Is there a smooth non-trivial action of $\mathrm{SU}(2)$ on a torus $\mathbb{T}^k$ for $k> 2$?

Question

For $k=2$ I have been able to show non-existence by using invariance of domain and that $\mathrm{SU}(2)$ does not have two-dimensional subgroups. What about $k>2$?

Answer 1

For $k=3$, there is no non-trivial action. I'm not sure what happens for higher $k$.

To see this, first recall that the connected closed subgroup of $SU(2)$ are given, up to conjugacy, as $\{I\}, \{diag(z,\overline{z}):z\in \mathbb{C}, |z| = 1\},$ and $SU(2)$ itself.

Now, pick a point $p\in T^3$ for which the orbit $SU(2)\cdot p$ is principal. I will write $P$ for the orbit. Since every orbit is diffeomorphic to $SU(2)$ mod the isotropy subgroup, $P$ has dimension $0$, $2$, or $3$. Of course, the case where it is $0$-dimensional, together with connectedness of $SU(2)$ will imply the action is trivial. So we need to deal with the other two cases.

Assume first that the $P$ has dimension $3$. By invariance of domain, the orbit must be all of $T^3$. In particular, in this case, the action is transitive. Counting dimensions, we find the isotropy subgroup of the action is $0$-dimensional, so is a discrete subgroup of $SU(2)$. It follows that the projection $SU(2)\rightarrow T^3$ is a covering. This is absurd since $\pi_1(T^3)$ is infinite.

Next assume that $P$ has dimension $2$. From the classification of subgroups of $SU(2)$, it follows that $P = S^2$ or $P = \mathbb{R}P^2$. Because $\dim T^3 - \dim P = 1$, a result of Mostert asserts that the orbit space $T^3/SU(2)$ is homeomorphic to either $[0,1]$ or $S^1$.

If the orbit space is $S^1$, Mostert shows that $T^3$ has the structure of a bundle over $S^1$ with fiber $P$. The long exact sequence of homotopy groups shows that $pi_1(P)$ is infinite, which is obviously absurd.

If the orbit space is $[0,1]$, Mostert shows there are precisely two non-principal orbits $S_1, S_2$ and that $P$ is a linear sphere bundle over both $S_1$ and $S_2$. In addition, he shows that $T^3$ can be obtained by gluing two disk bundles (obtained by filling in the fibers) along their common boundary $P$.

Since $\mathbb{R}P^2$ isn't a sphere bundle over anything, we must have $P = S^2$. From the classification of sphere bundles with total space $S^2$, then $S_1, S_2\in \{point, \mathbb{R}P^2\}$. If at least one $S_i$ is a point, a simple van Kampen argument shows that resulting space has fundamental group of order at most $2$, so it is not $\mathbb{T}^3$. If both $S_1, S_2$ are $\mathbb{R}P^2$, van Kampen shows $\pi_1$ is the free product $\mathbb{Z}/2\mathbb{Z} \ast \mathbb{Z}/2\mathbb{Z}$, so is not abelian. Thus, we don't get $T^3$ from this case either.

Answer 2

In fact, much more is true:

Suppose a compact Lie group $G$ acts non-trivially on a compact aspherical manifold $M$. Then $G$ is abelian.

A proof can apparently be found in

P. E. Conner and Frank Raymond, Actions of compact Lie groups on aspherical manifolds, Topology of Manifolds (Proc. Inst., Univ. of Georgia, Athens, Ga., 1969), Markham, Chicago, Ill., 1970, pp. 227–264. MR 0271958

but I am having trouble accessing it.

On the other hand, this article proves a slight generalization.

The idea of the proof is as follows. Writing $M = K(\pi,1)$, one finds an action of a larger group $G^\ast$ on the universal cover $\tilde{M}$ of $M$. This larger group contains $\pi$ (thought of as covering transformation) among other elements. (When $G = SU(2)$, the action on $T^k$ automatically lifts to $\mathbb{R}^k$; this lift plays the role of $G^\ast$. In this case, using discreteness of $\mathbb{Z}^k$ and connectedness of $SU(2)$, it's not too hard to see that $G^\ast \cong SU(2)\times \mathbb{Z}^k$).

Now, clearly $SU(2)$ contains many torsion elements. Hence, the following observation shows that $SU(2)$ cannot act non-trivially on $T^k$.

Observation: $G$ cannot contain any torsion elements. For, if this is false, pick a torsion element of prime order and consider the subgroup $H$ in $G$ generated by this element. Then one shows that $\pi_1(M/H)$ fits into a sequence $\pi_1(M)\rightarrow \pi_1(M/H)\rightarrow \pi_1(M)$ whose composition is the identity. (This part is least clear to me, but apparently enough is known about $\pi_1(M/H)$ for non-free actions to make this work.)

Because $M$ is a $K(\pi,1)$, this group level computation can be modeled via continous map $M\rightarrow M/H\rightarrow M$, where the composition $M\rightarrow M$ is homotopic to the identity. Then one argues that the map $H^{\dim M}(M/H)\rightarrow H^{\dim M}(M)$ cannot be surjective, giving a contradiction.