## A magic trick for defining obstruction bundles

Today I want to tell you about a magic trick for proving transversality and/or defining obstruction bundles in certain four-dimensional situations where the usual approach doesn’t work. This is helpful for defining cobordism maps between various kinds of contact homology in three dimensions (although to completely define cobordism maps you need much more than just this trick).

The question

Here is the setup. Let $(X,\omega)$ be a four-dimensional strong symplectic cobordism between nondegenerate contact three-manifolds $(Y_+,\lambda_+)$ and $(Y_-,\lambda_-)$. Let $\overline{X}$ denote the usual symplectization completion of $X$, and let $J$ be a generic almost complex structure on $\overline{X}$ satisfying the usual conditions.

Let $C$ be an immersed $J$-holomorphic curve in $\overline{X}$. Let $\pi:\widetilde{C}\to C$ be a degree $d$ branched cover of $C$ with $b\ge 0$ branch points.

When the number of branch points $b=0$, we would like a criterion which guarantees that $\widetilde{C}$ is cut out transversely. When the number of branch points $b>0$, it is more or less impossible for $C$ to be cut out transversely, but we would still like $\widetilde{C}$ to be “good” in the following sense.

Recall that there is a deformation operator

$D_C:L^2_1(N_C) \to L^2(T^{0,1}C\otimes N_C)$

where $N_C$ denotes the normal bundle to $C$. This operator is surjective if and only if $C$ is cut out transversely. The kernel of this operator consists of deformations of $C$ which are $J$-holomorphic to first order. The index of this operator is what we call the Fredholm index of $C$, denoted by $ind(C)$, which when $C$ is cut out transversely is the dimension of the moduli space of holomorphic curves near $C$. Now there is an induced operator

$D_{\widetilde{C}}:L^2_1(\pi^*N_C)\to L^2(T^{0,1}\widetilde{C}\otimes \pi^*N_C).$

To define this, choose a local complex trivialization of $N_C$. Then locally, in this trivialization,

$D_C = \overline{\partial} + \alpha$

where $\alpha$ is some $(0,1)$-form on $C$, determined by the derivative of the almost complex structure in directions normal to $C$. Using the same local trivialization for $\pi^*N_C$, we define

$D_{\widetilde{C}} = \overline{\partial} + \pi^*\alpha.$

Roughly speaking, $D_{\widetilde{C}}$ considers deformations of $\widetilde{C}$ in directions normal to $C$, so that the branch points don’t move.

Definition. The branched cover $\widetilde{C}$ is good if $Ker(D_{\widetilde{C}})=0$.

If there are no branch points and $ind(\widetilde{C})=0$, then $\widetilde{C}$ is good if and only if it is transverse. If there are branch points, then one needs all branched covers in the moduli space of branched covers of $C$ containing $\widetilde{C}$ to be good, in order to define an “obstruction bundle” over the moduli space of such branched covers in order to do gluing as in my joint papers with Taubes.

So the question is now, under what circumstances is $\widetilde{C}$ guaranteed to be good?

The usual approach

Recall that the Fredholm index of $\widetilde{C}$ is given in my notation by

$ind(\widetilde{C}) = -\chi(\widetilde{C}) + 2 c_\tau(\widetilde{C}) + CZ_\tau^{ind}(\widetilde{C}).$

By the Riemann-Hurwitz formula, we can rewrite this as

$ind(\widetilde{C}) = -d\chi(C) + b + 2d c_\tau(C) + CZ_\tau^{ind}(\widetilde{C}).$

Note that this is not the Fredholm index of the operator $D_{\widetilde{C}}$. Rather, we have

$ind(D_{\widetilde{C}}) = ind(\widetilde{C}) - 2b.$

Intuitively, this is because the operator $D_{\widetilde{C}}$ does not consider deformations of $\widetilde{C}$ that move the branch points, so the dimension of its domain is $2b$ too small.

Anyway, using the above notation, we can now state the result given by the usual approach. Let $g(\widetilde{C})$ denote the genus of $\widetilde{C}$. Let $h_+(\widetilde{C})$ denote the number of ends of $\widetilde{C}$ that are at positive hyperbolic orbits (including even covers of negative hyperbolic orbits).

Proposition 1. Suppose that $2g(\widetilde{C}) - 2 + ind(\widetilde{C}) + h_+(\widetilde{C}) - 2b < 0$. Then $\widetilde{C}$ is good.

