2
$\begingroup$

Let $\mathcal{A}$ be a unital simple C*-algebra. Show that $\mathfrak{Z}_{\mathcal{A}} = \mathbb{C}1_{\mathcal{A}}$, where $\mathfrak{Z}_{\mathcal{A}}$ denotes the center of $\mathcal{A}$ and $1_\mathcal{A}$ is identity (or unit) element in $\mathcal{A}$.

We have to show that $\mathfrak{Z}_{\mathcal{A}} = \mathbb{C}1_{\mathcal{A}}$. Clearly, $\mathbb{C}1_{\mathcal{A}}\subseteq \mathfrak{Z}_{\mathcal{A}}$. For the converse, suppose there exists $a\in \mathfrak{Z}_{\mathcal{A}}$ but not in $\mathbb{C}1_{\mathcal{A}}$. Define $\theta : \mathcal{A}\longrightarrow \mathcal{A}(a-\lambda 1_{\mathcal{A}})$ by $\theta(x) = x(a-\lambda 1_{\mathcal{A}})$, where $\lambda \in \mathbb{C}$. Clearly, $\theta$ is bijection and bounded (kernel($\theta$)=$\{0\}$, because $\mathcal{A}$ simple C*-algebra).

Above is clear. Please check now.

Therefore, $\mathcal{A} \cong \mathcal{A}(a-\lambda 1_{\mathcal{A}})$ (vector space isomorphism). This implies $ 1_{\mathcal{A}} = x (a-\lambda 1_{\mathcal{A}})$ for some $x\in\mathcal{A}$. It means that $(a-\lambda 1_{\mathcal{A}})$ is invertible for all $\lambda\in\mathbb{C}$. Therefore, $sp(a)$ (spectrum of $a$) is empty, which is a contradiction. Hence, $\mathfrak{Z}_{\mathcal{A}} = \mathbb{C}1_{\mathcal{A}}$.

Is the above correct? If yes, then please explain the following statement :

Therefore, $\mathcal{A} \cong \mathcal{A}(a-\lambda 1_{\mathcal{A}})$ (vector space isomorphism). This implies $ 1_{\mathcal{A}} = x (a-\lambda 1_{\mathcal{A}})$ for some $x\in\mathcal{A}$.

$\endgroup$
7
$\begingroup$

Another proof goes like this (I will denote the center by $Z(A)$):

Let $a \in Z(A)$ and pick some $\lambda \in \sigma(a)$ (the spectrum is always non-empty). Define $$ I := \overline{(a-\lambda)A}. $$ Then $I$ is a closed ideal in $A$ and for all $b \in A$ we have that $(a-\lambda)b$ is not invertible. Therefore $$ \lVert (a-\lambda)b - 1_A \rVert \geq 1 \qquad (b \in A). $$ In particular, $1_A \notin I$, so $I = \{0\}$ which implies that $a = \lambda 1_A$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Nice!$\ \ \ \ $ $\endgroup$ – Martin Argerami Oct 23 '16 at 17:26
  • $\begingroup$ I think the the result is still true, eventhough I don't see how to modify the above proof. However, the Dauns-Hofmann Theorem says that $C_b(Prim(A))$ is $*$-isomorphic to $Z(M(A))$. The first algebra is simply $\mathbb C$ so $M(A)$ has trivial center $\mathbb C$. Now, if $a \in Z(A)$, then $a \in Z(M(A))$. Indeed, for any $x \in M(A)$ you have $ax = s-\lim axe_\lambda = s-\lim xe_\lambda a = xa$, where we use that $xe_\lambda$ commutes with $a$ and where $(e_\lambda)$ is an a.u. for $A$. Hence $a = \lambda 1_{M(A)}$ for some $\lambda \in \mathbb C$. This is only possible for $\lambda = 0$. $\endgroup$ – user42761 Jan 12 '19 at 13:38
  • $\begingroup$ You mean a non-trivial one ? Then, no. If $A$ is simple we say that $Z(A) = \{0\}$ and $A$ contains no other ideals except $\{0\}$ and $A$ itself. $\endgroup$ – user42761 Jan 14 '19 at 8:53
1
$\begingroup$

I think the above is correct. But some details are missing. Note that $\theta$ is linear, but not multiplicative. One can still show that it is norm-continuous and that its kernel is a closed double-sided ideal: if $x(a-\lambda 1_{\mathcal A})=0$, then $yx(a-\lambda 1_{\mathcal A})=0$ and $xy(a-\lambda 1_{\mathcal A})=x(a-\lambda 1_{\mathcal A})y=0$. That makes $\theta$ injective. It is surjective by construction. Being bounded, it is then invertible by the open mapping theorem.

Once we know that $\mathcal A\simeq \mathcal A(a-\lambda 1)$, we deduce that $\mathcal A(a-\lambda 1)$ is closed. It is easy to show that it is an ideal. But then because $\mathcal A$ is simple, we get that $\mathcal A=\mathcal A(a-\lambda 1)$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ thanks a lot...... one thing i want to know..... what is the need of the following statement : Being bounded, it is then invertible by the open mapping theorem. $\endgroup$ – singh king Oct 22 '16 at 16:42
  • $\begingroup$ If you don't have that $\phi$ and $\phi^{-1}$ are bounded, you cannot conclude that $\mathcal A(a-\lambda 1)$ is closed. $\endgroup$ – Martin Argerami Oct 22 '16 at 17:50
  • $\begingroup$ can you please explain? $\endgroup$ – singh king Oct 22 '16 at 17:58
  • $\begingroup$ Since $\phi$ is bounded (or continuous) and $\phi(\mathcal{A}) = \mathcal{A}(a-\lambda1_\mathcal{A})$, $\mathcal{A}(a-\lambda1_\mathcal{A})$ is closed. $\endgroup$ – singh king Oct 22 '16 at 18:02
  • 1
    $\begingroup$ No. The range of a bounded operator is not necessarily closed. Functional Analysis 101. $\endgroup$ – Martin Argerami Oct 22 '16 at 18:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.