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.
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Sign up to join this communityIs 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.
There is a bit of a disconnect between the title and the actual question. Usually a semifree action is one in which the only isotropy groups are the trivial group and the whole group. The actions with only finite isotropy groups, in the body of your question, are often called ``pseudofree'' actions. If a torus were to act pseudofreely on a sphere then a subgroup of the form $G=\mathbf{Z}_{p}\times \mathbf{Z}_{p}$ for some prime $p$ would act freely on the sphere. It is well-known that the latter group cannot act freely on a sphere because such an action would imply that the group has periodic cohomology, which $G$ does not. See Bredon's book Introduction to Compact Transformation Groups, for example.
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.