User benjamin hayes - MathOverflow

Answer by Benjamin Hayes for spectra of sums in (Banach) algebras

2011-05-07T13:58:49Z

Here is a counterexample to $B/rad(B)$ being commutative. Let $S\colon l^{2}(\Bbb N)\to l^{2}(\Bbb N)$ be the unilateral shift. Then the spectrum of $S$ is $\overline{\Bbb D}={z\in \Bbb C:|z|\leq 1}.$ Now $S^{*}$ has the same spectrum. Consider the algebra they generate inside $B(l^{2}(\Bbb N)),$ it is a C $^{*}$ -algebra, hence its radical is zero. For any $\lambda,\mu \in \Bbb C$ we have $\|\lambda S+\mu S^{*}\|\leq |\lambda|+|\mu|.$ Thus the spectrum of $\lambda S+\mu S^{*}$ is contained in $(|\lambda|+|\mu|)\overline{\Bbb D}.$ It is straigthfoward to show that $(|\lambda|+|\mu|)\overline{\Bbb D}=\lambda\overline{\Bbb D}+\mu \overline{\Bbb D},$ so this does it.

In the case of a $C^{*}$ -algebra $A$ one can at say that the closed convex hull of the spectrum. of $a\in A$ is the set of $\phi(a),$ where $\phi$ is a state. Thus you always get that the closed convex hull of $a+b$ is contained in the sum of the closed convex hull of $a$ and $b.$ I don't know if this is true in arbitrary Banach algebras, but maybe there is a similar statement involving polynomial convex hulls.

EDIT: Okay this last statement about states may not be true, as I think you need $a$ to be normal for this statement about the closed convex hull of the spectrum of $a$ to be true.

Answer by Benjamin Hayes for a complex borel measure, whose Fourier transform goes to zero

2011-05-07T10:32:36Z

I'll try to write my suggestions assuming you are working on $\Bbb R^{n}.$

Here's what I would suggest, although this doesn't constitute a full solution. Write

$$\mu=\mu_{a}+\mu_{sc}+\mu_{d}$$

where $\mu_{a}$ is absolutely continuous with respect to Lebesgue measure, $\mu_{sc}$ is singular with respect to Lebesgue measure, but non-atomic and $\mu_{d}$ is purely atomic.

We have to argue that $\mu_{d}=0.$ By Riemann Lebesgue, we know that

$$\widehat{\mu_{a}}(\xi)\to 0$$ as $|\xi|\to \infty.$

Write $\mu_{d}=\sum_{n}c_{n}\delta_{x_{n}}$ with $x_{n}\ne x_{m}$ for $n\ne m.$

Then $$\widehat{\mu_{d}}(\xi)=\sum_{n}c_{n}e^{2\pi i \xi\cdot x_{n}}$$

and

$$\frac{1}{2^{d}T_{1}\cdots T_{d}}\int_{-T_{d}}^{T_{d}}\cdots\int_{-T_{1}}^{T_{1}}|\widehat{\mu_{d}}(\xi)|^{2}\prod_{j}d\xi_{j}=$$ $$\sum_{n}|c_{n}|^{2}+\frac{1}{2^{d}T_{1}\cdots T_{d}}\sum_{n\ne m}c_{n}\overline{c_{m}}\sin(2\pi (x_{n}-x_{m})T)$$

from here one can see that the above average tends to $\sum_{n}|c_{n}|^{2}$ as $T_{j}\to \infty$ for all $j,$ this makes it impossible for $\widehat{\mu_{d}}(\xi)\to 0$ as $|\xi|\to \infty.$

Now if one can do some analysis on $\widehat{\mu_{sc}}(\xi)$ as $|\xi|\to \infty,$ this may solve the problem.

In one dimension one can try to write $\mu_{sc}=dF,$ with $F$ a continuous function of bounded variation, it may be that some integration by parts trick works here but I'm not seeing it.

Answer by Benjamin Hayes for Examples of common false beliefs in mathematics.

2011-05-04T15:00:18Z

Here are mistakes I find surprisingly sharp people make about the weak$^{*}$ topology on the dual of $X,$ where $X$ is a Banach space.

-It is metrizable if $X$ is separable.

-It is locally compact by Banach-Alaoglu.

