New answers tagged

1

Given an abelian subalgebra $N$ of a von Neumann algebra $M$, such as the center of $M$, one can always write $N=L^\infty (X)$, for some measure space $X$. Furthermore one can decompose $M$ as a "direct integral" $$ M=\int_X^\oplus M_x\,dx $$ (sort of a continuous direct sum) of von Neumann algebras indexed by $X$. When $N$ is the center of $M$ ...


1

Use the following statement: If $A$ is a unital $C^*$-algebra and $x\in A$ there are unitaries $u_1,...,u_4$ and complex numbers $a_1,...,a_4$ with $x=\sum_i a_i u_i$. So: $$\varphi(x^*x)=\varphi\left(\sum_i(a_iu_i)^*\sum_j(a_ju_j)\right)= \sum_{ij} \overline{a_i}a_j\varphi(u_i^*u_j)$$ now note that $\varphi(u_i^*u_j)= \varphi\left(u_i(u_i^*u_j)u_i^*\right)...


3

Note that: $$\|(E-E_i)h\|^2 = \|Eh\|^2+\|E_ih\|^2 - \langle Eh, E_ih\rangle - \langle E_ih,Eh\rangle$$ As you have seen, $E_i\to E$ in WOT implies $\|E_ih\|^2\to \|Eh\|^2$, this is the only step that uses $E_i$ and $E$ being projections. From the definition WOT convergence it follows that $\langle Eh, E_ih\rangle\to\langle Eh,Eh\rangle = \|Eh\|^2$ and that $\...


0

So $f$ is a state. Lets first show that it is faithful, meaning that $f(a^*a)>0$ whenever $a\neq0$: Suppose $a\in A$ and $a\neq0$, then there is a state $\varphi$ with $\varphi(a^*a)>0$. Since $f_n$ is weak* dense you have a sub-sequence with $f_{n_k}\to \varphi$ in the weak* topology, but this topology is just so that you can recover $f_{n_k}(a^*a)\...


2

Something that in my opinion is not encouraged enough, is to try basic examples. You can produce a faithful state on $\mathbb C\oplus\mathbb C$ by $\tau(a,b)=\frac{a+b}2$. Then you can easily calculate $$\ker\tau=\{(a,-a):\ a\in\mathbb C\},$$ while $(\ker\tau)^+=\{0\}$. Also, note that linear functionals always have big kernels, as $\dim A/\ker\tau=1$. ...


3

You wrote as if $B$ is the finite-dimensional one, so I'll stick with that. Let $\gamma$ be a C$^*$-norm on $A\otimes M_n(\mathbb C)$ (one certainly exists, because we can represent $A\subset B(H)$ and then $A\otimes M_n(\mathbb C)$ can be represented in $B(H\otimes\mathbb C^n)$). For any $k$, the map $a\longmapsto \gamma(a\otimes E_{kk})$ defines a C$^*$-...


1

I know this a pretty old question, but I would like to give an incomplete answer, in the sense, that I will assume that $\mathcal{H}$ is finite-dim and that we are dealing with the anti-symmetrized Fock space so that everything is finite-dimensional. The generalization should be relatively straightforward, except that we would have to drown in details such ...


3

Recently, I wrote all this out in detail for myself, so here I share my notes with you. Note that the assumption that $X$ is Tychonoff can be ommitted. The construction works for every topological space. The Tychnoff assumption is only there to ensure that the canonical inclusion is injective. Recall that if $A$ is a commutative $C^*$-algebra, then we can ...


3

Consider a special set of characters of $C_b(X)$, for each $x\in X$ define: $$\delta_x: C_b(X)\to\Bbb C, \quad g\mapsto g(x)$$ Since the (non-zero) characters of $C_b(X)$ are the points of $\beta X$ this gives you a way of embedding $X$ into $\beta X$. Now if $f$ is some continuous function on $\beta X$ we may identify it also with an element $\tilde f\in ...


4

In the algebra of all bounded operators on $\ell^2$, consider the operator $a_n$ defined by $$ a_n(\xi ) = \frac 1{\sqrt n}\langle \xi , e_1\rangle e_n, \quad\forall \xi \in \ell^2, $$ where $\{e_n\}_{n\geq 1}$ is the canonical basis. I'll leave it up to you to verify that $$\sum_{n=1}^\infty a_na_n^*$$ converges in norm, but $$\sum_{n=1}^\infty a_n^*...


