User matthew daws - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T05:46:15Z http://mathoverflow.net/feeds/user/406 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/82871/reference-request-for-translating-from-top-to-c-alg Reference request for translating from Top to C*-alg Matthew Daws 2011-12-07T13:46:41Z 2013-03-09T13:24:56Z <p>Some recent questions on MO (for example, <a href="http://mathoverflow.net/questions/82708/do-subalgebras-of-cx-admit-a-description-in-terms-of-the-compact-hausdorff-spac" rel="nofollow">http://mathoverflow.net/questions/82708/do-subalgebras-of-cx-admit-a-description-in-terms-of-the-compact-hausdorff-spac</a>) have been about Gelfand duality-- namely, that the categories of compact Hausdorff spaces with continuous maps, and commutative unital C*-algebras with unital <code>$*$</code>-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 <a href="http://books.google.co.uk/books?id=f5cAlaptxd0C&amp;lpg=PP1&amp;dq=non-commutative%20geometry&amp;pg=PA3#v=onepage&amp;q=non-commutative%20geometry&amp;f=false" rel="nofollow">http://books.google.co.uk/books?id=f5cAlaptxd0C&amp;lpg=PP1&amp;dq=non-commutative%20geometry&amp;pg=PA3#v=onepage&amp;q=non-commutative%20geometry&amp;f=false</a>).</p> <blockquote> <p>Does anyone know a reasonably definitive reference for <strong>proofs</strong> of such dictionaries, in a self-contained form??</p> </blockquote> <p>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 <em>category</em> of compact spaces with continuous map, might there be a category theory book which is suitable?</p> <p>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<code>$^*$</code>-algebras $A$ and $B$ to be a non-degenerate <code>$*$</code>-homomorphism $\phi:A\rightarrow M(B)$ from $A$ to the multiplier algebra of $B$, where "non-degenerate" means that <code>$\{ \phi(a)b : a\in A,b\in B \}$</code> is linearly dense in $B$. Then the category of commutative C<code>$^*$</code>-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).</p> <blockquote> <p>Does anyone know a reasonably definitive reference in this more general setting?</p> </blockquote> http://mathoverflow.net/questions/118667/open-problems-in-the-theory-of-compact-quantum-groups/119152#119152 Answer by Matthew Daws for Open problems in the theory of compact quantum groups Matthew Daws 2013-01-17T09:07:56Z 2013-01-17T09:07:56Z <p>Probably not <em>important</em> in any sense, but something I thought about very briefly recently, and asked originally by Woronowicz in the Pseudo Group paper:</p> <blockquote> <p>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?</p> </blockquote> <p>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.</p> <p>Does anyone know any progress on this?</p> http://mathoverflow.net/questions/22174/regular-borel-measures-on-metric-spaces Regular borel measures on metric spaces Matthew Daws 2010-04-22T10:35:59Z 2012-12-02T18:46:54Z <p>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 <a href="http://en.wikipedia.org/wiki/Borel_hierarchy" rel="nofollow">Borel Hierarchy</a> and some transfinite induction. But, typically, I've lost the details.</p> <p>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 <a href="http://en.wikipedia.org/wiki/Lp_space#Dense_subspaces" rel="nofollow">http://en.wikipedia.org/wiki/Lp_space#Dense_subspaces</a> 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.</p> <p>(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).</p> <p>To clarify, some definitions (thanks Bill!):</p> <ul> <li>I guess by "Borel" I mean: the sigma-algebra generated by the open sets.</li> <li>A measure $\mu$ is "outer regular" if <code>$\mu(B) = \inf\{\mu(U) : B\subseteq U \text{ is open}\}$</code> for any Borel B.</li> <li>A measure $\mu$ is "inner regular" if <code>$\mu(B) = \sup\{\mu(K) : B\supseteq K \text{ is compact}\}$</code> for any Borel B.</li> <li>A measure $\mu$ is "Radon" if it's inner regular and locally finite (that is, all points have a neighbourhood of finite measure).</li> </ul> <p>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.</p> http://mathoverflow.net/questions/112468/ultrapowers-of-operators/112561#112561 Answer by Matthew Daws for Ultrapowers of operators Matthew Daws 2012-11-16T09:41:57Z 2012-11-16T09:41:57Z <p>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 <code>$(X)_{\mathcal U}^* = (X^*)_{\mathcal U}$</code> if and only if $X$ is super-reflexive, so there is <code>$\lambda \in (X)_{\mathcal U}^* \setminus (X^*)_{\mathcal U}$</code>. Choose <code>$y=(y_n)\in (X)_{\mathcal U}$</code>. Let <code>$T(x) = \lambda(x)y$</code> so $T$ is a rank-one map on <code>$(X)_{\mathcal U}$</code>. Suppose $T=(T_n)$. For each $n$ pick <code>$\mu_n\in X^*$</code> 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.</p> <p>If $X$ is super-reflexive, then I want to use some "co-ordinate" structure, so I need to think some more...