# Differential of a specific morphism to a Grassmannian

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

$$\alpha:\mathrm{Div}_{\geq0}(X)\to\mathrm{Pic}(X),\hspace{0.5cm}D\mapsto\mathcal{O}_X(D)$$

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

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?

• 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 Mar 7, 2023 at 22:26
• for any $Q \subset H^0(X,\mathcal{M}) = \mathbb{C}^{r+1}$ the map $\Phi$ can't be a local embedding. Mar 7, 2023 at 22:34
• @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. Mar 8, 2023 at 15:06

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.

• I have added LaTex formatting to your answer; please check if the result of my editing coincides with what you meant to write. Mar 17, 2023 at 8:52
• @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. Mar 17, 2023 at 20:06