# Dimension of isometry group of complete connected Riemannian manifold

Given an $n$-dimensional geodesically complete connected Riemannian manifold $M$, we want to prove that the dimension of its isometry group is $$\dim {\rm ISO}(M) \leq \frac{n(n+1)}2.$$

Does it suffice to say that, since Euclidean space $\mathbb{R}^n$ is expected to be maximally symmetric, and the number above is the dimension of its isometry group, namely translations plus rotations, then the bound must be true for any other manifold? Do you know a more rigorous proof?

## 1 Answer

The comparison with Euclidean space gives the correct intuition but to give further details one could mention that the orbit of a point $p\in M$ under the isometry group is a homogeneous space of the isometry group, whose dimension is at most that of $M$ itself, namely $n$. Now the action of the stabilizer at $p\in M$ is determined uniquely by the induced action on the tangent space at $p$, which can be identified with a subgroup of $SO(n)$. The latter is known to have dimension $n(n-1)/2$, so for the total dimension of the isometry group you get at most $n + n(n-1)/2 = n(n+1)/2$.

• the tricky part would be proving it is in fact a lie group using geodesics, right? – jj_p Jun 22 '13 at 8:27
• @MikhailKatz Can you please give a reference where it is discussed that the isometry group of a complete connected Riemannian manifold is a Lie group? Thanks. – caffeinemachine Dec 30 '17 at 17:55
• @caffeinemachine See math.stackexchange.com/questions/139402 (it's doesn't have to be complete) – punctured dusk Jul 1 '18 at 9:53