</p> http://mathoverflow.net/questions/50302/can-we-recover-a-von-neumann-algebra-from-its-predual/50731#50731 Answer by Matthew Daws for Can we recover a von Neumann algebra from its predual? Matthew Daws 2010-12-30T15:31:45Z 2012-08-14T15:38:06Z <blockquote> <p>In particular, can we dualize the product on a von Neumann algebra using some kind of tensor product?</p> </blockquote> <p>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.", <a href="http://www.ams.org/mathscinet-getitem?mr=806641" rel="nofollow">http://www.ams.org/mathscinet-getitem?mr=806641</a> Here he considered the Banach space projective tensor product, but this only works for subhomogeneous algebras.</p> <p>The general case was answered by Effros and Ruan in "Operator space tensor products and Hopf convolution algebras", <a href="http://www.ams.org/mathscinet-getitem?mr=2015023" rel="nofollow">http://www.ams.org/mathscinet-getitem?mr=2015023</a> If $M$ is a von Neumann algebra with predual <code>$M_*$</code> then there is a "tensor product", called the Extended Haagerup Tensor Product <code>$M_* \otimes_{eh} M_*$</code>, and a (complete) contraction <code>$\delta:M_* \rightarrow M_* \otimes_{eh} M_*$</code> 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 <code>$M_*\otimes M_*$</code> is not (norm) dense in <code>$M_* \otimes_{eh} M_*$</code>).</p> http://mathoverflow.net/questions/98448/fourier-transform-on-locally-compact-quantum-groups/98470#98470 Answer by Matthew Daws for Fourier Transform on Locally compact quantum groups Matthew Daws 2012-05-31T08:54:12Z 2012-05-31T08:54:12Z <p>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 $A(G)$).</p> <p>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).</p> <p>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.</p> <p>(I think here maybe I have computed things in the "dual" formalism to that of the original question).</p> <p>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...)</p> http://mathoverflow.net/questions/98225/separability-of-hilbert-spaces-from-gns-construction/98259#98259 Answer by Matthew Daws for Separability of Hilbert spaces from GNS construction. Matthew Daws 2012-05-29T08:08:05Z 2012-05-29T08:08:05Z <p>Some hints:</p> <ul> <li>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 $x\mapsto (x\xi|\xi)$.</li> <li>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.</li> </ul> <p>This doesn't use any special about $M$ or $f$.</p> http://mathoverflow.net/questions/97773/cake-cutting-and-amenable-groups/97835#97835 Answer by Matthew Daws for Cake-cutting and amenable groups Matthew Daws 2012-05-24T13:18:44Z 2012-05-24T19:33:13Z <p>Can't you just use the Lyapunov convexity theorem directly?</p> <p>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).</p> <p>(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 <code>$t\in (A\cup\{r\}) \cap (s^{-1}A\cup\{s^{-1}r\})$</code>, which implies that <code>$t=r\in s^{-1}A\cup\{s^{-1}r\}$</code>, 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.</p> <p>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.</p> <p>In particular, invariant means live in $M_c(\beta G)$, and so are atom-less, and so now we can just apply Lyapunov.</p> <p><strong>Edit:</strong> As Valerio points out, this shows that <code>$X=\{ (\mu_1(A),\cdots,\mu_n(A)) : A\subseteq\beta G \text{ is Borel}\}$</code> is a convex set in $[0,1]^n$. Now, each $A\subseteq G$ induces the clopen set <code>$O_A=\{ u\in\beta G: A\in u \}$</code>, 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 $\mu_i(C)-\mu_i(B)&lt;\epsilon$, 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 <code>$Y=\{ (\mu_1(A),\cdots,\mu_n(A)) : A\subseteq G\}$</code> is a subset of $X$, and is dense in $X$. <strong>I don't see right now why $Y$ need be convex.</strong></p> http://mathoverflow.net/questions/83336/induced-representations-of-topological-groups Induced representations of topological groups Matthew Daws 2011-12-13T14:31:59Z 2012-05-13T10:46:44Z <p>Sorry if this is a naive question-- I'm trying to learn this stuff (cross-posted from <a href="http://math.stackexchange.com/questions/89248/induced-representations-of-topological-groups" rel="nofollow">http://math.stackexchange.com/questions/89248/induced-representations-of-topological-groups</a>)</p> <p>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 $V$.</p> <p>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?</p> <blockquote> <p>In the topological setting, are $\operatorname{Ind}^G_H$ and $\operatorname{Res}$ in any sense adjoints?</p> </blockquote> <p>A slightly vague rider-- if (as I suspect) the answer is "no", can we be more precise about <em>why</em> the answer is no?