## Hind-Lisi and more

References for this post:

[HL] R. Hind and S. Lisi, Symplectic embeddings of polydisks

[CONCAVE] K. Choi, D. Cristofaro-Gardiner, D. Frenkel, M. Hutchings, and V. Ramos, Symplectic embeddings into four-dimensional concave toric domains

[T3] M. Hutchings and M. Sullivan, Rounding corners of polygons and the ECH of $T^3$

[CC2] M. Hutchings and C. H. Taubes, Proof of the Arnold chord conjecture in three dimensions II

[BUDAPEST] M. Hutchings, Lecture notes on ECH

0. Introduction

Hind and Lisi [HL] recently proved that if the polydisk $P(2,1)$ symplectically embeds into the ball $B(b)$ then $b\ge 3$. (I’ll review what this notation means below.) This bound is optimal, because the polydisk $P(a,1)$ symplectically embeds into the ball $B(1+a)$ by inclusion. Also, the $2$ is as large as possible for this sort of result, because whenever $a>2$, according to Schlenk’s book, one can use “symplectic folding” to symplectically embed $P(a,1)$ into $B(b)$ for certain $b < 1+a$.

So this result of Hind-Lisi is very nice. For a while I was wondering if one can reprove it using ECH methods; previously I couldn’t, but I was missing a couple of details, and I think I now know how to do this. Moreover the new method gives further obstructions to symplectically embedding polydisks into ellipsoids, and more generally symplectic embeddings of “convex toric domains”, which go beyond the obstructions coming from ECH capacities. Here are a couple of sample results.

First, we can extend the Hind-Lisi result to give some seemingly new obstructions to symplectic embeddings of skinnier polydisks into balls (I have no idea whether these are sharp).

Theorem 1. Let $a\in[1,8]$ and suppose that $P(a,1)$ symplectically embeds into $B(b)$. Then:

• If $1\le a\le 2$ then $b\ge 1+a$.
• If $2\le a\le 4$ then $b\ge (10+a)/4$.
• If $4\le a \le 9/2$ then $b\ge 7/2$.
• If $9/2 \le a \le 7$ then $b\ge (13+a)/5$.
• If $7\le a\le 8$ then $b\ge 4$.

Of course the third and fifth lines follow trivially from the lines above them. I have written the fifth line because the inequality $b\ge 4$ is significant until $a=8$, at which point it ties the volume constraint $b\ge\sqrt{2a}$.

We can also extend the Hind-Lisi result in a different direction by replacing balls with integral ellipsoids:

Theorem 2. Let $a\in[1,2]$ and let $b$ be a positive integer. Then $P(a,1)$ symplectically embeds into $E(bc,c)$ if and only if $bc \le a + b$.

Note that $P(a,1)$ includes into $E(bc,c)$ whenever $bc \ge a + b$. Thus Theorem 2 asserts that the inclusion is sharp when $b$ is an integer and $1\le a\le 2$. It is unclear why integrality of $b$ should be relevant here, but the method I am using works better for rational ellipsoids and best for integral ellipsoids.

I would now like to give an introduction to all of this, with a more formal writeup to come later.

1. Convex toric domains

Recall that if $\Omega$ is a domain in the (closed) first quadrant of the plane, we define the “toric domain”

$X_\Omega = \{z\in{\mathbb C}^2 \mid (\pi|z_1|^2,\pi|z_2|^2)\in\Omega\}.$

Let us now define a “convex toric domain” to be a domain $X_\Omega$ where $\Omega=\{(x,y)\mid 0\le x \le A, 0 \le y \le f(x)\}$ where $f$ is a convex function such that $f(0)=A$ and $f'(0)\le 0$. (Sorry for the confusing terminology, but the region below the graph of a concave function is convex!) For example, the polydisk

$P(a,b) = \{z\in{\mathbb C}^2\mid \pi|z_1|^2\le a, \pi|z_2|^2\le b\}$

is a convex toric domain where $A=a$ and $f\equiv b$ so that $\Omega$ is a rectangle. Also, the ellipsoid

$E(a,b) = \{z\in{\mathbb C}^2\mid \pi|z_1|^2/a + \pi|z_2|^2/b\}$

is a convex toric domain where $A=a$ and $f$ is linear so that $\Omega$ is a triangle.