1

Not all representations of $B(X)$ satisfy the continuity condition mentioned by the OP. To exhibit an example, let us first notice that $B(X)$ is a unital commutative C*-algebra, and hence it is isomorphic to $C(K)$, where $K$ is its spectrum. Given any Borel set $E\subseteq X$, we have that the characteristic function $\chi_E$ is an idempotent element in $...


1

The answer is negative and here is a counter-example. Let us begin with the following: Lemma. If $T$ and $S$ are positive operators on a Hilbert space, with $0\leq T\leq S\leq 1$, and if $\xi $ is any vector such that $T\xi =\xi $, then $S\xi =\xi $, as well. Proof. We have $$ \|\xi \|^2 = \langle \xi , \xi \rangle = \langle T\xi , \xi \rangle \leq \...


4

This is not true. Consider $A=M_2(\mathbb C)$, and $$B=\left\{\begin{bmatrix}b&0\\0&0\end{bmatrix}:b\in\mathbb C\right\}.$$ Note that $$\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}\begin{bmatrix}b&0\\0&0\end{bmatrix}=\begin{bmatrix}a_{11}b&0\\a_{21}b&0\end{bmatrix},$$ so clearly $AB$ cannot be all of $A$.


3

This is done in several places, for instance in chapter 8 in Kadison-Ringrose; it's done in more generality, as they show that any finite von Neumann algebras has a unique central valued trace. The "Murray-von Neumann way" is done in Sunder's An Invitation to von Neumann Algebras. Here is the (very rough) idea. Show that two projections with the ...


1

The functional $\varphi_+$ is positive, so (since $s(\varphi_+)\leq e$ and so $f\,s(\varphi_+)=0$), $$\|\varphi_+\|=\varphi_+(1)=\varphi_+(s(\varphi_+))=\varphi(e\,s(\varphi_+))=\varphi(e\,s(\varphi_+)-f\,s(\varphi_+))=\varphi(s(\varphi_+)).$$ You have that $\|\varphi_+\|\leq\|\varphi_1\|$, $\|\varphi_-\|\leq\|\varphi_2\|$, and $\|\varphi_+\|+\|\varphi_-\|=...


0

Here is an example. Consider the closure of graph $\Gamma$ of $$\sin\left(\frac{1}{|x|-1}\right), \qquad x\in(-1,1)$$ This is a compact space which looks sort of like this: Basically there are two intervals and that are connected via some line. Chose some homoemorphism of the line in the middle with $\Bbb R$ to embed the algebra $C(\overline\Gamma)$ into $...


4

From $u$ positive and $\|u\|\leq1$, you get $0\leq u\leq 1$. Also, being positive, $u$ has a (unique) positive square root. Then $$ u-u^2=u^{1/2}(1-u)u^{1/2}\geq0. $$


1

Without loss of generality, assume $A$ is unital. Then $C^*(u,1)\cong C(\sigma(u)$, the continuous functions on the spectrum of $u$, and this isomorphism takes $u$ to the inclusion function, $\sigma(u)\ni\lambda\mapsto\lambda\in\mathbb C$. The conditions that $u$ is positive and $\|u\|\leq1$ then imply that $0\leq \lambda\leq 1$ for all $\lambda\in\sigma$, ...


1

Given $u$ in $U(A/I)$, choose any $w$ in $A$ such that $\pi (w)=u$. One then has that $$ \pi (w^*w-1)=0, $$ so $w^*w-1\in I$. Given any $\varepsilon >0$, we then have by hypothesis that there exists a projection $p$ in $I$ (chosen e.g. from an approximate identity) such that $$ \varepsilon >\|(w^*w-1) - p(w^*w-1)\| = $$ $$ = \|(1-p)(w^*w-1)\|...


1

Note the second square bracket in $[0,\infty]$. What Takesaki is defining is a tracial weight. Many von Neumann algebras have no tracial states, but do have faithful tracial weights. These precisely those with no type III central summand, and where the identity is infinite (i.e., they have at least a component of type I$_\infty$ of II$_\infty$). The typical ...


2

To see that this inner product is well-defined, see theorem 6.3.1 in Murphy's book "$C^*$-algebras and operator theory". This has a very accessible exposition of the tensor product of Hilbert spaces. All you need for prerequisites is understanding how the algebraic tensor product of vector spaces is made. The $C^*$-algebraic case is much more ...