-The statement $X$ is weak $^{*}$ dense in the double dual of $X$ proves that the unit ball of $X$ is weak$^{*}$ dense in the unit ball of the double dual of $X.$

The first two are in fact never true if $X$ is infinite dimensional. While both statements in the third claim are true, the second one is significantly stronger, but a lot of people believe you can get it from the first by just "rescaling the elements" to have norm $\leq 1.$ (Although the proof of the statements in the third claim is not hard). The difficulty is that if $X$ is infinite dimensional then for any $\phi$ in the dual of $X,$ there exists a net $\phi_{i}$ in the dual of $X$ with $\|\phi_{i}\|\to \infty$ and $\phi_{i}\to \phi$ weak$^{*},$ so this rescaling trick cannot be uniformly applied. Really these all boil down to the following false belief:

-The dual of $X$ has a non-empty norm bounded weak$^{*}$ open set.

Again when $X$ is infinite dimensional this always fails.

Answer by Benjamin Hayes for Plessner's Theorem (1929)

2011-05-01T19:42:48Z

I know you said you were looking for Plessner's original, but you can give a short proof of this. Fix a compactly supported smooth function $\phi$ with integral one. Let $\phi_{t}(x)=t^{-1}\phi(x/t),$ and set $\nu_{t}=\mu*\phi_{t}.$ Note that $\nu_{t}$ is absolutely continuous with respect to Lebesgue with Radon-Nikodym derivative $\int\phi_{t}(x-y)\,d\mu(y).$ If $f$ is a $C_{0}(\Bbb R)$ function a straight-foward computation using Fubini's Theorem shows that $$\int fd\nu_{t}=\int\int f(y)\phi_{t}(y)d\tau_{x}\mu(y)dx$$ where $\tau_{x}\mu(E)=\mu(E-x).$ Because $\mu$ is continuous under translation the usual arguments show that $\nu_{t}$ converges to $\mu$ in norm, (here think of the total variation norm as the operator norm). Thus $\mu$ is a norm limit of $L^{1}(\Bbb R)$ functions and thus must be in $L^{1}(\Bbb R).$
(Note that the argument can be used to prove similar statements e.g. an $L^{\infty}$ function continuous under translations must agree almost everywhere with a uniformly continuous function).

Answer by Benjamin Hayes for Spatial isomorphisms of tensor product of factors

2011-04-30T19:47:23Z

Okay so I think I have an answer for existence of a spatial isomorphism in the case that the representation $\pi_{1},\pi_{2}$ are the identity representaion on $H.$ . Note that the condition $N\vee M'\cong N\overline{\otimes}M'$ spatially is independent of the way $M$ is represented. Indeed, suppose $M$ is represented on $H$ and $U$ is a unitary which conjugates $N\overline{\otimes}M'$ to $N\vee M'.$ If $K$ is another Hilbert spaces, and $M$ is represented as $M\otimes \Bbb C$ on $H\otimes K,$ then $U\otimes 1$ conjugates $N\overline{\otimes}(M'\overline{\otimes}B(K))$ to $N\vee (M'\overline{\otimes} B(K))$ (here $M'$ means the commutant in $H.$) Similarly, suppose we cut the representation by a projection $p$ in $M'.$ Since $U^{*}pU=1\otimes p$ by assumption, the unitary $U(p\otimes p):pH\otimes pH\to pH$ conjugates $N\overline{\otimes}pM'p$ and $N\vee pM'p.$ By the essential uniques of a normal representation of a Von Neumann Algebra, this proves the claim.

Let us first assume $M$ is type $II.$ Since $M$ is type $II,$ I will assume from here on that $M$ is represented on $L^{2}(M,\tau)$ with $\tau$ a (unique up to scalars) semifinite normal trace.

First off, let's note that $M$ cannot be type $II_{1}$ and this isomorphism be spatial, then we have $N'\cap M\cong (N\overline{\otimes} M')'=N'\overline{\otimes} M.$ Assume $M$ is type $II_{1},$ as noted in my last post, $N'\cap M$ cannot be finite dimensional, and being a subfactor of $M$ (being isomorhpic to $N'\overline{\otimes} M$) it must be a $II_{1}$-factor. But the assumption $N'\cap M$ is infinite dimensional implies that the inclusion $N\subseteq M$ is not finite index, i.e. $\dim_{N}(H)=\infty.$ Then, by definition, we know that $N'$ is infinite. (For more details see Section XIX.2 in Takesaki's Theory of Operator Algebras III).

