## First-Order Deformations

Today we actually get to some deformations. Let ${X_0}$ be a scheme of finite type over ${k}$. First, we’ll be working with the “ring of dual numbers” a lot, so we’ll just define it to be ${R=k[\epsilon]/(\epsilon^2)}$. Let’s recall a few useful properties first.

To give a map in ${k}$-schemes: Spec ${R\rightarrow X_0}$ is equivalent specifying a ${k}$-point and an element of the Zariski tangent space at that point.

Flatness is another important concept. An ${R}$-module, ${M}$, is flat if and only if the map ${M/\epsilon M\rightarrow M}$ by multiplication by ${\epsilon}$ is an injective. The proof is quite straightforward: Consider the exact sequence ${0\rightarrow k\stackrel{\epsilon}{\rightarrow} R\rightarrow k \rightarrow 0}$. If ${M}$ flat, tensor this with ${M}$ and you get ${0\rightarrow M/\epsilon M\stackrel{\epsilon}{\rightarrow} M\rightarrow M/\epsilon M\rightarrow 0}$, and hence injectivity. If the map is injective, then ${Tor^1(M,k)=0}$, so ${M}$ is flat.

Let’s move on now that those are out of the way. What exactly should these deformations be? Let’s say we have a nice family of schemes. This means that there is a map ${f:X\rightarrow T}$, and nice in our case means flat. ${T}$ parametrizes this family, since the fiber over any point gives a scheme ${X_t}$. There is a special fiber ${X_0}$, and deformations of ${X_0}$ are the schemes that occur in this family in a neighborhood of ${X_0}$. (Recall, this is still a “touchy-feely” idea of what a deformation is, don’t take this to be the definition).

For instance, you could parametrize some curves over ${\mathbb{A}^1_k}$ by ${Spec\left(k[x,y,t]/(xy-t)\right)\rightarrow Spec(k[t])}$. We could think about this family as hyperbolas $X_t=\{(x,y): xy=t\}$. As $t$ approaches 0, the hyperbolas degenerate into the coordinate axes. Deformations in this family of the special fiber ${X_0}$ are all irreducible, yet the special fiber is reducible.

Now for what we really care about in this post. A first-order deformation of ${X_0}$ is a scheme ${X'}$, flat over ${R}$ such that ${X'\otimes_R k\simeq X_0}$. Note the terminology comes from the fact that the family is over Spec ${R}$ which remembers “tangent” information. A second-order deformation would keep track of information via a family over Spec ${k[\epsilon]/(\epsilon^3)}$, etc (anyone see a completion happening in the near future?).

A first-order deformation ${X'}$ is something that completes the fiber diagram:

${\begin{matrix} X_0 & \rightarrow & X'\\ \downarrow & & \downarrow f\\ Spec k & \rightarrow & Spec R \end{matrix}}$

where ${f}$ is flat. Maybe a notation will be useful later: Def(${X_0/k}$).

Let’s use the last couple of posts to classify the first-order deformations of a non-singular scheme over an algebraically closed field. The claim is that Def(${X/k}$) (where these are up to isomorphism) is in bijective correspondence with ${H^1(X, \mathcal{T}_X)}$. So by the last post it is enough to show bijective correspondence with the infinitesimal extensions of ${X}$ by ${\mathcal{O}_X}$.

The proof is just adapting what we said at the start of the post. If ${X'\in Def(X/k)}$, then by flatness we can tensor with ${0\rightarrow k\rightarrow R\rightarrow k\rightarrow 0}$ and it remains exact: ${0\rightarrow \mathcal{O}_X\rightarrow \mathcal{O}_{X'}\rightarrow \mathcal{O}_X}$. Thus ${X'}$ is an infinitesimal extension of ${X}$ by ${\mathcal{O}_X}$. Conversely, any such extension is flat and hence a first-order deformation.

## Infinitesimal Extensions by Coherent Sheaves

Last time we looked at what the infinitesimal lifting property was and that a particularly nice class of schemes always satisfies it. A related concept is that of an infinitesimal extension of a scheme. Suppose ${X}$ is a scheme of finite type over ${k}$ and ${\mathcal{F}\in Coh(X)}$. A pair ${(X', \mathcal{I})}$ is an infinitesimal extension of ${X}$ by ${\mathcal{F}}$ if ${\mathcal{I}}$ is a sheaf of ideals satisfying ${\mathcal{I}^2=0}$ and as schemes ${(X', \mathcal{O}_X'/\mathcal{I})\simeq (X, \mathcal{O}_X)}$.

