Let $G$ be a compact Lie group, and $M$ be a smooth manifold on which $G$ acts smoothly and effectively. Denote by $F$ the $G$-fixed subset. In general, $F$ has finitely many connected components and each component is a submanifold of $M$. My question is: if $M$ admits an orientation-preserving $G$-action, and the fixed submanifold has positive dimension, then why the codimension of the fixed submanifold is even?

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    $\begingroup$ SO(3) acts linearly on $\mathbb R^3$ with fixed submanifold of codimension $3$, no? $\endgroup$ Oct 20, 2013 at 5:50

1 Answer 1


As Mariano's comment indicates, you need more conditions. The usual one is that $G$ is a torus.

Using $G$ compact, you can average a metric to get a $G$-invariant metric. Then the exponential map gives you a $G$-equivariant isomorphism of a neighborhood of a fixed point with a neighborhood of $0$ in the tangent space, reducing the problem to the real-linear case.

If $G$ is a torus, then this real representation breaks as a trivial representation (the tangent space to the component) and a bunch of $2$-dimensional real irreps. QED.


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