Let $X$ be a differentiable manifold and $G$ a finite group acting differentiably on $X$. The following formula for the Euler number $\text{e}(X/G)$ of the orbit space $X/G$ appears to be well-known: \begin{equation*}\tag{1} \text{e}(X/G) = \dfrac{1}{\lvert G \rvert} \sum_{g \in G} \text{e}(X^g) \end{equation*} where, for each $g \in G$, $X^g$ is the subspace of elements fixed by $g$. Instances of this formula can be found in [3] and [5], but without proof or link to such. A quite elaborate search on the web did not produce one, either.

Following the allusion to Lefschetz numbers in [3], a proof might be given, I suppose, along the following lines. Consider the canonical projection $\pi : X \rightarrow X/G$ and its induced map $\pi^* : \text{H}^*(X/G;\Bbb{Q}) \rightarrow \text{H}^*(X;\Bbb{Q})$ on rational cohomology. According to a celebrated result of Grothendieck in [4], this map is injective and maps $\text{H}^*(X/G;\Bbb{Q})$ isomorphically onto the $G$-invariants $\text{H}^*(X;\Bbb{Q})^G$ (this follows from the Leray SS for the map $\pi$, which collapses). Now for any representation of $G$ on a finite dimensional vector space $V$ over a field of characteristic 0, one has that \begin{equation*} P := \dfrac{1}{\lvert G \rvert} \sum_{g \in G} g : V \rightarrow V \end{equation*} is the projector onto the $G$-invariants $V^G$ and so, by taking traces, \begin{equation*} \dim V^G = \text{tr}\, P = \dfrac{1}{\lvert G \rvert} \sum_{g \in G} \text{tr}\, g. \end{equation*} Therefore, for each $i$, \begin{equation*} \dim \text{H}^i(X/G;\Bbb{Q}) = \dim \text{H}^i(X;\Bbb{Q})^G = \dfrac{1}{\lvert G \rvert} \sum_{g \in G} \text{tr}\, g^i, \end{equation*} where $g^i$ is the endomorphism induced by $g$ on $\text{H}^i(X;\Bbb{Q})$. Taking the alternating sum and interchanging the summation order leaves us with \begin{equation*} \begin{aligned} \text{e}(X/G) &= \sum_i (-1)^i \dim \text{H}^i(X/G;\Bbb{Q}) \\ &= \dfrac{1}{\lvert G \rvert} \sum_{g \in G} \sum_i (-1)^i \text{tr}\, g^i \end{aligned} \end{equation*} and so \begin{equation*}\tag{2} \text{e}(X/G) = \dfrac{1}{\lvert G \rvert} \sum_{g \in G} L(g), \end{equation*} where \begin{equation*} L(g) := \sum_i (-1)^i \text{tr}\, g^i \end{equation*} is the Lefschetz number of the endomorphism $g : X \rightarrow X$.

But now we are in business; (1) boils down to the sixty-four dollar question \begin{equation*}\tag{3} \forall g \in G: L(g) = \text{e}(X^g). \end{equation*} A high-brow reference for this is is the general Lefschetz fixed point theorem Theorem (2.12) of [2] (which is a synthesis of the main theorem Theorem (6.7) of [1] and the Localization Theorem (1.1) of [2]), which, after plodding through the quite intricate intricacies of the $K$-theory apparatus surrounding the Index Theorem, should grind out the desired answer (3), which is, in a sense a localized version of the Gauss-Bonnet Theorem. What makes me unhappy with this is that the derivation of (1) up to stage (3) is maybe on the level of a first year graduate course in Algebraic Topology, wherea the application of the full force of the Index Theorem in (3) appears like the proverbial cracking of a peanut with a sledgehammer. Now it is known that the Gauss-Bonnet Theorem is much more elementary than the Index Theorem (which just illustrates it but is not needed for its proof), and also localization results come in various flavours, so my question is whether there exists a much more elementary proof of (1) somewhere in the world.

[1] Atiyah, M.F. & Singer, I.M., The Index of Elliptic Operators: I, Ann.of Math, Sec.Ser., Vol. 87, No. 3 (May, 1968), pp. 484-530.

[2] Atiyah, M.F. & Segal, G.B., The Index of Elliptic Operators: II, Ann.of Math, Sec.Ser., Vol. 87, No. 3 (May, 1968), pp. 531-545.

[3] Bryan, J. & Fulman, J., Orbifold Euler characteristics and the number of commuting n-tuples in symmetric groups, [math/9712248] on $\mathtt{arxiv.org}$

[4] Grothendieck, A., Sur quelques points d'algèbre homologique, II, Tohoku Math. J. (2) Volume 9, Number 3 (1957), 119-221.

[5] Hirzebruch, F. & Höfer, T., On the Euler number of an orbifold, Math. Ann. 286, 255-260 (1990)


2 Answers 2


I like to think of it as a (kind of) categorification of Burnside's formula. If $X$ is a finite $G$-set, then Burnside's formula says that $$ |X/G|=\frac{1}{G} \sum_{g\in G} |X^g|. $$ If $X$ is (homeomorphic to the geometric realization of) a finite $G$-simplicial set, then you can apply Burnside's formula degree-wise, then take alternating sum to obtain your formula for Euler's number. Here you use that both strict orbits and fixed points commute with geometric realization.

For a general $G$-manifold you can realize it as the geometric realization by taking an equivariant triangulation (those exist at least for smooth $G$-manifolds).

  • $\begingroup$ Just the fundamentals: counting the vertices, edges, faces ... correctly - I like it. Many thanks for the answer. $\endgroup$ Jan 31, 2021 at 15:46

Here is a variant of Greg Arone's fine answer.

Following tom Dieck [p.227 of his book Transformation Groups], call a function $\chi$ from finite $G$-CW complexes to $\mathbb Q$ (or any abelian group) additive, if $\chi(X) = \chi(Y)$ if $X$ and $Y$ are $G$-homotopy equivalent, $\chi(\emptyset) = 0$, and $\chi(X \cup Y) = \chi(X) + \chi(Y) - \chi(X \cap Y)$. Because of how one builds $G$-CW complexes, an immediate lemma is that two additive functions agree on all $X$ if they agree on all the single orbit $G$-sets $G/H$.

Both the left and right side of your putative equation are easily seen to be additive functions. Thus they are equal if they are equal on finite $G$-sets, and this is Burnside's formula.

  • $\begingroup$ I think you can remove the dependency on $X$ being a $G$-CW complex by taking a cofibrant replacement in $G$-spaces for the genuine model structure (which should leave both sides of the equation unchanged) $\endgroup$ Jan 31, 2021 at 9:00
  • 1
    $\begingroup$ You need a finiteness condition for the Euler characteristic to be defined. $\endgroup$ Jan 31, 2021 at 10:34

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