## Problem with cylindrical cobordism maps fixed?

In a previous post, I discussed the problem of defining cylindrical contact homology for a contact three-manifold $(Y,\lambda)$ with no contractible Reeb orbits. (In that post I assumed for simplicity that there are no bad Reeb orbits, but I will allow bad Reeb orbits here. One can also somewhat relax the assumption of no contractible Reeb orbits, but let’s not try to solve all problems at once.) Here the differential is well-defined and has square zero for generic almost complex structures $J$ on ${\mathbb R}\times Y$, but there are transversality difficulties with defining cobordism maps. One can solve these transversality problems by using $S^1$-dependent almost complex structures, but this then introduces error terms into the chain map equation for the cobordism map. (You do get a cobordism map, but on a different theory: the “non-equivariant contact homology”, as detailed in this post.)

Now, I think I figured out how to fix the above difficulty and get a cobordism map on cylindrical contact homology of a contact three-manifold without contractible Reeb orbits. Warning: there is a good chance that this discussion, to the extent that it is correct at all, is converging to a convoluted alternate version of what Bourgeois and Oancea do in their paper on equivariant symplectic homology (which I really need to understand better). So apologies if this series of blog posts is turning into a diary of my rediscovery of things which are known in better ways to the experts. Still, I feel that when this blog asks questions, it should answer them if possible.

So, I will first review the problem with some updated notation, and then explain how I think it can be solved.

1. Review of nonequivariant contact homology (updated notation)

Let $(Y,\lambda)$ be a nondegenerate contact manifold of any dimension. Assume that either $Y$ is closed or we are in some other situation where Gromov compactness is applicable. We then define the nonequivariant contact homology complex $SC_*(Y,\lambda)$ as follows. This is defined over ${\mathbb Z}$. For each Reeb orbit $\gamma$, good or bad, there are two chain complex generators $\check{\gamma}$ and $\hat{\gamma}$.

To define the differential $\partial$, one chooses a generic one-parameter family $\{J_t\}_{t\in S^1}$ of almost complex structures $J_t$ on ${\mathbb R}\times Y$ each satisfying the usual conditions. One also chooses a base point $p_\gamma$ on the image of each Reeb orbit $\gamma$. If $\gamma_+$ and $\gamma_-$ are two distinct Reeb orbits, one defines ${\mathcal M}(\gamma_+,\gamma_-)$ to be the set of maps $u:{\mathbb R}\times S^1\to {\mathbb R}\times Y$ satisfying the equation $\partial_su + J_t\partial_tu=0$ such that $\lim_{s\to\pm\infty}\pi_Y(u(s,\cdot))$ is a reparametrization of $\gamma_\pm$. We mod out by ${\mathbb R}$ translation in the domain. This moduli space is cut out transversely if the family $\{J_t\}_{t\in S^1}$ is generic.

If $\alpha$ and $\beta$ are distinct Reeb orbits, the differential coefficient from $\check{\alpha}$ or $\hat{\alpha}$ to $\check{\beta}$ or $\hat{\beta}$ counts curves in ${\mathcal M}(\alpha,\beta)$, modulo ${\mathbb R}$ translation in the target. If $\alpha$ has a check on it, then we impose the further constraint $\lim_{s\to \infty} \pi_Y(u(s,0)) = p_\alpha$. If $\beta$ has a hat on it, then we impose the further constraint $\lim_{s\to-\infty} \pi_Y(u(s,0)) = p_\beta$. The differential also counts appropriate “Morse-Bott cascades”, see this post for details. Finally, if $\alpha=\beta$, then all differential coefficients between $\check{\alpha}$ and/or $\hat{\alpha}$ are zero, except that the coefficient $\langle\partial\hat{\alpha},\check{\alpha}\rangle = 2$ when $\alpha$ is a “bad” Reeb orbit.

We can write the differential as a block matrix in terms of the check and hat generators as

$\partial = \begin{pmatrix} \check{\partial} & \partial^+ \\ \partial^- & \hat{\partial}\end{pmatrix}.$

Here $\check{\partial}$ denotes the component going between check generators, and $\hat{\partial}$ denotes the component going between hat generators. These decrease the usual grading on contact homology by one. Next, $\partial^+$ denotes the component from hat to check generators; this preserves the usual grading on contact homology. The plus superscript is a reminder that this shifts the grading up more than usual. This is a new notation which is convenient for the calculations to come below. Finally, $\partial^-$ denotes the component from check to hat generators; this decreases the usual grading on contact homology by two, with the minus superscript indicating that the grading is shifted down more than usual.

2. Review of cylindrical contact homology (updated notation)

Continuing with the above setup, suppose that we can choose $J_t$ to be independent of $t\in S^1$ so that we still have the necessary transversality of the moduli spaces ${\mathcal M}(\gamma_+,\gamma_-)$. This is possible when $\dim(Y)=3$, although it is rarely possible in higher dimensions. We now define an operator $\delta$ on the free ${\mathbb Z}$-module generated by all Reeb orbits, good or bad, as follows. If $\alpha$ or $\beta$ is bad, then the coefficient $\langle\delta\alpha,\beta\rangle=0$. If $\alpha$ and $\beta$ are both good, then the coefficient $\langle\delta\alpha,\beta\rangle$ counts curves in ${\mathcal M}(\alpha,\beta)$, modulo $S^1$ translation in the domain and ${\mathbb R}$ translation in the target. We count each such curve $u$ with weight $\pm 1/m(u)$, where $m(u)$ denotes the covering multiplicity of $u$.

The operator $\delta$ now satisfies the equation

$\delta\kappa\delta = 0,$

where $\kappa$ is the operator that multiplies each Reeb orbit by its covering multiplicity. Thus we can define the cylindrical contact homology differential as either $\delta\kappa$ or $\kappa\delta$, and then it will have square zero. (And in the definition of the chain complex we throw out the bad orbits.)

While both of these differentials are defined over ${\mathbb Z}$, to get an invariant cylindrical contact homology we need to use ${\mathbb Q}$ coefficients. We will see one reason for this below.

Anyway, in the above situation the nonequviariant contact homology differential is given by

$\partial = \begin{pmatrix} \delta\kappa & \partial^+ \\ \partial^- & -\kappa\delta\end{pmatrix}$

where now $\partial^+(\hat{\alpha})$ equals $2\check{\alpha}$ when $\alpha$ is bad and $0$ when $\alpha$ is good. (When $J_t$ depends on $t$, the operator $\partial^+$ may include more terms.) Note that I’m guessing about the sign here, but it has to be something like this in order for the hat to check component of $\partial^2$ to equal zero. Note also that $\partial^2=0$ implies that  the coefficients of $\partial^-$ can be nonzero only between good Reeb orbits.

3. The problem with cobordism maps in the three dimensional case

Suppose now that we have an exact symplectic cobordism between two contact three-manifolds $(Y,\lambda)$ and $(Y',\lambda')$ as above, such that no Reeb orbit is contractible in the cobordism. Choose $S^1$-independent almost complex structures on the symplectizations of $Y$ and $Y'$ as above. I will use the same notation for the differentials on both contact three-manifolds. We would like to define a cobordism map between the cylindrical contact homology chain complexes.

To start, we can use an $S^1$-dependent almost complex structure on the completed cobordism to define a cobordism chain map $\phi_0$ between the nonequviariant contact homology chain complexes. This can be written in block form as

$\phi_0 = \begin{pmatrix} \check{\phi}_0 & \phi_0^+ \\ \phi_0^- & \hat{\phi}_0 \end{pmatrix}.$

Here the $0$ subscript is a reminder that this map preserves the grading on the nonequivariant contact homology. We will later look at maps that shift this grading by $k$, and these will have a $k$ subscript.

The check to check component of the chain map equation $\partial\phi_0=\phi_0\partial$ is

$\delta\kappa\check{\phi}_0 + \partial^+\phi_0^- = \check{\phi}_0\delta\kappa - \phi_0^+\partial^-.$

This means that if we had $\phi_0^+=0$, then we could use the good-to-good part of $\check{\phi}$ to define a chain map on cylindrical contact homology (using the convention $\delta\kappa$ for the differential). Likewise, we could use the good-to-good part of $\hat{\phi}$ to define a chain map on cylindrical contact homology, usuing the convention $\kappa\delta$ for the differential.

Actually, all we need is that $\phi_0^+=0$ going between good Reeb orbits, which will be important below.

If the almost complex structure on the cobordism were $S^1$-independent and we still had transversality, then we would indeed get $\phi_0^+=0$ (on all Reeb orbits). Unfortunately we usually cannnot get $S^1$-independence and transversality simultaneously on the cobordism.

In our situation where we have $S^1$-dependent $J$ and transversality, we would like to replace $\phi_0$ by a map of the form $\phi_0 - (\partial K_1 + K_1\partial)$ such that the hat to check component vanishes going between good Reeb orbits, where $K_1$ is natural up to an appropriate equivalence relation. But how are we going to cook up this map $K_1$?

4. Strategy

Let ${\mathcal J}$ denote the space of all families $\{J_t\}_{t\in S^1}$ on the completed cobordism, say with the smooth topology. There is an $S^1$-action on this space which shifts the parameter $t$. We know that for a generic point in ${\mathcal J}$, we get a chain map on the nonequivariant contact homology chain complex. Even though fixed points of the $S^1$ action are not sufficiently generic to give us chain maps, we would still like to detect something about the $S^1$-action and the fixed points in order to obtain the desired degree $1$ operation.

A generic path in ${\mathcal J}$ will give a chain homotopy between the chain maps determined by the endpoints of the path. In particular a generic loop in ${\mathcal J}$ will give a degree $1$ chain map. For example, a generic orbit of the $S^1$ action gives us a degree $1$ chain map. Also, since the space ${\mathcal J}$ is contractible, the resulting degree $1$ chain map is chain homotopic to zero. This is enough to get started, although we will also need to consider higher chain homotopies later… Let’s now explain the details.

5. Finding the chain homotopy: first step.

Choose a generic family $\{J_t\}_{t\in S^1}$ on the completed cobordism as needed to define the chain map $\phi_0$. The $S^1$-orbit in ${\mathcal J}$ of the family $\{J_t\}$ now defines a degree $1$ chain map $\phi_1$ on the nonequivariant contact homology chain complexes. What is this map $\phi_1$? It counts curves like in $\phi_0$, but which shift the grading up by $1$, that appear sometime in the orbit. Now as we rotate $\{J_t\}$ in its orbit, the holomorphic curves stay the same, except that we have to change the parametrization of the $S^1$ coordinate.

For example, suppose there is a map $u$ that contributes to $\langle\phi_0\hat{\alpha},\check{\beta}\rangle$. This means that it satisfies $\partial_su+J_t\partial_tu=0$ and $\lim_{s\to +\infty}\pi_Y(u(s,\cdot))$ is a reparametrization of $\alpha$ and $\lim_{s\to -\infty}\pi_{Y'}(u(s,\cdot))$ is a reparametrization of $\beta$. Now if we act on $\{J_t\}$ by $t_0\in S^1$, then the map $u$ gets replaced by $u(\cdot,\cdot - t_0)$. If $t_0$ is such that $\lim_{s\to +\infty}\pi_Y(u(s,t_0))=p_\alpha$, then we will pick up a contribution to $\langle\phi_1\check{\alpha},\check{\beta}\rangle$. That is, the curves that contribute to $\phi_0^+$ (which have no asymptotic marker constraints) also contribute to $\check{\phi}_1$ and $\hat{\phi}_1$, because as one reparametrizes the $S^1$ coordinate, sometimes the asymptotic marker constraint at the top or the bottom is satisfied. The upshot is that

$\phi_1 = \begin{pmatrix} \phi_0^+\kappa & 0 \\ \phi_1^- & -\kappa\phi_0^+\end{pmatrix}$

going between good Reeb orbits. (The diagonal entries involving bad Reeb orbits are more complicated because of the funny orientation issues with bad orbits.) Again, I am guessing about the sign, but it has to be something like this. The component $\phi_1^-$ is a bit trickier than the diagonal terms, but we don’t need to know what it is.

