# Rigid-Flexible Values for Embeddings of Four-Dimensional Ellipsoids into
Almost-Cubes

Cory H. Colbert  Washington and Lee University, Lexington, VA 24450
<EMAIL_ADDRESS>and  Andrew Lee  St. Thomas Aquinas College, Sparkill, NY
10976<EMAIL_ADDRESS>
###### Abstract.

We consider the embedding function $c_{b}(a)$ describing the problem of
symplectically embedding an ellipsoid $E(1,a)$ into the smallest possible
scaling by $\lambda>1$ of the polydisc $P(1,b)$. In particular, we calculate
rigid-flexible values, i.e. the minimum $a$ such that for $E(1,a^{\prime})$
with $a^{\prime}>a$, the embedding problem is determined only by volume. For
$1<b<2$ we find that these values vary piecewise smoothly outside a discrete
set of discontinuities at $b\in\left(\frac{n+1}{n}\right)^{2}$.

###### Key words and phrases:

Differential geometry, symplectic geometry MSC codes: 53D05, 53D22

###### Contents

  1. 1 Introduction, Statement of Results
    1. 1.1 Outline of Paper
    2. 1.2 Acknowledgments
  2. 2 Preliminaries: Two Methods for Finding Symplectic Embeddings
    1. 2.1 Obstructive Classes
    2. 2.2 Reduction at a Point
  3. 3 Classification of Exceptional Classes at $a=8.$
    1. 3.1 Proof of Lemma 3.1
    2. 3.2 Case I: $\mathbf{m}=(m^{\times 7},m-1)$
    3. 3.3 Case II: $\mathbf{m}=(m+1,m^{\times 7})$
  4. 4 No obstructive classes for $a\in(8,9]$
  5. 5 Reduction Calculations
    1. 5.1 Case I: $\lambda-1\geq d_{1}$
      1. 5.1.1 $a\in[9,10)$
      2. 5.1.2 $a\in[10,10.24)$
      3. 5.1.3 $a\in[10.24,12)$
      4. 5.1.4 $a\in[12,16)$
    2. 5.2 Case II: $d_{1}>\lambda-1$
      1. 5.2.1 $a\in[9,10)$
      2. 5.2.2 $a\in[10,11)$
      3. 5.2.3 $a\in[11,12)$
      4. 5.2.4 $a\in[12,13)$
      5. 5.2.5 $a\in[13,14)$
      6. 5.2.6 $a\in[14,15)$
      7. 5.2.7 $a\in[15,16)$

##  1\. Introduction, Statement of Results

The problem of embedding one symplectic manifold into another touches on a
wide variety of topics in symplectic geometry, and in this work we focus in
particular on embeddings of ellipsoids $E(a,b)$ into polydiscs $P(a,b)$ .
Here, a polydisc

$P(a,b):=\left\\{(z_{1},z_{2})\in\mathbb{C}^{2}\,|\,\pi\|z_{1}\|^{2}<a,\pi\|z_{2}\|^{2}<b\right\\}$

is the 4-dimensional open symplectic manifold $B_{2}(a)\times
B_{2}(b)\subset\mathbb{C}^{2}$, where each factor is a 2-disc of fixed radius
centered at $0\in\mathbb{C}$. Similarly the ellipsoid $E(a,b)$ is given by

$E(a,b):=\left\\{(z_{1},z_{2})\in\mathbb{C}^{2}\,|\,\frac{\pi\left|\left|z_{1}\right|\right|^{2}}{a}+\frac{\pi\left|\left|z_{2}\right|\right|^{2}}{b}<1\right\\}.$

We study the problem of finding embeddings of an ellipsoid into a polydisc.
This information is encoded by a function $c_{b}(a)$ whose value at $a$ is the
smallest $\lambda$ such that

