# Tangent Space of the Hodge bundle on the moduli space of curves

Let $$k$$ be an algebraically closed field and $$\mathcal M_g$$ denote the moduli space (stack) of smooth curves of genus $$g$$ over $$k$$. Using the universal curve $$\pi \colon \mathcal C_g \to \mathcal M_g$$, there is a natural vector bundle $$\pi_* \Omega_{\mathcal C_g / \mathcal M_g}$$ of rank $$g$$ called the Hodge bundle. The fiber over a point $$X \in \mathcal M_g$$ is $$H^0(X,\Omega_X)$$. Let $$\Omega \mathcal M_g$$ denote the total space of this bundle. I would like to understand the tangent space of $$\Omega \mathcal M_g$$ at a point $$(X, \omega)$$, where $$X$$ is a smooth curve and $$\omega$$ is a global section of $$\Omega_X$$ (for simplicity let us assume that $$g>1$$).

By definition a tangent vector at $$(X, \omega)$$ is a morphism from the dual numbers (i.e. $$\varepsilon^2=0$$) $$\operatorname{Spec}k[\varepsilon] \to \Omega \mathcal M_g$$ with image $$(X,\omega)$$. Since $$\Omega \mathcal M_g$$ is a vector bundle of rank $$g$$ this map is the same thing as two maps $$a \colon \operatorname{Spec}k[\varepsilon] \to \mathcal M_g$$ and $$b \colon \operatorname{Spec}k[\varepsilon] \to \mathbb A^g_k$$ with image $$X$$ and $$\omega$$ respectively. By the universal property of $$\mathcal M_g$$ the map $$a$$ corresponds to a first order deformation of $$X$$ and the map $$b$$ to a tangent vector of $$\mathbb A^g_k$$ at $$\omega$$, i.e. just a different point of $$\mathbb A^g_k$$ or another differential on $$X$$. I would like to arrive at the following description (that I will sketch below) of the tangent space that is used for example in Hubbard, John; Masur, Howard, Quadratic differentials and foliations, Acta Math. 142, 221-274 (1979). ZBL0415.30038. before proposition 4.5 (for quadratic differentials) or in Möller, Martin, Linear manifolds in the moduli space of one-forms, Duke Math. J. 144, No. 3, 447-487 (2008). ZBL1148.32007. in the proof of theorem 2.1.


I can see that this gives a vector space of dimension $$4g-3$$ and is therefore isomorphic to the tangent space at $$(X,\omega)$$ by an isomorphism that does not change the deformation of the curve $$X$$, i.e. the map $$a$$. Moreover given one deformation $$(\mathcal X, \tilde \omega)$$ of $$(X, \omega)$$ all other deformations of $$(X, \omega)$$ only differ by an element of $$H^0(X, \Omega_X)$$. But I can not make this last map canonical. I can not figure out what the deformations of $$(X, \omega)$$ are, that do not change the differential $$\omega$$.

• Yes, "deformations of $(X, \omega)$ that do not change $\omega$" do not make sense, especially since the Hodge bundle is not a flat bundle. Rather, there is a (non canonically split) short exact sequence $$0\to H^0(X, \Omega_X)\to T_{(X, \omega)} \Omega \mathcal{M}_g\to T_X \mathcal{M}_g\to 0.$$ – Piotr Achinger Apr 20 at 9:11

The universal property of the total space $$\mathbf{V}(E^\vee) = \operatorname{Spec}_M \operatorname{Sym} E^\vee$$ of a vector bundle (locally free sheaf) $$E$$ on some scheme $$M$$ is: giving a map $$T\to \mathbf{V}(E^\vee)$$ corresponds to giving a map $$f\colon T\to M$$ and a section $$\omega$$ of $$f^* E$$.
We apply this to $$T=\operatorname{Spec} k[\varepsilon]$$, $$M=\mathcal{M}_g$$ and $$E$$ the Hodge bundle, so that $$\mathbf{V}(E^\vee) = \Omega\mathcal{M}_g$$. Then a tangent vector at $$(X, \omega)$$ is the same as a first order deformation $$\tilde X = f^* \mathcal{C}_g$$ over $$T$$ (corresponding to an $$f\colon T\to \mathcal{M}_g$$, so that $$\tilde X$$ is the pullback of the universal family), together with a section $$\tilde\omega$$ of $$f^* E = f^* \pi_* \Omega_{\mathcal{C}_g/\mathcal{M}_g} = p_* \tilde f^* \Omega_{\mathcal{C}_g/\mathcal{M}_g} = p_* \Omega_{\tilde X/T}.$$ Here $$\tilde f\colon \tilde X = \mathcal{C}_g\times_{\mathcal{M}_g} T\to \mathcal{C}_g$$ and $$p\colon \tilde X\to T$$ are the projections. (For the first equality, one needs to check that the formation of $$\pi_* \Omega_{C/S}$$ for a smooth projective family of curves $$C\to S$$ commutes with base change along $$S'\to S$$.)
• @FabianRuoff yes, you trivialized the Hodge bundle (or its pullback to $T$), but there is no canonical way to do so. – Piotr Achinger Apr 20 at 12:26