User matthew daws - MathOverflow

Reference request for translating from Top to C*-alg

2011-12-07T13:46:41Z

Some recent questions on MO (for example, http://mathoverflow.net/questions/82708/do-subalgebras-of-cx-admit-a-description-in-terms-of-the-compact-hausdorff-spac) have been about Gelfand duality-- namely, that the categories of compact Hausdorff spaces with continuous maps, and commutative unital C*-algebras with unital $*$ -homomorphisms, are anti-equivalent. Thus one can "translate" properties about compact spaces over to C*-algebras. This can lead to a sort of "dictionary", see for example page 3 of Várilly's book on noncommutative geometry (which is http://books.google.co.uk/books?id=f5cAlaptxd0C&lpg=PP1&dq=non-commutative%20geometry&pg=PA3#v=onepage&q=non-commutative%20geometry&f=false).

Does anyone know a reasonably definitive reference for proofs of such dictionaries, in a self-contained form??

I'm guessing that perhaps such a thing doesn't exist, as these results are folklore (and are easy to prove really-- given the statement, the proofs often form nice exercises). As one is really just studying compact spaces via the category of compact spaces with continuous map, might there be a category theory book which is suitable?

Actually, I am more interested in the non-unital case. Rather than working with proper maps, I instead want to follow Woronowicz. Define a "morphism" between C $^*$ -algebras $A$ and $B$ to be a non-degenerate $*$ -homomorphism $\phi:A\rightarrow M(B)$ from $A$ to the multiplier algebra of $B$, where "non-degenerate" means that $\{ \phi(a)b : a\in A,b\in B \}$ is linearly dense in $B$. Then the category of commutative C $^*$ -algebras and morphisms is anti-equivalent to the category of locally compact spaces and continuous maps. One can then form a similar dictionary-- but here I think the proofs can be a bit trickier (or maybe just they use slightly less standard topology).

Does anyone know a reasonably definitive reference in this more general setting?

Answer by Matthew Daws for Open problems in the theory of compact quantum groups

2013-01-17T09:07:56Z

Probably not important in any sense, but something I thought about very briefly recently, and asked originally by Woronowicz in the Pseudo Group paper:

Let $(A,\Delta)$ be a CQG, and say that $a\in A$ is central if $\Delta(a)=\sigma\Delta(a)$ where $\sigma$ is the tensor swap map. You can show the all characters (traces of corepresentations) are central. Is the linear span of the characters norm dense in the central elements of A?

For a compact group, I'd use an averaging argument to prove this (approximate a by arbitrary matrix elements of coreps, and then average into characters). But I think the naive version of this works iff your Kac.

Does anyone know any progress on this?

Regular borel measures on metric spaces

2010-04-22T10:35:59Z

When teaching Measure Theory last year, I convinced myself that a finite measure defined on the Borel subsets of a (compact; separable complete?) metric space was automatically regular. I used the Borel Hierarchy and some transfinite induction. But, typically, I've lost the details.

So: is this true? Are related questions true? What are some good sources for this sort of questions? As motivation, a student pointed me to http://en.wikipedia.org/wiki/Lp_space#Dense_subspaces where it's claimed (without reference) that (up to a slight change of definition) the result is true for finite Borel measures on any metric space.

(I'm normally only interested in Locally Compact Hausdorff spaces, for which, e.g. Rudin's "Real and Complex Analysis" answers such questions to my satisfaction. But here I'm asking more about metric spaces).

To clarify, some definitions (thanks Bill!):

I guess by "Borel" I mean: the sigma-algebra generated by the open sets.
A measure $\mu$ is "outer regular" if $\mu(B) = \inf\{\mu(U) : B\subseteq U \text{ is open}\}$ for any Borel B.
A measure $\mu$ is "inner regular" if $\mu(B) = \sup\{\mu(K) : B\supseteq K \text{ is compact}\}$ for any Borel B.
A measure $\mu$ is "Radon" if it's inner regular and locally finite (that is, all points have a neighbourhood of finite measure).

So I don't think I'm quite interested in Radon measures (well, I am, but that doesn't completely answer my question): in particular, the original link to Wikipedia (about L^p spaces) seems to claim that any finite Borel measure on a metric space is automatically outer regular, and inner regular in the weaker sense with K being only closed.

Answer by Matthew Daws for Ultrapowers of operators

2012-11-16T09:41:57Z

As Andreas suggests, I shall fix $\mathcal U$ to be countably-incomplete. In fact, wlog, $\mathcal U$ will be over $\mathbb N$. If $X$ is not super-reflexive, then you don't even get all the rank-one operators. We know that $(X)_{\mathcal U}^* = (X^*)_{\mathcal U}$ if and only if $X$ is super-reflexive, so there is $\lambda \in (X)_{\mathcal U}^* \setminus (X^*)_{\mathcal U}$ . Choose $y=(y_n)\in (X)_{\mathcal U}$ . Let $T(x) = \lambda(x)y$ so $T$ is a rank-one map on $(X)_{\mathcal U}$ . Suppose $T=(T_n)$. For each $n$ pick $\mu_n\in X^*$ with $\|\mu_n\|\leq 1$ and with $\lim_n \mu_n(y_n)=\lim_n \|y_n\|$ (limits over $\mathcal U$ of course). Set $\mu=(\mu_n)$. Then $$ \mu(T(x)) = \lambda(x) \mu(y) = \lambda(x) = \mu((T_n)(x)) = \lim_n \mu_n(T_n(x_n)), $$ which holds for all $x$, so $\lambda = (\mu_n\circ T_n)\in (X^*)_{\mathcal U}$, contradiction.

If $X$ is super-reflexive, then I want to use some "co-ordinate" structure, so I need to think some more...

Answer by Matthew Daws for Can we recover a von Neumann algebra from its predual?

2010-12-30T15:31:45Z

In particular, can we dualize the product on a von Neumann algebra using some kind of tensor product?

The answer to this is explored in a number of papers. AFAIK, it was first considered by Quigg in "Approximately periodic functionals on C*-algebras and von Neumann algebras.", http://www.ams.org/mathscinet-getitem?mr=806641 Here he considered the Banach space projective tensor product, but this only works for subhomogeneous algebras.

The general case was answered by Effros and Ruan in "Operator space tensor products and Hopf convolution algebras", http://www.ams.org/mathscinet-getitem?mr=2015023 If $M$ is a von Neumann algebra with predual $M_*$ then there is a "tensor product", called the Extended Haagerup Tensor Product $M_* \otimes_{eh} M_*$ , and a (complete) contraction $\delta:M_* \rightarrow M_* \otimes_{eh} M_*$ which is the predual (in some slightly technical sense) of the multiplication map $M\otimes M\rightarrow M$. (My scare quotes are because the algebraic tensor product $M_*\otimes M_*$ is not (norm) dense in $M_* \otimes_{eh} M_*$ ).

Answer by Matthew Daws for Fourier Transform on Locally compact quantum groups

2012-05-31T08:54:12Z

My view is that one should treat LCQGs as a self-dual category; so there is no reason to prejudice, for a classical group $G$, the commutative case (leading to $L^1(\G)$) over the co-commutative (leading to $A(G)$).

The co-commutative is nice from the point of view of intrinsic groups-- this goes back to Takesaki and Tatsumma (and arguably Eymard, Herz etc.) where they showed that the intrinsic group of $VN(G)$ is just $G$ (with the same topology).

