The cylindrical contact homology differential

Sorry I’ve been too busy to post lately. There are a lot of things which I would like to write as soon as I have time. Anyway I wanted to make one brief remark now.

Cylindrical or linearized contact homology does not yet have a rigorous definition in most cases (although the heroic polyfold project of Hofer-Wysocki-Zehnder is supposed to fix this). However, for a contact three-manifold, if there are no contractible Reeb orbits, then there is little difficulty in defining the cylindrical contact homology differential for generic J. Since people keep asking me about this, I thought I should explain it here.

The most important point is that any cylinder which is a multiple cover of a nontrivial holomorphic cylinder has Fredholm index at least 1. To see this, assume that J is generic and let C be a nontrivial somewhere injective J-holomorphic cylinder. Then C has index at least 1.  Let \gamma_+ and \gamma_- denote the Reeb orbits at the positive and negative end of C, respectively. Choose trivializations of the contact plane field over \gamma_+ and \gamma_- which extend over C. Then

ind(C) = CZ(\gamma_+) - CZ(\gamma_-)

where CZ denotes the Conley-Zehnder index with respect to the above trivialization. Suppose C has index 1. Then we can choose the trivialization so that one of these Conley-Zehnder indices is 0. Then without loss of generality, CZ(\gamma_-)=0 and CZ(\gamma_+)=1. If C^k denotes the cylinder which is a k-fold cover of C, then

ind(C^k) = CZ(\gamma_+^k) - CZ(\gamma_-^k).

But \gamma_- is positive hyperbolic, and \gamma_+ is elliptic or negative hyperbolic, so it follows from the formulas for the Conley-Zehnder index in this post that CZ(\gamma_-^k)=0 and CZ(\gamma_+^k)\ge 1. Thus ind(C^k)\ge 1. We were assuming that ind(C)=1, but if ind(C)>1 then one can get even stronger inequalities.

Since there are no nontrivial multiply covered cylinders of index \le 0, there is no difficulty with the compactness to prove that the differential is defined and has square zero.

What about transversality for defining the differential? Note that an immersed cylinder of index 1 is automatically transverse for any J. This is because its “normal Chern number” 2g-2+ind+h_+, where g denotes the genus and h_+ denotes the number of ends at positive hyperbolic orbits, is equal to zero; and automatic transversality holds for immersed curves whenever the normal Chern number is less than or equal to zero. (Incidentally, while curves with nonpositive normal Chern number are good from this perspective, they are bad for ECH, because their multiple covers can have zero or negative ECH index, as I can explain another time.)

What about cylinders which are not immersed? The somewhere injective ones of course will be transverse for generic J. The only thing I’m not sure about is multiple covers of cylinders with singularities. I don’t know whether these are transverse for generic J or whether one needs to do something fancier to count them. (But I haven’t thought hard about this. Maybe one of the experts can weigh in here.)

So, modulo that last issue, defining the cylindrical contact differential is OK. However, negative index multiple covers become a problem if one tries to define cobordism maps on cylindrical contact homology, or if one allows contractible Reeb orbits and tries to define the contact homology algebra. However I think it is interesting to try to understand these objects without using too much technology, so I might discuss this some more later.

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6 Responses to The cylindrical contact homology differential

  1. Chris Wendl says:

    The problem with non-immersed index 1 multiply covered cylinders has a very simple solution: in the symplectization of a 3-manifold, there are none!

    Proof: As you observed, any index 1 cylinder has normal Chern number 0. Now, any curve u in a symplectization which is not a cover of a trivial cylinder satisfies the relation windpi(u) \le c_N(u), where windpi(u) is the integer introduced in HWZ “Properties… II” that counts points where u is not transverse to the Reeb vector field. By the similarity principle, those points are always isolated and count positively, so any index 1 cylinder therefore satisfies windpi(u) = 0, which means not only is it immersed, but it is everywhere transverse to the Reeb vector field. Then automatic transversality applies, even if it’s a multiple cover.

    Note that the above argument is VERY dependent on R-invariance and completely fails in a symplectic cobordism: morally, the reason why windpi(u) is nonnegative is that what it’s counting is the number of intersections between u and an infinitessimal R-translation of itself, and of course these intersections are positive if J is R-invariant. And of course it goes without saying that none of this works at all in higher dimensions.

    The relation I quoted between windpi and the normal Chern number is essentially due to Hofer-Wysocki-Zehnder (“Properites … II” in GAFA), but that was before anyone ever defined the normal Chern number… I think I rewrote it in these terms in one of my early papers, most likely the compactness paper in JEMS.

    And by the way, I believe I first learned the fact that automatic transversality can be used in this way to define cylindrical contact homology from Al Momin, and it should be explained in his thesis. I’ll point him to this post and see if he can confirm or deny that.

    • Great, thanks! (I think I should have remembered this, oh well.) I’ll be interested to hear more about what Al Momin did.

      This automatic transversality also implies that the cylindrical contact homology differential does not depend on J (because as you deform J, you will never see nontrivial index 0 curves). Is there any way to see what the differential is without using a J?

      • Chris Wendl says:

        I’m not quite with you there… why couldn’t you see index 0 curves as J deforms? I think you could, but only if both orbits are positive hyperbolic (otherwise the normal Chern number becomes negative and violates the windpi inequality). To make matters worse: index 0 cylinders with two positive hyperbolic orbits don’t satisfy automatic transversality, so I’m not sure how to deal with transversality for multiple covers in that case. We know at least that they are immersed (because windpi=0), so that’s nice, but is not sufficient for transversality. It’s not inconceivable that there could be some kind of “super-rigidity” phenomenon rescuing this situation, but there I’m out of my depth.

        About Al Momin: he could always chime in himself (I know he’s read this post), but since he hasn’t, most of what we’ve been talking about is explained in Proposition 2.10 of

      • Oops, I thought that 2g-2+ind+h_+\le0 implies automatic transversality, but I forgot that you also need h_+\le 1 (and I might be saying further dumb things regarding things that I haven’t thought about for a long time). So I retract my claim about the differential being independent of J. Thanks for the Momin regerence.

      • Chris Wendl says:

        This is a reply to Michael’s comment on my comment about automatic transversality for index 0 curves… for some reason the blog won’t let me reply to that comment directly.

        The correct statement of automatic transversality is that for immersed curves it holds if ind > 2g - 2 + h_+, or equivalently, the index is strictly greater than the normal Chern number. Notice the strict inequality; that’s why index 0 cylinders with h_+ = 2 don’t make it. (The “dual” of this criterion is the fact that a Cauchy Riemann operator on a line bundle is injective if the first Chern number is negative — also a strict inequality!) There is no general requirement that h_+ \le 1, but in practice the above inequality implies that this must hold if you’re interested in curves of index at most 2, which I believe is the case in every application where this has ever appeared (i.e. counting rigid curves, or constructing finite energy foliations).

  2. Al Momin says:

    I should have chimed in earlier and saved Chris the trouble (Chris pointed me here this morning) – I can at least say that I endorse his message. Should anyone wish a journal reference, I can perhaps save them a little trouble with this:

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