2

Let $M$ be a smooth manifold and Let $G$ be a finite group acting on $M$ by diffeomorphisms. Show that the set of fixed point $F=\{m \in M: g.m=m \}$ is a smooth manifold.

I am unable to deal with the set of fixed points. It doesn't seem to be an open set. It would have been trivial if it was open. Any hints?

zero2infinity
  • 753
  • 1
    Hm, does your text allow mixed-dimension manifolds (e.g., the disjoint union of a straight line and a plane)? – Hagen von Eitzen Apr 11 '21 at 07:34
  • Well, the set $F$ need not be an open subset, for example consider the plain $\mathbb R^2$ and the action of $\mathbb Z/2\mathbb Z={\overline 0,\overline 1}$ as $\overline 1\cdot (x,y)=(-x,-y)$, the unique fixed point is the origin $(0,0)$. – Alessandro Apr 11 '21 at 07:44
  • Suppose that $M$ is the real projective plane and $G$ is generated by a single involution $g$ acting as $g([x:y:z])=[-x:y:z]$. The fixed point set of $G$ is the disjoint union of a projective line and a singleton. Do you regard this as a manifold? – Moishe Kohan Apr 11 '21 at 09:59

2 Answers2

4

You may assume $M$ is a Riemannian manifold with some metric $g_0$. First, observe that $M$ admits a new Riemannian metric $g$ that is invariant under the group action: for example define $$ g=\sum_{\tau\in G} \tau^*g_0; $$ this can be defined since $G$ is a finite group.

In general, for any finitely many elements $\tau_1, \tau_2, ..., \tau_n$ assume $p$ is a fixed point of them, i.e. $\tau_j p=p$ for $j=1, ..., n$. Note that since $g$ is an invariant metric, the differential $d\tau_j$ is an orthogonal linear transformation on the tangent space $T_p$.

If there is no tangent vector $X$ at $p$ that is a common eigenvector of all $d\tau_j$ with eigenvalue $1$, then $p$ is an isolated fixed point: in fact, $\tau_j$ maps any geodesic from $p$ to a geodesic from $p$, so if a tangent vector $X$ is not an eigenvector with eigenvalue $1$ for some $d\tau_j$, then $\tau_j$ moves the geodesic $\exp_p(tX)$ (here $\exp_p$ is the exponential map from $p$) to a different geodesic, so there is no fixed point on $\exp_p(tX)$ for $0<t<\delta$.

Now assume the common eigenvectors of all $d\tau_j$ at $p$ with eigenvector $1$ form a linear subspace $V$ in $T_p$. Then as above we see the manifold $\exp_p V$ are all fixed points; on the other hand, any other point $q$ near $p$ can be written as $\exp_ptX$ so that $X\notin V$ and $t$ small, as above we see this point is moved to some other point by some $\tau_j$. So near $p$ the fixed point set is exactly $\exp_tV$.

Yuval
  • 3,439
  • 6
  • 20
  • 4
    This is a correct proof, of course. However, the more careful statement is that the fixed-point set is a disjoint union of closed subsets each of which is a smooth submanifold. The point is that different components of the fixed point set may have different dimensions. Some books allow manifolds of "nonconstant" dimension, for instance, Loring Tu does in his books and so does Lang. – Moishe Kohan Apr 11 '21 at 09:57
0

On the contrary, by its definition via an equality, $F$ will always by closed. You will need to show the properties of a smooth manifold step by step. For "locally Euclidean", you may want to use the implicit function theorem.

After that, the transition maps of the atlas may not "fit" the subsets a priori, but we haven't used finiteness of $G$ yet. Note how the action of $G$ on $M$ induces an action on transition maps near points $\in F$. By averaging over the finitely many transition maps in the $G$-orbit of an arbitrary transition map, you can obtain a $G$-invariant transition map. Being $G$-invariant, it respects $F$ and thus is a valid transition map for $F$.

Hagen von Eitzen
  • 374,180