Given ${(X, \mathcal{F})}$, as an example we always have the trivial infinitesimal extension by ${\mathcal{F}}$, which is ${(X, \mathcal{F})}$ where the structure sheaf on ${X}$ is given by ${\mathcal{O}_X'=\mathcal{O}_X\oplus\mathcal{F}}$ where multiplication is defined by ${(a\oplus f)\cdot (a'\oplus f')=aa'\oplus (af'+a'f)}$. Now notice that the way that ${\mathcal{F}}$ is a sheaf of ideals given this new structure sheaf, is ${0\oplus \mathcal{F}\subset \mathcal{O}_X'}$. Take ${(0,f), (0,f')\in\mathcal{F}}$ and see what happens ${(0,f)\cdot (0,f')=(0,0)}$. Thus ${\mathcal{F}^2=0}$ and clearly ${\mathcal{O}_X'/\mathcal{F}\simeq \mathcal{O}_X}$, so this is really an infinitesimal extension.

Let’s go back to our nice simple case from last time: Let ${X=}$ Spec ${A}$ be nonsingular. Then the claim is that any infinitesimal extension by a coherent sheaf ${\mathcal{F}}$ is isomorphic to the trivial one.

This immediately converts to an algebra problem. Let ${M}$ be the ${A}$-module such that ${\mathcal{F}=\tilde{M}}$. Suppose we have an infinitesimal extension ${(Spec A', \mathcal{I})}$. Then ${I}$ is an ideal of ${A'}$ with ${I^2=0}$, ${A'/I\simeq A}$, and ${I\simeq M}$ as an ${A}$-module.

This gives us an exact sequence ${0\rightarrow I\rightarrow A'\rightarrow A\rightarrow 0}$. Consider the isomorphism ${f: A'/I\rightarrow A}$, by last time there exists a lift ${A'/I\rightarrow A'}$. This is a retraction and hence the sequence splits giving ${A'\simeq A\oplus I\simeq A\oplus M}$ (we do need to check that the multiplication is the right one, but it is by computation). Taking the sheaves associated to this, we see that ${\mathcal{O}_X'\simeq \mathcal{O}_X\oplus \mathcal{F}}$, the trivial extension.

It’s good to notice that by what we did yesterday that these lifts are not actually unique. They are actually “parametrized” in some sense by ${Hom_A(\Omega_{A/k}, M)\simeq H^0(X, \mathcal{F}\otimes \mathcal{T})}$ where ${\mathcal{T}}$ is the tangent sheaf. So it is reasonable to guess that ${\mathcal{F}\otimes \mathcal{T}}$ plays a role here. Since we’ve classified the infinitesimal extensions of non-singular affine varieties, and non-singular varieties are locally of that form, let’s extend the result.

If ${X}$ is a non-singular variety over ${k=\overline{k}}$, and ${\mathcal{F}\in Coh(X)}$, then there is a one-to-one correspondence between the set of infinitesimal extensions of ${\mathcal{F}}$ by ${X}$ up to isomorphism and ${H^1(X, \mathcal{F}\otimes \mathcal{T})}$.

Take a finite affine cover of ${X}$, ${\{U_i=Spec(A_i)\}_1^n}$. Then we have a natural iso ${\check{H}^p(\mathcal{U}, \mathcal{F}\otimes\mathcal{T})\simeq H^p(X, \mathcal{F}\otimes\mathcal{T})}$, so we’ll think in terms of ${\check{C}}$ech cocycles.

If ${(X', \mathcal{I})}$ is an infinitesimal extension, then on each affine it must look trivial, so ${\mathcal{I}\big|_{U_i}\simeq \tilde{N_i}\simeq \mathcal{O}_{U_i}\oplus \mathcal{F}\big|_{U_i}}$. So ${A_i\oplus M_i=H^0(U_i, \mathcal{O}_X\oplus \mathcal{F})}$. Any two liftings differ by a section of ${Hom(\Omega_{X/k}, \mathcal{F})\simeq \mathcal{F}\otimes \Omega_{X/k}^* = \mathcal{F}\otimes\mathcal{T}}$. This is exactly saying that the pairwise intersections satisfy the cocycle condition and hence gives us an element of ${H^1(X, \mathcal{F}\otimes \mathcal{T})}$.

This basically reverses for the converse. Given an element of ${H^1(X, \mathcal{F}\otimes\mathcal{T})}$, then it satisfies the cocycle condition and hence patches by differences in the right thing to give an infinitesimal extension.

