Lecture notes on ECH 10: Why is the ECH differential a differential?

In the previous installment I outlined the proof that the differential \partial on the ECH chain complex is well-defined. This was not very complicated. However the proof that \partial^2=0 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 \partial^2=0, but it wouldn’t really be any shorter if we just proved that \partial^2=0.)

Basic strategy

The usual strategy for proving that \partial^2=0 is as follows. Given ECH generators \alpha_+ and \alpha_-, to prove that the coefficient \langle\partial^2\alpha_+,\alpha_-\rangle = 0, one should consider the moduli space of I=2 curves in {\mathcal M}(\alpha_+,\alpha_-), 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 (C_+,C_-) where C_+\in{\mathcal M}(\alpha_+,\alpha_0) and C_-\in{\mathcal M}(\alpha_0,\alpha_-) for some orbit set \alpha_0, and we have I(C_+)=I(C_-)=1. That is, if {\mathcal M}_k(\alpha,\beta) denotes the moduli space of I=k curves in {\mathcal M}(\alpha,\beta), modulo the usual equivalence relation, then {\mathcal M}_2(\alpha_+,\alpha_-) is a 1-manifold which has a compactification to a 1-manifold with boundary \overline{\mathcal M}_2(\alpha_+,\alpha_-), with a natural map

\partial\overline{\mathcal M}_2(\alpha_+,\alpha_-) \to \sqcup_{\alpha_0} {\mathcal M}_1(\alpha_+,\alpha_0) \times {\mathcal M}_1(\alpha_0,\alpha_-).

Now to complete the proof that \partial^2=0, we need to prove that for every orbit set \alpha_0, the following hold:

1) If \alpha_0 is an ECH generator, then the \alpha_0 component of the above map has degree 1.

2) If \alpha_0 is not an ECH generator (i.e. if \alpha_0 contains multiply covered hyperbolic orbits), then the \alpha_0 component of the above map has degree 0.

If we can prove both (1) and (2), then the number of points in \partial\overline{\mathcal M}_2(\alpha_+,\alpha_-) equals the coefficient \partial^2\alpha_+,\alpha_-\rangle, and since the number of points in the boundary of a compact 1-manifold is zero, we are done. (If we are proving \partial^2=0 over {\mathbb Z} instead of over {\mathbb Z}/2, 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 1 holomorphic curves C_+ and C_- in {\mathbb R}\times Y, and if there are (possibly multiply covered) Reeb orbits \gamma_1,\ldots,\gamma_n such that the negative ends of C_+, as well as the positive ends of C_-, are at \gamma_1,\ldots,\gamma_n, 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 (C_+,C_-). (If some of the Reeb orbits \gamma_i 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 C_+\in{\mathcal M}_1(\alpha_+,\alpha_0) and C_-\in{\mathcal M}(\alpha_0,\alpha_-), then the above picture usually does not apply. The reason is that if \gamma is an embedded Reeb orbit that appears with multiplicity m in \alpha_0, then we know that the total covering multiplicity of negative ends of C_+ at coverings of \gamma agrees with the total covering multiplicity of positive ends of C_- at coverings of \gamma, namely both totals equal m. However the individual covering multiplicities of the negative ends of C_+ and the positive ends of C_- at \gamma need not agree. In fact, we will see below that if {m}{>}{1}, then (in the absence of trivial cylinders) these multiplicities never agree when \gamma is elliptic, although they do agree when \gamma is hyperbolic.

At first sight, one might think that it is impossible to glue C_+ and C_- when \alpha_0 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 J-holomorphic curves with ECH index 1.

Partition conditions

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

If \gamma is hyperbolic, then p_\gamma^+(m)=p_\gamma^-(m). If \gamma is positive hyperbolic, then this partition is (1,\ldots,1). If \gamma is negative hyperbolic, then this partition is (2,\ldots,2) if m is even, and (2,\ldots,2,1) if m is odd.

If \gamma is elliptic with rotation angle \theta\in{\mathbb R}\setminus{\mathbb Q} for some trivialization of \xi|_\tau, then p_\gamma^\pm(m) = p_\theta^\pm(m), where p_\theta^\pm(m) depends only on m and the class of \theta in {\mathbb R}/{\mathbb Z} (which does not depend on \tau) and is defined as follows. To define p_\theta^+(m), let \Lambda_\theta^+(m) be the maximal concave polygonal path in the plane with vertices at lattice points which starts at the origin, ends at (m,\lfloor m\theta\rfloor), and lies below the line y=\theta x. Then p_\theta^+(m) consists of the horizontal displacements of the segments of \Lambda_\theta^+(m) connecting consecutive lattice points. The partition p_\theta^-(m) is defined analogously from the path \Lambda_\theta^-(m), which is the minimal convex polygonal path with vertices at lattice points which starts at the origin, ends at (m,\lceil m\theta\rceil), and lies above the line y=\theta x.

