# prove the dimension of invariant vectors set is constant

When I read the proof of “the fixed point set of an isometric map in Riemannian manifold is a totally geodesic submanifold”, there is a question about invariant vectors set.

Let $$\phi:M\to M$$ be an isometry provided M is a Riemannian manifold. For $$x\in M$$, defining invariant vectors set $$H:=\{v\in T_x M|(\text{d}\phi)_x(v)=v\}\,,$$ prove dim(H) is a constant.

I am not sure this question is definitely right, looking forward to some hints and ideas, thank you!

I think your statement is true on connected component of $$\mathrm{Fix}(\phi)$$, though I don't have a counter example for different components with different dimensions if $$M$$ is connected.

Suppose $$\phi(x)=x$$, and suppose its connected component in $$\mathrm{Fix}(\phi)$$ is not reduced to $$\{x\}$$. Let $$v \in T_xM$$ such that $$\mathrm{d}\phi(x)v = v$$. Let $$\gamma$$ be the unique geodesic with $$\gamma(0) = x$$ and $$\gamma'(0) = v$$. Then $$\phi\circ \gamma$$ is a geodesic (as $$\phi$$ maps geodesics to geodesics), with same initial data than $$\gamma$$. Thus, $$\phi\circ \gamma = \gamma$$.

Now, consider $$H_x \subset T_xM$$ the set of all fixed vectors of $$\mathrm{d}\phi(x)$$. It is a linear subspace. Let $$U\subset T_xM$$ be a convex neighbourhood of the origin on which the exponential map $$\exp_x$$ is a diffeomorphism onto its image. Suppose moreover that $$\exp_x(U)$$ is geodesically convex (one can assume this by reducing $$U$$). The idea is to prove that $$\exp_x(U\cap H_x) = \mathrm{Fix}(\phi)\cap \exp_x(U)$$. If this result is true, this shows that $$\mathrm{Fix} (\phi)$$ is a submanifold of $$M$$ with $$H_x = T_x \mathrm{Fix}(\phi)$$. Thus, $$\dim H_x$$ is locally constant as the rank of the tangent space of a submanifold.

The first inclusion $$\exp_x(H_x\cap U) \subset \mathrm{Fix}(\phi)\cap \exp_x(U)$$ is clear because of the first paragraph: the entire geodesic $$\gamma$$, with $$\gamma(0) = 0$$ and $$\gamma'(0) = v$$ is fixed by $$\phi$$, thus $$\phi(\gamma(t)) = \gamma(t)$$ for all $$t$$ and $$\exp_x(tv) \in \mathrm{Fix}(\phi)$$.

Conversly, suppose $$p \in \mathrm{Fix}(\phi)\cap \exp_x(U)$$. As $$\exp_x(U)$$ is geodesically convex, and $$\exp_x$$ is a diffeomorphism from $$U$$ to $$\exp_x(U)$$, let $$\gamma$$ be the unique minimizing geodesix from $$x$$ to $$p$$. Then $$\phi\circ \gamma$$ is also a minimizing geodesic from $$x$$ to $$p$$. By the assumptions on $$U$$, $$\gamma = \phi\circ \gamma$$, and as $$\gamma(0)$$ is a fixed point of $$\phi$$, this shows that $$\gamma'(0)$$ is fixed by $$\mathrm{d}\phi(x)$$ and $$v \in H_x$$.

We have shown that $$\exp_x\left(H_x \cap U \right)$$ is diffeomorphic to $$\mathrm{Fix}(\phi) \cap \exp_x(U)$$, and then it is a submanifold. The results follow.

Notice that we have shown more than that: we have shown that $$\mathrm{Fix}(\phi)$$ is a geodesic submanifold.

• The dimension of $Fix(\phi)$ can vary from component to component in connected manifolds. Consider for example $\Bbb{RP}^2$ and the map $[(x,y,z)]\mapsto [(-x,y,z)]$. You have as fixpointsets once the point $[(1,0,0)]$ but also the line $\{[(0,\cos(\varphi),\sin(\varphi))]\mid \varphi\}$. Commented Nov 13, 2020 at 19:14
• Thank you, it’s very helpful for me!
– ling
Commented Nov 14, 2020 at 1:34