Convex toric domains should be contrasted with the “concave toric domains” considered in [CONCAVE]. For a concave toric domain, $f$ is convex and $f(A)=0$. Note that an ellipsoid is both a convex toric domain and a concave toric domain; nothing else is.

Recall that ECH capacities give obstructions to symplectically embedding any symplectic four-manifold (typically with boundary) into another. McDuff showed that ECH capacities give a sharp obstruction to symplectically embedding one (four-dimensional) ellipsoid into another. More generally, at the Simons Center in June, Dan Cristofaro-Gardiner presented a proof of the remarkable result that ECH capacities give a sharp obstruction to symplectically embedding any concave toric domain into any convex toric domain.

On the other hand, ECH capacities do not give very good obstructions to symplectically embedding a convex toric domain into a concave toric domain, e.g. a polydisk into an ellipsoid. For example they only imply that if $P(2,1)$ symplectically embeds into $B(b)=E(b,b)$ then $b\ge 2$, although we actually want $b\ge 3$.

2. Introduction to the strategy.

To improve on the obstructions given by ECH capacities, the idea is a follows. First of all, the obstruction given by ECH capacities works as follows: if one symplectic four-manifold (with contact boundary) embeds into another, then we get a strong symplectic cobordism between the two contact manifolds, which induces a cobordism map on ECH. Whenever this cobordism map is nontrivial, a (possibly broken) holomorphic curve in the (completed) cobordism must exist, and the fact that this holomorphic curve has positive symplectic area gives us an inequality.

Now in general if we do not know very much about the holomorphic curve that will exist, then we will not get a very sharp inequality. The idea is to get restrictions on which holomorphic curves can exist in the cobordism, and thus improve the inequality. To do so we will use two facts:

(a) For the symplectic cobordism given by a symplectic embedding of one convex toric domain into another (the target of the embedding can also be more general), the holomorphic currents that one wants to count to define the ECH cobordism map are better behaved than usual (in particular there are no negative ECH index multiple covers).

(b) The quantity $J_0$ (a variant on the ECH index) bounds the topological complexity of holomorphic curves (when they are not multiply covered) and thus can be used to show that certain kinds of holomorphic curves cannot contribute to the ECH cobordism map (e.g. because the genus of a holomophic curve cannot be negative).

That was a bit vague, so let us now explain some details.

3. Convex ECH generators

Let $X_\Omega$ be a convex toric domain and let $Y$ denote its boundary. If $Y$ is smooth, then it has a natural contact form, given by the restriction of

$\lambda=\frac{1}{2}\sum_{i=1}^2(x_idy_i - y_idx_i).$

The claim is now that one can perturb $X_\Omega$ so that $Y$ is smooth, $\lambda$ is nondegenerate, and the generators of the ECH chain complex (up to large symplectic action) have the following combinatorial description.

Definition 3. A convex generator is a polygonal (i.e. comprised of finitely many line segments) path $\Lambda$ in the plane such that:

• $\Lambda$ starts at $(0,y)$ and ends at $(x,0)$ where $x$ and $y$ are nonnegative integers.
• The vertices of $\Lambda$ (the points where it changes direction) are at lattice points.
• The edges of $\Lambda$ (i.e. line segments between vertices) have nonpositive slope (possibly $-\infty$), and as one moves to the right, the slope of each edge is less than that of the preceding edge.
• Each edge of $\Lambda$ is labeled e’ or h’.
• Horizontal and vertical edges can only be labeled e’.

If $\Lambda$ is a convex generator, define its combinatorial ECH index by

$I(\Lambda) = 2(L(\Lambda)-1) - h(\Lambda)$

where $L(\Lambda)$ is the number of lattice points in the region bounded by $\Lambda$ and the axes (including lattice points on the boundary), and $h(\Lambda)$ is the number of edges labeled h’.

So far, the definition of a “convex generator” $\Lambda$ and its combinatorial ECH index do not depend on the convex domain $X_\Omega$. However there is also a combinatorial notion of symplectic action of $\Lambda$, which does depend on $\Omega$; it is given by

$A_\Omega(\Lambda) = \sum_{e\in Edges(\Lambda)}v_e\times p_e.$

Here $v_e$ is the vector determined by $e$, that is the difference between the final and initial endpoints; $\times$ denotes the cross product of vectors; and $p_e$ denotes a point on $\partial\Omega$ such that a tangent line to $\partial\Omega$ at $p_e$ is parallel to $e$. (If $p_e$ is a corner of $\partial\Omega$, this means that $\Omega$ is on the lower left of the line through $p_e$ parallel to $e$.)

