Let $M$ be a differentiable manifold, and $G$ a Lie group acting smoothly on $M$. Under which condition - if any - is the set of fixed points of the action a submanifold of $M$?

My thoughts so far: if we could determine this set as the preimage of some non-critical value of a diffeomorphism, we would be done. In the easy case $M=\mathbb R^n$, one can consider the function $g\cdot x - x$, with $g\in G$ fixed and $x\in M$ varying, and then pick the preimage of zero. In any case, this gives a larger set than those of all the fixed points: an intersection would be needed, and then who knows what may be happening?

I'm particularly interested in compact Lie groups, and I tried looking for counterexamples, but I can't find any. I have the feeling it shouldn't be too hard to find a set of fixed points self-intersecting in some point (I'm thinking of sth like the axes in $\mathbb R^2$), but I probably shouldn't be trusting feelings.

References, hints, generic intuitions are very welcome!

  • $\begingroup$ The preimage of a non-critical value of a diffeomorphism is a discrete set of points (and hence a submanifold, but not a very interesting one). Maybe you mean the preimage of a non-critical value of a submersion? $\endgroup$ – Jack Lee Apr 12 '16 at 20:44
  • $\begingroup$ @JackLee ouch, yes. I was thinking of sth like that, thanks $\endgroup$ – nelv Apr 13 '16 at 7:07

Let $x \in M$ be a fixed point of a $G$-action, $G$ acting properly on $M$ (meaning the map $G \times M \to M \times M$, $(g,m) \mapsto (m,gm)$, is proper). We'll use the following tubular neighborhood theorem:

If $M$ is a smooth proper $G$-manifold without boundary, and $N$ is a closed $G$-invariant submanifold, then there is an open neighborhood $U \supset N$ which is $G$-equivariantly diffeomorphic to the total space of a $G$-vector bundle over $N$ (that is, a vector bundle over $N$ such that the action of $G$ on $N$ extends to a fiberwise linear action of $G$ on the bundle.) See proposition 4.5 here.

Now, let's kill the fly. Since $x$ is a fixed point, it's a $G$-invariant submanifold. A $G$-vector bundle over it is the same thing as a vector space with a linear action of $G$. The fixed point set of this action is then necessarily a vector subspace. Transporting this back to the neighborhood of $x$, we see that locally near $x$, the fixed point set forms a submanifold, as desired.

(Indeed, the set of points $x$ such that the isotropy group of the action at $x$ contains some fixed closed subgroup $H \subset G$ forms a submanifold by the exact same argument; of course, the example above corresponds to the case $H=G$.)

  • $\begingroup$ thanks for the kind answer and the reference. So it's the properness - with some other machinery - really doing the job, avoiding the patologies of Prof. Lee's answer. Or uhm, yeah, at least that's a sufficient condition $\endgroup$ – nelv Apr 13 '16 at 7:10
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    $\begingroup$ @nelv I should clarify that talking about properness is a little disingenuous here. A noncompact Lie group that acts properly on $M$ has no fixed points: if it did, and $f: G \times M \to M \times M$ is the proper map, and $m$ is fixed, then $f^{-1}(m,m) = G \times \{m\}$, which is not compact. $\endgroup$ – user98602 Apr 13 '16 at 14:33

The fixed-point set can be extremely wild. For example, every closed subset of $\mathbb R^n$ is the fixed point set of some smooth $\mathbb R$-action. To see this, let $S\subseteq\mathbb R^n$ be closed, and let $f\colon \mathbb R^n\to\mathbb R$ be a smooth nonnegative function whose zero set is exactly $S$. (Such a function always exists; see, for example, Theorem 2.29 in my Introduction to Smooth Manifolds, 2nd ed.) Now let $V$ be the following vector field on $\mathbb R^n$: $$ V_x = \frac{f(x)}{1+f(x)} \,\left.\frac{\partial}{\partial x^1}\right|_x. $$ The fact that the coefficient of $V$ is bounded in absolute value by $1$ guarantees that $V$ is complete (exercise for the reader), so its flow is a smooth $\mathbb R$-action on $\mathbb R^n$. The fixed points of this action are exactly the zeros of $V$, which are the points of $S$.

  • $\begingroup$ yikes, I had not thought it could get that bad. Thank you for your answer : ) $\endgroup$ – nelv Apr 13 '16 at 7:11

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