# Let $G'\triangleleft G<\operatorname{Iso}(M)$ be a normal subgroup. A $G'$-stratum is the union of $G$-strata of lesser dimension

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?

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

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 $$\nu_pG(p)$$.