User dave penneys - MathOverflow

Examples of noncommutative analogs outside operator algebras?

2009-11-30T07:46:14Z

Theo's question made me wonder if there are other "noncommutative analogs" outside of operator algebras. Some noncommutative analogs from operator algebras include:

A $C^\ast$-algebra is a noncommutative topological space (cf. the Gelfand transform).
The multiplier algebra of a nonunital $C^\ast$-algebra is the noncommutative Stone-Cech compactification.
A spectral triple is a noncommutative manifold (add some extra data to the spectral triple to get a noncommutative Riemannian manifold cf. arXiv:0810.2088).
A von Neumann algebra is a noncommutative measure space.

Are there any other good examples? If you know more in operator algebras, that's great too.

EDIT: these algebras should be considered as various functions spaces for noncommutative spaces as per @Yemon's answer. I'm going to leave the above text as is unless there are requests for another edit.

Does such an infinite index subgroup exist?

2012-04-05T23:18:17Z

Notation: If $G$ is a countable group and $H$ is a subgroup, for $g\in G$, let $|\mathcal{O}_{gH}|$ be the size of the $H$-orbit of $gH$ in the $H$-set $G/H$.

Does there exist a countable group $G$ and a subgroup $H$ with $[G\colon H]=\infty$ such that:

There is a $g\in G$ with $|\mathcal{O} _{gH}|\neq |\mathcal{O} _{g^{-1}H}|$,
$|\mathcal{O} _{gH}|=\infty$ if and only if $|\mathcal{O} _{g^{-1}H}|=\infty$, and
There is a constant $M>0$ such that $\displaystyle \frac{|\mathcal{O} _{gH}|}{|\mathcal{O} _{g^{-1}H}|} \leq M$ for all $g\in G$ with $|\mathcal{O} _{gH}|,|\mathcal{O} _{g^{-1}H}|\neq \infty$.

For instance, 2 and 3 above would be satisfied if:

If $|\mathcal{O} _{gH}|\neq |\mathcal{O} _{g^{-1}H}|$, then $0< |\mathcal{O} _{gH}|,|\mathcal{O} _{g^{-1}H}|\leq M$ for some $M>1$ independent of $g$.

Answer by Dave Penneys for Does such an infinite index subgroup exist?

2012-04-10T17:07:53Z

I believe (1 and 2) and (3) are mutually exclusive. Here is a proof:

First, the commensurator $$ Comm_G(H) = \{g\in G : |\mathcal{O} _{gH}|, |\mathcal{O} _{g^{-1}H}|<\infty\} $$ is a group. We will show:

Lemma: $\varphi\colon Comm_G(H)\to \mathbb{Q}_{>0}$ by $g\mapsto \displaystyle\frac{[H\colon H\cap gHg^{-1}]}{[H\colon H\cap g^{-1}Hg]}$ is a homomorphism.

From the lemma, if we assume there is a $g\in Comm_G(H)$ such that $\varphi(g)=x>1$ (criteria 1 and 2), then the order of $g$ must be infinite, since $x>1$ implies $x^n>1$ for all $n\geq 1$. Since the order of $g$ is infinite, criterion 3 cannot hold since eventually $\varphi(g^n)=\varphi(g)^n=x^n>M$ for any $M>0$.

Proof of the lemma:

We must show $\varphi(g_1g_2)=\varphi(g_1)\varphi(g_2)$. Define the following constants:

For $i=1,2$, $a_i = [H\colon H\cap g_iHg_i^{-1}]$ and $b_i = [H\colon H\cap g_i^{-1}Hg_i]=[g_iHg_i^{-1}\colon H\cap g_iHg_i^{-1}]$
$a=[H\colon H\cap (g_1g_2)H(g_1g_2)^{-1}]$ and $b=[(g_1g_2)H(g_1g_2)^{-1}\colon H\cap (g_1g_2)H(g_1g_2)^{-1}]$

Note that since $x\mapsto g_1xg_1^{-1}$ is an automorphism of $G$, we have:

$a_2=[g_1Hg_1^{-1}\colon g_1Hg_1^{-1}\cap (g_1g_2)H(g_1g_2)^{-1}]$ and $b_2=[(g_1g_2)H(g_1g_2)^{-1}\colon g_1Hg_1^{-1}\cap (g_1g_2)H(g_1g_2)^{-1}]$

Now look at the subgroup $K=H\cap g_1 Hg_1^{-1}\cap (g_1g_2)H(g_1g_2)^{-1}$, and define

$a_1'=[H\cap (g_1g_2)H(g_1g_2)^{-1}\colon K]$
$a_2'=[H\cap g_1Hg_1^{-1}\colon K]$
$b_1'=[g_1Hg_1^{-1}\cap (g_1g_2)H(g_1g_2)^{-1}\colon K]$

which are all finite, since if we have a quadrilateral of groups $L_1\cap L_2\subset L_1,L_2\subset G$, we must have $[L_1\colon L_1\cap L_2] \leq [G\colon L_2]$. Now since index is multiplicative, we have

