From mboxrd@z Thu Jan 1 00:00:00 1970 X-Msuck: nntp://news.gmane.io/gmane.comp.tex.context/51089 Path: news.gmane.org!not-for-mail From: Xan Newsgroups: gmane.comp.tex.context Subject: Problems with bibtex and @incollection Date: Sat, 06 Jun 2009 21:06:29 +0200 Message-ID: <4A2ABE35.4070404@telefonica.net> Reply-To: mailing list for ConTeXt users NNTP-Posting-Host: lo.gmane.org Mime-Version: 1.0 Content-Type: multipart/mixed; boundary="------------060307080707070702010906" X-Trace: ger.gmane.org 1244315303 22830 80.91.229.12 (6 Jun 2009 19:08:23 GMT) X-Complaints-To: usenet@ger.gmane.org NNTP-Posting-Date: Sat, 6 Jun 2009 19:08:23 +0000 (UTC) Cc: Xan To: mailing list for ConTeXt users Original-X-From: ntg-context-bounces@ntg.nl Sat Jun 06 21:08:19 2009 Return-path: Envelope-to: gctc-ntg-context-518@m.gmane.org Original-Received: from ronja.vet.uu.nl ([131.211.172.88] helo=ronja.ntg.nl) by lo.gmane.org with esmtp (Exim 4.50) id 1MD1Fa-0003Eg-N4 for gctc-ntg-context-518@m.gmane.org; Sat, 06 Jun 2009 21:08:18 +0200 Original-Received: from localhost (localhost [127.0.0.1]) by ronja.ntg.nl (Postfix) with ESMTP id 5D9701FBDB; Sat, 6 Jun 2009 21:08:17 +0200 (CEST) Original-Received: from ronja.ntg.nl ([127.0.0.1]) by localhost (smtp.ntg.nl [127.0.0.1]) (amavisd-new, port 10024) with LMTP id 26050-02; Sat, 6 Jun 2009 21:07:00 +0200 (CEST) Original-Received: from ronja.vet.uu.nl (localhost [127.0.0.1]) by ronja.ntg.nl (Postfix) with ESMTP id 5B36B1FB89; Sat, 6 Jun 2009 21:07:00 +0200 (CEST) Original-Received: from localhost (localhost [127.0.0.1]) by ronja.ntg.nl (Postfix) with ESMTP id 4A06D1FB89 for ; Sat, 6 Jun 2009 21:06:58 +0200 (CEST) Original-Received: from ronja.ntg.nl ([127.0.0.1]) by localhost (smtp.ntg.nl [127.0.0.1]) (amavisd-new, port 10024) with LMTP id 26710-01 for ; Sat, 6 Jun 2009 21:06:40 +0200 (CEST) Original-Received: from filter1-til.mf.surf.net (filter1-til.mf.surf.net [194.171.167.217]) by ronja.ntg.nl (Postfix) with ESMTP id 720931FB7F for ; Sat, 6 Jun 2009 21:06:40 +0200 (CEST) Original-Received: from ctsmtpout1.frontal.correo (outmailhost.telefonica.net [213.4.149.242]) by filter1-til.mf.surf.net (8.13.8/8.13.8/Debian-3) with ESMTP id n56J6bPQ023466 for ; Sat, 6 Jun 2009 21:06:38 +0200 Original-Received: from [172.26.0.4] (83.58.163.247) by ctsmtpout1.frontal.correo (7.2.056.6) (authenticated as dxpublica) id 0000000000205273; Sat, 6 Jun 2009 21:06:31 +0200 User-Agent: Thunderbird 2.0.0.21 (X11/20090318) X-Bayes-Prob: 0.0001 (Score 0, tokens from: @@RPTN) X-CanIt-Geo: ip=213.4.149.242; country=ES; region=29; city=Madrid; latitude=40.4000; longitude=-3.6833; http://maps.google.com/maps?q=40.4000,-3.6833&z=6 X-CanItPRO-Stream: uu:ntg-context@ntg.nl (inherits from uu:default, base:default) X-Canit-Stats-ID: 238583274 - 512b4a2ee7a3 - 20090606 X-Scanned-By: CanIt (www . roaringpenguin . com) on 194.171.167.217 X-Virus-Scanned: amavisd-new at ntg.nl X-BeenThere: ntg-context@ntg.nl X-Mailman-Version: 2.1.11 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 X-Virus-Scanned: amavisd-new at ntg.nl Xref: news.gmane.org gmane.comp.tex.context:51089 Archived-At: This is a multi-part message in MIME format. --------------060307080707070702010906 Content-Type: text/plain; charset=ISO-8859-15; format=flowed Content-Transfer-Encoding: 7bit Hi, When I put one reference as @incollection Bibtex does not show me the title of the article in the book, but yes the booktitle. Why?. Anyone could help me please. I attached the files.... Xan. 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Despr=C3=A9s de par=C3=A8ntesis un '= .' 