So we have a degree $1$ chain map $\phi_1$. But remember that we are looking for a degree $1$ chain homotopy $K_1$, between $\phi_0$ and something with no hat to check component between good Reeb orbits. That is, we want to find a degree $1$ map

$K_1 = \begin{pmatrix} \check{K}_1 & K_1^+ \\ K_1^- & \hat{K}_1\end{pmatrix}$

such that

$\delta\kappa K_1^+ + \partial^+\hat{K}_1 + \check{K}_1\partial^+ - K_1^+\kappa\delta = \phi_0^+$

between good Reeb orbits. To find the chain homotopy $K_1$, note that since the orbit of $\{J_t\}$ in ${\mathcal J}$ is contractible, the map $\phi_1$ is chain homotopic to zero. That is, there is a map

$\phi_2 = \begin{pmatrix} \check{\phi}_2 & \phi_2^+ \\ \phi_2^- & \hat{\phi}_2 \end{pmatrix}$

with

$\partial\phi_2 - \phi_2\partial = \phi_1.$

The diagonal blocks of this equation, between good Reeb orbits, are

$\delta\kappa\check{\phi}_2 + \partial^+\phi_2^- - \check{\phi}_2\delta\kappa - \phi_2^+\partial^- = \phi_0^+\kappa,$

$\partial^-\phi_2^+ - \kappa\delta\hat{\phi}_2 - \phi_2^-\partial^+ + \hat{\phi}_2\kappa\delta = -\kappa\phi_0^+.$

Now the nice situation is where $\phi_2^+=0$ between good Reeb orbits. In this case, if we multiply the first equation on the right by $\kappa^{-1}$, multiply the second equation on the left by $\kappa^{-1}$, and subtract, we obtain

$\delta\kappa(\check{\phi}_2\kappa^{-1} + \kappa^{-1}\hat{\phi}_2) + \partial^+(\phi_2^-\kappa^{-1}) + (\kappa^{-1}\phi_2^-)\partial^+ - (\check{\phi}_2\kappa^{-1} + \kappa^{-1}\hat{\phi}_2)\kappa\delta = 2\phi_0^+$

between good Reeb orbits. It follows that we can solve the equation for $K_1$ by setting

$K_1^+ = \frac{1}{2}(\check{\phi}_2\kappa^{-1} + \kappa^{-1}\hat{\phi}_2),$

$\check{K}_1 = \frac{1}{2}\kappa^{-1}\phi_2^-,$

$\hat{K}_1 = \frac{1}{2}\phi_2^-\kappa^{-1},$

$K_1^- = 0.$

Note that we now have to use ${\mathbb Q}$ coefficients since we are taking the inverse of $\kappa$ and dividing by $2$.

But what if $\phi_2^+\neq 0$ between good Reeb orbits? It may look like we are back where we started, with a troublesome hat to check term, and have just made things more complicated!

6. Finding the chain homotopy: second step.

We would like to replace the map $\phi_2$ with a map of the form $\phi_2 - (\partial K_3 + K_3\partial)$ which has no hat to check term between good orbits. To find the map $K_3$, we play the same game again. We start with a disk in ${\mathcal J}$ which induces the chain homotopy $\phi_2$. The sweepout of the disk by the $S^1$-action then induces a degree $3$ chain map $\phi_3$. By the same argument as before, this map has the form

$\phi_3 = \begin{pmatrix} \phi_2^+\kappa & 0 \\ \phi_3^- & -\kappa\phi_2^+ \end{pmatrix}$

between good Reeb orbits. Since ${\mathcal J}$ is contractible, this degree $3$ map is chain homotopic to zero, so there is a degree $4$ map $\phi_4$ with $\partial\phi_4 - \phi_4\partial = \phi_3$. If $\phi_4^+=0$, then we can use the other blocks of $\phi_4$ to define $K_3$ as above. Otherwise, we have to continue this process.

Ultimately, we will obtain an infinite series for the desired chain map on cylindrical contact homology. However the $k^{th}$ term of this series will be a sum of compositions of maps, such that each composition of maps includes $k$ blocks from the nonequivariant contact homology differential. Consequently, by the symplectic action filtration, this infinite series will be well defined. That is, if we apply this infinite series to a Reeb orbit $\alpha$, and if there are only $k$ orbits in $Y$ or $Y'$ that have action less than $\alpha$, then we only need to add up the first $k$ terms of the series.

7. Higher dimensions?

One could now ask if we can use similar tricks to define the cylindrical contact homology of $(Y,\lambda)$ when $\dim(Y)>3$. Here, as explained before, we generally need an $S^1$-dependent almost complex structure to define the differential on the nonequivariant contact homology chain complex, which then has the form

$\partial = \begin{pmatrix} \check{\partial} & \partial^+ \\ \partial^- & \hat{\partial} \end{pmatrix}.$

We would like to perform a change of basis in the chain complex to arrange that $\partial^+$ vanishes, except for the part taking $\hat{\alpha}$ to $2\check{\alpha}$ when $\alpha$ is bad. If this has been done, then we can take $\check{\partial}$ or $\hat{\partial}$ going between good orbits to be the cylindrical contact homology differential.

How are we going to find the desired change of basis in the chain complex? As before, we can define ${\mathcal J}$ to be the space of smooth one-parameter families $\{J_t\}_{t\in S^1}$ of almost complex structures on ${\mathbb R}\times Y$ satisfying the usual conditions. The orbit of $\{J_t\}$ under the $S^1$ action on ${\mathcal J}$ now induces an isomorphism $\phi_0$ of chain complexes, and this is chain homotopic to the identity, and the disk in ${\mathcal J}$ inducing the chain homotopy can be swept out to induce a degree $2$ chain map $\phi_2$, which is then chain homotopic to zero, etc. This seems like a promising place to start, but now it is less straightforward to understand $\phi_0$ in terms of $\partial$, so I am not sure yet what is going on.

## Hidden branched covers of trivial cylinders

I would now like to explain another ECH-type trick, which I have been meaning to write about here for a while, and which may have applications to other kinds of contact homology in three dimensions and holomorphic curve counts in four dimensions.

1. The general situation.

Suppose we are trying to prove some kind of compactness for some kind of holomorphic curves in the symplectization of a contact three-manifold, or a symplectic cobordism between contact three-manifolds. We know from the general compactness result of Bourgeois-Eliashberg-Hofer-Wysocki-Zehnder that any sequence of holomorphic curves (of bounded symplectic area and genus between the same sets of Reeb orbits) has a subsequence which converges to a “holomorphic building”. A component of this building in a symplectization level may include a branched cover of a trivial cylinder. (Here, a “trivial cylinder” in the symplectization of a contact manifold $Y$ is a cylinder of the form ${\mathbb R}\times \gamma\subset{\mathbb R}\times Y$ where $\gamma$ is an embedded Reeb orbit in $Y$.)

For example, in the proof that $\partial^2=0$ in embedded contact homology, the limit of a sequence of holomorphic curves with ECH index $2$ may include branched covers of trivial cylinders in between curves with ECH index $1$. This is why the pair of papers with Taubes that prove that $\partial^2=0$ in ECH (as well as a more general gluing theorem) is 200 pages, instead of just a paragraph quoting previous gluing theorems.

Now a somewhat worse situation is when a branched cover of a trivial cylinder appears at the top or bottom of the holomorphic building. However I claim that this more or less never happens! In other words, branched covers of trivial cylinders stay “hidden” between other levels of the limiting building, and are never “exposed” at the top or bottom of the building.

I don’t have a general proof (or precise statement of the hypotheses) of this claim, but I can prove it in some cases, sometimes modulo some analysis which still needs to be worked out. I will now explain how to completely prove this claim in a very special case, which however is of some interest.

2. The special case.

Let $Y=S^3$, and let $\lambda$ be a nondegenerate, dynamically convex contact form on $S^3$. Recall that the term “dynamically convex”, going back to Hofer-Wysocki-Zehnder, means that each Reeb orbit $\gamma$ has Conley-Zehnder index $CZ(\gamma)\ge 3$. Here we define the Conley-Zehnder index of $\gamma$ using a trivialization of the restriction of the contact plane field to $\gamma$ that extends over a disk bounded by $\gamma$.

In this situation, one would like to define the cylindrical contact homology of $(S^3,\lambda)$, using dynamical convexity to rule out bubbling of holomorphic planes. That is, in the compactness arguments to prove that the cylindrical contact homology differential $\partial$ is well-defined and satisfies $\partial^2=0$, one has to worry about convergence to holomorphic buildings including holomorphic planes, together with some other holomorphic curves that are genus zero and have one positive but arbitrarily many negative ends. Dynamical convexity implies that every holomorphic plane has Fredholm index at least $2$.  If the other curves in the building are cut out transversely, then the total Fredholm index of the building will be too big, so this kind of degeneration will be ruled out.

3. The problem.

It may be possible to arrange transversality for the relevant holomorphic curves that are not branched covers of trivial cylinders. The paper by Bourgeois-Cieliebak-Ekholm arranges this transversality using $S^1$-dependent almost complex structures, but I don’t see how they get around the other problems with $S^1$-dependent almost complex structures that I have described in recent posts. In fact, there is some hope that the relevant transversality in the symplectization may work for a generic $S^1$-independent almost complex structure. Let’s suppose for the sake of argument that this works, or that we have arranged the necessary transversality some other way. We then get a well-defined differential $\partial$.

There is now a second problem, involving branched covers of trivial cylinders, that arises when one tries to prove that $\partial^2=0$. Specifically, a sequence of index $2$ cylinders may converge to a holomorphic building of the following type. There are two levels $u_1$ and $u_2$. The upper level $u_1$ is an index zero pair of pants with one positive end and two negative ends which is a branched cover of degree $m+1$ of a trivial cylinder ${\mathbb R}\times\gamma$, where $\gamma$ is an embedded elliptic Reeb orbit with monodromy angle $\theta\in(1,2)$ (with respect to the usual trivialization), and $m$ is a positive integer. The lower level $u_2$ has two components. One component is a cylinder which is a degree $m$ cover of ${\mathbb R}\times\gamma$. The other component is a somewhere injective index $2$ holomorphic plane with positive end at $\gamma$.

The above configuration would appear to mess up the proof that $\partial^2=0$, because the above configuration cannot be interpreted in any obvious way as contributing to $\partial^2=0$. However I claim that if $J$ is generic, then the above degeneration never happens.

The proof will use intersection theory, as in the definition of ECH and the work of Siefring.