$a a_1'=a_1a_2'$
$ba_1' = b_2 b_1'$
$a_2b_1'=b_1a_2'$

Solving for $a$ and $b$, we get $$ \frac{a}{b} = \frac{a_1a_2'}{a_1'}\frac{a_1'}{b_2b_1'}=\frac{a_1a_2'}{b_2b_1'}. $$ Now note that $\displaystyle \frac{a_2'}{b_1'}=\frac{a_2}{b_1}$, so we have $$ \varphi(g_1g_2)=\frac{a}{b} = \frac{a_1}{b_2}\frac{a_2}{b_1}= \frac{a_1}{b_1}\frac{a_2}{b_2}=\varphi(g_1)\varphi(g_2). $$

Answer by Dave Penneys for Restriction on the coefficients for an operator in the free group factor $ L(\mathbb{F}_2) $

2011-06-29T18:47:28Z

For convenience, let's identify $L(G)$ with its image in $\ell^2(G)$ as per @Matthew Daws' answer. For $f=\sum_{g\in G} \mu_g L_g\in\ell^2(G)$, we have $f\in L(G)$ if and only if $f* \xi\in \ell^2(G)$ for all $\xi\in \ell^2(G)$, where $*$ is convolution. Another way of saying this is that $L(G)$ is all $\ell^2$-sums which define bounded operators on $\ell^2(G)$ by convolution.

A good reference for this is Vaughan Jones' course notes/book on von Neumann algebras.

Answer by Dave Penneys for representing tensor-C*-categories in BIM

2011-06-13T15:08:10Z

I don't know about necessary conditions, but here are some results concerning sufficient conditions:

In MR1749868, Hayashi and Yamagami realize amenable $C^*$-tensor categories in the category of bifinite (Jones index) bimodules of the hyperfinite $II_1$-factor.-
In arXiv:0811.1764v4, Stefaan Vaes and Sébastien Falguières showed that "the representation category of any compact group is the [bifinite] bimodule category of a $II_1$-factor," i.e., given a compact group $G$ with representation category $C$, there is a $II_1$-factor $M$ whose category of bifinite bimodules is exactly $C$.-
Recently, Sven Raum and Sébastien Falguières showed that "all finite $C^*$-tensor categories are [bifinite] bimodule categories of a $II_1$-factor," i.e., given a finite $C^*$-tensor category $C$, there is a $II_1$-factor $M$ whose category of bifinite bimodules is exactly $C$. This paper has yet to appear on the arXiv. Here is the link to the conference at which the talk was given.-

I don't know of such results for type $III$ factors.

Is there a fusion category with an object which does not commute with its dual?

2011-05-26T18:17:02Z

Does there exist a fusion category with an object $X$ such that $XX^*\ncong X^*X$ (where the isomorphism need not be natural in any way)?

Feel free to add adjectives such as pivotal, spherical, unitary, etc.

Answer by Dave Penneys for What is the subfactor planar algebra of type $\tilde{A}_n$, of index 4?

2011-03-16T17:28:50Z

Yes, it is two dimensional, and this is allowed. It just means the planar algebra is not irreducible. I don't know of anyone that has thought about a presentation by generators and relations of this planar algebra yet.

One issue here is that since $d=[M\colon N]^{1/2}=2$ is not generic ($>2$), one has to be careful about the annular multiplicities of the subfactor (arXiv:math/0105071). So I don't know if the planar algebra qualifies as "annular multiplicities $*10$" like (extended) Haagerup.

Answer by Dave Penneys for Invertibility of the planar algebra-subfactor correspondence

2011-01-22T21:34:51Z

Yes, strongly amenable subfactors of the hyperfinite $II_1$-factor are completely classified by their standard invariant. The finite depth case was done by Popa's Classification of subfactors: the reduction to commuting squares (MR1055708), and the infinite depth case was finished by Popa's Classification of amenable subfactors of type $II$ (MR1278111). The reconstruction theorem in this case reproduces a hyperfinite $II_1$-subfactor.

The Guionnet-Jones-Shlyakhtenko construction reproduces an inclusion of interpolated free-group factors (arXiv:0911.4728), and there is a specific formula for which factors you get. So you need to start with the right factors (modulo the free-group factor isomorphism problem...).

EDIT: Noah's answer makes a really important point. I should point out that at index 6, Bisch, Nicoara, and Popa constructed an uncountable family of (non-amenable) subfactors of the hyperfinite $II_1$-factor with the same standard invariant with property (T) (MR2314611). As they say in the abstract:

We exploit the fact that property (T) groups have uncountably many non-cocycle conjugate cocycle actions on the hyperfinite $II_1$ factor.

For a discrete group, if you're amenable and you have property (T), then you're finite. For a subfactor, if you're amenable and you have property (T), then you're finite depth. So once you're in the infinite depth property (T) setting, there's no hope for a bijective correspondence between subfactors and planar algebras.

