# Tag Info

3

No. For instance, you can take the representation $\pi\oplus\pi$ on $H\oplus H$, and then $(\pi\oplus\pi)(x)$ will have twice the index of $\pi(x)$.

3

You can write the similarity as $NS=SM$. As $N$ and $M$ are normal, the Fuglede-Putnam theorem guarantees that $N^*S=SM^*$. Taking adjoints, $S^*N=MS^*$. Then $$S^*SM=S^*NS=MS^*S.$$ Using this identity repeteadly, $p (S^*S)M=Mp (S^*S )$ for all polynomials; taking limits, $f (S^*S)M=Mf (S^*S)$ for all continuous functions $f$. In particular, if ...

3

For an example of a non-proper $*$-morphism that isn't trivial, consider $A=B=\mathbb C\oplus \mathbb C$, and $\phi(a,b)=(a,0)$. For the existence of $\varphi_*$, the key property is that for any $*$-morphism $\varphi:C_0(X)\to C_0(Y)$, there exists $\varphi_*:Y\to X$, continuous, such that $$\tag{1}\varphi(f)(y)=f(\varphi_*(y))\ \ \ \ \text{ for all }y\in ... 3 Let A \subseteq B(H) be any AW*-algebra that is not a von Neumann algebra. (Actually, we don't need a full AW*-algebra, see below.) Let t \in A be any operator. By definition of AW*-algebra, every right-annihilator is generated by a projection. In particular, the right-annihilator of the singleton set \{t\} is generated by a projection q \in A. ... 2 You're right; C_0(X) is a C^*-algebra for any topological space X. However, for any X, we can canonically associate to it another space Y which is locally compact Hausdorff such that C_0(X)\cong C_0(Y), so in some sense we lose no generality by restricting X to be locally compact Hausdorff (and most interesting examples are). To construct ... 2 Here's one possible take on the question: K-theory for C^*-algebras is motivated by topological K-theory. Topological K_0 for a compact Hausdorff space X is the Grothendieck group of formal differences of isomorphism classes of locally trivial vector bundles over X, so we want the operator K_0-group of C(X) to match this. The Serre-Swan ... 2 Your two ideas would have been my first attempts; but I have no idea how to make them work (well, for the first one, I would try with T the flip, but I still wouldn't know how to do it). Let K\subset B(\ell^2) be the compact operators. On \overline {B(\ell^2)\odot B(\ell^2)}, consider the ideals \overline{K\odot B(\ell^2)} and \overline{K\odot ... 2 Define \phi:(B+I)/I\to B/B\cap I by$$\phi(b+j+I)=b+B\cap I,\ \ \ \ \ b\in B,\ j\in I.$$Of course we need to check that this is well-defined. If b_1+j_1=b_2+j_2, then$$ b_1-b_2=j_2-j_1\in B\cap I, $$so b_1+B\cap I=b_2+B\cap I. The map is obviously linear, multiplicative, *-preserving, and onto. As for injectivity, if b_1+B\cap I=b_2+B\cap I, ... 2 Given f and \epsilon, choose a polynomial p with \Vert f-p\Vert_{\infty,X}<\epsilon (where \Vert\cdot\Vert_{\infty,X} is the supremum norm oin X). Now see the corresponding polynomial function in \mathcal{A}, p:\mathcal{A}\to\mathcal{A}. (Remember: the functional calculus respects this notation, i.e., p(a), in the functional calculus, is ... 2 Note first that we may assume that all T_k are proper isometries; because if one of them, say T_1, is a unitary, we get \sum_{k=2}^NT_kT_k^*=0, which forces T_k=0 for all k\geq2. Since \sigma(T_k^*T_k)=\sigma(I)=\{1\}, using that \sigma(AB)\cup\{0\}=\sigma(BA)\cup\{0\} we deduce that \sigma(T_kT_k^*)=\{0,1\} (the zero has to be there ... 1 I will assume that the norm \|AB\| in the definition of \|A\|_{L(X)} is \|AB\|_{L(\mathbb C^n)}. The two norms are equal: you have, by definition,$$ \|A\|_{L(X)}\leq \|A\|_{L(\mathbb C^n)}, $$since \|AB\|_{L(\mathbb C^n)}\leq \|A\|_{L(\mathbb C^n)}\,\|B\|_{L(\mathbb C^n)}=\|A\|_{L(\mathbb C^n)}. Conversely,$$ ...