But in the commutative case, it's awful-- the intrinsic group of $L^1(G)$ is just the group of characters of $G$, which is rarely interesting outside of the abelian group case. Well, "interesting" is a bit extreme, giving maximal tori etc., but it certainly wouldn't give an injective Fourier transform.

(I think here maybe I have computed things in the "dual" formalism to that of the original question).

For a quantum example, I think Mehrdad showed that for $SU_\mu(2)$, you just get the maximal torus; so again the Fourier transform fails to be injective. That's not going to lead to an interesting theory (unless you have some specific application already in mind...)

Answer by Matthew Daws for Separability of Hilbert spaces from GNS construction.

2012-05-29T08:08:05Z

Some hints:

By replacing $H$ by $H\otimes\ell^2$, which doesn't change separability, you may suppose that every normal state $f$ is a vector state $x\mapsto (x\xi|\xi)$.
Then $H_f$ is the completion of $M$ for $(x|y) = f(y^*x) = (y^*x\xi|\xi) = (x\xi|y\xi)$. So the map $x\mapsto x\xi$ extends to an isometry from $H_f$ into $H$. So $H_f$ is separable.

This doesn't use any special about $M$ or $f$.

Answer by Matthew Daws for Cake-cutting and amenable groups

2012-05-24T13:18:44Z

Can't you just use the Lyapunov convexity theorem directly?

As usual, identify $\ell^\infty(G)$ with $C(\beta G)$, and work with $\beta G$ the Stone-Cech compactification. As this is a compact Hausdorff space, if $\mu$ is a regular measure on $\beta G$ then an atom of $\mu$ must be a point. So we can decompose $\mu$ as something in $\ell^1(\beta G)$ together with an atom-less measure, say a member of $M_c(\beta G)$ (continuous measures).

(Left) translation by members of $G$ give automorphisms of $\beta G$, and hence leave $\ell^1(\beta G)$ and $M_c(\beta G)$ invariant. I claim that nothing in $\ell^1(\beta G)$ can be left invariant. Let $\mu\in\ell^1(\beta G)$ be left invariant. Write $\beta G$ as the disjoint union of $G$-orbits, say $\bigcup_i G u_i$. Then $\mu$ must be supported on finite orbits (else we couldn't sum the coefficients, so we wouldn't be in $\ell^1$). If $u\in\beta G$ with $Gu$ finite, then there is $s\not=e$ in $G$ with $su=u$. Realise $u$ as an ultrafilter. Let $A\subseteq G$ be maximal with $A\cap s^{-1}A=\emptyset$. This means that if $r\not\in A$ then there is $t\in (A\cup\{r\}) \cap (s^{-1}A\cup\{s^{-1}r\})$ , which implies that $t=r\in s^{-1}A\cup\{s^{-1}r\}$ , that is, $sr\in A$. So $r\not\in A\implies sr\in A \implies r\in s^{-1}A$, so $G=A\cup s^{-1}A$. So Zorn implies there is $A\subseteq G$ with $A \cap s^{-1}A=\emptyset$ and $A\cup s^{-1}A=G$. Then either $A\in u$ so $A\in su$ so $s^{-1}A\in u$, contradiction; or $s^{-1}A\in u$ so $A\in su=u$ contradiction.

So I (hope!) I've shown that actually for any $u\in\beta G$, the orbit map $G\rightarrow\beta G; s\mapsto su$ is injective.

In particular, invariant means live in $M_c(\beta G)$, and so are atom-less, and so now we can just apply Lyapunov.

Edit: As Valerio points out, this shows that $X=\{ (\mu_1(A),\cdots,\mu_n(A)) : A\subseteq\beta G \text{ is Borel}\}$ is a convex set in $[0,1]^n$. Now, each $A\subseteq G$ induces the clopen set $O_A=\{ u\in\beta G: A\in u \}$ , and these sets $O_A$ form a base for the topology. Now each $\mu_i$ is regular, so given $\epsilon>0$ and $A\subseteq\beta G$ Borel, we can find $B,C\subseteq G$ with $O_B \subseteq A\subseteq O_C$ and with $\mu_i(C)-\mu_i(B)<\epsilon$, for all $i$ (under the obvious abuse of notation). (This follows as any open set is a union of sets of the form $O_C$, and then approximate with a finite union.) So $Y=\{ (\mu_1(A),\cdots,\mu_n(A)) : A\subseteq G\}$ is a subset of $X$, and is dense in $X$. I don't see right now why $Y$ need be convex.

Induced representations of topological groups

2011-12-13T14:31:59Z

Sorry if this is a naive question-- I'm trying to learn this stuff (cross-posted from http://math.stackexchange.com/questions/89248/induced-representations-of-topological-groups)

If $G$ is a group with subgroup $H$, then we have the restriction functor $\operatorname{Res}$ from $G-\operatorname{mod}$ to $H-\operatorname{mod}$. We also have this idea of induction, a functor $\operatorname{Ind}^G_H$ from $H-\operatorname{mod}$ to $G-\operatorname{mod}$. These are adjoints, which means (I think) that $\operatorname{Hom}_G(\operatorname{Ind}^G_H(V), U) \cong \operatorname{Hom}_H(V,\operatorname{Res}(U))$ naturally, for $G$-modules $U$ and $H$-modules $V$.

For locally compact groups, there is a theory worked out by MacKey and others. Actually, I have only read Rieffel's work on the subject (as I come from a functional Analysis background). For a locally compact $G$ and closed subgroup $H$, there is a very satisfactory notion of the functor $\operatorname{Ind}^G_H$ (where we consider "Hermitian modules", i.e. unitary representations on Hilbert spaces). What I don't see is how (or even if) this relates to the restriction functor?

In the topological setting, are $\operatorname{Ind}^G_H$ and $\operatorname{Res}$ in any sense adjoints?

A slightly vague rider-- if (as I suspect) the answer is "no", can we be more precise about why the answer is no?

Answer by Matthew Daws for If $A \subset X'$ annihilates only $0$, then $A$ is dense

2012-04-30T15:05:23Z

Let $\Phi\in X''$ be non-zero, let $\alpha_0\in X'$ be such that $\Phi(\alpha_0)=1$, and let $A$ be the collection of $\alpha\in X'$ with $\Phi(\alpha)=0$. Then $A$ is a subspace of $X'$, and any $\beta\in X'$ is equal to \[ \beta = \Phi(\beta)\alpha_0 + (\beta - \Phi(\beta)\alpha_0) \in \mathbb K\alpha_0 \oplus A, \] where $\mathbb K\alpha_0$ means the span of single element $\alpha_0$ (over whatever field you are using). In particular, $A$ is not norm dense.

Suppose now that $x\in X$ is non-zero, but with $\alpha(x)=0$ for all $\alpha\in A$. Then for any $\beta\in X'$ we have that \[ \beta(x) = \Phi(\beta)\alpha_0(x) + (\beta - \Phi(\beta)\alpha_0)(x) = \Phi(\beta) \alpha_0(x). \] Thus $x = \alpha_0(x) \Phi$, in particular, as $x$ is non-zero, $\alpha_0(x)\not=0$, and so $\Phi = \alpha_0(x)^{-1} x$.

Thus, if $X$ is not reflexive, I can choose $\Phi\in X''\setminus X$, and then I have found a suitable set $A$ which is not dense. So your condition is equivalent to reflexivity.