Answer by Dave Penneys for Subfactor summer reading list

2010-10-27T00:52:39Z

So it really depends on why you want to learn about subfactors. I'll try and give different reading lists based on different motivation.

The basics of $II_1$-subfactors:

If you're familiar with $II_1$-factors, then "Introduction to subfactors" (MR1473221) is a good place to start. Of course, Jones' original "Index for subfactors" (MR696688) is an enjoyable read as well. A more advanced and comprehensive treatment is Evans and Kawahigashi's "Quantum symmetries on operator algebras" (MR1642584).

Classification:

When mathematicians encounter a family of mathematical objects, we feel the need to classify them. Subfactors were first (and still!) classified by their principal graphs, which is the principal part of the Bratteli diagram of the standard invariant, or tower of relative commutants (see JS, EK, or Bisch's "Bimodules and higher relative commutants" MR1424954). The first classifications (index $\leq 4$) were completed by Jones, Ocneanu (MR996454) (very hard to find this source), and Popa (MR1278111). Popa showed that "amenable" subfactors of the hyperfinite $II_1$-factor (the index is the norm squared of the principal graph) are completely classified by their principal graphs.

Axiomatization:

There are several axiomatizations for the standard invariant of a subfactor: Ocneanu's paragroups (see EK), Popa's $\lambda$-lattices (MR1334479), and Jones' planar algebras (see section below on planar algebras). These three different axiomatizations play together nicely, and it's good to have an overview on what's really going on here. Unfortunately, there is no good unified source for this. Yet. However, you should think of it this way:

The standard invariant (or "representation theory") of a finite index $II_1$-subfactor $N\subset M$ is a unitary $2$-category with $2$ $0$-morphisms called $N$ and $M$, $1$-morphisms given by various bimodule summands of the basic constructions of $N\subset M$, and $2$-morphisms given by various intertwiner spaces. This is what Ocneanu calls a "paragroup" because it resembles the tensor category of representations of a finite group. This $2$-category is unitary, has nice duals, and satisfies Frobenius reciprocity, and other cool stuff as well. In particular, we can draw planar diagrams to represent different $2$-morphisms in the spaces. So this $2$-category has the structure of a planar algebra. For this category theory stuff, see Mueger's "From subfactors to categories and topology I" (MR1966524). In many cases, we can recover the special $2$-category from connections on a bipartite graph or on a commuting square (see EK, JS, Popa).

Reconstruction:

Popa proved that one can start with he standard invariant of a subfactor and reconstruct a subfactor with the same standard invariant (MR1334479). In the "strongly amenable" case (a bit more technical than "amenable"), you get a subfactor of the hyperfinite $II_1$-factor. You can also do the reconstruction completely planar algebraically. This result is due to Guionnet-Jones-Shlyakhtenko (arXiv:0712.2904), and a really easy version to understand was given by Jones-Shlyakhtenko-Walker (arXiv:0807.4146) (Kodiyalam and Sunder also have a version of this).

Planar Algebras:

If you want to know what a planar algebra is, see Peters' construction of the Haagerup subfactor planar algebra (arXiv:0902.1294), or Morrison-Peters-Snyder "Skein-theory for the $D_{2n}$ planar algebras" (MR2559686). If you want to know how a subfactor actually gives a planar algebra, see the first section of Jones-Penneys (arXiv:1007.3173), which relies on some proofs in Jones' "Planar Algebras I" (arXiv:math/9909027).

Examples:

A great class of examples is the Bisch-Haagerup subfactors (MR1386923) which are just slightly harder than group-subgroup examples (see JS). Some of the most important examples rely on the above reconstruction theorems. For example, there are the exotic Haagerup and Asaeda-Haagerup subfactors (MR1686551) and the composite Fuss-Catalan subfactors (MR1437496).

I'll keep updating this post. Right now I have office hours. Sections to come include:

Type III, Recent results

Answer by Dave Penneys for Can the minimal index of a subfactor take all values in {4cos^2(pi/n);n=3,4,5,...} u [4,infinity]?

2010-09-25T22:52:44Z

There is an irreducible Temperley-Lieb subfactor at every allowed index. For $n\geq 3$, it has index $4\cos^2(\pi/n)$ and principal graph $A_{n-1}$ (in fact all subfactors of index less than $4$ are irreducible), and for every $r\geq 4$, it has index $r$ and principal graph $A_\infty$. Doesn't that do the job by your remark?

Hochschild/Cyclic Homology of von Neumann Algebras: Useless?

2009-10-16T19:28:36Z

Hochschild homology gives invariants of (unital) $k$-algebras for $k$ a unital, commutative ring. If we let our algebra $A$ be the group ring $k[G]$ for $G$ a finite group, we get group homology. There are plenty of other connections to homological algebra. If we use cyclic homology, there are connections to geometry and topology involving the Chern character.

