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Let $X = \mathrm{Proj}\ \mathbb{C}[T_0,\dots,T_n]/I$ be a projective variety (Edit: I am mainly interested in the case of a curve), where $I$ is generated by polynomials with real coefficients. Then conjugation acts on $X$ (by conjugating coefficients of polynomials).

Question $(1)$: I am quite sure that conjugation is an automorphism of $X$ (automorphism of algebraic variety), is it right? Edit: This was answered by tjf below.

If it is an automorphism of $X$, then it also acts on the homology groups $H_k(X,\mathbb{Q})$ of $X$ by automorphisms. We can define in the obvious way invariant and alternating $k-$cycles: in the first case $\gamma = \overline{\gamma}$ and in the second case $\gamma = -\overline{\gamma}$. Not all $k-$cycles need to have one of this two forms, but we have that

$$ H_k(X,\mathbb{Q}) = H_k(X,\mathbb{Q})^+ \oplus H_k(X,\mathbb{Q})^-,$$

where $+$ and $-$ denote the fact of being invariant or alternating. The above is true because each $k-$cycle $\gamma$ can be decomposed uniquely as

$$ \gamma = \frac{\gamma + \overline{\gamma}}{2} + \frac{\gamma - \overline{\gamma}}{2},$$

a sum of an invariant and an alternating $k-$cycle.

Question $(2)$: are $H_k(X,\mathbb{Q})^+$ and $H_k(X,\mathbb{Q})^-$ isomorphic, or do they at least have the same dimension? I think this is some form of orbit-stabilizer, but I cannot formulate it properly. Edit: As Nicolas Hemelsoet points out, this is in general not true. So I would just like an answer for $X$ a curve and $k=1$.

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  • $\begingroup$ There are some obvious restrictions, e.g many times $H_k$ can be odd dimensional. So I guess your question is when $H_k$ is even dimensional, and it should be possible (I'm not super sure) to construct explicitely some counter-example using Hodge theory. Would you be interested ? $\endgroup$ – Nicolas Hemelsoet Feb 28 '18 at 10:31
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    $\begingroup$ Regarding Question (1), you must be careful. Conjugation is a scheme automorphism of $X$ but NOT an automorphism of algebraic variety. When the term 'algebraic variety' is used, there is an implicit base field $k$ over which this variety is defined; in this case it is $k=\mathbb{C}$. By definition, a morphism of algebraic varieties (over $k$) is a $k$-morphism. This is not the case for the complex conjugation. $\endgroup$ – tjf Feb 28 '18 at 12:21
  • $\begingroup$ @Nicolas Hemelsoet I have formulated a more general question than what I actually needed: I am just interested in the case where $X$ is a curve (Riemann surface) and $k=1$ (loops on $X$), so that $\mathrm{dim}_\mathbb{Q}(H_1(X,\mathbb{Q})) = 2g$, where $g$ is the genus of $X$. I do not know if I have enough background to understand counter-examples with Hodge theory, but if you have something for the special case I am interested in, I am all ears! $\endgroup$ – 57Jimmy Feb 28 '18 at 20:23
  • $\begingroup$ @tjf Thanks for the reminder, I had forgotten it! Still, I just need it to be a homeomorphism in order to induce an automorphism in homology, so all is well that ends well :) $\endgroup$ – 57Jimmy Feb 28 '18 at 20:23
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    $\begingroup$ If your curve is smooth then its real locus is a collection or circles so has Euler characteristic zero; thus you can apply Lefschetz fixed point formula to conclude that the dimension of subspace of H_1 on which conjugation is identity is half of the total dimension of H_1. If your curve is not smooth H_1 does not have to be even dimensional so the statement is false. $\endgroup$ – Max Mar 3 '18 at 9:11
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Here is one proof.

Lemma. Let $X$ be a cell complex, $G$ a finite group and $G\times X\to X$ is an action. Then for every field $F$ of characteristic zero, $$ H^*(X/G; F)\cong H^*(X;F)^{G}, $$
where the right hand-side is the ring of invariants under the $G$-action on $H^*(X)$.

This lemma is an application of the "transfer", you can find its proof for insatnce in Bredon's book "Compact Transformation Groups" or in these freely available notes by Alan Edmonds. (Edmonds treats only the case $F={\mathbb Q}$ but the general case is no different.)

Proposition. Let $G={\mathbb Z}_2$ acting as complex conjugation on an irreducible smooth complex-projective curve $X$ defined over ${\mathbb R}$. (More generally, one can allow a singular irreducible curve $X$ such that none of its singular points is real.) Then the dimension of the $G$-invariant subspace in $H^1(X; {\mathbb Q})$ is half of the dimension of $H^1(X; {\mathbb Q})$.

