2
$\begingroup$

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?

$\endgroup$
1
  • 5
    $\begingroup$ 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

6
$\begingroup$

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.

$\endgroup$

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.