## Rational SFT using only q variables IV

It is now time to explain how to use the q-variable version of rational SFT to define symplectic capacities and symplectic embedding obstructions. Before getting into details, I should tell you some good news and bad news. The bad news is that the rational SFT capacities are not very powerful. The good news is that rational SFT contains stronger embedding obstructions than the rational SFT capacities, and you can access these stronger obstructions if you can compute cobordism maps. This is a very interesting area for further exploration.

Spectral invariants

Let $(Y^{2n-1},\lambda)$ be a nondegenerate closed contact manifold as usual. If $\sigma\in HQ(Y,\lambda,0)$ is a nonzero class in the q-variable rational SFT, define $c_\sigma(Y,\lambda)\in{\mathbb R}$ to be the infimum over $L$ such that $\sigma$ is in the image of the map $HQ^L(Y,\lambda)\to HQ(Y,\lambda)$ induced by the inclusion of chain complexes.

If $(X^{2n},\omega)$ is a weakly exact symplectic cobordism from $(Y_+,\lambda_+)$ to $(Y_-,\lambda_-)$, recall that there is an induced map

$\Phi(X,\omega): HQ(Y_+,\lambda_+,0) \to HQ(Y_-,\lambda_-,0)$

which is the direct limit of maps

$\Phi^L(X,\omega): HQ^L(Y_+,\lambda_+,0)\to HQ^L(Y_-,\lambda_-,0)$

as $L\to\infty$. It follows as in ECH that if $\sigma_+\in HQ(Y_+,\lambda_+,0)$, if $\sigma_-=\Phi(X,\omega)(\sigma_+)\in HQ(Y_-,\lambda_-,0)$, and if $\sigma_-\neq 0$, then

$c_{\sigma_-}(Y_-,\lambda_-) \le c_{\sigma_+}(Y_+,\lambda_+).$

This is the basic inequality which leads to obstructions to symplectically embedding one Liouville domain into another.

As with the ECH spectrum, if $\lambda$ is degenerate, one can define $c_\sigma(Y,\lambda)$ as the limit of $c_\sigma(Y,\lambda_i)$ where $\{\lambda_i\}_{i=1,2,\ldots}$ is a sequence of nondegenerate contact forms converging in the $C^0$ topology to $\lambda$. The above inequality still holds for a weakly exact cobordism between possibly degenerate contact manifolds.

Definition of capacities

Let $(X^{2n},\omega)$ be a Liouville domain. What I mean by this is that $\omega$ is exact, and there is a contact form $\lambda$ on $Y=\partial X$ such that $d\lambda = \omega|_Y$. Note that the unit in the symmetric algebra on the polynomial algebra on the $q$ variables, which I denote by $1$, is a cycle in the chain complex computing $HQ(Y,\lambda,0)$. The homology class $[1]$ is nonzero, because the cobordism map $\Phi(X,\omega)$ maps it to $1$.

If $k$ is a nonnegative integer, we now define $c_k(X,\omega)\in[0,\infty]$ to be the minimum of $c_\sigma(Y,\lambda)$, where $\sigma$ ranges over classes in $HQ(Y,\lambda,0)$ such that $U^k\sigma=[1]$ whenever $U^k$ is a composition of $k$ of the $U$ maps associated to the components of $Y$ (possibly repeated). If no such class $\sigma$ exists, we define $c_k(X,\omega)=\infty$. The same or similar arguments to the definition of ECH capacities show the following:

• $c_k(X,\omega)$ does not depend on the choice of $\lambda$.
• If $(X',\omega')$ is another Liouville domain of the same dimension which symplectically embeds into $(X,\omega)$, then $c_k(X',\omega')\le c_k(X,\omega)$ for all $k$.
• We have the disjoint union property

$c_k((X_1,\omega_1)\sqcup (X_2,\omega_2)) = \max_{k_1+k_2=k}(c_{k_1}(X_1,\omega_1) + c_{k_2}(X_2,\omega_2)).$

Examples

I did some quick calculations to get an idea of how good these capacities are. I can explain the details of these calculations later, but for now here are the results (which you should regard as preliminary since I didn’t check every detail):

• The capacities of the four-dimensional ball $B^4(1)$, starting at $k=0$, are $0,1,1,2,2,2,3,3,3,\ldots$. That is, $c_k(B^4(1))=\lceil \frac{k+1}{3}\rceil$ for $k>0$.
• The capacities of the four-dimensinal polydisk $P(1,a)$ for $a\ge 1$ are given by $c_k(P(1,a)) = \min\{k,\lceil \frac{k-1}{2}\rceil + a\}$.
• If $2n>4$ then $c_k(B^{2n}(1)) = \lceil \frac{k}{2} \rceil$.

To test how powerful these capacities are, let us consider the question of when the disjoint union of $m$ four-dimensional balls of capacity $a$ be symplectically embedded into the four-ball of capacity $1$. (Recall that ECH capacities are sharp for this embedding problem.) McDuff-Polterovich showed that if such an embedding exists, then $a\le 1/2$ when $m=2$, $a\le 2/5$ when $m=5$, $a\le 3/8$ when $m=7$, and $a\le 6/17$ when $m=8$, and these bounds are optimal.