So we may assume $M$ is type $II_{\infty}.$

So if $M$ is a $II_{\infty}$ factor we may focus on the representation of $M$ on $L^{2}(M,\tau)$ with $\tau$ a fixed semifinite normal trace on $M$ (unique up to scalar multiplication). In this case we claim that this isomorphism is spatial if and only if $N'\cap M$ is infinite. Indeed if $N\vee M'\cong N\overline{\otimes}M'$ spatially then taking commutants implies that $N'\cap M\cong N'\overline{\otimes}M$ spatially, in particular since $M$ is infinite so is $N'\cap M.$

Conversely if $N'\cap M=(N\vee M')'$ is infinite, then since $(N\otimes M')'=N'\otimes M$ is $II_{\infty}$ we have two isomorphic Von Neumann algebras with properly infinite commutants. It is known that two such Von Neumann algebras must be spatially isomorphic (see Theorem V.3.1 in Takesaki's Theory of Operator Algebras I).

If $M$ is type $I$ i.e. isomorphic to $B(H)$ we may take the Hilbert space $M$ is represented on to be $H$ with the canonical action of $M.$ In this case since $M'=\Bbb C,$ it is easy to see that $N\cong N\otimes \Bbb C$ spatially.

Of course, one wouldn't really be able to regard this as explicit, because you would have to know how to write one representation of $M$ in terms of another by tensoring with the trivial representation and cutting by projections.

Answer by Benjamin Hayes for Spatial isomorphisms of tensor product of factors

2011-04-30T13:52:05Z

Potentially a silly question, but do you have an example where neither $M$ nor $N$ is finite type $I$ and one of $M$ and $N$ is not infinite type $I,$ and $N\vee M'\cong N\overline{\otimes} M',$ (by the map you defined) for $N$ and $M$ factors? It's not clear to me that there is a non-trivial case when this actually is an isomorphism, or that there is an example where these algebras are isomorphic by a different map. For example if $N=M$ this is impossible unless $M$ is type I, since $M\vee M'=B(H).$

Also if say $M$ or $N$ is type $II$ you need to have $N$ ``far'' from $M$ for $N\vee M\cong N\overline{\otimes} M,$ for example $N'\cap M$ cannot be finite dimensional (because then $N\vee M$ is type I begin a commutant of a type $I$ von Neumann algebra). So this will rule out finite index inclusions, for example.

Answer by Benjamin Hayes for What kind of completion is this?

2011-04-29T10:16:19Z

This is done in Conway's book on Functional Analysis, (at least as a Banach space but the proof should work as a Von Neumann algebra as well), although I don't have the book on me and don't know the exact chapter/section reference. Note that if $\mu$ and $\nu$ are measures on $X$ with $\mu <<\nu$ and $f=0$ $\nu$ almost everywhere then $f=0$ $\nu$ almost everywhere so we have a well-defined map $L^{\infty}(X,\nu)\to L^{\infty}(X,\mu).$ One can endow theinverse limit, call it Y, of $L^{\infty}(X,\mu)$ under these maps as a Banach space and show that it is isomorphic to $C(X)^{**}.$ The duality between $Y$ and $M(X)$ is somewhat clear, if $f=[f_{\mu}]$ is a compatible collection of functions (so $f_{\mu}=f_{\nu}$ $\mu$ almost everywhere if $\mu<<\nu$) then the integral of $f_{\mu}$ against $\mu$ is well-defined for each $\mu\in M(X)$ and gives the duality between $Y$ and $M(X).$

Comment by Benjamin Hayes

2011-10-13T17:05:07Z

Okay, thanks I think it see it in the compact case: converging at the function 1 on X gives you a bound on the norm so this probably reduces to the fact that weak$^{*}$ is metrizable on bounded sets.

Comment by Benjamin Hayes

2011-10-13T05:59:24Z

Why should it be obvious that $M^{+}(X)$ is metrizable? I think this fails for the weak$^{*}$ topology.

Comment by Benjamin Hayes

2011-10-12T03:47:20Z

$X^{∗}$ in the weak∗ topology is a countable union of $\{\phi\in X^{*}:\|\phi\|\leq N\}$, which have empty weak∗ interior. Hence, if the weak∗ topology were metrizable, we get a contradiction to the Baire Category Theorem. Are you sure you don't mean the weak∗ topology on the state space of $C_{0}(X)?

