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YCor
YCor
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fixed Fixed point algebra of a non amenable-amenable factor

Let $M$ be a non-amenable factor. Suppose $\alpha$ is an action of a group $\Bbb R$ on $M$.

Define $M^{\alpha}:=\{x\in M: \alpha_t(x)=x \text{ for any} \quad t\in \Bbb R\}$.$$M^{\alpha}:=\{x\in M: \alpha_t(x)=x \text{ for any} \; t\in \Bbb R\}.$$

If we know that there exits $t>0$ such that $\alpha_t=id$$\alpha_t=\mathrm{id}$, can we have $M^{\alpha}=\Bbb C 1 $?

fixed point algebra of a non amenable factor

Let $M$ be a non-amenable factor. Suppose $\alpha$ is an action of a group $\Bbb R$ on $M$.

Define $M^{\alpha}:=\{x\in M: \alpha_t(x)=x \text{ for any} \quad t\in \Bbb R\}$.

If we know that there exits $t>0$ such that $\alpha_t=id$, can we have $M^{\alpha}=\Bbb C 1 $?

Fixed point algebra of a non-amenable factor

Let $M$ be a non-amenable factor. Suppose $\alpha$ is an action of a group $\Bbb R$ on $M$.

Define $$M^{\alpha}:=\{x\in M: \alpha_t(x)=x \text{ for any} \; t\in \Bbb R\}.$$

If we know that there exits $t>0$ such that $\alpha_t=\mathrm{id}$, can we have $M^{\alpha}=\Bbb C 1 $?

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mathbeginner
mathbeginner
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fixed point algebra of a non amenable factor

Let $M$ be a non-amenable factor. Suppose $\alpha$ is an action of a group $\Bbb R$ on $M$.

Define $M^{\alpha}:=\{x\in M: \alpha_t(x)=x \text{ for any} \quad t\in \Bbb R\}$.

If we know that there exits $t>0$ such that $\alpha_t=id$, can we have $M^{\alpha}=\Bbb C 1 $?