(1.1) $E(1,a)\hookrightarrow P(\lambda,\lambda b).$

More precisely,

$c_{b}(a):=\inf\\{\lambda>0\,|\,E(1,a)\hookrightarrow P(\lambda,\lambda
b)\\}.$

Earlier work on embeddings of ellipsoids into polydiscs [2, 3] considered a
fixed value of $b$ and calculated $c_{b}(a)$ in its entirety for all $a$. For
example, $c_{1}(a)$ contains a so-called “infinite staircase,” a convergent
sequence of $a$-values $a_{n}\to a_{\infty}$ such that $c_{b}(a)$ is non-
decreasing and linear or constant on each $[a_{i},a_{i+1}]$. In addition, they
find that for sufficiently large values of $a$, $c_{b}(a)$ coincides with the
volume constraint $\sqrt{\frac{a}{2b}}$: for sufficiently elongated ellipsoids
the only obstruction to the existence of symplectic embeddings is the volume
constraint. In later work [2, 10] both the source and target in the embedding
problem were allowed to become elongated, with the goal of analyzing the
resulting changes in the embedding function $c_{b}(a)$.

To set notation, define

$RF(b):=\inf\left\\{A\,\,|\,\,c_{b}(a)=\sqrt{\frac{a}{2b}}\,\,\mbox{for
all}\,\,a\geq A\right\\}.$

We call this number the _rigid-flexible value_ , or the RF-value.

Varying this extra parameter $b$ uncovers delicate structure in the the
$b$-direction for functions $c_{b}(a)$. In [2] the authors find that for
$b\in\mathbb{N}$ with $b\geq 2$, $c_{b}(a)$ contains no infinite staircases
and can be computed explicitly for all $a$. On the other hand, [10] provides
examples of infinite sequences (some arbitrarily large and others converging
to 1) of irrational $b$ such that for each such $b$, $c_{b}(a)$ contains an
infinite staircase. [10] also computes the embedding functions $c_{b}(a)$ for
$b\in[1,\infty)$ and $a\in[1,3+\sqrt{2}]$. Previous joint work by the second
author computed $RF(b)$ in this problem for $b>2$, finding that it was
piecewise smooth in $b$ away from an infinite discrete set of discontinuities.

In this paper we examine the same problem but for $1<b<2$. We term these
polydiscs “almost-cubes” following [3], where cubes are polydiscs
$P(\lambda,b\lambda)$ with $b=1$. In this region we again see an infinite set
of discontinuities, this time accumulating at 1, but away from these points
$RF(b)$ is smooth.

######  Theorem 1.1.

Let $b\in(1,2)\setminus\left\\{\frac{(n+1)^{2}}{n^{2}}\right\\}$ for $n\geq
3$. Then for $k=\lfloor\frac{n}{2}\rfloor$,

(1.2)
$RF(b)=\begin{cases}2b\left(\frac{8}{2k^{2}+k+b(2k^{2}-k)}\right)^{2}&\mbox{when
$n$ is even}\\\
2b\left(\frac{8k^{2}+8k+1}{(2k+1)(k+1)+b(2k^{2}+k)}\right)^{2}&\mbox{when $n$
is odd}\\\ \end{cases}$

determined by the obstructive classes (for $k\geq 1$)

$\displaystyle S_{k}$
$\displaystyle:=((2k+1)(k+1),k(2k+1);(k^{2}+k+1),(k^{2}+k)^{\times 7}),$
$\displaystyle T_{k}$ $\displaystyle:=(2k^{2}+k,2k^{2}-k;(k^{2})^{\times
7},k^{2}-1).$

As an example, for $b\in\left(\frac{16}{9},2\right)$ we have $n=3$ so we
consider the $S_{k}$-classes with $k=1$. In this interval, the class
$S_{1}=(6,3;3,2^{\times 7})$ is obstructive at $a=8$. Then, for
$b\in\left(\frac{25}{16},\frac{16}{9}\right)$, we have $n=4$ so we consider
the $T_{k}$-classes with $k=2$. In this interval the class
$T_{2}=(10,6;4^{\times 7},3)$ is obstructive at $a=8$. Then for
$b\in\left(\frac{36}{25},\frac{25}{16}\right)$ we find
$S_{2}=(15,10;7,6^{\times 7})$ is obstructive at the same $a=8$.