We can compute the rational SFT capacities of $\sqcup_mB(a)$ using the disjoint property. We find that $c_2(\sqcup_2B(a))=2a$, while $c_2(B(1))$, so we recover the inequality $a\le 1/2$ when $m=2$. We also compute that $c_5(\sqcup_5B(a))=5a$ while $c_5(B(1))=2$, so we recover the inequality $a\le 2/5$ when $m=5$. Unfortunately, when $m=7$ or $m=8$ we do not even recover the volume constraint $a\le \sqrt{1/m}$.

The ultimate reason for this is that the first inequality comes from curves of degree $1$ in ${\mathbb C}P^2$, the second inequality comes from curves of degree $2$, the third inequality comes from curves of degree $3$, and the fourth inequality comes from curves of degree $6$. Since the latter two curves have positive genus, rational SFT does not see them. One might hope that rational SFT capacities see some non-embedded rational curves that ECH does not see, e.g. for embedding a polydisk into an ellipsoid where the ECH capacities do not give sharp obstructions, but so far I have not found an example where rational SFT capacities say anything more than ECH capacities.

When $2n>4$, rational SFT capacities say even less about packing a disjoint union of balls into a ball. All they tell us is that if $B^{2n}(a)\sqcup B^{2n}(b)$ symplectically embeds into $B^{2n}(1)$, then $a+b\le 1$, which was known to Gromov.

The good news

The good news is that rational SFT gives stronger obstructions to symplectic embeddings than rational SFT capacities, if you know something about the cobordism map.

In particular, rational SFT does recover the optimal obstructions to symplectically embedding the disjoint union of seven or eight four-balls of equal size into a four-ball! To see why, suppose there exists a symplectic embedding of $\sqcup_mB^4(a)$ into $B^4(1)$ where $m\in\{7,8\}$. Let $X$ be the symplectic cobordism obtained by removing the image of the embedding from $B^4(1)$. We can perturb this to obtain a cobordism between nearly-round ellipsoids with nondegenerate contact forms. Each ellipsoid has two embedded elliptic Reeb orbits which we denote by $\alpha$ and $\beta$, where $\alpha$ denotes the shorter orbit. The ECH generator differential vanishes because all generators have even grading.

We know from general properties of ECH cobordism maps that when $m=7$ (resp. $m=8$), the map induced by the cobordism has a nonzero coefficient from the ECH generator $\beta^3$ (resp. $\beta^6$) to the ECH generator consisting of $\alpha^2$ in one of the balls and $\alpha$ in the other six balls (resp. $\alpha^3$ in one of the balls and $\alpha^2$ in the other seven balls). Since the cobordism map decreases symplectic action, we conclude that $3\ge (2+6)a$ when $m=7$ and $6\ge (3+7\cdot 2)a$ when $m=8$, which are the optimal inequalities.

Now since the above coefficient of the ECH cobordism map is nonzero, there exists a (more precisely $1 \mod 2$) holomorphic curve between the corresponding ECH generators. In general one could get a broken and/or multiply covered curve, but (I think) one can rule that out in the present case because otherwise one would obtain a stronger embedding obstruction which cannot be true. Moreover, one can use the ECH partition conditions to see that each end of this holomorphic curve is at a singly covered Reeb orbit, so that it has $11$ ends when $m=7$ and $23$ ends when $m=8$. One can then use the relative adjunction formula to show that the holomorphic curve has $\chi=-9$ when $m=7$ and $\chi=-21$ when $m=8$. This implies that it is rational. So the cobordism map on rational SFT sees it!

More precisely, in rational SFT, the differential vanishes here because all holomorphic curves have even index, so we can identify elements of the chain complex with homology classes. If $\sigma_+$ denotes $\otimes_3\beta$ when $m=7$ and $\otimes_6\beta$ when $m=8$, and if $\Phi$ denotes the cobordism map on rational SFT, then $\Phi(\sigma_+)$ includes a monomial with $8$ of the $\alpha$ variables when $m=7$ and $17$ of the $\alpha$ variables when $m=8$. It follows that $c_{\Phi(\sigma_+)}\ge 8a$ when $m=7$ and $c_{\Phi(\sigma_+)}\ge 17a$ when $m=8$, so the inequality at the beginning of this post gives the optimal symplectic embedding obstruction.

In fact, as far as I know it it possible that all of the ECH obstructions to embedding a disjoint union of four-dimensional ellipsoids into an ellipsoid are seen by rational SFT cobordism maps.

The puzzle

In the above example of embedding seven or eight four-dimensional balls into a ball, it is a nontrivial exercise (which I haven’t really tried yet) to compute the relevant part of the rational SFT cobordism map without “cheating” and using information from ECH.

One can get some information about the cobordism map by using the fact that it commutes with the U maps, compare Lemma 3.2 in “The asymptotics of ECH capacities”. However in the present case one has to work harder because the U map is no longer injective on generators.

In dimension $2n>4$, if one can somehow compute the rational SFT cobordism map coming from an embedding of a disjoint union of balls into a ball, then one might get stronger obstructions to ball packing than the Gromov obstruction recalled above.

Likewise, for embedding one ellipsoid into another in higher dimensions, I suspect that rational SFT capacities say nothing more than Ekeland-Hofer capacities, but if one can compute the cobordism map one might be able to obtain stronger obstructions.