Proof. Suppose $\psi\in Ker(D_{\widetilde{C}})$ is not identically zero. We know, e.g. from the Carleman similarity principle, that every zero of $\psi$ is isolated and has positive multiplicity, so the signed count of zeroes of $\psi$ is nonnegative. On the other hand, we can bound the number of zeroes of $\psi$ similarly to my previous blog post “automatic transversality for dummies”. I’ll leave the calculation as an exercise; the result is

$\#\psi^{-1}(0) \le 2g(\widetilde{C}) - 2 + ind(\widetilde{C}) + h_+(\widetilde{C}) - 2b.$

If the right hand side is negative this is a contradiction. Thus if the right hand side is negative then $\widetilde{C}$ is good. QED.

Where the usual approach doesn’t work

Here are two examples where Proposition 1 is not applicable but we would still like to be able to prove that $\widetilde{C}$ is good.

Example 2. Suppose that $C$ is a cylinder with Fredholm index zero which has one positive end and one negative end, each at positive hyperbolic orbits. Suppose that $\widetilde{C}$ is also a cylinder, so that there are no branch points. If we are trying to define cobordism maps on cylindrical contact homology, we would like to be able to prove that $\widetilde{C}$ is cut out transversely, or equivalently good. Proposition 1 is not applicable here because the inequality that one needs is an equality in this case.

To motivate our next example, note that to define the cobordism maps needed to show that ECH does not depend on the almost complex structure, one needs to consider certain cases where $C$ and $\widetilde{C}$ both have ECH and Fredholm index equal to zero, with all ends at elliptic orbits. One wants to show that $\widetilde{C}$ is good so that one can do obstruction bundle gluing as in my paper with Taubes to prove the chain map equation. Here is a simple example of such a case where the inequality needed to apply Proposition 1 does not hold.

Example 3. Suppose that $C$ is embedded, has genus one, and has exactly one end, which is a positive end at a simple elliptic orbit $\gamma$ with monodromy angle (with respect to some trivialization $\tau$) $\theta\in(0,1/2)$. Suppose further that the Fredholm index $ind(C)=0$. Then the ECH index $I(C)=0$ also (this is true for any embedded curve whose ends are at distinct simple orbits by the relative adjunction formula). More precisely, $c_\tau(C)=-1$ and $Q_\tau(C)=0$. Now let $\widetilde{C}$ be a double cover of $C$ with one branch point, and then necessarily exactly one end, which is a positive end at the double cover of $\gamma$. Then it follows from the above calculations that $I(\widetilde{C})=ind(\widetilde{C})=0$. But $g(\widetilde{C})=2$, so the inequality needed to apply Proposition 1 is an equality in this case.

The magic trick

Now I can show you the magic trick. The statement is a bit complicated and not the most general possible, but this will illustrate the technique and is sufficient to handle the above two examples.

To proceed with the statement and the proof, we need to consider the normal bundle $N_C$. We can regard this as a kind of (completed) symplectic cobordism between the disjoint union over the ends of $C$ of the normal bundle of the corresponding (possibly multiply covered) Reeb orbit. Let me not try to define exactly what I mean by “symplectic cobordism” in this context. Suffice it to say that $C$, regarded as the zero section, defines an embedded holomorphic curve in $N_C$ whose ends are all at distinct simple Reeb orbits, even if none of the above is true for the original curve $C$.

There is now a well defined notion of the ECH index of $\widetilde{C}$ in the normal bundle $N_C$. This is defined by copying the usual formulas, in the normal bundle $N_C$. We can think of this as the ECH index “relative to $C$“, and we denote it by $I_C(\widetilde{C})$. If $C$ is embedded in $\overline{X}$ and all its ends are at distinct simple Reeb orbits, then the ECH index of $\widetilde{C}$ in $N_C$ agrees with the ECH index of $\widetilde{C}$ in $\overline{X}$, i.e. $I_C(\widetilde{C}) =I(\widetilde{C})$. If $C$ does not have these properties, then it is possible that $I_C(\widetilde{C})\neq I(\widetilde{C})$.

We also say that $\widetilde{C}$ satisfies the ECH partition conditions “relative to $C$” if it satisfies the usual ECH partition conditions in the normal bundle $N_C$. If all ends of $C$ are at distinct simple Reeb orbits then $\widetilde{C}$ satisfies the ECH partition conditions if and only if $\widetilde{C}$ satisfies the ECH partition conditions relative to $C$.

We can now state:

Proposition 4. Assume that $Ker(D_C)=0$. Suppose that either $ind(\widetilde{C})>I_C(\widetilde{C})$, or $ind(\widetilde{C})=I(\widetilde{C})$ and $\widetilde{C}$ does not satisfy the ECH partition conditions relative to $C$. Furthermore, if $\widetilde{C}\to C$ factors through a branched cover $\widehat{C}\to C$ whose degree is between $1$ and $d$, then assume that the above condition also holds with $\widetilde{C}$ replaced by $\widehat{C}$. Then $\widetilde{C}$ is good.

