Fun with ECH capacities 2

This is a continuation of the previous post about computing some examples of ECH capacities.  My goals in this post are (1) to recap some of the previous discussion using better notation, (2) to explain an effective way to compute the ECH capacities of triangles, and (3) to give an update on the status of the conjectures in the previous post.

(1) Let me briefly recall what we are doing.  For every domain $\Omega$ in the first quadrant of the plane we have a subset $X_\Omega$ of ${\mathbb C}^2$, namely the inverse image of $\Omega$ under the map sending $(z_1,z_2)\mapsto\pi(|z_1|^2,|z_2|^2)$.  We also have a sequence of numbers

$0 = c_0(\Omega) \le c_1(\Omega) \le c_2(\Omega) \le \cdots \le \infty$

which are the ECH capacities of $X_\Omega$.  We would like to calculate $c_k(\Omega)$, at least for some examples of interest.

Let’s recall what we know about the numbers $c_k(\Omega)$.  To do so it will help to introduce a bit of notation.  Let ${\mathcal S}$ denote the set of nondecreasing sequences $A=(A_k)_{k=0,1,\ldots}$ where each $A_k$ is a real number (or $\infty$) and $A_0=0$.  The ECH capacities of $X_\Omega$ give such a sequence which we denote by $c(\Omega)$.  Given two sequences $A,B\in{\mathcal S}$, define a new sequence $A+B\in{\mathcal S}$ by

$(A+B)_k = \max_{i+j=k}(A_i+B_j).$

Observe that the operation $+$ is associative and commutative.  Also, write $A\le B$ if $A_k\le B_k$ for all $k$.  Write $A < B$ if $A\le B$ and $\lim_{k\to\infty}(B_k-A_k)=\infty$.  (For this purpose we define $\infty-\infty=\infty$.  But that is not important for now because the examples of interest have finite ECH capacities.)  If $A < B$, define a new sequence $B-A\in{\mathcal S}$ by

$(B-A)_k = \min_{l\ge 0} (B_{k+l}-A_l).$

Note that the operation $-$ is not inverse to $+$.  In particular, it is easy to verify that $(B-A) + A \le B \le (B+A)-A$, but the reverse inequalities do not in general hold.

Now here are some properties of $c(\Omega)$:

• If $\Omega'$ and $\Omega$ are in the interior of the first quadrant, and “affine equivalent”, meaning that one can be obtained from the other by the action of $SL_2{\mathbb Z}$ and translation, then $c(\Omega)=c(\Omega')$.
• If $\Omega\subset \Omega'$, or more generally if $X_\Omega$ symplectically embeds into $X_{\Omega'}$, then $c(\Omega) \le c(\Omega')$.
• If $\Omega=\Omega_1\cup\Omega_2$ where $\Omega_1$ and $\Omega_2$ intersect only along their boundaries, then $c(\Omega_1)+c(\Omega_2)\le c(\Omega)$, and $c(\Omega_1) \le c(\Omega)-c(\Omega_2)$.
• If $\Omega$ is convex and intersects the axes in at most two points, then $c(\Omega) = m(\Omega)$, where $m_k(\Omega)$ is defined to be the minimum, over convex paths $\Lambda$ in the plane that enclose at least $k+1$ lattice points, of the length of $\Lambda$, as measured using the dual of a norm whose unit ball is a translate of $\Omega$.

Here the first property follows from the symplectomorphism invariance of ECH capacities.  The second property follows from the monotonicity axiom for ECH capacities.  In the third property, the first inequality follows from the monotonicity and disjoint union axioms for ECH capacities, and the second inequality is a consequence of the first.  These inequalities are not sharp in general, but they are sharp in some lucky cases.  The fourth property follows from Theorem 1.1 in my paper “Quantitative ECH”, as discussed in the previous post.  The assumption on intersecting the axes in at most two points can be dropped for rectangles and right triangles with legs on the axes, and quite possibly more generally.  In any case we always know using monotonicity that $c_k(\Omega) \ge m_k(\Omega)$.

(2) The formula for $c(\Omega)$ when $\Omega$ is convex, while combinatorial, is rather difficult to compute with.  I discussed some ways to simplify it in the previous post when $\Omega$ is a polygon, and now I would like to explain a way to simplify it even more when $\Omega$ is a triangle.  First recall that if $\Omega$ is affine equivalent to a right triangle with legs on the axes, then we can compute $c(\Omega)$ very easily:  If the vertices are $(0,0),(a,0),(0,b)$, then we simply write down all nonnegative integer linear combinations of $a$ and $b$ in nondecreasing order. Second, note that a general triangle can be approximated by a rational triangle, i.e. a triangle with vertices in ${\mathbb Q}^2$.  Third, observe that any rational triangle is affine equivalent to a triangle with all vertices on the axes. The ECH capacities of the latter kind of triangle can be effectively computed using the following proposition, which is a generalization of Lemma 2.6(ii) in Dusa McDuff’s paper “The Hofer conjecture on embedding symplectic ellipsoids”:

PROPOSITION. Suppose $\Omega_1$ is affine equivalent to a triangle with vertices $(a,0),(b,0),(0,c)$ where $a. Let $\Omega_2$ denote the triangle with vertices $(0,0),(a,0),(0,c)$, and let $\Omega_{12}$ denote the triangle with vertices $(0,0),(b,0),(0,c)$. Then

$c(\Omega_1) = c(\Omega_{12}) - c(\Omega_2).$

PROOF:  We know from the general properties above that $c(\Omega_1)\le c(\Omega_{12})-c(\Omega_2)$, so we just need to show that $c(\Omega_1)\ge c(\Omega_{12})-c(\Omega_2)$.  For this purpose it is enough to show that $m(\Omega_1)\ge m(\Omega_{12})-m(\Omega_2)$, since we know that $c(\Omega_1)\ge m(\Omega_1)$ while $c(\Omega_{12})=m(\Omega_{12})$ and $c(\Omega_2)=m(\Omega_2)$.

