# How is the smoothness of the space of deformations related to unobstructedness?

As a beginning differential geometer, I've been trying to learn about deformation theory. Other than Kodaira's book, I've found virtually no references from the point of view of differential geometry. As such, my understanding is hazy.

• I am asking this question in an attempt to clarify my vague understanding, especially in regard to the last bullet point below. Precise definitions of terms are welcome, as are references to books or introductory papers on the subject.

My vague understanding goes like this:

We are interested in studying the "deformations" of a geometric object $X$ (e.g., a complex manifold $X$). We can associate to $X$ its "deformation exact sequence," which amounts to a kernel-cokernel exact sequence in some cohomology theory (e.g., Cech cohomology) of some map $\varphi$ which I don't understand: $$0 \to \text{Ker}(\varphi_*) \to H^1_X \xrightarrow{\varphi_*} H^2_X \to \text{Coker}(\varphi_*) \to 0.$$ We interpret this as follows:

• $\text{Ker}(\varphi_*)$ is the space of infinitesimal (or "first-order") deformations. An object $X$ is rigid iff $X$ has no infinitesimal deformations -- i.e., $\text{Ker}(\varphi_*) = 0$.

• $\text{Coker}(\varphi_*)$ is the space of infinitesimal (or "first-order") obstructions. An object $X$ is unobstructed iff $X$ has no obstructions -- i.e, $\text{Coker}(\varphi_*) = 0$.

• Let $\mathcal{M}$ be the "moduli space of local deformations of $X$." The "formal tangent space" $T_X\mathcal{M}$ is the space of first-order deformations -- i.e., $T_X\mathcal{M} \cong \text{Ker}(\varphi_*)$.

• Somehow, in some cases, the smoothness of $\mathcal{M}$ is related to whether $X$ is unobstructed. In this case, $\dim(\mathcal{M}) = \dim(T_X\mathcal{M}) = \dim(\text{Ker}(\varphi_*))$.

More hazy thoughts: I think in the cases I'm interested in, the map $\varphi_*$ can be regarded as the differential of some map $\varphi \colon \text{Somewhere} \to \text{Somewhere Else}$, and $\mathcal{M} = \varphi^{-1}(\text{point})$, so $\mathcal{M}$ is smooth if $\varphi$ is a submersion (meaning $\varphi_*$ is surjective, so $\text{Coker}(\varphi_*) = 0$), in which case $T_X\mathcal{M} = \text{Ker}(\varphi_*)$. How any of this works precisely, I don't know.

• What is $\varphi$ here? – Qiaochu Yuan Feb 29 '16 at 7:19
• At least in related stories I know about, the relationship is really given by a form of the regular value theorem - $\varphi$ is the differential of some map, and unobstructedness means the preimage of a point - which should be the appropriate moduli space - is a smooth manifold. I can write down the story I know best (in which this is the moduli space of instantons) if desired. – user98602 Feb 29 '16 at 7:23
• @QiaochuYuan: I have no idea; all I know is this vague story. I'm looking for definitions / clarifications / anything, really. – Jesse Madnick Feb 29 '16 at 7:24
• @MikeMiller: I realized that there's some relationship to the Regular Value Theorem going on, too, and edited accordingly. But like I said, I'd be interested in any contribution: a story about the moduli space of instantons would be great, sure. – Jesse Madnick Feb 29 '16 at 7:28
• I think there is an explicit description of $\varphi :H^1(X,TX) \to H^2(X, TX)$ in D. Huybrechts, chapter 6. It's related to the Maurer-Cartan equation. – user99914 Feb 29 '16 at 8:30