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 [proposition] [lema] [text=3D{Proposici=C3=B3}] \defineenumeration [corollary] [lema] [text=3D{Corol=C2=B7lari}] \defineenumeration [theorem] [lema] [text=3D{Teorema}] %% Definici=C3=B3 \defineenumeration [definition] [lema] [text=3D{Definici=C3=B3},style=3Dtf,titlestyle=3Dbf,indenting=3Dyes] \defineenumeration [notation] [definition] [text=3D{Notaci=C3=B3},style=3Dtf,titlestyle=3Dbf,indenting=3Dyes] \defineenumeration [note] [definition] [text=3D{Nota},style=3Dtf,titlestyle=3Dbf,indenting=3Dyes] %% Demostraci=C3=B3 \defineenumeration[demo][text=3D{Demostraci=C3=B3.\space},number=3Dno,loc= ation=3Dserried,width=3Dfit,headstyle=3Ditalic,indentnext=3Dyes,between=3D= \blank,textdistance=3D.5em,closesymbol=3D{\mathematics{\Box}},style=3Dnor= mal,indenting=3Dyes] % Table of contents %% 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][style=3Dbold,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=3D10mm] %\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} % Other \setupunderbar[alternative=3Db] % Fix underline style % For putting underline with spaces: \underbar{\dorecurse{40}~} % Start the text \starttext \section{Preliminars} \startdefinition Siguin $G$ un grup, $A$ un conjunt finit de generadors d= e $G$ i ${\cal P} =3D \langle A \mid R \rangle$ una presentaci=C3=B3 fini= ta de $G$. Una paraula $w \in A^*$ es {\em nul-homot=C3=B2pica} per ${\ca= l P}$ si, i nom=C3=A9s si, $\pi(w) =3D 1 \in G$, o, equivalentment, si $w= $ forma un cicle dins el graf de Cayley $\Gamma_{G, A}$. \stopdefinition la defininci=C3=B3 d'=C3=A0rea posar que assumeixo que el conjunt de rela= cions cont=C3=A9 tamb=C3=A9 els sim=C3=A8trics. Totes les presentacions finites de $G$ tenen a $A$ com a conjunt finit de= generadors de $G$. \section{Millores directes de les fites de la funci=C3=B3 de Dehn per gru= ps seccionables} \startlema[area-concatenacio] Siguin $G$ un grup, ${\cal P} =3D \langle A= \mid R \rangle$ una presentaci=C3=B3 finita de $G$ i $u, v, w$ paraules = nul-homot=C3=B2piques per ${\cal P}$. Si $w =3D u v$ dins el grup lliure = $F(A)$, aleshores=20 \startformula \text{area} (w) \leq \text{area}(u) + \text{area}(v). \stopformula \stoplema \startdemo Si $\text{area}(u) =3D N$ i $\text{area}(v) =3D M$, 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 $x_i, y_j \in F(A)$, $r_i, s_j \in R$, on aquestes igualtats= s=C3=B3n dins el grup lliure $F(A)$. Com que $w =3D u v$ tamb=C3=A9 dins= el grup lliure, tenim que \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, $\text{area}(w) \leq N+M =3D \text{area}(u) = + \text{area}(v)$. \stopdemo \startlema[area-conjugats] Siguin $G$ un grup i ${\cal P} =3D \langle A \= mid R \rangle$ una presentaci=C3=B3 finita de $G$. Si $w$ =C3=A9s una par= aula nul-homot=C3=B2pica per ${\cal P}$ i $x \in F(A)$, llavors \startformula \text{area}(x^{-1}wx) \leq \text{area}(w). \stopformula \stoplema \startdemo Suposem que $\text{area}(w) =3D N$. Aleshores existeixen $x_i = \in F(A)$ 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 $F(A)$. 