Re: quantum logic

[Fred E.J. Linton <fejlinton@usa.net>]

I'll  address two of these questions.  The first:

> 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.

Actually, when the "underlying-set functor" for Banach spaces
is taken to be the unit disk functor, and the morphisms are
taken as the norm-decreasing maps, the situation is really great,
because the norm-decreasing maps DO constitute the unit disk
of the Banach space of bounded linear transformations, as you know.
And products and coproducts are as Banach spacists like to see them
(the familiar L-infinity style "full direct product" and and L-1 
style "weak direct product", respectively).

When the underlying-set functor is taken to be ALL the vectors 
of the Banach space, on the other hand, products and coproducts 
misbehave quite badly. 

As for the question,

> 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?

I'd answer by saying that unless the invertible bounded linear map
in the question IS an isometry I'd never dare call it one.

-- Fred (usually <FLinton@Wesleyan.edu>)

Toby Bartels <toby@math.ucr.edu> 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
> 
> 
> 
>