In the previous installment I outlined the proof that the differential on the ECH chain complex is well-defined. This was not very complicated. However the proof that is contained in a pair of papers by myself and Taubes totalling about 200 pages. Why is this so difficult? (The paper proves a more general gluing theorem than necessary to prove that , but it wouldn’t really be any shorter if we just proved that .)

**Basic strategy**

The usual strategy for proving that is as follows. Given ECH generators and , to prove that the coefficient , one should consider the moduli space of curves in , modulo the usual equivalence relation. By the proposition from installment 7, we know that this moduli space is a 1-manifold, and by the arguments from the previous installment, any sequence in this 1-manifold without a convergent subsequence converges to a broken pseudoholomorphic current where and for some orbit set , and we have . That is, if denotes the moduli space of curves in , modulo the usual equivalence relation, then is a 1-manifold which has a compactification to a 1-manifold with boundary , with a natural map

Now to complete the proof that , we need to prove that for every orbit set , the following hold:

1) If is an ECH generator, then the component of the above map has degree .

2) If is not an ECH generator (i.e. if contains multiply covered hyperbolic orbits), then the component of the above map has degree .

If we can prove both (1) and (2), then the number of points in equals the coefficient , and since the number of points in the boundary of a compact 1-manifold is zero, we are done. (If we are proving over instead of over , then we have to do all of this with orientations, and count degree with signs. Let’s not go there right now.)

**The difficulty**

In the more standard picture from symplectic field theory, if you have Fredholm index holomorphic curves and in , and if there are (possibly multiply covered) Reeb orbits such that the negative ends of , as well as the positive ends of , are at , then you can glue them to find an end of the moduli space of Fredholm index 2 curves that converges (in the sense of [BEHWZ]) to the broken curve . (If some of the Reeb orbits are repeated or multiply covered then there is more than one way to glue. To deal with this in SFT, one has to discard “bad” Reeb orbits, and one has to insert certain rational combinatorial factors in the differential. This issue also has implications for ECH as we will see below.)

In ECH, if we have and , then the above picture usually does not apply. The reason is that if is an embedded Reeb orbit that appears with multiplicity in , then we know that the *total* covering multiplicity of negative ends of at coverings of agrees with the total covering multiplicity of positive ends of at coverings of , namely both totals equal . However the *individual *covering multiplicities of the negative ends of and the positive ends of at need not agree. In fact, we will see below that if , then (in the absence of trivial cylinders) these multiplicities *never* agree when is elliptic, although they do agree when is hyperbolic.

At first sight, one might think that it is impossible to glue and when contains a multiply covered elliptic orbit. In fact, you might be thinking right now that I screwed up the definition of ECH, and I should be throwing out multiply covered elliptic orbits instead of multiply covered hyperbolic orbits!

To clarify what is going on here, we need to learn more about the covering multiplicities of ends of -holomorphic curves with ECH index .

**Partition conditions**

Let be a nondegenerate embedded Reeb orbit and let be a positive integer. We now define two partitions of , the “positive partition” and the “negative partition”, which we denote by and . (In my papers I called these the “outgoing partition” and the “incoming partition”. However that terminology seems to be confusing.)

If is hyperbolic, then . If is positive hyperbolic, then this partition is . If is negative hyperbolic, then this partition is if is even, and if is odd.

If is elliptic with rotation angle for some trivialization of , then , where depends only on and the class of in (which does not depend on ) and is defined as follows. To define , let be the maximal concave polygonal path in the plane with vertices at lattice points which starts at the origin, ends at , and lies below the line . Then consists of the horizontal displacements of the segments of connecting consecutive lattice points. The partition is defined analogously from the path , which is the minimal convex polygonal path with vertices at lattice points which starts at the origin, ends at , and lies above the line .

To give the simplest example, if , then and . In general, more interesting partitions are possible.

At this point I should probably give you some fun combinatorial exercises involving , for example hints to prove that and are *disjoint* whenever . But you’re probably wondering why I introduced these partitions. The reason why is the following:

**Proposition (equality condition in the writhe bound). **Let and be orbit sets. Let be a somewhere injective -holomorphic curve. Then equality holds in the writhe bound

and hence equality holds in the index inequality

,

only if the following hold:

1) For each , the covering multiplicities of the positive ends of at covers of are exactly the entries in the partition .

2) For each , the covering multiplicities of the negative ends of at covers of are exactly the entries in the partition .

