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4

The Boolean algebra of connected components is equivalent to the projections (the elements with $p^2=p$) in the algebra of functions, with multiplication of functions representing intersection and $(p_1,p_2) \to p_1 + p_2 - p_1p_2$ being the union of sets of components. I think there is a version of algebraic K-theory for topological algebras whose value ...


3

Because of the subspaces being selfadjoint, $B(H)V_1\subset V_2$ implies that $V_1B(H)\subset V_2$. If $V_1\ne0$, then $B(H)V_1B(H)\subset V_3$ contains all finite-rank operators, and thus $V_3$, being closed, contains the compact operators. If $V_3$ contains a non-compact operator, then the ideas in this answer show that $I\in V_7$ (I didn't count ...


2

Let $E=\{1,1/2,1/3,\cdots\}$ and let $\mu_{a}$ be the finite atomic meausre on $[0,1]$ that is supported on $E$ with $\mu\{1/n\}=1/n^{2}$. Let $\mu = \mu_{a}+m$ where $m$ is Lebesgue measure on $[0,1]$. Let $X=L^{2}_{\mu}[0,1]$. Let $\chi$ be the characteristic function of $E$, and define operators $A, B \in \mathcal{L}(X)$ by $$ Af = ...


1

Let $X$ be a compact Hausdorff space with more than one point. Then $C(X)$ is a natural, normal uniform algebra on $X$. This means that the character space of $C(X)$ is exactly $X$ and for each closed set $E\subseteq X$ and each closed set $F\subseteq X\setminus E$, there exists a function $f\in C(X)$ such that $f(y)=0$ for all $y\in E$ and $f(y)=1$ for each ...


1

Yes and you don't even have to take the closure. For instance if $A$ is a non-simple C*-algebra with trivial cetnre that has unique (faithful) trace (for instance $A=C^*(G)$ for a sufficiently non-commutative amenable group such as the group of permutations of integers that move at most finitely many entries), then $A\otimes \mathcal{Z}$, where $\mathcal{Z}$ ...


1

Kaplansky's density theorem lets you choose $a $ with $\|a\|\leq\|u\|$.


1

Since you have to relate the norm with the spectrum, I don't think this has an elementary algebraic proof. Facts needed (from the theory of Banach algebras): For any $A\in\mathcal S$, $\sigma(A)=\{\phi(A):\ \phi\in S(\mathcal A)\}$, where $S(\mathcal A)$ is the state space. For any $A\in\mathcal S$, $\|A\|=\text{spr}(A)=\max\{|\lambda|:\ ...


1

It is correct for positive operators, for more information see Gohberg, Krein, Introduction to the Theoryof Linear Non-Self-Adjoint Operators in Hilbert Space page 27


1

Let $L$ denote the matrix $$ L = \pmatrix{u \mathrm{I} + i S_3 && i S_- \\ i S_+ && u \mathrm{I} - i S_3} $$ My best guess is that whatever the author is getting at has something to do with the fact that $$ \operatorname{tr}(L^N) = 2I\,u^N + q_{2}u^{N-2} + \cdots + q_N $$ or, if $N$ is odd, $$ \operatorname{tr}(L^N) = 2I\,u^N + q_{2}u^{N-2} + ...



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