Von Neumann algebras are complex algebras, so we can take their Hochschild and cyclic homologies. When I have asked experts in the fields of von Neumann algebras and non-commutative geometry about what you get, I usually hear some approximation of the following: "There's also analysis in von Neumann algebras, so I wouldn't expect an algebraic invariant like Hochschild or cyclic homology to tell you anything useful."

Although this answer makes some sense, I find it very displeasing and cryptic. Why shouldn't it tell you something? Is there some way to make "it doesn't tell you anything" quantitative? Is there an example of a von Neumann algebra with nontrivial Hochschild or cyclic homology (different from that of the complex numbers)?

EDIT: After reading the responses so far, I should specify that I really want to know if there is a $II_1$-factor with nontrivial Hochschild or cyclic (co)homology.

When does a matrix define a convolution operator on a hypergroup?

2010-07-12T07:53:41Z

Let $H$ be a discrete hypergroup. Suppose I have a matrix $A=(A_{x,y})$ indexed over $H$ with nonnegative entries which defines a bounded operator on $\ell^2(H)$. When does there exist $f\in\ell^1(H)$ such that $A_{x,y}=\langle f*\delta_x,\delta_y\rangle$, i.e., $A$ is the matrix of transition probabilities for a random walk given by convolution with $f$?

A necessary condition is that $A$ commutes with $\ell^1(H)$ convolution on the right. Is this sufficient?

Operator Valued Weights

2009-10-17T21:54:29Z

One of the basic tools in subfactors is the conditional expectation. If $N\subset M$ is a $II_1$-subfactor (or an inclusion of finite factors), then there is a unique trace-preserving conditional expectation of $M$ onto $N$. This should be thought of as a (Banach space) projection of norm 1. In fact, it is the restriction of the Jones projection $e_N$ on $L^2(M)$ to $M$. In the finite index case, we get another conditional expectation (Jones projection...) from the basic construction $M_1=\langle M, e_N\rangle$ onto $M$.

In his thesis, Michael Burns showed that if we iterate the basic construction in the infinite index case, we only get half the conditional expectations (we only get the odd Jones projections). The other half of the time, we get a generalization of the conditional expectation called an operator valued weight, originally defined by Haagerup.

Given an inclusion of semifinite von Neumann algebras $(N, tr_N)\subset (M, tr_M)$, there is a unique normal, faithful, semi-finite trace-preserving operator valued weight $T\colon M_+\to \widehat{N_+}$, where we must take the "extended part" of the positive cone $N_+$ of $N$.

Edit as per @Dmitri's answer: Let $$ n_T=\{x\in M| T(x^\ast x)\in N_+\} $$ and set $$ m_T=n_T^\ast n_T=span\{x^\ast y| x,y\in n_T\}. $$ There is a natural extension of $T$ to $m_T$. Is there an example of a normal, faithful, semifinite operator valued weight such that

$N$ is not contained in $T(m_T)$, and/or
$1\notin T(M_+)$?

What about when $M$ and $N$ are factors ($M$ is $II_\infty$)?

Ideals in Factors

2009-10-15T15:58:51Z

One can easily prove that factors have no nontrivial ultraweakly closed 2-sided ideals as these are equivalent to nontrivial central projections. One can also show type $I_n$, type $II_1$, and type $III$ factors are algebraically simple (any 2-sided ideal must contain a projection. All projections are comparable in a factor, so you can show 1 is in the ideal). Ideals in $B(H)$ ($\dim(H)=\infty$, $H$ separable) have been studied extensively. What about ideals in $II_\infty$ factors?

One might think, since every $II_\infty$ factor $M$ can be written as $N\overline{\otimes} B(H)$ for $N$ a $II_1$ factor, if $I\subset B(H)$ is an ideal, then $N\otimes I$ is a 2-sided ideal. This is false. One needs to take the ideal generated by $N\otimes I$. What does that mean from a von Neumann algebra viewpoint? Is it the same as taking the norm closure?

We can also describe some ideals in terms of the trace. One has the equivalent of the Hilbert-Schmidt operators: $$I_2=\{x\in M | tr(x^\ast x)<\infty\}$$ and the trace class operators: $$I_1=\{x\in M | tr(|x|)<\infty\}=I_2^\ast I_2 =\left\{\sum^n_{i=1} x_i^\ast y_i | x_i, y_i\in I_2\right\}.$$ What is the relation of $I_j$ to $N\otimes L^j(H)$ for $j=1,2$ (where $L^2(H)$ is the Hilbert-Schmidt operators and $L^1(H)$ is the trace class operators in $B(H)$)? Is $I_j$ the norm closure of $N\otimes L^j(H)$?

How do you make a good math research poster for a non-mathematical audience?

2010-04-14T23:05:49Z

I have the opportunity to prepare a research poster for a non-mathematical, yet scientifically savvy audience, and I want to do it well. I have asked a few mathematicians, and I have heard the following sound advice:

Use interesting graphics.
Elaborate on possible applications to other scientific fields.

