# Tag Info

### 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 $S_2 = \{\lambda U \mid \lambda \in \mathbb{C}\}$ ...
• 1,446
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 ...
• 183k
1 vote
Accepted

### 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 ...
• 183k
1 vote
Accepted

### 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 $$E(pz)\leq\frac12.$$
• 183k
1 vote

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 $$\rho(g)u=u*\delta_{g^{-1}}.... • 22.2k 0 votes ### Center of a von Neumann algebra is properly contained in the maximal subalgebra a) For any a\in Z(A), b\in B you have ab=ba by definition of center. Then W^*(B,Z(A)) is an abelian von Neumann subalgebra of A that contains B. As B is maximal, W^*(B,Z(A))\subset B ... • 183k 0 votes ### Unitarily equivariant, linear, Hermitian maps on matrix algebras With respect, Martin's answer works too hard. The second condition says that L is an endomorphism of M_n(\mathbb{C}) as a complex representation of the unitary group U(n). So of course we should ... • 368k 2 votes Accepted ### Not-normal state on von Neumann algebra and finding a corner on which it is still faithful? As s.harp mentioned, the problem is that q can be the identity. For example, take \mathcal M=B(H) and \varphi any non-normal state that is zero on K(H) (these are common once you have the ... • 183k 1 vote Accepted ### Is the stabilization of a simple C^*-algebra simple? Let A,B be C^*-algebras. By a result of Takesaki, A\otimes B (minimal tensor product) is simple iff A and B are simple (reference: chapter IV in Takesaki's first book). Since the algebra of ... • 20.1k 1 vote Accepted ### Irreducible representations of C^* algebras and Commutants What you say is correct, but you don't even need to talk about von Neumann algebras. For any C^*-algebra A, you have$$ A=\overline{\operatorname{span}}\{a\in A:\ 0\leq a\leq 1\}. $$This a ... • 183k 4 votes Accepted ### C^*-algebra generated by the set Let's think about it for a moment. A C^*-subalgebra has to be closed under the algebra operations (sum, product and scalar multiplication), under conjugation, and since it has to be a Banach space ... • 33.3k 3 votes Accepted ### Von Neumann algebras are C^*-algebras A von Neumann algebra M lives in some B(H). The C^* relation$$\|T\|^2=\|T^*T\|$$holds in B(H). So all you need to check is that M is norm closed; and this comes for free since M is ... • 183k 2 votes Accepted ### Estimation for operators$$\|B\|^2A^*A-A^*B^*BA=A^*(\|B\|^2I-B^*B)A\ge 0$$• 7,336 1 vote ### Show that T^2\le 1 whenever 0\le T\le 1 in a Hilbert space Since you point out that 0\leq a\leq b doesn't necessarily imply a^2\leq b^2, it's worth mentioning that in an arbitrary unital C^*-algebra A, if a\in A and 0\leq a\leq 1, then a ... • 22.2k 5 votes Accepted ### Show that T^2\le 1 whenever 0\le T\le 1 in a Hilbert space There is another explanation which does not make use of the square root of T. The Cauchy-Schwarz inequality gives$$|\langle Tx,y\rangle |\le\langle Tx,x\rangle^{1/2}\langle Ty,y\rangle^{1/2}\le \|x\...
• 7,336

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