Let $G$ be a group of isometries acting effectively by isometries on a connected Riemannian manifold. And let $G'\triangleleft G$ be a normal subgroup. I am trying to prove that $\dim \operatorname{St}_G(p)\leq \dim \operatorname{St}_{G'}(p)$ for every $p\in M$; where $\operatorname{St}_X(p)$ stands for the stratum of the action of $X$ on $M$ through $p$.

We have the formula $$ \dim\operatorname{St}_G(p)=\dim G+\dim M_0^{G_p}-\dim N_G(G_p); $$ where $M_0^{G_p}$ is the connected component of the fixed-point set $M^{G_p}$ through $p$.

It is obvious that $\dim M_0^{G_p}\geq\dim M_0^{G'_p}$, since $G'_p\triangleleft G_p$. Thus, the result is true if we show that $\dim N_G(G_p)\geq \dim N_G'(G'_p)$, which I'm not sure if is true. If it is not true, is there any other way to prove my claim?

  • 1
    $\begingroup$ I'm pretty sure $\dim M_0^{G_p} > \dim M_0^{G'_p}$ should not be strict inequality. $\endgroup$
    – LSpice
    Commented Jul 5, 2020 at 22:44

1 Answer 1


We have to assume that the index $[G':G]$ is finite. In this case:

Let $G'$ be a normal subgroup of $G$ such that the quotient $\Gamma=G/G'$ is finite and acts by isometries in $X'=M/G'$, and $X=X'/\Gamma$. Thus, $(G')^0=G^0$ and, therefore the orbits $G'(p)$ and $G(p)$ have same connected components through $p$. This way, $\nu_pG(p)=\nu_pG'(p)$. Also, as $G_p^0=(G_p')^0$, the orbits associated to their respective slice representations, i.e., their infinitesimal actions on the normal space to the tangent space to their respective orbits, have same dimension. And $(\nu_pG(p))^{G_p}\subset (\nu_pG(p))^{G'_p}$. As the orbits of their associated slice representations have same dimension, the cohomogeneity of the action $(G'_p,(\nu_pG(p))^\dagger)$ can be no greater then the cohomogeneity then the action of $(G_p,(\nu_pG(p))^\dagger)$. Thus, $\dim\operatorname{St}_X(\pi(p))\leq \dim\operatorname{St}_{X'}(\pi'(p))$.

Here, $(\nu_pG(p)^\dagger)$ is the orthogonal complement of $(\nu_pG(p))^{G_p}$ in $\nu_pG(p)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.