Although easier for applied mathematicians, this will be ok as I study subfactors, which have connections to quantum physics and statistical mechanics, and planar algebras, which provide great graphics. But there are practical questions as well:

How can I use LaTeX to make a poster?
How can I avoid using mathematical symbols and technical language?

What makes a good math research poster? What are some good ways to target a non-mathematical audience? Does anyone have examples and/or templates using LaTeX? What other advice would you give a mathematician who has never made a research poster before?

A few asides:

I'm not sure if this is community wiki. I'm more than happy to click the box if requested.
Feel free to retag this question as you see fit.
I will answer my own question after the poster presentation, and I will have all of my materials available online.

Non-commutative geometry from von Neumann algebras?

2009-10-28T21:48:34Z

The Gelfand transform gives an equivalence of categories from the category of unital, commutative $C^*$-algebras with unital $*$-homomorphisms to the category of compact Hausdorff spaces with continuous maps. Hence the study of $C^*$-algebras is sometimes referred to as non-commutative topology.

All diffuse commutative von Neumann algebras acting on separable Hilbert space are isomorphic to $L^\infty[0,1]$. Hence the study of von Neumann algebras is sometimes referred to as non-commutative measure theory.

Connes proposed that the definition of a non-commutative manifold is a spectral triple $(A,H,D)$. From a $C^*$-algebra, we can recover the "differentiable elements" as those elements of the $C^*$-algebra $A$ that have bounded commutator with the Dirac operator $D$.

What happens if we start with a von Neumann algebra? Does the same definition give a "differentiable" structure? Is there a way of recovering a $C^*$-algebra from a von Neumann algebra that contains the "differentiable" structure on our non commutative measure space? This would be akin to our von Neumann algebra being $L^\infty(M)$ for $M$ a compact, orientable manifold (so we have a volume form). Or are von Neumann algebras just "too big" for this?

One of the reasons I am asking this question is Connes' spectral characterization of manifolds (arXiv:0810.2088v1) which shows we get a "Gelfand theory" for Riemannian manifolds if the spectral triples satisfy certain axioms. Connes starts with the von Neumann algebra $L^\infty(M)$ instead of the $C^*$-algebra $C(M)$.

Answer by Dave Penneys for Compact Hausdorff and C^*-algebra "objects" in a category.

2010-02-10T22:03:37Z

The "Bohrification" paper arXiv:0905.2275 may be relevant to Question 4. As I understand, they discuss the notion of $C^\ast$-algebra objects in a given topos.

Answer by Dave Penneys for Gelfand duality in NCG

2010-01-26T00:47:03Z

@Meyer has provided a very good answer, so I just want to add something from my notes from Professor Rieffel's $C^\ast$-algebra class about the primitive ideal space, which is a canonical topological space associated to a $C^\ast$-algebra (in fact, it is associated to any normed $*$-algebra, but let's keep it "easy"...).

Let $A$ be a $C^\ast$-algebra.

Definition: An ideal $I\subset A$ is called primitive if it is the kernel of an irreducible representation. Denote the set of all primitive ideals by $Prim(A)$.

Note that $Prim(A)$ is always a set as it is a subset of $P(A)$, the power set of $A$. An important theorem:

If $A$ is GCR, then the map $\widehat{A}\to Prim(A)$ by $(\pi, H)\mapsto \ker(\pi)$ is a bijection (where $\widehat{A}$ is the "set" of equivalence classes of irreducible representations).

This is far from the case if $A$ is NCR (there is some set theory here, such as the term "unclassifiable," that I don't want to get into as I am not a set theorist), but $Prim(A)$ is still a set, so we still get a topological space using the following fact:

All primitive ideals are prime.

The space of prime ideals of $A$ comes with the Jacobson, or "hull-kernel" topology, so we get the relative topology on $Prim(A)$.

A few facts:

In general, $Prim(A)$ is not Hausdorff, but it is $T_0$ (see @Meyer's comment for a counterexample).
$Prim(A)$ is locally compact.
If $A$ is separable, $Prim(A)$ has the Baire Category Property.

Once again, this is all from a course I took from Professor Rieffel. I hope it helps!

Do torsion-free groups give projectionless group ($C^\ast$) algebras?

2009-10-30T20:07:07Z

One of the reasons I study von Neumann algebras is that they always have plenty of projections. There are many projectionless $C^\ast$-algebras ($0$ and possibly $1$ are the only projections), but the von Neumann algebras they generate must have nontrivial projections (unless it's just the complex numbers, of course). A good example of this is the reduced group $C^\ast$-algebra of any free group $F_n$. If $n=1$, then $C_r^\ast(Z)\cong C(S^1)$ via the Gelfand transform, which is clearly projectionless. If $n\geq 2$, the proof is fairly complicated. See Davidson's book for a proof when $n=2$.

