# Vector field invariant under transitive action: restricts to free transitive action?

Thinking about how you can put vector fields on homogeneous spaces that respect the homogeneity, I'm interested in the following situation:

Let $V$ be a nonzero vector field on a manifold $M$, let the simply connected group $G$ act transitively on $M$, and let $V$ be invariant under this action, so $g_* V=V$ for all $g\in G$.

Now an obvious way to construct such a vector field is if $G$ acts freely and transitively, since you can pick a vector at any point and push it forward to any other point under the group action. If the action is not free, the ambiguity in choice of group element means that this may not work; the stabilizer need not map $V$ to itself. It is easy to come up with examples where this is not an obstruction, but in all those I've been able to think of, the action can be restricted to a free and transitive one. Must this always be the case? In other words, is the following true?

Proposition: Then $G$ has a subgroup acting freely and transitively.

• Is you question whether or not every transitive action of a simply connected Lie group restricts to a free transitive action on some subgroup? – Jason DeVito Dec 9 '13 at 2:18
• It is whether every transitive action that leaves some nonzero vector field invariant restricts to a free transitive action. – Holographer Dec 9 '13 at 8:34
• For what it's worth, I'm really enjoying your questions. I'm fairly slow to respond, but if you flag me, I'll see them eventually. And I can promise I'll think about them, even if I can't answer them ;-) – Jason DeVito Dec 10 '13 at 0:59

## 2 Answers

I think your question has a negative answer.

Let $G = SU(n)$ for $n\geq 3$. Then $G$ acts transitively on $S^{2n-1}$, preserving a nonzero vector field, and yet there is no subgroup of $SU(n)$ which acts freely and transitively on $S^{2n-1}$.

(We must exclude $n=2$ because the action of $SU(2)$ on $S^3$ is both transitive and free.)

To describe the action, we think of $S^{2n-1}\subseteq \mathbb{C}^n$ as $$S^{2n-1} = \{(z_1,...,z_n)\in\mathbb{C}^n: |z_1|^2 + \ldots + |z_n|^2 = 1\}.$$ For any $p\in S^{2n-1}$, we identify $T_p S^{2n-1}$ with the set of all vectors perpendicular to $p$.

Now, $SU(n)$ naturally acts on $\mathbb{C}^n$ just by multiplication and restricting the action to $S^{2n-1}$ gives the action we want.

Our vector field is the vector field tangent to the Hopf circles. More precisely, at the point $p\in S^{2n-1}$, $V(p) = ip$.

Then, because any element of $SU(n)$ is a complex linear map, we have $$A_\ast (V(p)) = A_\ast (ip) = A(ip) = i(Ap) = V(Ap)$$ (where we use the fact that the derivative of a linear map is itself in the second equality). Hence, $V(p)$ really is $SU(n)$ invariant.

Now, why doesn't any subgroup of $SU(n)$ acts freely and transitively on $S^{2n-1}$? One can copy the argument I used on your previous question: If a Lie group acts freely and transitively on a manifold, then the manifold is diffeomorphic to the Lie group, and in particular, carries a Lie group structure. But only $S^0$, $S^1$, and $S^3$ carry Lie group structures. (See, for example, an older answer of mine.

• One can carry out this same argument to prove that $G = Sp(n)$ for $n\geq 2$ acts freely on $S^{4n-1}$ preserving three independent vector fields, and yet no subgroup acts freely and transitively. Further, when $n\neq 2$ and $n\neq 4$, then no proper subgroup of $Sp(n)$ acts transitively at all. (And I'd be willing to bet, but don't immediately see how to prove, that $n=2$ and $n=4$ work as well.) – Jason DeVito Dec 9 '13 at 14:29
• Glad you like it. Thanks for pointing me to this question via the other question - I don't think I would have found this other wise. – Jason DeVito Dec 9 '13 at 21:18

It seems to me the answer is in the negative. Consider $SO(2)$ acting on $\mathbb{RP}^1$ by rotations. Then this action is transitive but not free, and no subgroup of $SO(2)$ gives a free transitive action because any neighborhood of the identity generates $SO(2)$.

Edit: ok so now $G$ must be simply connected. What about the action of $Spin(4)$ on $S^3$ through the homomorphism $Spin(4)\to SO(4)$? The action is certainly transitive and not free, and again for the same reason as before there is no free transitive restriction.

Edit2: On a second thought, I think my first example already can be easily tweaked to give a simply connected version of what you are asking. Just compose the original action of $SO(2)$ on $\mathbb{RP}^1$ with the universal cover $\mathbb{R}\to SO(2)$. Then the (contractible abelian) group $\mathbb{R}$ acts transitively on $\mathbb{RP}^1$ and no subgroup acts transitively and freely.

• Ok, you're certainly right, thanks. I've edited the question to ask something slightly different, with the same idea in mind, since for my application this sort of thing can be circumvented by passing to the covering group. – Holographer Dec 8 '13 at 17:53
• I don't understand the action of $SO(3)$ on $S^3$ you're describing, could you elaborate a bit? – Jason DeVito Dec 9 '13 at 2:17
• Oops sorry, embarrassing typo here. I meant $S^2$. I'm going to correct it. – johndoe Dec 9 '13 at 8:15
• The central object of interest is some vector field invariant under the action. There is no such vector field on $S^2$ for any choice of action. – Holographer Dec 9 '13 at 8:40
• You're right. I'm going to raise the dimensions involved and see what happens (trial and error is my motto...). – johndoe Dec 9 '13 at 8:52