## Friday, February 9, 2007

### Be wise, quantize!

Perhaps a better title for this series of posts would be Quantization and Noncommutative Geometry". This is a huge topic and certainly takes a lot of time and contributions by many people to do justice to the subject. In a nutshell I would say the revolution brought in physics by the advent of quantum mechanics in the hands of Heisenberg, Dirac, Schrodinger and others in the years 1925-1926 is in many ways echoed in mathematics through noncommutative geometry. It took almost 55 years (1925 to 1980, roughly, since Connes already in 1978 was talking about the foliation algebra of a foliation and proved an index theorem for them) to reach to the current phase of development of NCG (= noncommutative geometry). It is a long time and it is certainly interesting to know why it took so long, but that is another issue.

I would like to invite all those who are interested to contribute to the following issues or to a related topic of their choice.
1. Dirac quantization rules and NCG
2. No go theorems
3. Various quantization schemes: geometric quantization, deformation quantization, Berezin and Toeplitz quantization, etc....
3. Semiclassical limits
4. Applications to mathematics, e.g. to index theorems,
5. Quantum groups
6. Noncommutative geometry techniques, e.g. the role of groupoids, strict deformations
7. The role of operator algebras, and original ideas of von Neumann
8. Second quantization
9. Quantum field theory and NCG
As I said this list is incomplete, so feel free to add topics, and also discuss!

AC said...

This topic of "quantization and NCG" is very relevant. The word quantum', from the beginning, is not so much related to non-commutativity' but rather to integrality'. In the word quantum' there is really this discovery by Planck, of the formula
to be quantized in quanta of $\hbar \nu$. There is a confusion, created by people doing deformation theory who let one
believe that quantizing an algebra just means deforming it to a
non-commutative one. They take a commutative space and since they
deform the product into a non-commutative algebra, they believe they are quantizing. But this is wrong: you succeed in
quantizing a space only if you give a deformation into a very
specific algebra : the algebra of compact operators. And then, there
is an integrality, the integrality of the Fredholm index. The use of
the wrong vocabulary, creates confusion and does not help at all to understand. That's why I am so reluctant to use the word quantum' - instead of "non-commutative" and am against talking about "quantum spaces" or
"quantum geometry"....
this looks more flashy, perhaps, but the truth is that you are doing
something quantum only in very particular cases, otherwise you are
doing something non-commutative, that's all. Then this may be less
fashionable at the linguistic level, but never mind: it is much closer to reality.

Anonymous said...

Hi,

I am glad to see a blog like this here come into existence. Will be very interested in what is going on here.

I am from over at the n-category Cafe, where we like to think about what one might call "pull-push quantization" -- and its application in extended n-dimensional QFT.

(This seems to be particularly well adapted to "fundamental systems".)

John Baez is posting an entire lecture on Quantization and Cohomology to the Cafe.

While no true NCG experts, we are very interested in understanding relations to NCG, as for instance indicated in the thread Connes on Spectral Geometry of the Standard Model, II.

Best regards,
Urs

Anonymous said...

Since in the above entry it says

"I would like to invite all those who are interested to contribute to the following issues or to a related topic of their choice. [...] 7. The role of operator algebras"

I might maybe dare to audaciously go ahead and post a question belonging to that topic:

on Hilb -- the category of Hilbert spaces and isomorphism between them -- we have a functor to the category of C*-algebras, which sends every Hilbert space to the C*-algebra of its bounded endomorphisms.

The GNS construction, on the other hand, sends C*-algebras to Hilbert spaces, presumeably in a functorial way.

Question: are these two functors mutually adjoint?

masoud said...

Dear Anonymous,
Thanks for your question. In fact there is no such functor from C^*-algebras, per se, to Hilbert spaces. The GNS construction starts with a C^*-algebra and a STATE on it and produces a Hilbert space as well as a C^*-morphism from the given C^*-algebra into the algebra of bounded operators on that Hilbert space. This representation may very well fail to be faithful, but it is known that it is faithful iff the state is pure'. The GNS construction (in the unital case at least) in fact solves a universal problem in that the state becomes a
vacuum state' in that representation and it is universal for such a property. Being a universal construction, I guess it is pretty easy to guess that the GNS construction is adjoint to what functor from which category to which category. I can say more if needed, but a quick start would be to look at my recent lectures on
noncommutative geometry' (page 7) posted in the archive. You can also check Alain's 94 book which is available online. Please let me know if more is needed.

masoud said...

oops! in the above comment I meant to say irreducible' instead of faitthful' (thanks to Arup!)

urs said...

Hi again,

it was me who asked the above question (I did not mean to remain anonymous) -- many thanks for the reply!

There is a choice involved in the GNS construction, that of a state. This alone need not mean that we don't get an adjoint functor. It is quite common for adjoint functors to require us making lots of choices.

What happens if I choose a different state and do the construction again? By the universal property you mention, the two Hilbert spaces obtained this way should be isomorphic.

You wrote:

"I guess it is pretty easy to guess that the GNS construction is adjoint to what functor from which category to which category."

Yes, guessing is not the problem here! :-) I would just like an expert to confirm this.

