Why does the fixed point locus of a circle action always have even codimension

Question

So I'm studying equivariant cohomology, and I am in particular interested in the case where there is a $U(1)$ action on a Riemannian manifold $\mathcal{M}$. This is generated by some vector field $\xi\in\mathfrak{X}(\mathcal{M})$. The fixed points of this action are given by the set of points such that $\langle\xi,\xi\rangle=0$, where $\langle,\rangle$ is the inner product defined by the metric on $\mathcal M$. The claim is that this fixed point set is always of even codimension, but I do not understand why that must be the case.

Answer 1

This follows from the representations of the circle.

Since you are talking about the codimension of the fixed point set, I am going to assume that you already know that the fixed point set is an (embedded) submanifold of $M$, we will denote this submanifold by $M^{U(1)}$ [To prove this one constructs a $G$-invariant metric, and shows that the fixed points in a neighborhood of $p \in M^{U(1)}$ corresponds to a linear subspace of $T_pM$ under the exponential map arising out of the invariant metric.] So we just need to show that locally, in an adapted chart, the fixed point set is a linear subspace of even codimension.

First lets recall if a $G$-action on a manifold $M$ has a fixed point $p$, then this gives a representation of $G$ in $T_pM$ as follows: let $l_g: M \to M$ denote the action of $g\in G$ on $M$, $x \mapsto g\cdot x$. Then the derivative of $l_g$ at $p$, gives a linear map $(dl_g)_p: T_pM \to T_pM$. Since $l_g$ is a diffeomorphism, $(dl_g)_{p}$ is an invertible linear map. So, we have a representation, $G \to \operatorname{GL}(T_pM)$ given by $g \mapsto (dl_g)_{p}$.

Now recall the following fact about the representations of circle:

An irreducible real representation of $U(1)$ is either the trivial $1$-dimensional representation or a $2$-dimensional representation of the form $S^1 \to \operatorname{GL}(2, \mathbb{R})$, given by $$ e^{it} \mapsto \begin{bmatrix} \cos(mt) & -\sin(mt) \\ \sin(mt) & \cos(mt) \end{bmatrix} $$ where $m \in \Bbb Z_+$.

This means for any $p \in M^{U(1)}$, we can decompose representation $T_pM$ into irreducible components: $$ T_pM = V_1 \oplus \cdots \oplus V_k \oplus V'_1 \oplus \cdots \oplus V'_{n-2k}, $$ where each of the $V_i$ is a non-trivial two-dimensional representation and each of the $V_i'$ is a trivial one-dimensional representation. In other words, the subspace on which $U(1)$ acts trivially has even codimension $2k$.

Now suppose $p \in M^{U(1)}$ is a fixed point, then we can choose a small neighborhood $U$ in $M$ around $p$, such that the exponential map $T_pM \to U$ is a diffeomorphism. The map $l_g$ on the fixed point set and in particular on $U \cap M^{U(1)}$ is $l_gx = x$, so $(dl_g|_{U \cap M^{U(1)}})_p = \text{id}$. Therefore, under the exponential map, the points in $T_pM$ that correspond to the fixed points in $U$, are precisely $V'_1 \oplus \cdots \oplus V'_{n-2k}$, which has codimension $2k$.