Closed operators and duality

2012-04-26T15:48:12Z

Usually we would define a "densely defined, closed operator" on a Banach space $E$ to be a linear map $T:D(T)\rightarrow E$, where $D(T)$ is a dense subspace of $E$, and the graph of $T$, $G(T)=\{ (x,T(x)) : x\in D(T) \}$ is closed in $E\times E$. Then we can define an adjoint by setting \[ D(T^*) = \{ f\in E^* : \exists g\in E^*, f(Tx) = g(x) \ (x\in D(T)) \}. \] That $D(T)$ is dense means that if $f\in D(T^*)$ then the associated $g$ is unique, so we can define $T^*(f)=g$ . This level of generality seems rare-- e.g. Davies in his book "One-parameter semigroups" mentions this, notes that $D(T^*)$ can fail to be norm dense, and moves on to Hilbert spaces.

Indeed, most books seem to just start out working with Hilbert spaces (and then usually $T^*$ means the Hilbert space adjoint-- but this is essentially the same thing, up to twisting by some conjugation). Here you can apply Hilbert space techniques to show that $D(T^*)$ is dense etc.

It seems to me however that $D(T^*)$ will always at least be weak $^*$ -dense and that $G(T^*)$ will be weak $^*$ -closed in $E^*\times E^*$ . Moreover, the proofs don't seem to need Hilbert space techniques. Moreover, starting with such a "weak $^*$ -closed, densely defined operator" on $E^*$ , we can always find a densely-defined closed operator on $E$ which induces it. Applied to a reflexive Banach space, one builds a very satisfactory theory.

The only source I know which talks about "closed" operators in such generality is a paper by Ciorănescu and Zsidó, see MathSciNet or Project Euclid. Even they don't mention the duality result.

My question: Is there a good (or even bad) reference for all this? In particular, that a weak$^*$-closed operator is the adjoint of a closed operator?

Answer by Matthew Daws for Weak topology restricted to the unitary group of a von Neumann algebra

2012-04-25T20:14:57Z

No. Consider the algebra $\ell^\infty$. Let $w$ be an ultrafilter and define $\mu\in(\ell^\infty)^*$ by $\mu((x_n)) = \lim_{n\rightarrow w} x_n$. Now consider the sequence $x^{(n)}$ in the unitary group of $\ell^\infty$, defined by \[ x^{(n)}_m = \begin{cases} 1 &: m\leq n, \\ -1 &:m>n. \end{cases} \] So $\mu(x^{(n)}) = -1$ for all $n$; thus $x^{(n)}$ does not converge weakly to $1$.

However, for any $a=(a_n) \in \ell^1$, we have that $a(x^{(n)}) = \sum_{k\leq n} a_k - \sum_{k>n} a_k \rightarrow a(1) = \sum_k a_k$. So $x^{(n)} \rightarrow 1$ in the $\sigma$-weak topology.

Answer by Matthew Daws for Where is the error in this argument?

2012-04-13T08:05:47Z

A link to the literature: I think of $C^*(G)^*$ as being $B(G)$, the Fourier-Stieltjes algebra, realised as a (non-closed) algebra of continuous functions on $G$. Any member of $C^*(G)^*$ can be realised as the composition of a representation $\pi$ on $H$ with a vector functional $\omega_{\xi,\eta}$ on $H$. The resulting function in $B(G)$ is $g\mapsto (\pi(g)\xi|\eta)$.

Then $W^*(G)$ is $B(G)^*$ . As the tensor product of representations corresponds to the product in $B(G)$, it follows that $G_{\otimes}$ is actually just the collection of characters on $B(G)$, namely algebra homomorphisms $B(G)\rightarrow\mathbb C$. Such things were explored by Walter in his paper "On the structure of the Fourier-Stieltjes algebra"

It's shown that $G_{\otimes}$ is not a group, and that it contains proper partial isometries and projections; it is a semigroup though.

Separating vectors for C$^*$-algebras

2012-04-06T09:32:09Z

(I asked this on math.stackexchange, without response).

Let $A$ be a C$^*$-algebra, concretely acting on a Hilbert space $H$. Suppose that $\xi_0\in H$ is cyclic and separating for $A$ (that is, the map $A\rightarrow H, a\mapsto a(\xi_0)$ is injective with dense range). Let $M=A''$ the von Neumann algebra generated by $A$.

Need $\xi_0$ still be separating for $M$? That is, $x\in M, x(\xi_0)=0 \implies x=0$?

It is standard (and easy to prove) that this is equivalent to $\xi_0$ be cyclic for $M'$. However, the usual proof breaks down, and does not show this to be equivalent to $\xi_0$ being separating for $A$.

I think I can prove this using left Hilbert algebras. We turn $\mathfrak A = \{ a(\xi_0) : a\in A \}$ into a left Hilbert algebra algebra in the obvious way. Then run the Tomita-Takesaki machinery (actually not needed in full generality as we start with a state, not a weight). Then the von Neumann algebra generated by $\mathfrak A$ is nothing but $M$, and so the general theory tells us that $\varphi(x) = \|x\xi_0\|$ will be a faithful weight on $M$, which is what we need. Actually, it's not at all clear to me that this is correct-- I don't see why the map $S:\mathfrak A \rightarrow \mathfrak A; a\xi_0 \mapsto a^*\xi_0$ is preclosed. So now I suspect there might be a counter-example...

Almost orthogonal vectors

2010-05-16T06:46:00Z

This is to do with high dimensional geometry, which I'm always useless with. Suppose with have some large integer n and some small $\epsilon>0$. Working in the unit sphere of $\mathbb R^n$ or $\mathbb C^n$, I want to pick a large family of vectors $(u_i)_{i=1}^k$ which is almost orthogonal, in the sense that $|(u_i|u_j)| < \epsilon$ when $i\not=j$. I guess I'm interested in how the biggest choice of $k$ grows with $n$ and $\epsilon$.

For example, we can let ${u_1,\cdots,u_n}$ be the usual basis, and then choose $u_{n+1} = (1,1,\cdots,1)/n^{1/2}$, which works if $n^{-1/2} < \epsilon$. Then you can let $u_{n+2} = (1,\cdots,1,-1,\cdots,-1)/n^{1/2}$ and so forth, but it's not clear to me how far you can go.

Answer by Matthew Daws for Uniform $L_1$ convergence implies uniform convergence pointwise a.e.

2012-03-20T12:55:47Z

No. For each $n$ let $(A_{n,m})$ be a sequence of subsets of $\Omega$ each with measure less than $1/n$, but with $\bigcup_m A_{n,m} = \Omega$ (certainly you can do this if $\Omega=[0,1]$ with Lebesgue measure).

Now set $f^{(m)}_n = \chi_{A_{n,m}}$ . Then \[ \lim_n \ \sup_m \|f^{(m)}_n\|_1 = \lim_n \ \sup_m |A_{n,m}| < \lim_n \frac{1}{n} = 0. \] However, for any sequence $(k_n)$ and any $x\in\Omega$, \[ \sup_m |f^{(m)}_{k_n}(x)| = \sup_m \chi_{A_{k_n,m}}(x) = \chi_{\bigcup_m A_{k_n,m}}(x) = 1. \] Thus your conclusion cannot hold.

