From mboxrd@z Thu Jan 1 00:00:00 1970 X-Msuck: nntp://news.gmane.io/gmane.science.mathematics.categories/2478 Path: news.gmane.org!not-for-mail From: Toby Bartels Newsgroups: gmane.science.mathematics.categories Subject: Re: quantum logic Date: Mon, 20 Oct 2003 12:51:07 -0700 Message-ID: <20031020195106.GA2487@math-rs-n03.ucr.edu> References: <200310122208.h9CM8sf26075@math-cl-n01.ucr.edu> NNTP-Posting-Host: main.gmane.org Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii X-Trace: ger.gmane.org 1241018689 4341 80.91.229.2 (29 Apr 2009 15:24:49 GMT) X-Complaints-To: usenet@ger.gmane.org NNTP-Posting-Date: Wed, 29 Apr 2009 15:24:49 +0000 (UTC) To: categories Original-X-From: rrosebru@mta.ca Wed Oct 22 11:36:16 2003 -0300 Return-path: Envelope-to: categories-list@mta.ca Delivery-date: Wed, 22 Oct 2003 11:36:16 -0300 Original-Received: from Majordom by mailserv.mta.ca with local (Exim 4.10) id 1ACK0U-0002YF-00 for categories-list@mta.ca; Wed, 22 Oct 2003 11:30:38 -0300 Content-Disposition: inline In-Reply-To: Original-Sender: cat-dist@mta.ca Precedence: bulk X-Keywords: X-UID: 29 Original-Lines: 51 Xref: news.gmane.org gmane.science.mathematics.categories:2478 Archived-At: Michael Barr wrote in part: >After giving the matter some thought, I finally decided that the >category of Hilbert spaces should have as its morphisms norm-reducing >linear maps. At the very least that will ensure that an isomorphism is >an isometry. True, but are you begging the question by trying to ensure that? After all, an invertible bounded linear map is enough to deduce that Hilbert spaces are isomorphic (even in the sense of isometric), so why not count those maps as isomorphisms themselves? This matter is much bigger than Hilbert spaces, of course; moving to Banach spaces (a closed category even for arbitrary dimension), we can even see how, /as/ a closed category, it doesn't really matter! The question is, what is the forgetful functor from Ban to Set? Do we take the set of all vectors? or do we take the closed unit ball? The former corresponds to allowing all bounded linear maps as morphisms, while the latter corresponds to requiring norm-reducing linear maps. But in the closed category Ban, the Banach space of morphisms is, whatever your conventions, the space of all bounded linear maps. Still, this can be consistent with either choice of hom-SET, since the closed unit ball in the Banach space of bounded linear maps is none other than your preferred hom-set of norm-reducing maps. Jim Dolan (IIRC) suggested that Ban is more fundamentally a closed category than a category in the first place. We can do this on a more elementary level with metric spaces; is the hom-set the set of all Lipschitz continuous functions, or is it only the set of distance-reducing functions? But unlike with Banach (or Hilbert) spaces, this makes a difference even to the classification of metric spaces into isomorphism classes. The question becomes, is an isomorphism of metric spaces merely a relabelling of points keeping all distances the same, or does it also allow for a recalibration of ones ruler? Which is the correct interpretation may depend on the application, and how absolute -- rather than measured in some unit -- the distances are. (One can even recalibrate more generously to allow as morphisms all uniformly continuous maps, or even all continuous maps. Thus classically one speaks of variously "equivalent" metric spaces, such as "uniformly equivalent" or "topologically equivalent".) To get closed categories here, one must restrict to bounded metric spaces; the analysis is a little more fun than for Banach spaces, especially with the degeneracy surrounding the initial and terminal spaces. -- Toby