{n}}{\sqrt{2} \cdot 2^{n/2}}. \stopsplit \stopformula Si $n$ =C3=A9s senar, aleshores \startformula n!! =3D n \cdot (n-2)!! < n \cdot (n-1)!! =3D n \cdot 2^{(n-1)/2} \cdot (= (n-1)/2)!. \stopformula Per tant, operant \placeformula[-] \startformula \startsplit \NC \frac{(n!!)^2}{n! \cdot 2^{n/2}} \NC \leq \frac{n^2 \cdot 2^{n-1} \= cdot {((\frac{n-1}{2})!)}^2}{n! \cdot 2^{n/2}}\NR \NC \NC \sim \frac{\pi n^2 \cdot 2^{n/2 -1} \cdot n \cdot {(n/e)}^n \cd= ot 2^{1-n}}{\sqrt{2 \pi} \cdot \sqrt{n} \cdot {(n/e)}^n} \NR \NC \NC \sim \frac{\sqrt{\pi}}{\sqrt{2}} \cdot \frac{n^{5/2}}{2^{n/2}}.= \stopsplit \stopformula \indentation En tots dos casos, existeix una constant $C > 0$ tal que $F(= n)/n! \leq G(n)$ on \startformula G(n) \sim C \cdot \frac{n^{5/2}}{2^{n/2}}. \stopformula Arran d'aix=C3=B2 tenim que el quocient $F(n)/n!$ tendeix a zero quan $n$= tendeix a infinit. Per tant, $F(n) < n!$ per a $n$ suficientment gran. Vegem que $n! \not \preceq F(n)$. Per l'anterior, $n!/F(n) \geq 1/G(n) \s= im C \cdot 2^{n/2} \cdot n^{-5/2}$. Pel Lema~\in[thmi:lema-funcions-no-pr= eceq], basta veure que, per a tot $k > 2$, \startformula \lim_{x \to \infty} n!/(C \cdot 2^{n/2} \cdot n^{-5/2})=3D \infty \stopformula Per la f=C3=B3rmula de Stirling, tenim que \startformula \frac{(n!)^2}{C \cdot 2^{kn/2} \cdot (kn)^{-5/2}} \sim \frac{2 \pi n \cdo= t (n/e)^{2n} \cdot (kn)^{5/2}}{C \cdot 2^{kn/2}} \geq \frac{2\pi n \cdot = (kn)^{5/2}}{C} \cdot \left( \frac{n}{2^k e} \right)^{2n}, \stopformula que clarament tendeix a infinit quan $x$ tendeix a infinit. D'altra banda, $n$ suficientment gran, ${(n!!)}^2 \not \leq n!$, ja que, = usant $n!! =3D 2^{n/2} \cdot (n/2)!$ per a $n$ parell i la f=C3=B3rmula d= e Stirling, tenim que \placeformula[-] \startformula \startsplit \NC \frac{n!}{{(n!!)}^2} \NC \sim \frac{\sqrt{2 \pi n} \cdot {(n/e)}^n= }{2^n \cdot \pi n \cdot {(\frac{n/2}{e})}^n}\NR \NC \NC \sim \frac{\sqrt{2 \pi n}}{\pi n} \NR \NC \NC \sim \frac{\sqrt{2}}{\sqrt{\pi} \cdot n}. \stopsplit \stopformula Per tant, tenim una successi=C3=B3 de nombres naturals $(n_i)_{i \in \nat= uralnumbers}$ (tots els nombres parells) tal que \startformula \frac{n_i!}{{(n_i!!)}^2} \stopformula tendeix a zero quan $n_i$ tendeix a infinit. Per aix=C3=B2, $(n!!)^2 \not= \leq n!$ per a $n$ suficientment gran, ja que, en cas contrari, tendr=C3= =ADem que $n!/{(n!!)^2} \geq 1$. Per =C3=BAltim, vegem que $n! < e^{kn^3}$, per a tota constant $k > 0$ i = per a $n$ prou gran. Usant la f=C3=B3rmula de Stirling i operant, tenim q= ue \startformula \ln n! \sim \ln \bigl(\sqrt{2 \pi n} (n/e)^n\bigr) =3D \frac{1}{2} \ln (2= \pi) + \frac{1}{2} \ln n + n \ln n - n < kn^3, \stopformula del que es dedueix la desigualtat de forma directa. \stopmydemo \subsection{An=C3=A0lisi del cas asincr=C3=B2nic} Al contrari del cas en qu=C3=A8 un grup $G$ admet una secci=C3=B3 $\sigma= $ tal que la seva amplada sincr=C3=B2nica $\varphi_{\sigma} (n) < n-1$, p= er a $n$ suficientment gran, amb les t=C3=A8cniques que coneixem, no pode= m millorar, en general, la fita $\delta_G \preceq n!$ quan $\Phi_{\sigma}= (n) < n-1$, per a $n$ suficientment gran\footnote{Per demostrar que aque= sta fita superior =C3=A9s =C3=B2ptima, s'hauria de trobar un exemple d'un= grup $G$ tals que $\delta_G \simeq n!$ i $\Phi_{\sigma} (n) < n-1$, per = a $n$ suficientment gran, per qualque secci=C3=B3 $\sigma$ de $G$.}. Aque= st fet es posa de manifest si intentem obtenir resultats an=C3=A0legs als= obtinguts pel cas asincr=C3=B2nic: \startmylema[thmi:lema-cas-asincronic-geodesica] Siguin $G$ un grup, $X$ = un conjunt finit de generadors de $G$, ${\cal P} =3D \langle X \mid R \ra= ngle$ una presentaci=C3=B3 finita de $G$, $\sigma \colon G \rightarrow {(= X \cup X^{-1})}^*$ una secci=C3=B3 geod=C3=A8sica de $G$ respecte de $X$.= Aleshores, per a tota paraula $w \in {(X \cup X^{-1})}^*$ nul-homot=C3=B2= pica per ${\cal P}$, existeixen $u'_k$ paraules nul-homot=C3=B2piques per= ${\cal P}$, amb $k \in \{1, \ldots, {l(w)}^2 + l(w)\}$, de longitud $l(u= '_k) \leq 2\Phi({l(w)}/2)+2$ tals que \startformula \text{area}_{\cal P} (w) \leq \sum_{k=3D1}^{{l(w)}^2 + l(w)} \text{area}_= {\cal P}(u'_k). \stopformula \stopmylema \startmydemo La demostraci=C3=B3 =C3=A9s an=C3=A0loga al cas sincr=C3=B2nic. Sigui $w$= una paraula nul-homot=C3=B2pica per ${\cal P}$. Si $w =3D \varepsilon$, = aleshores el resultat =C3=A9s obvi. Si $w =3D x_1 \ldots x_r$, amb $r \ge= q 1$, $x_i \in X \cup X^{-1}$ i $i \in \{0, \ldots, r\}$. En aquest cas, = per a tot $i \in \{0, \ldots, l(w) -1\}$, considerem la paraula $v_i$ def= inida per la concatenaci=C3=B3 seg=C3=BCent: \startformula v_i =3D \sigma_{\pi(w(i))} \cdot x_{i+1} \cdot \sigma_{\pi(w(i+1))}^{-1}.= \stopformula Per a cada $i \in \{0, \ldots, l(w) -1\}$, $v_i$ =C3=A9s nul-homot=C3=B2p= ica per ${\cal P}$, ja que forma un cicle dins el graf de Cayley. I, a m=C3= =A9s, per construcci=C3=B3 \startformula w =3D v_0 \sharp (v_1 \sharp (\ldots, \sharp(v_{l(w) -1 })\ldots ). \stopformula Per tant, pel Lema~\in[thmi:lema-tecnic], \placeformula[form:desigualtat-w-vi-bis] \startformula \text{area}_{\cal P} (w) \leq \sum_{i=3D0}^{l(w)-1} \text{area}_{\cal P}= (v_i). \stopformula \indentation Per a tot $i \in \{0, \ldots, l(w)-1\}$, siguin $\rho_i$, $\= rho'_i$ reparametritzacions de $\naturalnumbers$ per a $\pi(w(i))$ i $\p= i(w(i+1))$. Pel Lema~\in[thmi:lema-distancia-menor-w-2], \placeformula[form:distancia-menor-que-lw-2-bis] \startformula d_{G, X} (\pi(w(i)), 1) \leq l(w)/2, \stopformula per a tot $i \in \{0, \ldots, l(w)\}$. Com que $\sigma$ =C3=A9s geod=C3=A8sica, aleshores, per a tot $i \in \{0,= \ldots, l(w)\}$, la longitud de $\sigma_i$ =C3=A9s menor o igual que $l(= w)/2$ i els conjunts \startformula \startalign \NC \{ \sigma_i (\rho_i (t)) \mid t \in \naturalnumbers \}, \NR \NC \{ \sigma_i (\rho'_i (t)) \mid t \in \naturalnumbers \} \NR \stopalign \stopformula tenen, com a m=C3=A0xim, $l(w)/2$ elements diferents. Per a tots $i \in \{0, \ldots, l(w)\}$ i $t \in \naturalnumbers$, siguin \startitemize[1] \item $x_i^{t)}$ la paraula (de com a m=C3=A0xim una lletra) corresponent= al cam=C3=AD geod=C3=A8sic que va des de $\sigma_i(\rho_i(t))$ a $\sigma= _i(\rho_i(t+1))$. \item $\overline{v}_{i, t}$ la paraula corresponent a un cam=C3=AD geod=C3= =A8sic que vagi des de $\sigma_i (\rho_i(t))$ fins a $\sigma_{i+1} (\rho'= _i (t))$. \item $u_{i, t}$ la paraula nul-homot=C3=B2pica per ${\cal P}$ formada pe= r la concatenaci=C3=B3 \startformula x_i^{t+1)} \cdot \overline{v}_{i, t+1} \cdot {\left(x_{i+1}^{t+1)}\right)= }^{-1} \cdot \overline{v}_{i, t}^{-1}, \stopformula \stopitemize \noindentation Com a m=C3=A0xim hi ha $l(\sigma_i) + l(\sigma_{i+1}) \leq= 2 l(w)/2 =3D l(w)$ paraules $u_{i, t}$, ja que quan \startformula \startcases \NC \sigma_i(\rho_i(m)) \MC =3D \sigma_i(\rho_i(m+1)), \NR \NC \sigma_i(\rho'_i(m)) \MC =3D \sigma_i(\rho'_i(m+1)), \NR \stopcases \stopformula per a algun $m \in \integers$, aleshores $u_{i, m} =3D u_{i, m+1}$. Per t= ant, per a tot $i \in \{0, \ldots, l(w)-1\}$, existeixen $t_1, \ldots, t_= {l(w)}$ tals que,=20 \startformula v_i =3D u_{i,t_1} \sharp (u_{i,t_2} \sharp (\ldots, \sharp(u_{i, t_{l(w)}= })\ldots ), \stopformula per la qual cosa, pel Lema~\in[thmi:lema-tecnic], \placeformula[form:desigualtat-vi-uij-bis] \startformula \text{area}_{\cal P} (v_i) \leq \sum_{j=3D0}^{l(w)} \text{area}_{\cal P}= (u_{i, t_j}). \stopformula \indentation Aleshores, combinant (\in [form:desigualtat-w-vi-bis]) i (\i= n[form:desigualtat-vi-uij-bis]), tenim que \startformula \text{area}_{\cal P} (w) \leq \sum_{i =3D 0}^{l(w) -1} \sum_{j =3D 0}^{l= (w)} \text{area}_{\cal P} (u_{i,t_j}).=20 \stopformula \indentation Per a tot $i \in \{0, \ldots, l(w)-1\}$ i $t \in \naturalnum= bers$, per definici=C3=B3 de $\Phi$ i per~{(\in[form:distancia-menor-que-= lw-2-bis])}, $l(\overline{v}_{i, t}) \leq \Phi(l(w)/2)$, per la qual cosa= $l(u_{i, t}) \leq 2 + 2 \varphi(l(w)/2)$. Per tant, reindexant el sumato= ri, tenim que existeixen com a m=C3=A0xim $l(w)(l(w)+1)$ paraules nul-hom= ot=C3=B2piques per ${\cal P}$, $u'_k$, tals que $l(u'_k) \leq 2 + 2 \Phi(= l(w)/2)$ i \startformula \text{area}_{\cal P} (w) \leq \sum_{k=3D1}^{{l(w)}^2+l(w)} \text{area}_{\= cal P}(u'_k). \stopformula \stopmydemo Amb una an=C3=A0lisi de la demostraci=C3=B3 es pot veure que la fita supe= rior del sumatori no es pot millorar (llevat de substraccions de constant= s) sense hip=C3=B2tesis suplement=C3=A0ries. \startmyproposition[thmi:recursio-delta-p] Siguin $G$ un grup, $X$ un con= junt finit de generadors de $G$, ${\cal P} =3D \langle X \mid R \rangle$ = una presentaci=C3=B3 finita de $G$ i $\sigma \colon G \to {(X \cup X^{-1}= )}^*$ una secci=C3=B3 geod=C3=A8sica tal que existeix un $n_0 \in \natura= lnumbers$ tal que $\Phi_{\sigma} (n) < n-1$ per a tot $n \geq n_0$. Alesh= ores existeix una constant $C \geq 1$, $C \in \naturalnumbers$, tal que \startformula \delta_{\cal P} (n) \leq C \cdot (n+1)!, \stopformula per a tot $n \in \naturalnumbers$. \stopmyproposition \startmydemo Com que $\Phi_{\sigma} (n) < n-1$ per a tot $n \geq n_0$ i $= \Phi_{\sigma}$ pren valors enters, aleshores $\Phi_{\sigma} (n) \leq n-2$= per a tot $n \geq n_0$. Pel Lema~\in[thmi:lema-cas-asincronic-geodesica], tenim que $\delta_{\cal= P}$ segueix la recursi=C3=B3 \startformula \delta_{\cal P} (n) \leq (n^2 + n) \cdot \delta_{\cal P} (n-2), \stopformula per a tot $n \geq n_0$. Sigui $f \colon \naturalnumbers \to \naturalnumbers\setminus\{0\}$ defini= da per la recursi=C3=B3 \startformula \startGROUP \NC f(n) =3D (n^2 + n) \cdot f(n-2), \NR \NC f(n_0) \geq \delta_{{\cal P}}(n_0) .