</p> http://mathoverflow.net/questions/95570/if-a-subset-x-annihilates-only-0-then-a-is-dense/95581#95581 Answer by Matthew Daws for If $A \subset X'$ annihilates only $0$, then $A$ is dense Matthew Daws 2012-04-30T15:05:23Z 2012-04-30T15:05:23Z <p>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 <code>$\beta = \Phi(\beta)\alpha_0 + (\beta - \Phi(\beta)\alpha_0) \in \mathbb K\alpha_0 \oplus A,$</code> 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.</p> <p>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 <code>$\beta(x) = \Phi(\beta)\alpha_0(x) + (\beta - \Phi(\beta)\alpha_0)(x) = \Phi(\beta) \alpha_0(x).$</code> Thus $x = \alpha_0(x) \Phi$, in particular, as $x$ is non-zero, $\alpha_0(x)\not=0$, and so $\Phi = \alpha_0(x)^{-1} x$.</p> <p>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.</p> http://mathoverflow.net/questions/95267/closed-operators-and-duality Closed operators and duality Matthew Daws 2012-04-26T15:48:12Z 2012-04-27T07:37:39Z <p>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 $T$, <code>$G(T)=\{ (x,T(x)) : x\in D(T) \}$</code> is closed in $E\times E$. Then we can define an adjoint by setting <code>$D(T^*) = \{ f\in E^* : \exists g\in E^*, f(Tx) = g(x) \ (x\in D(T)) \}.$</code> That $D(T)$ is dense means that if <code>$f\in D(T^*)$</code> then the associated $g$ is unique, so we can define <code>$T^*(f)=g$</code>. 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.</p> <p>Indeed, most books seem to just start out working with Hilbert spaces (and then usually <code>$T^*$</code> 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.</p> <p>It seems to me however that <code>$D(T^*)$</code> will always at least be weak<code>$^*$</code>-dense and that <code>$G(T^*)$</code> will be weak<code>$^*$</code>-closed in <code>$E^*\times E^*$</code>. Moreover, the proofs don't seem to need Hilbert space techniques. Moreover, starting with such a "weak<code>$^*$</code>-closed, densely defined operator" on <code>$E^*$</code>, 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.</p> <p>The only source I know which talks about "closed" operators in such generality is a paper by Ciorănescu and Zsidó, see <a href="http://www.ams.org/mathscinet-getitem?mr=0430867" rel="nofollow">MathSciNet</a> or <a href="http://projecteuclid.org/euclid.tmj/1178240775" rel="nofollow">Project Euclid</a>. Even they don't mention the duality result.</p> <blockquote> <p>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?</p> </blockquote> http://mathoverflow.net/questions/95188/weak-topology-restricted-to-the-unitary-group-of-a-von-neumann-algebra/95192#95192 Answer by Matthew Daws for Weak topology restricted to the unitary group of a von Neumann algebra Matthew Daws 2012-04-25T20:14:57Z 2012-04-25T20:14:57Z <p>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 <code>$x^{(n)}_m = \begin{cases} 1 &amp;: m\leq n, \\ -1 &amp;:m&gt;n. \end{cases}$</code> So $\mu(x^{(n)}) = -1$ for all $n$; thus $x^{(n)}$ does not converge weakly to $1$.</p> <p>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.</p> http://mathoverflow.net/questions/93873/where-is-the-error-in-this-argument/93941#93941 Answer by Matthew Daws for Where is the error in this argument? Matthew Daws 2012-04-13T08:05:47Z 2012-04-13T12:07:04Z <p>A link to the literature: I think of <code>$C^*(G)^*$</code> as being $B(G)$, the Fourier-Stieltjes algebra, realised as a (non-closed) algebra of continuous functions on $G$. Any member of <code>$C^*(G)^*$</code> 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 $g\mapsto (\pi(g)\xi|\eta)$.</p> <p>Then <code>$W^*(G)$</code> is <code>$B(G)^*$</code>. As the tensor product of representations corresponds to the product in $B(G)$, it follows that $G_{\otimes}$ is actually just the collection of <em>characters</em> on $B(G)$, namely algebra homomorphisms $B(G)\rightarrow\mathbb C$. Such things were explored by Walter in his paper <a href="http://projecteuclid.org/euclid.pjm/1102905857" rel="nofollow">"On the structure of the Fourier-Stieltjes algebra"</a></p> <p>It's shown that $G_{\otimes}$ is not a group, and that it contains proper partial isometries and projections; it is a semigroup though.</p> http://mathoverflow.net/questions/93295/separating-vectors-for-c-algebras Separating vectors for C$^*$-algebras Matthew Daws 2012-04-06T09:32:09Z 2012-04-07T06:32:12Z <p>(I asked this on <a href="http://math.stackexchange.com/questions/126023/separating-vectors-for-c-algebras" rel="nofollow">math.stackexchange</a>, without response).</p> <p>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 $A$.</p> <blockquote> <p>Need $\xi_0$ still be separating for $M$? That is, $x\in M, x(\xi_0)=0 \implies x=0$?</p> </blockquote> <p>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 <em>not</em> show this to be equivalent to $\xi_0$ being separating for $A$.