If $G$ is a torsion-free group, is the reduced group $C^\ast$-algebra of $G$ projectionless? This $C^\ast$-algebra always contains the group algebra $C[G]$, so a simpler question is whether $C[G]$ is projectionless if $G$ is torsion-free.

Note that torsion-free is a necessary condition as one gets a projection from summing up the elements in the cyclic group generated by a torsion element and dividing by the order of the element.

EDIT: changed typestting. still some bugs... help please?

Answer by Dave Penneys for What are some interesting sequences of functions for thinking about types of convergence?

2009-12-30T22:38:19Z

The sequence $(f_n)$ given by $f_n=n^{-1}\chi_{[-n,n]}$ converges uniformly, but not in $L^1$.

Pimsner-Popa Bases

2009-10-15T00:16:56Z

Let $N\subset M$ be a finite index $II_1$-subfactor. Let $B=\{b_i\}$ be a finite orthonormal (Pimsner-Popa) basis for $M$ over $N$. Let $d=[M\colon N]^{1/2}$. It is well known that $B_1=\{d b_{i_1} e_1 b_{i_2}\}$ is a (not necessarily orthonormal) basis for $M_1$ over $N$, where $M_1=\langle M, e_1\rangle$ is the basic construction of $N\subset M$ (for example, see Bisch, Bimodules and higher relative commutants, page 31). Under what conditions are we assured $B_1$ is orthonormal? orthogonal?

Integer index always works (see the Pimsner-Popa paper), and a variation always works in infinite index (with a definition of orthonormal basis due to Burns).

Answer by Dave Penneys for Examples of noncommutative analogs outside operator algebras?

2009-11-30T11:37:24Z

I remembered a few more from operator algebras:

a trace is like a noncommutative integral. This has lead to:

The Schatten $p$-ideals
If $M$ is a $II_1$-factor, then one forms the noncommutative $L^P$-spaces (see this link) $L^p(M)$ as the completion of $M$ with the norm $$tr(|x|^p)^{1/p}$$ for $1\leq p <\infty$ and as $M$ with the operator norm for $p=\infty$. Then we get the duality $L^1(M)^\ast=L^\infty(M)$, and since the trace is finite, we have, as in the finite measure space case, that $$M=L^\infty(M)\subset L^2(M)\subset L^1(M).$$ I believe we have $L^p(M)\subset L^q(M)$ if $p>q$, but I haven't checked it. Is it also true that $L^p(M)^*=L^q(M)$ if $p^{-1}+q^{-1}=1$? I have never checked this either.

Answer by Dave Penneys for Which is the correct ring of functions for a topological space?

2009-11-29T04:17:17Z

As an operator algebraist, I think the space of continuous functions to $\mathbb{C}$ which vanish at infinity is my preferred choice. Let me tell you why.

One of the basic ideas of noncommutative topology/geometry (and probably algebraic geometry, but I don't know much about that) is that we can trade the space for algebras of functions on that space. This is afforded by the Gelfand transform. The spectrum of a commutative $C^\ast$-algebra is the space of characters, i.e., $\ast$-algebra homomorphisms to $\mathbb{C}$.

If $X$ is compact Hausdorff, then the spectrum of $C(X)$ is $X$.
If $X$ is locally compact Hausdorff, but not compact, the spectrum of the non-unital $C^\ast$-algebra $C_0(X)$ is $X$. The spectrum of the unitalization of $C_0(X)$ ($C_0(X)\oplus \mathbb{C}$) is the one point compactification of $X$. The spectrum of the unital $C^\ast$-algebra $C_b(X)$ is $\beta X$, the Stone-Cech compactification of $X$. One should note that $C_b(X)$ is also the multiplier algebra of $C_0(X)$.
If $X$ is compact, but not Hausdorff, then $C(X)$ corresponds to some type of "Hausdorffization" of $X$.

Actually $C(X)$ and $C_0(X)$ are the same if $X$ is compact, but you want to denote it $C(X)$ to emphasize the fact that the algebra is already unital. Otherwise, when you add a unit, you take the one point compactification of a compact space which adds an extra point, which is not what you want.

Now let's suppose you have some additional structure, like $X$ is a compact manifold. Then you probably want the $C^\infty$-functions on $X$. However, these can be recovered from $C(X)$ as those operators whose iterated commutator with the Dirac operator is bounded. This inspired the notion of a spectral triple.

EDIT: In my haste to answer this question, I made some mistakes in the earlier answer as pointed out by @Jonas.

Answer by Dave Penneys for ADE type Dynkin diagrams

2009-11-25T13:05:52Z

They classify principal graphs of $II_1$-subfactors with index less than $4$. The principal graph can be $A_n$, $D_{2n}$, $E_6$, or $E_8$, but $D_{odd}$ and $E_7$ do not occur.

Do subgroups have "two sided bases"?

