Let $M$ be a smooth manifold, and let $f:M\to M$ be a smooth involution (i.e. $f^2=\text{id}$).

If we introduce a Riemannian metric on $M$ so that $f$ is isometry, we can prove easily that the fixed point set of $f$ makes a submanifold of $M$. (cf: [Klingenberg, Riemannian Geometry, Theorem 1.10.15]) My question is:

Can we prove this statement without using any Riemannian metric?

That is all of my question. Thank you.


1 Answer 1


Yes, by linearizing $f$ at a fixed point.

More explicitly, let $x_0\in M$ be a fixed point of $f$. Let $U$ be an open neigborhood of $x_0\in M$ for which there is a diffeomorphism with an open subset $V$ of $\mathbf R^n$. Replacing $U$ by $f^{-1}(U)\cap U$ if necessary, we may assume that $f(U)\subseteq U$.Then $f(U)=U$ since $f^2=\mathrm{id}$. Therefore, we may assume that $M=V$ is an open subset of $\mathbf R^n$. We may as well assume that $x_0$ is the origin in $\mathbf R^n$.

Now, consider the map $$ g\colon V\rightarrow \mathbf R^n $$ defined by $$ g(x)=x+D_0f(f^{-1}(x)), $$ where $D_0f$ is the differential of $f$ at $0$. The map $g$ is of course differentiable. Moreover, one has $$ g(f(x))=f(x)+D_0f(x)=D_0f(x+D_0f^{-1}(f(x)))=D_0f(x+D_0f(f^{-1}(x)))=D_0f(g(x)) $$ since $f^{-1}=f$. Observe that $$ D_0g=\mathrm{id}+D_0f\circ D_0f^{-1}=2\mathrm{id} $$ on $\mathbf R^n$. By the inverse function theorem, there is an open neighbohood $W$ of the origin in $\mathbf R^n$, contained in $V$, such that $$g\colon W\rightarrow \mathbf R^n$$ is a diffeomorphism onto its image. Replacing $W$ by $f^{-1}(W)$ if necessary, we may assume that $f(W)=W$ like before. Since $$ g(f(x))=D_0f(g(x)), $$ for all $x\in W$, we may assume that the action of $f$ on $V$ is the restriction of a linear map $L\colon\mathbf R^n\rightarrow\mathbf R^n$.

Since $f^2=\mathrm{id}$, one has $L^2=\mathrm{id}$. This means that we can diagonalize $L$ over $\mathbf R$, and we may assume that the matrix of $L$ in the standard basis is the diagonal matrix $\mathrm{diag}(1,\ldots,1,-1,\ldots,-1)$, say with $m$ diagonal entries equal to $-1$. Then, the set of fixed point of $f$ on $V$ is equal to $V\cap\mathbf R^m$, where $\mathbf R^m$ is identified with the subset $\mathbf R^m\times\{0\}^{n-m}$ of $\mathbf R^n$. This proves that the set of fixed points of $f$ is a smooth submanifold of $M$.

The argument applies more generally to any finite group action on $M$, or even, any action of a compact group on $M$, adapting the diagonizability part a little bit. The corresponding statement for topological manifolds is false. Note also that it is crucial here not to ask for a smooth manifold to be connected or even nonempty!

  • $\begingroup$ Great. Could you tell me some ideas or intuition behind the construction of the map $g$? $\endgroup$ Jun 6, 2016 at 10:11
  • $\begingroup$ Well, it is really a recurring construction of invariant elements when a finite group acts on an abelian group: if $G$ is a finite group acting on the left on an abelian group $A$ then $$\sum_{g\in G} g\cdot a$$ is a $G$-invariant element of $A$, for all $a\in A$. One applies that here to: $G=\mathbf Z/2$, acting on $V$ through $f$ and on $\mathbf R^n$ through $D_0f$. These actions induce an action by conjugation of $G$ on the abelian group $A$ of all differentiable maps from $V$ into $\mathbf R^n$... $\endgroup$ Jun 6, 2016 at 11:18
  • $\begingroup$ ...One takes $a$ to be the inclusion map of $V$ into $\mathbf R^n$ that we wanted to make $G$-invariant, and get a $G$-equivariant map from $V$ into $\mathbf R^n$. $\endgroup$ Jun 6, 2016 at 11:18
  • $\begingroup$ Very good. Now I have completely comprehended that, as you said on the answer, this argument runs more generally in the case of any finite group action. I'd like to give thanks to your beautiful answer and comments. $\endgroup$ Jun 7, 2016 at 3:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .