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

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?

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Isn't it E^2 = E? Btw, the projection on invariant subspace are usually also given by such an integral. – Marc Palm Apr 23 '11 at 12:53

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.
Regarding the second question, if $x$ is an element in the $\ast$-algebra generated algebraically by the isometries, then $x$ is in the linear span of elements of the form $y=S_{i_1}\cdots S_{i_n} S_{j_1}^\ast \cdots S_{j_m}^\ast$. Since $\alpha_t(y)=\exp((n-m)t)y$, it follows from the formula for integration of trigonometric polynomials that $E(y)=0$ unless $n=m$ (in which case $E(y)=y$). – Dima Shlyakhtenko Apr 24 '11 at 4:23
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.