RE: [math-fun] Re: Canonical 1-1 correspondences
-----Original Message----- From: John McCarthy [mailto:jmc@steam.Stanford.EDU] Sent: Tuesday, May 20, 2003 7:27 PM To: math-fun@mailman.xmission.com Cc: math-fun@mailman.xmission.com; jmclists@cs.Stanford.EDU Subject: Re: [math-fun] Re: Canonical 1-1 correspondences
John Conway wrote
But if you're going to allow infinite classes, then trivially the set of all correspondences is a canonical one.
Yes, but consider the correspondences between X and X* you get when you choose a basis for X. A basis element b of X corresponds to the functional that is 1 on b and 0 on the other basis elements. It seems to me that these correspondences should be the canonical set of correspondences. However, I don't know of a general criterion for preferring them.
Aren't these exactly the linear bijections between X and X*? Andy Latto andy.latto@pobox.com
On Wed, 21 May 2003, Andy Latto wrote:
-----Original Message----- From: John McCarthy [mailto:jmc@steam.Stanford.EDU] Sent: Tuesday, May 20, 2003 7:27 PM To: math-fun@mailman.xmission.com Cc: math-fun@mailman.xmission.com; jmclists@cs.Stanford.EDU Subject: Re: [math-fun] Re: Canonical 1-1 correspondences
John Conway wrote
But if you're going to allow infinite classes, then trivially the set of all correspondences is a canonical one.
Yes, but consider the correspondences between X and X* you get when you choose a basis for X. A basis element b of X corresponds to the functional that is 1 on b and 0 on the other basis elements. It seems to me that these correspondences should be the canonical set of correspondences. However, I don't know of a general criterion for preferring them.
Aren't these exactly the linear bijections between X and X*?
Yes, they are. JHC
I see that in my previous posts I only hinted at what I wanted to say. 1. It always seemed to me that the natural way to prove two sets equipollent was to give a bijection. Proofs not involving a bijection strike me as less desirable and less informative. 2. Sometimes there is a canonical bijection between the sets, and that's nice. 3. However, when there isn't a canonical bijection, there may be a canonical set of bijections no single one of which is canonical. 4. This is the case with a space and its dual space. The set of linear maps is indeed canonical. I suppose the example is too easy to consider the set of linear maps as providing the good proof of equipollence. 5. I suppose there are cases when there is neither a canonical 1-1 correspondence or a canonical set of them. A criterion is wanted for excluding the set of all 1-1 correspondences; it's unlikely to be constructed as a part of the proof of equipollence.
Come on, people. Epsilon more consideration. A linear isomorphism between a vector space and its dual is equivalent to a bilinear form on the vector space. (mapping X to the set of maps from X to Y is the same as giving a map of X x X to Y). When you map a basis to its dual basis, this gives a particular kind of isomorphism of X to X* (provided X is finite dimensional) that is described by the identity matrix. The bilinear form is symmetric, so not all isomorphisms are represented in this way. Any orthogonal matrix describes a basis that determines the same isomorphism. On Wednesday, May 21, 2003, at 09:50 AM, John Conway wrote:
On Wed, 21 May 2003, Andy Latto wrote:
-----Original Message----- From: John McCarthy [mailto:jmc@steam.Stanford.EDU] Sent: Tuesday, May 20, 2003 7:27 PM To: math-fun@mailman.xmission.com Cc: math-fun@mailman.xmission.com; jmclists@cs.Stanford.EDU Subject: Re: [math-fun] Re: Canonical 1-1 correspondences
John Conway wrote
But if you're going to allow infinite classes, then trivially the set of all correspondences is a canonical one.
Yes, but consider the correspondences between X and X* you get when you choose a basis for X. A basis element b of X corresponds to the functional that is 1 on b and 0 on the other basis elements. It seems to me that these correspondences should be the canonical set of correspondences. However, I don't know of a general criterion for preferring them.
Aren't these exactly the linear bijections between X and X*?
Yes, they are. JHC
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participants (4)
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Andy Latto -
John Conway -
John McCarthy -
William Thurston