Consequently, the behavior of $RF(b)$ on $1<b<2$ differs from the left
continuity which appears for $b>2$, calculated in [12]. In this case $RF(b)$
is neither left nor right continuous. Experimental evidence suggests that
$RF(b)$ at these discontinuities is determined by obstructive classes at
$a=7$. This resembles the case $b=1$, for which [3] shows that
$RF(1)=7\frac{1}{32}$ whereas [2] shows that $RF(2)=8\frac{1}{36}$.

###  1.1. Outline of Paper

In Section 3 we first describe the exceptional classes which determine the
$RF$-value for $b\not\in\left\\{\left(\frac{n+1}{n}\right)^{2}\right\\}$.
Then, in Section 4, we describe the computer-assisted search establishing that
no more obstructive classes can appear when $a\in(8,9]$. The remainder of the
paper is devoted to applying the reduction method for $a$ in the interval
$[9,\infty)$, to establish that there are no obstructions to embeddings other
than the volume constraint. This argument splits into many cases, according to
the ordering of the terms.

###  1.2. Acknowledgments

The second author thanks Dan Cristofaro-Gardiner for suggesting this problem
and providing support and encouragement along the way.

This work used the NCSA Delta CPU at the University of Illinois Urbana-
Champaign through allocation MTH230018 from the Advanced Cyberinfrastructure
Coordination Ecosystem: Services & Support (ACCESS) program, which is
supported by National Science Foundation grants #2138259, #2138286, #2138307,
#2137603, and #2138296.

##  2\. Preliminaries: Two Methods for Finding Symplectic Embeddings

Here, we review the methods we use for detecting symplectic embeddings. This
is not an exhaustive list, and more detailed expositions are in [2, 9], so we
review only what we use. The following definition is central to both methods.
Fix $b\geq 1$. Since the function $c_{b}(a)$ is continuous in $a$, it suffices
to compute it for $a\geq 1$ rational.

######  Definition 2.1.

The weight expansion $\textbf{w}(a,b)$ of such an $a$ is the finite decreasing
sequence

$\textbf{w}(a)=(1^{\times\ell_{0}},w_{1}^{\times\ell_{1}},...,w_{n}^{\times\ell_{n}}),$

where $w_{1}=a-\ell_{0}<1,w_{2}=1-\ell_{1}w_{1}<w_{1}$, and so on.

We write $\textbf{w}(a)$ to mean $\textbf{w}(1,a)$, and we write $w(a_{i})$ to
mean the $i$th term of the weight expansion $\textbf{w}(a)$.

In [7, Thm. 1.1], it is shown that there is a canonical decomposition of any
ellipsoid $E(1,a)$ into a collection of balls

$E(1,a):=\coprod_{i}B(w_{i})$

where the $w_{i}$ are the terms in the weight expansion $\textbf{w}(a)$.

######  Theorem 2.2 ([3]).

Let $a,b\in\mathbb{Q}$ be positive. There exists a symplectic embedding
$E(1,a)\hookrightarrow P(\lambda,\lambda b)$ if and only if there is a
symplectic embedding

$E(1,a)\coprod B(\lambda)\coprod B(\lambda b)\hookrightarrow B(\lambda(b+1)).$

This reduces the polydisc problem to a ball-packing problem of embedding balls
of capacity $e_{i}$ into a ball of capacity $\mu$:

(2.1) $\coprod_{i}B(e_{i})\hookrightarrow B(\mu).$

The purpose of introducing these constructions is that a collection of balls
can be embedded symplectically precisely when the associated multiple blow-up
of $\mathbb{C}P^{2}$ carries a symplectic form in a certain cohomology class.
We denote by $\overline{\mathcal{C}}_{K}(X_{n})$ the set of cohomology classes
represented by symplectic forms for which the anticanonical class is
$K=-3L+\sum_{i}E_{i}$, where $L$ is Poincaré dual to a line in
$\mathbb{C}P^{2}$ and each $E_{i}$ is dual to the $i$th exceptional sphere. By
[8], the embedding (2.1) exists when the following cohomology class is in the
symplectic cone:

