Classification of first order deformations of n-pointed non-singular variety

Question

Why is the set of first order deformations equal to $H^1(X,T_{X}(-p_1 -p_2 ... -p_k))$? And what is the motivation for studying this? (I know the proof for unpointed curves, which can be found in a few sources: Hartshorne's deformation theory, Arbarello, Cornalba, Griffiths' Geometry of Algebraic Curves II etc.)

Answer 1

It shouldn't be hard to adapt the proof for unpointed curves. The idea is the following. If $\newcommand \G {\mathcal G} \G$ is a sheaf of groups on $X$, say a topological space, then $H^1(X,\G)$ always classifies isomorphism classes of "locally trivial things over $X$ such that their sheaf of automorphisms over $X$ coincides with $\G$". Example: if $G$ is a finite group, then $H^1(X,G)$ classifies $G$-torsors over $X$. The proof is purely formal and is easiest to carry out in Cech cohomology.

Let's say deformations = 1st order deformations to avoid repeating myself. All deformations of smooth affine varieties are trivial. This means that deformations of any smooth variety are locally trivial, and so can be classified by their sheaf of automorphisms as above. An automorphism of a trivial deformation is the same thing as an infinitesimal automorphism which (on a smooth variety) is the same thing as a vector field (a section of $T_X$). So $H^1(X,T_X)$ classifies isomorphism classes of deformations of $X$.

Now do the same thing for infinitesimal automorphisms that fix the $p_i$. This is the same thing as a vector field which vanishes at each $p_i$, which is the same as a section of $T_X(-p_1-\ldots -p_n)$. Intuitively (differential-geometrically) this is clear: a vector field on a manifold is an infinitesimal automorphism because you can flow along it, and this flow fixes precisely those points where the vector field vanishes. This assertion is enough to finish the proof. So I would look at how Hartshorne identifies automorphisms of $X \times_k k[\varepsilon]$ over $X$ with sections of $T_X = \mathrm{Hom}(\Omega^1,\mathcal O_X) = \mathrm{Der}(\mathcal O_X,\mathcal O_X)$ and modify that proof to take into account that the automorphism should fix a number of points.

Answer 2

$\newcommand{\e}{\epsilon}$ I just wanted to add the following to Dan Petersen's answer.

While it's certainly true that deformations of a smooth variety are locally trivial, a priori one might expect that the additional data of $n$ marked points might allow for locally nontrivial deformations of the variety-with-marking. However, it turns out that the additional data of the marking doesn't add any nontriviality. Intuitively, automorphisms of a trivial first order deformation of $X$ are given by sections of the tangent sheaf. Deformations of a point viewed as a closed subscheme of $X$ are basically just tangent vectors at the point. Thus, it is reasonable to wonder if every deformation of the point can be sent to the trivial deformation by an automorphism.

Mostly for my own benefit, I'll give a proof here:

It will suffice to consider the case where there is just 1 marked point $p$.

Let $k$ be a field, and $B$ a smooth $k$-algebra. Then, every deformation of $B$ over the dual numbers is trivial. Let's view the point $p$ as a section $\text{Spec }k\rightarrow\text{Spec }B$, with ideal $I\subset B$. Thus, we want to show that for every extension of $I$ to an ideal $I'\subset B[\e]$, flat over $k[\e]$, there is an automorphism $\sigma$ of $B[\e]$ which is $k[\e]$-linear and induces the identity on $B[\e]/\e B\cong B$ (so that it's an automorphism of $B[\e]$ as a deformation), such that $\sigma(I') = I[\e] := I\otimes_k k[\e]$.

In Hartshorne's book Deformation theory, Proposition 2.3 shows that the choies of ideals $I'$ described above are in bijection with homomorphisms $\varphi\in Hom_B(I,B/I)$, where given a homomorphism $\varphi$, one defines: $$I' := I_\varphi := \{i + \e b : i\in I, b\in B, \text{and }(b\text{ mod } I) = \varphi(i)\}$$ Here, we view $i+\e b$ as an element of $B'$ via the splitting $B[\e] = B\oplus\e B$.

Next, note that any automorphism $\sigma$ of $B[\e]$ ($k[\e]$-linear, inducing the identity on $B$) can be written as: $$\sigma(x) = x + \e d_\sigma x$$ The basic properties of automorphisms shows that $d_\sigma$ gives a $k[\e]$-linear derivation $B[\e]\rightarrow B$, which actually kills $\e$, and hence we have $$d_\sigma\in\text{Der}_k(B,B)$$ Conversely, the sum of $\text{Id}_{B[\e]}$ with any $k$-linear derivation $B\rightarrow B$ is also an automorphism ($k[\epsilon]$-linear, inducing the identity on $B$).

Now, given an element $\varphi\in Hom_B(I,B/I)$, $\varphi$ clearly factors through $I/I^2 = \Omega_{B/k}\otimes_B (B/I)$ (here using the factor that $p$ is a point, hence etale over $B/I \cong k$), thus $$\varphi\in\text{Hom}_{B/I}(\Omega_{B/k}\otimes_B(B/I),B/I) = \text{Hom}_B(\Omega_{B/k},B/I) = \text{Der}_k(B,B/I)$$ where the first equality comes from adjointness of pushforward and pullback of sheaves of modules.

Let $r : B/I\rightarrow B$ be the composition $B/I\stackrel{\sim}{\rightarrow} k\rightarrow B$.

Then, given such a $\varphi$ now viewed as an element of $\text{Der}_k(B,B/I)$, one can define the automorphism $$\sigma_{\varphi} : B[\e]\rightarrow B[\e] \qquad x\mapsto x - \e (r\circ\varphi)(\overline{x})$$ where $\overline{x}$ is the image of $x$ in $B[\e]/(\e) \cong B$. Since $r\circ\varphi$ is a $k$-linear derivation, this yields an automorphism, and for any $i + \e b\in I_\varphi$, we have: $$\sigma_\varphi(i+\e b) = i+\e b - \e\cdot r(\varphi(i+\e b)) = i+\e b - \e\cdot r(\varphi(i)) = i+\e(b-r(\varphi(i)))$$ However, by definition of $I_\varphi$, we must have $\varphi(i) = (b\text{ mod }I)$, which is to say that $b-r(\varphi(i))\in I$, so $\sigma_\varphi(I_\varphi) \subset I[\e]$. The fact that $\sigma_\varphi$ is an automorphism and $I_\varphi$ was flat over $k[\e]$ implies that $\sigma_\varphi(I_\varphi) = I[\e]$, which is the desired result.

Answer 3

Here's a "quick and dirty" answer, far from being precise. Take the case of curves that you mention: the idea is that , the more points that you add to your moduli problem, the more parameters you need. The dimension of the moduli space gets bigger. In fact, by serre duality your space is iso to $H^0(X,2K_X +p_1+p_2+ ... + p_k)$ and this more positive than the mere def space $H^0(X,2K_X)$ hence it is very likely to have more sections, i.e. more directions in which you can deform your pointed curve. Work out yourself the example of $X=P^1$: it is very instructive. You will see straight away that the rough dimension count of the moduli space corresponds to the dimension of the first order deformation space.