\NR \stopGROUP \stopformula Aleshores tenim que $\delta_{\cal P}(n) \leq f(n)$ per a tot $n \geq n_0$= =2E Considerem la funci=C3=B3 $F \colon \naturalnumbers \to \reals$ definida = per $F(n) =3D \ln f(n)$. Tenim que $F(n)$ satisf=C3=A0 la recursi=C3=B3 \startformula \startGROUP \NC F(n) =3D \ln n + \ln (n+1) + F(n-2), \NR \NC F(n_0) =3D \ln f(n_0).\NR \stopGROUP \stopformula Es pot veure que $F$ =C3=A9s de la forma \startformula F(n) =3D \startdisplaycases \NC F(0) +\ln (n+1)! \MC \text{si } n \text{ parell} \NR \NC F(1) + \ln (n+1)! \MC \text{si } n \text{ senar} \NR \stopdisplaycases \stopformula Per tant, prenent $K =3D \max \{F(0), F(1)\}$, tenim que $F(n) \leq K + \= ln (n+1)!$, i, per tant, \startformula \delta_{\cal P} (n) \leq f(n) \leq e^{F(n)} \leq e^K \cdot (n+1)!, \stopformula per a tot $n \geq n_0$. Ara b=C3=A9, podem prendre $C \geq e^K$, $C \in \= naturalnumbers$, prou gran tal que $\delta_{\cal P} (n) \leq C \cdot (n+1= )!$ per a tot $n \in \naturalnumbers$. \stopmydemo \startmylema[thmi:lema-preceq-necessari] Siguin $f, g \colon \naturalnumb= ers \to \naturalnumbers$ i $C \geq 1$, $C \in \naturalnumbers$, una const= ant. \startitemize[a][left=3D(,right=3D),stopper=3D] \item Si $g$ =C3=A9s creixent i $f \preceq Cg$, aleshores $f \preceq g$. \item $(n+1)! \simeq n!$. \stopitemize \stopmylema \startmydemo \startitemize[a][left=3D(,right=3D),stopper=3D] \item Si $f \preceq C g$, aleshores existeix $k > 0$ tal que $f(x) \leq k= C g(kx + k) + kx + k$. Com que $g$ =C3=A9s creixent i $C \geq 1$, \placeformula[-] \startformula \startsplit \NC f(x) \NC \leq k C g(kx + k) + kx + k\NR \NC \NC \leq k C g(kC x + kC) + kC x + kC \NR \NC \NC \leq (k C) g((kC) x + (kC)) + (kC) x + (kC). \stopsplit \stopformula Per tant, $f \preceq g$. \item De forma obvia, tenim que $(n+1)! \leq k (kn + k)! + kn + k$ prenen= t $k =3D1$. Per tant, $(n+1)! \preceq n!$. D'altra banda, com que $n! \l= eq (n+1)!$, tenim que $n! \preceq (n+1)!$. Per tant, $n! \simeq (n+1)!$. \stopitemize \stopmydemo \startmyproposition Sigui $G$ un grup finitament generat. Si existeix una= secci=C3=B3 geod=C3=A8sica $\sigma$, respecte d'algun conjunt de generad= ors finit de $G$, tal que $\Phi (n) < n-1$ per a $n$ suficientment gran, = aleshores $G$ =C3=A9s finitament presentat i la funci=C3=B3 de Dehn de $G= $, $\delta_G$, satisf=C3=A0 que $\delta_G (n) \preceq (n+1)! \simeq n!$. \stopmyproposition \startmydemo Pel Teorema~\in[thme:bridson-finitament-presentat], $G$ =C3=A9s finitamen= t presentat. Si $\sigma$ =C3=A9s una secci=C3=B3 respecte d'un conjunt de= generadors $X$, sigui ${\cal P} =3D \langle X \mid R \rangle$ una presen= taci=C3=B3 finita de $G$ (tota presentaci=C3=B3 finita ${\cal P}' =3D \la= ngle Y \mid S\rangle$ en tendr=C3=A0 una d'equivalent d'aquesta forma). Sigui $n_0$ tal que $\Phi_{\sigma}(n) < n-1$ per a tot $n \geq n_0$. Per = la Proposici=C3=B3~\in[thmi:recursio-delta-p] existeix una constant $C \g= eq 1$ tal que=20 \startformula \delta_{{\cal P}} (n) \leq C \cdot (n+1)!. \stopformula per a tot $n \in \naturalnumbers$. Com que $\delta_G$ =C3=A9s la classe d'equival=C3=A8ncia de les funcions = de Dehn de les presentacions finites de $G$ m=C3=B2dul $\simeq$, tenim qu= e existeix $C' \geq 1$ tal que \startformula \delta_{G} (n) \preceq C' \cdot (n+1)!. \stopformula Ara b=C3=A9, pel Lema~\in[thmi:lema-preceq-necessari], \startformula \delta_{G} (n) \preceq (n+1)! \simeq n!. \stopformula \stopmydemo Aquesta proposici=C3=B3 realment confirma que, amb les t=C3=A8cniques emp= rades en el cas sincr=C3=B2nic, no podem millorar la fita de la funci=C3=B3= de Dehn per a grups que admeten una secci=C3=B3 $\sigma$ tal que $\Phi_{= \sigma} (n) < n-1$ per a $n$ suficientment gran. \section{L'amplada mitjana respecte de dos valors} En aquesta secci=C3=B3 definirem l'amplada mitjana d'una secci=C3=B3 $\si= gma$ respecte de dos valors, $\lambda_{\sigma, s, k}$, i veurem que si un= grup $G$ admet una secci=C3=B3 tal que la seva amplada mitjana no =C3=A9= s {\em molt gran}, aleshores $G$ =C3=A9s finitament presentat i t=C3=A9 e= l problema de la paraula resoluble. \startmynotation Siguin $G$ un grup, $X$ un conjunt finit de generadors d= e $G$ i $n \in \naturalnumbers$. Indicarem amb $K_{G, X} (n)$\mysymbol{$K= _{G, X}(n)$} el conjunt \startformula K_{G, X} (n) =3D \{ (g, h) \in G \times G \mid d_{G, X} (1, g), d_{G, X} = (1, h) \leq n, \; d_{G, X} (g, h) =3D1 \}. \stopformula Quan $G$ i $X$ siguin clars pel context, els podrem ometre i escriure, si= mplement, $K(n)$\mysymbol{$K(n)$}. \stopmynotation \startmynotation Siguin $G$ un grup, $X$ un conjunt finit de generadors d= e $G$, $\sigma \colon G \to {(X \cup X^{-1})}^*$ una secci=C3=B3, $g, h \= in G$ i $t \in \naturalnumbers$. Indicarem amb $D_{\sigma, g, h} (t)$\mys= ymbol{$D_{\sigma, g, h} (t)$} el nombre \startformula D_{\sigma, g, h} (t) =3D d_{G, X} (\sigma_g (t), \sigma_h (t)). \stopformula \stopmynotation \startmydefinition Siguin $G$ un grup, $X$ un conjunt finit de generadors= de $G$, $\sigma \colon G \to {(X \cup X^{-1})}^*$ una secci=C3=B3, $s, k= \in \naturalnumbers$. La {\em $(s, k)$-amplada mitjana de \sigma}\index[= (s,k)-amplada mitjana+d'una secci=C3=B3]{$(s, k)$-amplada mitjana+d'una s= ecci=C3=B3}, o {\em amplada mitjana de $\sigma$ respecte de $s$ i $k$}\in= dex{amplada+mitjana respecte de dos valors}, =C3=A9s la funci=C3=B3 $\lam= bda_{\sigma, s, k} \colon \naturalnumbers \to \naturalnumbers$ definida p= er $\lambda_{\sigma, s, k} (0) =3D 0$ i, per a tot $n > 0$,=20 \startformula \lambda_{\sigma, s, k} (n) =3D \max \{( D_{\sigma, g, h} (t+s) + D_{\sig= ma, g, h} (t+k) )/2 \mid t \in \naturalnumbers, (g, h) \in K_{G, X} (n) \= }. \stopformula Escriurem simplement $\lambda_{s, k} (n)$\mysymbol{$\lambda_{\sigma, s, k= }$} quan $\sigma$ sigui clara pel context o quan no existeixi confusi=C3=B3= possible. \stopmydefinition Estemdrem aquesta funci=C3=B3 als nombres reals mitjan=C3=A7ant $\lambda_= {\sigma, s, k} (x) =3D \lambda_{\sigma, s, k} (\lfloor x \rfloor)$ si $x = > 0$ i $\lambda_{\sigma, s, k} (x) =3D \lambda_{\sigma, s, k} (0)$ si $x = < 0$. Per simetria, tenim que $\lambda_{\sigma, s, k} =3D \lambda_{\sigma, k, s= }$, per a tots $s, k \in \naturalnumbers$. Per tant, a partir d'ara, supo= sarem que $s \leq k$. \startmylema[thmi:minoracio-lambda-phi] Siguin $G$ un grup, $X$ un conjun= t finit de generadors de $G$ i $\sigma \colon G \to {(X \cup X^{-1})}^*$.= Aleshores $\lambda_{\sigma, s, k} (n) \leq \varphi_{\sigma} (n)$, per a = tots $s \leq k \in \naturalnumbers$. \stopmylema \startmydemo Per a tot $s \in \naturalnumbers$, indiquem amb \startformula M_s(n) =3D \max \{d_{G, X}(\sigma_g (t+s), \sigma_h (t+s)) \mid t \in \na= turalnumbers, (g, h) \in K_{G, X} (n) \}. \stopformula De forma clara tenim que, per a tot $n \in \naturalnumbers$, $M_s(n) \leq= \varphi_{\sigma}(n)$ i $2\lambda_{\sigma, s, k} (n) \leq M_s (n) + M_k = (n)$. Per tant, $\lambda_{\sigma, s, k} (n) \leq \varphi_{\sigma} (n)$. \stopmydemo \startmycorollary Siguin $s \leq k \in \naturalnumbers$. Tot grup finitam= ent generat admet una secci=C3=B3 $\sigma$, respecte d'algun conjunt fini= t de generadors, tal que $L_{\sigma}(n) =3D n$ i $\lambda_{\sigma, s, k}(= n) \leq n$, per a tot $n \in \naturalnumbers$. \stopmycorollary \startmydemo Pel Teorema~\in[thme:bridson-finitament-presentat], tot grup= finitament generat admet una secci=C3=B3 $\sigma$, respecte d'algun conj= unt finit de generadors, tal que $L_{\sigma} (n) =3D n$ i $\varphi_{\sigm= a} (n) \leq n$, per a tot $n \in \naturalnumbers$. Aplicant el lema anter= ior, tenim que $\lambda_{\sigma, s, k} (n) \leq n$, per a tot $n \in \nat= uralnumbers$. \stopmydemo Volem veure que si un grup finitament generat $G$ admet una secci=C3=B3 $= \sigma$ tal que, per a qualques $s \leq k \in \naturalnumbers$, $\lambda_= {\sigma, s, k} (n) < n-1$, per a $n$ suficientment gran, aleshores $G$ =C3= =A9s finitament presentat i $G$ t=C3=A9 el problema de la paraula resolub= le. En comptes de demostrar aquest fet per a tots $s, k$, ho demostrarem = nom=C3=A9s per a $s =3D 0$ i $k=3D1$, reduint els altres casos a aquest. \subsection{Reducci=C3=B3 de $\lambda_{s, k}$ a $\lambda_{0, 1}$} \startmylema[thmi:reduccio-lambdask-a-lambda-0k] Siguin $G$ un grup, $X$ = un conjunt finit de generadors de $G$, $\sigma \colon G \to {(X \cup X^{-= 1})}$ una secci=C3=B3, $0 < s \leq k \in \naturalnumbers$ i $f \colon \na= turalnumbers \to \naturalnumbers$ una funci=C3=B3 qualsevol. Aleshores: \startitemize[a][left=3D(,right=3D),stopper=3D] \item Si $\lambda_{\sigma, 0, k-s} (n) < f(n)$ per a $n$ suficientment gr= an, aleshores $\lambda_{\sigma, s, k} (n) < f(n)$ per a $n$ suficientment= gran. \item Si $f$ =C3=A9s estrictament creixent, aleshores si $\lambda_{\sigma= , s, k} (n) < f(n)$ per a $n$ suficientment gran, llavors $\lambda_{\sigm= a, 0, k-s} (n) < f(n)$ per a $n$ suficientment gran. \stopitemize \stopmylema \startmydemo \startitemize[a][left=3D(,right=3D),stopper=3D] \item Fent el canvi de variable $t' =3D t+s$, tenim que \placeformula[-] \startformula \startsplit \NC 2 \lambda_{\sigma, s, k} (n) \NC =3D \max \{D_{\sigma, g, h} (t+s) = + D_{\sigma, g, h} (t+k) \mid t \in \naturalnumbers, (g, h) \in K_{G, X} = (n) \}\NR \NC \NC =3D \max \{D_{\sigma, g, h} (t') + D_{\sigma, g, h} (t'+k-s) \m= id t' \geq s, (g, h) \in K_{G, X} (n)\} \NR \NC \NC \leq 2 \lambda_{\sigma, 0, k-s} (n), \stopsplit \stopformula per a tot $n \in \naturalnumbers$, Per tant, $\lambda_{\sigma, s, k} (n) \leq \lambda_{\sigma, 0, k-s} (n)$ = per a tot $n \in \naturalnumbers$. Per tant, per hip=C3=B2tesi, $\lambda_= {\sigma, s, k} (n) < f(n)$ per a $n$ suficientment gran. \item Per a tot $n \in \naturalnumbers$, indiquem amb $N_{s, k} (n)$ el m= =C3=A0xim \startformula \max \{D_{\sigma, g, h} (t) + D_{\sigma, g, h} (t+k-s) \mid t =3D 0, \ldo= ts, s-1, (g, h) \in K_{G, X} (n) \}. \stopformula Tenim que \startformula \startsplit \NC 2 \lambda_{\sigma, 0, k-s} (n) \NC =3D \max \{D_{\sigma, g, h} (t) + = D_{\sigma, g, h} (t+k-s) \mid t \in \naturalnumbers, (g, h) \in K_{G, X} = (n) \} \NR \NC \NC =3D \max \big\{\max \{D_{\sigma, g, h} (t) + D_{\sigma, g, h} (t+= k-s) \mid t \geq s, (g, h) \in K_{G, X} (n) \}, \NR \NC \NC \quad \qquad \quad N_{s, k} (n) \big\} \NR \NC \NC =3D \max \{ 2 \lambda_{\sigma, s, k} (n), N_{s, k} (n) \} \stopsplit \stopformula (el darrer pas s'aconsegueix fent el canvi de variable $t' =3D t-s$; vege= u l'apartat anterior). $N_{s, k} (n)$ satisf=C3=A0 que \startformula \startsplit \NC N_{s, k} (n) \NC \leq \max \{D_{\sigma, g, h} (t) \mid t=3D0, \ldots,= s-1, (g, h) \in K_{G, X} (n) \} \NR \NC \NC \quad + \max \{D_{\sigma, g, h} (t+s) \mid t=3D0, \ldots, s-1, (g= , h) \in K_{G, X} (n) \}\NR \NC \NC \leq 2 \max \{D_{\sigma, g, h} (t) \mid t=3D0, \ldots, s-1, (g, h= ) \in K_{G, X} (n) \}, \NR \NC \NC \leq 2 \max \{d_{G, X} (g, h) \mid (g, h) \in B_{G, X} (1, s-1)\}= =2E \stopsplit \stopformula La darrera desigualtat =C3=A9s perqu=C3=A8 la dist=C3=A0ncia entre $\sigm= a_g(t)$ i $\sigma_h(t)$ =C3=A9s menor que la m=C3=A0xima dist=C3=A0ncia e= ntre dos elements de la bolla $B_{G, X} (1, s-1)$, ja que $\sigma_g(t)$ i= $\sigma_h(t)$ pertanyen a aquesta bolla, per a qualssevol $g, h \in K_{G= , X} (n)$. Per tant, com que $f$ =C3=A9s estrictament creixent, existeix = $n_1 \in \naturalnumbers$, independent de $n$ ($n_1$ nom=C3=A9s dep=C3=A8= n de $s$), tal que $f(n_1) > N_{s, k} (n)$, per a tot $n \in \naturalnumb= ers$. D'altra banda, sigui $n_0 \in \naturalnumbers$ tal que $\lambda_{\sigma, = s, k} (n) < f(n)$ per a tot $n \geq n_0$, el qual existeix per hip=C3=B2t= esi, i sigui $N =3D \max \{n_0, n_1\}$. Aleshores, per a tot $n \geq N$, \startformula \startsplit \NC 2 \lambda_{\sigma, 0, k-s} (n) \NC =3D \max \{2 \lambda_{\sigma, s, k= } (n), N_{s, k} (n)\} \NR \NC \NC \leq \max \{f(n_1), 2 \lambda_{\sigma, s, k} (n) \} \NR \NC \NC \leq \max \{f(N), 2f(n) \} \NR \NC \NC \leq 2 f(n). \stopsplit \stopformula Per tant, $\lambda_{\sigma, 0, k-s} (n) < f(n)$ per a $n$ suficientment g= ran. \stopitemize \stopmydemo \startmylema[thmi:reduccio-lambda-0k-a-lambda-01] Siguin $G$ un grup, $X$= un conjunt finit de generadors de $G$, $\sigma \colon G \to {(X \cup X^{= -1})}^*$ una secci=C3=B3 i $k > 1$. Si $\lambda_{\sigma, 0, k} (n) < n-k$= per a $n$ suficientment gran, aleshores $\lambda_{\sigma,0,1} (n) < n-1$= per a $n$ suficientment gran. \stopmylema \startmydemo Sigui $n_0 \in \naturalnumbers$ tal que $\lambda_{\sigma, 0,= k} (n) < n-k$ per a tot $n \geq n_0$. Per a tots $n \in \naturalnumbers$, $g, h \in K_{G, X} (n)$ i $t \in \nat= uralnumbers$, aplicant la desigualtat triangular, tenim que \startformula \startsplit \NC d_{G, X} (\sigma_g (t+1), \sigma_h (t+1)) \NC \leq d_{G, X} (\sigma_g= (t+1), \sigma_g (t+k))\NR \NC \NC \quad + d_{G, X} (\sigma_g (t+k), \sigma_h (t+k))\NR \NC \NC \quad + d_{G, X} (\sigma_h (t+k), \sigma_h (t+1))\NR \NC \NC \leq 2 (k-1) + d_{G, X} (\sigma_g (t+k), \sigma_h (t+k)). \stopsplit \stopformula Per tant, per a tot $n \geq n_0$, \startformula \startsplit \NC 2 \lambda_{\sigma, 0, 1} (n) \NC =3D \max \{D_{\sigma, g, h} (t) + D_= {\sigma, g, h} (t+1) \mid t \in \naturalnumbers, (g, h) \in K_{G, X} (n)\= }\NR \NC \NC \leq \max \{D_{\sigma, g, h} (t) + 2(k-1) + D_{\sigma, g, h} (t+k= ) \mid t \in \naturalnumbers, (g, h) \in K_{G, X} (n)\}\NR \NC \NC \leq \max \{D_{\sigma, g, h} (t) + D_{\sigma, g, h} (t+k) \mid t = \in \naturalnumbers, (g, h) \in K_{G, X} (n)\}+ 2 (k-1)\NR \NC \NC \leq 2\lambda_{\sigma, 0, k} (n) + 2(k-1) \NR \NC \NC < 2 (n-k) + 2(k-1) \NR \NC \NC =3D 2 (n-1). \stopsplit \stopformula Aleshores $\lambda_{\sigma, 0, 1} (n) < n-1$ per a tot $n \geq n_0$. \stopmydemo \startmynotation Sigui $f \colon \naturalnumbers \to \naturalnumbers$ una= funci=C3=B3 qualsevol i $s \leq k \in \naturalnumbers$. Indicarem amb: \startitemize[a][left=3D(,right=3D),stopper=3D] \item ${\cal S}(\varphi, f)$ la classe dels grups finitament generats $G$= tal que existeix una secci=C3=B3 $\sigma$ de $G$ respecte d'un conjunt f= init de generadors de $G$ tal que $\varphi_{\sigma} (n) < f(n)$ per a $n$= suficientment gran. \item ${\cal S} (\lambda_{s, k}, f)$ la classe dels grups finitament gene= rats $G$ tal que existeix una secci=C3=B3 $\sigma$ de $G$ respecte d'un c= onjunt finit de generadors de $G$ tal que $\lambda_{\sigma, s, k} (n) < f= (n)$ per a $n$ suficientment gran. \stopitemize \stopmynotation Amb aquesta notaci=C3=B3, tenim que la classe dels grups sincr=C3=B2nicam= ent seccionables coincideix amb la uni=C3=B3 de les classes ${\cal S}(\va= rphi, f(n) \equiv k)$ amb $k \in \naturalnumbers$ constant, i que els teo= remes~\in[thme:bridson-fita-sincronica] i \in[thme:bridson-finitament-pre= sentat] estableixen que si $G \in {\cal S}(\varphi, f(n) =3D n-1)$, alesh= ores $G$ =C3=A9s finitament presentat i $\delta_{G} \preceq e^{kn^3}$ per= alguna constant $k > 0$. \startmyproposition[thmi:proposicio-reduccio-lambdes] Per a tot $r \geq 1= $, sigui $f_r \colon \naturalnumbers \to \naturalnumbers$ la funci=C3=B3 = definida per $f_r (n) =3D n-r$. Aleshores: \startformula \startgather {\cal S}(\varphi, f_1) \subseteq {\cal S} (\lambda_{0, 1}, f_1),\NR \bigcup_{0 \leq s \leq k} {\cal S} (\lambda_{s, k}, f_{k-s}) =3D \bigcup_= {0 \leq k} {\cal S} (\lambda_{0, k}, f_k) \subseteq {\cal S} (\lambda_{0,= 1}, f_1). \NR \stopgather \stopformula \stopmyproposition \startmydemo Pel Lema~\in[thmi:minoracio-lambda-phi], $\lambda_{\sigma, 0= , 1} \leq \varphi_{\sigma}$. Per tant, clarament, ${\cal S} (\varphi, f_1= ) \subseteq {\cal S} (\lambda_{0, 1}, f_1)$. D'altra banda, pel Lema~\in[thmi:reduccio-lambdask-a-lambda-0k], per a to= ts $0 < s \leq k$, ${\cal S} (\lambda_{s, k}, f_{k-s}) =3D {\cal S} (\lam= bda_{0, k-s}, f_{k-s})$. Aquest fet tamb=C3=A9 es compleix per a $s =3D 0= $ i $0 \leq k$ de forma trivial. Per tant, \startformula \bigcup_{0 \leq s \leq k} {\cal S} (\lambda_{s, k}, f_{k-s}) =3D \bigcup_= {0 \leq k} {\cal S} (\lambda_{0, k}, f_k) \stopformula \indentation Finalment, pel Lema~\in[thmi:reduccio-lambda-0k-a-lambda-01]= , ${\cal S} (\lambda_{0, k}, f_k) \subseteq {\cal S} (\lambda_{0,1}, f_1)= $. \stopmydemo Arran d'aquesta proposici=C3=B3 tenim que la classe dels grups que admete= n una secci=C3=B3 tal que, per a qualques $s \leq k$, $\lambda_{s, k} (n)= < n-1$ per a $n$ suficientment gran =C3=A9s, exactament, la classe dels = grups que admeten una secci=C3=B3 tal que $\lambda_{0,k'} (n) < n-1$ per = a $n$ suficientment gran, per algun $k'\geq 0$. Per tant, a partir d'ara,= podrem ocupar-nos, nom=C3=A9s, d'aquesta darrera classe de grups. \subsection{Els grups de ${\cal S}(\lambda_{0,1}, f(n)=3Dn-1)$ s=C3=B3n f= initament presentats} \startmylema[thmi:lema-dels-conjugats-de-conjugats]Siguin $A$ un conjunt = qualsevol, $X, Y \subseteq F(A)$. Si $w$ es pot expressar com producte de= conjugats de paraules de $X$ i cada $x \in X$ es pot expressar com produ= cte de conjugats de $Y$, aleshores $w$ es pot expressar com producte de c= onjugats de $Y$. \stopmylema \startmydemo Per hip=C3=B2tesi, tenim que \startformula \startmathalignment \NC w \NC =3D \prod_{i=3D1}^N u_i^{-1} x_i u_i, \NR \NC x_i \NC =3D \prod_{i=3D1}^{N_i} v_{ij}^{-1} y_{ij} v_{ij}, \NR \stopmathalignment \stopformula on $u_i, v_{ij} \in F(A)$, $x_i \in X$, $y_{ij} \in Y$. Llavors \startformula w =3D \prod_{i=3D1}^N u_i^{-1} x_i u_i =3D \prod_{i=3D1}^N u_i^{-1} \Big(= \prod_{i=3D1}^{N_i} v_{ij}^{-1} y_{ij} v_{ij}\Big) u_i =3D \prod_{i,j} (v= _{ij} u_i)^{-1} y_{ij} (v_{ij} u_i). \stopformula Per tant, $w$ es pot expressar com a producte de conjugats de $Y$. \stopmydemo \startmytheorem[thmi:lambda-0-1-finitament-presentat] Qualsevol grup $G \= in {\cal S}(\lambda_{0,1}, f(n)=3Dn-1)$ =C3=A9s finitament presentat. M=C3= =A9s concretament, si $X$ =C3=A9s un conjunt finit de generadors de $G$, = $n_0 \in \naturalnumbers$ i $\sigma \colon G \to {(X \cup X^{-1})}^*$ una= secci=C3=B3 tal que $\lambda_{0,1} (n) < n-1$ per a $n \geq n_0$, alesho= res existeix una presentaci=C3=B3 finita ${\cal P}$ de $G$ de