If I get around to explaining the proof of the writhe bound, then I will explain the proof of this proposition at the same time. If is generic, then (modulo a technical point which has not been proved but which I think is doable), the partition conditions are not only necessary but also sufficient for equality in the writhe bound.

**Why no multiply covered hyperbolic orbits? Reason #3**

Using the above proposition, let us see if we can understand why in the special case where all Reeb orbits are hyperbolic. Let us try to prove points (1) and (2) from the section “Basic strategy” above.

Point (1) (at least with coefficients; coefficients involve additional issues) follows from the standard gluing theory mentioned above: if is an ECH generator, then it contains no multiply covered orbits, so the negative ends of and the positive ends of are at the same, distinct, embedded Reeb orbits.

For point (2), suppose that contains a positive hyperbolic orbit with multiplicity . Since , we know that has negative ends at , and has positive ends at . To glue and , we need to choose a bijection between the negative ends of at and the positive ends of at . There are ways to do this. Note that is even.

Likewise, if contains a negative hyperbolic orbit with multiplicity , then has negative ends at the double cover of (plus one end at if is odd), while has positive ends at covers of with these same multiplicities. To glue and , we need to choose a bijection between the negative ends of and the positive ends of at the double cover of ; and for each pair of positive and negative ends identified by this bijection, we need to choose one of two ways to identify the sheets of the cover. The number of ways to do this is where . Note that this number of choices is again even.

For each hyperbolic orbit that appears in with multiplicity , we need to make choices as above. The total number of choices is even. This immediately implies point (2) if we are working with coefficients. If we use coefficients instead, then work of Bourgeois-Mohnke implies that whenever one composes the bijection for a positive hyperbolic orbit with a transposition, or switches a sheet identification for a negative hyperbolic orbit, the sign of the gluing changes. It follows that half of the gluing choices have one sign and half of the gluing choices have the other sign, which proves point (2) again.

**Gluing along multiply covered elliptic orbits**

The above hopefully explains why we throw out multiply covered hyperbolic orbits. But how can we glue along multiply covered elliptic orbits, for which the positive and negative partitions are disjoint? Why don’t we have to also throw out multiply covered elliptic orbits?

To simplify the notation, consider the case where consists of just a single pair , where is elliptic and . Let and be somewhere injective curves of Fredholm index 1, but let us not yet assume that they have ECH index 1. To simplify the discussion let us assume that and do not contain trivial cylinders or covers thereof. The covering multiplicities of the negative ends of determine a partition of , and the covering multiplicities of the positive ends of determine a partition of . Since we are not assuming equality in the index inequality, we do not necessarily have .

If , can we glue and to a curve of Fredholm index ? The answer is sometimes, but not directly. What we really do is glue a triple , where is a Fredholm index zero degree branched cover of , so that the covering multiplicities of the positive ends of are the entries in the partition , and the covering multiplicities of the negative ends of are the entries in the partition . (Whether such a Fredholm index zero branched cover exists depends on the combinatorics of the partitions; if it exists, it is necessarily a union of genus zero curves.) It now makes sense to try to glue these, but there is an obstruction to gluing because is not transverse (as the dimension of its component in the moduli space of holomorphic curves is twice the number of branch points, while the Fredholm index is zero). That is, one can glue “up to” an element of the cokernel of the linearized Cauchy-Riemann operator for , and this element of the cokernel is the gluing obstruction. If you haven’t seen a gluing proof before you then you probably do not understand what I am talking about here. Maybe I’ll explain it later.

Anyway there is a finite set of branched covers that can be glued and these comprise the zeros of a section of an “obstruction bundle” over the moduli space of branched covers. The number of gluings is the signed count of zeroes of this section. The 200-page pair of gluing papers with Taubes computed this number of gluings, which we can denote by . It turns out that is always nonnegative (for appropriate orientation conventions); sometimes it is zero (when no Fredholm index zero branched cover exists), and sometimes it is bigger than . But it turns out that we have the following:

**Amazing fact. ** *if and only if *.

That is, the number of gluings equals exactly in the situation that is needed for ECH!

The details of this story would take a very long time to explain. However there are “big picture” reasons to believe that it will work out before going through 200 pages of analysis. The big picture reasons involve the symmetric product picture which I hinted at in installment 3 of these notes. In fact I think this may lead to a simpler proof that , which was my original plan for proving that before switching to the branched cover approach for reasons explained here. I will try to explain this symmetric product picture later.