</p> <p><strike>I <em>think</em> I can prove this using left Hilbert algebras. We turn <code>$\mathfrak A = \{ a(\xi_0) : a\in A \}$</code> 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.</strike> 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...</p> http://mathoverflow.net/questions/24864/almost-orthogonal-vectors Almost orthogonal vectors Matthew Daws 2010-05-16T06:46:00Z 2012-04-03T00:17:47Z <p>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 $|(u_i|u_j)| &lt; \epsilon$ when $i\not=j$. I guess I'm interested in how the biggest choice of $k$ grows with $n$ and $\epsilon$.</p> <p>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 $n^{-1/2} &lt; \epsilon$. 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.</p> http://mathoverflow.net/questions/91708/uniform-l-1-convergence-implies-uniform-convergence-pointwise-a-e/91709#91709 Answer by Matthew Daws for Uniform $L_1$ convergence implies uniform convergence pointwise a.e. Matthew Daws 2012-03-20T12:55:47Z 2012-03-20T12:55:47Z <p>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).</p> <p>Now set <code>$f^{(m)}_n = \chi_{A_{n,m}}$</code>. Then <code>$\lim_n \ \sup_m \|f^{(m)}_n\|_1 = \lim_n \ \sup_m |A_{n,m}| &lt; \lim_n \frac{1}{n} = 0.$</code> However, for any sequence $(k_n)$ and any $x\in\Omega$, <code>$\sup_m |f^{(m)}_{k_n}(x)| = \sup_m \chi_{A_{k_n,m}}(x) = \chi_{\bigcup_m A_{k_n,m}}(x) = 1.$</code> Thus your conclusion cannot hold.</p> http://mathoverflow.net/questions/90146/stone-ech-compactification-of-mathbb-r/90185#90185 Answer by Matthew Daws for Stone-Čech compactification of $\mathbb R$ Matthew Daws 2012-03-04T08:33:59Z 2012-03-04T14:55:27Z <p>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 $\beta X$.</p> <p>(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.)</p> <p>I'll now use that $\beta X$ is the character space of $C^b(X)$. Let $U\subseteq X$ be open.</p> <blockquote> <p>Lemma: Assume that $U$ is relatively compact. Under the isomorphism $C(\beta X)=C^b(X)$, we identify the ideal <code>$\{ f\in C(\beta X) : f(x)=0 \ (x\not\in U) \}$</code> with <code>$\{ F\in C^b(X) : f(x)=0 \ (x\not\in U) \}$</code></p> <p>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 $U$.</p> </blockquote> <p>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 <code>$C(\beta X) / \{ f\in C(\beta X) : f(x)=0 \ (x\not\in U) \}$</code>. So by the above, we identify $C(\beta X \setminus U)$ with <code>$C^b(X) / \{ F\in C^b(X) : F(x)=0 \ (x\not\in U) \}$</code>. 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 <code>$C^b(X) / \{ F\in C^b(X) : F(x)=0 \ (x\not\in U) \}$</code> 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.</p> <p>(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.)</p> http://mathoverflow.net/questions/90062/algebraic-vs-topological-ergodicity/90078#90078 Answer by Matthew Daws for algebraic VS topological ergodicity Matthew Daws 2012-03-02T21:49:45Z 2012-03-03T14:09:41Z <p>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 <code>$\|f\circ\phi_{g_i^{-1}} - f\|_\infty \rightarrow 0$</code>. 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 $U$).</p> <p>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.</p> <p>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.</p> <p>For (2): let $G=\mathbb Z$, so $\phi$ is generated by a single homeomorphism of $X$. Let <code>$X=\{ z\in\mathbb C : |z|\leq 1\}$</code> 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 $r&lt;1$, so by continuity $f=0$. Thus the action $\alpha$ is "ergodic", but $\phi$ has non-trivial invariant open and closed sets.</p> <p><strong>Edit:</strong> 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.</p> http://mathoverflow.net/questions/89839/do-signed-measures-on-sigma-rings-always-have-a-hahn-decomposition/89848#89848 Answer by Matthew Daws for Do signed measures on sigma-rings always have a Hahn decomposition? Matthew Daws 2012-02-29T09:11:51Z 2012-02-29T17:39:37Z <p>Your axioms are:</p> <ol> <li>$\emptyset\in\mathcal R$</li> <li>$A,B\in\mathcal R \implies (A\cup B)\setminus (A\cap B)\in\mathcal R$</li> <li>$A_n\in\mathcal R \implies \bigcap_n A_n\in\mathcal R$</li> </ol> <p>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 <a href="http://en.wikipedia.org/wiki/Delta-ring" rel="nofollow">http://en.wikipedia.org/wiki/Delta-ring</a> ) and not a $\sigma$-ring.</p> <p>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 <code>$\phi(A) = |A\cap\{\text{evens}\}| - |A\cap\{\text{odds}\}|.$</code> 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 $\mathcal R$.</p> <p>For a positive result, note that for any $A\in\mathcal R$, the collection <code>$\mathcal R_A = \{ A\cap B : B\in\mathcal R \}$</code> 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!)</p> http://mathoverflow.net/questions/89423/peter-weyl-theorem-as-proven-in-cartiers-primer/89428#89428 Answer by Matthew Daws for Peter-Weyl theorem as proven in Cartier's Primer Matthew Daws 2012-02-24T17:55:20Z 2012-02-24T18:10:49Z <p>(Same as pm's answer, with details.)