la forma $\= langle X \mid R\rangle$, amb \startformula R =3D \{w =3D 1 \mid w \in {(X \cup X^{-1})}^*, \pi(w) =3D 1, l(w) \leq 2= n_0, \}, \stopformula amb $\pi {(X \cup X^{-1})}^* \to G$ el morfisme exhaustiu definit anterio= rment. \stopmytheorem \startmydemo Com que $G \in {\cal S}(\lambda_{0,1}, f(n)=3Dn-1)$, aleshor= es existeixen $X$ un conjunt finit de generadors de $G$ i $\sigma \colon = G \to {(X \cup X^{-1})}^*$ una secci=C3=B3 tal que $\lambda_{0,1} (n) < n= -1$ per a $n$ suficientment gran. Sigui $n_0 \in \naturalnumbers$ tal que= $\lambda_{0,1} (n) < n-1$ per a tot $n \geq n_0$. Com que $X$ =C3=A9s un conjunt finit de generadors, tenim el morfisme exh= austiu $\pi \colon {(X \cup X^{-1})}^* \to G$. Siguin el conjunt de relac= ions \startformula R =3D \{w =3D 1 \mid w \in {(X \cup X^{-1})}^*, \pi(w) =3D 1, l(w) \leq 2= n_0, \}, \stopformula el qual =C3=A9s sim=C3=A8tric per ser $\pi$ un morfisme, i la presentaci=C3= =B3 finita ${\cal P} =3D \langle X \mid R \rangle$. Volem veure que ${\cal P}$ =C3=A9s una presentaci=C3=B3 (finita) de $G$. = A l'igual que en la demostraci=C3=B3 de la Proposici=C3=B3~\in[thmi:propo= sicio-fites-funcions-Dehn-asincronics-longitud], hem de veure que tota pa= raula $w \in {(X \cup X^{-1})}^*$ tal que $\pi(w) =3D 1$ =C3=A9s nul-homo= t=C3=B2pica per ${\cal P}$. Geom=C3=A8tricament, aquest fet =C3=A9s equiv= alent a provar l'exist=C3=A8ncia d'un diagrama de van Kampen ${\cal D}_w$= per a $w$ respecte de ${\cal P}$. Ho veurem per inducci=C3=B3 sobre $l(w= )$. \startitemize[1] \item Si $l(w) \leq 2n_0$, aleshores podem prendre el diagrama de van Kam= pen que t=C3=A9 frontera $w$ i una sola cara. Aquest diagrama existeix pe= rqu=C3=A8, en aquest cas, $w$ =C3=A9s una paraula de $R$. \item Suposem-ho cert fins a $l(w) =3D n$ i provem-ho per a $l(w)=3D n+1$= =2E Siguin \startformula=20 m =3D \max \{l(\sigma_{\pi(w(i))}) \mid i=3D 0, \ldots, l(w)\} \leq L_{\s= igma} (l(w)/2) \stopformula i $I =3D \{0, \ldots, l(w)\} \times \{0, \ldots, m \}$. Considerem $\Lambda$ el CW-complex planar de dimensi=C3=B3 $2$ format per= $m+1$ segments horitzontals i $l(w)+1$ segments verticals el qual divide= ix el rectangle de llarg=C3=A0ria $l(w)$ i altura $m$ en rectangles d'=C3= =A0rea unitat i que t=C3=A9 $I$ com a conjunt de v=C3=A8rtexos. Transform= em $\Lambda$ en un CW-complex planar i dirigit $\tilde{\Lambda}$ de dimen= si=C3=B3 $2$ amb els v=C3=A8rtexos i els arcs etiquetats mitjan=C3=A7ant = els passos seg=C3=BCents (el qual est=C3=A0 representat, esquem=C3=A0tica= ment, a la Figura~\in[fig:diagrama-van-Kampen-per-a-lambda-0-1]): \startitemize[1] \item L'assignaci=C3=B3 de l'etiqueta $\sigma_{\pi(w(i))}(j)$ al v=C3=A8r= tex $(i, j) \in I$. En particular, $(i, m)$ t=C3=A9 $\pi(w(i))$ com a eti= queta (els v=C3=A8rtexos $(i, j)$ amb $j < m$ poden tenir, tamb=C3=A9, aq= uesta etiqueta). \item Per a tot v=C3=A8rtex $(i, j)$ amb $i \not \in \{0, l(w)\}$, l'etiq= uetatge de l'aresta que va des de $(i, j)$ a $ (i,j+1)$ amb la lletra $(j= +1)$-=C3=A8ssima de la paraula $\sigma_{\pi(w(i))}$, i l'assignaci=C3=B3 = del sentit d'aquesta aresta cap a $(i,j+1)$. =C3=89s a dir, les etiquetes= de les arestes del segment vertical $i$-=C3=A8ssim, llegides d'abaix a d= alt, formen la paraula $\sigma_{\pi(w(i))}$. D'altra banda, etiquetarem t= otes les arestes dels segments $0$-=C3=A8ssim i $l(w)$-=C3=A8ssim amb la = paraula $\varepsilon$ i els dirigirem cap a dalt. \item Per a cada $(i, j), (i+1, j) \in I$ amb $j \neq 0$, considerem la p= araula $w_{i, j} \in {(X \cup X^{-1})}^*$ tal que $\gamma(w_{i, j})$ =C3=A9= s un cam=C3=AD geod=C3=A8sic de $\sigma_{\pi(w(i))}(j)$ a $\sigma_{\pi(w(= i+1))}(j)$ dins el graf de Cayley $\Gamma_{G, X}$. A m=C3=A9s, dividim l'= aresta que va de $(i, j)$ a $(i+1, j)$ dins $\Lambda$ en $l(w_{i, j})$ se= gments dirigits, els quals determinaran $l(w_{i,j})+1$ punts. Dos d'aques= ts punts seran $\sigma_{\pi(w(i))}(j)$ i $\sigma_{\pi(w(i+1))}(j)$ (els e= xtrems). Etiquetarem cadascun d'aquests punts, successivament des de $(i,= j)$ fins a $(i+1,j)$, amb les lletres de $w_{i, j}$ i dirigirem tots els= segments cap a $(i+1, j)$. En particular, d'aquesta manera, les etiquete= s dels arcs del segment horitzontal $m$-=C3=A8ssim de $\tilde{\Lambda}$, = llegides de de $(0,m)$ fins a $(l(w), m)$ formen la paraula $w$. Quan $j =3D 0$, prendrem $w_{i, j} =3D \varepsilon$ en sentit cap a $(i, = j)$. Aix=C3=AD etiquetarem l'aresta que va des de $(i, j)$ a $(i, j+1)$ a= mb $\varepsilon$ i sentit de $(i, j+1)$ cap a $(i,j)$. \stopitemize \placefigure [here] [fig:diagrama-van-Kampen-per-a-lambda-0-1] {El diagrama $\tilde{\Lambda}$ contru=C3=AFt a partir de $\Lambda$ i $\= sigma$.} {\startcombination[1*1] { \starttikzpicture[scale=3D1] % punts \filldraw[color=3Dblue!50] (0,13) circle (2pt); \filldraw[color=3Dblue!50] (1,13) circle (2pt); \filldraw[color=3Dblue!50] (4,13) circle (2pt); \filldraw[color=3Dblue!50] (5,13) circle (2pt); \filldraw[color=3Dblue!50] (8,13) circle (2pt); \filldraw[color=3Dblue!50] (9,13) circle (2pt); \filldraw[color=3Dblue!50] (0,12) circle (2pt); \filldraw[color=3Dblue!50] (1,12) circle (2pt); \filldraw[color=3Dblue!50] (8,12) circle (2pt); \filldraw[color=3Dblue!50] (9,12) circle (2pt); \filldraw[color=3Dblue!50] (0,11) circle (2pt); \filldraw[color=3Dblue!50] (1,11) circle (2pt); \filldraw[color=3Dblue!50] (8,11) circle (2pt); \filldraw[color=3Dblue!50] (9,11) circle (2pt); \filldraw[color=3Dblue!50] (0,6) circle (2pt); \filldraw[color=3Dblue!50] (0,7) circle (2pt); \filldraw[color=3Dblue!50] (4,6) circle (2pt); \filldraw[color=3Dblue!50] (4,7) circle (2pt); \filldraw[color=3Dblue!50] (5,6) circle (2pt); \filldraw[color=3Dblue!50] (5,7) circle (2pt); \filldraw[color=3Dblue!50] (8,6) circle (2pt); \filldraw[color=3Dblue!50] (8,7) circle (2pt); \filldraw[color=3Dblue!50] (9,6) circle (2pt); \filldraw[color=3Dblue!50] (9,7) circle (2pt); \filldraw[color=3Dblue!50] (0,2) circle (2pt); \filldraw[color=3Dblue!50] (1,2) circle (2pt); \filldraw[color=3Dblue!50] (8,2) circle (2pt); \filldraw[color=3Dblue!50] (9,2) circle (2pt); \filldraw[color=3Dblue!50] (0,1) circle (2pt); \filldraw[color=3Dblue!50] (1,1) circle (2pt); \filldraw[color=3Dblue!50] (8,1) circle (2pt); \filldraw[color=3Dblue!50] (9,1) circle (2pt); \filldraw[color=3Dblue!50] (0,0) circle (2pt); \filldraw[color=3Dblue!50] (1,0) circle (2pt); \filldraw[color=3Dblue!50] (4,0) circle (2pt); \filldraw[color=3Dblue!50] (5,0) circle (2pt); \filldraw[color=3Dblue!50] (8,0) circle (2pt); \filldraw[color=3Dblue!50] (9,0) circle (2pt); % Noms \draw (0, 11.5) node[left] {$\varepsilon$}; \draw (0, 6.5) node[left] {$\varepsilon$}; \draw (0, 1.5) node[left] {$\varepsilon$}; \draw (0, 0.5) node[left] {$\varepsilon$}; \draw (9, 0.5) node[right] {$\varepsilon$}; \draw (9, 1.5) node[right] {$\varepsilon$}; \draw (9, 6.5) node[right] {$\varepsilon$}; \draw (9, 11.5) node[right] {$\varepsilon$}; \draw (0.5, 0) node[below] {$\varepsilon$}; \draw (4.5, 0) node[below] {$\varepsilon$}; \draw (8.5, 0) node[below] {$\varepsilon$}; \draw (0, 13) node[above] {$\pi(w(l(w)))=3D1$}; \draw (3.5, 13) node[above] {$\pi(w(i+1))$}; \draw (5.5, 13) node[above] {$\pi(w(i))$}; \draw (9, 13) node[above] {$\pi(w(0))=3D1$}; \draw (4, 4) node[left] {$\sigma_{\pi(w(i+1))}$}; \draw (5, 4) node[right] {$\sigma_{\pi(w(i))}$}; %\draw (5, 5.7) node[right] {$\sigma_{\pi(w(i))}(j)$}; %\draw (5, 7.3) node[right] {$\sigma_{\pi(w(i))}(j+1)$}; %\draw (4, 5.7) node[left] {$\sigma_{\pi(w(i+1))}(j)$}; %\draw (4, 7.3) node[left] {$\sigma_{\pi(w(i+1))}(j+1)$}; \draw (4.5, 6) node[below] {$w_{i,j}$}; \draw (4.5, 7) node[above] {$w_{i,j+1}$}; \draw (2.5, 10) node {$\vdots$}; \draw (2.5, 3) node {$\vdots$}; \draw (6.5, 10) node {$\vdots$}; \draw (6.5, 3) node {$\vdots$}; % L=C3=ADnies \draw (0,0) -- (1,0) -- (2,0) -- (3,0) -- (4,0) -- (5,0) -- (6,0) -- (7,0= ) -- (8, 0) -- (9, 0); \draw (0,13) -- (1,13) -- (2,13) -- (3,13) -- (4,13) -- (5,13) -- (6,13) = -- (7,13) -- (8, 13) -- (9, 13); \foreach \x in {0,1,...,12} {\draw (0, \x) -- (0,\x+1);} \foreach \x in {0,1,...,12} {\draw (9, \x) -- (9,\x+1);} \foreach \x in {1,2,11,12} {\draw (0,\x) -- (1,\x); \draw (1,\x) -- (8,\x); %\draw (7,\x) -- (8,\x); \draw (8,\x) -- (9,\x);} \foreach \x in {6,7} {\draw (0,\x) -- (1,\x); \draw (1,\x) -- (3,\x); \draw (3,\x) -- (4,\x) -- (5,\x) -- (6,\x); \draw (6,\x) -- (8,\x); \draw (8,\x) -- (9,\x);} \foreach \y in {0,1,...,12} {\draw (1,\y) -- (1,\y+1);} \foreach \y in {0,1,...,12} {\draw (4,\y) -- (4,\y+1);} \foreach \y in {0,1,...,12} {\draw (5,\y) -- (5,\y+1);} \foreach \y in {0,1,...,12} {\draw (8,\y) -- (8,\y+1);} % Sentit \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green!50,line width=3D1mm]{>}}}] (1,0) -- (4,0); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green!50,line width=3D1mm]{>}}}] (5,0) -- (8,0); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green!50,line width=3D1mm]{>}}}] (9,2) -- (9,6); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green!50,line width=3D1mm]{>}}}] (9,7) -- (9,11); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green!50,line width=3D1mm]{>}}}] (0,2) -- (0,6); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green!50,line