Center of unital simple C*-algebra is $\mathbb{C}$.

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 $x\in\mathcal{A}$.

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 $a = \lambda 1_A$.

• Nice!$\ \ \ \$ – Martin Argerami Oct 23 '16 at 17:26
• 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 $\lambda = 0$. – user42761 Jan 12 '19 at 13:38
• 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. – user42761 Jan 14 '19 at 8:53

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 $$\mathcal A=\mathcal A(a-\lambda 1)$$.

• 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. – singh king Oct 22 '16 at 16:42
• If you don't have that $\phi$ and $\phi^{-1}$ are bounded, you cannot conclude that $\mathcal A(a-\lambda 1)$ is closed. – Martin Argerami Oct 22 '16 at 17:50
• can you please explain? – singh king Oct 22 '16 at 17:58
• Since $\phi$ is bounded (or continuous) and $\phi(\mathcal{A}) = \mathcal{A}(a-\lambda1_\mathcal{A})$, $\mathcal{A}(a-\lambda1_\mathcal{A})$ is closed. – singh king Oct 22 '16 at 18:02
• No. The range of a bounded operator is not necessarily closed. Functional Analysis 101. – Martin Argerami Oct 22 '16 at 18:05