</p> <p>Well, if $$R_f(\varphi)(h) = \int_G \varphi(g) f(g^{-1}h) \ dg$$ then the adjoint satisfies <code>\begin{align*} (R_f^*(\varphi)|\psi) &amp;= (\varphi|R_f(\psi)) = \int_{G\times G} \varphi(h) \overline{ \psi(g) f(g^{-1}h) } \ dg \ dh \\ &amp;= \int_{G\times G} \varphi(h) \tilde f(h^{-1}g) \overline{\psi(g)} \ dh \ dg = (R_{\tilde f}(\varphi) | \psi) \end{align*}</code> where $\tilde f(g) = \overline{ f(g^{-1}) }$. So $R_f^* = R_{\tilde f}$ is again a right translation operator.</p> http://mathoverflow.net/questions/89274/the-dual-space-of-cx-x-is-noncompact-metric-space/89278#89278 Answer by Matthew Daws for the dual space of C(X) (X is noncompact metric space) Matthew Daws 2012-02-23T12:57:46Z 2012-02-23T12:57:46Z <p>What you state in the first paragraph is the Riesz Representation Theorem (see <a href="http://en.wikipedia.org/wiki/Riesz_representation_theorem#The_representation_theorem_for_linear_functionals_on_Cc.28X.29" rel="nofollow">http://en.wikipedia.org/wiki/Riesz_representation_theorem#The_representation_theorem_for_linear_functionals_on_Cc.28X.29</a>) This is valid for all locally compact Hausdorff spaces; so in particular for $\mathbb R$ (ah, I guess, if you look at $C_0(\mathbb R)^*$).</p> <p>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 <a href="http://en.wikipedia.org/wiki/Stone_cech_compactification" rel="nofollow">http://en.wikipedia.org/wiki/Stone_cech_compactification</a> ) 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 $X$).</p> <p>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!)</p> http://mathoverflow.net/questions/87224/matrices-with-entries-in-a-c-algebra/87251#87251 Answer by Matthew Daws for Matrices with entries in a $C^*$-algebra Matthew Daws 2012-02-01T16:33:17Z 2012-02-01T16:33:17Z <p>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 $\|x\| = \|(x,x)\|^{1/2}$.</p> <blockquote> <p>Lemma: We have that $(x,y)^* (x,y) \leq \|x\|^2 (y,y)$ the order in the C<code>$^*$</code>-algebra sense.</p> <p>Proof: (Copied from Lance's Hilbert C<code>$^*$</code>-module book). Wlog $\|x\|=1$. For $a\in A$ let $a\cdot x = (ax_i)$. Then <code>\begin{align*} 0 &amp;\leq (a\cdot x-y, a\cdot x-y) \\ &amp;= a (x,x) a^* - (y,x)a^* - a(x,y) + (y,y) \\ &amp;\leq aa^*- (y,x)a^* - a(x,y) + (y,y) \end{align*}</code> The claim that <code>$a(x,x)a^* \leq aa^*$</code> follows as if $c\in A^+$ then always <code>$aca^* \leq \|c\|aa^*$</code>. Now set $a=(x,y)^*=(y,x)$ and the claim follows.</p> </blockquote> <p>In particular, $\|(x,y)\|^2 \leq \|x\|^2 \|y\|^2$ and so <code>$\sup\{ \|(x,y)\| : \|x\| \leq 1 \} = \|y\|$</code>.</p> <p>So you define <code>$\| a \| = \sup \{ (x, ay) : \|x\|\leq 1, \|y\|\leq 1 \}$</code> where $(ay)_i = \sum_j y_j a_{ij}^*$. Then from the observation above, <code>\begin{align*} \|a\|^2 &amp;= \sup \{ \|ay\|^2 : \|y\|\leq 1 \} = \sup\{ (ay,ay) : \|y\|\leq 1 \} \\ &amp;= \sup\Big\{ \Big\| \sum y_j a_{ij}^* a_{ik} y_k^* \Big\| : \|y\|\leq 1 \Big\} \\ &amp;= \sup\{ \|(y, (a^*a)y)\| : \|y\|\leq 1 \} \\ &amp;\leq \sup\{ \|(a^*a)y\| : \|y\|\leq 1 \} \\ &amp;= \sup\{ \|(z,(a^*a)y)\| : \|y\|\leq 1, \|z\|\leq 1 \} = \|a^*a\|. \end{align*}</code> However, for any $z$, <code>$\|az\| = \sup\{ \|(x,az)\| : \|x\|\leq 1 \} \leq \sup\{ \|(x,ay)\| : \|x\|\leq 1, \|y\|\leq \|z\| \} = \|a\|\|z\|.$</code> 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: <code>$\|a\|^2 \leq \|a^*a\| \leq \|a^*\| \|a\| = \|a\|^2$</code> and so we have equality throughout, establishing the C<code>$^*$</code>-identity for $M_n(A)$.</p> <p>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<code>$^*$</code>-algebra.</p> http://mathoverflow.net/questions/85591/liftings-of-l-infty-functions "Liftings" of L^\infty functions Matthew Daws 2012-01-13T16:13:26Z 2012-01-13T16:23:15Z <p>This is motivated by this question: <a href="http://mathoverflow.net/questions/85411/is-there-an-inclusion-of-l-inftyg-into-c-0g" rel="nofollow">http://mathoverflow.net/questions/85411/is-there-an-inclusion-of-l-inftyg-into-c-0g</a> and Bill Johnson's comments there.</p> <p>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:</p> <ol> <li>$\rho(f)=f$ $\mu$-a.e.</li> <li>if $f=g$ $\mu$-a.e. then $\rho(f) = \rho(g)$</li> <li>if $f$ is continuous, then $\rho(f)=f$.</li> </ol> <p>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...</p> <p>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.</p> <p>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 <em>bounded</em> Radon measure. So my question is:</p> <blockquote> <p>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?</p> </blockquote> <p>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?