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Let $X\overset{f}\to S$ be a flat morphism of schemes, and $S=\operatorname{Spec}(k)$ for an algebraically closed field $k=\bar{k}$. I'm just interested, for the moment, with schemes $X$ corresponding to smooth projective curves.

Assume the functors $\operatorname{Div}_{X/S}^d$, and $\operatorname{Pic}_{X/S}^d$ are representable, and let $X_d$ and $W_d$ denote the respective representing schemes (for a definition of the functors see for example "Kleiman - The Picard scheme" at pages 17 and 23, but I don't think it's crucial for the following).

I derived (using known facts) a description of the tangent sheaf of the $d$-component of the Picard scheme $W_d$. However, I don't feel very confident with the technical machinery of algebraic geometry, so my proof is probably not completely correct.

Could you please give it a look?

Any help/comment/suggestion would be incredibly appreciated!

Proposition: Let $\pi:X\times X_d \to X_d$ the natural projection and $u:X_d\to W_d$ the Abel Jacobi map on degree $d$. Then the sheaves over $X_d$ $$ u^* T_\bullet W_d \quad\text{and}\quad R^1 \pi_* \mathcal{O}_{X\times X_d} $$ are isomorphic, and they coincide with the constant sheaf over $X_d$ with fibers $H^1(\mathcal{O}_X)$.

References used in the proof (for basic results you probably already know):

  • $[BLR]$ $=$ Bosch, Lutkebohmert, and Raynaud - Neron Models
  • $[HART]$ $=$ Robin Hartshorne - Algebraic Geometry


First of all we notice that we have isomorphisms $$ T_0 \operatorname{Pic}_{X/S} \cong R^1f_* \mathcal{O}_X \cong H^1(X, \mathcal{O}_X),$$ where the first one is the content of Theorem 1 of $[BLR]$, and the second one follows from the obvious identity of functors $f_*\square \equiv \Gamma(X,\square)$.

Since $\operatorname{Pic}_{X/S}$ is a group variety its tangent sheaf is constant with fibers $T_0\operatorname{Pic}_{X/S} \cong H^1(X, \mathcal{O}_X)$. The same is true if we restrict to the subscheme $W_d$: The tangent sheaf $ T_\bullet W_d $ is the constant sheaf with fibers $H^1(X,\mathcal{O}_X)$ over $W_d$.

Actually, also $R^1 \pi_* \mathcal{O}_{X\times X_d}$ is constant with fibers $H^1(X, \mathcal{O}_X)$. To see this consider the fiber diagram

enter image description here

and apply Proposition 9.3 (page 255) of $[HART]$ to the structure sheaf $\mathcal{O}_X$. We get $$ R^1 \pi_* \mathcal{O}_{X\times X_d} \cong g^* \left( R^1f_*\mathcal{O}_X \right) \cong g^* H^1(X,\mathcal{O}_X), $$ from which we deduce that $R^1 \pi_* \mathcal{O}_{X\times X_d}$ is the constant sheaf on $X_d$ with fibers $H^1(X,\mathcal{O}_X)$.

To conclude it's enough to notice that $u: X_d \to W_d$ is a surjective map of schemes, so from the above description of $T_\bullet W_d$ and $R^1 \pi_* \mathcal{O}_{X\times X_d}$ as constant sheaves it follows that $$ u^* T_\bullet W_d \cong R^1 \pi_* \mathcal{O}_{X\times X_d}. $$

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up vote 0 down vote accepted

I thought more about it, and I'm now sure this is almost correct. The only mistake is in the final part, when I say

notice that $u:X_d \to W_d$ is a surjective map of schemes

In fact, $u$ is not always surjective, and anyway we don't need its surjectivity to conclude that pulling back the constant sheaf $\underline{H^1(\mathcal{O}_X)}$ on $W_d$ we get the constant sheaf $\underline{H^1(\mathcal{O}_X)}$ on $X_d$.

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