## Rational SFT using only q variables

I have talked a lot about ECH capacities in this blog and elsewhere. These give obstructions to symplectically embedding one symplectic four-manifold with boundary into another. This obstruction is sharp for embedding one ellipsoid into another, as shown by McDuff. However is not so good for say embedding a polydisk into a ball, as shown by simple examples.

A separate point which I would like to reiterate here is that one can use formalism similar to the definition of ECH capacities to define capacities using other versions of contact homology. For example, the analogue in linearized contact homology of the “full ECH capacities” conjecturally agrees with the Ekeland-Hofer capacities, at least for convex domains in ${\mathbb R}^4$.

I would now like to explain how to define an analogue of ECH capacities for rational SFT. Work of Hind-Kerman and Hind-Lisi (obtaining some powerful results about embeddings by studying rational curves) suggests that one might obtain some useful new embedding obstructions this way. To do so, we first need to modify the definition of rational SFT to make it more suitable for defining capacities.

The usual definition of rational SFT

Let us first briefly review the usual definition of rational SFT due to Eliashberg, Givental, and Hofer. Let $Y$ be a closed contact manifold of dimension $2n-1$ and let $\lambda$ be a nondegenerate contact form on $Y$. For each (possibly multiply covered) “good” Reeb orbit $\gamma$, one introduces two formal variables $p_\gamma$ and $q_\gamma$. (A Reeb orbit is “bad” if it is the double cover of a Reeb orbit whose Conley-Zehnder index has opposite sign, otherwise it is “good”.) One defines an algebra ${\mathcal A}$ which is a suitable Novikov completion of the supercommutative algebra over ${\mathbb Q}$ generated by the variables $p_\gamma$ and $q_\gamma$. (In the supercommutativity, the parity of the grading of a variable $p_\gamma$ or $q_\gamma$ is the parity of the Conley-Zehnder index of $\gamma$, plus the parity of $n+1$.) The algebra ${\mathbb A}$ has a Poisson bracket $\{\cdot,\cdot\}$ defined via the symplectic form $\omega=\sum_\gamma k_\gamma^{-1}p_\gamma\wedge q_\gamma$. Here $k_\gamma$ denotes the covering multiplicity of the Reeb orbit $\gamma$ (which is $1$ if and only if $\gamma$ is embedded). What this means is that the Poisson bracket of two monomials is the sum of all ways of annihilating a $p_\gamma$ factor in one monomial with a $q_\gamma$ factor in the other monomial, counted with appropriate signs and combinatorial factors.

To define the differential on ${\mathcal A}$, one chooses a generic almost complex structure $J$ on ${\mathbb R}\times Y$ satisfying the usual conditions to define any kind of contact homology. Moreover, one needs to make some abstract perturbation of the $J$-holomorphic curve equation to arrange transversality of moduli spaces of (abstractly perturbed) $J$-holomorphic curves. Hopefully polyfold theory will soon make this possible; let us assume that this has been done and pretend that all relevant moduli spaces of holomorphic curves are cut out transversely, so that we can focus on trying to understand the formalism.

One now defines the Hamiltonian $H$ to be a sum over all index $1$ holomorphic curves modulo ${\mathbb R}$ translation. A holomorphic curve with positive ends at Reeb orbits $\alpha_1,\ldots,\alpha_k$ and negative ends at Reeb orbits $\beta_1,\ldots,\beta_l$ contributes a term of $q_{\beta_1}\cdots q_{\beta_l}p_{\alpha_1}\cdots p_{\alpha_k}$, times an appropriate sign and combinatorial factor, to the Hamiltonian. Looking at ends of index 2 moduli spaces of genus zero curves shows that

$\{H,H\}=0.$

Here the Poisson bracket formalism cleverly keeps track of the fact that if a connected genus zero curve splits into two pieces, then it does so along a single circle. This is because as mentioned above, when you take the Poisson bracket of two monomials, only one $p$ variable can annihilate a corresponding $q$ variable.

One now defines a differential $d$ on the algebra ${\mathcal A}$ by $d = \{H,\cdot\}$. The Jacobi identity for the Poisson bracket (with some funny signs coming from the supercommutativity) together with the identity $\{H,H\}=0$ imply that $d^2=0$. Then rational SFT is the homology of this differential. Also, the Poisson bracket makes it into an algebra, which is very nice.

The problem

Unfortunately this is not very well suited to defining capacities. To define capacities we first want to define a filtered theory, considering only generators of symplectic action less than $L$. However for the differential to decrease symplectic action, we should define the symplectic action of $p_\gamma$ to be minus the symplectic action of $\gamma$; but then there would be infinitely many generators of action less than $L$, involving arbitrarily long Reeb orbits, so this would not work so well.

Instead, we need a version of rational SFT involving only the $q$ variables. While doing so, we need to somehow keep track of the fact that we are never allowed to glue together the same pair of genus zero curves more than once to obtain a higher genus curve. Here is how I propose to do this.

