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Let $T=S^{1}\times S^{1}\times ...\times S^{1}$ ($n$ times) be $n$ dimensional torus and $X$ be a $T$-space.

Lemma: If $X$ has finitely many connected orbit type, then there is a subcircle $L=S^{1}\subseteq T$ such that $X^{L}=X^{T}$ (fixed point sets of the action).

I want to prove this lemma.

In some books, The hypotesis finitely many connected orbit type is not necessary. That is, There is always a subcircle $L=S^{1}\subseteq T$ such that $X^{L}=X^{T}$. Is this true?

If $G$ is a compact connected Lie group, then there is a subcircle $% L=S^{1}\subseteq G$ such that $X^{L}=X^{G}$? That is, there is subtorus $T\subseteq G$ such that $X^{T}=X^{G}$?

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When they are finitely orbit types, the result is true. $x$ and $y$ are in the same orbit type if and only the satbilizer $Stab(x)$ of $x$ is conjugated to $Stab(y)$, since $T$ is commutative, it implies that $Stab(x)=Stab(y)$, thus there exists a finite number of subgroups of $T$, $H_1$,...,$H_n$ such that $H_i$ is a stabilizer of a point. Suppose that $H_n=T$ (the stabilizer of fixed points) then you can find a circle $L$ which is not contained in $H_i$ for $i<n$, $X^L=X^T$.

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The answer to the last question is "no". There exist smooth actions of non-abelian compact Lie groups on euclidean space with empty fixed point set. See Bredon's book, Introduction to Compact Transformation Groups, for example. But Smith Theory shows that any circle or torus group action on euclidean space has non-empty fixed point set.

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