The more precise statement is that for any $L>0$ large and $\epsilon>0$ small, one can perturb $Y$ by a perturbation of size less than $\epsilon$ in an appropriate sense, such that there is a bijection between ECH generators of (actual) action less than $L$ with convex generators of (combinatorial) action less than $L$, such that the actual ECH index agrees with the combinatorial ECH index, and the action agrees with the combinatorial action up to $\epsilon$ error. This is analogous to Lemma 3.3 in [CONCAVE] and proved similarly. (If you actually read this proof and try to compare, note that here we are first perturbing $\Omega$ so that its boundary is nearly horizontal at the beginning and nearly vertical at the end.)

4. The chain complex

Now let us consider the ECH of the boundary of a convex toric domain with ${\mathbb Z}/2$ coefficients. We know that the homology of the chain complex has one generator in each nonnegative even degree. What do the homology generators actually look like? And before discussing this question we should maybe first ask if there is a combinatorial formula for the differential.

I conjecture that if $\Lambda$ is any path in which all edges are labeled e’, then $\Lambda$ is a cycle which represents the homology generator of grading $I(\Lambda)$. This would follow from a more general conjecture about the differential. Namely, extend each convex path to a closed polygon by appending “virtual” edges along the axes, with all of the virtual edges labeled e’. The conjecture is then that the differential acts on these extended generators by rounding corners and locally losing one h’ (possibly plus some relatively unimportant “double rounding” terms), just as in [T3]. Keon Choi may know how to prove this, related to his thesis work.

Anyway, to study symplectic embeddings into ellipsoids, we don’t need to know all of that. We just need to know the following fact, which can be proved more quickly:

Lemma 4. Suppose $\Omega$ is the triangle with vertices $(0,0)$, $(a,0)$, and $(0,b)$, so that $X_\Omega = E(a,b)$. Let $P$ be a line in the plane of slope $-b/a$ which passes through at least one lattice point in the first quadrant. Let $\Lambda$ be the maximal convex generator to the left of $P$, with all edges labeled e’. Then $\Lambda$ is a cycle which represents a nontrivial homology class in the ECH of (perturbed) $\partial E(a,b)$.

Proof. Let $k=I(\Lambda)/2$. One can check that $\Lambda$ uniquely minimizes the combinatorial action among generators of index $2k$, and its combinatorial action agrees with the kth ECH capacity of $E(a,b)$. This is similar to Example 1.23 in [CONCAVE]. The claim follows immediately (because the degree $2k$ homology class is represented by a cycle which is a sum of generators with action less than or equal to the kth ECH capacity, and $\Lambda$ is the only such generator.)

5. The cobordism map

Suppose now that a convex toric domain $X_\Omega$ with boundary $Y$ symplectically embeds into (the interior of) another convex toric domain $X_{\Omega'}$ with boundary $Y'$. Then $X_\Omega$ minus the interior of the image of $X_{\Omega'}$ is a strong symplectic cobordism from $Y'$ to $Y$. We further get a strong symplectic cobordism between the perturbed contact forms as above. This induces a map $ECH(Y')\to ECH(Y)$, which is an isomorphism, because the cobordism is diffeomorphic to the product $[0,1]\times S^3$. We know from my work with Taubes [CC2] using Seiberg-Witten theory that given an appropriate almost complex structure $J$ on the cobordism, this ECH cobordism map is induced by a chain map, such that whenever a coefficient of the chain map is nonzero, there is a (possibly broken) $J$-holomorphic current with ECH index zero between the corresponding ECH generators.

In particular, the chain map decreases the symplectic action. All of our symplectic embedding obstructions come from this fact.

Now the key observation which gives us some control over this chain map is:

Lemma 5. The chain map (up to any given symplectic action $L$) can be chosen so that it only counts $I=0$ unbroken $J$-holomophic currents, such that the components are disjoint and have $I=0$, and the somewhere injective curve underlying each component is embedded and also has $I=0$.

