Suppose $M$ is a compact oriented smooth manifold and $G$ is a finite group acting on it. Then it is well-known, although I have yet to find a proof or derivation of it, that the (normal topological) Euler characteristic of the orbifold $M/G$ is $\chi(M/G) = \frac1{|G|}\sum_i\sum_{g\in G} (-1)^i\mathrm{Tr}_{H^i(M)}(g^*)$, where $H^i(M)$ is the $i$-th (De Rham) cohomology group and $g^*$ is the map on it induced by $g$.

Everywhere where I have looked, this is then said to equal $\frac1{|G|}\sum_{g\in G}\chi(M^g)$ (where $M^g$ is the set of points that $g$ leaves fixed) because of the Lefschetz formula. Not being familiar with this formula, I've looked up several versions of it. Especially the one for compact oriented manifolds seems very useful, but it (along with a number of other versions) holds only when all the fixed points are isolated. In particular, the set of fixed points should be countable. However, I have seen this formula being used used in situations in which this does not hold. For example, let the permuation group $S_n$ act on $M\times\dots\times M = M^n$ by permuting the factors. The set of fixed points of this action is certainly not countable.

So how does this work? Is this Lefschetz formula applicable to this situation after all, or is there another usable version of it that should be used here? Also, is there perhaps a book or arXiv document that shows how to calculate the first expression of the Euler characteristic?


As far as I understand your question, you want to see a derivation of the formula for $\chi(M/G)$. Here it is:

  1. The difficult part of the argument os to show that there is an isomorphism $H^* (M/G; \mathbb{Q}) \cong H^* (M; \mathbb{Q})^G$ (the $G$-invariants). It is induced from the quotient map $M \to M/G$, but that is not so important. In the following, all cohomology have rational coefficients.

  2. If that is done, the argument is easy. By elementary representation theory of finite groups:

$$ dim (H^i (M)^G) = \frac{1}{|G|} \sum_{g \in G} Tr_{H^i (M)} (g). $$

  1. We show the difficult part in two steps. Consider the Borel construction $EG \times_G M$. There is a fibre bundle $\pi:EG \times_G M \to BG$ with fibre $M$ and a map $f: EG \times_G M \to M/G$.

  2. The Leray-Serre spectral sequence of $\pi$ begins with $E_{2}^{pq}=H^p (G;H^q (M))$. If $p > 0$, this group is zero and hence $H^n(EG \times_G M) \cong H^0 (G, H^n(M))=H^n (M)^G$.

  3. $f$ induces an isomorphism in cohomology: Let $x \in M$ and let $H$ be the stabilizer subgroup at $x$. Pick a $G$-equivariant Riemann metric on $M$. Consider $V=\bigcup_{g \in G} B_{\epsilon}(x)$. If $\epsilon$ is small enough, then $V$ is a disc bundle a $G$-equivariant vector bundle on $G/H$. Clearly, $EG \times_G V \simeq EG \times_G G/H \simeq BH$. Moreover, $V/G\simeq \mathbb{R}/H$ is contractible. The consequence of this discussion is that there exists a finite cover of $M/G$ by open contractible sets, such that all intersections are again contractible or empty. Also, the preimages of the covering sets and their intersections have trivial rational cohomology, because the cohomology of $BH$ is trivial. By the Mayer-Vietoris sequence, induction on the number of covering sets and repeated application of the 5-lemma, it follows that $f^* :H^* (M/G) \to H^* (EG \times_G M)$ is an isomorphism in rational cohomology.


Just to add to Johannes answer, when you have a map $f:M\to M$ and you want to compute $\sum_i (-1)^i \textrm{Tr}_{H_i(M)}(f_\ast)$, one may proceed as follows : imagine $i$-cycles in $M$, i.e., elements in $H_i(M)$, arising as either $i$-cycles in $M^f$ which are fixed by $f_\ast$ or they are not fixed by $f_\ast$. One can do this rigorously by choosing an appropriate triangulation of $M$, homotoping $f$ to be a simplicial map and choosing a basis for $H_i(M)$ for all $i$ simultaneously. It then follows that in this basis, the trace picks up $1$ for each element in the basis of $H^i(M^f)$, which when summed up with alternating signs produces $\chi(M^f)$. The classical Lefschetz fixed point formula can be proven in exactly the same way.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.