# Euler characteristic expression in terms the number of fixed points of an $\mathbb{S}^1$ action

I have found in a paper* that I am reading that

Given $$(M,J)$$ compact (smooth) manifold with an almost complex structure $$J$$, if we have an $$\mathbb{S}^1$$ action with isolated fixed points then $$\chi(M) = |M^{\mathbb{S}^1}|$$ where $$\chi(M)$$ is the Euler characteristic and $$M^{\mathbb{S}^1}$$ is the set of fixed points of the action.

How can this be proven? Can we use Poincarè-Hopf theorem is some way or some kind of Moment map as a Morse function?

*https://arxiv.org/abs/1604.00277 footnote on page 20.

• How does $J$ interact with the action? – Thomas Rot Feb 6 '19 at 11:06
• It wasn't said anything in the footnote where I have found the above claim. In the rest of the paper they use symplectic (even Hamiltonian) actions thus I assume that the action preserves the almost complex structure. – Warlock of Firetop Mountain Feb 6 '19 at 14:03
• It's been a while since I thought about such things. But in the hamiltonian setting I think you can prove a standard form for the action around a fixed point (Theorem of Franks). This allows you then to compute the index of the fixed points and apply the Poincare Hopf theorem. Maybe J suffices to prove the standard form? I think you can also use the Atiyah-Bott-Berline-Vergne localization theorem in equivariant cohomology (applied to the top chern class, which computes the euler characteristic when evaluated against the fundamental class of the manifold). – Thomas Rot Feb 6 '19 at 14:11
• But that may be not helpful depending on your background. But as I said, It has been a while since I thought about such things. – Thomas Rot Feb 6 '19 at 14:17
• Thanks, how would you extend the top Chern class to the Borel construction $X_G$ though? I have only seen a method to extend the symplectic form. – Warlock of Firetop Mountain Feb 6 '19 at 14:41

This is true for any smooth circle action on a closed manifold. There is no need to have an almost complex structure.

In fact, more generally we have the following.

Suppose $$S^1$$ acts smoothly on a closed manifold $$M$$. Then the fixed point set $$M^{S^1}$$ has the property that $$\chi(M) = \chi(M^{S^1})$$.

(So, in particular, the result holds even if the fixed point set is not isolated).

Proof: By averaging an arbitrary Riemannian metric on $$M$$, we may assume the action is isometric. The fixed point set of an isometric action is always a totally geodesic submanifold, but it may have several components. (It can only have finitely many components since it is compact). In particular, asking about $$\chi(M^{S^1})$$ makes sense.

We also point out that in this situation, the number of isotropy groups is finite. (This is true, up to conjugacy, for any compact Lie group action on any closed manifold, as a consequence of the slice theorem).

Let $$N$$ be a component of $$M^{S^1}$$ and let $$\nu N$$ be an embedding of the normal bundle of $$N$$ into some $$\epsilon$$-neighborhood of $$N$$. By shrinking $$\epsilon$$, we may assume that $$\nu N_1 \cap \nu N_2 = \emptyset$$ for disjoint components $$N_1,N_2\subseteq M^{S^1}$$. I will use the notation $$\nu M^{S^1}$$ to denote the the union of the $$\nu N_i$$

Note that the $$S^1$$ action preseves $$\nu M^{S^1}$$ since is is characterized as the set of points a distance $$< \epsilon$$ away from $$M^{S^1}$$ and the action is isometric.

It follows that the $$S^1$$ action also preserves $$M\setminus \nu M^{S^1}$$. Since we've removed the points with isotropy group $$S^1$$, and all other closed subgroups of $$S^1$$ are finite (and there are only finitely many of them) there is a neighborhood $$U$$ of the identity $$1\in S^1$$ with the property that any $$p\in U$$ acts with no fixed points. In particular, the action field ( $$\frac{d}{dt}|_{t=0} e^{it} \cdot m$$, for $$m\in M$$) has no zeros. By Poincare-Hopf, $$M\setminus \nu M^{S^1}$$ has zero Euler characteristic.

In a similar fashion, the $$S^1$$ action preserves the boundary $$\partial \nu M^{S^1}$$ since $$\partial \nu M^{S^1}$$ consists of all points in $$M$$ a distance $$\epsilon$$ from $$N$$. Repeating the argument in the previous paragraph, we deduce $$\partial \nu N$$ also has zero Euler characteristic.

Now we are basically done. Write $$M = (M\setminus \nu M^{S^1}) \cup \nu M^{S^1}$$. Using the fact that each $$\nu N_i$$ deformation retracts to $$N_i$$ (so, in particular, $$\chi(\nu M^{S^1}) = \chi(M^{S^1})$$, we compute \begin{align*} \chi(M) &= \chi(M\setminus \nu M^{S^1}) + \chi(\nu M^{S^1}) - \chi(\partial \nu M^{S^1})\\&= 0 + \chi(\nu M^{S^1}) + 0 \\ &= \chi(M^{S^1}). \end{align*}

• It's been a long time, but I believe I learned this argument from Bredon's Transformation groups book. – Jason DeVito Feb 6 '19 at 15:27
• Nice. What version of the Poincaré-hopf theorem are you using, since $M\setminus \nu M^S^1$ is non-compact? – Thomas Rot Feb 6 '19 at 16:22
• @Thomas: That's a good question! How about this for an alternative argument of this statement: if $S^1$ acts on $X$ with no fixed points and has finitely many finite isotropy groups, then $\chi(X) = 0$. The point is that for any sufficiently large prime $p$, the $p$-th roots of unity $\langle \mu_p\rangle$ act freely on $X$. (We just need $p$ to not divide the orders of the finitely many isotropy groups). This gives a covering $X \rightarrow X/\langle \mu_p\rangle$. Thus, $\chi(X) = p\chi( X/\langle \mu_p\rangle)$ so all sufficiently large primes divide $\chi(X)$. Thus, $\chi(X) = 0$. – Jason DeVito Feb 6 '19 at 16:28
• Thanks Jason, exquisite answer I really appreciate also the key reference. So it seems that we don't need to make it act by isometries right? Because the slice theorem aready provides us with an invariant neighborhood and the action it must send the boundary $\partial \nu F$ to the boundary since it acts by diffeos. – Warlock of Firetop Mountain Feb 6 '19 at 16:55
• @Warlock: We don't need to make it act by isometries, but we can assume it does without loss of generality. This tells us, e.g., that the fixed point sets are embedded submanifolds, etc. Also, I believe the proof of the slice theorem is significantly easier in the isometric case, etc. – Jason DeVito Feb 6 '19 at 18:51