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edited Dec 14, 2019 at 18:45

user108998

Here´s an attempt at a more topological approach, perhaps someone can tell me where it goes wrong as I´m slightly suspicious of both argument and conclusion, nonetheless I´ll post it in case something can be gained from it.

Let $M$ be a compact smooth oriented manifold and $G$ a finite group acting in an oriented manner on $M$. I claim that if there is a finite set $M^{G}$ of fixed points then $\chi(M)$ is divisible by the order of $G$.

To see this let $\nu$ be a vector field on $M$, chosen sufficiently generically that it has finitely many singularities and none of these lie in $M^{G}$. Now by averaging $\nu$ over $G$ we may assume that $\nu$ is $G$-equivariant, and still has singularity set disjoint from $M^{G}$. Now $G$-equivariance implies that $G$ permutes the singularity set of $\nu$ and for all $g$ in $G$ we should have $Ind_{x}(\nu)=Ind_{gx}(\nu)$ because $G$ preserves orientations, from which the theorem follows from the Hopf Index theorem.

Now for the question, $\sigma$ acts complex analytically, and so preserves orientations. By above we deduce that $\chi(M)$ is even. $\chi(M\setminus{M^{\sigma}})$ is even as $M\setminus{M^{\sigma}}:=U$ has a free action of $C_{2}$. Now one observes that $\chi(U)=\chi_{c}(U)$ as $U$ is an complex variety, and further that $\chi_{c}(U)=\chi(M)-\vert M^{\sigma}\vert$ and concludes.

Edit: the claim is false as abx notes in the comments. I'll leave this in case there's something to be salvaged by the Hopf index approach to the problem, but I'm not optimistic.

Here´s an attempt at a more topological approach, perhaps someone can tell me where it goes wrong as I´m slightly suspicious of both argument and conclusion, nonetheless I´ll post it in case something can be gained from it.

Let $M$ be a compact smooth oriented manifold and $G$ a finite group acting in an oriented manner on $M$. I claim that if there is a finite set $M^{G}$ of fixed points then $\chi(M)$ is divisible by the order of $G$.

To see this let $\nu$ be a vector field on $M$, chosen sufficiently generically that it has finitely many singularities and none of these lie in $M^{G}$. Now by averaging $\nu$ over $G$ we may assume that $\nu$ is $G$-equivariant, and still has singularity set disjoint from $M^{G}$. Now $G$-equivariance implies that $G$ permutes the singularity set of $\nu$ and for all $g$ in $G$ we should have $Ind_{x}(\nu)=Ind_{gx}(\nu)$ because $G$ preserves orientations, from which the theorem follows from the Hopf Index theorem.

Now for the question, $\sigma$ acts complex analytically, and so preserves orientations. By above we deduce that $\chi(M)$ is even. $\chi(M\setminus{M^{\sigma}})$ is even as $M\setminus{M^{\sigma}}:=U$ has a free action of $C_{2}$. Now one observes that $\chi(U)=\chi_{c}(U)$ as $U$ is an complex variety, and further that $\chi_{c}(U)=\chi(M)-\vert M^{\sigma}\vert$ and concludes.

Here´s an attempt at a more topological approach, perhaps someone can tell me where it goes wrong as I´m slightly suspicious of both argument and conclusion, nonetheless I´ll post it in case something can be gained from it.

Let $M$ be a compact smooth oriented manifold and $G$ a finite group acting in an oriented manner on $M$. I claim that if there is a finite set $M^{G}$ of fixed points then $\chi(M)$ is divisible by the order of $G$.

To see this let $\nu$ be a vector field on $M$, chosen sufficiently generically that it has finitely many singularities and none of these lie in $M^{G}$. Now by averaging $\nu$ over $G$ we may assume that $\nu$ is $G$-equivariant, and still has singularity set disjoint from $M^{G}$. Now $G$-equivariance implies that $G$ permutes the singularity set of $\nu$ and for all $g$ in $G$ we should have $Ind_{x}(\nu)=Ind_{gx}(\nu)$ because $G$ preserves orientations, from which the theorem follows from the Hopf Index theorem.

Now for the question, $\sigma$ acts complex analytically, and so preserves orientations. By above we deduce that $\chi(M)$ is even. $\chi(M\setminus{M^{\sigma}})$ is even as $M\setminus{M^{\sigma}}:=U$ has a free action of $C_{2}$. Now one observes that $\chi(U)=\chi_{c}(U)$ as $U$ is an complex variety, and further that $\chi_{c}(U)=\chi(M)-\vert M^{\sigma}\vert$ and concludes.

Edit: the claim is false as abx notes in the comments. I'll leave this in case there's something to be salvaged by the Hopf index approach to the problem, but I'm not optimistic.

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created Dec 14, 2019 at 10:09

user108998

Here´s an attempt at a more topological approach, perhaps someone can tell me where it goes wrong as I´m slightly suspicious of both argument and conclusion, nonetheless I´ll post it in case something can be gained from it.

Let $M$ be a compact smooth oriented manifold and $G$ a finite group acting in an oriented manner on $M$. I claim that if there is a finite set $M^{G}$ of fixed points then $\chi(M)$ is divisible by the order of $G$.

To see this let $\nu$ be a vector field on $M$, chosen sufficiently generically that it has finitely many singularities and none of these lie in $M^{G}$. Now by averaging $\nu$ over $G$ we may assume that $\nu$ is $G$-equivariant, and still has singularity set disjoint from $M^{G}$. Now $G$-equivariance implies that $G$ permutes the singularity set of $\nu$ and for all $g$ in $G$ we should have $Ind_{x}(\nu)=Ind_{gx}(\nu)$ because $G$ preserves orientations, from which the theorem follows from the Hopf Index theorem.

Now for the question, $\sigma$ acts complex analytically, and so preserves orientations. By above we deduce that $\chi(M)$ is even. $\chi(M\setminus{M^{\sigma}})$ is even as $M\setminus{M^{\sigma}}:=U$ has a free action of $C_{2}$. Now one observes that $\chi(U)=\chi_{c}(U)$ as $U$ is an complex variety, and further that $\chi_{c}(U)=\chi(M)-\vert M^{\sigma}\vert$ and concludes.