$\mu L-\sum_{i}e_{i}E_{i}\in\overline{\mathcal{C}}_{K}(X_{n}).\newline
\indent$

The above fact gives a sufficient criterion for a class to lie in
$\overline{\mathcal{C}}_{K}(X_{n})$. If there is a symplectic form in a given
class, it must have non-negative intersection with certain holomorphic curves.
By [6], this is also sufficient: if $\mathcal{E}_{K}(X_{n}):=\\{e\in
H_{2}(X_{n})\,|\,\langle e,e\rangle=-1,\,\,\langle K,e\rangle=-1\\}$ is the
set of exceptional classes, then we may characterize the symplectic cone
referenced above as

(2.2) $\overline{\mathcal{C}}_{K}(X_{n})=\\{\omega\in
H^{2}(X_{n})\,|\,\langle\omega,e\rangle\geq 0\,\,\forall
e\in\mathcal{E}_{k}(X_{n})\\}.$

In summary, an embedding of the form (2.1) exists when the associated weight
expansion represents a cohomology class in the symplectic cone
$\overline{\mathcal{C}}_{K}(X_{n})$, and the equality (2.2) characterizes
elements of the symplectic cone as those which pair non-negatively with so-
called exceptional classes $\mathcal{E}_{k}(X_{n})$.

We now require a condition for determining when homology classes are
exceptional. With respect to the basis of
$H^{2}(X_{n};\mathbb{R})\simeq\mathbb{R}^{n+1}$ given above, a _Cremona
transform_ is the map given by

$(d;m_{1},...m_{n})\mapsto(2d-m_{1}-m_{2}-m_{3};d-m_{2}-m_{3},d-m_{1}-m_{3},d-m_{1}-m_{2},m_{4},...,m_{n})$

We will use Cremona transforms in two different contexts below. First, we can
state the condition for an integral homology class to lie in
$\mathcal{E}_{k}$, which is proven in [9] based on work of [5, 6].

######  Theorem 2.3.

A class $(d;m_{1},...,m_{n})\in H_{2}(X_{n};\mathbb{Z})$ is in
$\mathcal{E}_{K}(X_{n})$ if and only if its entries satisfy the Diophantine
equations

(2.3) $\displaystyle 3d-1=\sum_{i}m_{i}\newline $ $\displaystyle
d^{2}+1=\sum_{i}m_{i}^{2}$

and $(d;m_{1},...,m_{n})$ reduces to $(0;-1,0,\ldots,0)$ after a sequence of
Cremona transforms.

For the problem of embedding ellipsoids into polydiscs, the natural
compactification of $P(a,b)$ adds a single point to each disc, yielding
$S^{2}\times S^{2}$. This manifold is in fact diffeomorphic to the 2-fold
blow-up of $\mathbb{C}P^{2}$, so the $n$-fold blow-up of $S^{2}\times S^{2}$
(denoted $Y_{n}$) can be identified with $X_{n+1}$. The induced isomorphism on
cohomology $\psi:H^{2}(Y_{n})\to H^{2}(X_{n+1})$ is given by

$(d,e;m_{1},m_{2},m_{3},...,m_{n})\mapsto(d+e-m_{1};d-m_{1},e-m_{1},m_{2},...,m_{n})$

Using this isomorphism we may translate the Diophantine equations (2.3) into
the following conditions.

(2.4) $\displaystyle\sum_{i}m_{i}$ $\displaystyle=$ $\displaystyle
2(d+e)-1\newline $ (2.5) $\displaystyle\sum_{i}m_{i}^{2}$ $\displaystyle=$
$\displaystyle 2de+1$

###  2.1. Obstructive Classes

We can now describe the first of the two methods we employ to find symplectic
embeddings. If $A:=(d,e;\textbf{m})\in\mathcal{E}_{k}$ with the basis given
previously, the condition

(2.6) $\lambda=\frac{\langle\textbf{w}(a),\textbf{m}\rangle}{d+be}>0$

is equivalent to

(2.7) $\lambda(d+be)-\langle\textbf{w}(a),\textbf{m}\rangle>0.$

The following statement is [2, Method 1’].

