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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

If you have a linear map $F:V\to V((x))$, then the coefficient of $x^i$ in $F(v)$ depends linearly on $v$. Hence you can write $F(v)=\sum_{i\in\mathbb Z} f_i(v)x^i$ where each $f_i:V\to V$ is linear. (In addition, you know that for each $v\in V$, there is some $i_0$ such that $f_i(v)=0$ for $i<i_0$, but this does not play a role here.) Now you simply ...


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

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

You can write $(\Bbb C^n, \lVert\cdot\rVert_p)=L^p(\Bbb Z/n\Bbb Z) = L^p(G)$, the inner group is compact, hence when $p\ge 1$ you have a Banach space structure and the usual convolution on functions over a compact group give you an algebra structure, given as $$(f\star g)(x)=\int_G f(y)g(x-y)\,dy.$$ Here "-" means the subtraction in the group structure ...


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

This is not true. Let $$ T=\begin{bmatrix}1&2\\2&4\end{bmatrix},\ \ P=\begin{bmatrix}1&0\\0&0\end{bmatrix}. $$ Then $T$ is positive (selfadjoint, with eigenvalues $0$ and $5$), but $$ T-PTP=\begin{bmatrix}0&2\\2&4\end{bmatrix} $$ is not positive (selfadjoint with eigenvalues $2\pm2\sqrt2$).


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

Question (1) seems to be missing from your post. For the identification (2), the unitary $U : \ell^2(\Gamma \times \Gamma) \to \ell^2(\Gamma) \otimes \ell^2(\Gamma)$ defined by $\delta_{(g,h)} \mapsto \delta_g \otimes \delta_h$ intertwines $\lambda'$ and $\lambda \otimes \lambda$. Let $F : H \otimes \ell^2(\Gamma) \to \ell^2(\Gamma) \otimes H$ be the flip ...


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 ...


2

I don't know why you say that $f(\sigma(x))=F_1$. A point in $\sigma(x)$ is either in $F_1$ or in $F_2$, and so $f(x)$ is either $0$ or $1$; and then $f(\sigma(x))=\{0,1\}$.


1

If $\|\varphi(A)\|\leq c\,\|A\|$ when $A$ is selfadjoint, then for arbitrary $A$ you have $$ \|\varphi(A)\|=\|\varphi(\text{Re}\,A)+i\varphi(\text{Im}\,A)\| \leq\|\varphi(\text{Re}\,A)\|+\|\varphi(\text{Im}\,A)\|\\ \leq c\,(\|\text{Re}\,A\|+\|\text{Im}\,A)\|\leq 2c\|A\|, $$ since $$ \|\text{Re}A\|=\frac12\,\|A+A^*\|\leq\frac12\,(\|A\|+\|A^*\|)=\|A\|. $$


1

I post here the answer I received for the same question in mathoverflow: Here I list some facts that may be useful for building your intuition: 1. Two commutative Morita equivalent $C^*$-algebra are in fact $*$-isomorphic. 2 If $A$ is $C^*$-algebra and you take $B=M_n(A)$ then $A$ and $B$ are Morita equivalent. 3 Many invariants for $C^*$-algebras such as ...


1

The inequality does not hold in general. Let $$ T=\begin{bmatrix}1&1\\0&0\end{bmatrix},\ \ Q_1=\begin{bmatrix}0&0\\0&1\end{bmatrix}, \ \ Q_2=\begin{bmatrix}1&0\\0&0\end{bmatrix}. $$ Note that $$ ...


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

They are not equivalent on an infinite-dimensional Hilbert space. The weak convergence for operators in this case is in the usually called "weak operator topology": $$ A_n\xrightarrow{wot} A\ \ \iff\ \ \langle A_nx,y\rangle\to\langle Ax,y\rangle,\ \ \forall x,y\in H. $$ The weak operator topology is known to be coarser than the $\sigma$-weak operator ...



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