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

Question

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 $x\in\mathcal{A}$.

Answer 1

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 $a = \lambda 1_A$.

Answer 2

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