4. Why this degeneration never happens.

Suppose that $u$ is a holomorphic cylinder which is “close to breaking” into the holomorphic building $(u_1,u_2)$. To explain what I mean by this, let us not mod out by ${\mathbb R}$ translation on ${\mathbb R}\times Y$. Then there is some large positive real number $R$ and some small $\epsilon>0$ such that the intersection of $u$ with $[R,\infty)\times Y$ is within distance $\epsilon$ (pick your favorite norm) of the translate of $u_1$ by $+2R$, the intersection of $u$ with $[-R,R]\times Y$ is within distance $\epsilon$ of the $m$-fold cover of ${\mathbb R}\times\gamma$ union ${\mathbb R}\times\gamma$, and the intersection of $u$ with $(-\infty,R]\times Y$ is within distance of $\epsilon$ of the translate of $u_2$ by $-2R$.

Now let $C$ denote the intersection of $u$ with $[-2R+T,2R+T]\times Y$ where $T$ is a large constant which is independent of $R$. The curve $u$ is necessarily somewhere injective (because $u_2$ is), so $C$ is a surface which is embedded except for finitely many singularities. Let $\delta\ge 0$ denote the count of singularities of $C$ with the usual positive integer weights. Let $\zeta_+$ denote the intersection of $u$ with $\{2R+T\}\times Y$, regarded as a braid with $m+1$ strands in a three-dimensional tubular neighborhood of the Reeb orbit $\gamma$. Likewise let $\zeta_-$ denote the braid obtained by intersecting $u$ with $\{-2R+T\}\times Y$.

A version of the relative adjunction formula, cf. my lecture notes on ECH, section 3.3, implies that

$0 = -1 + w(\zeta_+) - w(\zeta_-) -2\delta.$

Here $w(\zeta_\pm)$ denotes the writhe of the braid $\zeta_\pm$ with respect to the usual trivialization. In particular, it follows that we have a strict inequality

$w(\zeta_+) > w(\zeta_-)$.

We are now going to obtain some independent bounds on the writhes $w(\zeta_\pm)$ and get a contradiction.

5. Writhe bounds.

First, the positive asymptotics of $u$ give the writhe bound

$w(\zeta_+) \le m\lfloor (m+1)\theta\rfloor,$

cf. the lecture notes on ECH, Lemma 5.5(a). This bound can be improved when $m+1$ and $\lfloor(m+1)\theta\rfloor$ have a common factor; one can then subtract $gcd(m+1,\lfloor (m+1)\theta\rfloor) - 1$ from the right hand side. I think this is proved in Siefring’s paper on intersection theory. However we will not need that improvement here.

Now the braid $\zeta_-$ has two components: a component $\zeta_-^1$ with one strand, and a component $\zeta_-^m$ with $m$ strands. Because the negative end of $u$ corresponding to $\zeta_-^m$ decays exponentially for time $4R-T$ before becoming $\zeta_-^m$, while the positive end of $u_2$ corresponding to $\zeta_-^1$ exponentially decays only for time $T$ before becoming $\zeta_-^1$, it follows that if $R$ is sufficiently large then the braid $\zeta_-^1$ wraps around the braid $\zeta_-^m$. Therefore

$w(\zeta_-) = w(\zeta_-^m) + 2m\eta(\zeta_-^1),$

where $\eta(\zeta_-^1)$ denotes the winding number of $\zeta_-^1$ around $\gamma$.

Again, as in Lemma 5.5(a) of the ECH lecture notes, the negative asymptotics of $u$ imply that

$w(\zeta_-^m)\ge (m-1)\lceil m\theta\rceil.$

(Again, this inequality can be improved when $m$ and $\lceil m\theta\rceil$ have a common factor, but we do not need this.) Finally, similarly to Proposition 3.2 in my second gluing paper with Taubes, if $J$ is generic then

$\eta(\zeta_-^1)=1.$

Putting this all together, we get

$w(\zeta_+) - w(\zeta_-) \le m\lfloor (m+1)\theta\rfloor - (m-1)\lceil m\theta\rceil - 2m.$

6. One last step.

To complete the proof, we need to use the assumption that $u_1$ has Fredholm index zero. The Fredholm index of $u_1$ is

$ind(u_1) = 1 + (2\lfloor (m+1)\theta \rfloor + 1) - (2\lfloor m\theta\rfloor + 1) - (2\lfloor\theta\rfloor + 1)$

$= 2(\lfloor (m+1)\theta\rfloor - \lceil m\theta\rceil).$

Putting the fact that this is zero into the previous inequality, we get

$w(\zeta_+) - w(\zeta_-) \le \lceil m\theta\rceil - 2m.$

Since $\theta < 2$, we have $\lceil m\theta\rceil \le 2m$. Thus $w(\zeta_+) - w(\zeta_-)\le 0$, which is the desired contradiction.

## Non-equivariant contact homology

In a previous post, I explained how that the transversality difficulties in defining cylindrical contact homology can be resolved by using $S^1$-dependent almost complex structures, but at the expense of obtaining the wrong theory. The theory that one naturally obtains this way is what one might call “non-$S^1$-equivariant cylindrical contact homology” (which is related to “positive symplectic homology”). It is then a nontrivial matter to extract the desired cylindrical contact homology from this.

In the present post I would like to start over and spell this out in greater generality. (The previous post considered a situation where cylindrical contact homology can be defined, but there is a difficulty defining cobordism maps.) I spoke about this recently at the IAS and Columbia; thanks are due to a number of people in the audiences there who helped clear up some of my many confusions on this topic.

I claim that the theory I am about to describe can be made completely rigorous by doing the following: (1) write down some details about Morse-Bott theory a la Bourgeois in the transverse case (there may already be a reference for this, I don’t know); (2) check some details about orientations, and (3) fix any conceptual errors I have made.

1. Beginning the definition of the chain complex

Let $(Y,\lambda)$ be a nondegenerate contact manifold of any dimension, and assume that there are no contractible Reeb orbits. Assume also that either $Y$ is closed, or we are in some other situation where Gromov compactness is applicable. For example, $Y$ could be a tubular neighborhood of a (possibly degenerate) Reeb orbit in some larger contact manifold $(Y_0,\lambda_0)$, and $\lambda$ could be a nondegenerate perturbation of $\lambda_0$ in this tubular neighborhood. This is the situation of “local contact homology” as in the preprint of Hryniewicz and Macarini.

We now define a chain complex over ${\mathbb Z}$ as follows. For each Reeb orbit $\gamma$ (good or bad), there are two generators, which we will denote by $\hat{\gamma}$ and $\check{\gamma}$. (This notation is borrowed from the paper of Bourgeois-Ekholm-Eliashberg. What I am doing is very similar to things that they and Bourgeois-Oancea have done. I make no claim to any originality here, I am just trying to understand what is going on.) The grading of $\hat{\gamma}$ is one greater than the grading of $\check{\gamma}$.

To define the differential on the chain complex, choose a generic family $\{J_t\}_{t\in S^1}$ of almost complex structures on ${\mathbb R}\times Y$, each satisfying the usual conditions (derivative of the ${\mathbb R}$ direction is sent to the Reeb vector field, contact hyperplane sent to itself compatibly with $d\lambda$, invariant under ${\mathbb R}$ translation). If $\gamma_+$ and $\gamma_-$ are Reeb orbits, define ${\mathcal M}(\gamma_+,\gamma_-)$ to be the set of maps $u:{\mathbb R}\times S^1\to{\mathbb R}\times Y$ such that

$\partial_su + J_t\partial_tu = 0$

and $\lim_{s\to\pm\infty}u(s,\cdot)$ is a reparametrization of $\gamma_\pm$. Here $s$ denotes the $S^1$ coordinate and $t$ denotes the ${\mathbb R}$ coordinate. Also, in the moduli space ${\mathcal M}(\gamma_+,\gamma_-)$, we mod out by ${\mathbb R}$ translation in the domain.

Note that we cannot mod out by $S^1$ translation in the domain, because the equation above is not $S^1$-invariant. Consequently, the expected dimension of this moduli space is one greater than usual. That is, we expect that

$\dim {\mathcal M}(\gamma_+,\gamma_-) = |\gamma_+| - |\gamma_-| + 1$

where $|\gamma_+| - |\gamma_-|$ denotes the usual grading difference (which in general depends on a choice of relative homology class of cylinders connecting $\gamma_+$ and $\gamma_-)$.

I claim that if the family $\{J_t\}_{t\in S^1}$ is generic, then the moduli space ${\mathcal M}(\gamma_+,\gamma_-)$ is cut out transversely in an appropriate sense, and is a manifold of the above dimension. This should follow similarly to the proof of the analogous result for Hamiltonian Floer homology.

2. Morse-Bott moduli spaces

To continue, for each Reeb orbit $\gamma$, choose a base point $p_\gamma$ in the image of $\gamma$ in $Y$. If $\alpha$ and $\beta$ are distinct Reeb orbits, we define “Morse-Bott moduli spaces” ${\mathcal M}(\hat{\alpha},\hat{\beta})$, ${\mathcal M}(\hat{\alpha},\check{\beta})$, ${\mathcal M}(\check{\alpha},\hat{\beta})$ and ${\mathcal M}(\check{\alpha},\check{\beta})$ as follows. Each of these spaces is, as a set, a disjoint union of subsets ${\mathcal M}^k(\cdot,\cdot)$ indexed by nonnegative integers $k$, which (at the risk of some confusion) I will call “levels”. The “primary” levels ${\mathcal M}^0(\cdots,\cdots)$ are given as follows:

${\mathcal M}^0(\hat{\alpha},\hat{\beta}) = \{u\in{\mathcal M}(\alpha,\beta) \mid \lim_{s\to -\infty} \pi_Y(u(s,0)) = p_\beta\}/{\mathbb R},$

${\mathcal M}^0(\hat{\alpha},\check{\beta}) = {\mathcal M}(\alpha,\beta)/{\mathbb R},$

${\mathcal M}^0(\check{\alpha},\hat{\beta}) = \{u\in{\mathcal M}(\alpha,\beta) \mid \lim_{s\to +\infty} \pi_Y(u(s,0)) = p_\alpha, \lim_{s\to -\infty} \pi_Y(u(s,0)) = p_\beta\}/{\mathbb R},$

${\mathcal M}^0(\check{\alpha},\check{\beta}) = \{u\in{\mathcal M}(\alpha,\beta) \mid \lim_{s\to +\infty} \pi_Y(u(s,0)) = p_\alpha\}/{\mathbb R}.$

Here $\pi_Y$ denotes the projection ${\mathbb R}\times Y\to Y$, and we are modding out by ${\mathbb R}$ translation in the target. In short, ${\mathcal M}^0(\cdot,\cdot)$ is just the moduli space from Section 1 modulo ${\mathbb R}$ translation in the target, where a check on a Reeb orbit indicates an asymptotic point constraint on the positive end, and a hat on a Reeb orbit indicates an asymptotic point constraint on the negative end.

The “higher” levels ${\mathcal M}^k(\cdot,\cdot)$ for $k>0$ consist of tuples $(u_0,\ldots,u_k)$ satisfying the following conditions. First, there are distinct Reeb orbits $\alpha=\gamma_0,\gamma_1,\ldots,\gamma_{k+1}=\beta$ such that $u_i\in {\mathcal M}(\gamma_i,\gamma_{i+1})/{\mathbb R}$. Second, the positive end of $u_0$ has a point constraint if $\alpha$ has a check on it. Third, the negative end of $u_k$ has a point constraint if $\beta$ has a hat on it. Fourth, if $1\le i\le k$, then the three points $p_{\gamma_i}$, $\lim_{s\to-\infty}\pi_Y(u_{i-1}(s,0))$, and $\lim_{s\to+\infty}\pi_Y(u_i(s,0))$ are cyclically ordered around the image of the Reeb orbit $\gamma_i$, with respect to the orientation given by the Reeb flow.