Rational SFT with only q variables

Define a new algebra ${\mathcal A}'$ over ${\mathbb Q}$ as follows. A generator of ${\mathcal A}'$ is an ordered $m$-tuple $(x_1,\ldots,x_m)$, for $m\ge 0$, where each $x_i$ is a nonconstant monomial in the $q$ variables. We mod out by the usual supercommutativity for changing the order of the $q$ variables in one of the monomials $x_i$. Also, we can re-order the $m$ elements in the $m$-tuple $(x_1,\ldots,x_m)$ and multiply by $-1$ whenever we commute two odd degree elements.

One can alternately think of a generator of ${\mathcal A}'$ as a monomial in the $q$ variables, in which the factors have been partitioned into $m$ equivalence classes. The idea is that this is a list of Reeb orbits, where Reeb orbits in the same equivalence class are boundary components of the same connected genus zero curve above the Reeb orbits, and so are not allowed to be glued to the same component of a holomorphic curve below the Reeb orbits.

We next define a product ${\mathcal A}\times {\mathcal A}'\to {\mathcal A}'$ as follows. Let $y=q_{\alpha_1}\ldots q_{\alpha_k}p_{\beta_1}\cdots p_{\beta_l}$ be a monomial in ${\mathcal A}$, and let $x=(x_1,\ldots,x_m)$ be a generator of ${\mathcal A}'$. The idea is to think of $y$ as a connected genus zero curve with $l$ positive ends and $k$ negative ends, and sum over all ways of attaching the positive ends to $x$ without increasing the genus. More precisely, $y\cdot x$ is the sum over all ways of annihilating each of the $p_{\beta_i}$ factors in $y$ with a $q_{\beta_i}$ factor in $x$, such that the $q_{\beta_i}$ factors in $x$ that are annihilated are in $l$ distinct equivalence classes. The resulting new generator of ${\mathcal A}'$ has $m-l+1$ equivalence classes. One equivalence class consists of the variables $q_{\alpha_1},\ldots,q_{\alpha_k}$, together with the remaining variables in the $l$ equivalence classes from which $q_{\beta_i}$ variables were annihilated. The remaining $m-l$ equivalence classes are the equivalence classes in $x$ from which no variable was annihilated. (This is easier to explain by drawing a picture…) One multiplies by the same sign and combinatorial factor as in the definition of the Poisson bracket on ${\mathcal A}$.

If $y_1,y_2$ are odd degree monomials in ${\mathcal A}$, and if $x$ is a generator of ${\mathcal A}'$, then we have

$y_1\cdot(y_2\cdot x) + y_2\cdot(y_1\cdot x) = \{y_1,y_2\}\cdot x.$

The idea is that both sides of the equation sum over all ways of gluing a positive end of $y_1$ or $y_2$ to a negative end of $y_2$ or $y_1$, and then gluing all remaining positive ends to distinct equivalence classes in $x$.

We now define a differential $d'$ on ${\mathcal A}'$ by $d'(x)=H\cdot x$. The above identity together with $\{H,\cdot H\}=0$ imply that $(d')^2=0$. The homology of $d'$ is the “q-variable only” version of rational SFT, which can be used to define capacities. In particular it has a symplectic action filtration, in which the symplectic action of a monomial in the $q$ variables is the sum of the symplectic actions of the Reeb orbits corresponding to the variables.

Did that make any sense at all? If so, I’ll explain in a subsequent post how to use the above to define some new symplectic capacities (which may or may not be interesting, we’ll see).

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### 5 Responses to Rational SFT using only q variables

1. Chris Wendl says:

I haven’t had a chance to digest this fully yet, but this post makes me very happy and I am immensely looking forward to the next one. Here are the questions that occur to me immediately:

(1) Is there a nice “purely algebraic” way to define the action of A on A’, without appealing to our intuition about holomorphic curves?

(2) What do we get from cobordisms? (Presumably your next post will address this anyway…)

• Janko Latschev says:

In fancy words, A’ is the symmetric algebra on A. The rational curves in the symplectization give you an L_\infty structure on A, with curves with \ell inputs giving you the \ell-to-one operation. What Michael describes is the “packed up” version of this which assembles everything into a coderivation of square 0 on A’=S(A). This was in fact the way Kai and I first thought of (rational) SFT when we tried to find the relation to string topology. From what I remember, there were two drawbacks (maybe more):
1) this is a huge complex algebraically, making things somewhat unwieldy
2) I never found a nice way to formulate (algebraically, in terms of the usual SFT algebra stuff) the morphisms you expect to get from cobordisms, at least not nice enough to be presented in public. In fact, non-exact cobordisms will only give you something like a pairing, I suppose…

In the linearizable world you can drop everything down to a certain version of an infinity Lie bialgebra structure on the vector space generated by the Reeb orbits, which is much smaller and more managable. Again, I never worked out properly how to describe the morphisms in that context (nicely), in part because they were simple enough to formulate in the general SFT picture,.But that is just my ignorance.

• OK, thanks. It doesn’t surprise me that you have already thought of this. Hopefully I’ll have something new to say when we get to capacities.