From mboxrd@z Thu Jan 1 00:00:00 1970 X-Msuck: nntp://news.gmane.io/gmane.science.mathematics.categories/3387 Path: news.gmane.org!not-for-mail From: "George Janelidze" Newsgroups: gmane.science.mathematics.categories Subject: Re: Linear--structure or property? Date: Fri, 11 Aug 2006 16:53:04 +0200 Message-ID: NNTP-Posting-Host: main.gmane.org Mime-Version: 1.0 Content-Type: text/plain;charset="iso-8859-1" Content-Transfer-Encoding: 7bit X-Trace: ger.gmane.org 1241019274 8536 80.91.229.2 (29 Apr 2009 15:34:34 GMT) X-Complaints-To: usenet@ger.gmane.org NNTP-Posting-Date: Wed, 29 Apr 2009 15:34:34 +0000 (UTC) To: "Categories list" Original-X-From: rrosebru@mta.ca Fri Aug 11 14:04:15 2006 -0300 Return-path: Envelope-to: categories-list@mta.ca Delivery-date: Fri, 11 Aug 2006 14:04:15 -0300 Original-Received: from Majordom by mailserv.mta.ca with local (Exim 4.61) (envelope-from ) id 1GBaNp-0003UN-UA for categories-list@mta.ca; Fri, 11 Aug 2006 14:01:17 -0300 Original-Sender: cat-dist@mta.ca Precedence: bulk X-Keywords: X-UID: 20 Original-Lines: 76 Xref: news.gmane.org gmane.science.mathematics.categories:3387 Archived-At: Dear Steve, It is true that constructing such examples with more than one object is even easier in a sense. However your example needs a minor correction: What is 1+1 in C(x,x) and in C(y,y)? If it is 0, them M must be a Z/2Z-module, since for every element u in M we have u+u = 1u+1u = (1+1)u = 0u = 0. If it is 1, them M must be idempotent (as before: u+u = 1u+1u = (1+1)u = 1u = u). If it is 1 in C(x,x) and 0 in C(y,y), then M becomes trivial, which destroys the example. And... why did not you and I just take the monoid {0,1}, which becomes a commutative semiring for both additions?! More generally, take any non-degenerated Boolean algebra. It has multiplication=intersection=meet, and at least two additions (symmetric difference and union=join) both good for that multiplication. George ----- Original Message ----- From: "Stephen Lack" To: "Categories list" Sent: Friday, August 11, 2006 12:49 PM Subject: categories: RE: Linear--structure or property? It's a structure. Consider the following category C. Two objects x and y, with hom-categories C(x,x)=C(y,y)={0,1} C(y,x)={0} C(x,y)=M with composition defined so that each 1 is an identity morphism and each 0 a zero morphism, and with M an arbitrary set. Any commutative monoid structure on M makes C into a linear category. Steve. -----Original Message----- From: cat-dist@mta.ca on behalf of Michael Barr Sent: Fri 8/11/2006 6:14 AM To: Categories list Subject: categories: Linear--structure or property? Bill Lawvere uses "linear" for a category enriched over commutative semigroups. Obviously, if the category has finite products, this is a property. What about in the absence of finite products (or sums)? Could you have two (semi)ring structures on the same set with the same associative multiplication? Robin Houston's startling (to me, anyway) proof that a compact *-autonomous category with finite products is linear starts by proving that 0 = 1. Suppose the category has only binary products? Well, I have an example of one that is not linear: Lawvere's category that is the ordered set of real numbers has a compact *-autonomous structure. Tensor is + and internal hom is -. Product is inf and sum is sup, but there are no initial or terminal objects and the category is not linear.