In the above we assumed that the Reeb orbits $\alpha$ and $\beta$ are distinct. When they are the same, we define ${\mathcal M}(\hat{\alpha},\hat{\alpha}) = {\mathcal M}(\check{\alpha},\hat{\alpha}) = {\mathcal M}(\check{\alpha},\check{\alpha}) = \emptyset$. Finally, we define ${\mathcal M}(\hat{\alpha},\check{\alpha})$ to be the empty set if $\alpha$ is good, and two (positively oriented) points if $\alpha$ is bad.

It follows from the dimension formula in Section 1 (and a bit more transversality which I think is not hard to arrange) that, ignoring for now the points where different levels come together, the Morse-Bott moduli spaces are manifolds of dimensions

$\dim{\mathcal M}(\hat{\alpha},\check{\beta}) = |\alpha| - |\beta|$,

$\dim{\mathcal M}(\check{\alpha},\check{\beta}) = \dim{\mathcal M}(\hat{\alpha},\hat{\beta}) = |\alpha|-|\beta|-1$,

$\dim{\mathcal M}(\check{\alpha},\hat{\beta}) = |\alpha| - |\beta| - 2$.

3. Morse-Bott gluing

Let $x,y,z$ denote chain complex generators, i.e. Reeb orbits with hats or checks on them. I claim (and this is where some “Morse-Bott gluing” is required) that if $\dim {\mathcal M}(x,y)=1$, then this entire moduli space is a $1$-manifold, and these $1$-manifolds, as well as the $0$-dimensional moduli spaces, can be oriented so that

$\partial{\mathcal M}(x,y) = \coprod_z{\mathcal M}(x,z) \times {\mathcal M}(z,y)$

as oriented manifolds.  (Here by the boundary, I really mean the boundary of a compactification obtained by adding one point for each end.)

For example, suppose $x=\hat{\alpha}$ and $y=\check{\beta}$ where $|\alpha|-|\beta|=1$. Consider the primary level ${\mathcal M}^0(x,y) = {\mathcal M}(\alpha,\beta)/{\mathbb R}$. This is a one-dimensional manifold, but it has boundary because curves in this moduli space can break. A curve can break into a pair $(u_0,u_1)$ where $u_0\in {\mathcal M}(\alpha,\gamma)/{\mathbb R}$ and $u_1\in{\mathcal M}(\gamma,\beta)/{\mathbb R}$ for a third Reeb orbit $\gamma$. For dimension reasons we must have either $|\gamma|=|\alpha|$ or $|\gamma|=|\beta|$. Without much loss of generality, suppose $|\gamma|=|\alpha|$. Then $u_1$ lives in the interior of a one-dimensional moduli space ${\mathcal M}(\gamma,\beta)/{\mathbb R}$. Instead of regarding the pair $(u_0,u_1)$ as a boundary point of the moduli space ${\mathcal M}(x,y)$, we extend the moduli space ${\mathcal M}(x,y)$ by including part of $\{u_0\}\times{\mathcal M}(\gamma,\beta)/{\mathbb R}$, namely the “half” in which the cylic ordering constraint is satisfied along $\gamma$. This is why we need to introduce the level ${\mathcal M}^1(x,y)$. We now continue this process. The curve $u_1$ may itself break, leading us to the next level ${\mathcal M}^2(x,y)$, and so forth.

Now what happens if the cylic ordering constraint at one of the intermediate levels ceases to hold? There are three ways this can happen. First, it could happen that the points $\lim_{s\to-\infty}\pi_Y(u_{i-1}(s,0))$ and $\lim_{s\to+\infty}\pi_Y(u_i(s,0))$ coincide for some $i\in\{1,\ldots,k\}$. In this case, we can glue $u_{i-1}$ and $u_i$ to jump down to a lower level of the moduli space where $k$ decreases by one. Second, the point $\lim_{s\to+\infty}\pi_Y(u_i(s,0))$ could coincide with $p_{\gamma_i}$ for some $i\ge 1$. Now we are truly stuck and cannot extend the moduli space further. But we are at a point in the product ${\mathcal M}(x,\check{\gamma_i})\times {\mathcal M}(\check{\gamma_i},y)$, so that’s OK. Third, the point $\lim_{s\to-\infty}\pi_Y(u_i(s,0))$ can coincide with $p_{\gamma_{i+1}}$ for some $i\le k$. In this case we see a boundary point in ${\mathcal M}(x,\hat{\gamma}_{i+1})\times {\mathcal M}(\hat{\gamma}_{i+1},y)$.

In studying the boundary of the moduli space ${\mathcal M}(x,y)$, there is one remaining issue. Suppose for example that $x=\check{\alpha}$ and $y=\check{\beta}$ where $\beta$ is bad. The primary level is then

${\mathcal M}^0(\check{\alpha},\check{\beta}) = \{u\in{\mathcal M}(\alpha,\beta) \mid \lim_{u\to +\infty}\pi_Y(u(s,0)) = p_\alpha\}/{\mathbb R}.$

Now, because there is no point constraint on the negative end at the bad orbit, there is a problem orienting this moduli space by the usual method of “coherient orientations” (which is what one needs to get the boundary orientations to come out right). The upshot is that we can coherently orient the part of this moduli space where $\lim_{s\to -\infty}\pi_Y(u(s,0))$ does not equal the base point $p_\beta$. When we hit the base point, we have to stop. We interpret this stopping point as a boundary point in ${\mathcal M}(\check{\alpha},\hat{\beta}) \times {\mathcal M}(\hat{\beta},\check{\beta})$. This is why we want ${\mathcal M}(\hat{\beta},\check{\beta})$ to be nonempty when $\beta$ is bad. And the reason why we want this set to contain two points is that if we start with an element of ${\mathcal M}(\check{\alpha},\hat{\beta})$, then there are two ways to “glue” it to obtain an end of the moduli space ${\mathcal M}(\check{\alpha},\check{\beta})$, because there are two directions in which the point $\lim_{s\to -\infty}\pi_Y(u(s,0))$ can move along the Reeb orbit $\beta$.

5. Non-equivariant contact homology

If you believe all of that, we now have a chain complex $SC_*(Y,\lambda)$, which depends on the additional choice of the generic family of almost complex structures $\{J_t\}_{t\in S^1}$ and the base points $p_\gamma$. The differential, which we will denote by $\partial_{SH}$, counts points in the zero-dimensional Morse-Bott moduli spaces with signs given by a coherent orientation. The gluing theory outlined above then implies that $\partial_{SH}^2=0$. We denote the homology of this chain complex $SH_*(Y,\lambda)$ and claim that it does not depend on the additional choices.

It is useful to write the differential in block matrix form as

$\partial_{SH} = \begin{pmatrix} \check{\partial}_1 & \partial_0 \\ \partial_2 & \hat{\partial}_1 \end{pmatrix}.$

Here $\check{\partial}_1$ denotes the component of the differential going between checked Reeb orbits with usual grading difference one; $\hat{\partial}_1$ denotes the component between hatted orbits with usual grading difference one; $\partial_0$ denotes the component from hatted orbits to checked orbits with usual grading difference zero; and $\partial_2$ denotes the component from checked orbits to hatted orbits with usual grading difference two.

6. Example: S^1-independent J

Suppose that we can choose the family of almost complex structures $\{J_t\}_{t\in S^1}$ to all agree with a single almost complex structure $J$ on ${\mathbb R}\times Y$ so that the necessary transversality holds to define cylindrical contact homology. This is rarely possible, but it is possible when $\dim Y = 3$.

There are now two combinatorial conventions for defining the cylindrical contact homology differential, which I denote by $\partial_{CH}^+$ and $\partial_{CH}^-$. If $\alpha$ and $\beta$ are good Reeb orbits, then the coefficient of $\partial_{CH}^+$ from $\alpha$ to $\beta$ counts holomorphic cylinders $u$ from $\alpha$ to $\beta$, multiplied by $m(\alpha)/m(u)$, where $m$ denotes the covering multiplicity. With the other convention $\partial_{CH}^-$, one instead multiplies by $m(\beta)/m(u)$. One needs combinatorial factors such as these to account for the fact that there are multiple ways to glue holomorphic cylinders along multiply covered Reeb orbits. The two conventions give rise to isomorphic chain complexes over ${\mathbb Q}$; the isomorphism from the first convention to the second multiplies each Reeb orbit by its covering multiplicity. (On the other hand, cylindrical contact homology is not invariant if one uses ${\mathbb Z}$ coefficients; we will discuss this in another post.)

I claim now that for our $S^1$-independent $J$, we have

$\partial_{SH} = \begin{pmatrix} \partial_{CH}^+ & \partial_0^{good} \\ \partial_2 & \partial_{CH}^- \end{pmatrix}.$

Here $\partial_0^{good}(\hat{\alpha})$ is $0$ if $\alpha$ is good, and $2\check{\alpha}$ if $\alpha$ is bad. Terms $\check{\beta}$ with $\beta\neq\alpha$ cannot appear in $\partial_0^{good}(\hat{\alpha})$ for dimensional reasons. It is an exercise to check that $\partial_{CH}^+$ and $\partial_{CH}^-$ appear on the diagonal, because of the way the point constraints work.

The chain complex is now filtered by the usual grading of Reeb orbits, so we can compute its homology by a spectral sequence. The first term of the spectral sequence is the homology of $\partial_0^{good}$. This kills the bad orbits, if we are using rational coefficients. Let us do this (use rational coefficients), even though the chain complex is defined over the integers. The second term in the spectral sequence is then two copies of cylindrical contact homology. The third term now computes the homology of a map on the second term induced by $\partial_2$, so that we get an exact triangle

$CH\stackrel{(\partial_2)_*}{\longrightarrow} CH \longrightarrow SH\otimes{\mathbb R}\longrightarrow \cdots$

as in Bourgeois-Oancea.

7. What about the general case?

It is not clear how to generalize the above example to the case where the almost complex structure is $S^1$-dependent. The chain complex $SC_*$ is still filtered by the usual grading of Reeb orbits. However the first term in the spectral sequence no longer has an obvious interpretation in terms of cylindrical contact homology, because $\partial_0$ may include coefficients from $\hat{\alpha}$ to $\check{\beta}$ where the Reeb orbits $\alpha$ and $\beta$ are different. Also, cobordism maps will not respect this filtration, but may shift it up by $1$.

In conclusion, I still don’t see how to define cylindrical contact homology in general using $S^1$-dependent almost complex structures. But we haven’t yet squeezed every drop of information out of the above moduli spaces. And maybe the non-equivariant contact homology also has some uses.

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## What is the geometric meaning of ECH capacities?

