2
$\begingroup$

Let $f : Y \longrightarrow X := \mathbb{P}^1\setminus\{0,1,\infty\}$ be the smooth proper morphism associated to the Legendre family, which is an elliptic fibration of the punctured line, with fibre above $t$ being the elliptic curve $E_t := y^2 = x(x-1)(x-t)$.

Let $\mathbb{H} := R^1f_*\mathbb{Q}$ be the derived pushforward of the constant sheaf $\mathbb{Q}$ on $Y$, its stalk at a point $t$ is $H^1_{dR}(E_t, \mathbb{Q})$.

Let $x_0\in X$, we may consider the (topological) monodromy representation $\rho: \pi_1(X,x_0)\longrightarrow GL(\mathbb{H}_{x_0})$. Apriori the image of $\rho$ is a subset of $GL_2(\mathbb{Q})$, however, it is well known that the Zariski closure is $SL_2(\mathbb{Q})$.

Denote by $\mathcal{H}$ the $C^{\infty}$-local system on $X$, given by $\mathcal{H} := C^{\infty}(X)\otimes \mathbb{H}$, where $C^{\infty}(X)$ is the sheaf of smooth functions on $X$. Consider the de-Rham cohomology on $X$ with coefficients in $\mathcal{H}$, i.e. $H^1_{dR}(X,\mathcal{H})$. I read somewhere that there is an action of the monodromy group, $SL_2$, on $H^1_{dR}(X,\mathcal{H})$. I imagine this should be standard. How is this action defined?

The reason I find this confusing, is that the cocycle classes in $H^1_{dR}(X,\mathcal{H})$ correspond to smooth 1-forms on $X$, with coefficients in $\mathcal{H}$, which are closed and $\pi_1(X,x_0)$-invariant. I.e. a 1-cocycle has the form $\nu(z)dz\otimes \omega$, where $\nu(z)$ is holomorphic on $X$, and $\nu(z)dz\otimes \omega$ is a global section of $\Omega^1_X\otimes \mathcal{H}$, or alternatively $\overline{\nu(z)dz}\otimes \omega$, with $\overline{\nu(z)}$ antiholomorphic.

Then it is clear how $\pi_1(X,x_0)$ acts on $\Omega_X^1$. It also acts on $\mathcal{H}$ via $\rho$. However, it isn't clear how $SL_2$ acts on $\Omega^1_X$, and if it doesn't, and its action on a 1-cocycle $\nu(z)dz\otimes \omega$ is only given by its action on $\mathbb{H}$, then it isn't clear anymore that the result remains a global section.

Any clarification would be appreciated! Thanks!

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.