To give the simplest example, if \theta\in(0,1/m), then p_\theta^+(m) = (1,\ldots,1) and p_\theta^-(m)=(m). In general, more interesting partitions are possible.

At this point I should probably give you some fun combinatorial exercises involving p_\theta^\pm(m), for example hints to prove that p_\theta^+(m) and p_\theta^-(m) are disjoint whenever {m}{>}{1}. But you’re probably wondering why I introduced these partitions. The reason why is the following:

Proposition (equality condition in the writhe bound). Let \alpha=\{(\alpha_i,m_i)\} and \beta=\{(\beta_j,n_j)\} be orbit sets. Let C\in{\mathcal M}(\alpha,\beta) be a somewhere injective J-holomorphic curve. Then equality holds in the writhe bound

w_\tau(C) \le CZ_\tau^I(\alpha,\beta) - CZ_\tau^{ind}(C),

and hence equality holds in the index inequality

ind(C) \le I(C) - 2\delta(C),

only if the following hold:

1) For each i, the covering multiplicities of the positive ends of C at covers of \alpha_i are exactly the entries in the partition p_{\alpha_i}^+(m_i).

2) For each j, the covering multiplicities of the negative ends of C at covers of \beta_j are exactly the entries in the partition p_{\beta_j}^-(n_j).

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 J 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 \partial^2=0 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 {\mathbb Z}/2 coefficients; {\mathbb Z} coefficients involve additional issues) follows from the standard gluing theory mentioned above: if \alpha_0 is an ECH generator, then it contains no multiply covered orbits, so the negative ends of C_+ and the positive ends of C_- are at the same, distinct, embedded Reeb orbits.

For point (2), suppose that \alpha_0 contains a positive hyperbolic orbit \gamma with multiplicity m{>}{1}. Since p_\gamma^\pm(m)=(1,\ldots,1), we know that C_+ has m negative ends at \gamma, and C_- has m positive ends at \gamma. To glue C_+ and C_-, we need to choose a bijection between the negative ends of C_+ at \gamma and the positive ends of C_- at \gamma. There are m! ways to do this. Note that m! is even.

Likewise, if \alpha_0 contains a negative hyperbolic orbit \gamma with multiplicity m{>}{1}, then C_+ has \lfloor m/2\rfloor negative ends at the double cover of \gamma (plus one end at \gamma if m is odd), while C_- has positive ends at covers of \gamma with these same multiplicities. To glue C_+ and C_-, we need to choose a bijection between the \lfloor m/2\rfloor negative ends of C_+ and the \lfloor m/2\rfloor positive ends of C_- at the double cover of \gamma; 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 k!2^k where k=\lfloor m/2\rfloor \ge 1. Note that this number of choices is again even.

For each hyperbolic orbit that appears in \alpha_0 with multiplicity {>}1, we need to make choices as above. The total number of choices is even. This immediately implies point (2) if we are working with {\mathbb Z}/2 coefficients. If we use {\mathbb Z} 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 \alpha_0 consists of just a single pair (\gamma,m), where \gamma is elliptic and {m}{>}{1}. Let C_+\in{\mathcal M}(\alpha_+,\alpha_0) and C_-\in{\mathcal M}(\alpha_0,\alpha_-) 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 C_+ and C_- do not contain trivial cylinders or covers thereof. The covering multiplicities of the negative ends of C_+ determine a partition p_- of m, and the covering multiplicities of the positive ends of C_- determine a partition p_+ of m. Since we are not assuming equality in the index inequality, we do not necessarily have p_\pm = p_\gamma^\pm(m).

If p_+\neq p_-, can we glue C_+ and C_- to a curve of Fredholm index 2? The answer is sometimes, but not directly. What we really do is glue a triple (C_+,C_0,C_-), where C_0 is a Fredholm index zero degree m branched cover of {\mathbb R}\times\gamma, so that the covering multiplicities of the positive ends of C_0 are the entries in the partition p_-, and the covering multiplicities of the negative ends of C_0 are the entries in the partition p_+. (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 C_0 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 C_0, 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 G_\gamma(p_-,p_+). It turns out that G_\gamma(p_-,p_+) 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 1. But it turns out that we have the following:

Amazing fact. G_\gamma(p_-,p_+)=1 if and only if p_\pm = p_\gamma^\pm(m).

That is, the number of gluings equals 1 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 \partial^2=0, which was my original plan for proving that \partial^2=0 before switching to the branched cover approach for reasons explained here.  I will try to explain this symmetric product picture later.

This entry was posted in ECH. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s