In my last post and in the comments I misremembered and garbled various things about automatic transversality. Let me try to clean things up here.

As Chris Wendl pointed out, a correct statement is the following. Let be a (completed) four-dimensional strong symplectic cobordism between contact manifolds, and let be an almost complex structure on satisfying the usual conditions. If is a -holomorphic curve in , let denote the genus of , let denote the number of ends of at positive hyperbolic orbits, and let denote the Fredholm index of .

**Theorem.** Let be an immersed -holomorphic curve in with ends at nondegenerate Reeb orbits. Suppose that . Then is automatically transverse (regardless of whether or not is generic).

Chris Wendl has a paper proving more general versions of this in which does not have to be immersed and can have ends at Morse-Bott Reeb orbits. (I have used the Morse-Bott version before.) But let me just explain the proof of the above, in my notation.

**Proof. **To simplify notation, assume all ends of are positive; it is easy to fix the notation when this is not the case.

Since is immersed, it has a well-defined normal bundle , and we can regard the linearized Cauchy Riemann operator as a map from sections of to sections of . Suppose that is not transverse. Then the formal adjoint has a nonzero kernel. Let be a section of which is an element of . It is a standard fact that every zero of is isolated and has negative multiplicity. (This follows for example from the Carleman similarity principle.) Also the zero set of is compact (this follows from stuff about the asymptotics of which I can review later). So the algebraic count of zeros of is negative:

Now the algebraic count of zeroes of is given in terms of the relative first Chern class of by the formula

Here is a trivialization of over the ends of , and denotes the sum of the winding numbers of around the ends with respect to the trivialization . Indeed, one can regard this formula as the definition of the relative first Chern class .

We can take the trivialization of over the ends to be induced by a trivialization of the contact plane field over the ends, which we also denote by , together with the canonical trivialization of over the ends. Then

It also follows from the asymptotics of which I didn’t explain here that

Here indexes the ends of , denotes the Reeb orbit at the end of , and denotes the Conley-Zehnder index with respect to . Note that is even if and only if is positive hyperbolic. So if denotes the total number of ends of , we can rewrite the above inequality as

Next recall that the Fredholm index is given by the formula

We now put everything together as follows:

So if is not transverse. QED.

It is easy to see that automatic transversality can fail for an index cylinder with both ends at positive hyperbolic orbits (contrary to a foolish claim that I made in the comments to the previous post). Just think about the symplectic fixed point Floer homology for the indentity map on a surface, perturbed by a Morse function. (This is not quite the contact geometry setting but it is close.) It is possible to have a flow line between two index critical points (e.g. for the height function on a torus standing on its side), and this gives rise to a holomorphic cylinder of index zero whose ends are at positive hyperbolic orbits.

Sorry, I still have a minor quibble. 🙂

I don’t think the example at the end shows that automatic transversality fails for index 0 cylinders with two positive hyperbolic orbits. In fact, there’s no situation I can think of in which automatic transversality implies the non-existence of certain holomorphic curves: for the cylinders in question, the necessary criterion is not satisfied so there is simply no conclusion.

But I like the example anyway because it does give a good hint of the fact that hyperbolic orbits cause problems while elliptic orbits do not: there can be no index 0 gradient flow lines between two critical points of index 0 or 2, and if one didn’t already see that from the fact that those critical points are either maxima or minima, one could PROVE it using the fact that index 0 cylinders with two elliptic orbits DO satisfy automatic transversality, i.e. if such a gradient line existed, then the corresponding cylinder would have to be Fredholm regular, which it clearly is not since it is not isolated due to R-invariance.

I’m not sure I understand your quibble, but I think we’re saying the same thing. What I am saying is that nontrivial index 0 cylinders with both ends at positive hyperbolic orbits are not automatically transverse (because sometimes they exist in which case they are necessarily not transverse). Of course the automatic transversality theorem that I reviewed does not apply here because .

Ah, I see… yes, quite right, I withdraw my quibble!

Hi Michael,

Could you explain why the count of zeroes of is negative? I know that elements of the cokernel are solution of the Cauchy-Riemann type equation on , so shouldn’t the count of zeroes be positive? Sorry for being stupid (or ignorant) here.

In a local complex coordinate and a local trivialization of , the operator has the form where is a matrix. The formal adjoint then has the local form . The symbol of this operator is that of a conjugate Cauchy-Riemann operator, so its zeroes have negative multiplicity instead of positive. That is, if the zeroth order term were not there, then would be in the kernel of if and only if is holomorphic; and the Carleman similarity principle basically says that you can locally change the coordinates and trivializations to make the zeroth order term disappear, so that this special case implies the general case.