Question

Suppose I have a smooth hypersurface $X$ in $\mathbb{P}^n$ which is invariant under a (say finite) group $G$ of projective transformations. What can be said about the action of $G$ on the deformation space $H^1(X,T_X)$? I could imagine that many examples (especially with $dim(X)=1$ or $dim(X)=2$) have been worked out, if this is of any interest at all.

Answer 1

Let's assume that we are working over $\mathbb{C}$.

First of all, hypersurfaces in $\mathbb{P}^n$ are unobstructed, so their first-order deformations always correspond to small deformations (deformations over a disk).

As a general fact, when you consider a smooth variety $X$ with a finite group $G$ acting $holomorphically$ on it, the invariant subspace $H^1(X, T_X)^G$, it parametrizes those first-order deformations that preserve the holomorphic $G$-action. This essentially comes from the fact that, being the action of $G$ holomorphic, if you take $\sigma \in G$, then $\sigma_*$ commutes with $\bar{\partial}$ and the Green operator $\boldsymbol{G}$, so if $\varphi(t)$ solves the Kuranishi equation

$\varphi(t)=t + \frac{1}{2}\bar{\partial}^* \boldsymbol{G}[\varphi(t), \varphi(t)]$

for $t$, then $\sigma_*\varphi(t)$ solves the Kuranishi equation for $\sigma_*t$, and $\sigma_{*} \varphi(t) = \varphi(\sigma_*t)$.

Example Let us consider a quintic Fermat surface $X \subset \mathbb{P}^3$ of equation

$x^5+y^5+z^5+w^5=0$.

It admits a free action of the cyclic group $\mathbb{Z}_5$ given as follows: if $\xi$ is a primitive $5$-th root of unity, then

$\xi \cdot (x,y,z,w)=(x, \xi y, \xi^2 z, \xi^3 w) $.

The quotient $Y := X/\mathbb{Z}_5$ is a Godeaux surface (i.e. a surface of general type with $p_g=q=0, K^2=1$ ) with fundamental group $\mathbb{Z}_5$. M. Reid proved that, conversely, every Godeaux surface with fundamental group $\mathbb{Z}_5$ arises in this way and that, moreover, the corresponding moduli space is generically smooth of dimension $8$. Then in this case we have

$\dim H^1(X, T_X)=40$

$\dim H^1(X, T_X)^G=H^1(Y, T_Y)=8$,

since the number of moduli of quintics keeping the free $G$-action equals the number of moduli of the Godeaux surface $Y$ (well, Horikawa showed that the deformations of quintic surfaces are complicated enough, anyway $40$ is the right number).

Actually, one can say more and check that for every irreducible character $\chi$ of $G$ one has

$\dim H^1(X, T_X)^{\chi} = 8$,

but I do not know any easy interpretation of these eigenspaces in terms of the deformations of the quintic.

Answer 2

Suppose that the action of $G$ on the smooth hypersurface $X$ of degree $d$ in $\mathbb P^n$ comes from a linear action of $G$ on $k^{n+1}$, as in Francesco's example. Assume also that $n \geq 3$, and $d \geq 2$. From the conormal sequence $$ 0 \longrightarrow \mathrm T_X \longrightarrow \mathrm T_{\mathbb P^n}{\mid}_X \longrightarrow \mathcal O_X(d) \longrightarrow 0 $$ we get an exact sequence $$ \mathrm H^0(X, \mathrm T_{\mathbb P^n}{\mid}_X)\longrightarrow \mathrm H^0(X, \mathcal O_X(d)) \longrightarrow \mathrm H^1(X, \mathrm T_X)\ . $$ One can show that the homomorphism $\mathrm H^0(X, \mathcal O_X(d)) \longrightarrow \mathrm H^1(X, \mathrm T_X)$ is surjective, except in the single case $n = 3$, $d = 4$, where there is a 1-dimensional cokernel. Let us exclude this particular case (which can also be treated).

For each $i \geq 0$, let $V_i$ be the space of forms of degree $i$ in $n+1$ variables; there is a natural action of $G$ on $V_i$. In representation theoretic terms, $V_i = \mathop{\rm Sym}^i(k^{n+1})^\vee$. Let $f \in V_d$ an equation for $X$ and $L$ the substspace generated by $f$. From the Euler sequence $$ 0 \longrightarrow \mathcal O_X \longrightarrow \mathcal O_X(1)^{n+1} \longrightarrow \mathrm T_{\mathbb P^n}{\mid}_X \longrightarrow 0 $$ we get a surjection $\mathrm H^0(X, \mathcal O_X(1))^{n+1} \to \mathrm H^0(X, \mathrm T_{\mathbb P^n}{\mid}_X)$; thus $\mathrm H^1(X, T_X)$ can be interpreted as the cokernel of a map $\phi \colon \mathrm H^0(X, \mathcal O_X(1))^{n+1} \to \mathrm H^0(X, \mathcal O_X(d))$. We have $\mathrm H^0(X, \mathcal O_X(1)) = V_1$ and $\mathrm H^0(X, \mathcal O_X(d)) = V_d/L$; furthermore, by unwinding the definitions one can show that the map $\phi \colon V_1^{n+1} \to V_d/L$ sends $(\ell_0, \dots, \ell_n)$ into the class of $\ell_0f_{x_0} + \cdots + \ell_n f_{x_n}$.

So, one can describe $\mathrm H^1(X, \mathrm T_X)$ as the quotient of $V_d$ modulo the subspace generated by $f$ and by the classes of the form $\ell_0f_{x_0} + \cdots + \ell_n f_{x_n}$, where the $\ell_i$ are homogeneous of degree 1. If the characteristic of $k$ does not divide $d$, then from Euler's formula we see that $f$ is of the form $\ell_0f_{x_0} + \cdots + \ell_n f_{x_n}$, so we don't need to add it.

This gives a description of the action of $G$ on $H^1(X, \mathrm T_X)$, which allows to compute it, at least in simple cases (the calculations could become unwieldy, particularly in the non-abelian case).