## Infinitesimal Lifting Property

I’ve basically recuperated from my test and I’m trying to get back into the AG frame of mind. I have about 5 posts half written, so I’m going to actually try to finish this one and start up a nice little series. I’m taking a class on deformation theory this quarter (which hasn’t actually started yet), so this series will review some of the very, very small amount of deformation theory scattered throughout the exercises of Hartshorne.

Let’s start with some basics on the infinitesimal lifting property. First assume ${k}$ an algebraically closed field and ${A}$ a finitely generated ${k}$-algebra with Spec ${A}$ a nonsingular variety (over ${k}$). Suppose ${0\rightarrow I\rightarrow B'\rightarrow B\rightarrow 0}$ is exact with ${B'}$ a ${k}$-algebra and ${I}$ an ideal with ${I^2=0}$. Then ${A}$ satisfies the infinitesimal lifting property: whenever there is a ${k}$-algebra hom ${f:A\rightarrow B}$, there is a lift ${g: A\rightarrow B'}$ making the obvious diagram commute.

First note that if ${g, g':A\rightarrow B'}$ are two such lifts then, ${\theta=g-g'}$ is a ${k}$-derivation of ${A}$ into ${I}$. A quick subtlety is that a “${k}$-derivation” is an ${A}$-module map that is a derivation and evaluates to zero on ${k}$. So we need to understand how ${I}$ is an ${A}$-module. But ${I^2=0}$, so it is a ${B}$-module, which in turn is an ${A}$-module (via ${g}$ and ${g'}$ which will be used). The reason ${im \theta\subset I}$ is that ${\theta}$ is a lift of the zero map since ${g}$ and ${g'}$ both lift ${f}$. Since the sequence is exact and ${\theta}$ lands in the kernel, it is in the image of the one before it, i.e. ${I}$.

Derivation:

$\displaystyle \begin{array}{rcl} \theta(ab) & = & g(a)g(b)-g'(a)g'(b) \\ & = & g(a)g(b)-g(a)g'(b)+g(a)g'(b)-g'(a)g'(b)\\ & = & g(a)(g(b)-g'(b))+(g(a)-g'(a))g'(b)\\ & = & g(a)\theta(b) + \theta(a)g'(b)\\ & = & a\cdot\theta(b)+b\cdot\theta(a) \end{array}$

Evaluates to 0 on ${k}$: Since ${g(1)=g'(1)=1}$, ${\theta(1)=0}$. Thus ${\theta(k)=k\cdot 0 = 0}$.

Since ${\Omega_{A/k}}$ is a universal object, we can consider ${\theta\in Hom_A(\Omega_{A/k}, I)}$. Conversely, given any ${\theta\in Hom_A(\Omega_{A/k}, I)}$, we can compose with the universal map ${d}$ to get ${\theta'=\theta\circ d: A\rightarrow I}$ is a ${k}$-derivation. Compose this with the inclusion ${I\rightarrow B'}$, call this ${\psi: A\rightarrow B'}$. Since composing again with ${B'\rightarrow B}$ gives ${0}$, ${\psi}$ is a lift of ${0}$ and hence ${g'=\psi + g}$ is a lift of ${f}$ (note we’ve only guaranteed ${k}$-linear so far, not algebra hom). Finally let’s check it preserves multiplication:

$\displaystyle \begin{array}{rcl} \psi(ab)+g(ab) & = & \theta'(ab)+g(ab) \\ & = & \theta'(a)g(b)+g(a)\theta'(b)+g(ab)\\ & = & \theta'(a)\theta'(b) + \theta'(a)g(b)+g(a)\theta'(b)+g(a)g(b)\\ & = & (\theta'(a)+g(a))(\theta'(b)+g(b))\\ & = & g'(a)g'(b) \end{array}$

Now let ${P=k[x_1, \ldots , x_n]}$ for which ${A=P/J}$ for some ${J}$. So we get another exact sequence ${0\rightarrow J\rightarrow P\rightarrow A\rightarrow 0}$. We now check that there is a map ${h:P\rightarrow B'}$ such that the square ${\begin{matrix} P & \stackrel{h}{\longrightarrow} & B' \\ \downarrow & & \downarrow \\ A & \stackrel{f}{\longrightarrow} & B \end{matrix}}$ commutes and this induces an ${A}$-linear map ${\overline{h}: J/J^2\rightarrow I}$.