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 $\text{area}(x^{-1}wx) \leq N =3D \text{area}(= w)$, que =C3=A9s el que vol=C3=ADem veure. \stopdemo \startlema[lema-tecnic] Siguin $G$ un grup, $A$ un conjunt finit de gener= adors de $G$ i ${\cal P} =3D \langle A \mid R \rangle$ una presentaci=C3=B3= finita de $G$, $g_1, g_2, g_3, g_4, g_5, g_6 \in G$ i els camins $a$, $b= $, $c$, $d$, $e$, $f$ i $g$ dins el graf de Cayley $\Gamma_{G, A}$ que un= eixen 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] % 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=3D1mm]{>}}}] (0,0) -- (2,0); \draw [decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow= [blue,line width=3D1mm]{>}}}] (2,0) -- (4,0); \draw [decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow= [blue,line width=3D1mm]{>}}}] (4,0) -- (4,2); \draw [decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow= [blue,line width=3D1mm]{>}}}] (4,2) -- (2,2); \draw [decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow= [blue,line width=3D1mm]{>}}}] (2,2) -- (0,2); \draw [decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow= [blue,line width=3D1mm]{>}}}] (0,2) -- (0,0); \draw [decorate,decoration=3D{markings,mark=3Dat position .5 with {\arrow= [blue,line width=3D1mm]{>}}}] (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, 0.3) node {$a$}; \draw (3, 0.3) node {$b$}; \draw (3.7, 1) node {$c$}; \draw (3, 1.7) node {$d$}; \draw (1, 1.7) node {$e$}; \draw (0.3, 1) node {$f$}; \draw (2.3, 1) node {$g$}; % \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} D'altra banda, siguin $w_1, w_2$ i $w \in A^*$ les paraules formades per = la composici=C3=B3 de les paraules corresponents a aquests camins (que se= guirem indicant de la mateixa manera) definides com: \startformula \startmathalignment \NC w_1 \NC =3D f^{-1}e^{-1}g^{-1}a^{-1}, \NR[+] \NC w_2 \NC =3D d^{-1} c^{-1} b^{-1} g,\NR[+] \NC w \NC =3D f^{-1} e^{-1} d^{-1} c^{-1} b^{-1} a^{-1}.\NR[+] \stopmathalignment \stopformula Aleshores $w, w_1, w_2$ s=C3=B3n nul-homot=C3=B2piques per ${\cal P}$ i a= m=C3=A9s, \startformula \text{area}(w) \leq \text{area}(w_1) + \text{area}(w_2). \stopformula \stoplema \startdemo De forma =C3=B2bvia tenim que $w_1$, $w_2$ i $w$ s=C3=B3n nul-= homot=C3=B2piques per ${\cal P}$, ja el seus camins dins el graf de Cayle= y formen cicles (per exemple el cam=C3=AD corresponent a $w_1$ forma un c= icle 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, fin= alment, de $g_2$ a $g_1$). D'altra banda, dins el grup lliure $F(A)$ tenim que \startformula w =3D (f^{-1}e^{-1}g^{-1}a^{-1}) a g (d^{-1}c^{-1}b^{-1}g)g^{-1}a^{-1} =3D= w_1 ag w_2 (ag)^{-1}. \stopformula Per tant, dins $F(A)$, $w =3D w_1 x^{-1} w_2 x$ amb $x \in F(A)$. Llavors= , aplicant el lemes \in[area-concatenacio] i \in[area-conjugats], tenim q= ue \startformula \text{area}(w) =3D \text{area}(w_1 x^{-1}w_2 x) \leq \text{area}(w_1) + \= text{area}(x^{-1} w_2 x) \leq \text{area}(w_1) + \text{area}(w_2). \stopformula \stopdemo \startdefinition{Paraules congruents} Siguin $G$ un grup, $A$ un conjunt = finit de generadors de $G$ i ${\cal P} =3D \langle A \mid R\rangle$ una p= resentaci=C3=B3 finita de $G$. Dues paraules $w_1, w_2 \in A^*$ nul-homot= =C3=B2piques per ${\cal P}$ s=C3=B3n {\em congruents} si, i nom=C3=A9s si= , dins el graf de Cayley $\Gamma_{G, A}$ existeixen punts $g_1, g_2, g_3,= g_4, g_5, g_6 \in G$ i camins $a, b, c, d, e, f, g$ que uneixen els pare= lls 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 parau= les corresponents a aquests camins (que seguirem indicant de la mateixa m= anera) satisfan \startformula \startmathalignment \NC w_1 \NC =3D f^{-1}e^{-1}g^{-1}a^{-1}, \NR[+] \NC w_2 \NC =3D d^{-1} c^{-1} b^{-1} g.