Answer by Matthew Daws for Stone-Čech compactification of $\mathbb R$

2012-03-04T08:33:59Z

I can show the following (which Anton was asking about in comments). Let $X$ be locally compact and Hausdorff, and $U\subseteq X$ open. Let $X_\infty$ be the one-point compactication, so $U$ is still open in $X_\infty$. By the universal property of the Stone-Cech compactification, there is a continuous map $\phi:\beta X\rightarrow X_\infty$ which is the identity on $X$. Then $\phi^{-1}(U)$ is open in $\beta X$, and is just the canonical image of $U$ in $\beta X$. So $U$ open in $X$ shows that $U$ is open in $\beta X$.

(This fails for general closed sets. If $F\subseteq X$ is closed, then $F$ is only closed in $X_\infty$ if $F$ is also compact.)

I'll now use that $\beta X$ is the character space of $C^b(X)$. Let $U\subseteq X$ be open.

Lemma: Assume that $U$ is relatively compact. Under the isomorphism $C(\beta X)=C^b(X)$, we identify the ideal $\{ f\in C(\beta X) : f(x)=0 \ (x\not\in U) \}$ with $\{ F\in C^b(X) : f(x)=0 \ (x\not\in U) \}$

Proof: $X$ is itself open in $\beta X$, and the image of $C_0(X)$ in $C(\beta X)$ is just the functions vanishing off $X$. If $F\in C^b(X)$ vanishes off $U$ then $F\in C_0(X)$ (as $U$ is relatively compact) and so the associated $f$ in $C(\beta X)$ vanishes off $U$. Conversely, if $f\in C(\beta X)$ vanishes off $U$ then the associated $F\in C^b(X)$ is just the restriction of $f$ to $X$, and so vanishes off $U$.

By the Tietze theorem, the restriction map $C(\beta X) \rightarrow C(\beta X \setminus U)$ is a surjection. So we can identify $C(\beta X\setminus U)$ with the quotient $C(\beta X) / \{ f\in C(\beta X) : f(x)=0 \ (x\not\in U) \}$ . So by the above, we identify $C(\beta X \setminus U)$ with $C^b(X) / \{ F\in C^b(X) : F(x)=0 \ (x\not\in U) \}$ . If $X$ is normal, then we can again use Tietze to extend any $F\in C^b(X\setminus U)$ to all of $X$. It follows that $C^b(X) / \{ F\in C^b(X) : F(x)=0 \ (x\not\in U) \}$ is isomorphic to $C^b(X\setminus U) = C(\beta(X\setminus U))$. So $\beta X \setminus U = \beta (X\setminus U)$ (in a fairly canonical way) under the hypotheses that $X$ is normal and $U$ is relatively compact.

(I'm not sure what happens for non-normal $X$. For $X=\mathbb R$ and $U$ an open interval, we obviously don't need Tietze.)

Answer by Matthew Daws for algebraic VS topological ergodicity

2012-03-02T21:49:45Z

Given there are no other answers, let me say something about (1). We need only check continuity at the identity of $G$. Let $(g_i)$ be a net converging to $e_G$. That $\alpha_{g_i}\rightarrow I$ means that for each $f\in C(X)$, we have $\|f\circ\phi_{g_i^{-1}} - f\|_\infty \rightarrow 0$ . That $\phi_{g_i^{-1}} \rightarrow I$ means that whenever $K\subseteq X, U\subseteq X$ are compact (=closed) and open, respectively, with $K\subseteq U$, we have that $\phi_{g_i^{-1}}(K) \subseteq U$ for large $i$. (This is a basic open set about $I$ in the compact-open topology). To show that $\alpha$ continuous implies that $\phi$ is continuous, you can use a simple Urysohn's lemma argument (to find a function $1$ on $K$ and $0$ off $U$).

Conversely, suppose $\phi$ is continuous, and let $f\in C(X)$. For $\epsilon>0$ we can cover $f(X)\subseteq\mathbb C$ by a finite number of closed discs of radius $\epsilon/2$, say $(L_k)$. Then cover each $L_k$ by an open disc of slightly larger radius, say $V_k$. Then $K_k=F^{-1}(L_k)$ is closed in $X$, and $U_k=f^{-1}(V_k)$ is open, and contains $K_k$. So for $i$ large, by continuity of $\phi$, we have that $\phi_{g_i^{-1}}(K_k) \subseteq U_k$. Thus $f(x) \in L_k \implies x\in K_k \implies \phi_{g_i^{-1}}(x) \in U_k \implies \alpha_{g_i}(f)(x) \in V_k$. Hence $\|f - \alpha_{g_i}(f)\|_\infty$ is small. So $\alpha$ is continuous.

I think a good reference for the locally compact case is Dana Williams's book "Crossed Products of $C*$-Algebras". It nicely dots all the is and crossed all the ts.

For (2): let $G=\mathbb Z$, so $\phi$ is generated by a single homeomorphism of $X$. Let $X=\{ z\in\mathbb C : |z|\leq 1\}$ and define $\phi(re^{i\theta}) = r^2 e^{i\theta}$. Then $0$ and all points on the circle are fixed; the orbit of any other point $re^{i\theta}$ has accumulation points $0$ and $e^{i\theta}$. Let $f\in C(X)$ be invariant; translate so $f(0)=0$. Then $f(re^{i\theta}) = \lim_n f(r^{2n}e^{i\theta})=0$ for all $r<1$, so by continuity $f=0$. Thus the action $\alpha$ is "ergodic", but $\phi$ has non-trivial invariant open and closed sets.

Edit: An easier example has $X=[0,1]$ and $\phi(s)=s^2$. Then ${0}, {1}$ are non-trivial fixed closed sets, and $(0,1)$ is an invariant open set; but again $\alpha$ only leaves the constant functions invariant.

Answer by Matthew Daws for Do signed measures on sigma-rings always have a Hahn decomposition?

2012-02-29T09:11:51Z

Your axioms are:

$\emptyset\in\mathcal R$
$A,B\in\mathcal R \implies (A\cup B)\setminus (A\cap B)\in\mathcal R$
$A_n\in\mathcal R \implies \bigcap_n A_n\in\mathcal R$

Then 2 and 3 show that $A,B\in\mathcal R \implies A\setminus B\in\mathcal R$. Then 2 shows that $\mathcal R$ is closed under disjoint unions, so combined with 3, we see that $A,B\in\mathcal R \implies A\cup B = (A\setminus B) \cup (B\setminus A) \cup (A\cap B) \in \mathcal R$. So actually your axioms describe a "$\delta$-ring" (see http://en.wikipedia.org/wiki/Delta-ring ) and not a $\sigma$-ring.

Here then is a counter-example to your question. Let $X=\mathbb N$ with $\mathcal R$ being the collection of all finite subsets of $X$. Define $\phi$ by \[ \phi(A) = |A\cap\{\text{evens}\}| - |A\cap\{\text{odds}\}|. \] This is a "measure" in your sense. But then the only choice for $P$ would be the set of even numbers, and that's not in $\mathcal R$.

