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6

One can prove that $p,q$ are Murray -von-Neumann equivalent if they fulfill $$ \|aa^*-p\|<\frac 12,\quad \|a^*a-q\|<\frac 12. $$ The proof is standard (for example it is outlined in the comment of @ougao), I was just lucky to find $\|(1/2-p)^{-1}\|=2$. We first assume that the $C^*$-algebra $A$ is unital. First note that $(1-2p)^2=1$, hence ...


3

I had forgotten about this question until Jonas Meyer put a bounty on it. In hindsight, I think I now know how to solve parts of it... I'll write down what I've got and maybe someone can give a complete argument building on it. I'll address the case $1 \in A$ first since it is somewhat simpler. The argument does not seem to be overly difficult, but it is a ...


3

Inspired by George Lowther's solution, here is another more elementary example of a (nonunital) C*-algebra $A$ such that $A$ has no nonzero projections, but $M_2(A)$ does have nonzero projections. Let $D$ be the sub algebra of $M_2(\mathbb{C})$ consisting of scalar multiples of the identity. Let $C$ be the sub algebra of $M_2(\mathbb{C})$ consisting of ...


3

The answer to question 2 is negative. It is possible that there are projections in $M_\infty(A)$ which are not equivalent to any projection in $A$, even when $A$ is a $C^*$-algebra (so, representable) and spacious. To do this, I will construct a (non-unitial) $C^*$-algebra which has no non-zero projections, so is trivially spacious. However, $M_2(A)$ will ...


3

Yes, $$ \mathbb C\,I=\{\lambda\,I:\ \lambda\in\mathbb C\}. $$ It is coherent with the notation $$ AB=\{ab:\ a\in A,\ b\in B\}, $$ and with $$ A+B=\{a+b:\ a\in A,\ b\in B\}, $$ etc.


3

Oh well, how silly of me not to have thought of the following argument sooner. In what follows, $ (e_{i})_{i \in I} $ is a self-adjoint approximate identity of $ A $ that is bounded in norm by $ 1 $. It is $ C^{*} $-folklore that such an approximate identity exists. Lemma. If $ (x_{i})_{i \in I} $ is a convergent net in $ A $ whose limit is $ x $, ...


3

Let me list some facts about the algebra $A=C^\ast(\mathbb{Z}_2\ast\mathbb{Z}_2)$. Denote by $s,t\in\mathbb{Z}_2\ast\mathbb{Z}_2$ the canonical generators. $\bullet$ $A$ is the universal algebra generated by two reflections: if $S,T\in B(H)$ are two reflections, i.e. $S=S^*,T=T^*,S^2=I,T^2=I,$ then there exists a unique homomorphism from $A$ onto the ...


2

Yes, a morphism in this category is epic iff it is surjective. Obviously surjective morphisms are epic, so it suffices to show that if $f: X\to Y$ is not surjective, there is a locally compact Hausdorff space $Z$ and distinct $g_1, g_2 \in \hom(Y,Z)$ such that $g_1 \circ f = g_2 \circ f$. I will use the notion of a perfect map as defined in Engelking's ...


2

Here is my own attempt on the problem, which turns out to be essentially San’s argument above (I only realized this when writing my response). I would therefore like to focus on demystifying the continuous functional calculus that takes place in the argument and on making portions of it more transparent. All credit is due to San, of course, for posting the ...


1

If $x\geq0$, let $f$ be a state on $B(H_2)$; then the functional $f\circ\phi$ is unital and contractive. It is well-known that a unital contractive functional is positive (I know of proofs in Paulsen and Davidson's books). So $f(\phi(x))\geq0$ for all states on $B(H_2)$, which means that $\phi(x)\geq0$. Finally, this can be done for each amplification of ...


1

The following categories are contravariantly equivalent: locally compact Hausdorff spaces with proper continuous maps commutative C$^*$-algebras with non-degenerate $*$-homomorphisms Here, a $*$-homomorphism $f : A \to B$ is non-degenerate if the following equivalent conditions are satisfied: The ideal generated by the set-theoretic image of $f$ is ...


1

The Axiom Schema of Separation tells us that if $ A $ is a set and $ \varphi $ is a formula with a free variable, then the collection $$ \{ x \mid x \in A ~ \text{and} ~ \varphi[x] \} $$ is a set. Now, fix both $ (G,A,\alpha) $ and $ f \in {C_{c}}(G,A) $, and let $ \varphi[r] $ denote the sentence There exists a covariant representation $ (\pi,U) $ of $ ...


1

The whole point of the multiplier algebra is that you should be able to multiply. So $\varphi(a)b$ has to make sense. If you look at page 39 in Murphy's book, you see that the definition of $M(I)$ includes a canonical identification of $I$ as an ideal on $M(I)$. As Phoenix mentioned, $\varphi$ extends the inclusion $I\to M(I)$. So ...


1

The answer is no. By Takesaki I, Theorem V.1.41, the von Neumann algebra generated by P and Q is the direct sum of an abelian part and a type $I_2$ part, in which the projections P and Q have the form $$ P = \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix} , \quad Q = \begin{pmatrix} c^2 & cs \\ cs & s^2 \\ \end{pmatrix}, $$ where $c$ and ...


1

Group algebras are not my thing, but here is an attempt. You have, according to Wikipedia, that $\mathbb Z_2*\mathbb Z_2\simeq \mathbb Z\rtimes\mathbb Z_2$, so $$ C(\mathbb T)\times_\alpha\mathbb Z_2\simeq C^*(\mathbb Z\rtimes\mathbb Z_2)\simeq C^*(\mathbb Z_2*\mathbb Z_2). $$ The first isomorphism holds due to the fact that $\mathbb Z\rtimes\mathbb Z_2$ ...


1

Let $\mathcal{A}$ be unital and $\omega$ be a state. Then $$ 0 \leq \omega(|a - \omega(a){\bf 1}_\mathcal{A}|^2) = \omega(a^*a - a\omega(a^*) - a^* \omega(a) + |\omega(a)|^2{\bf 1}_\mathcal{A}) = \omega(a^*a) - |\omega(a)|^2. $$


1

If $A$ is non-unital, then by definition $$ \sigma_A(a) = \sigma_{A_1}(a) $$ where $A_1$ is the unitization of $A$, and necessarily one has that $0\in \sigma_A(a)$. One can then define $$ I = \{f \in C(\sigma_A(a)) : f(0) = 0\} $$ and obtain a homomorphism $\varphi : I \to A$ as you have mentioned. However, $I \neq C_0(\sigma_A(a))$ in the sense of ...


1

By continuity of $p, q :X \rightarrow M_n \mathbb(C)$, and (ii), and the total disconnectedess of $X$, find a partition of X into clopen sets $X_1,\cdots, X_k$ and complex matrix $v_1,\cdots,v_k$ such that $\|v_i^*v_i - p(x)\|<1$ and $\|v_iv_i^* - q(x)\|<1$ for all $x \in X_i$. Now, define the map $f: X \rightarrow M_n (C)$, $f(x) = v_i$, $x \in ...


1

As you say, an ultrapower of $R $ contains $L (\mathbb F_2) $, which is not hyperfinite, while every subfactor of the hyperfinite II $_1$ is hyperfinite. Another way to see it is that $R $ is separable, while its ultrapowers aren't.



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