</p> http://mathoverflow.net/questions/85411/is-there-an-inclusion-of-l-inftyg-into-c-0g/85412#85412 Answer by Matthew Daws for Is there an inclusion of $L_\infty(G)$ into $C_0(G)^{**}$? Matthew Daws 2012-01-11T13:28:19Z 2012-01-11T14:19:57Z <p><strong>Undeleted:</strong> 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"...</p> <p>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...</p> <p><strong>Edit:</strong> 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...</p> http://mathoverflow.net/questions/83504/relation-between-sot-convergence-of-t-and-t/83511#83511 Answer by Matthew Daws for relation between SOT-convergence of T and T' Matthew Daws 2011-12-15T12:10:58Z 2011-12-15T12:10:58Z <p>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.</p> <p>To deal with the $1/n$ term, instead use a weighted shift. So something like <code>$T\xi = T(\xi_1,\xi_2,\xi_3,\cdots) = (2\xi_2,\frac{3}{2}\xi_3,\frac{4}{3}\xi_4,\cdots).$</code> Then <code>$T^2\xi = (3\xi_3, \frac{4}{2}\xi_4, \frac{5}{3}\xi_5, \cdots),$</code> and so forth. So $\frac{1}{n}T^n$ will be SOT null. But <code>$T^*\xi = (0,2\xi_1,\frac{3}{2}\xi_2,\frac{4}{3}\xi_3,\cdots).$</code> 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.</p> http://mathoverflow.net/questions/82962/reasonable-crossnorm-on-banach-algebra-tensor-product-constructed-from-isometric/82966#82966 Answer by Matthew Daws for Reasonable crossnorm on Banach algebra tensor product constructed from isometric non-degenerate representations Matthew Daws 2011-12-08T14:18:40Z 2011-12-08T14:25:08Z <p>Let <code>$D_A\subseteq A^*$</code> be the functionals of the form $\mu(\pi(\cdot)x)$, for <code>$x\in X, \mu\in X^*, \|x\|\leq 1, \|\mu\|\leq 1$</code>. As $\pi$ is an isometry, Hahn-Banach shows that the convex hull of $X$ is weak<code>$^*$</code>-dense in the closed ball of <code>$A^*$</code>, say <code>$A^*_{[1]}$</code>.</p> <p>Similarly for $D_B$ using $\rho$. It's clear (*) that <code>$|(\mu_A\otimes\mu_B)\tau| \leq \| (\pi\otimes\rho)(\tau) \|_{op} \qquad (\tau\in A\otimes B, \mu_A\in D_A, \mu_B\in D_B).$</code> 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 <code>$|(\mu_A\otimes\mu_B)\tau| \leq \| (\pi\otimes\rho)(\tau) \|_{op} \qquad (\tau\in A\otimes B, \mu_A\in A^*_{[1]}, \mu_B\in B^*_{[1]}),$</code> and that's all you need to show, I think?</p> <p>Why is <code>(*)</code> true? Given <code>$x\in X,\mu\in X^*,y\in Y,\lambda\in Y^*$</code>, we have that $x\otimes y\in X\widehat\otimes Y$ of norm $\|x\|\|y\|$, and also <code>$\mu\otimes\lambda \in (X\widehat\otimes Y)^*$</code> (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.</p> http://mathoverflow.net/questions/80864/a-left-inverse-for-the-comultiplication-on-a-hopf-von-neumann-algebra/81060#81060 Answer by Matthew Daws for A left inverse for the comultiplication on a Hopf von Neumann algebra Matthew Daws 2011-11-16T12:19:01Z 2011-11-30T09:41:10Z <p>This is far from a full answer (but maybe it will inspire other answers).</p> <p>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.</p> <p>Firstly, if $M=L^\infty(G)$ with $\Delta(f)(s,t) = f(st)$, then define <code>$T_*:L^1(G) \rightarrow L^1(G\times G)$</code> by <code>$T_*(f)(s,t) = f(st) f_0(s),$</code> where $f_0\in L^1(G)$ is some fixed, positive function with <code>$\|f_0\|_1=1$</code>. This is bounded, as <code>\begin{align} \|T_*(f)\|_1 &amp;= \int_G \int_G |f(st) f_0(s)| \ dt \ ds = \int_G \int_G |f(ss^{-1}t)| \ dt \ |f_0(s)| \ ds \\ &amp;= \int_G \|f\|_1 |f_0(s)| \ ds = \|f\|_1, \end{align}</code> using the left-invariance of the Haar measure. Then the pre-adjoint <code>$\Delta_*:L^1(G\times G)\rightarrow L^1(G)$</code> is <code>$\Delta_*(f)(t) = \int_G f(s,s^{-1}t) \ ds,$</code> i.e. convolution (if you set $f=a\otimes b$). So then <code>$\Delta_* T_*(f) (t) = \int_G T_*(f)(s,s^{-1}t) \ ds = \int_G f(ss^{-1}t) f_0(s) \ ds = f(t).$</code> So <code>$\Delta_* T_*$</code> is the identity, as required.</p> <p>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.</p> <p>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 $VN(G)$.</p> <p><strong>Edit (29 Nov):</strong> (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 $\psi$.</p> <p>So in particular, this gives a positive answer (even with the "normal") condition in the separable SIN case, say.</p> http://mathoverflow.net/questions/74387/100-bounty-ended-do-invariant-measures-maximize-the-integral/81260#81260 Answer by Matthew Daws for 100€ bounty ended: Do invariant measures maximize the integral? Matthew Daws 2011-11-18T16:27:18Z 2011-11-19T08:53:43Z <p><strong>Edit:</strong> Here is what I think is a counter-example.</p> <p>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 <code>$\int f(x) \ d\mu(x) = \lim_{x\in\mathcal V} f(x).$</code> 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 $y\in\mathbb Z$, <code>$\int \phi(x+y) \ d\nu(x) = \nu(2\mathbb N-y) = \begin{cases} \nu(2\mathbb N) &amp;: y\in 2\mathbb Z, \\ \nu(2\mathbb N+1) &amp;: y\in 2\mathbb Z+1, \end{cases} = \alpha.