# Tag Info

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-...

2

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 $\sum|\lambda_n|<\infty$.

2

The fact that $\overline{J}$ is an ideal follows from the continuity of the algebraic operations. For example, if $x \in \overline{J}$ and $y \in A$ then we can choose $x_n \in J$ such that $x_n \rightarrow x$ and then $yx_n \rightarrow yx$ and since $J$ is an ideal, $yx_n \in J$ and so $yx \in \overline{J}$. Similarly for right multiplication and addition. ...

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 ...

1

Seems to me the only example is $\Bbb C$. Say $\hat a$ is the Gelfand transform. Since $||\hat a||_\infty\le||a||$, similarly for $a^{-1}$, and $||\hat a||_\infty||1/\hat a||_\infty\ge1$ it follows that $|\hat a|$ is constant whenever $a$ is invertible. Hence, whether $a$ is invertible or not, $|\hat a-\lambda|$ is constant for every $\lambda$ not in the ...

1


Only top voted, non community-wiki answers of a minimum length are eligible