Start from a connected closed Riemann surface $\Sigma\_g,$
obtained as the (symmetric) covering of an open and/or unoriented surface 
$\Sigma,$ namely $\Sigma=\Sigma\_g/\Omega,$ where $\Omega$ is an antiholomorphic involution.


We have on $\Sigma\_g$ $2g$ one cycles $\delta\_i$ and we can choose them such that
$\delta\_i$ intersects positively $\delta\_{i+g}$ and no other
(this gives the so-called A and B-cycles);
we have a basis of $g$ holomorphic one-forms $\omega\_k.$
Construct a $g \times 2g$ matrix of complex numbers


$$ \begin{pmatrix} \int\_{\delta\_1} \omega\_1 & \cdots & \int\_{\delta\_{2g}} \omega\_1 \\
\vdots & & \vdots\\
\int\_{\delta\_1} \omega\_g & \cdots & \int\_{\delta\_{2g}} \omega\_g \end{pmatrix}. $$


One can show (see e.g. Griffiths-Harris p. 228) that the $g$ period vectors
$\Pi\_i=\left( \int\_{\delta\_1} \omega\_i , \ldots , \int\_{\delta\_{2g}} \omega\_i \right)$
are linearly independent over $\mathbb{R},$
and so their linear combinations with integer coefficients
span a lattice
$\Lambda$ inside $\mathbb{C}^{g}.$
Define the Jacobian of $\Sigma\_g$ as $Jac(\Sigma\_g)=\mathbb{C}^{g}/\Lambda.$


My QUESTION is: how does $\Omega$
act on (or lifts to) (the cohomology of) $Jac(\Sigma\_g)$ ?
Namely, if we pick coordinates $z\_i$ over the torus,
does it amount, inside cohomology, to something like $\mathrm{d}z\_i \to \mathrm{d}\overline{z}\_i$ ?
Is there some good reference about this story?


Bonus QUESTION: How does $\Omega$ lift to the moduli space $\mathcal{M}\_{g,n}$ ?


(Note: this has a double on MSE, as soon as one gets an answer, the other is canceled) 



lift of antiholomorphic involution of Riemann surface to its Jacobian's cohomology
The cohomology of $Jac(\Sigma\_g)$ is the exterior algebra of $H^1$, which is canonically isomorphic to $H^1(\Sigma \_g,\Bbb{C})$. So what you are really asking is how $\Omega $ acts on $H^1(\Sigma \_g,\Bbb{C})$. We have the Hodge decomposition $H^1(\Sigma \_g,\Bbb{C})=H^{1,0}\oplus H^{0,1}$, and $\Omega $ maps one summand into the other. I don't think one can say much more.