width=3D1mm]{>}}}] (0,7) -- (0,11); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green!50,line width=3D1mm]{>}}}] (4,2) -- (4,6); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green!50,line width=3D1mm]{>}}}] (4,7) -- (4,11); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green!50,line width=3D1mm]{>}}}] (5,2) -- (5,6); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green!50,line width=3D1mm]{>}}}] (5,7) -- (5,11); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green!50,line width=3D1mm]{>}}}] (4,13) -- (1,13); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green!50,line width=3D1mm]{>}}}] (8,13) -- (5,13); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green!50,line width=3D1mm]{>}}}] (5,6) -- (4,6); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green!50,line width=3D1mm]{>}}}] (5,7) -- (4,7); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green!50,line width=3D1mm]{>}}}] (4,1) -- (1,1); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green!50,line width=3D1mm]{>}}}] (8,1) -- (5,1); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green!50,line width=3D1mm]{>}}}] (4,2) -- (1,2); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green!50,line width=3D1mm]{>}}}] (8,2) -- (5,2); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green!50,line width=3D1mm]{>}}}] (4,11) -- (1,11); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green!50,line width=3D1mm]{>}}}] (8,11) -- (5,11); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green!50,line width=3D1mm]{>}}}] (4,12) -- (1,12); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green!50,line width=3D1mm]{>}}}] (8,12) -- (5,12); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green!50,line width=3D1mm]{>}}}] (4,6) -- (1,6); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green!50,line width=3D1mm]{>}}}] (8,6) -- (5,6); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green!50,line width=3D1mm]{>}}}] (4,7) -- (1,7); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green!50,line width=3D1mm]{>}}}] (8,7) -- (5,7); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green!50,line width=3D1mm]{>}}}] (1,2) -- (1,6); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green!50,line width=3D1mm]{>}}}] (1,7) -- (1,11); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green!50,line width=3D1mm]{>}}}] (8,2) -- (8,6); \draw[decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow[= green!50,line width=3D1mm]{>}}}] (8,7) -- (8,11); \draw[loosely dashed] (6,14.3)-- (3,14.3); \draw[loosely dashed] (6,-1.3)-- (3,-1.3); \draw[loosely dashed] (-1.3,5)-- (-1.3,8); \draw[loosely dashed] (10.3,5)-- (10.3,8); \draw[decorate,decoration=3D{markings,mark=3Dat position .45 with {\arrow= [red!50,line width=3D0.6mm]{angle 60}}}] (6,14.3) -- (3,14.3); \draw[decorate,decoration=3D{markings,mark=3Dat position .55 with {\arrow= [red!50,line width=3D0.6mm]{angle 60}}}] (6,14.3) -- (3,14.3); \draw[decorate,decoration=3D{markings,mark=3Dat position .45 with {\arrow= [red!50,line width=3D0.6mm]{angle 60}}}] (3,-1.3)-- (6,-1.3); \draw[decorate,decoration=3D{markings,mark=3Dat position .55 with {\arrow= [red!50,line width=3D0.6mm]{angle 60}}}] (3,-1.3)-- (6,-1.3); \draw[decorate,decoration=3D{markings,mark=3Dat position .45 with {\arrow= [red!50,line width=3D0.6mm]{angle 60}}}] (-1.3,5)-- (-1.3,8); \draw[decorate,decoration=3D{markings,mark=3Dat position .55 with {\arrow= [red!50,line width=3D0.6mm]{angle 60}}}] (-1.3,5)-- (-1.3,8); \draw[decorate,decoration=3D{markings,mark=3Dat position .45 with {\arrow= [red!50,line width=3D0.6mm]{angle 60}}}] (10.3,5)-- (10.3,8); \draw[decorate,decoration=3D{markings,mark=3Dat position .55 with {\arrow= [red!50,line width=3D0.6mm]{angle 60}}}] (10.3,5)-- (10.3,8); \draw (4.5, 14.6) node {$w$}; \draw (4.5, -1.6) node {$\varepsilon$}; \draw (-1.6, 6.5) node {$\varepsilon$}; \draw (10.6, 6.5) node {$\varepsilon$}; \stoptikzpicture} { } \stopcombination} \indentation Per a cada cara de $\tilde{\Lambda}$ determinada pels v=C3=A8= rtexos $(i, j)$, $(i,j+1)$, $(i+1,j+1)$ i $(i+1,j) \in I$, sigui $u_{i, j= }$ la paraula formada llegint les etiquetes de la seva frontera des de $(= i,j)$ en el sentit contrari a les agulles del rellotge. Pel Lema~\in[thmi= :lema-distancia-menor-w-2] i perqu=C3=A8 els camins corresponents a $w_{i= ,j}$ i $w_{i,j+1}$ s=C3=B3n geod=C3=A8sics, \startformula l(w_{i,j}) + l(w_{i,j+1}) \leq 2 \lambda_{0, 1} (l(w)/2), \stopformula per a tots $(i, j), (i, j+1) \in I$. Per tant, com que $\lambda_{0, 1} (n= ) < n-1$ per a tot $n \geq n_0$, \startformula l(u_{i, j}) \leq 2+2 \lambda_{0, 1} (l(w)/2) < l(w). \stopformula \noindentation D'altra banda, $\pi(u_{i, j}) =3D 1$ (ja que, per construc= ci=C3=B3, forma un cicle dins el graf de Cayley $\Gamma_{G, X}$). Per tan= t, $\tilde{\Lambda}$ =C3=A9s un diagrama de van Kampen per a $w$ respecte= de ${\cal Q} =3D \langle X \mid S\rangle$ amb $S$ el conjunt de relacion= s \startformula S =3D \{ u =3D 1 \mid u \in {(X \cup X^{-1})}^*, \pi(u) =3D 1, l(u) < l(w= )\}. \stopformula \noindentation Per tant, pels lemes~\in[thmi:lema-de-van-Kampen] i \in[th= mi:w-nul-homotopica-producte-conjugats], $w$ es pot posar com a producte = de conjugats de $u_{i, j}$ i les seves inverses. Com que $l(u_{i, j}) < l= (w)$, per hip=C3=B2tesi d'inducci=C3=B3, tenim que existeix un diagrama d= e van Kampen $D_{i, j}$ de frontera $u_{i, j}$ respecte de ${\cal P}$. Pe= r tant, cada $u_{i, j}$ es pot posar com a producte de conjugats de parau= les de $R$ (els conjugats de les inverses de paraules de $R$ tamb=C3=A9 s= =C3=B3n paraules de $R$, ja que $R$ =C3=A9s sim=C3=A8tric). Combinant aqu= ests dos fets, pel Lema~\in[thmi:lema-dels-conjugats-de-conjugats], $w$ e= s pot posar com a producte de conjugats de paraules de $R$, per la qual c= osa existeix un diagrama de van Kampen $D_w$ per a $w$ sobre ${\cal P}$. \stopitemize \stopmydemo \startmycorollary Siguin $s \leq k \in \naturalnumbers$ i un grup $G$. Si= $G \in {\cal S} (\lambda_{s, k}, f(n) =3D n-(k-s))$, aleshores $G$ =C3=A9= s finitament presentat. \stopmycorollary \startmydemo =C3=89s conseq=C3=BC=C3=A8ncia directa de la Proposici=C3=B3= ~\in[thmi:proposicio-reduccio-lambdes] i del Teorema~\in[thmi:lambda-0-1-= finitament-presentat]. \stopmydemo \subsection{L'ordre de la funci=C3=B3 de Dehn dels grups de ${\cal S}(\la= mbda_{0,1}, f(n)=3Dn-1)$} \startmylema[thmi:lema-tecnic-nombre-de-subdiagrames-lambda-01] Siguin $G= $ un grup, $X$ un conjunt finit de generadors de $G$ tal que $X =3D X^{-1= }$, ${\cal P} =3D \langle X \mid R\rangle$ una presentaci=C3=B3 finita de= $G$ fixada i $\sigma \colon G \to X^*$ una secci=C3=B3. Per a qualsevol = paraula $w \in X^*$ nul-homot=C3=B2pica per ${\cal P}$, existeix un diagr= ama de van Kampen ${\cal D}_w$ per a $w$ respecte de ${\cal P}$ el qual e= s pot expressar com la uni=C3=B3 de com a m=C3=A0xim \startformula l(w) \cdot ({\lvert X \rvert}^{l(w) \lambda_{0,1} (l(w)/2)} +1) \stopformula subdiagrames 1-connectats, cadascun dels quals t=C3=A9 per=C3=ADmetre com= a m=C3=A0xim \startformula 2 \lambda_{0,1} (l(w)/2) +2. \stopformula \stopmylema \startmydemo Sigui $w$ una paraula nul-homot=C3=B2pica per ${\cal P}$. Co= nsiderem $\tilde{\Lambda}$ el diagrama planar dirigit i etiquetat constru= =C3=AFt a partir de $w$ i $\sigma$ com en el Teorema~\in[thmi:lambda-0-1-= finitament-presentat] (Figura~\in[fig:diagrama-van-Kampen-per-a-lambda-0-= 1]). Cada segment de $\tilde{\Lambda}$ t=C3=A9 associada una paraula sobr= e $X$, la qual correspon a llegir les seves etiquetes de forma consecutiv= a a partir d'un dels seus extrems. Si indiquem amb $h(j), v(i)$ les parau= les associades al segment horitzontal $j$-=C3=A8ssim i al segment vertica= l $i$-=C3=A8ssim, respectivament (amb la convenci=C3=B3 que comencem a co= mptar des dels segments inferior i dret), aleshores $h(0) =3D v(0) =3D v(= l(w)) =3D \varepsilon$ i $v(i) =3D \sigma_{\pi(w(i))}$, per a tot $i \in = \{1, \ldots, l(w)-1\}$. D'altra banda, per a qualsevol $j \neq 0$, $h(j)$= est=C3=A0 formada per la concatenaci=C3=B3 de $l(w)$ paraules, $w_{i, j}= $, corresponents a camins geod=C3=A8sics entre $(i, j)$ i $(i+1,j)$. Aque= stes paraules s=C3=B3n tals que \startformula l(w_{i, j}) + l(w_{i, j+1}) \leq 2 \lambda_{0, 1} (l(w)/2), \stopformula per a qualsevol $i \in \{0, \ldots, l(w)\}$. Per tant, \startformula l(h(j)) + l(h(j+1)) \leq 2 l(w) \cdot \lambda_{0, 1} (l(w)/2). \stopformula Aix=C3=B2 vol dir que, parell a parell, la suma de les longituds de les p= araules associades als segments horitzontals =C3=A9s menor o igual que $2= l(w) \cdot \lambda_{0, 1} (l(w)/2)$. Aix=C3=B2 implica que una de les pa= raules t=C3=A9 longitud menor que $l(w) \cdot \lambda_{0, 1} (l(w)/2)$: s= i $l(h(j)) \leq l(w) \cdot \lambda_{0, 1} (l(w)/2)$, aleshores ho tenim. = Si $l(h(j)) > l(w) \cdot \lambda_{0, 1} (l(w)/2)$, aleshores \startformula \startsplit \NC l(h(j+1)) \NC \leq 2l(w) \cdot \lambda_{0, 1} (l(w)/2) - l(w) \cdot \= lambda_{0, 1} (l(w)/2)\NR \NC \NC \leq l(w) \cdot \lambda_{0, 1} (l(w)/2). \stopsplit \stopformula Per tant, com a m=C3=A0xim hi pot haver ${\lvert X \rvert}^{l(w) \cdot \l= ambda_{0, 1} (l(w)/2)}$ parells amb les dues paraules diferents. Per tant= , com a m=C3=A0xim hi ha $1+{\lvert X \rvert}^{l(w) \cdot \lambda_{0, 1} = (l(w)/2)}$ paraules diferents corresponents a segments horitzontals (Figu= ra~\in[fig:raonament-parells-maxim-paraules]). \placefigure [here] [fig:raonament-parells-maxim-paraules] {Esquema dels parells associats als segments horitzontals de $\tilde{\L= ambda}$.