# Tag Info

3

You cannot define type I von Neumann algebras in terms of minimal projections, since many don't have them; $L^\infty[0,1]$, for example. What you probably saw is that some texts only care about factors, and in a factor the abelian projections are precisely the minimal ones. Rank one projections (not "operators") are an example of a minimal projection. But ...

3

If $A$ is any unital C*-algebra, the hypotheses are satisfied with $C=A$ and $B=\mathbb C1$.

2

Let's first check that $P^\perp$ has invariant range. Let $a \in A$ and let $\eta \in [A\xi_1]^\perp = (A\xi_1)^\perp$. Then, for any $b \in A$, $$\langle a\eta, b\xi_1 \rangle = \langle \eta, a^\ast b \xi_1 \rangle = 0,$$ where $a^\ast b \in A$ and hence $a^\ast b \xi_1 \in A\xi_1$ precisely because $A$ is a self-adjoint algebra of operators in $B(H)$, ...

2

1: On a quick look, I don't immediately see how to show that $\psi$ is contractive in general. But note that when $A$ is unital so is $\psi$, which together with positivity makes it contractive. I think this idea can be extended to the non-unital case. 2: When you prove that $\psi$ is positive, you don't need to use that $A$ is $A$, just that it is a ...

2

The usual definition of a $W^*$-algebra is a weak operator closed $*$-subalgebra of $B(H)$, that it's equivalent to the one you gave is a theorem of Sakai. More directly, the predual to $B(H)$ is the space of trace class operators.

2

Let $M_1=M_2=M_2(\mathbb C)$, and $M_3=\mathbb C\,I_2$. Then $$vN(M_1,M_2)\cap M_3=\mathbb C\,I_2,$$ $$vN(M_1,M_3)\cap vN(M_2,M_3)=M_1\cap M_2=M_2(\mathbb C).$$

2

If you have a bounded operator $A$, then the holomorphic functional calculus is always an option, and it is based on Cauchy's integral representation: $$f(A) = \frac{1}{2\pi i} \oint_{C} f(\lambda)\frac{1}{\lambda I-A}"\,d\lambda = \frac{1}{2\pi i} \oint_{C} f(\lambda)(\lambda I-A)^{-1}\,d\lambda.$$ The contour $C$ is any simple ...

2

Hint as rhetorical question: If it's isometric, and $\hat{x} = 0$, what does that tell you about $\lVert x\rVert$?

1

Injectivity of the Gelfand transform is equivalent to the assertion that characters separate points. This can be verified by using that in the commutative case characters are precisely the pure states and so they have to separate points since the states do.

1

This looks like an "associativity" property. Both algebras live in $B(H_B\otimes H_A\otimes \ell^2(\Gamma))$, so the topology is the same. At the pre-closure level, you should convince yourself that $C_c(\Gamma,B\odot A)$ and $B\odot C_c(\Gamma,A)$ are equal, and that their closures give the two algebras you want to consider.

1

We can say that $\sigma(p) \subseteq \{0,1\}$. $p - p^2 = 0$, and hence, by the spectral theorem $$\{0\} = \sigma(0) = \{\lambda - \lambda^2 \mid \lambda \in \sigma(p) \}$$

1

Martini's answer is the canonical one. But in this case one can check things explicitly: We can assume that $A$ is unital, as we can always work on the unitization (we actually need it to define the spectrum). If $p=I$ of $p=0$, it is straightforward to check that the spectra are respectively $\{1\}$ and $\{0\}$. For non-trivial $p$: If $\lambda=0$, $p$ ...

1

The fact that the range of $P$ is invariant under $A$ can be written as $PaP=aP$ for all $a\in A$. Now fix $a\in A$; since $a^*\in A$, $Pa^*P=a^*P$. Take adjoints and you get $PaP=Pa$. So we have shown that $Pa=aP$ for all $a\in A$.

1

I think it is simply this fact: take $x\in\ker\pi$. From the relation in the first paragraph of the proof, $$x=\lim_n\Psi_n(\pi(x))=\lim_n\Psi(0)=0.$$ So $\pi$ is an isometric epimorphism, and so an isomorphism.

1

Your proof shows that $||A^*A|| \leq ||A||^2$. You still have to show $||A||^2 \leq ||A^*A||$. Now $sup_{||x||=1}||(Ax,Ax)|| \leq ||A^*A||$ by the first equality in your post. The result thus follows.

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