Why is this a conditional expectation into the fixed point algebra?

Question

Let $A$ be a C*-algebra and let $\alpha$ be an action of the circle group $S_1$ on $A$ (Gauge action). We define the following map: $$E:A\rightarrow A;\quad E(a):=\int\alpha_t(a)\textrm{d} t.$$ My question is why $E$ is a conditional expectation into the fixed point algebra for the gauge action?

For example if we take the Cuntz algebra $O_n$ and the gauge action as $$\alpha_t(S_j)=e^{it}S_j\quad \forall j=1\dots n,$$ where $ S_j $ is a generating set of isometries for $O_n$, and extending the action to the whole of $O_n$.

What is the value of the integration after restricting to the *-algebra generated algebraically by the isometries?

Answer 1

For your first question, you can check that $E(a)$ is invariant with respect to the action, by using that the Haar measure is left-invariant. The other properties are also straightforward to check.

With respect to your second question, note that $\alpha_t(S_j^*) = e^{-it} S_j^*$. Consider an element $S_{i_1} \cdots S_{i_k} S_{j_1}^* \cdots S_{j_l}$ of the Cuntz algebra. Using the relation above you can see that this is invariant with respect to the gauge action if and only if $k=l$. These operators generate the fixed point algebra. I don't know if it's easy to see, for example, what the image of $S_i$ is under this map.

Answer 2

Let $G$ be a compact group with Haar measure $\mu$ normalised such that $\mu (G)=1$. Let $A$ be a C*-algebra and $\alpha : G\rightarrow \text{Aut} (A)$ a continuous action of $G$ on $A$. Define $$ E(x)=\int_{G}\alpha_{t}(x)d\mu(t). $$ We claim that this is a faithful conditional expectation of $A$ onto the fixed-point algebra $A^{\alpha}$ of $\alpha$. It is norm-decreasing since $$ \|E(x)\|=\|\int_{G}\alpha_{t}(x)d\mu (t)\|\leq\int_{G}\|\alpha_{t}(x)\|d\mu (t) =\|x\|, $$ and if $x\in A^{\alpha}$, then $E(x)=\int_{G}\alpha_{t}(x)d\mu (t)=\int_{G}x d\mu (t)=x$. We need to show that $E$ is also a projection onto the fixed-point algebra. But for fixed $t'\in G$, we get from linearity of integration and left invariance of the Haar measure that $$ \alpha_{t'}(E(x))=\int_{G}\alpha_{t'}(\alpha_{t}(x))d\mu (t)=\int_{G}\alpha_{t't}(x)d\mu (t)=E(x). $$ Now to show faithfulness, let $x\in A$ be a non-zero positive element and let $\omega$ be a state on $A$ such that $\omega (x)=\|x\|$. Since $t\mapsto\omega (\alpha_{t}(x))$ is continuous, we can find a non-empty, open subset $V$ of $G$ such that $\omega (\alpha_{t}(x))\geq \|x\| / 2$ for $t\in V$. It follows that $$ \omega (E(x))=\int_{G}\omega(\alpha_{t}(x))d\mu (t)\geq \int_{V}\omega (\alpha_{t}(x))d\mu (t)\geq\frac{\|x\|}{2}\mu (V), $$ which is non-zero since the Haar measure of a non-empty open set is non-zero.