To prove this, one needs to do a calculation, using the special nature of the contact forms in question, to show that if $J$ is generic, then multiply covered curves with negative ECH index cannot arise here. I will explain this some other time. I will just remark for now that this lemma also holds for symplectic embeddings of convex or concave toric domains into concave toric domains. (It doesn’t work for concave toric domains into convex toric domains, but this is the case where Dan showed that ECH capacities already give a sharp symplectic embedding obstruction.)

6. Controlling topological complexity

For the above cobordism arising from a symplectic embedding of one convex toric domain into another, we now want to control the topological complexity of $J$-holomorphic currents with ECH index index $I=0$. To do so, we use the quantity $J_0$ (this is an integer similar to the ECH index, not an almost complex structure), see e.g. Section 5.2 of the lecture notes [BUDAPEST].

In our situation, $J_0$ can be computed combinatorially as follows. If $\Lambda$ is a convex generator, going from $(0,y)$ to $(x,0)$, then

$J_0(\Lambda) = I(\Lambda) - 2x - 2y - e(\Lambda).$

Here $e(\Lambda)$ denotes the number of edges of $\Lambda$ that are labeled e’, plus the number of edges that are labeled h’ and contain at least one interior lattice point. (In other words, $e(\Lambda)$ is the number of elliptic orbits in the orbit set corresponding to $\Lambda$.)

Now the significance of $J_0$ is that if $C$ is an $I=0$ holomophic curve from $\Lambda'$ to $\Lambda$ which does not have any multiply covered components, then

$J_0(\Lambda') - J_0(\Lambda) = -\chi(C) + O(C).$

Here $O(C)$ is the sum, over all Reeb orbits in $\Lambda$ or $\Lambda'$, of the number of ends of $C$ at covers of that orbit minus one. This is proved as in Exercise 5.9 of [BUDAPEST].

We can be a bit more specific as follows. By the partition conditions, we know that all positive ends of $C$ are at simple orbits, and $C$ can have at most one negative end at covers of any given orbit. To make use of this fact, let $n(\Lambda)$ denote the total number of Reeb orbits in $\Lambda$, namely the number of edges, plus the number of h’ edges with at least one interior lattice point. Let $m(\Lambda)$ denote the total multiplicity of all Reeb orbits in $\Lambda$, namely the sum over all edges of the number of interior lattice points minus one. Then

$O(C) = m(\Lambda') - n(\Lambda').$

Also, if $C$ is connected with genus $g$, then

$\chi(C) = 2 - 2g - m(\Lambda') - n(\Lambda)$.

Putting the above together, we get

$2x(\Lambda) + 2y(\Lambda) = 2g - 2 + 2x(\Lambda') + 2y(\Lambda') + 2m(\Lambda') + h(\Lambda) - h(\Lambda').$

Since $g\ge 0$ and $h(\Lambda)\ge 0$, we conclude the following:

Lemma 6. Suppose there exists an $I=0$, connected, embedded curve from $\Lambda'$ to $\Lambda$. Suppose that all edges of $\Lambda'$ are labeled e’. Then

$x(\Lambda) + y(\Lambda) \ge x(\Lambda') + y(\Lambda') + m(\Lambda') - 1.$

(OK, I hate the writing style where the proof comes before the statement, sorry.)

Anyway, to prove Theorems 1 and 2, we will apply Lemma 6 repeatedly. Let’s see now how this works.

7. Proof of the first part of Theorem 1.

To start, let us prove the first part of Theorem 1, namely that if $P(a,1)$ symplectically embeds into $B(b)$, and if $1\le a \le 2$, then $b\ge 1 + a$.

Assume that $P(a,1)$ symplectically embeds into $B(b)$, assume that $a\ge 1$, and assume that $b < 1+a$. We will show that $b\ge 3$.

Step 1. Consider the convex generator $\Lambda'$ consisting of the straight line from $(0,1)$ to $(1,0)$, labeled e’. Then $I(\Lambda') = 4$ and $A_{B(b)}(\Lambda') = b$. By Lemmas 4 and 5, there is an $I=0$ current from $\Lambda'$ to some convex generator $\Lambda$ with $I(\Lambda)=4$ and $A_{P(a,1)}(\Lambda)\le b$.

