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2

I don't know about "geometrical view", but the condition essentially follows from assuming that multiplication is continuous. Note the "essentially"; we'll see below what I mean by that. Say we have an algebra with a norm, and multiplication is (jointly) continuous. Continuity at $(0,0)$ shows that there exists $\delta>0$ so that $$||xy||\le1\quad(||x||,|...


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I could solve the problems: We show that $\pi$ is open onto its image $\pi(S)$. Let $\{s\} \subseteq S$, then we have $\{\delta_s\} = \pi(s) \cap (\operatorname{ev}_{1_{\{s\}}})^{-1}(\mathbb C \setminus \{0\})$, where $\operatorname{ev}_{1_{\{s\}}}(\varphi) = \varphi(1_{\{s\}})$ for all $\varphi \in G_S$. Hence $\{\delta_s\}$ is open in $\pi(S)$ and $\pi(S)...


2

Any $*$-homomorphism between C$^*$-algebras is contractive. This is standard (i.e., it appears in every book on the subject) and is due to three things: The C $^*$-identity $\|a\|^2=\|a^*a\|$, which reduces the problem to norms of positives; The equality $\|a\|=\text {spr}\, (a) $ for $a $ positive; The fact that a $*$-homomorphism reduces the spectral ...


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The only example is $\Bbb C$. Say $A$ is a Banach algebra with the given property and say the identity is $e$. First suppose that $a$ is invertible and $\lambda\in\sigma(a)$, the spectrum of $a$. Then $|\lambda|\le||a||$. And $\lambda^{-1}\in\sigma(a^{-1})$, so $|\lambda^{-1}|\le||a^{-1}||=||a||^{-1}$. So $|\lambda|=||a||$. Now if $a$ is any element of $A$,...


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I recently wrote up a solution to this very problem, which I've copied below. Note that in this context, $\mathfrak A$ is the space of bounded operators on $B$. I hope you find this helpful. $ \newcommand{\f}{\mathfrak} \DeclareMathOperator{\rad}{r} \newcommand{\eps}{\varepsilon} \DeclareMathOperator{\spec}{spec} \newcommand{\lp}{\left(} \newcommand{\rp}{\...


3

If your example $\mathcal{B}$ were a real Banach algebra instead of a complex Banach algebra, then you would be right that there are four connected components, since $\mathcal{B}$ can be identified with $\mathbb{R}^2$ and the invertible elements split into four quadrants. But over $\mathbb{C}$, you have the complement in $\mathbb{C}^2$ of two (complex) one-...


1

Well, $$||x||^{m^2}=(||x||^m)^m\le(K||x^m||)^m=K^m||x^m||^m\le K^mK||(x^m)^m||=K^{m+1}||x^{m^2}||.$$Now your argument for $n\le m$ also works for $n\le m^2$...


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Your definition of $C^*$-algebra is missing the property $(xy)^*=y^*x^*$. Without that requirement, a more natural (to me) example where 1-2-3 are true and 4 is false is to take $M_2(\mathbb C)$ and define the adjoint as conjugation entrywise. Condition 1 cannot be deduced from 2-3-4. For example, let $A=\ell^\infty(\mathbb N)$, and let $$x^*=(\overline{\...


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No. For example, say $$Ax=\sum\lambda_n\langle x,e_n\rangle e_n,$$where $(e_n)$ are orthomormal and $\lambda_n$ are scalars. Then $A$ is compact if $\lambda_n\to0$, while $A$ is trace class requires $\sum|\lambda_n|<\infty$.


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No. Compact, positive definite operators, for example, are trace class if and only if their eigenvalues are summable



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