You furthermore wrote:

"but a quick start would be to look at my recent lectures on
noncommutative geometry' (page 7) posted in the archive. You can also check Alain's 94 book which is available online"

Okay, thanks. I will see if I can track these down. Thanks a lot!

masoud said...

hi again,
I think before discussing the functoriality property of the GNS construction it is important to realize that the Hilbert space that one defines through GNS, in general, depends on the state. For different states \phi on the same C*-algebras you can get Hilbert spaces of different dimensions. Example: start with a commutative algebra A =C(X), or A =C[0, 1] for that matter. Then a state on A is simply a probability measure on X. If you choose the Dirac mass at a point, the associated GNS Hilbert space is H=C (complex numbers). For a measure supported at a finite number of points, you get a finite dimensional Hilbert space, while if you start with the Lebesgue measure you obviously obtain L^2[0, 1]. There are simple noncommutative examples too. For A= M_n(C), and the tracial state, the GNS Hilbert space is H= M_n(C) with Hilbert-Schmidt inner product, while if your initial state is \phi (a_ij)= a_00, you obtain C^n, an n-dimensional Hilbert space.

So, the initial data for the GNS is really a pair (A, \phi) of a C*-algebra and a state on it and the role of \phi is by no means auxiliary.

urs said...

You write:

"So, the initial data for the GNS is really a pair (A, \phi) of a C*-algebra and a state on it and the role of \phi is by no means auxiliary."

Ah, thanks. I did not appreciate the role played by the states, properly.

So, let me see if I can come up with a well-formed version of my question then.

I presume we have a category of pairs (A,\phi), whose morphisms are pairs consisting of an algebra isomorphism and of the respective action on the state.

On this category the GNS construction has better chances to extend to a functor to Hilb. Does it?

And then I would want a functor from Hilb back to that category of pairs, which is weakly inverse, or at least adjoint, to the former one.

So I'd need to send any given Hilbert space to the pair consisting of its algebra of bounded operators, together with a state on that.

Does that exist?

masoud said...

That is good; at least now we have a common starting point. I don't think an interesting functor from Hilb back to the category of pairs really exits (but see below). In fact Hilb is not the target category for the GNS (unless you are willing to forget a lot by composing it with a forgetful functor to Hilb). I can now elaborate on my first response to your question. Given a pair (A, \phi), and assuming A is unital for simplicity, the GNS construction gives us a triple (H, \pi, v), where H is a Hilbert space, \pi is a representation of A in L(H), and v is a vector in H of length one. The following two conditions are satisfied: v is a cyclic vector for \pi and \phi (a) = < \pi(a)v, v> for all a in A. One checks that any other triple with these conditions is unitarily equivalent (in an obvious sense) to the GNS triple. Now if you insist on seeing the GNS construction as a functor, something like this can be done: I suppose if you consider a category consisting of quadruples (A, H, \pi, v)as above with morphisms consisting of unitary equivalences and another category consisting of pairs (A, \phi), then the GNS construction defines an equivalence of these two categories. The quasi inverse is given by
the functor ( A, H, \pi, v) ------->(A, \phi) where \phi (a)= <\pi (a) v, v>.

urs said...

Hi,

okay, good. Thanks for you patience with me!

Hopefully without straining that patience too much once again, let me try to suggest a slight modification of the category of quadruples which you mentioned, a modification that comes closer to the original motivation I had:

You suggest a category whose objects are quadruples consisting of

H -- a Hilbert space

A -- a (unital, say) C*-algebra

\pi -- a rep of A on H by bounded operators (you mean B(H) instead of L(H), right?)

v -- a unit vector in H (required to by cyclic with respect to pi(A), as far as I understand) .

Now, what if I passed instead to a category of pairs (other pairs than those we discussed above), by borrowing the idea of your quadruples, but taking the algebra A always to be B(H) and hence \pi the identity?

So I am now thinking of objects being pairs

(H,v)

with H a Hilbert space and v a cyclic vector (wrt B(H)).

I would want to send this pair

(Hilbert space, vector)

to a pair

(C*-algebra,state)

the way you indicated, only that in this setup \pi would not be additonal information so that I just get

(H,v) |-->(A, \phi)

where

A = B(H)

and where

\phi (a)= <a v, v>.

Does that make sense?

yemon said...

Several years ago I had a go at formulating the GNS embedding that arises by taking the direct product of all GNS reps over all states as a left adjoint, but couldn't get the details to work. (I think this is what Urs' original post was aiming at.)

One problem, if I recall correctly, is that the uniqueness of factorization needed in the definition of left adjoint doesn't seem quite to work.

By the way, for general Banach spaces E the operation `given E, form the Banach algebra B(E)' is not functorial -- so the Hilbert space case, if it works, has to be a little subtler.

masoud said...

This is in response to Urs's last question. Yes, it makes sense, but one has to be a bit careful about the class of states on L(H) (=B(H), in your notation). For any positive trace class operator p with Tr (p)=1,
\phi (a)=Tr (ap) is a state one L(H). It is pure iff p is a rank one projection which in this case then is identified with a unit vector v. It is however not true that all states on L(H) are of this type. Only normal (meaning continuous in the ultra weak topology of L(H)) states on L(H) are of this type. Thus the two categories you want to look at are the category of pairs (H, v) and the category of pairs (L(H), \phi), where \phi is a pure normal state on L(H) and the GNS construction sets an equivalence between the two. Admittedly this is a rather degenerate case of the GNS construction but that is what it is....
As for the last comment, the association H \to L(H) is functorial if you restrict to the class of unitary equivalences, as the intention was.