Last week, Keon Choi, Dan Cristofaro-Gardiner, David Frenkel, Vinicius Ramos, and I posted a new paper, Symplectic embeddings into four-dimensional concave toric domains. To explain what concave toric domains are, recall that if $\Omega$ is a domain in the first quadrant of the plane, we define a (four-dimensional) “toric domain”

$X_\Omega = \{z\in{\mathbb C}^2 \mid \pi(|z_1|^2,|z_2|^2)\in\Omega\}.$

For example, if $\Omega$ is a right triangle with vertices $(0,0), (a,0), (0,b)$, then $X_\Omega$ is an ellipsoid, which we denote by $E(a,b)$. We also define the ball $B(a)=E(a,a)$. If $\Omega$ is a rectangle with vertices $(0,0), (a,0), (0,b), (a,b)$, then $X_\Omega$ is the polydisk which we denote by $P(a,b)$. Define a “concave toric domain” to be a domain $X_\Omega$ where $\Omega$ is bounded by the axes and the graph of a convex function $[0,a]\to [0,b]$ for some positive real numbers $a,b$. So an ellipsoid is a concave toric domain, but a polydisk is not. The (nondisjoint) union of two ellipsoids is also an example of a concave toric domain.

In the paper we computed the ECH capacities of all concave toric domains, and obtained some applications to symplectic embedding problems. For now I will skip the discussion of the symplectic embedding problems, and focus on the following question: What is the geometric meaning of ECH capacities?

For a concave toric domain $X_\Omega$, we showed that there is a nonincreasing sequence of nonnegative real numbers $a_1,a_2,\ldots$ such that the $k^{th}$ ECH capacity of $X_\Omega$ is given by

$c_k(X_\Omega) = c_k\left(\coprod_iB(a_i)\right).$

It turns out after the fact that the numbers $a_i$ can be described geometrically as follows. First, $a_1$ is the largest real number $a$ such that there exists a symplectic embedding $B(a)\to X_\Omega$. Then $a_2$ is, roughly speaking, the largest real number $a$ such that there exists a symplectic embedding of $int (B(a))$ into the remaining space. And so forth. That is the numbers $a_1,a_2,\ldots$ measure the size of the balls in a “greedy ball packing” of $X_\Omega$.

To make this precise, if $X$ is any domain, we can define $a_1(X)$ to be the supremum over $a$ such that there exists a symplectic embedding $B(a)\to X$.  We can then inductively define $a_{k+1}(X)$ to be the limit, as $a_1',\ldots,a_k'$ approach $a_1(X),\ldots,a_k(X)$ from below, of the maximum of $a_1(X\setminus \phi(\coprod_{i=1}^k B(a_i')))$, where the maximum is over symplectic embeddings $\phi:\coprod_{i=1}^kB(a_i')\to X$.

We now know that if $X$ is a concave toric domain, then

$c_k(X) = c_k\left(\coprod_{i=1}^\infty B(a_i(X))\right).$

For which other domains $X$ is this true? Note that the left hand side of the above equation is always greater than or equal to the right hand side, by the monotonicity of ECH capacities. However the reverse inequality need not be true.

The first counterexample I could find is the polydisk $P(2,3)$. Here, if I am not mistaken, the sequence of numbers $a_i$ is $(2,2,1,1,1,1,0,\ldots)$. For $k=4$, we have $c_4(P(2,3))=7$, but $c_4(\coprod_iB(a_i))=6$.

So I guess we have more work to do to understand what ECH capacities are measuring in general.

## An obstruction to cylindrical contact homology?

Last week I was at a very nice conference at IMPA, and one of the topics of informal discussion was to what extent cylindrical contact homology can be defined, in the absence of contractible Reeb orbits, but without using polyfolds. At the conference I was mostly discussing this with Umberto Hryniewicz, and I also had some previous discussions on related topics with Jo Nelson, whose thesis considers under what conditions one can define linearized contact homology with pre-polyfold technology. Anyway something interesting came up in the discussion last week: namely, it appears that there is a homological obstruction to this working. While this obstruction is probably (hopefully) an artifact of my ignorance, I would like to discuss it here, in case others have useful wisdom or suggestions. So here is what I am going to do in this post:

1. I am going to consider the simplest situation where a difficulty arises, namely cylindrical contact homology on a contact three-manifold with no contractible Reeb orbits. While examples of closed contact manifolds with this property may be uncommon, another natural situation in which such a setup arises is the “local contact homology” in a neighborhood of a Reeb orbit introduced in this preprint by Macarini and Hryniewicz. Or on a closed three-manifold one might work below a symplectic action level in which there are no contractible orbits. Anyway, in three dimensions when there are no contractible Reeb orbits, thanks to automatic transversality results of Wendl and others, one can define the differential using a generic almost complex structure. However there are transversality difficulties in defining cobordism maps which you need for example to show that the homology is an invariant of contact structures.
2. I’ll review how by using $S^1$-dependent almost complex structures, one can fix the transversality difficulties, but at the expense of messing up the chain map equation.
3. I’ll explain how one can add correction terms to fix the chain map equation, but at the expense of ultimately obtaining the wrong theory: an analogue of reduced symplectic homology.
4. I will then identify a homological obstruction to defining cobordism maps on cylindrical contact homology using this approach.
5. Finally I will speculate on how this obstruction might result from my ignorance rather than an actual problem.

Somehow this is much easier to explain by drawing pictures on a blackboard, but I’ll still give it a try here and hopefully it will be understandable.

1. Cylindrical contact homology

Let $(Y^3,\lambda)$ be a closed nondegenerate contact three-manifold with no contractible Reeb orbits. For each embedded Reeb orbit $\gamma$, choose a base point $p_\gamma$ on the image of $\gamma$.  Choose a generic almost complex structure $J$ on ${\mathbb R}\times Y$ satisfying the usual conditions. We can then define the cylindrical contact homology chain complex $CC_*(Y,\lambda,J)$ as follows. It is freely generated over ${\mathbb Q}$ by “good” Reeb orbits. In this situation a Reeb orbit is “bad” if it is an even multiple cover of a negative hyperbolic orbit, otherwise it is “good”. If $\gamma_+$ and $\gamma_-$ are two good Reeb orbits, the differential coefficient $\langle\partial\gamma_+,\gamma_-\rangle$ counts, with appropriate signs and combinatorial factors, maps

$u:{\mathbb R}\times S^1\to{\mathbb R}\times Y$

satisfying the holomorphic curve equation

$\partial_su + J(u)\partial_tu=0$

such that $\lim_{s\to \pm\infty}\pi_Y(u(s,\cdot))$ is a reparametrization of $\gamma_\pm$, and $\lim_{s\to +\infty}\pi_Y(u(s,0))$ is the base point $p_{\widehat{\gamma_+}}$. Here $\pi_Y$ denotes the projection ${\mathbb R}\times Y\to Y$, and $\widehat{\gamma}$ denotes the embedded Reeb orbit underlying $\gamma$. We count such holomorphic maps in index one moduli spaces, modulo ${\mathbb R}$ translation.

If $J$ is generic, then all such $u$ that are somewhere injective are cut out transversely. Furthermore, in this four-dimensional situation, automatic transversality results imply that even the multiply covered $u$ are also cut out transversely. Thus “classical” transversality arguments suffice to define the differential, and also to prove that $\partial^2=0$. (I reviewed this in a couple of earlier postings, and if I recall correctly this was originally published in a paper by Al Momin.)

Let us denote the homology of the chain complex by $CH_*(Y,\lambda,J)$. One can use arguments which I will outline below to show that this is independent of $J$. One would also like to show that this depends only on the contact structure $\xi=Ker(\lambda)$ and not on $\lambda$, but here we run into some difficulties which I will describe below.

2. Cobordism map difficulties

Let $(X^4,\omega)$ be an exact symplectic cobordism from $(Y_+,\lambda_+)$ to $(Y_-,\lambda_-)$. One would like to show that this induces a map

$CH_*(X,\omega): CH_*(Y_+,\lambda_+,J_+) \to CH_*(Y_-,\lambda_-,J_-)$.

In the simplest case when $X=[0,1]\times Y$ one would like to use these maps to show that $CH_*(Y,\lambda,J)$ depends only on the pair $(Y,\xi)$.

One can try to define the desired cobordism map as follows. Let $\overline{X}$ denote the “completion” of $X$ obtained by attaching symplectization ends to its boundary components. Let $J$ be an almost complex structure on $\overline{X}$ with agrees with $J_\pm$ on the ends and which is $\omega$-compatible on $X$. We would now like to define a chain map

$\phi: CC_*(Y_+,\lambda_+,J_+) \to CC_*(Y_-,\lambda_-,J_-)$

by counting index zero holomorphic maps

$u:{\mathbb R}\times S^1 \to \overline{X}$

satisfying the same conditions as before. If $J$ is generic, then the somewhere injective solutions are cut out transversely. However it is sometimes (maybe most of the time) impossible to obtain transversality for the multiply covered solutions for any $J$. I can show you explicit counterexamples.

One could try to define a map just using the somewhere injective curves, together with some contributions from their multiple covers, as in Taubes’s Gromov invariant of closed symplectic four-manifolds. However this doesn’t work, due to two related difficulties. First, there are examples showing that such a count cannot give the correct map. Second, there is a compactness problem: a sequence of somewhere injective index zero solutions may converge to a “broken holomorphic curve”, or “holomorphic building”, containing a negative index multiple cover in the cobordism level, together with some positive index solutions in the symplectization levels. The cobordism map needs to consider contributions from all moduli spaces of such buildings. It’s not trivial to see how to count these so as to get the chain map equation to hold, but this might be possible using obstruction bundles, as I may discuss later.

3. Fixing transversality

To simplify the discussion, let us assume henceforth that there are no bad Reeb orbits. If there are bad Reeb orbits then we have to say some things more carefully, but we an already see the main difficulties in the absence of bad Reeb orbits.

To continue the discussion, there is a simple way to perturb the holomorphic curve equation to obtain transversality. Umberto Hryniewicz was telling me about this last week, and Kai Cieliebak also told me about this last summer. Namely one chooses a family $\{J_t\}_{t\in S^1}$ of almost complex structures on $\overline{X}$ such that each almost complex structure $J_t$ in the family satisfies the usual conditions. Given Reeb orbits $\gamma_+$ of $\lambda_+$ and $\gamma_-$ of $\lambda_-$, we consider maps

$u:{\mathbb R}\times S^1\to\overline{X}$

satisfying the equation

$\partial_su + J_t(u)\partial_tu=0$

such that $\lim_{s\to\pm\infty}\pi_{Y_\pm}(u(s,\cdot))$ is a reparametrization of $\gamma_\pm$. Let ${\mathcal M}(\gamma_+,\gamma_-)$ denote the moduli space of such maps $u$. If the family $\{J_t\}$ of almost complex structures is generic, then this moduli space is cut out transversely, and has dimension

$dim({\mathcal M}(\gamma_+,\gamma_-)) = |\gamma_+| - |\gamma_-| + 1$

where $|\gamma_+| - |\gamma_-|$ denotes the difference in Conley-Zehnder indices relative to the homotopy class of $u$. The $S^1$-dendence of the almost complex structure gets rid of the usual difficulties with multiple covers.

There are evaluation maps

$e_\pm: {\mathcal M}(\gamma_+,\gamma_-) \to \widehat{\gamma_\pm}$

defined by $e_\pm(u) = \lim_{s\to\pm\infty}u(s,0)$. We can arrange that these evaluation maps are transverse to the base points $p_{\widehat{\gamma_\pm}}$, e.g. by choosing the base points generically, or probably also by choosing the family $\{J_t\}$ generically. One might then try to define the cobordism chain map $\phi$ by

$\langle\phi\gamma_+,\gamma_-\rangle = \#\{u\in{\mathcal M}(\gamma_+,\gamma_-) \mid e_+(u)=p_{\widehat{\gamma_+}}\}.$

The motivation for this definition is as follows: if the family $\{J_t\}$ were $S^1$-independent, and if by some miracle we still had transversality (for example if we were working below a symplectic action level in which all Reeb orbits are simple), then this would agree with the definition we attempted before, and everything would work.