2009-11-24T03:34:32Z

Let $H\leq G$ be an inclusion of finite groups. Define a map $E\colon \mathbb{C}[G]\to \mathbb{C}[H]$ to be the $\mathbb{C}$-linear extension of $$ E(g)=\begin{cases} g &\text{if } g\in H\\ 0 &\text{else,} \end{cases} $$ i.e., $E$ is the projection onto $\mathbb{C}[H]$. A finite subset $B\subset \mathbb{C}[G]$ will be called a left basis for $G$ over $H$ if $$ x=\sum\limits_{b\in B} b E(b^\ast x) $$ for all $x\in \mathbb{C}[G]$, where $\ast$ is the anti-linear extension of the map $g\mapsto g^{-1}$. For an example, take $B$ to be a set of left-coset representatives. Similarly, we can define a right basis to be a finite subset $B\subset \mathbb{C}[G]$ such that $$ x=\sum\limits_{b\in B} E(x b^\ast)b $$ for all $x\in\mathbb{C}[G]$.

Note that there exist groups for which there is a basis which is both a left and right basis, but $H$ is not a normal subgroup of $G$. One can take the subgroup of the symmetric group $S_n$ ($n\geq 3$) which fixes $1$. Then a set of left and right coset representatives is given by $$ \{ (1 j)|j=1,\dots,n\}. $$ Does there always exist a basis which is both a left and right basis, or are there inclusions of groups for which there is no simultaneous left and right basis?

The motivation for this question is another question from subfactor theory: if $N\subset M$ is a finite index, extremal $II_1$-subfactor, does there always exist a Pimsner-Popa basis which is both a left and right basis? The subgroup subfactor is an example of such a subfactor, and the question posed above is a watered-down version of the subfactor question, where perhaps an answer is already known or more easily obtainable.

Why are fusion categories interesting?

2009-11-19T21:52:58Z

In the same vein as Kate and Scott's questions, why are fusion categories interesting? I know that given a "suitably nice" fusion category (which probably means adding adjectives such as "unitary," "spherical," and "pivotal"), we get a subfactor planar algebra which, in turn, gives us a subfactor. Also, I vaguely understand that these categories give us Turaev Viro TQFTs.

What else do fusion categories do? What's a good reference for the Turaev Viro stuff?

Answer by Dave Penneys for What is to Quantize something?

2009-11-20T02:00:00Z

In mathematics, quantization often refers to some kind of deformation of a classical object. The Heisenberg Uncertainty Principle says that the position and momentum operators do not commute. In fact, $[X,P]=i\hbar$. In the limit as $\hbar\to 0$, these operators commute once again. Technically speaking, this is nonsense as $\hbar$ is a universal constant, but in mathematics, we are free to play with parameters. A couple of examples include:

the noncommutative torus, the universal $C^\ast$-algebra generated by two unitaries satisfying $uv=e^{i\theta} vu$. As $\theta\to 0$, we get $C(\mathbb{T}^2)$, the continuous functions on the $2$-torus. We usually think of the deformed algebra as a quantization of the commutative one.
some quantum groups are deformations of universal enveloping algebras, i.e., we get the universal enveloping algebra as $q\to 1$.

Answer by Dave Penneys for Measurable functions and unbounded operators in von Neumann algebras

2009-11-16T06:26:57Z

I think your question should be as follows:

Given a von Neumann algebra $M$, can we define a canonical set $S$ such that

if $M=L^\infty(X)$ acting on $L^2(X)$, then $S$ is (isomorphic to) the set of a.e. defined measurable functions, and
if $M=B(H)$ acting on $H$, then $S$ is the densely defined, closed unbounded operators?

If we take this slight alteration of the question, then I believe the answer is the closed affiliated operators. Here's a sketch of a proof which I think should work, but you should check the details just to make sure.

Let $A=L^\infty(X)$. If $f$ is an a.e. defined measurable function, define as in my other answer $$ D(M_f)={ \xi\in L^2(X) | f\xi\in L^2(X)}. $$ Then $M_f\colon D(M_f)\to L^2(X)$ is closed and affiliated with $A$.

Now suppose $T$ is a closed, densely defined operator affiliated to $A$ acting in $L^2(X)$ (abbreviated $T\eta A$). Then we can do polar decomposition to get $T=U|T|$ where $U\in A$ and $|T|\eta A$. Hence we have reduced to the case where $T$ is positive and self adjoint. Since $T\eta A$, we must have that $f(T)\in A$ for all bounded Borel functions $f$. In particular, for $n\geq 1$, $T_n=\chi_{[0,n]}(T)\in A$, and $T_n$ increases to $T$. It should be clear how to proceed now to get that $T$ is multiplication by an a.e. defined measurable function.

Answer by Dave Penneys for ubiquity, importance of path algebras

2009-11-14T19:01:28Z

One idea comes to mind for number two: Bases! If we have a unital inclusion of finite dimensional semi-simple complex algebras, the isomorphisms with the path algebras fix bases and make our lives easier.

