# Isotropy groups on maximal symmetric manifolds

The questions originally arises while I’m reading a book on General Relativity. There was the following statement.

Let $Iso(M,g)$ the group of isometries on the maximal symmetric Riemannian manifold $M$. Consider the isotropy group of a certain $p \in M$, which is a subgroup of the isometry group. Then the group action of the isotropy group on $T_p M$ is identical to the action of $SO(n)$ on $\mathbb{R}^n$ for every manifold $M$.

The question that remains is why this true? When it’s technically to prove this, I’m also fine with the argument.

• What means "maximal symmetric"? – Moishe Kohan Oct 17 '17 at 12:35
• This just means that the isometry group is of dimension $\frac{n(n+1)}{2}$ – Hamilcar Oct 17 '17 at 12:36

There are several problems with the statement you quoted.

First, "the maximal symmetric Riemannian manifold $M$" doesn't really make sense. This terminology would be meaningful if among all symmetric Riemannian manifolds, there's one that's maximal (in some sense -- it contains all the others, perhaps?), and it's unique (implied by the use of "the"). I think what you want to say here is "a maximally symmetric Riemannian manifold."

Second (probably a typo), it makes no sense to say "identical to the action of $SO(n)$ on $\mathbb R$," because $T_pM$ is an $n$-dimensional vector space (where $n$ is the dimension of $M$) while $\mathbb R$ is $1$-dimensional. Probably they meant to write $\mathbb R^n$.

Third, it's absolutely not correct to say "the group action of the isotropy group on $T_pM$ is identical to the action of $SO(n)$ on $\mathbb R^n$. It can't be "identical," because $\mathbb R^n$ and $T_pM$ are not identical vector spaces. What would make sense is to say that the action is equivalent to the action of $SO(n)$ on $\mathbb R^n$ in the sense of representation theory.

Even with these corrections, the statement is false. For example, the unit sphere $\mathbb S^n\subseteq \mathbb R^{n+1}$ is maximally symmetric, but its isotropy action at the north pole is equivalent to the action of $O(n)$ on $\mathbb R^n$, not $SO(n)$.

Anthony Carapetis's answer shows that the isotropy group contains $SO(T_pM)$. But it will only be contained in $SO(T_pM)$ if you restrict attention to orientation-preserving isometries.

First note that since the action of the group on $M$ is by isometries, the induced action of the isotropy subgroup $\mathrm{Stab}(p)$ on $T_p M$ must be by linear isometries; so it is a subgroup of $SO(T_p M).$ (So far this is true for any Riemannian manifold.)

In order to show that in this case it is the entirety of $SO(T_p M)$, we can make a dimensional argument. The orbit of $p$ under the action of the isometry group has dimension at most $n$ (since the manifold itself does), and can be identified with $\mathrm{Isom}(M,g)/\mathrm{Stab}(p)$; so we have $\frac12n(n+1) - \mathrm{dim}(\mathrm{Stab}(p))\le n$ and thus $\mathrm{dim}(\mathrm{Stab}(p))\ge \frac12 n(n-1).$ Since this is exactly the dimension of $SO(T_p M)$, we are done.

Edit: As Jack points out, I really should be writing orientation-preserving isometries everywhere if we're talking about $SO$.