User tobias fritz - MathOverflow

Tannaka duality for C*-algebras?

2013-04-21T17:00:59Z

Tannaka-Krein duality shows how to recover a group $G$ from its category $\mathbf{Rep}(G)$ of finite-dimensional complex representations and the forgetful functor $F:\mathbf{Rep}(G)\to \mathbf{Vect}_{\mathbb{C}}$.

On the other hand, there is Takesaki's theorem which shows how to recover a (separable) $C^*$-algebra $A$ from its representation theory. We have just started looking into the details of this and trying to reformulate this in categorical terms. It seems tantalizingly similar to the Tannaka-Krein reconstruction theorem. In particular, it seems that Takesaki secretly also considers natural transformations from the forgetful functor $F:\mathbf{Rep}(A)\to\mathbf{Hilb}$ to itself.

So, the question is:

Can Takesaki's duality theorem indeed be formulated in categorical terms similar to Tannaka-Krein duality? Where can we read about it?

We're mostly interested in the unital case. Follow-up question:

Consider the category of unital $C^*$-algebras and unital completely positive maps. Is there a "nice" description of its opposite as a concrete category?

Answer by Tobias Fritz for Realizing universal C-algebras as concrete C-algebras

2013-01-28T17:42:45Z

Edit: as pointed out in the comments, the following answers the question for unital C*-algebras presented in terms of generators and relations. When I say C*-algebra, I really mean unital C*-algebra.

It may depend on what exactly you mean by "concrete", but I highly doubt that there is a general solution to this; finding a concrete realization of a universal C*-algebra requires classifying all the representations of the given generators and relations on Hilbert space, and this is an extremely difficult problem in general. For a more rigorous argument, see below.

But in your four examples, the universal C*-algebras are all commutative, and simple answers are possible:

$C([-1,1])$.
$C([0,1])$.
$C(\mathbb{D})$ where $\mathbb{D}$ is the unit disk.
$C(\{0,1\})=\mathbb{C}^2$ .

In each case, the generator is the identity function, just as in your $C(\mathbb{T})$ example. It is a good exercise to verify the required universal property in each of these cases.

Another good example is the C*-algebra freely generated by two projections. This turns out to be the group C*-algebra $C^*(\mathbb{Z}_2\ast \mathbb{Z}_2)=C^*(\mathbb{Z}\rtimes\mathbb{Z}_2)$ and can be realized concretely as the subalgebra of $C([0,1],M_2(\mathbb{C}))$ containing those matrix-valued functions which are diagonal on the endpoints $0$ and $1$. See this paper of Raeburn and Sinclair.

So why do I think that a general solution is impossible? Consider the word problem for groups: there are groups given in terms of generators and relations for which there is no algorithm that can decide whether a given word in the generators represents the unit element. Now we can look at the maximal group C*-algebra of such a group. This C*-algebra is itself given by the same generators and relations together with additional relations requiring the generators to be unitary. If your intended meaning of a "concrete representation" comprises the existence of an algorithm that decides whether a given formal combination of generators represents $0$, then it follows that such a concrete representation cannot exist.

graphs with independence number = Shannon capacity

2012-10-04T19:31:38Z

For $G$ a graph, let $\alpha(G)$ be its independence number and $\Theta(G)=\lim_n \sqrt[n]{\alpha(G^{\boxtimes})}$ its Shannon capacity, where $\boxtimes$ denotes strong product.

Consider graphs $G$ and $H$ satisfying $\alpha(G)=\Theta(G)$ and $\alpha(H)=\Theta(H)$. For example, $G$ and $H$ could be perfect, but the more interesting situations arise when neither of them is perfect.

Question: Does this assumption imply
(1) $\alpha(G\boxtimes H) = \alpha(G)\alpha(H)$ ?
(2) $\Theta(G\boxtimes H) = \Theta(G)\Theta(H)$ ?
(3) $\Theta(G + H) = \Theta(G) + \Theta(H)$ ?

Here, $G+H$ stands for the disjoint union of $G$ and $H$.

If my reasoning is correct, then (1) and (2) are equivalent and imply (3).

As far as I can see, neither the work of Haemers nor the results of Alon have anything directly to say about these questions. But then again, I am not an expert on this, so I might have missed something obvious.

