# Can we get fixed points sets of principal isotropy groups of orthogonal representations via iteration of involutions?

Consider an orthogonal representation of a compact Lie group $G$ on an Euclidean space $V$. Denote by $H$ a fixed principal isotropy group, which we assume to be non-trivial. Consider the fixed point set $V^H$. As a motivation to my question, recall that $V^H$ contains the normal spaces to the principal orbits it meets. The normalizer of $H$ in $G$, $N(H)$, acts on $V^H$. Let $\bar N = N(H)/H$. The inclusion $V^H\to V$ induces an isometry between quotient metric spaces $V^H/\bar N\to V/G$. There are very many features of the representation $(G,V)$ that can be recovered from the simpler representation $(\bar N,V^H)$, more than I would like to mention now (e.g. Luna-Richardson theorem).

My question is whether it is true that there always exists a subgroup $L\subset H$ which is a finitely iterated $\mathbf Z_2$-extension of the identity and gives the same fixed point set, $V^L=V^H$.

(This was an assumption in a result I once proved with G. Thorbergsson to get a sufficient condition for the orbits of a representation to be taut in the sense of Morse theory. At that time, we just thought it easier to check this condition directly in the cases we wanted to use our result.)

Application: Let $M$ be any $G$-orbit in $V$. Then the fixed point set $M^H$ is an orbit of $\bar N$ in $V^H$. If the result above is true, then $M^H$ can be obtained from $M$ by successively taking fixed points of involutions. It then follows from a theorem of Floyd that the sum of the Betti numbers of $M^H$ with respect to $\mathbf Z_2$-coefficients is less than or equal to the corresponding sum for $M$. In a sense, the topology of $M^H$ is not more complicated than that of $M$.

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