## Rational SFT using only q variables III

Let $(Y,\lambda)$ be a nondegenerate closed contact manifold of dimension $2n-1$. In the previous two posts, I defined (modulo the usual transversality issues) a modified version of rational SFT, which we are temporarily denoting by $HQ(Y,\lambda)$. I now want to define a $U$ map on this theory, by analogy with the $U$ map on ECH.

Definition of the U map

Recall that ${\mathcal A}$ denotes the (completed) algebra generated by the $p$ and $q$ variables. We choose a generic almost complex structure $J$ on ${\mathbb R}\times Y$ and perturbation of the holomorphic curve equation, and then $H\in{\mathcal A}$ denotes the Hamiltonian counting connected index one rational curves.

Now pick a generic point $y\in Y$. We can then consider connected rational curves of index $2n-2$ in ${\mathbb R}\times Y$ with a marked point mapping to $(0,y)\in{\mathbb R}\times Y$. Counting these curves with the usual signs and combinatorial factors, assuming transversality as usual, defines an element $U_y\in{\mathcal A}$. Looking at ends of moduli spaces of index $2n-1$ rational curves with a marked point mapping to $(0,y)$ shows that the Poisson bracket $\{U_y,H\}=0$. This is enough to show that $U_y$ induces a map on the usual rational SFT (which has been studied in the SFT literature).

Now recall from the first post in this series that ${\mathcal A}'$ denotes the symmetric algebra on the polynomial algebra on the $q$ variables, and we have a multiplication ${\mathcal A}\times {\mathcal A}'\to{\mathcal A}'$. This is related to the Poisson bracket as follows: if $a_1,a_2$ are monomials in ${\mathcal A}$ and $b\in {\mathcal A}'$ then

$\{a_1,a_2\}\cdot b = a_1\cdot(a_2\cdot b) - (-1)^{|a_1||a_2|} a_2\cdot (a_1\cdot b),$

where $|a|$ denotes the grading parity of $a$. It follows from the above identity and $\{H,H\}=0$ that multiplication by $H$ defines a differential $d'$ on ${\mathcal A}'$, whose homology we are calling $HQ(Y,\lambda)$. It also follows from the above identity and $\{U_y,H\}=0$ that $U_y$ is a chain map from $({\mathcal A}',d')$ to itself. The induced map on $HQ(Y,\lambda)$ is the $U$ map, which we denote by $U$.

The $U$ map respects the symplectic action filtration, by the usual Stokes’s theorem argument.

When $Y$ is connected, $U$ does not depend on the choice of base point $y$. To see this, let $y'$ be another base point and let $\gamma:[0,1]\times Y$ be a generic path from $y$ to $y'$. We then consider pairs $(t,u)$ where $t\in[0,1]$ and $u$ is an index $2n-3$ rational curve with a marked point mapping to $\gamma(t)$. Counting these with appropriate signs and combinatorial factors defines an element $K\in{\mathcal A}$ such that $\{K,H\}=U_y - U_{y'}$. It follows from this equation and the above identity that $U_y$ and $U_{y'}$ induce the same map on $HQ(Y,\lambda)$.

If $Y$ is disconnected, then there is a different $U$ map for each component of $Y$. If $(Y_i,\lambda_i)$ are the components of $Y$, then it follows from the definition that $HQ(Y,\lambda)=\otimes_i HQ(Y_i,\lambda_i)$, and the $U$ map associated to the $i^{th}$ component of $Y$ is the tensor product of the $U$ map on $(Y_i,\lambda_i)$ with the identity on the other factors. Just like in ECH. [Correction 5/1/13: It is not quite right that $HQ(Y,\lambda) = \otimes_i HQ(Y_i,\lambda_i)$. Rather, there is a map $\otimes_i HQ(Y_i,\lambda_i) \to HQ(Y,\lambda)$, induced by the obvious inclusion of chain complexes, and this map is injective because it has a right inverse induced by a chain map which "disconnects" Reeb orbits in different components of $Y$. Under these maps, the $U$ map on $Y$ associated to the $i^{th}$ component agrees with the tensor product of the $U$ map on $(Y_i,\lambda_i)$ with the identity on the other factors.]

Similar arguments show that the $U$ map (or maps when $Y$ is disconnected) do not depend on $J$ (and the perturbation of the holomorphic curve equation). Rather than explain this, let us explain a more general phenomenon which implies it, namely how the $U$ map behaves under cobordisms.

The U map and cobordism maps

Let $(X,\omega)$ be an exact symplectic cobordism from $(Y_+,\lambda_+)$ to $(Y_-,\lambda_-)$. In the previous post we defined a cobordism map $\Phi(X,\omega): HQ(Y_+,\lambda_+) \to HQ(Y_-,\lambda_-)$ which respects the symplectic action filtrations. I claim if $U_+$ denotes the $U$ map on $HQ(Y_+,\lambda_+)$ determined by one of the components of $Y_+$, and if $U_-$ denotes the $U$ map on $HQ(Y_-,\lambda_-)$ determined by one of the components of $Y_-$, and if these components of $Y_+$ and $Y_-$ are contained in the same component of $X$, then

$U_-\circ \Phi(X,\omega) = \Phi(X,\omega)\circ U_+: HQ(Y_+,\lambda_+) \to HQ(Y_-,\lambda_-).$

To see this, recall from the previous post that counting connected index zero rational curves in the completed cobordism defines an element $\phi_1\in {\mathcal A}_{-,+}$, and $\phi = \exp(\phi_1)\in{\mathcal A}'_{-,+}$ (see the previous post for explanation of this notation) induces the cobordism map $\Phi(X,\omega)$.