\NR[+] \stopmathalignment \stopformula En aquest cas, indicarem amb $w_1 \sharp w_2$ a la paraula definida com \startformula w_1 \sharp w_2 =3D f^{-1} e^{-1} d^{-1} c^{-1} b^{-1} a^{-1}. \stopformula \stopdefinition 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 $\text{area}(u\sharp v) \leq \text{area}(u) + \text{area}= (v)$, o sigui, tenim que la funci=C3=B3 $\text{area} \colon \{w \in A^* \= mid \text{nul-homot=C3=B2pica per } {\cal P} \} \to \naturalnumbers$ =C3=A9= s subadditiva per a paraules congruents, o sigui, subadditiva per l'opera= ci=C3=B3 $\sharp$. \startlema[lema0]Siguin $G$ un grup, $A$ un conjunt finit de generadors d= e $G$ i ${\cal P} =3D \langle A \mid R\rangle$ una presentaci=C3=B3 finit= a de $G$. Si $w$ =C3=A9s una paraula nul-homot=C3=B2pica per ${\cal P}$, = aleshores $d(1, w(t)) \leq {\lvert w\rvert}/2$, per a tot $t \in \natural= numbers$. \stoplema \startdemo La dist=C3=A0ncia $d(1, w(t))$ =C3=A9s menor o igual que la lo= ngitud del menor cam=C3=AD des d'$1$ a $w(t)$ que passa per sobre la corb= a. Si $t \leq {\lvert w\rvert }/2$, aleshores aquesta longitud =C3=A9s me= nor o igual que ${\lvert w\rvert}/2$, ja que podem seguir el cam=C3=AD so= bre la corba que va de $w(t)$ a $w(0)$. Si $t > {\lvert w\rvert }/2$, lla= vors aquesta longitud tamb=C3=A9 =C3=A9s menor igual que ${\lvert w\rvert= }/2$, ja que podem seguir el cam=C3=AD sobre la corba de $w(t)$ a $w({\l= vert w\rvert})$. En tot cas, $d(1, w(t)) \leq {\lvert w\rvert}/2$. \stopdemo \startlema[subadditivitat-area] Siguin $G$ un grup, $A$ un conjunt finit = de generadors de $G$, ${\cal P} =3D \langle A \mid R \rangle$ una present= aci=C3=B3 finita de $G$, $\sigma \colon G \rightarrow A^*$ una secci=C3=B3= d'amplada $\varphi$. Aleshores, per a tota paraula $w \in A^*$ nul-homot= =C3=B2pica per ${\cal P}$, existeixen $u_k$ paraules nul-homot=C3=B2pique= s per ${\cal P}$, amb $k \in \{1, \ldots, {\lvert w \rvert}^2/2\}$, de lo= ngitud $l(u_k) \leq 2\varphi({\lvert w \rvert}/2)+2$ tals que \startformula \text{area}(w) \leq \sum_{k=3D1}^{{{\lvert w \rvert}^2}/2} \text{area}(u_= k). \stopformula \stoplema \startdemo Sigui $w$ una paraula nul-homot=C3=B2pica per ${\cal P}$. Si $= w =3D \varepsilon$, aleshores el resultat =C3=A9s obvi, ja que basta agaf= ar $w_i =3D \varepsilon$. Si $\lvert w \rvert \geq 1$, considerem els seg= =C3=BCents objectes (tal com es mostra a la Figura \in[figura-area]): \startitemize[2] \item El conjunt d'=C3=ADndexos $I =3D \{(i,j) \mid i =3D 0, \ldots, {\lv= ert w \rvert}, j =3D 0, \ldots, {{\lvert w \rvert}^2}/2\}$. \item Per a tot $(i, j) \in I$, els punts $\sigma_{\pi(w(i))}(j)$ del gra= f de Cayley $\Gamma_{G, A}$, que indicarem amb $\sigma_{i}(j)$. \item Per a tots $(i, j), (i',j') \in I$, agafem un cam=C3=AD geod=C3=A8s= sic $\rho_{(i,j)(i',j')}$ entre $\sigma_i(j)$ i $\sigma_{i'}(j')$. Escoll= im $\rho_{(i,{\lvert w \rvert}/2)(i+1,{\lvert w \rvert}/2)}$ de manera qu= e sigui igual al cam=C3=AD geod=C3=A8ssic (de longitud unitat) sobre $w$ = que uneix $\sigma_i(j)$ i $\sigma_{i+1}(j)$. \item Per a tot $(i, j) \in I$ tal que $(i+1,j), (i,j+1), (i+1,j+1) \in I= $, sigui $\theta_{i, j}$ el cam=C3=AD determinat per la composici=C3=B3 (= en aquest ordre) dels camins geod=C3=A8ssics $\rho_{(i,j)(i,j+1)}$, $\rho= _{(i,j+1)(i+1,j+1)}$, $\rho_{(i+1,j+1)(i+1,j)}$, $\rho_{(i+1,j)(i,j)}$, i= $u_{i, j}$ la paraula corresponent a $\theta_{i,j}$. \item Per a tot $i \in \{0, \ldots, {\lvert w\rvert } -1\}$, sigui $\tau_= i$ el cam=C3=AD determinat pels v=C3=A8rtexs $\sigma_i(0) =3D \sigma_{i+1= }(0) =3D 1$, $\sigma_i ({\lvert w \rvert}/2)$ i $\sigma_{i+1} ({\lvert w = \rvert}/2)$ i per les composicions de camins \startformula \startmathalignment \NC \rho_{(i,0)(i,1)}\rho_{(i,1)(i,2)}\ldots \rho_{(i,{\lvert w \rvert}/= 2-1)(i,{\lvert w \rvert}/2)}, \NR[+] \NC \rho_{(i,{\lvert w \rvert}/2)(i+1,{\lvert w \rvert}/2)},\NR[+] \NC \rho_{(i+1,{\lvert w \rvert}/2)(i+1,{\lvert w \rvert}/2-1)}\ldots \= rho_{(i+1,1)(i+1,0)}\rho_{(i+1,1)(i+1,0)}.