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3 votes

The extension of injective map is also injective?

The answer to both is no. Take $S_1 = \{\lambda S\ \mid \lambda \in \mathbb{C}\}$ where $S \in B(\ell^2(\mathbb{N}))$ is the unilateral shift and $S_2 = \{\lambda U \mid \lambda \in \mathbb{C}\}$ ...
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1 vote

$C^*$-ideals in tensor products

It is always non-trivial. Since the ideal $J$ is a closed subspace of $A$, by Hahn-Banach we can find $\varphi\in A^*$ with $\|\varphi\|=1$, $\varphi|_J=0$ and $\varphi(a)=1$. Fix $k_0\in K(H)$ and ...
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1 vote
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Trace in a finite dimensional $C^*$-Algebra

You are missing a crucial part of the statement from the book, which is that $r$ is rational. If $r$ is irrational, then the embedding you are looking for does not exist; the reason is that a matrix ...
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1 vote
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Want to show that $pz$ is a projection in a $C^*$-algebra with $q \le \frac{1}2$

Since $z$ is in the centre, $$ E(pz)=E(p)z=u\,1_{(0,\frac12]}(u). $$ Since $t\,1{(0,\frac12]}(t)\leq\frac12$ for all nonnegative $t$, functional calculus gives you $$ E(pz)\leq\frac12. $$
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1 vote

Relation between right regular representation and convolution of group algebras

The issue is in your formula for the right regular representation in terms of the convolution product: You state that $$\rho(g)u=u*\delta_g.$$ But the correct formula is $$\rho(g)u=u*\delta_{g^{-1}}.$...
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  • 22.2k
0 votes

Center of a von Neumann algebra is properly contained in the maximal subalgebra

a) For any $a\in Z(A)$, $b\in B$ you have $ab=ba$ by definition of center. Then $W^*(B,Z(A))$ is an abelian von Neumann subalgebra of $A$ that contains $B$. As $B$ is maximal, $W^*(B,Z(A))\subset B$ ...
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0 votes

Unitarily equivariant, linear, Hermitian maps on matrix algebras

With respect, Martin's answer works too hard. The second condition says that $L$ is an endomorphism of $M_n(\mathbb{C})$ as a complex representation of the unitary group $U(n)$. So of course we should ...
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2 votes
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Not-normal state on von Neumann algebra and finding a corner on which it is still faithful?

As s.harp mentioned, the problem is that $q$ can be the identity. For example, take $\mathcal M=B(H)$ and $\varphi$ any non-normal state that is zero on $K(H)$ (these are common once you have the ...
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1 vote
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Is the stabilization of a simple $C^*$-algebra simple?

Let $A,B$ be $C^*$-algebras. By a result of Takesaki, $A\otimes B$ (minimal tensor product) is simple iff $A$ and $B$ are simple (reference: chapter IV in Takesaki's first book). Since the algebra of ...
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1 vote
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Irreducible representations of $C^*$ algebras and Commutants

What you say is correct, but you don't even need to talk about von Neumann algebras. For any C$^*$-algebra $A$, you have $$ A=\overline{\operatorname{span}}\{a\in A:\ 0\leq a\leq 1\}. $$ This a ...
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4 votes
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$C^*$-algebra generated by the set

Let's think about it for a moment. A $C^*$-subalgebra has to be closed under the algebra operations (sum, product and scalar multiplication), under conjugation, and since it has to be a Banach space ...
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3 votes
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Von Neumann algebras are $C^*$-algebras

A von Neumann algebra $M$ lives in some $B(H)$. The C$^*$ relation $$\|T\|^2=\|T^*T\|$$ holds in $B(H)$. So all you need to check is that $M$ is norm closed; and this comes for free since $M$ is ...
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2 votes
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Estimation for operators

$$\|B\|^2A^*A-A^*B^*BA=A^*(\|B\|^2I-B^*B)A\ge 0$$
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1 vote

Show that $T^2\le 1$ whenever $0\le T\le 1$ in a Hilbert space

Since you point out that $0\leq a\leq b$ doesn't necessarily imply $a^2\leq b^2$, it's worth mentioning that in an arbitrary unital $C^*$-algebra $A$, if $a\in A$ and $0\leq a\leq 1$, then a ...
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5 votes
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Show that $T^2\le 1$ whenever $0\le T\le 1$ in a Hilbert space

There is another explanation which does not make use of the square root of $T.$ The Cauchy-Schwarz inequality gives $$|\langle Tx,y\rangle |\le\langle Tx,x\rangle^{1/2}\langle Ty,y\rangle^{1/2}\le \|x\...
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