This is a problem that's been bugging me for some time, and therefore I've decided to ask it here. Let $X$ be a smooth projective (irreducible) variety over an algebraically closed field of characteristic 0, let $\mathcal{L},\mathcal{M}\in\mathrm{Pic}(X)$ be two line bundles on $X$, let $Q\subseteq|\mathcal{M}|$ be a linear system of dimension $n\geq1$, let $\mathrm{Div}_{\geq0}(X)$ be the scheme of effective divisors on $X$, let $\mathrm{Aut}^0(X)$ denote the component of the automorphism group scheme of $X$ that contains the identity, and consider the maps


$$\beta:\mathrm{Aut}^0(X)\to\mathrm{Pic}(X),\hspace{0.5cm} \theta\mapsto \mathcal{L}\otimes\theta^*\mathcal{M}^{-1}.$$

Now consider the map (where the fiber product incorporates the above maps)

$$\Phi:\mathrm{Div}_{\geq0}(X)\times _{\mathrm{Pic}(X)}\mathrm{Aut}^0(X)\to\mathbb{G}(n,|\mathcal{L}|)$$ $$(D, \theta)\mapsto D+\theta^*Q$$

Question: Assuming the fiber product is not empty, what is the differential of $\Phi$ at a point?

  • $\begingroup$ I think it would be helpful you'd me slightly more specific about the example you hae in mind. In general, there is no chance that this map is a local ismorphism. Take $X = \mathbb{P}^r$, $\mathcal{M} = \mathcal{L} = \mathcal{O}_{\mathbb{P}^r}(1)$. We have (up to a finite quotient whose kernel acts trivially on $X$) $Aut^0(X) = \mathrm{SL}_{r+1}$ which is of dimension $(r+1)^2-1 = r^2+2r$. Since $\mathcal{M}$ is $\mathrm{SL}_{r+1}$-equivariant, one checks that $\{0\} \times \mathrm{SL}_{r+1} \subset \mathrm{Div}_{\geq 0}(X) \times_{\mathrm{Pic}(X)} Aut^0(X)$. An easy dimension count shows that $\endgroup$
    – Libli
    Mar 7, 2023 at 22:26
  • $\begingroup$ for any $Q \subset H^0(X,\mathcal{M}) = \mathbb{C}^{r+1}$ the map $\Phi$ can't be a local embedding. $\endgroup$
    – Libli
    Mar 7, 2023 at 22:34
  • $\begingroup$ @Libli Thanks for your nice example. The specific situation I have is the following: let $G\leq\mathrm{Aut}(X)$ be a finite group such that $X/G\simeq\mathbb{P}^d$, and let $Q$ be the linear system $\pi^*|\mathcal{O}(1)|$ where $\pi:X\to\mathbb{P}^d$ is a quotient map. $\endgroup$
    – rfauffar
    Mar 8, 2023 at 15:06

1 Answer 1


The differential of $\Phi$ at a point $(D,\theta)$ in the fiber product is given by the induced map on tangent spaces:

$$T(\mathrm{Div} _{\ge 0}(X)) \times T(\mathrm{Pic} (X)) \times T(\mathrm{Aut} ^0 (X)) → T(\mathbb G (n,|\mathcal L|))$$

where $T$ denotes the tangent space.

To compute this differential, we can use the fact that the tangent space of $\mathrm{Div} _{\ge 0}(X)$ at a point $D$ is isomorphic to $H^0(X,\mathcal O _D)$, the space of global sections of the structure sheaf of $X$ twisted by $D$. Similarly, the tangent space of $\mathrm {Pic} (X)$ at a point $\mathcal L$ is isomorphic to $H^1 (X, \mathcal O _{\mathcal L})$, the space of global sections of the sheaf of holomorphic sections of $\mathcal L$. Finally, the tangent space of $\mathrm{Aut} ^0 (X)$ at the identity is isomorphic to $H^0 (X,TX)$, the space of vector fields on $X$.

Using these identifications, we can write down the differential of $\alpha$ and $\beta$, and the map induced by $Q \subseteq |\mathcal M|$ on the space of global sections of $\mathcal L$. The differential of $\alpha$ is given by the map

$$T(\mathrm{Div} _{\ge 0}(X)) \ni f \mapsto \mathcal O _X (f+D) \otimes \mathcal O _X (D)^{-1} \in T(\mathrm{Pic} (X))$$

while the differential of $\beta$ is given by the map

$$T(\mathrm {Aut} ^0 (X)) \ni V \mapsto \mathcal L \otimes V^* \mathcal M ^{-1} \otimes \mathcal M \otimes {\mathcal L} ^{-1} \in T(\mathrm{Pic} (X)) \ . $$

To compute the differential of the map induced by $Q$, we need to consider the linear system $Q$ as a subspace of $H^0 (X, \mathcal M)$. Let $s_1, \dots, s_n$ be a basis for $Q$, and let $t_1, \dots, t_m$ be a basis for $H^0 (X, \mathcal M)$ such that $s_1, \dots, s_n, t_1, \dots, t_m$ is a basis for $H^0 (X, \mathcal M)$. Then the map induced by $Q$ is given by the matrix

$$[ s_1, \dots, s_n, t_1, \dots, t_m ] \ .$$

To compute the differential of $\Phi$, we use the product rule for differentials:

$$\mathrm d \Phi (D, \theta) = \mathrm d (D + \theta^* Q) + (\mathrm d \alpha(D), \mathrm d \beta(\theta)) + (\mathrm d (Q),0) \ ,$$

where $\mathrm d (Q)$ is the differential of the map induced by $Q$. The first term on the right-hand side is just the differential of the translation map $T_{\mathrm {Div} _{\ge 0} (X)}$ given by $\mathrm d (D + \theta \wedge Q) = \mathrm d D + \theta \wedge \mathrm d Q$, while the second term is the product of the differentials of $\alpha$ and $\beta$. The last term is zero, since the differential of $Q$ does not depend on $\theta$.

Putting everything together, we obtain the differential of $\Phi$ at a point $(D, \theta)$ in the fiber product.

  • $\begingroup$ I have added LaTex formatting to your answer; please check if the result of my editing coincides with what you meant to write. $\endgroup$
    – Alex M.
    Mar 17, 2023 at 8:52
  • $\begingroup$ @Rina Thanks for your answer. There are a few details that I think are left out and I would like to understand better, so I'll have to think about it a bit. $\endgroup$
    – rfauffar
    Mar 17, 2023 at 20:06

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