Consider the following very basic example. The affine scheme $\text{Spec } k[x, y]/xy$ is singular at the origin. This singularity can be detected as follows: the Zariski tangent space at the origin, which we think of as the space of $k$-algebra maps $k[x, y]/xy \to k[e]/e^2$ which reduce to $x, y \mapsto 0 \bmod e$, is 2-dimensional, since we can take

$$x \mapsto ae, y \mapsto be$$

for any $a, e$. But most of these tangent vectors don't lift to 2-jets. The space of 2-jets at the origin is the space of $k$-algebra maps $k[x, y]/xy \to k[e]/e^3$ which reduce to $x, y \mapsto 0 \bmod e$, and if $x \mapsto ae$ with $a \neq 0$ then we must have $y \mapsto be^2$ for some $b$, and vice versa. So the only tangent vectors which lift to 2-jets are those with $a = 0$ or $b = 0$. (Geometrically this should make sense since this variety is the union of the $x$- and $y$-axis and these are the tangent vectors pointing along those axes.)

This example can be used to motivate the definition of formal smoothness in algebraic geometry. This is a condition weaker than being smooth, and it's usually stated in much more generality than we need here: all we need is that formal smoothness of a space $M$ (variety, scheme, stack, whatever) implies that any tangent vector $\text{Spec } k[e]/e^2 \to M$ lifts to a 2-jet $\text{Spec } k[e]/e^3 \to M$.

Now take $M$ to be a moduli space of whatever sort of objects you're trying to deform. A point of $M$ is an object of that sort. A tangent vector to a point is a first-order deformation of that object. And a 2-jet is a second-order deformation. So if $M$ is smooth at a point, then we expect any tangent vector to lift to a 2-jet, which is precisely the statement that first-order deformations lift to second-order deformations.

If $X$ is an object describing a point of the moduli space $M$, a further question is what any of this business has to do with, say, the tangent bundle of $X$. The short story is that the correct definition of "tangent space of $M$ at $X$" should return an object called the tangent complex of $X$ (shifted by $1$), which is a derived version of the tangent bundle of $X$, and "$M$ is smooth at $X$" means that the tangent complex is concentrated in degree $\pm 1$ (it's one of these, not either, but don't ask me to figure out which), where it is just the usual tangent bundle. Obstructions live in the tangent complex in other degrees, so if there aren't any other degrees then there aren't any other obstructions.

Here's a fairy tale.

Let $M$ be an oriented Riemannian 4-manifold with a $G$-bundle $E \to M$ over it. (Often we take $G=SU(2)$.) Given a connection $A$ on it, I can define its curvature $F_A$, and its self-dual part $F_A^+ = (*F_A + F_A)/2$. The instanton equation is $F_A^+ = 0$.

The space of connections is affine over $\Omega^1(\mathfrak g_E)$, 1-forms with values in the endomorphism bundle of $E$. Because $F_{A+a} = F_A + d_A a + a \wedge a$, the map $a \mapsto F_{A+a}^+$ is given by $F_A^+ + d_A^+ a + (a \wedge a)^+$. The derivative is this map is $d_A^+: \Omega^1(\mathfrak g_E) \to \Omega^2_+(\mathfrak g_E)$.

Now, the space I'm interested in is not the space of instantons (solutions to $F_A^+ = 0$), but rather the moduli space thereof, where I mod out by the gauge group $\mathcal G = \text{Aut}(E)$ of fiber-preserving bundle automorphisms. The Lie algebra of this is $\Omega^0(\mathfrak g_E)$, and the differential of the action of $\mathcal G$ on $\mathcal A$ at $A$ is $\sigma \mapsto d_A\sigma$.

This is precisely the setup you've given above, with $d_A^+ = \varphi$ in this setting. Actually, I prefer to write my complex a little larger: $$\Omega^0(\mathfrak g_E) \xrightarrow{d_A} \Omega^1(\mathfrak g_E) \xrightarrow{d_A^+} \Omega^2_+(\mathfrak g_E).$$ Now what I should really do so that $\varphi$ is actually correct is consider instead the map $\varphi = d_A^* \oplus d_A^+: \Omega^1 \to \Omega^0 \oplus \Omega^2_+$. At least infinitesimally, this is the solutions to the equations living in a certain slice of the gauge group action; so when I mod out by the gauge group, we've actually found what my moduli space looks like infinitesimally. (At least, this is true provided $d_A$ does not have kernel. Assume it doesn't, for the cases where it does are important but special and I do not think illuminating in the general deformation theory framework.)