For a positive result, note that for any $A\in\mathcal R$, the collection $\mathcal R_A = \{ A\cap B : B\in\mathcal R \}$ is a $\sigma$-algebra on $A$, and $\phi$ restricted to $A$ is a signed measure in the usual sense. So there is a Hahn-Decomposition for $A$. You might think one could glue the Hahn decompositions together. Perhaps getting $P\subseteq X$ such that for all $B\in\mathcal R$, we did have $B\cap P, B\setminus P \in\mathcal R$ and $\phi(B\setminus P) \leq 0 \leq \phi(B\cap P)$. (This is true in my example). But my gut reaction suggests that this probably can't be done in general (I await a good counter-example from someone else!)

Answer by Matthew Daws for Peter-Weyl theorem as proven in Cartier's Primer

2012-02-24T17:55:20Z

(Same as pm's answer, with details.)

Well, if $$ R_f(\varphi)(h) = \int_G \varphi(g) f(g^{-1}h) \ dg $$ then the adjoint satisfies \begin{align*} (R_f^*(\varphi)|\psi) &= (\varphi|R_f(\psi)) = \int_{G\times G} \varphi(h) \overline{ \psi(g) f(g^{-1}h) } \ dg \ dh \\ &= \int_{G\times G} \varphi(h) \tilde f(h^{-1}g) \overline{\psi(g)} \ dh \ dg = (R_{\tilde f}(\varphi) | \psi) \end{align*} where $\tilde f(g) = \overline{ f(g^{-1}) }$. So $R_f^* = R_{\tilde f}$ is again a right translation operator.

Answer by Matthew Daws for the dual space of C(X) (X is noncompact metric space)

2012-02-23T12:57:46Z

What you state in the first paragraph is the Riesz Representation Theorem (see http://en.wikipedia.org/wiki/Riesz_representation_theorem#The_representation_theorem_for_linear_functionals_on_Cc.28X.29) This is valid for all locally compact Hausdorff spaces; so in particular for $\mathbb R$ (ah, I guess, if you look at $C_0(\mathbb R)^*$).

If $X$ is any topological space, then of course we can talk of $C^b(X)$ (the bounded continuous functions on $X$). This is still a commutative C$^*$-algebra, and so is isomorphic to $C(K)$, where $K$ is some compact Hausdorff space. The process of moving from $X$ to $K$ is functorial; purely at the topological level it corresponds to constructing the Stone-Cech compactification (see http://en.wikipedia.org/wiki/Stone_cech_compactification ) Point evaluation at $x\in X$ induces a character on $C^b(X) = C(K)$ and hence a point $k$ of $K$; we thus get a (continuous) map $X\rightarrow K$. This is injective if $X$ is completely regular; but it can fail to be injective (basically, we might lack enough continuous functions to separate points of $X$).

Back to your question: $C^b(X)^* = C(K)^* = M(K)$. For $\mathbb R$, we find that $K$ is nothing but $\beta\mathbb R$ the Stone-Cech compactification (quite a large space!)

Answer by Matthew Daws for Matrices with entries in a $C^*$-algebra

2012-02-01T16:33:17Z

For $x=(x_i)_{i=1}^n, y=(y_i)_{i=1}^n \subseteq A$ define $(x,y) = \sum_i x_i y_i^* \in A$, and set $\|x\| = \|(x,x)\|^{1/2}$.

Lemma: We have that $(x,y)^* (x,y) \leq \|x\|^2 (y,y)$ the order in the C $^*$ -algebra sense.

Proof: (Copied from Lance's Hilbert C $^*$ -module book). Wlog $\|x\|=1$. For $a\in A$ let $a\cdot x = (ax_i)$. Then \begin{align*} 0 &\leq (a\cdot x-y, a\cdot x-y) \\ &= a (x,x) a^* - (y,x)a^* - a(x,y) + (y,y) \\ &\leq aa^*- (y,x)a^* - a(x,y) + (y,y) \end{align*} The claim that $a(x,x)a^* \leq aa^*$ follows as if $c\in A^+$ then always $aca^* \leq \|c\|aa^*$ . Now set $a=(x,y)^*=(y,x)$ and the claim follows.

In particular, $\|(x,y)\|^2 \leq \|x\|^2 \|y\|^2$ and so $\sup\{ \|(x,y)\| : \|x\| \leq 1 \} = \|y\|$ .

So you define \[ \| a \| = \sup \{ (x, ay) : \|x\|\leq 1, \|y\|\leq 1 \} \] where $(ay)_i = \sum_j y_j a_{ij}^*$. Then from the observation above, \begin{align*} \|a\|^2 &= \sup \{ \|ay\|^2 : \|y\|\leq 1 \} = \sup\{ (ay,ay) : \|y\|\leq 1 \} \\ &= \sup\Big\{ \Big\| \sum y_j a_{ij}^* a_{ik} y_k^* \Big\| : \|y\|\leq 1 \Big\} \\ &= \sup\{ \|(y, (a^*a)y)\| : \|y\|\leq 1 \} \\ &\leq \sup\{ \|(a^*a)y\| : \|y\|\leq 1 \} \\ &= \sup\{ \|(z,(a^*a)y)\| : \|y\|\leq 1, \|z\|\leq 1 \} = \|a^*a\|. \end{align*} However, for any $z$, \[ \|az\| = \sup\{ \|(x,az)\| : \|x\|\leq 1 \} \leq \sup\{ \|(x,ay)\| : \|x\|\leq 1, \|y\|\leq \|z\| \} = \|a\|\|z\|. \] It follows from this that the norm on $M_n(A)$ is an algebra norm (i.e. submultiplicative). From the very definition, the involution is an isometry on $M_n(A)$. So we have the usual trick: \[ \|a\|^2 \leq \|a^*a\| \leq \|a^*\| \|a\| = \|a\|^2 \] and so we have equality throughout, establishing the C $^*$ -identity for $M_n(A)$.

The idea is to define a generalised Hilbert space of rows of $A$ with an $A$-valued inner-product, and then copy the usual proof that operators on a Hilbert space are a C $^*$ -algebra.

"Liftings" of L^\infty functions

2012-01-13T16:13:26Z

This is motivated by this question: http://mathoverflow.net/questions/85411/is-there-an-inclusion-of-l-inftyg-into-c-0g and Bill Johnson's comments there.

Let $X$ be a locally compact Hausdorff space and $\mu$ a Radon measure on $X$. (We can simplify things to make $X$ compact if we like, so "Radon" becomes much the same a "Borel"; but see below). Suppose $\mu$ has full support. Let $M^\infty(X,\mu)$ be the space of all bounded $\mu$-measurable functions on $X$; so $L^\infty(X,\mu)$ is the quotient of $M^\infty(X,\mu)$ by the subspace of $\mu$-null functions. A "strong lifting" $\rho$ is a map $M^\infty(X,\mu) \rightarrow M^\infty(X,\mu)$ which is unital, linear, multiplicative, and such that:

$\rho(f)=f$ $\mu$-a.e.
if $f=g$ $\mu$-a.e. then $\rho(f) = \rho(g)$
if $f$ is continuous, then $\rho(f)=f$.

This means basically that $\rho$ picks out a representative of each equivalence class in $L^\infty(X,\mu)$ (and respects continuous functions) and so allows us to genuinely think of $L^\infty(X,\mu)$ as a space of functions. Very nice...

However, it seems that there is a hidden technicality. We might ask that each function $\rho(f)$ actually be Borel. Apparently this is open, even for $X=[0,1]$ with Lebesgue measure.