\NR[+] \stopmathalignment \stopformula que uneixen, respectivament, $1$ amb $\sigma_i({\lvert w \rvert}/2)$, $\s= igma_i({\lvert w \rvert}/2)$ amb $\sigma_{i+1}({\lvert w \rvert}/2)$, i $= \sigma_{i+1}({\lvert w \rvert}/2)$ amb $1$. I indiquem amb $v_i$ la paraula corresponent a $\tau_i$. \stopitemize \placefigure [here] [figura-area] {Camins sobre $w$} {\startcombination[1*1] { \starttikzpicture[scale=3D1] % Els punts \filldraw (0,-4) circle (2pt); \filldraw (0.4216,3.9603) circle (2pt); % primer punt: avaluo ({3*sin(\t = r)},{4*cos(\t r)}); a t =3D 0.141 \filldraw (-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 (0,-4) circle (2pt); % perqu=C3=A8 me quedi el punt damunt. % Els combings \draw plot[domain=3D0:0.4216,smooth,variable=3D\t] ({-\t + \t* (\t - 0.42= 16)*sin(rand r)},{18.8812*\t -4 }); % el sentit d'omega \draw[decorate,decoration=3D{markings,mark=3Dat position .9 with {\arrow[= blue,line width=3D1mm]{<}}}] plot[domain=3D-3.141:3.141,smooth,variable=3D= \t] ({3*sin(\t r)},{4*cos(\t r)}); % Els noms \draw (0, -4.3) node {$1 \in G$}; %\draw (0.9, 8.3) node {$\pi(w(i))$}; %\draw (-0.9, 8.3) node {$\pi(w(i+1))$}; \draw (2.5, -3) node {$w$}; % Els noms dels camins %\draw (1, 0.3) node {$a$}; %\draw (3, 0.3) node {$b$}; %\draw (3.7, 1) node {$c$}; %\draw (3, 1.7) node {$d$}; %\draw (1, 1.7) node {$e$}; %\draw (0.3, 1) node {$f$}; %\draw (2.3, 1) node {$g$}; % PROVES %\draw[out=3D45,in=3D-45] (0,0) to (0.5,8); %\draw[color=3Dblue,->] (0,0) .. controls (0.1,2) .. (0.2,3) .. controls = (0.3,4) and (0.4,6) .. (0.5,8); %\draw (0,0) arc (-90:90:3 and 4); %\draw (0,0) arc (270:90:3 and 4); %\draw[color=3Dgreen] plot[domain=3D-3.141:3.141,smooth,variable=3D\t] ({= 4*sin(\t + (.1 * rand) r)},{4*cos(\t r)}); %\draw (0,0) arc (-90:81.82:2 and 4); %\draw[decorate,decoration=3D{random steps,segment length=3D2mm, amplitud= e=3D2pt}] (0,0) arc (-90:97.18:3.5 and 4); % \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} \indentation Cada paraula $v_i$ =C3=A9s nul-homot=C3=B2pica per ${\cal P}= $, ja que el seu cam=C3=AD corresponent, $\tau_i$, forma un cicle dins el= graf de Cayley $\Gamma_{G, A}$ (que t=C3=A9 punt inicial i final $\sigma= _i(0) =3D 1$). Per contrucci=C3=B3, \startformula w =3D v_0 \sharp (v_1 \sharp (\ldots, \sharp(v_{{\lvert w \rvert}-1})\ldo= ts ), \stopformula llavors, per aplicaci=C3=B3 reiterada del Lema \in[lema-tecnic], tenim qu= e \placeformula[vi] \startformula \text{area}(w) \leq \sum_{i=3D0}^{{\lvert w \rvert}-1} \text{area}(v_i). \stopformula \indentation D'altra banda, cada $u_{i,j}$ =C3=A9s nul-homot=C3=B2pica pe= r ${\cal P}$, ja que el seu cam=C3=AD corresponent, $\tau_{i,j}$, forma u= n cicle dins el graf de Cayley amb punt inicial i punt final $\sigma_i(j)= $. A m=C3=A9s, tamb=C3=A9 per construcci=C3=B3,=20 \startformula v_i =3D u_{i,0} \sharp (u_{i,1} \sharp (\ldots, \sharp(u_{i,{\lvert w \rv= ert}/2-1})\ldots ), \stopformula per la qual cosa, pel Lema \in[lema-tecnic], tenim que \placeformula[uij] \startformula \text{area}(v_i) \leq \sum_{j=3D0}^{{\lvert w \rvert}/2-1} \text{area}(u= _{i,j}). \stopformula Aleshores, combinant (\in[vi]) i (\in[uij]) tenim que \placeformula[-] \startformula \text{area}(w) \leq \sum_{i=3D0}^{{\lvert w \rvert}-1} \sum_{j=3D0}^{{\lv= ert w \rvert}/2-1} \text{area}(u_{i,j}). \stopformula \indentation Pel Lema \in[lema0], tenim que cada $u_{i,j}$ t=C3=A9 longit= ud $l(u_{i,j}) \leq 2 \varphi({\lvert w \rvert}/2)+2$. Aleshores, si rein= dexam aquest sumatori amb la bijecci=C3=B3 \startformula \{u_{i, j} \mid i =3D 0, \ldots, {\lvert w \rvert}-1, j=3D 0, \ldots, {\l= vert w \rvert}/2-1\} \longleftrightarrow \{u_k \mid k =3D 1, \ldots, {{\l= vert w \rvert}^2}/2\}, \stopformula tenim que \startformula \text{area}(w) \leq \sum_{k=3D0}^{{{\lvert w \rvert}}^2/2} \text{area}(u_= k) \stopformula amb $u_k$ paraules nul-homot=C3=B2piques per ${\cal P}$ tals que $l(u_k) = \leq 2 \varphi({\lvert w \rvert}/2)+2$, que =C3=A9s el que vol=C3=ADem ve= ure. \stopdemo \startproposition[desigualtat-dehn] Siguin $G$ un grup, $A$ un conjunt fi= nit de generadors de $G$, ${\cal P} =3D \langle A \mid R \rangle$ una pre= sentaci=C3=B3 finita de $G$ i una secci=C3=B3 $\sigma \colon G \to A^*$ a= mb amplada $\varphi$. Aleshores \startformula \delta_{{\cal P}} (n) \leq \frac{1}{2} \delta_{{\cal P}} (2 \varphi(n/2)+= 2) \cdot n^2. \stopformula \stopproposition \startdemo Sigui $w$ una paraula nul-homot=C3=B2pica per ${\cal P}$. Pel = Lema \in[subadditivitat-area], tenim que existeixen $u_k$ paraules nul-ho= mot=C3=B2piques per ${\cal P}$, amb $k \in \{1, \ldots, {\lvert w \rvert}= /2\}$, de longitud $l(u_k) \leq 2 \varphi({\lvert w \rvert}/2) +2$ tals q= ue \startformula \text{area}(w) \leq \sum_{i=3D1}^{{{\lvert w \rvert}^2}/2} \text{area}(u_= k). \stopformula Com que $l(u_k) \leq 2 \varphi({\lvert w \rvert}/2) +2$, aleshores $\text= {area}(u_k) \leq \delta_{{\cal P}}(2 \varphi({\lvert w \rvert}/2) +2)$. A= m=C3=A9s, de forma =C3=B2bvia, ${\lvert w \rvert} \leq l(w)$. Per tot ai= x=C3=B2, tenim que \placeformula[-] \startformula \startsplit \NC \text{area}(w) \NC \leq \sum_{k=3D1}^{{{\lvert w \rvert}^2}/2} \tex= t{area}(u_k) \NR \NC \NC \leq \sum_{k=3D1}^{{{\lvert w \rvert}^2}/2} \delta_{{\cal P}} = (2 \varphi({\lvert w \rvert}/2) +2) \NR \NC \NC \leq \frac{1}{2} {\lvert w \rvert}^2 \cdot \delta_{{\cal P}} (= 2 \varphi({\lvert w \rvert}/2) +2) \NR \NC \NC \leq \frac{1}{2} {l(w)}^2 \cdot \delta_{{\cal P}} (2 \varphi({= l(w)}/2) +2). \stopsplit \stopformula \indentation Llavors \placeformula[-] \startformula \startsplit \NC \delta_{{\cal P}} (n) \NC =3D \max \{ \text{area}(w) \mid w \text{ n= ul-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({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(n/2) +2) \stopsplit \stopformula que =C3=A9s el que vol=C3=ADem veure. \stopdemo \definemathcases[displaycases][style=3D\displaystyle] \startlema[equacio-funcional] Sigui $F\colon \naturalnumbers \to \reals $= una funci=C3=B3 que cumpleix la recursi=C3=B3 $F(n) =3D F(n-2) + 2 \ln n= + \ln \frac{1}{2}$. 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 $1!! =3D 1$= , $0!! =3D 1$, $n!! =3D n \cdot (n-2)!!$. \stoplema \startdemo Com que la recursi=C3=B3 $F(n) =3D F(n-2) + 2 \ln n + \ln \fra= c{1}{2}$ =C3=A9s d'ordre $2$, per la Teoria d'equacions en difer=C3=A8nci= es, la soluci=C3=B3 d'aquesta recursi=C3=B3 =C3=A9s =C3=BAnica si es cone= ixen les condicions inicials $F(1)$ i $F(0)$. Per tant, basta comprovar q= ue 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. \stopdemo \starttheorem[Theorema-n!!-Presentacions] Siguin $G$ un grup, $A$ un conj= unt finit de generadors de $G$, ${\cal P} =3D \langle A \mid R \rangle$ u= na presentaci=C3=B3 finita de $G$ i $\sigma \colon G \to A^*$ una secci=C3= =B3 d'amplada $\varphi$ tal que existeix un $n_0 \in \naturalnumbers$ tal= que $\varphi(n) < n-1$ per a tot $n \geq n_0$. Aleshores existeix $C$ co= nstant, 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, $C$ compleix que \startformula C \geq \frac{(\delta_{{\cal P}}(n_0)+1)\cdot 2^{\frac{n_0+1}{2}}}{(n_0!!)