# Fixed point set for a subcircle of torus actions

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 $X^{T}=X^{G}$?

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 $i<n$, $X^L=X^T$.