So the problem is that $\rho(f)$ might genuinely be only $\mu$-measurable. In the question I link to above, the hope was that we could use $\rho$ to embed $L^\infty(X,\mu)$ into $C_0(X)^{**}$, but that would require us to be able to integrate $\rho(f)$ against any bounded Radon measure. So my question is:

Which measures can we integrate $\rho(f)$ against (for all $f$)? Assuming we cannot integrate against all measures, is there a good counter-example to illuminate things?

My reference for all of this is the book "Topics in the theory of lifting" by A. and C. Ionescu Tulcea. This is an old book; I am not an expert. Has any progress been made on e.g. the Borel measurability question?

Answer by Matthew Daws for Is there an inclusion of $L_\infty(G)$ into $C_0(G)^{**}$?

2012-01-11T13:28:19Z

Undeleted: This is perhaps a little more tangential to the original question than I'd hoped. But maybe it gives some hints as to why the conclusions aren't that "worrying"...

I don't think this is silly. For example, if $G=\mathbb R$ then for each $t\in\mathbb R$ consider $$ f_{t,n} = \frac{n}{2} \chi_{[t-1/n,t+1/n]} \in L^1(\mathbb R). $$ Then each $f_{t,n}$ is a unit vector in $L^1(\mathbb R)$. Then given $F\in L^\infty(G)$, we define $$\tilde F(t) = \lim_n \langle F,f_{t,n} \rangle = \lim_n \frac{n}{2} \int_{t-1/n}^{t+1/n} F(s) \ ds$$ Maybe this limit doesn't actually exist-- but the sequence is bounded, so just force it to converge via an ultra-filter limit, or similar. I think you have just described an abstract version of this construction. The point is that $\tilde F$ is little more than a function $\mathbb R\rightarrow\mathbb C$ which is bounded; I don't see why it need have any continuity or measurability properties...

Edit: Actually, maybe a better argument is the following. If you convolve an $L^1(G)$ function by an $L^\infty(G)$ function, then you get a (left or right, depending on taste) uniformly continuous function, which you can then integrate against a bounded measure. So if $(e_\alpha)$ is a bai for $L^1(G)$, then define $T:L^\infty(G)\rightarrow M(G)^*$ by taking an ultrafilter limit: $\langle T(F),\mu\rangle = \lim_\alpha \int_G e_\alpha * F \ d\mu$. This gives the identity on $C_0(G)$ (indeed, on left/right uniformly continuous functions). I think this construction would be well-known to Banach algebraists...

Answer by Matthew Daws for relation between SOT-convergence of T and T'

2011-12-15T12:10:58Z

If we didn't have the $1/n$ term, what's the standard example here? Let $T$ be the left shift on $\ell^2$, so $T^n\rightarrow 0$ strongly, but $T^*$ is the right shift, an isometry.

To deal with the $1/n$ term, instead use a weighted shift. So something like \[ T\xi = T(\xi_1,\xi_2,\xi_3,\cdots) = (2\xi_2,\frac{3}{2}\xi_3,\frac{4}{3}\xi_4,\cdots). \] Then \[ T^2\xi = (3\xi_3, \frac{4}{2}\xi_4, \frac{5}{3}\xi_5, \cdots), \] and so forth. So $\frac{1}{n}T^n$ will be SOT null. But \[ T^*\xi = (0,2\xi_1,\frac{3}{2}\xi_2,\frac{4}{3}\xi_3,\cdots). \] So I haven't quite got the numbers right so that $\frac{1}{n}(T^*)^n$ is an isometry, but "asymptotically" it will be; it certainly isn't SOT null.

Answer by Matthew Daws for Reasonable crossnorm on Banach algebra tensor product constructed from isometric non-degenerate representations

2011-12-08T14:18:40Z

Let $D_A\subseteq A^*$ be the functionals of the form $\mu(\pi(\cdot)x)$, for $x\in X, \mu\in X^*, \|x\|\leq 1, \|\mu\|\leq 1$ . As $\pi$ is an isometry, Hahn-Banach shows that the convex hull of $X$ is weak $^*$ -dense in the closed ball of $A^*$ , say $A^*_{[1]}$ .

Similarly for $D_B$ using $\rho$. It's clear (*) that \[ |(\mu_A\otimes\mu_B)\tau| \leq \| (\pi\otimes\rho)(\tau) \|_{op} \qquad (\tau\in A\otimes B, \mu_A\in D_A, \mu_B\in D_B). \] However, this inequality is preserved by taking convex combinations of the elements of $D_A$ and $D_B$, and by taking weak$^*$-closures. But that would then show that \[ |(\mu_A\otimes\mu_B)\tau| \leq \| (\pi\otimes\rho)(\tau) \|_{op} \qquad (\tau\in A\otimes B, \mu_A\in A^*_{[1]}, \mu_B\in B^*_{[1]}), \] and that's all you need to show, I think?

Why is (*) true? Given $x\in X,\mu\in X^*,y\in Y,\lambda\in Y^*$ , we have that $x\otimes y\in X\widehat\otimes Y$ of norm $\|x\|\|y\|$, and also $\mu\otimes\lambda \in (X\widehat\otimes Y)^*$ (this is $w\otimes z\mapsto \mu(w)\lambda(z)$) has norm $\|\mu\| \|\lambda\|$. Then $(\mu_A\otimes\mu_B)\tau = (\mu\otimes\lambda)((\pi\otimes\rho)(\tau)(x\otimes y))$. Actually, this shows that the result would remain true if you replaced the projective tensor norm on $X\otimes Y$ be any reasonable cross-norm.

Answer by Matthew Daws for A left inverse for the comultiplication on a Hopf von Neumann algebra

2011-11-16T12:19:01Z

This is far from a full answer (but maybe it will inspire other answers).

I think all the cases Yemon gives, we only get a cb map, but we don't get normality. However, in some special cases, you do get a normal map.

Firstly, if $M=L^\infty(G)$ with $\Delta(f)(s,t) = f(st)$, then define $T_*:L^1(G) \rightarrow L^1(G\times G)$ by \[ T_*(f)(s,t) = f(st) f_0(s), \] where $f_0\in L^1(G)$ is some fixed, positive function with $\|f_0\|_1=1$ . This is bounded, as \begin{align} \|T_*(f)\|_1 &= \int_G \int_G |f(st) f_0(s)| \ dt \ ds = \int_G \int_G |f(ss^{-1}t)| \ dt \ |f_0(s)| \ ds \\ &= \int_G \|f\|_1 |f_0(s)| \ ds = \|f\|_1, \end{align} using the left-invariance of the Haar measure. Then the pre-adjoint $\Delta_*:L^1(G\times G)\rightarrow L^1(G)$ is \[ \Delta_*(f)(t) = \int_G f(s,s^{-1}t) \ ds, \] i.e. convolution (if you set $f=a\otimes b$). So then \[ \Delta_* T_*(f) (t) = \int_G T_*(f)(s,s^{-1}t) \ ds = \int_G f(ss^{-1}t) f_0(s) \ ds = f(t). \] So $\Delta_* T_*$ is the identity, as required.

More generally, and as Yemon alludes to in a comment, if $(M,\Delta)$ is a compact Kac algebra, then it's operator biprojective (see Aristov's paper "Amenability and compact type for Hopf-von Neumann algebras from the homological point of view"). So we can choose $T$ to be cb and normal (and with various $M_*$ module properties). If we can choose $T$ with these module properties (but not necessarily normal) then $M$ is said to be operator biflat. For $VN(G)$ this was investigated by Aristov, Runde and Spronk, "Operator biflatness of the Fourier algebra and approximate indicators for subgroups". It seems to be unknown if $VN(G)$ ever fails to be operator biflat. Of course, this is a much stronger condition that Yemon asks for.

