# lift of antiholomorphic involution of Riemann surface to its Jacobian's cohomology

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)

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.

• @Piotr Achinger: The involution is anti-holomorphic.
– abx
Dec 22 '13 at 9:20
• @jj_p: Be careful that $\Omega$ does not extend naturally to the Jacobian. $Jac(\Sigma)$ is $H^{0,1}/H^1(\Sigma,\Bbb{Z})$, and $\Omega$ does not preserve $H^{0,1}$.
– abx
Dec 22 '13 at 9:23
• For a complex torus $T=V/L$, there is a natural identification $H_1(T,\mathbb{Z})=L$, hence $H^1(T,\mathbb{C})=(L\otimes \mathbb{C})^*$. For $T=Jac(\Sigma _g)$, $L=H_1(C,\mathbb{Z})$, hence $H^1(T,\mathbb{C})=H^1(\Sigma _g,\mathbb{C})$.
– abx
Dec 22 '13 at 12:51
• Go through the calculations of the following short paper, depending on genus and number of fixed curves, you will be able to find a classification. V. A. Kransov, On theta characteristics of real algebraic curves, Mathematical Notes, Vol. 64, No. 3, 1998. Dec 23 '13 at 1:47
• In some cases, the involution you get on the Jac has no fixed point, while in some other cases, it has so many components. About you question on $\mathcal{M}_{g,n}$: the involution lifts to an involution on this space. Space of fixed curves is a "submanifold" of half dimension; some of its components correspond to curves with real structure but some others do not. Moreover, two real curves with underlying isomorphic complex structures might not be isomorphic as real curves. Dec 23 '13 at 1:52