} {\startcombination[1*1] { \starttikzpicture[scale=3D1] % punts \foreach \x in {0,1,2,3,4,7,8,9} {\draw (\x,0) -- (\x,0.5);} \draw (0, 0.5) node[above] {$h(0)$}; \draw (1, 0.5) node[above] {$h(1)$}; \draw (2, 0.5) node[above] {$h(2)$}; \draw (3, 0.5) node[above] {$\ldots$}; \draw (5.5, 0) node {$\ldots$}; \draw (0.5, 0) node[below] {$\underbrace{\; }_{ \startsubstack \text{parell} \NR \text{n=C3=BAm. 0}\NR \stopsubstack}$}; \draw (1.5, 0) node[below] {$\underbrace{\; }_{ \startsubstack \text{parell} \NR \text{n=C3=BAm. 1}\NR \stopsubstack}$}; \draw (2.5, 0) node[below] {$\underbrace{\; }_{ \startsubstack \text{parell} \NR \text{n=C3=BAm. 2}\NR \stopsubstack}$}; \draw (3.5, 0) node[below] {$\underbrace{\; }_{\;}$}; \draw (7.5, 0) node[below] {$\underbrace{\; }_{\;}$}; \draw (8.5, 0) node[below] {$\underbrace{\; }_{\;}$}; \draw (3.5, -1) node[above] {$\ldots$}; % Noms %\draw (1.5, 1.2) node {$x$}; %\draw (0.7, 1.4) node {$y$}; %\draw (0.7, 3) node {$1$}; \stoptikzpicture} { } \stopcombination} En el cas en qu=C3=A8 dues paraules corresponents a segments horitzontals= diferents siguin iguals, podem eliminar la part de $\tilde{\Lambda}$ que= est=C3=A0 entre elles. D'aquesta manera, podem construir un diagrama $\D= elta$ que t=C3=A9 $l(w)+1$ l=C3=ADnies verticals (les mateixes que $\tild= e{\Lambda}$) i com a m=C3=A0xim $1+{\lvert X \rvert}^{l(w) \cdot \lambda_= {0, 1} (l(w)/2)}$ l=C3=ADnies horitzontals\footnote{Aquestes t=C3=A8cniqu= es de {\em cirugia} es poden trobar a diverses refer=C3=A8ncies \cite[bri= dson, epstein].}. Com que $\tilde{\Lambda}$ =C3=A9s un diagrama de van Kampen per a $w$ amb= les paraules de la frontera de les seves cares formant paraules de longi= tud menor que $l(w)$, aleshores, per inducci=C3=B3, existeix un diagram d= e van Kampen respecte de ${\cal P}$ per a la paraula corresponent a cada = cara de $\tilde{\Lambda}$. Per tant, podem encastar aquests diagrames a l= es cares de $\tilde{\Lambda}$. Com que $\Delta$ =C3=A9s un subdiagrama de= $\tilde{\Lambda}$, llavors tamb=C3=A9 podem encastar aquests diagrames a= $\Delta$. Per tant, $\Delta$ =C3=A9s un diagrama de van Kampen per a $w$= respecte de ${\cal P}$ format per, com a m=C3=A0xim, \startformula l(w) \cdot ({\lvert X \rvert}^{l(w) \cdot \lambda_{0, 1} (l(w)/2)} +1 ) \stopformula subdiagrames 1-connectats (els diagrames de van Kampen de les cares de $\= Delta$) de per=C3=ADmetre com a m=C3=A0xim $2 \lambda_{0,1} (l(w)/2) +2$.= \stopmydemo \startmytheorem[thmi:funcio-isoperimetrica-lambda01] Sigui $G$ un grup, $= X$ un conjunt finit de generadors de $G$ tal que $X =3D X^{-1}$, $\sigma = \colon G \to X^*$ una secci=C3=B3 tal que $\lambda_{0,1} (n) < n-1$ per a= $n$ suficientment gran, i $F \colon \naturalnumbers \to \naturalnumbers$= tal que $F(n) \geq 1$, per a tot $n \geq 1$. Si \startformula F(n) \geq n ({\lvert X \rvert}^{n \lambda_{0, 1} (n/2)}+1 ) F\big(2 \lamb= da_{0, 1} (n/2) + 2 \big), \stopformula per a $n$ suficientment gran, aleshores $F$ =C3=A9s una funci=C3=B3 isope= rim=C3=A8trica per a qualque presentaci=C3=B3 finita de $G$. \stopmytheorem \startmydemo Sigui $n_1 \in \naturalnumbers$ tal que, per a tot $n \geq n= _1$, \startformula F(n) \geq n ({\lvert X \rvert}^{n \lambda_{0, 1} (n/2)}+1 ) F\big(2 \lamb= da_{0, 1} (n/2) + 2 \big) \stopformula i $\lambda_{0,1} (n) < n-1$. Pel Teorema~\in[thmi:lambda-0-1-finitament-p= resentat], ${\cal P} =3D \langle X \mid R\rangle$ amb \startformula R =3D \{w =3D 1 \mid w \in {(X \cup X^{-1})}^*, \pi(w) =3D 1, l(w) \leq 2= n_1, \} \stopformula =C3=A9s una presentaci=C3=B3 finita de $G$. Vegem que $F$ =C3=A9s una fun= ci=C3=B3 isoperim=C3=A8trica per a ${\cal P}$, =C3=A9s a dir, que, per a = tota paraula $w \in {(X \cup X^{-1})}^*$ nul-homot=C3=B2pica per ${\cal P= }$ tal que $l(w) \leq n$, tenim que $\text{area}_{\cal P} (w) \leq F(n)$.= Demostrem-ho per inducci=C3=B3 sobre $l(w)$: \startitemize[1] \item Si $w$ =C3=A9s una paraula nul-homot=C3=B2pica per ${\cal P}$ tal q= ue $l(w) \leq 2n_1$, aleshores existeix un diagrama de van Kampen per a $= w$ amb una =C3=BAnica cara, ja que $w$ =C3=A9s una paraula de $R$. Per ta= nt, $\text{area}_{\cal P} (w) =3D 1$, que =C3=A9s menor o igual que $F(2n= _1)$ per hip=C3=B2tesi. \item Suposem que $w$ =C3=A9s una paraula nul-homot=C3=B2pica per ${\cal = P}$ tal que $l(w) =3D n > 2n_1$ i que qualsevol paraula nul-homot=C3=B2pi= ca per ${\cal P}$ de longitud $r < n$ =C3=A9s frontera d'un diagrama de v= an Kampen respecte de ${\cal P}$ que t=C3=A9 com a m=C3=A0xim $F(r)$ care= s. Aleshores, pel Lema~\in[thmi:lema-tecnic-nombre-de-subdiagrames-lambda= -01], existeix un diagrama de van Kampen per a $w$ respecte de ${\cal P}$= el qual t=C3=A9 com a m=C3=A0xim \startformula n ({\lvert X \rvert}^{n \lambda_{0, 1} (n/2)}+1 ) F\big(2 \lambda_{0, 1} = (n/2) + 2 \big) \stopformula cares. Per tant, aplicant la hip=C3=B2tesi sobre $F$, tenim que aquest no= mbre =C3=A9s menor o igual que $F(n)$. \stopitemize \stopmydemo \startmytheorem[thmi:funcio-Dehn-lambda-0-1] Sigui $G$ un grup. Si $G \in= {\cal S} (\lambda_{0,1},f(n) =3D n-1)$, aleshores existeix $k > 0$ tal q= ue \startformula \delta_G(n) \preceq e^{kn^3}. \stopformula \stopmytheorem \startmydemo Com que $G \in {\cal S} (\lambda_{0,1},f(n) =3D n-1)$, alesh= ores existeixen $X$ un conjunt finit de generadors de $G$ tal que $X =3D = X^{-1}$ i $\sigma \colon G \to X^*$ tal que $\lambda_{0,1} (n) < n-1$ per= a $n$ suficientment gran. Vegem que existeix $k > 0$ tal que la funci=C3=B3 $n \mapsto e^{kn^3}$ =C3= =A9s isoperim=C3=A8trica per a qualque presentaci=C3=B3 de $G$. Pel Teore= ma~\in[thmi:funcio-isoperimetrica-lambda01], basta veure que existeix $k = > 0$ tal que, per a $n$ suficientment gran, \startformula kn^3 \geq \ln n + \ln ({\lvert X \rvert}^{n \lambda_{0, 1} (n/2)}+1 ) + = k (2 \lambda_{0, 1} (n/2) + 2)^3. \stopformula Ara b=C3=A9, per a $n$ suficientment gran, tenim que \startformula \startsplit \NC \ln n \NC + \ln ({\lvert X \rvert}^{n \lambda_{0, 1} (n/2)}+1 ) + k = (2 \lambda_{0, 1} (n/2) + 2)^3 \NR \NC \NC \leq \ln n + \ln (2{\lvert X \rvert}^{n \lambda_{0, 1} (n/2)}) + = k (2 \lambda_{0, 1} (n/2) + 2)^3 \NR \NC \NC \leq \ln n + \ln 2 + n \lambda_{0, 1} (n/2)\ln \lvert X \rvert + = k (2 \lambda_{0, 1} (n/2) + 2)^3\NR \NC \NC \leq \ln n + \ln 2 + n (n/2-2)\ln \lvert X \rvert + k (n-2)^3\NR \NC \NC \leq kn^3 + n^2 (\frac{1}{2} \ln \lvert X \rvert - 6k) + o(n^2). \stopsplit \stopformula Per tant, prenent $k > \frac{1}{12} \ln \lvert X \rvert$, tenim que aix=C3= =B2 =C3=A9s menor o igual que $kn^3$ per a $n$ suficientment gran. Aleshores $n \mapsto e^{kn^3}$ =C3=A9s una funci=C3=B3 isoperim=C3=A8tric= a per a qualque presentaci=C3=B3. Per tant, la funci=C3=B3 de Dehn d'aque= sta presentaci=C3=B3 =C3=A9s menor o igual que $e^{kn^3}$. Per tant, la f= unci=C3=B3 de Dehn de $G$ =C3=A9s $\preceq e^{kn^3}$. \stopmydemo \startmycorollary Siguin $s \leq k \in \naturalnumbers$ i $G$ un grup. Si= $G \in {\cal S} (\lambda_{s, k}, f(n) =3D n-(k-s))$, aleshores existeix = $k > 0$ tal que $\delta_G \preceq e^{kn^3}$. En particular, $G$ t=C3=A9 e= l problema de la paraula resoluble. \stopmycorollary \startmydemo De la Proposici=C3=B3~\in[thmi:proposicio-reduccio-lambdes] = i del Teorema~\in[thmi:funcio-Dehn-lambda-0-1], s'estableix directament a= quest fet. \stopmydemo \subsubsection{Els grups de ${\cal S} (\lambda_{0, k}, f(n) =3D n-1)$ amb= seccions geod=C3=A8siques} En aquest apartat, trobarem una fita superior per a les funcions de Dehn = dels grups que admeten una secci=C3=B3 geod=C3=A8sica $\sigma$ tal que, p= er a qualque $k \geq 0$, $\lambda_{\sigma, 0, k} (n) < n-k$. per a $n$ su= ficientment gran. Quan $k > 1$, aquesta fita superior =C3=A9s estrictamen= t m=C3=A9s petita que la fita superior corresponent als grups que admeten= una secci=C3=B3 geod=C3=A8sica $\kappa$ tal que $\varphi_{\kappa} (n) < = n-1$ per a $n$ suficientment gran. \startmyproposition[thmi:desigualtat-dehn-p-lambda-0-k] Siguin $G$ un gru= p, $X$ un conjunt finit de generadors de $G$, ${\cal P} =3D \langle X \mi= d R \rangle$ una presentaci=C3=B3 finita de $G$, $k \geq 0$ i $\sigma \co= lon G \to {(X \cup X^{-1})}^*$ una secci=C3=B3 geod=C3=A8sica. Si $\sigma= $ =C3=A9s tal que $\lambda_{0, k} (n) < n-k$ per a $n$ suficientment gran= , aleshores \startformula \delta_{\cal P} (n) \leq \frac{n^2}{2k} \delta_{\cal P} (n-2). \stopformula \stopmyproposition \startmydemo Sigui $w$ una paraula nul-homot=C3=B2pica per ${\cal P}$. Sigui $m$ l'ent= er m=C3=A9s gran tal que $l(w)/2 \geq m \cdot k$, el qual coincideix amb = la part entera $\lfloor (l(w)/2)/k \rfloor$. Considerem $u_{i, j}$ les pa= raules nul-homot=C3=B2piques per ${\cal P}$ definides per: \startitemize[1] \item Per a tots $i \in \{0, \ldots, l(w)-1\}$ i $j \in \{1, 2, \ldots, m= \}$, la paraula $u_{i,j}$ est=C3=A0 formada per la concatenaci=C3=B3, en = aquest ordre, de les paraules seg=C3=BCents: \startitemize[2] \item La paraula corresponent a un cam=C3=AD geod=C3=A8sic des de $\sigma= _{\pi(w(i))} (mj)$ a $\sigma_{\pi(w(i+1))} (mj)$. \item La subparaula de $\sigma_{\pi(w(i+1))}$ tal que el seu corresponent= cam=C3=AD va des de $\sigma_{\pi(w(i+1))} (mj)$ a $\sigma_{\pi(w(i+1))} = (m(j-1))$. \item La paraula corresponent a un cam=C3=AD geod=C3=A8sic des de $\sigma= _{\pi(w(i+1))} (m(j-1))$ a $\sigma_{\pi(w(i))} (m(j-1))$. \item La subparaula de $\sigma_{\pi(w(i))}$ tal que el seu corresponent c= am=C3=AD va des de $\sigma_{\pi(w(i))} (m(j-1))$ fins a $\sigma_{\pi(w(i)= )} (mj)$. \stopitemize \item Per a tots $i \in \{0, \ldots, l(w)-1\}$ i $j =3D m+1$, $u_{i,j}$ =C3= =A9s la concatenaci=C3=B3, en aquest ordre, de: \startitemize[2] \item La lletra $(i+1)$-=C3=A8ssima de $w$. \item La subparaula de $\sigma_{\pi(w(i+1))}$ tal que el seu corresponent= cam=C3=AD va de $\sigma_{\pi(w(i+1))} (l(\sigma_{\pi(w(i+1))}))$ fins a = $\sigma_{\pi(w(i+1))} (mk)$. \item La paraula corresponent a un cam=C3=AD geod=C3=A8sic entre $\sigma_= {\pi(w(i+1))} (mk)$ i $\sigma_{\pi(w(i))} (mk)$. \item La subparaula de $\sigma_{\pi(w(i))}$ que va des de $\sigma_{\pi(w(= i))} (mk)$ fins a $\sigma_{\pi(w(i))} (l(\sigma_{\pi(w(i))}))$. \stopitemize \stopitemize Per construcci=C3=B3, tenim que $u_{i, j}$ s=C3=B3n paraules congruents i= {\em formen} $w$, =C3=A9s a dir, existeixen paraules $w_i$, $i \in \{0, = \ldots, l(w)-1\}$, tals que $w =3D w_1 \sharp ( w_2 \sharp (\ldots (w_{l(= w)-1})\ldots ))$ i cada $w_i =3D u_{i, 0} \sharp( \ldots \sharp (u_{i, m+= 1})\ldots )$. Per tant, \startformula \text{area}_{\cal P} (w) \leq \sum_{i=3D0}^{l(w)-1} \sum_{j=3D1}^{m+1} \t= ext{area}_{\cal P} (u_{i, j}). \stopformula A m=C3=A9s, per definici=C3=B3 de $\lambda_{0,k}$, la longitud de cada $u= _{i, j}$ =C3=A9s menor o igual que $2k + 2 \lambda_{0, k} (l(w)/2)$. Com que $\lambda_{0,k} (n) < n-k$ per a $n$ suficientment gran, llavors l= a funci=C3=B3 de Dehn satisf=C3=A0 que \startformula \startsplit \NC \delta_{\cal P} (n) \NC \leq n \cdot \frac{n/2}{k} \delta_{\cal P} (2= k + 2 \lambda_{0, k} (n/2))\NR \NC \NC \leq \frac{n^2}{2k} \delta_{\cal P} (n-2), \stopsplit \stopformula per a $n$ suficientment gran. \stopmydemo Per t=C3=A8cniques similars a les del Lema~\in[thmi:equacio-funcional] es= pot veure la proposici=C3=B3 seg=C3=BCent: \startmyproposition Sigui $F \colon \naturalnumbers \to \naturalnumbers$ = una funci=C3=B3 que satisf=C3=A0 la recursi=C3=B3 \startformula F(n) =3D 2 \ln n - \ln 2k + F(n-2) \stopformula per a tot $n \geq 2$. Aleshores $F$ =C3=A9s de la forma \placeformula[-] \startformula F(n) =3D \startdisplaycases \NC F(0) + 2 \ln n!! - \frac{n}{2} \ln 2k \MC \text{si } n \text{ p= arell} \NR \NC F(1) + 2 \ln n!! - \frac{n+1}{2} \ln 2k \MC \text{si } n \text{ s= enar} \NR \stopdisplaycases \stopformula \stopmyproposition \startmytheorem Siguin $k \geq 0$ i $G$ un grup tal que admet una secci=C3= =B3 $\sigma$ tal que $\lambda_{0, k} (n) < n-k$ per a $n$ suficientment g= ran. Aleshores \startformula \delta_G (n) \preceq \frac{(n!!)^2}{{(2k)}^{n/2}}. \stopformula \stopmytheorem \startmydemo Pel Teorema~\in[thmi:lambda-0-1-finitament-presentat], $G$ =C3= =A9s finitament presentat. Sigui ${\cal P}$ una presentaci=C3=B3 finita d= e $G$ la forma ${\cal P} =3D \langle X \mid R\rangle$, on $X$ =C3=A9s el = conjunt finit de generadors de $G$ respecte del qual est=C3=A0 definida $= \sigma$ (Si $G$ t=C3=A9 una presentaci=C3=B3 finita ${\cal Q} =3D \langle= Y \mid S\rangle$, amb $Y \neq X$, sempre en podem trobar una isomorfa a = ${\cal Q}$ amb $X$ com a conjunt de generadors). Sigui $n_0$ tal que $\lambda_{0, k} (n) < n-k$ per a tot $n \geq n_0$. Pe= r la Proposici=C3=B3~\in[thmi:desigualtat-dehn-p-lambda-0-k], $\delta_{\c= al P} (n) \leq (n^2/2k) \cdot \delta_{\cal P} (n-2)$. Si considerem $f \c= olon \naturalnumbers \to \naturalnumbers\setminus\{0\}$ tal que $f(n) =3D= (n^2/2k) \cdot f(n-2)$ i $f(n_0) \geq \delta_{\cal P} (n_0)$, tenim que= $\delta_{\cal P} (n) \leq f(n)$, per a tot $n \geq n_0$. Sigui $F(n) =3D= \ln f(n)$. Tenim que $F$ compleix la recursi=C3=B3 \startformula F(n) =3D 2\ln n - ln 2k + F(n-2). \stopformula Pel lema anterior, existeix una constant $C > 1$ tal que $F(n) \leq C + 2= \ln n!! - (n/2) \ln 2k$. A m=C3=A9s, podem prendre $C$ suficientment gra= n tal que, per a tot $n \in \naturalnumbers$, \startformula \delta_{\cal P} (n) \leq f(n) \leq e^{F(n)} \leq e^{C} \cdot \frac{(n!!)^= 2}{{(2k)}^{n/2}}. \stopformula Per tant \startformula \delta_G (n) \preceq e^{C} \cdot \frac{(n!!)^2}{{(2k)}^{n/2}}, \stopformula i, pel Lema~\in[thmi:lema-preceq-necessari], \startformula \delta_G (n) \preceq \frac{(n!!)^2}{{(2k)}^{n/2}}. \stopformula \stopmydemo \startmyproposition Per a tot $k \geq 1$, sigui $F_k (n) =3D \frac{(n!!)^= 2}{{(2k)}^{n/2}}$. Si $k > 1$, aleshores $F_k \preceq F_1$ per=C3=B2 $F_1= \not \preceq F_k $. \stopmyproposition \startmydemo De forma evident, si $k > 1$, aleshores $F_k (n) < F_1 (n)$ = per a tot $n \in \naturalnumbers$. Per tant, $F_k \preceq F_1$. D'altra banda, per a tot $k > 1$, el quocient $F_1(n)/F_k(n) =3D k^{n/2}$= =2E Per tant, pel Lema~\in[thmi:lema-funcions-no-preceq], per demostrar q= ue $F_1 \not \preceq F_k$, basta veure que $\lim_{n \to \infty} (F_1 (n))= ^2 / k^{Cn/2} =3D \infty$, per a tota constant $C > 2$ o, el que =C3=A9s = el mateix, $\lim_{n \to \infty} (F_1 (n))^2 / k^{Cn} =3D \infty$, per a t= ota constant $C > 0$. Si $n$ =C3=A9s parell, aleshores $n!! =3D 2^{n/2} \cdot (n/2)!$, i si $n$= =C3=A9s senar, aleshores $n!! \geq (n-1)!!$. Per tant, aplicant la f=C3=B3= rmula de Stirling, tenim que, per a tot $n \in \naturalnumbers$, \startformula (n!!)^2 \geq 2^{n-1} \cdot \pi \cdot (n-1) \cdot \left(\frac{n-1}{2e}\rig= ht)^{n-1}. \stopformula Per tant, per a tot $n \geq 2$, \startformula \startsplit \NC \frac{(F_1(n))^2}{k^{Cn}} \NC \geq \frac{((n-1)!!)^4 / 2^n}{k^{Cn}}\N= R \NC \NC \sim \frac{2^{2n-2} \cdot \pi^2 \cdot (n-1)^2 \cdot ((n-1)/2e)^{= 2n-2}}{k^{Cn} \cdot 2^n}\NR \NC \NC =3D \frac{\pi^2 (n-1)^2 \cdot ((n-1)/e)^{2 (n-1)}}{k^{Cn} \cdot 2= ^n}\NR \NC \NC \geq \frac{\pi^2 (n-1)^2 \cdot ((n-1)/e)^{2 (n-1)}}{k^{2C(n-1)} \= cdot 2^{2(n-1)}}\NR \NC \NC =3D \pi^2 (n-1)^2 \cdot \left(\frac{n-1}{2ek^C}\right)^{2 (n-1)}.= \stopsplit \stopformula Aquesta expressi=C3=B3 tendeix a infinit quan $n$ tendeix a infinit. Per = tant, tamb=C3=A9 ho fa el quocient $(F_1(n))^2/k^{Cn}$. \stopmydemo \subsection{Difer=C3=A8ncies entre $\varphi$ i $\lambda_{0, k}$} En aquesta secci=C3=B3 reflexionarem sobre si la classe de grups que, per= a qualque $k \geq 1$, admeten una secci=C3=B3 $\sigma$ tal que $\lambda_= {\sigma, 0, k} (n) < n-1$, per a $n$ suficientment gran, =C3=A9s m=C3=A9s= general que la classe de grups que admeten $\sigma$ tal que $\varphi_{\s= igma} (n) < n-1$, per a $n$ prou gran. Tamb=C3=A9 farem aquesta reflexi=C3= =B3 quan $\sigma$ sigui geod=C3=A8sica. Si s'observen els teoremes~\in[thmi:lambda-0-1-finitament-presentat] i \i= n[thmi:funcio-Dehn-lambda-0-1] pot par=C3=A8ixer que aquests no milloren = els resultats obtinguts per Bridson per al cas en qu=C3=A8 els grups adme= ten una secci=C3=B3 tal que $\varphi (n) < n-1$ per a $n$ suficientment g= ran (Teorema~\in[thme:bridson-finitament-presentat] i Proposici=C3=B3~\in= [thmi:proposicio-resum-ordre-dehn-bridson]). Aquesta impressi=C3=B3 =C3=A9= s deguda a qu=C3=A8 s'obtenen les mateixes conclusions per a $\lambda_{0,= 1}$ que per a $\varphi$: $G$ =C3=A9s finitament presentat i existeix $k >= 0$ tal que $\delta_{G} (n) \preceq e^{kn^3}$, per a tot grup $G$ tal que= admet una secci=C3=B3 $\sigma$ tal que, per a $n$ suficientment gran, $\= lambda_{\sigma, 0,1} (n) < n-1$ o $\varphi_{\sigma} (n) < n-1$. Creiem que aquesta impressi=C3=B3 =C3=A9s falsa. Concretament, pensam que= el rang d'aplicaci=C3=B3 dels Teoremes \in[thmi:lambda-0-1-finitament-pr= esentat] i \in[thmi:funcio-Dehn-lambda-0-1] =C3=A9s major, estrictament, = que el corresponent al Teorema~\in[thme:bridson-finitament-presentat] i l= a Proposici=C3=B3~\in[thmi:proposicio-resum-ordre-dehn-bridson] o, en alt= res paraules, que existeix un grup $G_0$ el qual admet una secci=C3=B3 $\= sigma$ tal que $\lambda_{\sigma, 0, 1} (n) < n-1$, per a $n$ suficientmen= t gran, per=C3=B2, per a tota secci=C3=B3 $\kappa$, $\varphi_{\kappa} (n)= \not < n-1$ per a $n$ suficientment gran. No hem pogut establir aquest fet perqu=C3=A8 no hem pogut trobar un exemp= le expl=C3=ADcit d'aquest grup, encara que existeixen diverses raons plau= sibles per a aquesta exist=C3=A8ncia: \startitemize[1] \item En general, $\lambda_{0,1} (n)$ =C3=A9s menor, estrictament, que $\= varphi(n)$, ja que =C3=A9s una mitjana de valors (la mitjana de valors =C3= =A9s menor que el m=C3=A0xim d'aquests valors). \item Si $G$ =C3=A9s un grup, $X$ =C3=A9s un conjunt de generadors finit = de $G$ i $\sigma \colon G \to {(X \cup X^{-1})}^*$ =C3=A9s una secci=C3=B3= tal que $\varphi_{\sigma}(n) \not < n-1$, per a $n$ suficientment gran, = aleshores existeix una successi=C3=B3 $(n_i)_{i \in \naturalnumbers}$ tal= que $\varphi_{\sigma} (n_i) \in \{n_i-1, n_i\}$. Per tant, existeixen $g= _i, h_i \in G$ tals que $d_{G, X} (g_i, h_i) =3D 