It seems surprising to me that we can just write down a suitable (normal) map $T$ for $L^\infty(G)$, but that it seems that Yemon's question is open for $VN(G)$.

Edit (29 Nov): (This is technical; I hope I got all the details correct). Suppose $G$ is a locally compact group such that $VN(G)$ admits a faithful normal trace (I think is true if $G$ is a separable SIN group, for example). Then let $\varphi$ be the Plancheral weight on $VN(G)$; let $\omega$ be a normal tracial state. Then $\psi=\varphi\otimes\omega$ is a semifinite trace on $VN(G\times G)=VN(G)\overline\otimes VN(G)$, and for $x\in VN(G)_+$, we have that $$ \psi(\Delta(x)) = \varphi((\iota\otimes\omega)\Delta(x)) = \omega(1) \varphi(x) = \varphi(x) $$ as $\varphi$ is invariant for $\Delta$ (I know this from the quantum group setting; see papers of Kustermans and Vaes). Let $(\sigma_t)$ be the modular automorphism group for $\varphi$; then $(\sigma_t\otimes\iota)\Delta(x) = \Delta(\sigma_t(x))$ for $x\in VN(G)$, and so as $\omega$ is a trace, we see that $\Delta(VN(G))$ is invariant for the modular automorphism group of $\psi$. By a theorem of Takesaki there is a normal conditional expectation $\epsilon:VN(G\times G)\rightarrow \Delta(VN(G))$ with $\psi(x) = \psi(\epsilon(x))$ for all $x$ in the definition ideal of $\psi$.

So in particular, this gives a positive answer (even with the "normal") condition in the separable SIN case, say.

Answer by Matthew Daws for 100€ bounty ended: Do invariant measures maximize the integral?

2011-11-18T16:27:18Z

Edit: Here is what I think is a counter-example.

Let $\phi$ be the indication function of the even natural numbers, let $\mathcal U$ be an ultrafilter supported on the even naturals, and let $\mathcal V$ be an ultrafilter defined on the even negative integers. Define $\mu\in M(\mathbb Z)$ by \[ \int f(x) \ d\mu(x) = \lim_{x\in\mathcal V} f(x). \] Then $\mu\in I_\phi$ (as, in fact, $\int \phi(x+y) \ d\mu(y)=0$ for all $x$). If $\nu\in I_\phi$ then $\nu$ must assign the same measure to $2\mathbb N$ and $2\mathbb N+1$, say $\alpha\leq 1/2$. You also need to argue that $\nu$ must assign zero measure to any finite set (else it won't be $\phi$-invariant). So for any $y\in\mathbb Z$, \[ \int \phi(x+y) \ d\nu(x) = \nu(2\mathbb N-y) = \begin{cases} \nu(2\mathbb N) &: y\in 2\mathbb Z, \\ \nu(2\mathbb N+1) &: y\in 2\mathbb Z+1, \end{cases} = \alpha. \] Thus \[ \int \int \phi(x+y) \ d\nu(x) \ d\mu(y) = \int \alpha \ d\mu(y) = \alpha, \] as $\mu$ is a probability measure. By contrast, let $\nu$ be defined by \[ \int f(x) \ d\nu(x) = \lim_{y\in\mathcal U} f(y). \] Then \[ \int \phi(x+y) \ d\nu(x) = \begin{cases} 1 &: y\in 2\mathbb Z, \\ 0 &: y \in 2\mathbb Z+1, \end{cases} \] and so \[ \int \int \phi(x+y) \ d\nu(x) \ d\mu(y) = \int \chi_{2\mathbb Z}(y) \ d\mu(y) = 1. \] So $F$ is not maximised on $I_\phi$.

In fact, by replacing $2\mathbb N$ by $k\mathbb N$, I think you get that $F$ has norm one, but $F(\nu)\leq 1/k$ for any $\nu\in I_\phi$.

But somehow, to my mind, what's wrong is that the $\mu\in I_\phi$ you choose is very poor. So here's a revised conjecture:

Let $\mu\in I_\phi$ maximise the integral $\int \phi(x) \ d\mu(x)$. Then $F$ attains its maximum on $I_\phi$.

Old post: (Explains my thinking).

I think of these questions using the Arens products, from abstract Banach algebra theory. So I work over the complex numbers; but this is not a problem.

Consider $A=\ell^1(\mathbb Z)$ with the convolution product, so $A$ is commutative. Then $A^*=\ell^\infty(\mathbb Z) = C(\beta\mathbb Z)$ is an $A$-module: $(a\cdot f)(b) = f(ba)$ for $a,b\in A,f\in A^*$ . Then $A^{**}=M(\beta\mathbb Z)$ the space of finite Borel measures on the Stone-Cech compactification $\beta\mathbb Z$. Your space $M(\mathbb Z)$ is just the positive measures $\mu\in A^{**}$ with $\mu(1)=1$.

We try to extend the product of $A$ to $A^{**}$. Firstly we define a bilinear map $A^{**}\times A^*\rightarrow A^*$ by \[ (\mu\cdot f)(a) = \mu(a\cdot f) \qquad (\mu\in A^{**}, f\in A^*, a\in A). \] But then we have two choices for the product on $A^{**}$ : \[ (\mu \Box \lambda)(f) = \mu(\lambda\cdot f), \quad (\mu\diamond\lambda)(f) = \lambda(\mu\cdot f) \qquad (\mu,\lambda\in A^{**}, f\in A^*). \] A little thought shows that $\mu\diamond\lambda = \lambda\Box\mu$.

So if $\phi\in A^*$ if positive then $\mu\in I_\phi$ if and only if $\mu\cdot\phi = \mu(\phi) 1$. This follows, as writing $\delta_x\in A=\ell^1(\mathbb Z)$ for the point mass at $x\in\mathbb Z$, we have \[ (\phi\cdot\delta_x)(\delta_y) = \phi(\delta_{x+y}) \implies (\mu\cdot\phi)(\delta_x) = \mu(\phi\cdot\delta_x) = \int \phi(x+y) \ d\mu(y). \] So the condition that $\mu\in I_\phi$ becomes that $(\mu\cdot\phi)(\delta_x)$ is constant in $x$, which is seen to be equivalent to $\mu\cdot\phi = \mu(\phi) 1$.

Similarly, your map $F$ is just $F(\nu) = (\mu\Box\nu)(\phi)$.

As you allude to, it's known that $\lambda\Box\mu \not= \mu\Box\lambda$ for arbitrary $\lambda,\mu$. However, we say that $f\in A^*$ is "weakly almost periodic" (WAP) if $(\lambda\Box\mu)(f) = (\mu\Box\lambda)(f)$ for all $\mu,\lambda\in A^{**}$ . So if $\phi$ is WAP and $\mu\in I_\phi$ then for any $\nu\in M(\mathbb Z)$, \[ F(\nu) = (\mu\Box\nu)(\phi) = (\nu\Box\mu)(\phi) = \nu(\mu\cdot\phi) = \nu(1) \mu(\phi) = \mu(\phi), \] as $\nu$ is a probability measure. So actually $F$ is constant on $M(\mathbb Z)$ and so certainly attains its maximum at a point of $I_\phi$.

