Let $G$ be a compact Lie group and $M$ be a smooth manifold on which $G$ acts smoothly and effectively. Then the orbit space $M/G$ admits an Whitney stratification as follows $$M/G=\bigsqcup_{H<G}(M_{(H)}/G).$$ Where $H$ is a closed subgroup of $G$, and $M_{(H)}$ is the set of the points in $M$ such that the isotroy groups are conjugate to $H$. A result is if $M$ admit an orientation preserved by the $G-$action then $$codimM_{(H)}\geq2,$$ when $H$ is not trivial subgroup of $G$. Why?

## 2 Answers

Maybe I add something to Peter Michor's answer. The slice theorem says, that for $y\in M_{(H)}$ with $G_y=H$ there exists a $G$-invariant open neighbourhood $U_y$, such that $$ U_y\cong G\times_H V_y$$ and in particular $$M_{(H)}\cap U_y= G\times_H V_y^H$$ where $V_y= T_y M/T_y (G\cdot y)$. So $M_{(H)}$ can only have codimension 1 in $M$ if $\dim V_y^H=\dim V_y-1$. This means that $H$ has to act only via one reflection on $V_y$, but this cannot be orientation preserving!

A codimension 1 stratum needs a reflection in the isotropy group; better, in the isotropy representation $G_x:T_xM\to T_xM$ for x over a codimension 1 stratum.