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.


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


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.

  • $\begingroup$ That is an interesting example, as it both shows that the action may be nontrivial, and gives an interpretation of the fixed part. What would you say is the best way to 'check' the decomposition of $H^1(X,T_X)$ as a $G$-representation? In the given case, I would try and write that as the quotient of $H^0(X,\mathcal{O}(5))$ by $H^0(X,T_{\mathbb{P}^3}$, but this does no seem to depend on the given equation (except for its degree), and I wouldn't know the action on the latter. $\endgroup$
    – bellini
    Aug 18 '10 at 19:26
  • $\begingroup$ Once you know that the invariant part is $8$-dimensional, the fact that also all the other eigenspaces are $8$-dimensional comes almost immediately by symmetry considerations. I never tried the computation you are suggesting, anyway probably the action on the latter group should be not too difficult to understand: write a basis for $H^0(P^3, T_{P^3})$ by using Euler sequence and restrict it to $X$ (here you use the particular form of the equation). $\endgroup$ Aug 18 '10 at 19:59
  • $\begingroup$ The contribution of the eigenspaces, not just for a hypersurface but in general when the variety you're working with is (say) smooth, is that they can be used to defined so-called "natural" deformations to which the group action doesn't extend, at least when $G$ is abelian. See Pardini's paper Abelian Covers (Crelle early nineties) for the definition of natural deformation. $\endgroup$
    – Barbara
    Aug 20 '10 at 13:17

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).

  • $\begingroup$ Thanks, Angelo. Now I see where the equation $f$ comes in. I agree that it looks uninviting, but if no one else comes up with a better idea, this is the way to go. I would guess that calculations like this have been done 150 years ago in invariant theory. $\endgroup$
    – bellini
    Aug 21 '10 at 8:06

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