Here is why Proposition 4 is applicable to Examples 2 and 3. In both of these examples, $ind(\widetilde{C})=I_C(\widetilde{C})=0$. However neither of these examples satisfies the ECH partition conditions relative to $C$. (In Example 2, the partition conditions relative to $C$ would require that $\widetilde{C}$ has $d$ positive ends and $d$ negative ends. In Example 3, the partition conditions relative to $C$ would require that $\widetilde{C}$ has two positive ends.)

Proof of Proposition 4. First observe that there is a unique almost complex structure on the normal bundle $N_C$ (regarded here as four-manifold, not a bundle) whose restriction to the fibers agrees with the almost complex structure $J$ on $X$, such that a local section $\psi$ is in the kernel of the operator $D_C$ if and only if $\psi$ is a holomorphic map from a neighborhood in $C$ to $N_C$.

Now suppose $\widetilde{\psi}$ is a nonzero element of $Ker(D_{\widetilde{C}})$. Let $\psi$ denote the image of $\widetilde{\psi}$ under the projection $\pi^*N_C\to N_C$. Then $\psi$ is a holomorphic curve in $N_C$.

By the assumption that $Ker(D_C)=0$ and the assumption about intermediate branched covers in Proposition 4, we can assume without loss of generality that $\psi$ is somewhere injective.

Now a version of the ECH index inequality tells us that $ind(\psi) \le I_C(\psi)$, with equality only if $\psi$ satisfies the ECH partition conditions relative to $C$. This is proved just like the usual ECH index inequality, except that in this case one does not need Siefring’s nonlinear analysis; instead one can use the much easier, linear analysis in my paper “An index inequality for pseudoholomorphic curves in symplectizations”.

Now in the first sentence of the above paragraph, one can replace $\psi$ everywhere by $\widetilde{C}$ without changing anything. We then have a contradiction to the assumptions of Proposition 4, which means that $\psi$ couldn’t exist, so $\widetilde{C}$ is good. QED.

Remark. There is a variant of Example 3 in which $C$ is the same as before, but now there are no branch points, so that $\widetilde{C}$ has genus one and two ends. Here neither Proposition 1 nor Proposition 4 is applicable to show that $\widetilde{C}$ is good, i.e. transverse. However I don’t think one needs to consider this case to define ECH cobordism maps. The reason is that the curves you would want to glue this to would have just one negative end (by the ECH partition conditions), so you would want to add a branch point so that $\widetilde{C}$ has just one positive end as in Example 3.

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### 5 Responses to A magic trick for defining obstruction bundles

1. Chris Wendl says:

A discussion reminded me of this post that I had intended to read 7 months ago but not gotten around to until now. I have two questions:

(1) Do I understand correctly that if C is an index 0 immersed cylinder in a 4-dimensional cobordism, then this trick implies that whenever C is regular, all of its unbranched multiple covers also are? (In other words, one need not make J “even more generic” in order to achieve regularity for the covers?)

(2) Do I also understand correctly that this trick does NOT apply to the situation that arises in defining the Gromov invariant, i.e. showing that for generic J, all the unbranched multiple covers of embedded index 0 tori are regular? Unless I’m very confused, it seems that in that case, the partition conditions are trivially satisfied but ind and I_C are always equal. (I believe also that in that case, it should not suffice merely to achieve transversality for the embedded curves, as one may really have to make J more generic in order to handle the covers. At any rate that’s how I understand Taubes’s argument, though my understanding of Taubes is fragile.)

2. Don’t worry about the 7 months; there’s too much stuff to read, and I am happy to discuss blog postings arbitrarily far in the future.

(1) I don’t think the trick always implies this. It depends on the relation between the ECH and Fredholm indices and on the partition conditions. Note also the key word “not” in Proposition 4.

(2) Right. J needs to be more generic in order for the covers to be regular. Indeed, as one varies J in a generic one-parameter family, at some times double covers may not be regular, leading to “torus doubling bifurcations”. (I should go back and read Taubes’s proof that the covers are regular for generic J, because I think I never really understood this.)

• Chris Wendl says:

My question (1) was a bit sloppy: I meant to also assume the asymptotic orbits are positive hyperbolic (which guarantees among other things that all the unbranched covers also have index 0). Is what I said true then?

(Automatic transversality makes me less worried about the case with elliptic orbits… of course I am conveniently forgetting that there could also be negative hyperbolic orbits, as I sometimes do.)

• OK, then yes, what you said is true: If you have an immersed index zero cylinder with ends at positive hyperbolic orbits, and if it is regular, then its unbranched covers are automatically regular. This was Example 2 above (except that in Example 2 I said that one of the ends is on the negative side of the cobordism, but this is not relevant to anything).