TITLE: Pseudofree $T^2$ actions on spheres
QUESTION [7 upvotes]: Is it possible to construct a smooth action of $S^1\times S^1$ on $S^{2n+1}$ ($n\ge 2$) such that no point on $S^{2n+1}$ has an infinite stabilizer? 
Note that if such an action exists, it can not be linear.

REPLY [4 votes]: This should only be considered as a comment to the answer above, unfortunately it does not fit in a comment.
One can show, that $\mathbb{Z}_p\times \mathbb{Z}_p$ cannot act freely on $S^n$ also via "elementary" topological methods as follows (I learned this in the book of Hatcher, but dont ask me where):
Assume $n\geq 2$ (all other cases are trivial)
Suppose there exists a free $\mathbb{Z}_p\times \mathbb{Z}_p$ action on $S^n$. Then the quotient $X:=S^n/(\mathbb{Z}_p\times \mathbb{Z}_p)$ can be made into a CW-complex. The long exact sequence for fibrations yiels
$$\pi_k(X)=\begin{cases}0 & 2\leq k\leq n \\ \mathbb{Z}_p\times \mathbb{Z}_p  &k=1\end{cases}$$ 
Now we construct a $K(\mathbb{Z}_p\times \mathbb{Z}_p,1)$ as follows. Via the quotient map $S^n\to X$ we attach one (n+1) cell, to form a space $X'$, and the attach cells of dimension greater than $n+1$ to kill all  homotopy groups higher equals $n+1$. Call the resulting space $Y$. By definition $Y$ is a $K(\mathbb{Z}_p\times \mathbb{Z}_p,1)$ with one (n+1) cell. This means
$$H_{n+1}(Y,\mathbb{Z}_p)\in \{0,\mathbb{Z}_p\}$$
On the other hand: It is well known that the infinite dimensional lens spaces $S^\infty/\mathbb{Z}_p$ is a $K(\mathbb{Z}_p,1)$,  hence $S^{\infty}/\mathbb{Z}_p\times S^{\infty}/\mathbb{Z}_p$ is a $K(\mathbb{Z}_p\times \mathbb{Z}_p,1)$. Now using Künneth we see that 
$$H^{n+1}(S^{\infty}/\mathbb{Z}_p\times S^{\infty}/\mathbb{Z}_p,\mathbb{Z}_p)\notin \{\mathbb{Z}_p,1\}$$
Since $K(\mathbb{Z}_p\times \mathbb{Z}_p,1)$ is unique up to homotopy this shows that such an action cannot exist.