$</code> Thus <code>$\int \int \phi(x+y) \ d\nu(x) \ d\mu(y) = \int \alpha \ d\mu(y) = \alpha,$</code> as $\mu$ is a probability measure. By contrast, let $\nu$ be defined by <code>$\int f(x) \ d\nu(x) = \lim_{y\in\mathcal U} f(y).$</code> Then <code>$\int \phi(x+y) \ d\nu(x) = \begin{cases} 1 &amp;: y\in 2\mathbb Z, \\ 0 &amp;: y \in 2\mathbb Z+1, \end{cases}$</code> and so <code>$\int \int \phi(x+y) \ d\nu(x) \ d\mu(y) = \int \chi_{2\mathbb Z}(y) \ d\mu(y) = 1.$</code> So $F$ is not maximised on $I_\phi$.</p> <p>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 $\nu\in I_\phi$.</p> <p>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:</p> <blockquote> <p>Let $\mu\in I_\phi$ maximise the integral $\int \phi(x) \ d\mu(x)$. Then $F$ attains its maximum on $I_\phi$.</p> </blockquote> <p><strong>Old post:</strong> (Explains my thinking).</p> <p>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.</p> <p>Consider $A=\ell^1(\mathbb Z)$ with the convolution product, so $A$ is commutative. Then <code>$A^*=\ell^\infty(\mathbb Z) = C(\beta\mathbb Z)$</code> is an $A$-module: $(a\cdot f)(b) = f(ba)$ for <code>$a,b\in A,f\in A^*$</code>. Then <code>$A^{**}=M(\beta\mathbb Z)$</code> the space of finite Borel measures on the Stone-Cech compactification $\beta\mathbb Z$. Your space $M(\mathbb Z)$ is just the positive measures <code>$\mu\in A^{**}$</code> with $\mu(1)=1$.</p> <p>We try to extend the product of $A$ to $A^{**}$. Firstly we define a bilinear map <code>$A^{**}\times A^*\rightarrow A^*$</code> by <code>$(\mu\cdot f)(a) = \mu(a\cdot f) \qquad (\mu\in A^{**}, f\in A^*, a\in A).$</code> But then we have two choices for the product on <code>$A^{**}$</code>: <code>$(\mu \Box \lambda)(f) = \mu(\lambda\cdot f), \quad (\mu\diamond\lambda)(f) = \lambda(\mu\cdot f) \qquad (\mu,\lambda\in A^{**}, f\in A^*).$</code> A little thought shows that $\mu\diamond\lambda = \lambda\Box\mu$.</p> <p>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 <code>$(\phi\cdot\delta_x)(\delta_y) = \phi(\delta_{x+y}) \implies (\mu\cdot\phi)(\delta_x) = \mu(\phi\cdot\delta_x) = \int \phi(x+y) \ d\mu(y).$</code> 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 $\mu\cdot\phi = \mu(\phi) 1$.</p> <p>Similarly, your map $F$ is just $F(\nu) = (\mu\Box\nu)(\phi)$.</p> <p>As you allude to, it's known that $\lambda\Box\mu \not= \mu\Box\lambda$ for arbitrary $\lambda,\mu$. However, we say that <code>$f\in A^*$</code> is "weakly almost periodic" (WAP) if $(\lambda\Box\mu)(f) = (\mu\Box\lambda)(f)$ for all <code>$\mu,\lambda\in A^{**}$</code>. So if $\phi$ is WAP and $\mu\in I_\phi$ then for any $\nu\in M(\mathbb Z)$, <code>$F(\nu) = (\mu\Box\nu)(\phi) = (\nu\Box\mu)(\phi) = \nu(\mu\cdot\phi) = \nu(1) \mu(\phi) = \mu(\phi),$</code> 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 $I_\phi$.</p> <p>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 <code>$\lim_n \lim_m \phi(x_n+y_m) = \lim_m \lim_n \phi(x_n+y_m)$</code> 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.</p> <p>We may as well assume that <code>$\|\phi\|_\infty=1$</code>. 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 $\mathbb N$.</p> <blockquote> <p>So a test case is to find $\phi$ not in WAP and with <code>$\nu(\phi)&lt;\|\phi\|_\infty$</code> 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 $\phi$?</p> </blockquote> <p>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.</p> http://mathoverflow.net/questions/79821/conditional-expectations-onto-masas-in-type-iii-factors/79856#79856 Answer by Matthew Daws for Conditional expectations onto masas in type III factors Matthew Daws 2011-11-02T18:28:26Z 2011-11-02T20:33:52Z <p>Takesaki showed in section 6 of:</p> <p>MR0303307 (46 #2445) Takesaki, Masamichi Conditional expectations in von Neumann algebras. J. Functional Analysis 9 (1972), 306–321. <a href="http://www.sciencedirect.com/science/article/pii/0022123672900043" rel="nofollow">http://www.sciencedirect.com/science/article/pii/0022123672900043</a></p> <p>that the following are equivalent for a von Neumann algebra M (not necessarily a factor):</p> <ul> <li>M is finite</li> <li>Every MASA in M admits a conditional expectation (i.e. norm one normal projection) onto it.</li> </ul> <p><strong>Edit:</strong> 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!</p> http://mathoverflow.net/questions/78988/when-does-mathscinet-review-a-paper/78992#78992 Answer by Matthew Daws for When does mathscinet review a paper? Matthew Daws 2011-10-24T16:08:40Z 2011-10-24T16:08:40Z <p>You can see the "official line" here: <a href="http://www.ams.org/publications/math-reviews/mr-edit" rel="nofollow">http://www.ams.org/publications/math-reviews/mr-edit</a> In particular,</p> <blockquote> <p>Elementary articles or books, or articles that have not been refereed are ordinarily not listed.