Now let's look at your formalism. What does it mean when I'm asking that $\varphi$ is unobstructed? It means precisely that my instanton equation $F_A^+$ has a local inverse in the slice we chose (provided you believe the regular value theorem in this setting; one has to pass to a Sobolev space completion to get it to really work): both that $d_A$ is injective (= $d_A^*$ is surjective = my connection is irreducible; don't worry about this) and that $d_A^+$ is surjective (that $F_A^+$ has a local inverse globally). That is, I've shown that my moduli space is locally smooth near $A$ a solution to the instanton equation. This is the only way I ever actually show smoothness. In this situation, the actual honest to god tangent space inside the slice (and hence to the moduli space) at $A$ is precisely the same as $\text{ker}(d_A^* \oplus d_A^+)$

And what does rigidity mean? It means that this tangent space is zero, and hence that I cannot automatically find other solutions locally.

One last note is that I don't seem to have actually mentioned cohomology here. The point is that when $\varphi$ is unobstructed, the only cohomology of my complex is in the middle degree. I can calculate this, then, as the Euler characteristic of the whole complex - $H^0 - H^1 + H^2_+ = \dim \mathcal M$. Using the Atiyah-Singer index theorem I calculate the left hand side to be $8c_2(E) - 3(1-b_1(M) + b_2^+(M))$, or something like that.

This fairy tale should bear some resemblance to reality. I've thought about this much less, but here's an idea of what the idea should look like for deformations of complex structures. A complex structure on $M$, by Newlander-Nirenberg, is a connection $TM \xrightarrow{\bar \partial} TM \otimes T^*M$ with $(\bar \partial)^2 = 0$. A deformation of this is given by adding a $TM$-valued 1-form $A$; and the linearization of $(\bar \partial + A)^2 = 0$ is $\bar \partial A = 0$. So you're interested in the map $\varphi: \Omega^1(TM) \to \Omega^2(TM)$ given by $\bar \partial$. And I want to mod out by the automorphisms of the local coordinate system, which is infinitesimally the image of $\Omega^0(TM) \to \Omega^1(TM)$, I think but don't remember. I also have the vague impression this problem is unobstructed for some reason? But I do not remember why. So what you care about is precisely $H^1_{\bar \partial}(TM)$.

Let's specialize to the Riemann surface case (which is why I believe what I'm saying above, because it gives the results I know are true). There, I actually just have a complex $\Omega^0(TM) \to \Omega^1(TM) \to \Omega^2(TM)$. Riemann Roch should tell us that the Euler characteristic of this complex is $3-3g$, and I have for some reason convinced myself that the last map is surjective, so $3-3g = H^0(TM) - H^1(TM)$. When $M$ is genus 0, $H^0(TM) = 3$ and $H^1(TM) = 0$; when $M$ is genus 1, $H^0(TM) = 1$, so $H^1(TM) = 1$; and when $M$ is genus $g \geq 2$, the automorphism group (whose Lie algebra, note, is holomorphic vector fields = $H^0(TM)$) is 0-dimensional and $H^1(TM) = 3g-3$, as we expect.

• As a note, the reason one wants to add further conditions should be related to achieving ellipticity of your operators. That's why I need to consider $d_A^+ \oplus d_A^*$; that's why I need to mod out by the image of $\bar \partial$ below. – user98602 Feb 29 '16 at 8:19
• By unobstructed do you mean $\varphi = 0$? – user99914 Feb 29 '16 at 8:47
• @JohnMa: I mean that it's surjective, as in the OP. – user98602 Feb 29 '16 at 8:49