Note a map out of ${P}$ is completely determined by where the ${x_i}$ go. Since ${B'\rightarrow B}$ surjective, choose any ${b_i\in B'}$ such that ${b_i\mapsto f(\overline{x_i})}$. Extend this to get ${h}$. By definition ${h}$ makes the square commute. Chasing around exactness, we get that if ${a\in J}$, then considering ${a\in P}$ gives ${h(a)\in I}$. Thus restricting gives ${\overline{h}: J\rightarrow I}$. Since ${I^2=0}$ we have ${h(a^2)=h(a)^2=0}$, so this descends to a map ${\overline{h}: J/J^2\rightarrow I}$. It is clearly ${A}$-linear.

Let ${X=}$ Spec ${P}$ and ${Y=}$ Spec ${A}$. The sheaf of ideals ${\mathcal{J}=\tilde{J}}$ defines ${Y}$ as a subscheme. Then by nonsingularity (Theorem 8.17 of Hartshorne) we have an exact sequence ${0\rightarrow \mathcal{J}/\mathcal{J}^2\rightarrow \Omega_{X/k}\otimes \mathcal{O}_Y\rightarrow \Omega_{Y/k}\rightarrow 0}$. Take global sections of this sequence to get the exact sequence ${0\rightarrow J/J^2\rightarrow \Omega_{P/k}\otimes A\rightarrow \Omega_{A/k}\rightarrow 0}$ (${H^1}$ vanishes by Serre).

Now apply the functor ${Hom_A(\cdot, I)}$ to get the exact sequence ${0\rightarrow Hom_A(\Omega_{A/k}, I)\rightarrow Hom_P(\Omega_{P/k}, I)\rightarrow Hom_A(J/J^2, I)\rightarrow 0}$. Exactness on the right is due to ${\Omega_{A/k}}$ being locally free and hence projective, so ${Ext^1}$ vanishes.

That surjectivity is exactly what we needed to say a lift exists. Take ${\overline{h}\in Hom_A(J/J^2, I)}$ as constructed before. Then choose ${\theta\in Hom_P(\Omega_{P/k}, I)}$ that maps to it. Compose with the universal map and inclusion to get a derivation ${P\rightarrow B'}$ (we’ll just relabel this ${\theta}$). Set ${h'=h-\theta}$. Since if ${j\in J}$ we have ${h'(j)=h(j)-\theta(j)=\overline{h}(j)-\overline{h}(j)=0}$ it descends to a map ${g:A\rightarrow B'}$ which is the desired lift.

Next time we’ll move on to infinitesimal extensions and rephrase what we just did in those terms.

## Naturality of Flows

This is something I always forget exists and has a name, so I end up reproving it. Since this sequence of posts is a hodge-podge of things to help me take a differential geometry test, hopefully this will lodge the result in my brain and save me time if it comes up.

I’m not sure whether to call it a lemma or not, but the setup is you have a smooth map ${F:M\rightarrow N}$ and a vector field on ${M}$, say ${X}$ and a vector field on ${N}$ say ${Y}$ such that ${X}$ and ${Y}$ are ${F}$-related. Define ${M_t}$ and ${N_t}$ to be the image of flowing for time ${t}$ and let ${\theta}$ and ${\eta}$ be the flows of ${X}$ and ${Y}$ respectively. Then the lemma says for all ${t}$ we have ${F(M_t)\subset N_t}$ and ${\eta_t\circ F=F\circ \theta_t}$ on ${M_t}$.

This is a “naturality” condition because all it really says is that the following diagram commutes:

${\begin{matrix} M_t & \stackrel{F}{\longrightarrow} & N_t \\ \theta_t \downarrow & & \downarrow \eta_t \\ M_{-t} & \stackrel{\longrightarrow}{F} & N_{-t} \end{matrix}}$

Proof: Let ${p\in M}$, then ${F\circ \theta^p: \mathbb{R}\rightarrow N}$ is a curve that satisfies the property $\displaystyle {\frac{d}{dt}\Big|_{t=t_0}(F\circ \theta^p)(t)=DF_{\theta^p(t_0)}(\frac{d}{dt}\theta^p (t)\Big|_{t=t_0})=DF_{\theta^p(t_0)}(X_{\theta^p(t_0)})=Y_{F\circ \theta^p(t_0)}}$. Since ${F\circ \theta^p(0)=F(p)}$, and integral curves are unique, we get that ${F\circ\theta^p(t)=\eta^{F(p)}(t)}$ at least on the domain of ${\theta^p}$.

