Automatic transversality for dummies

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 $X$ be a (completed) four-dimensional strong symplectic cobordism between contact manifolds, and let $J$ be an almost complex structure on $X$ satisfying the usual conditions. If $C$ is a $J$ -holomorphic curve in $X$ , let $g$ denote the genus of $C$ , let $h_+$ denote the number of ends of $C$ at positive hyperbolic orbits, and let $ind$ denote the Fredholm index of $C$ .

Theorem. Let $C$ be an immersed $J$ -holomorphic curve in $X$ with ends at nondegenerate Reeb orbits. Suppose that $2g-2+h_+<ind$ . Then $C$ is automatically transverse (regardless of whether or not $J$ is generic).

Chris Wendl has a paper proving more general versions of this in which $C$ 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 $C$ are positive; it is easy to fix the notation when this is not the case.

Since $C$ is immersed, it has a well-defined normal bundle $N$ , and we can regard the linearized Cauchy Riemann operator $D$ as a map from sections of $N$ to sections of $T^{0,1}C\otimes N$ . Suppose that $C$ is not transverse. Then the formal adjoint $D^*$ has a nonzero kernel. Let $\psi$ be a section of $T^{0,1}C\otimes N$ which is an element of $Ker(D^*)$ . It is a standard fact that every zero of $\psi$ is isolated and has negative multiplicity. (This follows for example from the Carleman similarity principle.) Also the zero set of $\psi$ is compact (this follows from stuff about the asymptotics of $\psi$ which I can review later). So the algebraic count of zeros of $\psi$ is negative:

$\#\psi^{-1}(0) \le 0.$

Now the algebraic count of zeroes of $\psi$ is given in terms of the relative first Chern class of $T^{0,1}C\otimes N$ by the formula

$\#\psi^{-1}(0) = c_1(T^{0,1}C\otimes N,\tau) + wind_\tau(\psi).$

Here $\tau$ is a trivialization of $T^{0,1}C\otimes N$ over the ends of $C$ , and $wind_\tau(\psi)$ denotes the sum of the winding numbers of $\psi$ around the ends with respect to the trivialization $\tau$ . Indeed, one can regard this formula as the definition of the relative first Chern class $c_1(\cdot,\tau)$ .

We can take the trivialization $\tau$ of $T^{0,1}C\otimes N$ over the ends to be induced by a trivialization of the contact plane field $\xi$ over the ends, which we also denote by $\tau$ , together with the canonical trivialization of $T^{0,1}C$ over the ends. Then

$c_1(T^{0,1}\Sigma\otimes N,\tau) = \chi(C) + c_1(N,\tau) = c_1(TX|_C,\tau).$

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

$wind_\tau(\psi) \ge \sum_i \lceil CZ_\tau(\gamma_i)/2\rceil.$

Here $i$ indexes the ends of $C$ , $\gamma_i$ denotes the Reeb orbit at the $i^{th}$ end of $C$ , and $CZ_\tau$ denotes the Conley-Zehnder index with respect to $\tau$ . Note that $CZ_\tau(\gamma_i)$ is even if and only if $\gamma_i$ is positive hyperbolic. So if $k$ denotes the total number of ends of $C$ , we can rewrite the above inequality as

$2 wind_\tau(\psi) \ge \sum_i CZ_\tau(\gamma_i) + k - h_+.$

Next recall that the Fredholm index is given by the formula

$ind = -\chi(C)+ 2c_1(TX|_C,\tau) + \sum_i CZ_\tau(\gamma_i).$

We now put everything together as follows:

$0 \ge 2\#\psi^{-1}(0)$

$= 2c_1(T^{0,1}C\otimes N,\tau) + 2wind_\tau(\psi)$

$= 2c_1(TX|_C,\tau) + 2wind_\tau(\psi)$

$\ge 2c_1(TX|_C,\tau) + \sum_i CZ_\tau(\gamma_i) + k - h_+$

$= ind + 2 - 2g - h_+.$

So $0 \ge ind + 2 - 2g - h_+$ if $C$ is not transverse. QED.

It is easy to see that automatic transversality can fail for an index $0$ 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 $1$ 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.

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5 Responses to Automatic transversality for dummies

Chris Wendl says:

September 22, 2012 at 5:30 pm

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.

- floerhomology says:
  
  September 22, 2012 at 11:15 pm
  
  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 $2g-2+h_+ = ind$ .
  
  - Chris Wendl says:
    
    September 22, 2012 at 11:38 pm
    
    Ah, I see… yes, quite right, I withdraw my quibble!
Paolo says:

October 15, 2012 at 5:19 pm

Hi Michael,

Could you explain why the count of zeroes of $\psi$ is negative? I know that elements of the cokernel are solution of the Cauchy-Riemann type equation on $T^{0,1} C \otimes N$ , so shouldn’t the count of zeroes be positive? Sorry for being stupid (or ignorant) here.

- floerhomology says:
  
  October 15, 2012 at 5:27 pm
  
  In a local complex coordinate $z=s+it$ and a local trivialization of $N$ , the operator $D$ has the form $\partial_s + i \partial_t + A(s,t)$ where $A(s,t)$ is a $2\times 2$ matrix. The formal adjoint $D^*$ then has the local form $-\partial_s + i\partial_t + A(s,t)^*$ . 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 $A(s,t)$ were not there, then $\psi$ would be in the kernel of $D^*$ if and only if $\overline{\psi}$ 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.