Answer by Dave Penneys for Measurable functions and unbounded operators in von Neumann algebras

2009-11-13T18:25:59Z

It is not always the case that the composite of two operators is nice. In general, the composite of $S$ and $T$ is defined by $$ D(ST)={\xi\in D(T) | T\xi\in D(S)}. $$ There is no reason to expect this space to be dense.

I agree with @Yemon. You want the affiliated operators. Here's an interesting parallel for finite measure spaces that I heard in a mini-course from Ozawa. Just as $$L^\infty(X,\mu)\subset L^2(X,\mu)$$ which is a subset of all measurable functions, if $(M,tr)$ is a finite von Neumann algebra, $$M=L^\infty(M,tr)\subset L^2(M,tr)\subset \eta(M),$$ the affiliated operators. The inclusion for $L^2(M)$ into $\eta(M)$ is given by $$ \xi\mapsto (L_\xi\colon m\mapsto \xi m), $$ (be careful - you need to take the closure of $L_\xi$. Clearly $L_\xi$ commutes with right multiplication by $U(M)$) and the image of $L^2(M)$ is the set of all $T\in \eta(M)$ such that if $$ |T|=\int t dE(t) $$ is the usual spectral measure, then $$ \int t^2 d tr(E(t)) <\infty. $$

About your comment in response to @Yemon: Indeed the trace works. If you have a positive function, use indicator functons to show the trace is semi-finite and faithful (see Takesaki I). Normal is clear. If you have a measurable function $f$ on $(X,\mu)$, construct a multiplication operator by letting $$ D(M_f)={\xi\in L^2(X,\mu)| f\xi \in L^2(X,\mu)}. $$ If $\mu$ is finite, and $f$ is a measurable, real valued, a.e. finite function, then $M_f$ is self-adjoint, and the spectrum is the essential range of $f$.

Reed and Simon have a good introduction to unbounded operators, in particular the spectral theorem.

Somehow the brackets aren't showing up for the sets above, but I hope it's legible.

Comment by Dave Penneys

2012-04-09T18:12:25Z

And aren't the indices the same as $[H\colon U]=(10)(12)=120=(24)(5)=[H\colon V]$?

Comment by Dave Penneys

2012-04-09T14:53:09Z

Thanks for your answer Mark! I've been working through the example, and 2 and 3 still seem non-trivial (especially 3). For example, $[H\colon H\cap t^2 H t^{-2}]$ seems highly dependent on how $t$ moves $U\cap V$ (and thus depends on the choice of isomorphism $U\cong V$). Are there any good references for the general properties of HNN extensions and why 2 and 3 should hold? Thanks again!

Comment by Dave Penneys

2011-07-01T09:11:31Z

It's not like the third one, which only converges almost uniformly, not uniformly.

Comment by Dave Penneys

2010-11-23T19:12:26Z

+1 for the title.

Comment by Dave Penneys

2010-10-27T16:37:11Z

do you mean "non-commutative" in paragraph 1?

Comment by Dave Penneys

2010-09-26T20:16:40Z

Every standard invariant that arises from a finite index type $II_1$ subfactor also arises as the standard invariant of a type $III$ subfactor, and vice versa. See Izumi's paper "On type II and type III principal graphs of subfactors."

Comment by Dave Penneys

2010-04-01T17:19:48Z

Temperley-Lieb is a pretty famous counterexample. And why is Perron-Frobenius not Frobenius-Perron?

Comment by Dave Penneys

2010-01-27T00:18:52Z

thanks! i will edit the answer to refer to your comment as a counterexample.

Comment by Dave Penneys

2010-01-26T23:36:31Z

Doesn't the primitive ideal space of $B_0(H)$ consist of two points $\\{\\{0\\},B_0(H)\\}$, as the trivial representation is irreducible as well?

Comment by Dave Penneys

2009-11-30T17:35:48Z

@Andrew - most of the non locally compact spaces that I deal with on a day to day basis are infinite dimensional topological vector spaces as you point out. But in this case, wouldn't the "correct" ring of functions be the continuous linear maps? What are some other examples?

Comment by Dave Penneys

2009-11-30T07:52:41Z

somehow the hyperlinks have a mind of their own... help please?

Comment by Dave Penneys

2009-11-29T06:45:40Z

@Jonas - you have every reason to harp away. somehow my brain is not working clearly right now. i edited again. i think i need to get off of MO right now and get some sleep. hopefully i haven't made any more ridiculous errors. please point them out if you see them.

Comment by Dave Penneys

2009-11-29T06:13:26Z

@Jonas - thank you. don't know what i was thinking...

Comment by Dave Penneys

2009-11-29T05:29:10Z

Can you supply a reference? Is Aharoni's proof constructive? Can you give an example of such an embedding?

Comment by Dave Penneys

2009-11-24T18:40:09Z

yes, thanks. i fixed it.