</p> </blockquote> <p>In my experience, (and in addition to Thierry's and Andreas's list) this includes:</p> <ol> <li>Books reviews</li> <li>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.</li> </ol> <p>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).</p> <p>As to my case (2.) above-- yes, perhaps this <em>is</em> 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...</p> http://mathoverflow.net/questions/118667/open-problems-in-the-theory-of-compact-quantum-groups/119060#119060 Comment by Matthew Daws Matthew Daws 2013-01-17T09:04:05Z 2013-01-17T09:04:05Z As for (b), it's trivial for compact, shown by Reiji for discrete, and I <i>think</i> still wide open for Kac...? http://mathoverflow.net/questions/112468/ultrapowers-of-operators/112561#112561 Comment by Matthew Daws Matthew Daws 2012-11-16T12:38:35Z 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 $(\mathcal B(X))_{\mathcal U}$ <i>will</i> 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. http://mathoverflow.net/questions/112468/ultrapowers-of-operators/112561#112561 Comment by Matthew Daws Matthew Daws 2012-11-16T12:35:49Z 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 $\mu_i(y_i)&gt; \|y_i\|-1/n$; 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... http://mathoverflow.net/questions/100368/idempotent-homomorphisms-of-von-neumann-algebras Comment by Matthew Daws Matthew Daws 2012-06-22T20:31:58Z 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.?) http://mathoverflow.net/questions/99451/is-the-set-of-all-probability-measures-weak-closed/99456#99456 Comment by Matthew Daws Matthew Daws 2012-06-13T19:25:40Z 2012-06-13T19:25:40Z Annoyingly, this is exactly the same as an answer I gave over at math.stackexchange: <a href="http://math.stackexchange.com/questions/157795/is-the-set-of-all-probability-measures-weak-closed" rel="nofollow" title="is the set of all probability measures weak closed">math.stackexchange.com/questions/157795/&hellip;</a> 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... http://mathoverflow.net/questions/99410/when-does-the-adjoint-operator-map-closed-convex-subsets-to-closed-convex-subset Comment by Matthew Daws Matthew Daws 2012-06-13T12:27:23Z 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. http://mathoverflow.net/questions/99410/when-does-the-adjoint-operator-map-closed-convex-subsets-to-closed-convex-subset Comment by Matthew Daws Matthew Daws 2012-06-13T12:10:02Z 2012-06-13T12:10:02Z This is very similar to a question here: <a href="http://math.stackexchange.com/questions/157069/linear-image-of-closed-convex-set" rel="nofollow" title="linear image of closed convex set">math.stackexchange.com/questions/157069/&hellip;</a> http://mathoverflow.net/questions/99091/spectrum-of-l-inftyx-mu Comment by Matthew Daws Matthew Daws 2012-06-08T15:47:30Z 2012-06-08T15:47:30Z A related question on math.se: <a href="http://math.stackexchange.com/questions/81324/a-problem-on-c-ast-algebras-and-w-ast-algebras" rel="nofollow" title="a problem on c ast algebras and w ast algebras">math.stackexchange.com/questions/81324/&hellip;</a> http://mathoverflow.net/questions/99091/spectrum-of-l-inftyx-mu Comment by Matthew Daws Matthew Daws 2012-06-08T11:58:03Z 2012-06-08T11:58:03Z @Andr&#233;: Okay, we agree to disagree I guess. +1 your answer. http://mathoverflow.net/questions/99091/spectrum-of-l-inftyx-mu Comment by Matthew Daws Matthew Daws 2012-06-08T08:41:47Z 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. http://mathoverflow.net/questions/96373/group-c-algebras-of-finite-groups Comment by Matthew Daws Matthew Daws 2012-05-09T08:20:32Z 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... http://mathoverflow.net/questions/96084/interpolation-space-between-l-2-and-l-infty Comment by Matthew Daws Matthew Daws 2012-05-06T07:39:54Z 2012-05-06T07:39:54Z I would <i>guess</i> 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 &quot;interpolation&quot; space? Why, in fact, can I not take $Z=Y$? http://mathoverflow.net/questions/95564/ideals-of-cx-with-only-finitely-many-number-of-zerosets Comment by Matthew Daws Matthew Daws 2012-05-01T20:38:58Z 2012-05-01T20:38:58Z Is actually $Z[I]$ equal to the intersection of all the sets $Z[f]$ with $f\in I$? http://mathoverflow.net/questions/95267/closed-operators-and-duality Comment by Matthew Daws Matthew Daws 2012-04-27T14:16:30Z 2012-04-27T14:16:30Z @Bill: Thanks for those suggestions. Kato, Chapter III Section 5 gets very close to what I had in mind. http://mathoverflow.net/questions/95267/closed-operators-and-duality/95327#95327 Comment by Matthew Daws Matthew Daws 2012-04-27T10:50:31Z 2012-04-27T10:50:31Z This almost does it in too much detail; but it's a pretty good reference. Thanks!