However, when $\{J_t\}$ is $S^1$-dependent, the above definition of $\phi$ might not satisfy the chain map equation. Here is why. Suppose that $|\gamma_+|-|\gamma_-|=1$. To try to prove that $\langle(\partial\phi-\phi\partial)\gamma_+,\gamma_-\rangle=0$, we want to consider the ends of the one-dimensional moduli space

${\mathcal M} = \{u\in{\mathcal M}(\gamma_+,\gamma_-) \mid e_+(\gamma_+)=p_{\widehat{\gamma_+}}\}.$

We expect that this has a compactification $\overline{\mathcal M}$ whose boundary is given by fiber products over evaluation maps,

$\partial\overline{\mathcal M} = \coprod_{|\gamma_+'|=|\gamma_+|-1} \{u\in\mathcal{M}(\gamma_+,\gamma_+') \mid e_+(\gamma_+) = p_{\widehat{\gamma_+}}\}/{\mathbb R} \times_{\widehat{\gamma_+'}} {\mathcal M}(\gamma_+',\gamma_-)$

$-\sqcup \coprod_{|\gamma_-'|=|\gamma_-|+1} \{u\in {\mathcal M}(\gamma_+,\gamma_-')\mid e_+(u)=p_{\widehat{\gamma_+}}\} \times_{\widehat{\gamma_-'}} {\mathcal M}(\gamma_-',\gamma_-).$

Here ${\mathcal M}(\gamma_+,\gamma_+')$ denotes the moduli space of solutions to our equation in ${\mathbb R}\times Y_+$; since $J_+$ is $S^1$-independent, if we impose the “asymptotic marker” constraint $e_+=p_{\widehat{\gamma_+}}$ then we recover the differential coefficient $\langle\partial\gamma_+,\gamma_+'\rangle$. Likewise for ${\mathcal M}(\gamma_-',\gamma_-)$. Thus, in the above boundary equation, the number of points in the second line agrees with the coefficient $\langle\partial\phi\gamma_+,\gamma_-\rangle$. However the number of points in the first line might not agree with the coefficient $\langle\phi\partial\gamma_+,\gamma_-\rangle$. The reason is that the holomorphic maps in ${\mathcal M}(\gamma_+',\gamma_-)$ that appear do not necessarily satisfy the asymptotic marker constraint $e_+=p_{\widehat{\gamma_+'}}$.

To clarify what is going on here, for each generic $t\in S^1$ we can define an alternate map

$\phi_t:CC_*(Y_+,\lambda_+,J_+) \to CC_*(Y_-,\lambda_-,J_-)$

as follows. Define

${\mathcal M}_t(\gamma_+,\gamma_-) = \{u\in{\mathcal M}(\gamma_+,\gamma_-) \mid \lim_{s\to +\infty} u(s,t) = p_{\widehat{\gamma_+}}\}.$

We then define the coefficent $\langle\phi_t\gamma_+,\gamma_-\rangle$ to count points in the moduli space ${\mathcal M}_t(\gamma_+,\gamma_-)$ with appropriate signs and combinatorial factors. In particular $\phi_0=\phi$. Now in the boundary equation, as we said before, the number of points in the second line agrees with the coefficient $\langle\partial\phi_0\gamma_+,\gamma_-\rangle$. However each point in the first line is counted by $\langle\phi_t\partial\gamma_+,\gamma_-\rangle$ for some $t\in S^1$.

If $\phi_t$ were independent of $t$ then everything would be OK and we would obtain the chain map equation $\partial\phi=\phi\partial$. However $\phi_t$ might not be independent of $t$. Here is why. Consider an interval $[t_0,t_1]\subset S^1$. Suppose that $|\gamma_+|=|\gamma_-|$ and consider the one-dimensional moduli space

${\mathcal M} = \bigcup_{t\in [t_0,t_1]}{\mathcal M}_t(\gamma_+,\gamma_-).$

We expect that this has a compactification $\overline{\mathcal M}$ whose boundary is given by

$\partial\overline{\mathcal M} = {\mathcal M}_{t_1}(\gamma_+,\gamma_-) - {\mathcal M}_{t_0}(\gamma_+,\gamma_-)$

$+ \bigcup_{t\in[t_0,t_1]} \coprod_{|\gamma_+'| = |\gamma_+| - 1} {\mathcal M}_t(\gamma_+,\gamma_+') \times_{\widehat{\gamma_+'}} {\mathcal M}(\gamma_+',\gamma_-)$

$+\bigcup_{t\in[t_0,t_1]} \coprod_{|\gamma_-'|=|\gamma_-|+1} {\mathcal M}_t(\gamma_+,\gamma_-')\times_{\widehat{\gamma_-'}} {\mathcal M}(\gamma_-',\gamma_-).$

In other words, we can define a map

$\psi_{[t_0,t_1]}: CC_*(Y_+,\lambda_+,J_+) \to CC_{*+1}(Y_-,\lambda_-,J_-)$

by defining $\langle\psi_{[t_0,t_1]}\gamma_+,\gamma_-\rangle$ to be a count of points in $\bigcup_{t\in[t_0,t_1]}{\mathcal M}_t(\gamma_+,\gamma_-)$. We then have

$\phi_{t_1} - \phi_{t_0} = \partial\psi_{[t_0,t_1]} + \psi_{[t_0,t_1]}\partial.$

To summarize, the difficulty with proving the chain map equation comes from holomorphic maps in ${\mathcal M}(\gamma_+,\gamma_-)$ where $|\gamma_-|=|\gamma_+|+1$. If $J_t$ where $S^1$-independent and if we had transversality, then such maps would not exist. However I do not see any obvious way to rule out the existence of such maps when $J_t$ is $S^1$-dependent.

Since the map $\phi$ is not a chain map, we can try to add some correction terms to fix this. If $|\gamma_+|=|\gamma_-|$, a natural correction term to add to the coefficient $\langle\phi\gamma_+,\gamma_-\rangle$ would be a count of elements of

$\coprod_{|\gamma_+'|=|\gamma_+|-1}\{(u_+,u_0)\in {\mathcal M}_0(\gamma_+,\gamma_+') \times {\mathcal M}(\gamma_+',\gamma_-) \mid p_{\widehat{\gamma_+'}} < e_-(u_+) < e_+(u_0)\}.$

Here the condition

$p_{\widehat{\gamma_+'}} < e_-(u_+) < e_+(u_0)$

means that the three points in question are positively cyclically ordered in the circle $\widehat{\gamma_+'}$, with its orientation given by the Reeb vector field. If you add these correction terms, then at first it looks like the chain map equation will now work, but there is an additional error term. Let $\phi_+$ denote the sum of the original map $\phi_0$ plus the above correction terms. One then finds, assuming the signs work out, that

$\partial\phi_+ - \phi_+\partial = \phi_{-1}U.$

Here

$\phi_{-1}:CC_*(Y_+,\lambda_+,J_+) \to CC_{*+1}(Y_-,\lambda_-,J_-)$

is defined by setting  $\langle\phi_{-1}\gamma_+,\gamma_-\rangle$ to be a count of elements of the “bad” moduli space ${\mathcal M}(\gamma_+,\gamma_-)$ with appropriate signs and combinatorial factors. Meanwhile

$U:CC_*(Y_+,\lambda_+,J_+) \to CC_{*-2}(Y_+,\lambda_+,J_+)$

is a well-known chain map in contact homology which appears (under different names) in work of Bourgeois-Oancea and Bourgeois-Ekholm-Eliashberg. The coefficent $\langle U\gamma_+,\gamma_+'\rangle$ is a sum of two terms. The first term counts curves $u\in{\mathcal M}(\gamma_+,\gamma_+')$ satisfying the two asymptotic marker constraints $e_+(u) = p_{\widehat{\gamma_+}}$ and $e_-(u) = p_{\widehat{\gamma_-}}$. The second term (which one needs to add to the first term to obtain a chain map) counts pairs $(u_+,u_+')$ where $u_+\in{\mathcal M}(\gamma_+,\gamma_+'')$ for some Reeb orbit $\gamma_+''$ with $|\gamma_+''|=|\gamma_+|-1$, the curve $u_+$ satisfies the positive asymptotic marker constraint $e_+(u_+) = p_{\widehat{\gamma_+}}$, the curve $u_-$ satisfies the negative asmptotic marker constraint $e_-(u_-)=p_{\widehat{\gamma_+'}}$, and we have the cyclic ordering condition

$p_{\gamma_+''} < e_-(u_+) < e_+(u_+').$

Instead of explaining the details of the above equation, let us now put it in a more general framework which will make it appear almost obvious.

4b. Symplectic homology

Let $(Y^3,\lambda)$ be a nondegenerate contact three-manifold with no contractible Reeb orbits, and let $J$ be a generic almost complex structure as needed to define its cylindrical contact homology. Following section 3.2 of the paper Effect of Legendrian surgery by Bourgeois-Ekholm-Eliashberg,  we can define a “symplectic homology” chain complex $SC_*(Y,\lambda,J)$ as follows. There are two generators for each (good or bad) Reeb orbit $\gamma$. One can think of these as the minima and maxima of a Morse function on the underlying embedded Reeb orbit $\widehat{\gamma}$, whose minima and maxima are both very close to the base point $p_{\widehat{\gamma}}$. We can write the differential as a $2\times 2$ block matrices, where the blocks correspond to minima versus maxima, as

$\partial_{SH} = \begin{pmatrix}\partial & \partial_{Morse} \\ U & -\partial\end{pmatrix}.$

Here $\partial$ is the differential on the cylindrical contact homology chain complex (I think we should interpret its coefficients between bad Reeb orbits to be zero), $U$ is the map defined above, and $\partial_{Morse}$ is the twisted Morse differential on the underlying embedded Reeb orbits; this is zero for good Reeb orbits and $2$ for bad Reeb orbits. If we continue to make the simplifying assumption that there are no bad Reeb orbits, then this is just zero. We denote the homology of this complex by $SH_*(Y,\lambda,J)$. This is what Bourgeois-Ekholm-Eliashberg call “reduced symplectic homology”, except that there they have a symplectic filling (and allow for contractible Reeb orbits), and we don’t have a symplectic filling (and assume that there are no contractible Reeb orbits).

Now return to our exact symplectic cobordism from $(Y_+,\lambda_+)$ to $(Y_-,\lambda_-)$. We can then use our generic $S^1$-family of almost complex structures $\{J_t\}$ on $\overline{X}$ to define a chain map

$\Phi_{SH}:SC_*(Y_+,\lambda_+,J_+) \to SC_*(Y_-,\lambda_-,J_-).$

The idea of this map is as follows. From a maximum to a minimum it counts curves with no asymptotic marker constraint. From a minimum to a minimum it counts curves with a positive asymptotic marker constraint. From a maximum to a maximum it counts curves with a negative asymptotic marker constraint. From a minimum to a maximum it counts curves with both positive and negative asymptotic marker constraints. These “curves” are not only elements of the moduli spaces ${\mathcal M}(\gamma_+,\gamma_-)$, but also “cascades” consisting of sequences of such curves satisfying cyclic ordering constraints at the intermediate levels. The result has the block matrix form

$\Phi_{SH} = \begin{pmatrix} \phi_+ & \phi_{-1} \\ \phi_1 & \phi_-\end{pmatrix}.$

Here $\phi_+$ is the map we defined before which consisted of the first attempt at a chain map plus correction terms; $\phi_{-1}$ is the map counting “bad” curves which we defined above; $\phi_-$ is similar to $\phi_+$ but uses negative asymptotic marker constraints instead of positive ones; and the details of $\phi_1$ are not important for our discussion.