Edit (see Will Traves' answer): Actually, I am specifically interested in those $G$ and $H$ which are well-covered.

Edit: The paper is here.

Answer by Tobias Fritz for Upper bound on Shannon capacity based on independence number

2012-11-26T23:51:33Z

An inequality as simple as $\Theta(G)\leq \alpha(G)+1$ can certainly not hold for all $G$: take some $G$ with $\Theta(G)>\alpha(G)$ and consider the disjoint union $G+G$. Since $\alpha$ is additive under disjoint union while $\Theta$ is superadditive, this $G+G$ will have a gap between $\Theta$ and $\alpha$ which is at least twice as big as $G$'s. Now repeat this process if necessary.

This paper of Alon and Lubetzky seems highly relevant. After proving several negative results (which I don't fully grasp), they conjecture that $$ \Theta(G) \leq 2\max_{k=1,\ldots,|G|} \sqrt[k]{\alpha(G^k)} , $$ where $|G|$ is the number of vertices.

Answer by Tobias Fritz for Skew fraction fields of *-algebras

2012-11-01T21:28:23Z

If I'm not mistaken, it follows immediately from abstract nonsense that the $*$-operation extends uniquely from $R$ to the field of fractions.

The notion of localization is more general than that of field of fractions, and hence it is sufficient to prove that if $S\subseteq R$ is any set with $S^*=S$ , then $RS^{-1}$ inherits a unique $*$-operation.

But this is an immediate consequence of the universal property of $RS^{-1}$: the $*$ -operation is a homomorphism $R^{op}\to R$, which can be composed with $R\to RS^{-1}$ to give $R^{op}\to RS^{-1}$. By $S^*=S$ , this homomorphism maps every element of $S$ to an invertible element, and hence this induces $*:(RS^{-1})^{op}\to RS^{-1}$ which is compatible with the original $*:R\to R$ . In particular, this compatibility guarantees also that the extended $*$ -operation is antilinear.

The universal property should also guarantee that the extended $*$ is an involution as well.

Edit (see comments): The following is a proof sketch showing that $(RS^{-1})^{op}$ and $R^{op}S^{-1}$ are canonically isomorphic.

I think the easiest way to see this is to show that $(RS^{-1})^{op}$ has the universal property of $R^{op}S^{-1}$, i.e. that the canonical morphism $R^{op}\to (RS^{-1})^{op}$, which is defined simply as the opposite of $R\to RS^{-1}$, is the universal morphism with domain $R^{op}$ which maps $S$ to invertibles.

To see this, consider any $R^{op}\to T$ which maps $S$ to invertibles. Then its opposite $R\to T^{op}$ also maps $S$ to invertibles, and hence factors through $RS^{-1} \to T^{op}$. Taking opposites again gives the desired $(RS^{-1})^{op}\to T$.

making a graph well-covered without changing its Shannon capacity

2012-10-12T20:17:00Z

This strongly relates to an earlier question of mine.

Let $G$ be a graph, $\alpha(G)$ its independence number and $\Theta(G)$ its Shannon capacity.

Question: can one 'add new vertices' to $G$ such that $G\subseteq G'$ becomes an induced subgraph of some well-covered $G'$ with $\alpha(G')=\alpha(G)$ and $\Theta(G') = \Theta(G)$?

If so, this would be a way to 'uniformize' a graph without changing its Shannon capacity. Has anyone considered this question before?

Thanks!

Comment by Tobias Fritz

2013-04-21T21:15:54Z

Thanks for this, that's a good start. Those Yoneda lemma applications are incredibly sneaky!

Comment by Tobias Fritz

2013-02-10T21:51:32Z

What do you mean exactly by "solving" a linear system? Deciding whether it has a solution? Interesting semirings are $\mathbb{R}_{\geq 0}$ and $\mathbb{Q}_{\geq 0}$, over which deciding whether a solution exists is linear programming and hence in $P$. On the other hand, finding a minimal set of generating solutions such that every other solution is a linear combination of the given ones is very difficult (vertex enumeration problem, e.g. for polytopes).