4

For projections $p,q \in A$ we have $0 \le p,q \le 1$ so $$-1 \le -q \le p-q \le p \le 1$$ which implies $\sigma(p-q) \subseteq [-1,1]$ and therefore $\|p-q\| \le 1$.


1

Consider the following set: $$S_p:=\left\{ J \subseteq P_0(M)\ \middle|\ \sum_{P\in J}P \le p,\quad P\cdot Q = 0 \text{ for all $P,Q$ in $J$} \right\}$$ ie $S_p$ consists of all sets of mutually orthogonal support projections that sum up to something less than $p$ (this sum is to be understood to converge in the ultra-weak topology or in SOT or in whatever ...


2

I believe $C_0(\mathbb R)$ is an example. If $C_0(\mathbb R)$ is an essential ideal in some unital algebra $A$, then $A$ embedds in the multiplier algebra of $C_0(\mathbb R)$, so $A$ must be commutative. The spectrum of $A$ will then be a compact space containing an open dense copy of $\mathbb R$. Since the closure of a connected set is connected, we ...


2

The answer is no. To exhibit a counter-example, consider the space $$ X=\{p\in M_2(\mathbb C): p^*=p=p^2, \ \text{tr}(p)=1\}, $$ with the topology inherited from $M_2(\mathbb C)$. Since orthogonal projections correspond bijectively to their range, $X$ may be seen as the space formed by the one-dimensional subspaces of $\mathbb C^2$, namely the ...


1

$\newcommand{\Asa}{A_{\text{sa}}}\newcommand{\tt}{\tilde \tau }$There are few strong results on positivity that apply to general $^*$-algebras, so let me restrict to C*-algebras here. Among the crucial tools at our disposal is the fact that every self-adjoint element decomposes as a difference os positive elements. Definition. Given a C$^*$-algebra $A$, ...


1

The characters of $C_0(X)$ are the maps $\{ ev_x\mid x\in X\}$ where $ev_x: C_0(X)\to\Bbb C$ is given by $f\mapsto f(x)$. As such the associated semi-definite inner-product $(,)_{ev_x}$ is given by: $$(f,g)_{ev_x}:= ev_x(f^* g) =\overline{f(x)}\cdot g(x)$$ And clearly $C_0(X) /N_{ev_x}\cong\Bbb C$ where $N_{ev_x}$ is the null space of $(,)_{ev_x}$. The ...


1

If $p=u^*u$ and $q=uu^*$, then $$u=uu^*u=qu=pqu=pu=pup\in pMp.$$ Since $p$ is abelian, it now follows that $$p=u^*u=uu^*=q.$$


3

Here's a simple example. Let $R=\mathbb{Z}[x,y]/(xy,y^2)$ and let $f:R\to R$ be given by $f(x)=x+y$ and $f(y)=y$. This restricts to the identity on the ideal $I=(y)$ and induces the identity $R/I\to R/I$, but is not the identity. One simple obstruction is that if a (left) ideal $I\subseteq R$ contains an element $a$ that is not a (right) zero divisor, then ...


2

Besides the non-zero eigenvalues of $T$, let us make sure to include $\lambda_0=0$, and let $P_0$ be the orthonormal projection onto the null space of $T$. Then clearly $$ T = \sum_{i=1}^{\infty}\lambda_i P_i = \sum_{i=0}^{\infty}\lambda_i P_i, $$ and also $$ H = \bigoplus_{i=0}^{\infty}P_i(H). $$ Notice that an operator $S$ commutes with $T$ iff $S$...


0

Here is a proof which works more generally for $x,y\ge 0$ in a $C^*$-algebra, where the spectral radius satisfies $r(z)\le \|z\|=\sqrt{\|z^*z\|}$ for every element $z$, and $r(t)=\|t\|$ for every normal element $t$. The main difference with @user1551's argument is that we will use the invertibility of $y$. Other than that, the idea is essentially the same. ...


1

Yes, the book is wrong. It was supposed to be $ub \in A$ for all $b \in B$. See also Farah's book Corollary 1.6.13.


1

We also have that $\omega_n$ is faithful on $eAe$ by Lemma 3.6, and that $e_0=ee_0e\in eAe$. Since $e_0$ is nonzero, it then follows that $\omega_n(e_0)>0$. However, $$0=\omega(e_0)\geq\omega_n(e_0)>0,$$ a contradiction.


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