Assuming the analysis needed to define Morse-Bott theory a la Bourgeois (I don’t know a reference for the full details of this), but without any apparent transversality difficulties, we then have the chain map equation

$\partial_{SH}\Phi_{SH} - \Phi_{SH}\partial_{SH}=0.$

Expanding this as a block matrix, we obtain

$0 = \begin{pmatrix} \partial\phi_+ - \phi_+\partial - \phi_{-1}U & \partial\phi_{-1} + \phi_{-1}\partial \\ U\phi_+ - \partial\phi_1-\phi_1\partial-\phi_-U & U\phi_{-1} - \partial\phi_-+\phi_-\partial \end{pmatrix}.$

The upper left entry is the equation we wrote down before, expressing the failure of $\phi_+$ to be a chain map on the cylindrical contact homology chain complex. The lower right entry expresses a similar failure of $\phi_-$ to be a chain map. The upper right entry says that the map $\phi_{-1}$ counting “bad” holomorphic curves is a chain map. In particular it induces a map

$(\phi_{-1})_*: CH_*(Y_+,\lambda_+,J_+) \to CH_{*+1}(Y_-,\lambda_-,J_-).$

5. The obstruction

I claim that the above map $(\phi_{-1})_*$ is an obstruction to defining a cobordism map on cylindrical contact homology (at least using this approach). One way to think of this is that the chain map $\Phi_{SH}$ depends on some choices. Different choices will lead to chain homotopic chain maps. One could ask if there are some choices for which this chain map has $\phi_{-1}=0$. But one can calculate using the above matrices that chain homotopic maps $\Phi_{SH}$ give rise to chain homotopic maps $\phi_{-1}$.

So is this obstruction zero or not?

1. If we can perform some kind of abstract perturbation of the space of holomorphic maps ${\mathbb R}\times Y \to \overline{X}$ so as to obtain transversality while preserving $S^1$-symmetry, then we will have $\phi_{-1}=0$. I don’t know enough about polyfolds to be able to tell whether or not the theory is supposed to be able to do this. The difficulty being that cylinders are not stable in the Deligne-Mumford sense.
2. One could look at a sequence of $S^1$-dependent almost complex structures that converges to an $S^1$-independent almost complex structure and see what happens to the moduli space of curves counted by $\phi_{-1}$. The limit should be the zero set of a section of an obstruction bundle over the moduli space of index $-1$ holomorphic buildings. Maybe there is some way to use $S^1$ symmetry to show that the count of zeroes of this section vanishes.
3. I haven’t yet digested the preprint $S^1$-equivariant symplectic homology and linearized contact homology by Bourgeois-Oancea, but maybe this suggests a way to modify the definition of contact homology to resolve the above problems.

The answers to the above questions may be well known to some experts, so if you know any answers or have any suggestions please comment, thanks.

Posted in Contact homology, Open questions | 7 Comments

## A magic trick for defining obstruction bundles

Today I want to tell you about a magic trick for proving transversality and/or defining obstruction bundles in certain four-dimensional situations where the usual approach doesn’t work. This is helpful for defining cobordism maps between various kinds of contact homology in three dimensions (although to completely define cobordism maps you need much more than just this trick).

The question

Here is the setup. Let $(X,\omega)$ be a four-dimensional strong symplectic cobordism between nondegenerate contact three-manifolds $(Y_+,\lambda_+)$ and $(Y_-,\lambda_-)$. Let $\overline{X}$ denote the usual symplectization completion of $X$, and let $J$ be a generic almost complex structure on $\overline{X}$ satisfying the usual conditions.

Let $C$ be an immersed $J$-holomorphic curve in $\overline{X}$. Let $\pi:\widetilde{C}\to C$ be a degree $d$ branched cover of $C$ with $b\ge 0$ branch points.

When the number of branch points $b=0$, we would like a criterion which guarantees that $\widetilde{C}$ is cut out transversely. When the number of branch points $b>0$, it is more or less impossible for $C$ to be cut out transversely, but we would still like $\widetilde{C}$ to be “good” in the following sense.

Recall that there is a deformation operator

$D_C:L^2_1(N_C) \to L^2(T^{0,1}C\otimes N_C)$

where $N_C$ denotes the normal bundle to $C$. This operator is surjective if and only if $C$ is cut out transversely. The kernel of this operator consists of deformations of $C$ which are $J$-holomorphic to first order. The index of this operator is what we call the Fredholm index of $C$, denoted by $ind(C)$, which when $C$ is cut out transversely is the dimension of the moduli space of holomorphic curves near $C$. Now there is an induced operator

$D_{\widetilde{C}}:L^2_1(\pi^*N_C)\to L^2(T^{0,1}\widetilde{C}\otimes \pi^*N_C).$

To define this, choose a local complex trivialization of $N_C$. Then locally, in this trivialization,

$D_C = \overline{\partial} + \alpha$

where $\alpha$ is some $(0,1)$-form on $C$, determined by the derivative of the almost complex structure in directions normal to $C$. Using the same local trivialization for $\pi^*N_C$, we define

$D_{\widetilde{C}} = \overline{\partial} + \pi^*\alpha.$

Roughly speaking, $D_{\widetilde{C}}$ considers deformations of $\widetilde{C}$ in directions normal to $C$, so that the branch points don’t move.

Definition. The branched cover $\widetilde{C}$ is good if $Ker(D_{\widetilde{C}})=0$.

If there are no branch points and $ind(\widetilde{C})=0$, then $\widetilde{C}$ is good if and only if it is transverse. If there are branch points, then one needs all branched covers in the moduli space of branched covers of $C$ containing $\widetilde{C}$ to be good, in order to define an “obstruction bundle” over the moduli space of such branched covers in order to do gluing as in my joint papers with Taubes.

So the question is now, under what circumstances is $\widetilde{C}$ guaranteed to be good?

The usual approach

Recall that the Fredholm index of $\widetilde{C}$ is given in my notation by

$ind(\widetilde{C}) = -\chi(\widetilde{C}) + 2 c_\tau(\widetilde{C}) + CZ_\tau^{ind}(\widetilde{C}).$

By the Riemann-Hurwitz formula, we can rewrite this as

$ind(\widetilde{C}) = -d\chi(C) + b + 2d c_\tau(C) + CZ_\tau^{ind}(\widetilde{C}).$

Note that this is not the Fredholm index of the operator $D_{\widetilde{C}}$. Rather, we have

$ind(D_{\widetilde{C}}) = ind(\widetilde{C}) - 2b.$

Intuitively, this is because the operator $D_{\widetilde{C}}$ does not consider deformations of $\widetilde{C}$ that move the branch points, so the dimension of its domain is $2b$ too small.

Anyway, using the above notation, we can now state the result given by the usual approach. Let $g(\widetilde{C})$ denote the genus of $\widetilde{C}$. Let $h_+(\widetilde{C})$ denote the number of ends of $\widetilde{C}$ that are at positive hyperbolic orbits (including even covers of negative hyperbolic orbits).

Proposition 1. Suppose that $2g(\widetilde{C}) - 2 + ind(\widetilde{C}) + h_+(\widetilde{C}) - 2b < 0$. Then $\widetilde{C}$ is good.

Proof. Suppose $\psi\in Ker(D_{\widetilde{C}})$ is not identically zero. We know, e.g. from the Carleman similarity principle, that every zero of $\psi$ is isolated and has positive multiplicity, so the signed count of zeroes of $\psi$ is nonnegative. On the other hand, we can bound the number of zeroes of $\psi$ similarly to my previous blog post “automatic transversality for dummies”. I’ll leave the calculation as an exercise; the result is

$\#\psi^{-1}(0) \le 2g(\widetilde{C}) - 2 + ind(\widetilde{C}) + h_+(\widetilde{C}) - 2b.$

If the right hand side is negative this is a contradiction. Thus if the right hand side is negative then $\widetilde{C}$ is good. QED.

Where the usual approach doesn’t work

Here are two examples where Proposition 1 is not applicable but we would still like to be able to prove that $\widetilde{C}$ is good.

Example 2. Suppose that $C$ is a cylinder with Fredholm index zero which has one positive end and one negative end, each at positive hyperbolic orbits. Suppose that $\widetilde{C}$ is also a cylinder, so that there are no branch points. If we are trying to define cobordism maps on cylindrical contact homology, we would like to be able to prove that $\widetilde{C}$ is cut out transversely, or equivalently good. Proposition 1 is not applicable here because the inequality that one needs is an equality in this case.

To motivate our next example, note that to define the cobordism maps needed to show that ECH does not depend on the almost complex structure, one needs to consider certain cases where $C$ and $\widetilde{C}$ both have ECH and Fredholm index equal to zero, with all ends at elliptic orbits. One wants to show that $\widetilde{C}$ is good so that one can do obstruction bundle gluing as in my paper with Taubes to prove the chain map equation. Here is a simple example of such a case where the inequality needed to apply Proposition 1 does not hold.

Example 3. Suppose that $C$ is embedded, has genus one, and has exactly one end, which is a positive end at a simple elliptic orbit $\gamma$ with monodromy angle (with respect to some trivialization $\tau$) $\theta\in(0,1/2)$. Suppose further that the Fredholm index $ind(C)=0$. Then the ECH index $I(C)=0$ also (this is true for any embedded curve whose ends are at distinct simple orbits by the relative adjunction formula). More precisely, $c_\tau(C)=-1$ and $Q_\tau(C)=0$. Now let $\widetilde{C}$ be a double cover of $C$ with one branch point, and then necessarily exactly one end, which is a positive end at the double cover of $\gamma$. Then it follows from the above calculations that $I(\widetilde{C})=ind(\widetilde{C})=0$. But $g(\widetilde{C})=2$, so the inequality needed to apply Proposition 1 is an equality in this case.

The magic trick

Now I can show you the magic trick. The statement is a bit complicated and not the most general possible, but this will illustrate the technique and is sufficient to handle the above two examples.

To proceed with the statement and the proof, we need to consider the normal bundle $N_C$. We can regard this as a kind of (completed) symplectic cobordism between the disjoint union over the ends of $C$ of the normal bundle of the corresponding (possibly multiply covered) Reeb orbit. Let me not try to define exactly what I mean by “symplectic cobordism” in this context. Suffice it to say that $C$, regarded as the zero section, defines an embedded holomorphic curve in $N_C$ whose ends are all at distinct simple Reeb orbits, even if none of the above is true for the original curve $C$.

There is now a well defined notion of the ECH index of $\widetilde{C}$ in the normal bundle $N_C$. This is defined by copying the usual formulas, in the normal bundle $N_C$. We can think of this as the ECH index “relative to $C$“, and we denote it by $I_C(\widetilde{C})$. If $C$ is embedded in $\overline{X}$ and all its ends are at distinct simple Reeb orbits, then the ECH index of $\widetilde{C}$ in $N_C$ agrees with the ECH index of $\widetilde{C}$ in $\overline{X}$, i.e. $I_C(\widetilde{C}) =I(\widetilde{C})$. If $C$ does not have these properties, then it is possible that $I_C(\widetilde{C})\neq I(\widetilde{C})$.

