Re: quantum logic

[Michael Barr <barr@barrs.org>]

I will stick to my perception that if you dealing with Hilbert or Banach
spaces isomorphisms should be just that.  It makes no difference to the
*-autonomous structure anyway.

For Banach spaces, if you take as underlying functor the closed unit ball,
it has an adjoint.  It is not tripleable, however, but C^*-algebras are
(with the unit ball underlying functor).

On Mon, 20 Oct 2003, Toby Bartels wrote:

> 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
>
>
>