Conjugation and homology on projective variety 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 $k=1$.

 A: 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). 


*

*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.   

*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.) 

*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. 

*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, $H^2(Y; {\mathbb Q})=0$. 

*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 $X$). 