1$, $d_{G, X}(1, g_i), d= _{G, X}(1, h_i) \leq n_i$ i $(t_i)_{i \in \naturalnumbers}$ tals que \startformula d_{G, X} (\sigma_{g_i} (t_i), \sigma_{h_i} (t_i)) \in \{n_i -1, n_i\} \stopformula (=C3=A9s a $t_i$ que s'agafa el m=C3=A0xim de les dist=C3=A0ncies entre $= \sigma_{g_i}$ i $\sigma_{h_i}$). Pareix probable que existeixin (molts) g= rups tals que els valors precedents i consecutius de $t_i$ siguin prou pe= tits, =C3=A9s a dir, que \startformula d_{G, X} (\sigma_{g_i} (t_i \pm 1), \sigma_{h_i} (t_i \pm 1)) < n_i. \stopformula Si \startformula d_{G, X} (\sigma_{g_i} (t_i \pm 1), \sigma_{h_i} (t_i \pm 1)) < n_i -2, \stopformula aleshores $\lambda_{0,1} (n) < n -1$. Per tant, l'exist=C3=A8ncia de $G_0= $ pot venir com a conseq=C3=BC=C3=A8ncia de l'exist=C3=A8ncia d'un grup (= infinit) tal que les seves dist=C3=A0ncies entre elements {\em oscil=C2=B7= lin}. \stopitemize \noindentation De fet, ni tan sols hem pogut construir un grup $G_1$ amb = la funci=C3=B3 de Dehn $\preceq e^{kn^3}$, per qualque $k > 0$, i tal que= no admet=C3=A9s una secci=C3=B3 amb amplada $\varphi(n) < n-1$ per a $n$= suficientment gran. Al marge de l'exist=C3=A8ncia de $G_1$ (l'exist=C3=A8ncia de $G_0$ implic= a l'exist=C3=A8ncia de $G_1$), una de les dificultats amb les quals hem t= opat de manera m=C3=A9s freq=C3=BCent quan hem intentat demostrar l'exist= =C3=A8ncia de $G_0$ =C3=A9s la coincid=C3=A8ncia de valor entre $\lambda_= {0,1} (n)$ i $\varphi(n)$. Per exemple, si en $\integers \oplus \integers= $, prenem la secci=C3=B3 $\sigma((i, j)) =3Da^i b^j$, amb $a =3D (1, 0)$ = i $b=3D (0, 1)$ (exemple \in[exemple-Z+Z-grups-seccionables], p=C3=A0gina= \at[exemple-Z+Z-grups-seccionables]), aleshores tenim que $\varphi_{\sig= ma} (n) =3D 2 =3D \lambda_{0,1} (n)$. Per a l'exist=C3=A8ncia de $G_0$, h= a d'existir un grup $G_2$ (que pot coincidir amb $G_0$) tal que $\lambda_= {0,1} (n) < \varphi(n)$ assimpt=C3=B2ticament o, equivalentment, $\liminf= _{n \to \infty} {\varphi(n)}/{\lambda_{0,1}} (n) > 1$. Tampoc hem pogut e= stablir l'exist=C3=A8ncia de $G_2$. Notem que $1 \leq \varphi/\lambda_{0,= 1} \leq 2$, ja que $\varphi \leq 2 \lambda_{0, 1}$. Tot fa pensar que nec= essitam un invariant geom=C3=A8tric associat al quocient $\varphi/{\lambd= a_{0,1}}$ (o a la seva difer=C3=A8ncia) per demostrar aquest fet. Per tot aix=C3=B2, enunciem la conjectura seg=C3=BCent: \startmyconjecture Existeixen grups finitament presentats $G_0$, $G_1$ i = $G_2$ tals que \startitemize[1] \item $G_0$ admet una secci=C3=B3 $\sigma$ tal que $\lambda_{\sigma, 0, 1= } (n) < n-1$, per a $n$ suficientment gran, per=C3=B2, per a tota secci=C3= =B3 $\kappa$, $\varphi_{\kappa} (n) \not < n-1$ per a $n$ suficientment g= ran. \item $G_1$ no admet cap secci=C3=B3 $\sigma$ amb amplada $\varphi_{\sigm= a} (n) < n-1$ per a $n$ suficientment gran i $\delta_{G} (n) \preceq e^{k= n^3}$ per qualque $k > 0$. \item Per a tota secci=C3=B3 $\sigma$ de $G_2$ (respecte d'algun conjunt = finit de generadors de $G_2$), \startformula \liminf_{n \to \infty} {\varphi_{\sigma}(n)}/{\lambda_{\sigma, 0,1}} (n) = > 1. \stopformula \stopitemize \stopmyconjecture L'exist=C3=A8ncia de $G_0$ implicaria, per definici=C3=B3, que ${\cal S} = (\varphi, f(n) =3D n-1) \subsetneq {\cal S} (\lambda_{0,1}, f(n) =3D n-1)= $. De fet, creiem que ${\cal S} (\lambda_{0, k}, f(n) =3D n-k)$ =C3=A9s u= na classe incomparable amb ${\cal S}(\lambda_{0, k'}, f(n) =3D n-k')$ qua= n $k \neq k'$. Per =C3=BAltim, en el cas geod=C3=A8sic, tenim que la classe de grups tal= s que admeten una secci=C3=B3 geod=C3=A8sica tal que $\varphi(n) < n-1$ p= er a $n$ suficientment gran tenen funci=C3=B3 de Dehn $\preceq (n!!)^2/2^= {n/2}$ mentre que la classe de grups que admeten una secci=C3=B3 geod=C3=A8= sica tal que, per qualque $k > 1$, $\lambda_{0, k}(n) < n-k$ tenen funci=C3= =B3 de Dehn $\preceq (n!!)^2/(2k)^{n/2}$. Hem vist que $(n!!)^2/(2k)^{n/2= }$ =C3=A9s estrictament menor, m=C3=B2dul $\simeq$, que $(n!!)^2/2^{n/2}$= =2E El Teorema de Sapir-Birget-Rips (Teorema~\in[thme:teorema-sapir-birge= t-rips]) i l'{\em abund=C3=A0ncia} dels grups finitament presentats sugge= reixen que aix=C3=B2 =C3=A9s una ra=C3=B3 m=C3=A9s per conjecturar que aq= uestes classes de grups s=C3=B3n diferents: \startmyconjecture Existeix un grup finitament presentat $G_3$ tal que ad= met una secci=C3=B3 geod=C3=A8sica tal que, per qualque $k > 1$, $\lambda= _{0, k} (n) < n-k$, per a $n$ suficientment gran, per=C3=B2 no admet cap = secci=C3=B3 geod=C3=A8sica tal que $\varphi (n) < n-1$ per a $n$ suficien= tment gran. \stopmyconjecture \section{$\varphi_k$} Una altra possible generalitzaci=C3=B3 de $\varphi$ =C3=A9s, en comptes d= e realitzar la mitjana de les dist=C3=A0ncies de dos valors, com f=C3=A9i= em amb $\lambda_{s, k}$, fer la mitjana de $k$ valors consecutius. \startmydefinition Siguin $G$ un grup, $X$ un conjunt finit de generadors= de $G$, $\sigma \colon G \to {(X \cup X^{-1})}^*$ una secci=C3=B3, $k \g= eq 0$. La {\em amplada mitjana $k$-=C3=A8ssima de \sigma}\index[amplada+m= itjana k-=C3=A8ssima d'una secci=C3=B3]{amplada+mitjana $k$-=C3=A8ssima d= 'una secci=C3=B3}, o {\em amplada mitjana de $k+1$ valors de $\sigma$}\in= dex[amplada+mitjana respecte de k+1 valors]{amplada+mitjan respecte de $k= +1$ valors}, =C3=A9s la funci=C3=B3 $\varphi_{\sigma, k} \colon \naturaln= umbers \to \naturalnumbers$\mysymbol{$\varphi_{\sigma}$} definida per $\v= arphi_{\sigma, k} (0) =3D 0$ i, per a tot $n > 0$, \startformula \varphi_{\sigma, k} (n)=3D \max \{ \sum_{i=3D0}^k D_{\sigma, g, h} (t+i)= \mid t \in \naturalnumbers, (g, h) \in K_{G, X} (n) \}. \stopformula Quan $\sigma$ sigui clara pel context i sigui una secci=C3=B3 gen=C3=A8ri= ca, escriurem simplement $\varphi_k (n)$\mysymbol{$\varphi_k$}. \stopmydefinition Estendrem $\varphi_k$ als nombres reals mitjan=C3=A7ant $\varphi_k (x) =3D= \varphi_k (\lfloor x \rfloor)$ si $x > 0$ i $\varphi_k (x) =3D \varphi_k= (0)$ si $x < 0$. \startmylema Per a tot $k \geq 0$, tenim que $\varphi_k (n) \leq \varphi = (n)$. \stopmylema \startmydemo Clarament, per a tot $n \in \naturalnumbers$, \startformula \startsplit \NC \varphi_k (n) \NC =3D \max \{ \frac{1}{k+1} \sum_{i=3D0}^k D_{\sigma,= g, h} (t+i) \mid t \in \naturalnumbers, (g, h) \in K_{G, X} (n)\}\NR \NC \NC \leq \frac{1}{k+1} \sum_{i=3D0}^k \max \{D_{\sigma, g, h} (t+i) \= mid t \in \naturalnumbers, (g, h) \in K_{G, X} (n)\}\NR \NC \NC \leq \frac{k+1}{k+1} \varphi (n)\NR \NC \NC =3D \varphi(n). \stopsplit \stopformula \stopmydemo \startmycorollary Tot grup finitament generat admet una secci=C3=B3 $\sig= ma$, respecte d'algun conjunt finit de generadors, tal que $L_{\sigma} (n= ) =3D n$ i $\varphi_k (n) \leq n$, per a tot $n \in \naturalnumbers$. \stopmycorollary \startmydemo =C3=89s conseq=C3=BC=C3=A8ncia directa del lema anterior i d= el Teorema~\in[thme:bridson-finitament-presentat]. \stopmydemo \bigskip Conjectura: $\varphi_k (n) < n-1$ aleshores $G$ t=C3=A9 el problema de la= paraula resoluble. $\varphi_k$ el que fa =C3=A9s regularitzar les dist=C3= =A0ncies entre els punts. Aix=C3=B2 concorda amb veure $G$, com a espai, = m=C3=A8tric molt alluny. --> connexi=C3=B3 amb "asymptotic connes" (\cite= [riley-tesi] i altres per exemple meier). \section{M=C3=A9s d'una secci=C3=B3} \section{$p_w$} \section{altres} Generalitzar generalitzar $\lambda_{s, k}$ i dem=C3=A9s al cas asincr=C3=B2= nic. Conjecturem que existeixen grups que no s=C3=B3n sincr=C3=B2nics d'a= questa classe. \completepublications[criterium=3Dall] %all per tots \title{=C3=8Dndex alfab=C3=A8tic} \placeindex \stoptext =20 \startmydefinition Siguin $G$ un grup, $X$ un conjunt finit de generadors= de $G$, $\sigma \colon G \to {(X \cup X^{-1})}^*$ una secci=C3=B3, $k \g= eq 0$. L'{\em amplada mitjana $k$-=C3=A8ssima} o {\em amplada mitjana res= pecte de $k+1$ valors}\index[amplada+mitjana k-=C3=A8ssima]{amplada+mitja= na $k$-=C3=A8ssima} \index[amplada+mitjana respecte de k+1 valors]{amplada+mitjana respecte d= e $k+1$ valors} =C3=A9s la funci=C3=B3 $\varphi_{\sigma, k} \colon \naturalnumbers \to \= naturalnumbers$ definida per $\varphi_{\sigma, k} (0) =3D 0$\mysymbol{$\v= arphi_{\sigma, k}$} i, per a tot $n > 0$, \startformula \varphi_{\sigma, k} (n) =3D \max \{ \frac{1}{k+1} \sum_{i=3D0}^k D_{\sigm= a, g, h} (t+i) \mid t \in \naturalnumbers, (g, h) \in K_{G, X} (n)\}. \stopformula Quan $\sigma$ sigui clara pel context o quan $\sigma$ sigui una secci=C3=B3= gen=C3=A8rica, escriurem simplement $\varphi_k$. De forma trivial, tenim= que $\varphi_0 =3D \varphi$.=20 \stopmydefinition --------------030701000305020104030602 Content-Type: text/plain; charset="us-ascii" MIME-Version: 1.0 Content-Transfer-Encoding: 7bit Content-Disposition: inline ___________________________________________________________________________________ If your question is of interest to others as well, please add an entry to the Wiki! maillist : ntg-context@ntg.nl / http://www.ntg.nl/mailman/listinfo/ntg-context webpage : http://www.pragma-ade.nl / http://tex.aanhet.net archive : https://foundry.supelec.fr/projects/contextrev/ wiki : http://contextgarden.net ___________________________________________________________________________________ --------------030701000305020104030602--