1

Consider $M_2(\mathbb C)$. As vectorspace it is isomorphic to $\mathbb C ^4$. But $K_0(M_2(\mathbb C)) = K_0(\mathbb C) = \mathbb Z \neq \mathbb Z^4 = K_0(\mathbb C)^4.$

1

ou have shown that the sequence of positive operators $$(b+1/n)^{1/2}-(a+1/n)^{1/2}$$ converges to $b^{1/2}-a^{1/2}$. So now all you have to show is that a limit of positives is positive. The two ways that come to mind are: by representing $A\subset B (H)$ (and then using wot convergence); or, if you want to stay abstract, by looking at states. Edit: ...

1

This is a corollary of the fact that images of $\ast$-homomorphism of $C^\ast$-algebras are closed. More precisely, denote by $\pi$ the canonical projection $A\to A/I$. Then $\pi(B)\subset A/I$ is closed and so is $B+I=\pi^{-1}(\pi(B))\subset A$ as the preimage of a closed set under a continuous map.

1

You're on the right track: you want to look at indicator functions. If $f$ is a function such that $f^2=f$, can you prove it must be an indicator function? Now of course, you're right that most indicator functions aren't continuous. Can you describe the indicator functions which are? And of those, which vanish at infinity? Answers to these questions can ...

1

It's true that any finite number of maps at the beginning don't affect the direct limit, so you may ignore the first two. Then the direct limit is the stabilization of the image of your $3\times 3$-matrix. Explicitly, if you call $A$ this matrix, then the image of $A^n$ stabilizes for $n$ big enough, and the limit is this image. Now just looking at its ...

1

You are right: When taking a direct limit like $$A_1\overset{T_1}{\longrightarrow}A_2\overset{T_2}{\longrightarrow}\cdots,$$ the first elements of your direct sequence do not matter. In fact, you can take any subsequence $(n_k)$ (and compose the maps appropriatelly to get a directed sequence $A_{n_1}\to A_{n_2}\to\cdots$) and you still get the same direct ...

1

There is something off with that argument as it is written. Nothing prevents, for instance, that $x-\Phi_i(x_i')\leq0$ and nonzero. In that case, $$f(x-\Phi_i(x_i'))=0,$$ and then $$0<\|x-\Phi_i(x_i)\|,\ \ \ \ \|f(x-\Phi_i(x_i'))\|=0.$$ This is how I think that argument can be saved. If $x\geq0$, $y=y^*$, and $\|x-y\|<\varepsilon$, then ...

1

In your attempt to prove that $\phi$ is pure you are mixing the definition you gave ($\psi\leq\phi$ implies $\psi=t\phi$) with being an extreme point (the two things are equivalent, but I think you need to make up your mind). For the converse, you need to show that any measure that has support with at least two points does not satisfy the definition you ...

1

I think you make it unncessarily complicated. By the minimality of $E$, you have $P_{\lambda_i}=P_{\lambda_j}$ for all $i,j$. But each $\lambda_i$ is the particular projection corresponding to $\lambda_i$; so $\lambda_i=\lambda_j$ for all $i,j$ (projections corresponding to different eigenvalues are orthogonal to each other). Thus $T=\lambda_1\,E$.

1

Your isometries are not such. Consider $$\rho_1=\begin{bmatrix}1/2&0\\0&1/2\end{bmatrix},\ \ \rho_2=\begin{bmatrix}1&0\\0&0\end{bmatrix}.$$ Then $$\|\rho_1\|=1/2,\ \ \|\rho_2\|=1,$$ while $$\|\rho_1\|_1=\|\rho_2\|_1=1.\ \$$ For the proof, if $\rho=\sum_k\lambda_kP_k$ with $\sum_k\lambda_k=1$ and $0\leq\lambda_k$ for all $k$ and at ...

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