= ^2}. \stopformula \stoptheorem \startdemo Per la Proposici=C3=B3 \in[desigualtat-dehn], tenim que \startformula \delta_{{\cal P}} (n) \leq \frac{1}{2} \delta_{{\cal P}} (2 \varphi(n/2)+= 2) \cdot n^2. \stopformula Com que $\varphi(n) < n-1$ per a tot $n \geq n_0$, llavors $\varphi(n) \l= eq n-2$, ja que la funci=C3=B3 $\varphi$ nom=C3=A9s pren valors naturals.= Per aix=C3=B2, per a tot $n \geq n_0$, tenim que $2 \varphi(n/2) +2 \leq= n-2$. Per tant, $\delta_{{\cal P}}$ compleix 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 \startmathalignment \NC f(n) \NC =3D \frac{1}{2} f(n-2) \cdot n^2, \NR[+] \NC f(n_0) \NC \geq \delta_{{\cal P}}(n_0) \NR \stopmathalignment \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 $F(n) =3D \ln f(n)$. Per (\in[desigualtat-f]) prenent lo= garitmes i operant, tenim que $F$ compleix que \placeformula \startformula \startmathalignment \NC F(n) \NC =3D F(n-2) + 2 \ln n + \ln \frac{1}{2}, \NR[+] \NC F(n_0) \NC =3D \ln f(n_0)\NR \stopmathalignment \stopformula Pel Lema \in[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 Agafant $f(n_0) \geq \delta_{{\cal P}}(n_0) + 1 > 0$, $F(0) =3D \ln C_1$ = i $F(1) =3D \ln C_2$ amb $C_1$ i $C_2$ constants, que nom=C3=A9s depenen = de $n_0$ i de $\delta_{{\cal P}}$, 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 = agafar $f(n_0) > 0$ 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. \stopdemo \starttheorem Siguin $G$ un grup, $A$ un conjunt finit de generadors de $= G$ i $\sigma \colon G \to A^*$ una secci=C3=B3 d'amplada $\varphi$ tal qu= e existeix $n_0 \in \naturalnumbers$ tal que $\varphi(n) < n-1$ per a tot= $n \geq n_0$. Aleshores la funci=C3=B3 de Dehn de $G$, $\delta_G$, safis= f=C3=A0 \startformula \delta_{G} (n) \simeq \frac{(n!!)^2}{2^{n/2}}, \stopformula per a tot $n \geq n_0$. \stoptheorem \startdemo Sigui ${\cal P} =3D \langle A \mid R\rangle$ una presentaci=C3= =B3 qualsevol de $G$. Pel Teorema \in[Theorema-n!!-Presentacions] existei= x una constant $C_{{\cal P}, n_0}$, que dep=C3=A8n de ${\cal P}$ i de $n_= 0$, tal que=20 \startformula \delta_{{\cal P}} (n) \leq C_{{\cal P},n_0} \cdot \frac{(n!!)^2}{2^{n/2}}= . \stopformula per a tot $n \geq n_0$. Com que les funcions de Dehn de dues presentacion= s de $G$ s=C3=B3n $\simeq$-equivalents \cite[extras=3D{, Proposici=C3=B3~= 1.3.3}][bridson-tutorial], llavors la funci=C3=B3 de Dehn, $\delta_G$, =C3= =A9s $\simeq$-equivalent a $\delta_{\cal P}$ i, per tant, \startformula \delta_G \simeq \frac{(n!!)^2}{2^{n/2}}, \stopformula per a tot $n \geq n_0$. \stopdemo \startnote \startitemize[1] \item Aquest teorema redueix de forma notable les fites superiors de les = funcions de Dehn conegudes per grups seccionables. Bridson prov=C3=A0 que= si $G$ =C3=A9s un grup tal que admet una secci=C3=B3 d'amplada $\varphi$= tal que $\varphi(n) < n-1$ assimpt=C3=B2ticament, aleshores la seva func= i=C3=B3 de Dehn $\delta_G$ t=C3=A9 ordre $e^{kn^3}$ \cite[extras=3D{, Teo= rema~4.3}][bridson], i Riley demostr=C3=A0 que si $G$ admet una secci=C3=B3= {\em geod=C3=A8ssica} tal que $\varphi(n) < n-1$ de forma assimpt=C3=B2t= ica, aleshores $\delta_{G}(n)$ =C3=A9s equivalent linealment a $n!$ \cite= [extras=3D{, Teorema~2}][riley]). Com que $(n!!)^2$ =C3=A9s una fita supe= rior m=C3=A9s baixa i no =C3=A9s equivalent linealment a cap de les dues = fites anteriors, aleshores aquest teorema millora les fites. Record que $f$ =C3=A9s equivalent linealment a $g$ si, i nom=C3=A9s si, $= f \preceq g$ i $g \preceq f$, on $f \preceq g$ significa que existeix una= constant $k > 0$ tal que $f(x) \leq k g(kx + k) + kx + k$ per a tot $x \= geq 0$. \item Tenim molt m=C3=A9s que $\delta_G (n) < (n!!)