So, to be interesting, we need to ask the question for $\phi$ which are not WAP. An alternative characterisation of $\phi$ being in WAP is that the set of translates of $\phi$ in $\ell^\infty(\mathbb Z)$ forms a relatively weakly compact set. A nice characterisation of Grothendieck shows that this is equivalent to \[ \lim_n \lim_m \phi(x_n+y_m) = \lim_m \lim_n \phi(x_n+y_m) \] whenever all the limits exist, for sequences $(x_n),(y_m)$ in $\mathbb Z$. If $\phi$ is the indicator function of $\mathbb N$, then it's not in WAP.

We may as well assume that $\|\phi\|_\infty=1$ . Another "easy" case is when we can find $\nu\in I_\phi$ with $\nu(\phi)=1$. Then $F(\nu) = \mu(\nu\cdot\phi) = \mu(1) \nu(\phi) = 1$; while for any $\lambda\in M(\mathbb Z)$, clearly $|F(\lambda)| = |\mu(\lambda\cdot\phi)| \leq 1$ as $\mu$ is a probability measure, and $\lambda\cdot\phi$ is bounded by $1$ (again, as $\lambda$ is a probability measure and $\phi$ is bounded by $1$). Notice that this case covers your example of when $\phi$ is the indicator function of $\mathbb N$.

So a test case is to find $\phi$ not in WAP and with $\nu(\phi)<\|\phi\|_\infty$ for all $\nu\in I_\phi$ (notice that $I_\phi$ is always non-empty, as $\mathbb Z$ is amenable). Do you have an example of such a $\phi$?

Actually, if $\phi$ is the indicator function of the even natural numbers, then that's an example. And that leads to my (hopeful) counter-example.

Answer by Matthew Daws for Conditional expectations onto masas in type III factors

2011-11-02T18:28:26Z

Takesaki showed in section 6 of:

MR0303307 (46 #2445) Takesaki, Masamichi Conditional expectations in von Neumann algebras. J. Functional Analysis 9 (1972), 306–321. http://www.sciencedirect.com/science/article/pii/0022123672900043

that the following are equivalent for a von Neumann algebra M (not necessarily a factor):

M is finite
Every MASA in M admits a conditional expectation (i.e. norm one normal projection) onto it.

Edit: As Jon Bannon helpfully points out, the original question asked "when does a MASA admit a conditional expectation onto it", and so this answer only says "not always" which isn't really a full answer!

Answer by Matthew Daws for When does mathscinet review a paper?

2011-10-24T16:08:40Z

You can see the "official line" here: http://www.ams.org/publications/math-reviews/mr-edit In particular,

Elementary articles or books, or articles that have not been refereed are ordinarily not listed.

In my experience, (and in addition to Thierry's and Andreas's list) this includes:

Books reviews
More problematic is the case of certain authors and journals who are, shall we say, excessively "prolific"-- sometimes one sees papers which are clearly "new mathematics" (or least, claim to be) but which do not get a review. For obvious reasons I won't give an example, but it's not hard to find using the MathSciNet search tool.

You ask "In particular does it at all imply that the paper has little or no merit from a mathematical viewpoint?". Well, certainly not, in some sense. A book review, or an elementary survey article, might well contain interesting mathematics, and might well be useful to read (which also one can clearly understand why they wouldn't get a review-- a "review of a review" would be quite silly).

As to my case (2.) above-- yes, perhaps this is Math Reviews (or an editor, or someone who was sent the paper to review) making that judgement. This does seem to be a grey area, as it's not covered by the "Editorial Statement" I linked to (except maybe in the word "elementary"). It would be interesting to get more information about this...

Comment by Matthew Daws

2013-01-17T09:04:05Z

As for (b), it's trivial for compact, shown by Reiji for discrete, and I think still wide open for Kac...?

Comment by Matthew Daws

2012-11-16T12:38:35Z

@Bill: Not sure I see this. If $(X)_{\mathcal U}$ is reflexive and has the approximation property then the image of $(\mathcal B(X))_{\mathcal U}$ will contain all compacts. So you need to start building general operators in $\mathcal B((X)_{\mathcal U})$. If e.g. $X$ is Hilbert then I can do this, but I'm don't see how just having a copy of a Hilbert space in $(X)_{\mathcal U}$ is going to help.

Comment by Matthew Daws

2012-11-16T12:35:49Z

Sorry-- sloppy. If $\mathcal U$ is countably incomplete, then let $(A_n)$ be a decreasing sequence of elements of $\mathcal U$ with $\bigcap_n A_n=\emptyset$. Then for example you'd choose $(\mu_i)$ as follows: if $i\in A_n\setminus A_{n+1}$ pick $\mu_i\in X^*$ with $\mu_i(y_i)\geq0$ and with $\mu_i(y_i)> \|y_i\|-1/n$; otherwise choose $\mu_i$ arbitrary (as this won't affect the equivalence class $(\mu_i)$ in $(X^*)_{\mathcal U}$. Hopefully you now see how to make the rest of the argument run...

Comment by Matthew Daws

2012-06-22T20:31:58Z

So can I ask a naive question: is it correct that $F$ is just an algebra homomorphism (not assumed normal, or a $*$-map, etc.?)

Comment by Matthew Daws

2012-06-13T19:25:40Z

Annoyingly, this is exactly the same as an answer I gave over at math.stackexchange: math.stackexchange.com/questions/157795/… I had already suggested to Peter (who may or may not also be Andy) that it was polite to point out when you are cross-posting...

Comment by Matthew Daws

2012-06-13T12:27:23Z

@Peter: No; but if you look on this site, you'll see that it's generally considered good to cross-link. You probably wouldn't know that as a new user, so I thought I'd highlight this for you.

Comment by Matthew Daws

2012-06-13T12:10:02Z

This is very similar to a question here: math.stackexchange.com/questions/157069/…

Comment by Matthew Daws

2012-06-08T15:47:30Z

A related question on math.se: math.stackexchange.com/questions/81324/…

Comment by Matthew Daws

2012-06-08T11:58:03Z

@André: Okay, we agree to disagree I guess. +1 your answer.

Comment by Matthew Daws

2012-06-08T08:41:47Z

Unless I'm missing something, this is a very standard fact about abelian von Neumann algebras which can be found in standard texts, just as Amin and Yulia say. So voting to close, as not research level.

Comment by Matthew Daws

2012-05-09T08:20:32Z

It's an old book, but I still think Dixmier treats this very well-- he does the compact group case, but of course finite is a special case of this...

Comment by Matthew Daws

2012-05-06T07:39:54Z

I would guess that people are voting to close, and downvoting, because you haven't given nearly enough information to all us to answer this. What is an "interpolation" space? Why, in fact, can I not take $Z=Y$?

Comment by Matthew Daws

2012-05-01T20:38:58Z

Is actually $Z[I]$ equal to the intersection of all the sets $Z[f]$ with $f\in I$?

Comment by Matthew Daws

2012-04-27T14:16:30Z

@Bill: Thanks for those suggestions. Kato, Chapter III Section 5 gets very close to what I had in mind.

Comment by Matthew Daws

2012-04-27T10:50:31Z

This almost does it in too much detail; but it's a pretty good reference. Thanks!