Comment by Benjamin Hayes

2011-10-12T03:42:27Z

I think $M(T)$ is not metrizable in the weak$^{*}$ topology, and in fact my claim that this fails for every infinite dimensional Banach space i also think is true. The rough outline of the proof I saw was this: 1. If $X^{*}$ is weak$^{*}$ metrizable, then a first countabliity at the origin argument implies that $X^{*}$ has a translation invariant metric given the weak$^{*}$ topology. 2. One can characterize completeness topologically for translation-invariant metrics, and see directly that if $X^{*}$ had a translation-invariant metric given the weak$^{*}$ topology it would be complete.

Comment by Benjamin Hayes

2011-09-27T20:54:32Z

Preserving norms isn't trivial.

Comment by Benjamin Hayes

2011-06-15T03:20:17Z

In Takesaki II Problem IX.7 there is a outline of a proof to the statment that $\tau(f(S))\leq \tau(f(T))$ for any $f\geq 0$ continuous with $f(0)=0.$ My guess is that one can work with this a little bit to show that the same inequality holds for $f=\chi_{(s,\infty)}$ which at least handles the factor case. The general case you may be able to do by working with the extended center-valued trace but I don't really know.

Comment by Benjamin Hayes

2011-06-13T06:11:17Z

I think the precise statement is Takesaki's duality theorem. If $\mathbb{R}$ acts on $M,$ then there is a ``dual action'' of $\mathbb{R}$ (or more precisely $\widehat{\mathbb{R}}$) and Takesaki's duality theorem shows that $(M\rtimes \mathbb{R})\rtimes \mathbb{R}\cong M\overline{\otimes} B(L^{2}(\mathbb{R}).$ This is done in Takesaki II Theorem $X.2.3 $ (iii) (and actually the more general case of crossed products by locally compact abelian groups is considered). If I recall, the proof is just playing around a lot with Fourier transforms.

Comment by Benjamin Hayes

2011-05-23T07:31:50Z

I believe the standard definition is no closed $G$-invariant subspaces.

Comment by Benjamin Hayes

2011-05-17T19:35:21Z

If you take unitary representations on infinite dimensional Hilbert spaces you can reformulate some of this stuff. In this case, $V\otimes V^{*}$ is isomorphic to Hilbert-Schmidt operators from $V$ to $V,$ and has a copy of the trivial representation if and only if $V$ is finite dimensional. Also, one can reformulate Schur's Lemma to say that a unitary representation is irreducible if and only if there are no nontrivial bounded operators commuting with $\pi(g)$ for all $g\in G.$ (Think of commuting operators as $G$-modular maps). This requires some tools though, namely the spectral theorem.

Comment by Benjamin Hayes

2011-05-08T08:26:46Z

I amenability should be defined by either F$\'o$lner sequences or invariant means. As an operator algebraist, I like the C`$^{∗}$ and W$^{∗}$ variants as well, but more recall these as really useful facts to remember. I like the F$\'o$lner sequence definition, its the most intuitive and hands on (how do you find an invariant mean? usually its not constructive), and for many approximation and dynamical purposes (entropy) it's nicest to use. The nice thing about the almost invariant vectors/pdf definition is it has may generalizations the others don't (haggerup/weakly amenable)

Comment by Benjamin Hayes

2011-05-08T07:31:28Z

Okay, thanks for your correction and clarification. Can one use this to get some sort of statement about the spectrum of $a+b.$ (I'm not familiar with the definition of numerical range as in an arbitrary unital Banach algebra).

Comment by Benjamin Hayes

2011-05-07T16:08:36Z

Nice! I vaguely remembered something like this when typing my last post, but couldn't quite remember what the exact statement was.

Comment by Benjamin Hayes

2011-05-07T15:03:42Z

This sounds highly related to Geometric Group Theory. I would check in books on that area, in particular De La Harpe's Geometric Group Theory, where he actually computes some of these if I recall correctly.

Comment by Benjamin Hayes

2011-05-07T14:16:07Z

Can't you get the statement that compact Lie groups are linear more or less from Peter-Weyl? This is at least writes your compact Lie group as an inverse limit of Lie groups $G_{n}.$ It seems that once $\dim G_{n}=\dim G_{m}$ for $m\geq n,$ the maps $G_{m+1}--->G_{m}--->G_{m-1}--->...--->G_{n},$ will have to be covers and this process can't be nontrivial for very long.

Comment by Benjamin Hayes

2011-05-07T14:01:43Z

What is a graded $C^{*}$ -algebra? And what are its odd elements?