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Nik Weaver
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Suppose $(G,\alpha, C(X))$ is a triple for which this is true. Then let $\beta$ be the action of $G\times \mathbb{Z}$ on $C(X)$ which agrees with $\alpha$ on $G$ and is trivial on $\mathbb{Z}$. The statement will no longer be true for this action because $X$ hasn't changed but the crossed product has been tensored with $C(\mathbb{T})$ so it has many more tracial states.

(Answer to the original question which omitted the hypothesis that $X$ is infinite: I don't think this is true for $G = \mathbb{Z}$. Let $X$ be a singleton and let $\alpha$ be the trivial action of $\mathbb{Z}$. Then $C(X)\rtimes \mathbb{Z} \cong C^*(\mathbb{Z}) \cong C(\mathbb{T})$, which is abelian so it has lots of tracial states, but there is only one probability measure on $X$.)

I don't think this is true for $G = \mathbb{Z}$. Let $X$ be a singleton and let $\alpha$ be the trivial action of $\mathbb{Z}$. Then $C(X)\rtimes \mathbb{Z} \cong C^*(\mathbb{Z}) \cong C(\mathbb{T})$, which is abelian so it has lots of tracial states, but there is only one probability measure on $X$.

Suppose $(G,\alpha, C(X))$ is a triple for which this is true. Then let $\beta$ be the action of $G\times \mathbb{Z}$ on $C(X)$ which agrees with $\alpha$ on $G$ and is trivial on $\mathbb{Z}$. The statement will no longer be true for this action because $X$ hasn't changed but the crossed product has been tensored with $C(\mathbb{T})$ so it has many more tracial states.

(Answer to the original question which omitted the hypothesis that $X$ is infinite: I don't think this is true for $G = \mathbb{Z}$. Let $X$ be a singleton and let $\alpha$ be the trivial action of $\mathbb{Z}$. Then $C(X)\rtimes \mathbb{Z} \cong C^*(\mathbb{Z}) \cong C(\mathbb{T})$, which is abelian so it has lots of tracial states, but there is only one probability measure on $X$.)

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Nik Weaver
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  • 213

I don't think this is true for $G = \mathbb{Z}$. Let $X$ be a singleton and let $\alpha$ be the trivial action of $\mathbb{Z}$. Then $C(X)\rtimes \mathbb{Z} \cong C^*(\mathbb{Z}) \cong C(\mathbb{T})$, which is abelian so it has lots of tracial states, but there is only one probability measure on $X$.