Comment by Tobias Fritz

2013-01-28T21:38:38Z

@Benjamin: you certainly know this, but just to make things clear: the universal representation is the direct sum of the GNS representations associated to all states. In particular, it is also a faithful representation on a Hilbert space, just like the left regular rep. I agree that the latter is much more "concrete" than the universal rep, but the difference seems to lie in how explicit the rep itself and its underlying Hilbert space are defined. My (possibly naive) idea was to try and make the notion of concreteness precise by relating it to decidability.

Comment by Tobias Fritz

2013-01-28T19:37:30Z

@Benjamin: the reduced C*-algebra of a discrete group coincides with the universal one if and only if the group is amenable. And yes, there are amenable groups with undecidable word problem! (Kharlampovich constructed a solvable fp group with undecidable wp.) Hence my question about what the OP exactly means by "concrete": if it entails decidability, then the answer to the OP's main question is negative. On the other hand, if e.g. the left regular representation of a group still counts as concrete, then why not regard the universal representation as concrete as well?

Comment by Tobias Fritz

2013-01-28T18:46:47Z

@Nik: thanks, I just clarified this. I haven't thought about the non-unital case.

Comment by Tobias Fritz

2013-01-28T03:54:26Z

If you're willing to consider pseudo-Riemannian manifolds as well, then any solution to the Einstein field equations of general relativity will give you a perfectly nice example of a Ricci-flat manifold. Many of these, for example the Schwarzschild or Kerr solutions, are not flat.

Comment by Tobias Fritz

2013-01-27T02:21:48Z

See also geometric measure theory and Hadwiger's theorem, which is very nicely explained in a series of posts on the n-Category Café starting here: golem.ph.utexas.edu/category/2011/06/hadwigers_theorem_part_1.html

Comment by Tobias Fritz

2013-01-14T23:48:23Z

A neat application of linear codes also arises in network coding: en.wikipedia.org/wiki/…

Comment by Tobias Fritz

2012-12-30T19:27:11Z

duplicate: mathoverflow.net/questions/115421/…

Comment by Tobias Fritz

2012-12-29T20:12:15Z

Yes, if I recall correctly, this one can also be admired on the glass front of the MPIM's reception desk.

Comment by Tobias Fritz

2012-12-25T08:22:53Z

Is a free product of Hopfian groups again Hopfian?

Comment by Tobias Fritz

2012-12-12T18:22:27Z

@Graphth: thanks a bunch, that's very helpful! We didn't know about that paper of Haemers. It will take us some time now to revise our paper now that we have this new information. Leave me a PM if we can use your real name in the acknowledgments.

Comment by Tobias Fritz

2012-12-10T18:24:03Z

What do you mean by factoring in Groupoids? Are you referring to the category of groupoids or about your $\mathcal{C}$ being a groupoid? I assume that you don't mean the latter, because the only groupoids that have products are the empty one and those equivalent to the trivial group... Also, your question may need to make a bit more precise. Even with fixed $\mathcal{C}$, the answer may depend on how you represent the objects as data structures. In many cases, even isomorphism is not decidable, e.g. in the category of finitely presented groups when a group is given in terms of a presentation.

Comment by Tobias Fritz

2012-12-08T17:32:57Z

One can also ask whether $\alpha(G)=\Theta(G)$ implies $\alpha(G)=\vartheta(G)$; if this is true, then the other conjectures immediately follow from the basic properties of the Lovász number $\vartheta$. Alas, we haven't been able to answer any of these four questions yet; but since our work lies mainly in relating these questions to problems in the mathematical foundations of quantum mechanics, we also haven't tried really hard. Our paper should be on the arXiv within a few days; I will announce it here.

Comment by Tobias Fritz

2012-12-08T15:35:34Z

A brief search reveals this, which might be of relevance: research.microsoft.com/pubs/102435/…

User tobias fritz - MathOverflow

Tannaka duality for C*-algebras?

Answer by Tobias Fritz for Realizing universal C*-algebras as concrete C*-algebras

graphs with independence number = Shannon capacity

Answer by Tobias Fritz for Upper bound on Shannon capacity based on independence number

Answer by Tobias Fritz for Skew fraction fields of *-algebras

making a graph well-covered without changing its Shannon capacity

Comment by Tobias Fritz

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Answer by Tobias Fritz for Realizing universal C-algebras as concrete C-algebras