There are in fact only three convex generators $\Lambda$ with $I(\Lambda)=4$: the horizontal line from $(0,0)$ to $(2,0)$, labeled e’, which has action $2$; the line from $(0,1)$ to $(1,0)$, labeled e’, which has action $1+a$; and the vertical line from $(0,2)$ to $(0,0)$, labeled e’, which has action $2a$. If there is a holomorphic curve from $\Lambda'$ to either of the latter two generators, then we immediately get $b\ge 1+a$ or $b\ge 2a$, which contradicts our assumption that $b< 1+a$. Thus $I=0$ curves from $\Lambda'$ can only go to the horizontal line.

Step 2. Consider the convex generator $\Lambda'$ consisting of the straight line from $(0,2)$ to $(2,0)$, labeled e’. Then $I(\Lambda')=10$ and $A_{B(b)}(\Lambda')=2b$. By Lemmas 4 and 5, there is an $I=0$ current $C$ from $\Lambda'$ to some convex generator $\Lambda$ with $I(\Lambda)=10$ and $A_{P(a,1)}(\Lambda)\le 2b$.

Now the current $C$ must actually be a connected, embedded holomorphic curve. Why? Well, if $C$ were disconnected, then each component would have $I=0$, and so by the end of Step 1, each component can only go to the horizontal line of length 2. Hence $\Lambda$ is a horizontal line of length 4. But this only has $I=8$ which is not big enough. Likewise, if $C$ is multiply covered, then again by the end of Step 1, the embedded curve underlying $C$ can only map to the horizontal line of length 2, so again $\Lambda$ is the horizontal line of length 4 which is impossible.

Since $C$ is connected and embedded, we can apply Lemma 6 to conclude that

$x(\Lambda) + y(\Lambda) \ge 5$.

Now the action of $\Lambda$ is given by

$A_{P(a,1)}(\Lambda) = x(\Lambda) + a y(\Lambda).$

If $y(\Lambda)>0$, then $A_{P(a,1)}(\Lambda)\ge 4+a$, so $4+a\le 2b$. Combining this with our assumption $b < 1+a$ then gives $b > 3$.

Assume now that $2b < 4+a$. Then $\Lambda'$ can only be the horizontal line of length 5, labeled e’. In particular this gives $b\ge 5/2$. We need one more step to obtain $b\ge 3$.

Step 3. We are now assuming $b<1+a$ and $2b<4+a$. Consider the convex generator $\Lambda'$ given by the straight line from $(0,3)$ to $(3,0)$, labeled e’. This has $I=18$ and $A_{B(b)}=3b$. By Lemmas 4 and 5, there is a holomorphic current $C$ from $\Lambda'$ to a convex generator $\Lambda$ with $I(\Lambda) = 18$ and $A_{P(a,1)}(\Lambda) \le 3b$.

By Steps 1 and 2, $C$ is connected and embedded. (Otherwise $\Lambda$ is a horizontal line of length at most $7$, which only has $I=14$, which is too small.) Thus Lemma 6 applies to give

$x(\Lambda) + y(\Lambda) \ge 8$.

If $y(\Lambda)>0$, then $A_{P(a,1)}(\Lambda)\ge 7+a$, so $7+a\le 3b$. Combining this with our assumption that $b<1+a$ then gives $b>3$.

If $y(\Lambda)=0$, then $\Lambda$ must be a horizontal line of length $9$, since its ECH index is $18$. Thus $A_{P(a,1)}(\Lambda)=9$, so $b\ge 3$.

QED

Which holomorphic curve is giving the obstruction? If you analyze the above proof, what it is saying is that if $a\ge 1$ and if $P(a,1)$ symplectically embeds into $B(b)$, then at least one of the following holomorphic curves must exist:

• A pair of pants from the “diagonal” Reeb orbit of action $b$ to the “horizontal” and “vertical” simple orbits for the polydisk, implying that $b\ge 1+a$.
• A cylinder from $\Lambda'$ as in Step 1 to the vertical line of length 2, implying that $b\ge 2a$.
• A holomorphic curve with two positive ends at the “diagonal” orbit of action $b$, and negative ends with total action at least $4+a$, implying that $2b\ge 4+a$.
• A holomorphic curve with three positive ends at the “diagonal” orbit of action $b$, and either one negative end with action $9$, or with negative ends of total action at least $7+a$.

Either way, we obtain $b\ge 1+a$ if $a\le 2$. In any case, this looks quite different from the Hind-Lisi argument, which studies curves of “degree” $d$ and takes the limit as $d\to\infty$.