Thus if ${p\in M_t}$ then ${F(p)\in N_t}$, or equivalently ${F(M_t)\subset N_t}$. But we just wrote that ${F(\theta^p(t))=\eta^{F(p)}(t)}$ where defined, which is just a different form of the equation ${\eta_t\circ F=F\circ \theta_t(p)}$.

We get a nice corollary out of this. If our function ${F:M\rightarrow N}$ was actually a diffeo, then take ${Y=F_*X}$ the pushforward, and we get that the flow of the pushforward is ${\eta_t=F\circ \theta_t\circ F^{-1}}$ and the flow domain is actually equal ${N_t=F(M_t)}$.

In algebraic geometry we care a lot about families of things. In the differentiable world, the nicest case of this would be when you have a smooth submersion: ${F: M\rightarrow N}$, where ${M}$ is compact and both are connected. Then since all values are regular, ${F^{-1}(n_0)}$ is smooth embedded submanifold. If ${N}$ were say ${\mathbb{R}}$ (of course, ${M}$ couldn’t be compact in this case), then we would have a nice 1-dimensional family of manifolds that are parametrized in a nice way.

It turns out to be quite easy to prove that in the above circumstance all fibers are diffeomorphic. In AG we often call this an “iso-trivial” family, although I’m not sure that is the best analogy. The proof basically comes down to the naturality of flows. Given any vector field ${Y}$ on ${N}$, we can lift it to a vector field ${X}$ on ${M}$ that is ${F}$-related. I won’t do the details, but it can be done clearly in nice choice of coordinates ${(x^1, \ldots, x^n)\mapsto (x^1, \ldots, x^{n-k})}$ and then just patch together with a partition of unity.

Let ${M_x}$ be the notation for ${F^{-1}(x)}$. Fix an ${x\in N}$, then by the above naturality lemma ${\theta_t\Big|_{M_x} : M_x\rightarrow M_{\eta_t(x)}}$ is well-defined and hence a diffeomorphism since it has smooth inverse ${\theta_{-t}}$. Let ${y\in N}$. Then as long as there is a vector field on ${N}$ which flows ${x}$ to ${y}$, then we’ve shown that ${M_x\simeq M_y}$, so since ${x}$, ${y}$ were arbitrary, all fibers are diffeomorphic. But there is such a vector field, since ${N}$ is connected.

## Lie groups have abelian fundamental group

Last year I wrote up how to prove that the fundamental group of a (connected) topological group was abelian. Since Lie groups are topological groups, they also have abelian fundamental groups, but I think there is a much neater way to prove this fact using smooth things. Here it is:

Lemma 1: A connected Lie group that acts smoothly on a discrete space is must be a trivial action.

Proof: Suppose our action is ${\theta: G\times M \rightarrow M}$ by ${\theta(g,x)=g\cdot x}$. Consider a point ${q}$ with non-trivial orbit. Then ${\text{im}\theta_{q}=\{q\}\cup A}$ where ${A}$ is non-empty. Thus ${G=\theta_{q}^{-1}(q)\cup \theta_q^{-1}(A)}$, a disconnection of ${G}$. Thus all orbits are trivial.

Lemma 2: A discrete normal subgroup of a Lie group (from now on always assumed connected) is central.

Proof: Denote the subgroup ${H}$. Then ${\theta(g,h)=ghg^{-1}}$ is a smooth action on ${H}$ (a discrete set). Thus by Lemma 1, ${ghg^{-1}=h}$ for all ${g\in G}$. Thus ${H}$ is central, and in particular abelian.

Now for the main theorem. Let ${G}$ be a connected Lie group. Let ${U}$ be the universal cover of ${G}$. Then the covering map ${p: U\rightarrow G}$ is a group homomorphism. Since ${U}$ is simply connected, the covering is normal and hence ${Aut_p(U)\simeq \pi_1(G, e)}$. By virtue of being normal, we also get that ${Aut_p(U)}$ acts transitively on the fibers of ${p}$. In particular, on the set ${p^{-1}(e)}$, which is discrete being the fiber of a discrete bundle. But this set is ${\ker p}$, which is a normal subgroup. I.e. a discrete normal subgroup, which by Lemma 2 is abelian.

Fix ${q\in \ker p}$. Then we get an ismorphism ${\ker p \simeq Aut_p(U)}$ by ${x\mapsto \phi_x}$ where ${\phi_x}$ is the unique covering automorphism that takes ${q}$ to ${x}$. Thus ${Aut_p(U)}$ is abelian which means ${\pi_1(G,e)}$ is abelian.