^2$, tal com posa de m= anifest l'apartat $a$ del teorema. \item La Proposici=C3=B3 \in[desigualtat-dehn] tamb=C3=A9 implica fites i= nferiors, molt grolleres per cert. \stopitemize Podem confirmar aquest teorema amb els resultats seg=C3=BCents: \starttheorem[teorema4.2millorat]Siguin $G$ un grup, $A$ un conjunt finit= de generadors de $G$, $\sigma \colon G\rightarrow A^*$ una secci=C3=B3 d= e $G$ respecte de $A$ i $F\colon \naturalnumbers \rightarrow \naturalnumb= ers$ una funci=C3=B3 qualsevol tal que $F(n) \geq 1$ per a tot $n > 0$. S= i $\varphi(n) < n-1$ per a $n$ suficientment gran i $F$ compleix que \startformula F(n) \geq \frac{1}{2} n^2 F\big(2 \varphi({n}/{2}) + 2\big) \stopformula per a $n$ suficientment gran, aleshores $F$ =C3=A9s una funci=C3=B3 isope= rim=C3=A8trica de $G$. \stoptheorem \startdemo Basta veure que $F$ =C3=A9s una funci=C3=B3 isoperim=C3=A8tric= a per a alguna presentaci=C3=B3 finita de $G$, =C3=A9s a dir, que existei= x una presentaci=C3=B3 finita ${\cal P}$ de $G$ tal que qualsevol paraula= nul-homot=C3=B2pica $w$ de longitud $l(w) \leq n$ satisf=C3=A0 que $\tex= t{area}(w) \leq F(n)$. Sigui $N$ tal que $\varphi(n) < n-1$ i $F(n) \geq \frac{1}{2} n^2 F(2 \va= rphi(n/2) +2)$ per a tot $n \geq N$. Tenim que ${\cal P} =3D \langle A \m= id R_0 \rangle$ =C3=A9s una presentaci=C3=B3 finita de $G$, on $R_0 =3D \= {w \in A^* \mid w \text{ nul-homot=C3=B2pica, } l(w) \leq 2N\}$ \cite[ext= ras=3D{, Proposici=C3=B3~3.1}][bridson]. Sigui $w$ una paraula nul-homot=C3=B2pica per ${\cal P}$. Provem per indu= cci=C3=B3 sobre $l(w) $ que $F$ =C3=A9s una funci=C3=B3 isoperim=C3=A8tri= ca per aquesta presentaci=C3=B3. \startitemize[1] \item Si $w$ =C3=A9s una paraula nul-homot=C3=B2pica tal que $l(w)\leq 2N= $, aleshores $\text{area}(w) =3D 1$ (ja que, per estar $w \in R_0$, exist= eix un diagrama de van Kampen ${\cal D}$ sobre ${\cal P}$ que t=C3=A9 $w$= com a frontera i com a =C3=BAnica cara. A m=C3=A9s, $C({\cal D}) =3D \te= xt{area}(w)=3D 1$). Com que $F(n) \geq 1$ per a tot $n \geq 1$, tenim que= $\text{area}(w) =3D 1 \leq F(2N)$. \item Suposem que $l(w) =3D n > 2N$ i que qualsevol paraula nul-homot=C3=B2= pica de longitud $r < n$, $w_r$, =C3=A9s tal que $\text{area}(w_r) \leq F= (r)$. Vegem que $\text{area}(w) \leq F(n)$. Pel Lema \in[subadditivitat-area], existeixen $u_k$ paraules nul-homot=C3= =B2piques per ${\cal P}$, amb $k =3D \{1, \ldots, n^2/2\}$, tals que \startformula \text{area}(w) \leq \sum_{i=3D1}^{n^2/2} \text{area}(u_k) \stopformula i $l(u_k) \leq 2 \varphi(n/2) + 2$. Com que $\varphi(n) < n-1$, tenim que $l(u_k) \leq n-2$ i per tant podem = aplicar hip=C3=B2tesi d'inducci=C3=B3. Aplicant la desigualtat sobre $F$ = que suposam per hip=C3=B2tesi, tenim que \placeformula[-] \startformula \startsplit \NC \text{area}(w) \NC \leq \sum_{i=3D1}^{n^2/2} \text{area}(u_k) \NR \NC \NC \leq \sum_{i=3D1}^{n^2/2} F(2 \varphi(n/2) + 2) \NR \NC \NC \leq \frac{1}{2} n^2 F(2 \varphi(n/2) + 2) \NR \NC \NC \leq F(n). \stopsplit \stopformula \stopitemize \stopdemo \startcorollary $(n!!)^2$ =C3=A9s una funci=C3=B3 isoperim=C3=A8trica. \stopcorollary \startdemo Pel Teorema \in[teorema4.2millorat] basta veure que la funci=C3= =B3 $n \mapsto (n!!)^2$ compleix \stopdemo \completepublications[criterium=3Dcite] %all per tots \stoptext --------------060307080707070702010906 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 ___________________________________________________________________________________ --------------060307080707070702010906--