8. Proof of the rest of Theorem 1.

To prove the rest of Theorem 1, we continue to play the above game. Namely:

Step 4. Assume $a\ge 2$, $b<1+a$, $2b<4+a$, and $3b<7+a$. (We can make this last assumption without loss of generality because $(7+a)/3 > (10+a)/4$ when $a\ge 2$.) Let $\Lambda'$ be the straight line from $(0,4)$ to $(4,0)$, labeled e’. This has $I=28$ and $A_{B(b)}=4b$. Then there is a holomorphic current $C$ from $\Lambda'$ to a convex generator $\Lambda$ with $I(\Lambda)=28$ and $A_{P(a,1)}(\Lambda) \le 4b$. The above assumptions imply that $C$ is connected and embedded, so we can apply Lemma 6 to deduce that $x(\Lambda) + y(\Lambda) \ge 11$. Thus

$4b\ge \min(10+a,14).$

This gives the lower bound on $b$ in Theorem 1 for $2\le a \le 4$.

Step 5. Now assume $a\ge 4$, $b<1+a$, $2b<4+a$, $3b<7+a$, and $4b<10+a$. Let $\Lambda'$ be the straight line from $(0,5)$ to $(5,0)$. Then we similarly obtain

$5b\ge \min(13+a,20).$

This gives the lower bound on $b$ in Theorem 1 for $9/2\le a \le 7$.

QED.

9. Proof of Theorem 2.

We now prove that if $1\le a\le 2$ and $b$ is a positive integer, and if $P(a,1)$ symplectically embeds into $E(bc,c)$, then $bc \le a + b$.

Theorem 1 covered the case where $b=1$.

To prove the rest, suppose that $b>1$ is an integer and that $P(a,1)$ symplectically embeds into $E(bc,c)$. Assume that $a\ge 1$ and $a+b>bc$. We need to show that $a>2$. This is actually a little simpler than the proof of Theorem 1 and only requires two steps.

Step 1. Let $\Lambda'$ be the convex generator given by the straight line from $(0,1)$ to $(b,0)$, labeled e’. Then $I(\Lambda') = 2b$ and $A_{E(bc,c)}(\Lambda') = bc$. By Lemmas 4 and 5, there is a holomorphic current $C$ from $\Lambda'$ to a convex generator $\Lambda$ with $I(\Lambda) = 2b$ and $A_{P(a,1)}(\Lambda) \le bc$. Now $C$ must be connected and embedded, since $\Lambda'$ consists only of a single, simple Reeb orbit. Thus Lemma 6 implies that $x(\Lambda) + y(\Lambda) \ge b+1$. If $y(\Lambda)>0$ then $A_{P(a,1)}(\Lambda) = x(\Lambda) + ay(\Lambda) \ge a+b$, contradicting our assumption that $a+b>bc$. The conclusion is that $\Lambda$ can only be the horizontal line of length $b+1$, labeled e’.

Step 2. Now let $\Lambda'$ be the convex generator given by the straight line from $(0,2)$ to $(2b,0)$, labeled `e’. Then $I(\Lambda')= 6b+4$ and $A_{E(bc,c)}(\Lambda')=2bc$. By Lemmas 4 and 5, there is a holomorphic current $C$ from $\Lambda'$ to a convex generator $\Lambda$ with $I(\Lambda)= 6b+4$ and $A_{P(a,1)}(\Lambda)\le 2bc$. Now $C$ must be connected and embedded, since otherwise, by Step 1, $\Lambda$ is a horizontal line of length $2b+2$, which has index $4b+4<6b+4$. Thus Lemma 6 applies to give

$x(\Lambda) + y(\Lambda) \ge 2b+3$.

If $y(\Lambda)>0$ then $A_{P(a,1)}(\Lambda) = x(\Lambda) + a y(\Lambda) \ge 2b+2+a$. Thus $2b+2+a\le 2bc$. By assumption $2bc < 2a+2b$, so $2 < a$ and we are done.

If on the other hand $y(\Lambda)=0$, then $x(\Lambda)=3b+2$, so $3b+2\le 2bc$. By assumption $2bc<2a+2b$, so we get $b+2 < 2a$. Since we are assuming $b\ge 2$, it follows that $a>2$ so we are done again.

QED