By an approximation argument, we may assume that $a/c$ and $b/c$ are irrational.

Recall from the previous post that if $\Omega$ is any triangle, then $m_k(\Omega)$ is the minimum of $2A(\Omega)\lambda$ over triangles $\Lambda$ that enclose at least $k+1$ lattice points and are obtained from $\Omega$ by scaling by $\lambda$ and translating.  Also $A(\Omega)$ denotes the area of $\Omega$.  (In the previous post I also rotated $\Omega$ by 90 degress, as that was natural in the context of the discussion there, but since this does not change the answer I will not do it here.)  Let us call a triangle “minimal” if it achieves this minimum.  Note that if a triangle is minimal, then every edge must contain a lattice point;  otherwise we could move an edge towards an opposite vertex so as to decrease $\lambda$.  If $\Omega$ is affine equivalent to a right triangle with legs on the axes, then the converse is also true, i.e. $\Lambda$ is minimal if and only if every edge of $\Lambda$ encloses a lattice point.

Now fix $k$ and let $\Delta_1$ be a minimal triangle for computing $m_k(\Omega_1)$.  Since $\Delta_1$ is minimal, it follows as above that the bottom, horizontal edge of $\Delta_1$ must have integer $y$ coordinate. So by translating we may assume that the bottom edge of $\Delta_1$ is on the $x$ axis, and by further translating we can assume that the top vertex of $\Delta_1$ has $x$ coordinate in the interval $(-1,0]$.  Now let $\Delta_2$ denote the triangle formed by the left edge of $\Delta_1$ together with the axes, and let $l+1$ denote the number of lattice points enclosed by $\Delta_2$.  Let $\Delta_{12}$ denote the triangle formed by the right edge of $\Delta_1$ and the axes.  Note that each of the nonhorizontal edges of $\Delta_1$ contains exactly one lattice point (because $\Delta_1$ is minimal and $a/c$ and $b/c$ are irrational).  Consequently, $\Delta_{12}$ encloses $k+l+1$ lattice points,  $\Delta_2$ is minimal for computing $m_l(\Omega_2)$, and $\Delta_{12}$ is minimal for computing $m_{k+l}(\Omega_{12})$.  Now write the bottom vertices of $\Delta_1$ as $(a',0),(b',0)$ with $a'.  Then it follows from the above that

$m_k(\Omega_1) = (b'-a')c,$

$m_l(\Omega_2) = a'c,$

$m_{k+1}(\Omega_{12}) = b'c.$

Thus $m_k(\Omega_1) = m_{k+l}(\Omega_{12})-m_l(\Omega_2) \ge (m(\Omega_{12})-m(\Omega_2))_k$ as desired.

QED

(3) Enough for now about computing $c(\Omega)$ when $\Omega$ is convex (although there are still some interesting mysteries which I plan to discuss later).  If $\Omega$ is a nonconvex polygon, can we compute $c(\Omega)$ combinatorially using the above properties?  Or do we have to go back to the definition of ECH capacities and count holomorphic curves?  I don’t know.

In the previous post I considered the example where $\Omega$ is a quadrilateral with vertices $(0,0),(A,0),(1-a,a),(0,1)$ where $a<1.  One can get a lower bound on $c_k(\Omega)$ by cutting $\Omega$ into two triangles (each affine equivalent to a right triangle with legs on the axes).  And one can get an upper bound on $c_k(\Omega)$ by adding a triangle (whose ECH capacities are determined by the above proposition) to get a right triangle with legs on the axes.  I noticed that in some examples I worked out, the lower and upper bounds are equal.

Since then, I tried a couple more examples.  I found one case where the upper and lower bounds are not equal.  This case is where $a=2/5, A=3, k=5$.  In this case the lower bound on $c_5$ is $14/5$, while the upper bound on $c_5$ is $3$.  However, in all other examples that I computed (which admittedly were not very many), the upper and lower bounds were equal. So I don’t know what’s going on here.

Anyway, this does not disprove Conjecture 2 in the previous post, although it does cast some doubt on it. Recall that Conjecture 2 in the previous post was introduced because it implies Conjecture 1 in the previous post.  However Conjecture 2 is much more than is needed to prove Conjecture 1.  To prove Conjecture 1, one does not need $c(\Omega)$ to equal the lower bound; one just needs good enough upper bounds on $c(\Omega)$.

Anyway, I think a lot more calculation and/or intelligent thought is needed to sort all this out.  Meanwhile, if anyone is still reading this and has questions about how the calculations work, please post a comment or email me.  I am happy to clarify.  Also, since these are just informal notes, I may well have made mistakes.