We also say that $\widetilde{C}$ satisfies the ECH partition conditions “relative to $C$” if it satisfies the usual ECH partition conditions in the normal bundle $N_C$. If all ends of $C$ are at distinct simple Reeb orbits then $\widetilde{C}$ satisfies the ECH partition conditions if and only if $\widetilde{C}$ satisfies the ECH partition conditions relative to $C$.

We can now state:

Proposition 4. Assume that $Ker(D_C)=0$. Suppose that either $ind(\widetilde{C})>I_C(\widetilde{C})$, or $ind(\widetilde{C})=I(\widetilde{C})$ and $\widetilde{C}$ does not satisfy the ECH partition conditions relative to $C$. Furthermore, if $\widetilde{C}\to C$ factors through a branched cover $\widehat{C}\to C$ whose degree is between $1$ and $d$, then assume that the above condition also holds with $\widetilde{C}$ replaced by $\widehat{C}$. Then $\widetilde{C}$ is good.

Here is why Proposition 4 is applicable to Examples 2 and 3. In both of these examples, $ind(\widetilde{C})=I_C(\widetilde{C})=0$. However neither of these examples satisfies the ECH partition conditions relative to $C$. (In Example 2, the partition conditions relative to $C$ would require that $\widetilde{C}$ has $d$ positive ends and $d$ negative ends. In Example 3, the partition conditions relative to $C$ would require that $\widetilde{C}$ has two positive ends.)

Proof of Proposition 4. First observe that there is a unique almost complex structure on the normal bundle $N_C$ (regarded here as four-manifold, not a bundle) whose restriction to the fibers agrees with the almost complex structure $J$ on $X$, such that a local section $\psi$ is in the kernel of the operator $D_C$ if and only if $\psi$ is a holomorphic map from a neighborhood in $C$ to $N_C$.

Now suppose $\widetilde{\psi}$ is a nonzero element of $Ker(D_{\widetilde{C}})$. Let $\psi$ denote the image of $\widetilde{\psi}$ under the projection $\pi^*N_C\to N_C$. Then $\psi$ is a holomorphic curve in $N_C$.

By the assumption that $Ker(D_C)=0$ and the assumption about intermediate branched covers in Proposition 4, we can assume without loss of generality that $\psi$ is somewhere injective.

Now a version of the ECH index inequality tells us that $ind(\psi) \le I_C(\psi)$, with equality only if $\psi$ satisfies the ECH partition conditions relative to $C$. This is proved just like the usual ECH index inequality, except that in this case one does not need Siefring’s nonlinear analysis; instead one can use the much easier, linear analysis in my paper “An index inequality for pseudoholomorphic curves in symplectizations”.

Now in the first sentence of the above paragraph, one can replace $\psi$ everywhere by $\widetilde{C}$ without changing anything. We then have a contradiction to the assumptions of Proposition 4, which means that $\psi$ couldn’t exist, so $\widetilde{C}$ is good. QED.

Remark. There is a variant of Example 3 in which $C$ is the same as before, but now there are no branch points, so that $\widetilde{C}$ has genus one and two ends. Here neither Proposition 1 nor Proposition 4 is applicable to show that $\widetilde{C}$ is good, i.e. transverse. However I don’t think one needs to consider this case to define ECH cobordism maps. The reason is that the curves you would want to glue this to would have just one negative end (by the ECH partition conditions), so you would want to add a branch point so that $\widetilde{C}$ has just one positive end as in Example 3.

## The blind leading the blind: adventures in online education at duolingo

We interrupt our regular programming for a bit of amusement. The background for this is that I am considering developing an online calculus class (not a MOOC for now, rather something more modest), and so I have been thinking a lot about online education lately. In particular I am curious to see what is already out there.

Now in general I have some qualms about excessively computerized education. I think that education is about more than just optimizing the process of loading the student’s mind with a set collection of facts and rote skills. Rather, there is a certain give and take between the teacher and the student (which works best face to face), and both learn together.

That said, there are some areas where one cannot get around memorizing a certain number of facts and rote skills. A prime example of such an area is elementary language learning. Thus I was quite interested when I heard about a relatively new website called duolingo.com. The idea of this site is that students learn languages for free, and in the process of learning they translate websites in a crowdsourced manner (and the site is funded by translation fees). Now this sounds a bit preposterous, and I can’t find any details about how it works, but after playing with it for a bit, here is what I think is going on.

The language learner sees a series of quizzes. It is like a game and potentially addictive. Quizzes can either review what you are already supposed to know or introduce new material. Quiz questions ask you to translate a short phrase or sentence either from the language you are learning to your native language or vice versa. It is very engaging, constantly asking you to produce output with instant feedback, which I think is an excellent feature for any kind of online education.

That is the good part. Now the bad part is that it is full of errors and inconsistencies, especially at the higher levels. I’ll show you some examples in a minute. Here is my speculation on what is going on. There is some website that duolingo is being paid to translate. The software breaks up the texts into little bits and quizzes language learners on how to translate these bits. (Never mind that it is questionable whether one can get good translations by breaking up sentences into little bits. Just try this with a long German sentence in which a verb is split into two pieces three lines apart…) The quiz answers are initially graded using some software along the lines of Google Translate (and maybe also some algorithm which compares passages in two different languages and decides whether or not they are saying the same thing). Thus, when a question is first introduced into the quiz system, it may have errors. After a while, the grading is adjusted depending on what the users do. For example, if the same supposedly wrong answer is given some number of times, then this answer is reclassified as correct. There is also a button where you can complain if you think you were graded incorrectly (I pressed this button many times when I was trying it out).

I tested this out saying that I am a native English speaker trying to learn German. (I know enough German to feel competent to evaluate the program in this area.) I tested out of the basic levels and then did some review quizzes. Here are some of the errors I found. (If I am the one who was wrong in any of these cases, please correct me.)

• It asked me to translate a sentence which began “If we didn’t exist…” I started with “Wenn wir nicht existierten…”. It marked this wrong and said that I should have written “Wenn wir nicht existieren würden…” As far as I know these mean exactly the same thing, but mine is less clumsy. In general it doesn’t deal well with the fact that there can be different equivalent ways to say the same thing, especially in longer sentences.
• It asked me to translate “I drank wine.” I wrote “Ich habe Wein getrunken”. It marked this wrong and wanted “Ich trank Wein.” Another case of equivalent ways to say the same thing (although here you prefer one or the other depending on whether you are speaking or writing).
• It asked me to translate “Mit zwölf Jahren, ist ein Hund alt”. I answered “A dog is old when it is twelve”. It wanted “a dog is old with twelve” or “with twelve years, a dog is old.” Neither of these makes much sense in English. It seemed to be insisting on the word for word translation of “mit” as “with”, but you can’t just replace German prepositions by similar English prepositions or you will often get nonsense.
• It asked me to translate “I go every year”. I answered “Ich gehe jedes Jahr hin”. It marked that wrong and wanted “Ich gehe jedes Jahr”. Another example where word for word translation gives you bad results if you don’t have an idea of the overall meaning.
• It asked me to translate “He hears notes”. I wrote “Er hört Töne”. It marked this wrong and wanted “Er hört Noten”. While “Note” means “note” in some cases, according to my German dictionary the things that you here are “Töne”, not “Noten”. Correct me if I’m wrong.
• It asked me to translate “Grandmother is making tea”. I wrote “Großmutter kocht Tee”. It marked this wrong and wanted “Großmutter macht Tee”. I wasn’t sure about this, and since I didn’t have a native German speaker handy to ask, I used a little trick: a Google search for “Tee kochen” (in quotes) gives three times as many hits as “Tee machen”. I conclude from this that mine is probaby better, and in any case acceptable.
• It asked me to translate “Das ist bei uns ganz klar der Diesel”. It was multiple choice, and the only option remotely close to this was “That is for us clearly the diesel”, which is not something a native English speaker would ever say.
• It asked me to translate “Wir bleiben in Kontakt”. I answered “We will stay in touch”. It marked this wrong wanted “We stay in touch.” First of all, it doesn’t deal well with the fact that there is not a bijection between verb tenses in the two languages; the present tense in German can correspond to the present or future tense in English. Second, “We stay in touch” is somewhat unnatural English. I can think of situations in which one might say this, but in general the program has a bad tendency to translate things into the “habitual present” tense in English when this does not make sense. For example a correct answer to another question was “The elephant eats an apple.” This is (usually) not a correct English sentence. (I guess you could say it while narrating a story in the present tense, but that is uncommon.) You could say “The elephant is eating an apple” (to refer to what the elephant is doing right now) or “The elephant eats apples” (to indicate that apples are part of the elephant’s diet). But “The elephant eats an apple” does not make sense as a standalone sentence. It could make sense as part of a sentence, e.g. “If the elephant eats an apple, it will get indigestion”. But it looked like we were supposed to translate a complete sentence… The program also seemed to have a lot of trouble with the Konjunktiv, but I don’t have examples handy.
• It asked me to translate “Ein Auftrag aus der Zukunft”. Ignoring the fact that this is a weird thing to say, I translated it as “An assignment from the future”. It marked this wrong and wanted “A assignment from the future”. Really? Surely it would be trivial to program it to know when to use “a” versus “an”.
• It asked me to translate “I do not go on vacation without a pocket book”. It was multiple choice. The only close choice was “Ich fahre nicht in den Urlaub nicht ohne Taschenbuch”. There is one “nicht” too many there.
• It asked me to translate “Does he have his insurances?”. I don’t think that’s correct English, nor do I know what it is supposed to mean.

There is more like this (and there were some hilarious examples much worse than these on the discussion forum), but you get the idea. I would estimate that in the more advanced levels of the German section, about 20 percent of the translation questions have significant issues.

The sentences that look like garbage are obviously the output of computer translation software which we are supposed to be correcting. This is not good for the language learner. We can recognize when our own language is wrong, but if we are learning a new language we don’t know. There are discussion forums where you can ask about questions you have, but it might take a while to get an answer, and the person answering your question might not know any more than you.

The worst part of this is that after a while you are trained to give answers which you know are wrong but the computer will accept in order to avoid losing points. (It is like a game and if you lose too many points you have to backtrack to the beginning of the quiz and get some of the same annoying wrong questions again.) In this way, flaws in the translation software are reinforced and amplified into systematic errors in the program.

Despite all of the above criticism, it is a fun and valuable program. If you need a little break from work, go try it and see what you think. (Hopefully some of the above problems will be fixed. “A” versus “an” should take five minutes.) And then think about how we can teach math online in an engaging manner without propagating errors.

P.S. I don’t know if they are actually doing real translation work yet or if it the system is still in an experimental stage. As I said I couldn’t find any details about what they are doing. There are also some texts that you can try translating yourself, or edit other people’s translations of. However I suspect that at least some of this is not for pay but rather for data gathering purposes. For example some of the texts are classic works (e.g. fairy tales) which have been translated many times before.

P.P.S. Another amusing English sentence which turned up was “You are swimming better than he does.” At least the foreign language sentences seem to have fewer errors, or maybe I just don’t notice them.

Posted in Uncategorized | 8 Comments