Now pick a base point $x\in X$. Then counting index $2n-2$ connected rational curves in the completed cobordism with a marked point mapping to $x$ defines an element $U_x\in{\mathcal A}_{-,+}$. Define

$U_x' = U_x\otimes\phi\in{\mathcal A}'_{-,+}.$

This counts possibly disconnected index $2n-2$ rational curves in the completed cobordism with a marked point mapping to $x$. Considering ends of moduli spaces of possibly disconnected index $2n-1$ rational curves in the completed cobordism with a marked point mapping to $x$ shows that

$H_-\cdot U_x' = U_x'\cdot H_+.$

Hence $U_x'$ induces a map $HQ(Y_+,\lambda_+)\to HQ(Y_-,\lambda_-)$.

Now this induced map is equal to $\Phi(X,\omega)\circ U_+$ if $U_+$ is the $U$ map on $HQ(Y_+,\lambda_+)$ corresponding to a component of $Y_+$ in the same component of $X$ as $x$. It is also equal to $U_-\circ \Phi(X,\omega)$ if $U_-$ is the $U$ map on $HQ(Y_-,\lambda_-)$ corresponding to a component of $Y_-$ in the same component of $X$ as $x$. One defines a chain homotopy by taking a path $\gamma:[0,\infty)\to X$ which starts at $x$ and goes off to infinity on one of the ends and counting index $2n-3$ rational curves in the completed cobordism with  a marked point mapping to a point on $\gamma$.

If $(X,\omega)$ is a weakly exact cobordism from $(Y_+,\lambda_+)$ to $(Y_-,\lambda_-)$, then the cobordism map $\Phi(X,\omega):HQ(Y_+,\lambda_+,0)\to HQ(Y_-,\lambda_-,0)$ commutes with the $U$ maps in the same way, by the same argument.

Now we are ready to define capacities! I will do this in the next post.

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### 7 Responses to Rational SFT using only q variables III

1. Chris Wendl says:

Other than wanting to have a well-behaved action filtration and to avoid Novikov rings, was it really necessary to assume $$(X,\omega)$$ was (weakly) exact in this discussion?

2. No, the weakly exact assumption is not really necessary. Analogously to the ECH case as I described here last May and August (a paper about this in progress), a general symplectic cobordism $(X,\omega)$ from $(Y_+,\lambda_+)$ to $(Y_-,\lambda_-)$ induces maps $\Phi(X,\omega,A):HQ(Y_+,\lambda_+,\partial_+A)\to HQ(Y_-,\lambda_-,\partial_-A)$ for $A\in H_2(X,\partial X)$ which respect the symplectic action filtration up to a shift by $\rho(A)$, and maps $\overline{\Phi}(X,\omega)$ on the Novikov completions.

By the way, if you want to use latex in a comment, put your math in between single dollar signs, and after the first dollar sign insert the word latex’ (with no space between the first dollar sign and the word latex’). Unfortunately I don’t know how to see a latex preview for comments.

• Chris Wendl says:

Thanks for the tip! This is a test: $E=mc^2$

3. Janko Latschev says:

I thought I was subscribed to the blog, but I guess I wasn’t, so I missed installments II and III until today. Anyway, very nice job. I guess the definition of the capacities will involve the special cycle in $HQ(Y,\lambda)$ corresponding to the empty monomial as a generator of $\mathcal A'$, say $(\emptyset)$. To test my understanding of things so far: Suppose $(Q_1\cdots Q_s)$ is a monomial of monomials, i.e. an element of $\mathcal A'$. A configuration of curves in a cobordism $(X,\omega)$ in relative homology class $A$ which has no negative ends contributes $(\emptyset)^r$ to $\Phi(X,\omega,A)(Q_1\dots Q_s)$ where $r$ is the number of connected components one gets by imagining that each $Q_j$ corresponds to a kind of formal rational curve with only negative ends, and then performing the formal gluing the way you described it in your first post (i.e. only allowing configurtions where each piece ends up having genus 0)?

• That’s close, except that $\emptyset^r=\emptyset$. It might help to think of ${\mathcal A}'$ as the symmetric algebra on polynomials in the $q$ variables as you suggest, and write $Q_1\otimes\cdot\otimes Q_s$ instead of $(Q_1,\ldots,Q_s)$. One can then write $1$ instead of $\emptyset$.

• Janko Latschev says:

I actually meant what I said, i.e. that the empty monomial $\emptyset$ is one of the generators of the symmetric algebra and not the empty “monomial of monomials”, and then $\emptyset^r \ne \emptyset$, because they are monomials of different degree. Probably your approach is a lot better for what you want to do, since it ignores information irrelevant for your purpose, but one could dream of cases where the number of components without negative ends ($r$ in the example) has some interesting meaning as well. (Needless to say, I don’t have specific examples in mind.)
I will stop rambling now.

4. OK, you could also do it that way. This would be useful for example if you were trying to count index zero genus zero holomomorphic curves in a closed symplectic manifold by cutting along a contact type hypersurface, in which case the exponent of the emptyset would keep track of the number of components of a closed holomorphic curve. To simplify what I am currently doing, I will not do that and will instead declare that $\emptyset^r=\emptyset$ and call this $1$.