Symplectic folding is sometimes optimal

Reference: [BEYOND] = “Beyond ECH capacities”

I have played with a few more calculations using the methods in [BEYOND]. Here is the most interesting thing I have found so far.

In [BEYOND, Thm. 1.2], it was shown, among other things, that if the polydisk $P(a,1)$ symplectically embeds into the four-dimensional ball $B(c)$, and if $2 \le a \le 4$, then $c\ge (10+a)/4$. On the other hand, Felix Schlenk showed using symplectic folding that if $2\le a\le 4$, then $P(a,1)$ symplectically embeds into $B(c)$ whenever $c> (4+a)/2$. These bounds agree for $a=2$ and disagree for $a>2$. One might ask whether one can improve one or both of these bounds to get them to agree. In fact, it turns out that symplectic folding is sometimes optimal, in the following sense:

Theorem. If $2\le a\le 12/5$, and if $P(a,1)$ symplectically embeds into $B(c)$, then $c\ge (4+a)/2$.

This is a direct application of [BEYOND, Thm. 1.18], and I will assume the statement of the latter theorem below. (In other words, the following is basically an addendum to be added in the next version of [BEYOND], unless I discover some way to improve it first.)

Proof of Theorem. Suppose that $2\le a\le 12/5$, that $P(a,1)$ symplectically embeds into $B(c)$, and that $c < (4+a)/2$. We will obtain a contradiction in four steps. Below, the symbol $\le$ between convex generators means $\le_{P(a,1),B(c)}$.

Step 1. We first show that if $\Lambda \le e_{1,1}^d$ with $d\le 9$, then $y(\Lambda) \le 1$.

If $y(\Lambda)\ge 2$, then as in Step 1 of the proof of [BEYOND, Thm. 1.2], we have

$3d-3+2a \le dc$.

Combining this with our assumption that $c < (4+a)/2$ gives

$(d-4)a > 2d-6$.

If $d<4$ then it follows that $a < 4/3$; if $d=4$ then it follows that $2<0$; and if $5\le d\le 9$ then it follows that $a>12/5$. Either way this contradicts our hypothesis that $2\le a \le 12/5$.

Step 2. We now show that if $\Lambda\le e_{d,d}$, and if $y(\Lambda)\le 1$, then $\Lambda$ includes a factor of $e_{1,0}$.

If not, then the only possibility for $\Lambda$ with the correct ECH index is

$\Lambda = e_{(d^2+3d-2)/2,1}$.

The action inequality in the definition of $\le_{P(a,1),B(c)}$ then implies that

$(d^2+3d-2)/2 + a \le dc.$

Combining this with our assumption that $c < (4+a)/2$ gives

$(d-2)a > d^2-d-2$.

If $d=1$ then it follows that $a<2$; if $d=2$ then it follows that $0<0$; and if $d\ge 3$ then it follows that $a > d+1$. Either way this contradicts our hypothesis that $2 \le a \le 12/5$.

Step 3. We now show that there does not exist any convex generator $\Lambda$ with $\Lambda \le e_{1,1}^9$.

If $\Lambda$ is such a generator, then we know from Step 1 that $y(\Lambda) \le 1$.

If $y(\Lambda) = 0$, then the only possibility for $\Lambda$ with the correct ECH index is $\Lambda = e_{1,0}^{54}$. Then $54 \le 9c$, which combined with our assumption that $c < (4+a)/2$ implies that $a > 8$, contradicting our hypotheses.

If $y(\Lambda) = 1$, then $x(\Lambda) \ge 27$, or else we would have $I(\Lambda) \le 106$, contradicting the fact that $I(\Lambda) = 108$. Since $x(\Lambda) \ge 27$, it follows that

$27 + a \le 9c$.

Combining this with our assumption that $c < (4+a)/2$ gives $a > 18/7$, contradicting our hypothesis that $a \le 12/5$.

Step 4. We now apply [BEYOND, Thm. 1.18] to $\Lambda'=e_{1,1}^9$, to obtain a convex generator $\Lambda$, and factorizations $\Lambda=\Lambda_1\cdots\Lambda_n$ and $\Lambda'=\Lambda_1'\cdots\Lambda_n'$, satisfying the three bullet points in [BEYOND, Thm. 1.18].

By Step 3 and the first bullet point, we must have $n>1$.

By Step 2 and the first two bullet points, all of the $\Lambda_i$ must be equal, and all of the $\Lambda_i'$ must be equal. Thus either $n=9$ and $\Lambda_i'=e_{1,1}$ for each $i$, or $n=3$ and $\Lambda_i'=e_{3,3}$ for each $i$.

If $n=9$, then by Steps 1 and 2, we have $\Lambda=e_{1,0}^2$ for each $i$. But then $I(\Lambda)=36$, contradicting the fact that $I(\Lambda)=108$.

If $n=3$, then by Step 1, and the facts that $I(\Lambda_i)=18$ and $x(\Lambda_i) + y(\Lambda_i) \ge 8$, the only possibilities are that $\Lambda_i = e_{1,0}^9$ for each $i$, or $\Lambda_i = e_{1,0}e_{6,1}$ for each $i$. In the former case we have $I(\Lambda) = 54$, and in the latter case we have $I(\Lambda)=102$. Either way, this contradicts the fact that $I(\Lambda)=108$.

QED

Remark. It is conceivable that with more work, the hypothesis $a\le 12/5$ could be weakened to $a \le (\sqrt{7}-1)/(\sqrt{7}-2) = 2.54858\cdots$. The significance of the latter number is that if $a$ is less than it, and if $d$ is sufficiently large with respect to $a$, then there does not exist any convex generator $\Lambda$ with $\Lambda \le e_{1,1}^d$. We might then be able to use arguments similar to the above to get a contradiction.

More generally, one can maybe get more information by considering all of the holomorphic curves that exist in the cobordism coming from a symplectic embedding. We know that certain curves must exist in order to give a chain map on ECH satisfying the required properties. However the existence of certain curves excludes the existence of others, for example when their intersection number would be negative. When you write down all of these conditions, it is like a giant logic puzzle, and the challenge is to extract significant information from it using a manageable amount of computation.