# Free circle action from a torus

So these lecture notes contain this statement (Exercise 3.3.5):

If $T$ is a torus acting on a compact manifold $M$ such that every isotropy subgroup has codimension greater than one, then there exists a circle inside $T$ that acts freely on $M$.

Could someone point me to a proof? This doesn't seem to be very straightforward. I would imagine this has a lot to do with finiteness of number of orbit types on $M$ (perhaps we don't even need $M$ to be a manifold, if we a priori know that there are only finitely many orbit types?). Also, can we weaken the assumption about codimension of isotropy groups to the action being fixed point free?

Suppose that a compact group $G$ acts on a compact manifold $M$, We say that the orbits of $x$ and $y$ have the same type if and only if $G_x$ is conjugated to $G_y$ where $G_x$ is the stabilizer of $x$. There exists a finite number of type since $M$ is compact. Suppose now that $G$ is the torus $T^n$, a type of an orbit is characterized by a subgroup $H_i$ of $T^n$ since $H_i=gH_ig^{-1}$. It the dimension of $H_i< n$, there exists a morphism $f:S^1\rightarrow T^n$ whose image does not intersects a type of an orbit since the number of type is finite. $S^1$ acts freely on $M$ via $f$.
• I'm afraid I don't understand. How do I produce such a morphism $S^1 \to T^n$ e.g. if there exists a point with isotropy $H=T^{n-1} \times \mathbb{Z}_n$? Also, I skimmed the book you referred to, but can't find the relevant part. Could you please be more specific? – phdstudent Nov 17 '15 at 13:30
• The reference whose for the definition of orbit type and the fact that the number of orbit type is finite if $M$ is compact. The idea of the proof is very simple: take a. circle in $T^n$ which avoids all the stabilizers, it exists since the cardinality of the type is finite and the dimension of a stabilizer<n – Tsemo Aristide Nov 17 '15 at 13:33
• OK, let me more specific: suppose that $n=2$, i.e. $T^2=S^1\times S^1$, and there are two only stabilizers: $S^1 \times \{1\}$ and $\{1\}\times S^1$. I see that I can embed a circle that avoids those, but how do I know the emedding is homomorphic? – phdstudent Nov 17 '15 at 14:15
• There is a dense family of $S^1$ sitting inside $T^2$. To see this remark that $T^2$ is the quotient of $R^2$ by $t_{e_1},t_{e_2}$ where $t_{e_1}$ is the translation of direction $e_1$, basically every translation $t_{ae_1+be_2}, a,b\in Q$ yields to a circle sitting in $T^2$. – Tsemo Aristide Nov 17 '15 at 14:20