Proof. Let $Y:= X/G$. I first consider the case when $G$ has nonempty fixed-point set in $X$. Then $X$ (as a topological space) is obtained by gluing two copies of $Y$ along a disjoint union of circles $F\subset Y$ (the projection of the fixed-point set of the action of $G$ on $X$). Then, since $\chi(F)=0$,
$$ \chi(X)=2\chi(Y). $$ Since $H^2(Y; {\mathbb Q})=0$, we obtain that $$\chi(X)= 2- dim H^1(X; {\mathbb Q})= 2- 2 dim H^1(Y; {\mathbb Q})$$ and, hence (by Lemma), $$ dim H^1(X; {\mathbb Q})= 2 dim H^1(Y; {\mathbb Q})= 2 dim H^1(X; {\mathbb Q})^G. $$ If $G$ acts freely on $X$, the proof is essentially the same. Since the Euler characteristic is multiplicative under covering maps and taking into account that $Y=X/G$ is nonorientable and, hence, has $H^2(Y; {\mathbb Q})=0$, again obtain $$ 2- dim H^1(X; {\mathbb Q})= \chi(X)= 2\chi(Y)= 2- 2 dim H^1(Y; {\mathbb Q}). $$ Then, proceed as before. qed

If $X$ has odd number of singular points, then $dim H^1(X; {\mathbb Q})$ is odd and, hence, the proposition cannot be true. I leave it to somebody else to sort out the case of general complex-projective curves.

Edit. Here are answers to your questions. Almost all of this one learns in a graduate course in algebraic or differential topology (plus a complex analysis class).

  1. First of all, this is a general fact of differential topology that if $G$ is a compact group acting smoothly on a manifold $M$ then the fixed-point set $Fix_M(G)$ of the action is a smooth submanifold (this was discussed several times at MSE, for instance here). If $M$ is compact then $Fix_M(G)$ is also compact. (The latter is actually a fact of general topology: a closed subset of a compact space is compact.) In the case when $M$ is a Riemann surface $X$ and $G$ is cyclic orientation-reversing (antiholomorphic) then by linearizing the action at its fixed-points you see that the generator of $g$ (in holomorphic coordinates near any fixed point) has the form $z\mapsto \bar{z}$. Then $Fix_M(G)$ is a 1-dimensional manifold. By the classification of 1-dimensional manifolds, every connected 1-dimensional manifold is either empty or is a circle. Every compact 1-dimensional manifold is a finite union of circles.

  2. If $Fix_X(G)$ is nonempty and $X$ is connected (which is always the case if $X$ is an irreducible smooth complex projective curve) then $Y=X/G$ is a connected manifold with nonempty boundary (projection of the fixed point set of $G$). Hence, $H^2(Y; {\mathbb Q})=0$. You can see this, for instance, by observing that $Y$ is homotopy-equivalent to a bouquet of circles. (One can derive this for instance from the classification of compact surfaces with boundary.)

  3. If $G$ acts freely then $Y=X/G$ is a connected manifold. (This is a general fact about properly discontinuous free group actions on manifolds: The quotient is always a manifold and the quotient map is a covering.) The fact that the image of a connected space under a continuous map is connected is a fact of general topology which one typically learns in a general topology class. In particular, $X/G$ cannot be a disjoint union of two homeomorphic copies of anything.

  4. If $X$ is an orientable connected manifold and $X\to X/G$ is a quotient by a free properly discontinuous action which does not preserve orientation then $Y=X/G$ is a nonorientable connected manifold. (This was discussed many times at MSE, e.g. here, here, here,....) In particular, $H^2(Y; {\mathbb Q})=0$.

  5. Lastly, consider the complex manifold $M={\mathbb C}P^n$ and let $g: M\to M$ be the complex conjugation $$ (z_0: z_1:...:z_n)\mapsto (\bar{z}_0: \bar{z}_1:...:\bar{z}_n). $$ Then for every complex line $L$ in $T_pM$ the map $dg_p: L\to dg_p(L)\subset T_pM$ is orientation-reversing. This is a fact of linear algebra. (Consider the lift of $g$ to ${\mathbb C}^{n+1}$.) In particular, if $X\subset M$ is a Riemann surface and $g(X)= X$, then $g: X\to X$ reverses orientation on $X$ (the orientation is induced by the complex structure of $X$).

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  • $\begingroup$ Thank you very much for your answer! The smooth case is actually enough for me. I still have a couple of questions, being rather new in this field: $\endgroup$ – 57Jimmy Mar 5 '18 at 19:16
  • $\begingroup$ 1) Why is $F$ a disjoint union of circles? 2) Why in the second case don't we get a disjoint union of two copies of $Y$ with the same reasoning as in the first case? 3) Why is $H^2(Y,\mathbb{Q})=0$? 4) Why is $Y$ non-orientable in the second case? $\endgroup$ – 57Jimmy Mar 5 '18 at 19:29
  • $\begingroup$ P.S. concise answers are enough, I just need to know if they are standard facts of some sort that I can look up somewhere or if they are specific to this problem $\endgroup$ – 57Jimmy Mar 5 '18 at 19:37
  • $\begingroup$ @57Jimmy All this is very standard. For instance if you have a connected surface with nonempty boundary then it has zero 2nd cohomology; same for nonorientable connected surfaces. Complex conjugation reverses orientation, hence quotient by a free action given by complex conjugation will need nonorientable. $\endgroup$ – Moishe Kohan Mar 5 '18 at 21:53
  • $\begingroup$ Thanks again a lot for your answer. Please allow me one last question: you say that in the case where $G$ acts freely, we still have $H_2(X/G,\mathbb{Q})=0$. This seems to be in contrast to a statement in the notes of Edmonds you have linked, just after Corollary 9.2: "If $G$ acts freely on a closed oriented $n$-manifold $M$, then $H_n(M, \mathbb{Z}) = \mathbb{Z} = H_n(M/G, \mathbb{Z})$". What am I missing? $\endgroup$ – 57Jimmy Mar 9 '18 at 10:13

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