######  Theorem 2.4.

An embedding (1.1) exists iff $\lambda\geq\sqrt{\frac{a}{2b}}$ and

$\mu_{b}(d,e;m):=\frac{\sum_{i}m_{i}\cdot w_{i}(a)}{d+be}\leq\lambda$

for every $(d,e;m_{1},...,m_{n})\in H_{2}(Y_{n};\mathbb{Z})$ which satisfies
equations (2.4, 2.5) and reduces to $(0;-1)$ after some sequence of Cremona
transforms.

######  Definition 2.5.

We say that a class $A$ is obstructive at $a>0$ if the quotient in (2.6) is
larger than the volume constraint $\sqrt{\frac{a}{2b}}$.

To use this method, then, we must find all obstructive classes. In practice,
this method proceeds first by finding a discrete subset of $a$ which have a
special relationship to the obstructive classes, described as follows.

If the graph of $c_{b}(a)$ does not follow the volume constraint, then locally
it must be given by the obstruction function of some class $(d,e;\textbf{m})$.
Restricting to the interval where this class determines the graph, [9] Lemma
2.1.3 states that there is a particular $a$-value on the interval whose weight
expansion coincides with the number of positive entries of m, the tail of the
obstructive class.

######  Lemma 2.6.

Let $C=(d,e;\textbf{m})$ be an exceptional class, and let $I$ be the maximal
nonempty open set on which $\mu_{a,b}(C)>\sqrt{\frac{a}{2b}}$ for all $a\in
I$. Then, there is a unique element $a_{0}\in I$ such that the length of the
weight expansion for $a_{0}$, denoted $\ell(a_{0})$, satisfies
$\ell(a_{0})=\ell(\textbf{m})$, and moreover $\ell(a)>\ell(\textbf{m})$ if
$a\in I\setminus\\{a_{0}\\}$.

The following quantities will be relevant going forward: let $\ell_{0}$ denote
the number of $1$’s in the weight expansion of $a$ and subsequently let
$\ell_{i}$ denote the lengths of subsequent blocks. When a class
$(d,e;\textbf{w}(a))$ is obstructive, we have a vector of error terms
$\epsilon$ defined as

$\textbf{m}=\frac{d+be}{\sqrt{2a}}\textbf{w}(a)+\epsilon.$

Each contribution to this error, thought of as the difference between m and
$\frac{d+be}{\sqrt{2a}}\textbf{w}(a)$, will be important, so we define
$v_{i}=\frac{d+be}{\sqrt{2a}}w_{i}$ for $i=0,...,M$ where $M$ denotes the
length of the weight expansion of $a$.
Also let

$\sigma=\sum_{\ell_{0}+1}^{M}\epsilon_{i}$

With these terms established, the following result (proven in [2, 9]) will be
used repeatedly.

######  Lemma 2.7.

For all $(d,e;\textbf{m})\in\mathcal{E}$, suppose that $a\in\mathbb{Q}$ and
$b\in\mathbb{R}$ with $a,b>1$.Then we have

  1. (1)

If $\mu_{b}(d,e;\textbf{m})(a)>\sqrt{\frac{a}{2b}}$ then $d=be+h$ where
$|h|<\sqrt{2b}$ and $\langle\epsilon,\epsilon\rangle<1-\frac{h^{2}}{2b}$

The second step of the obstructive class method establishes that in fact only
finitely many classes must be checked at the values of $a$ characterized in
Lemma 2.6. This is the purpose of the inequality (2.9) below. Proofs of this
fact appear in [9, 2]. A proof of the inequality (2.9) appears in [12].

######  Lemma 2.8.

Let $(d,e;\textbf{m})$ be an exceptional class such that $\ell(a)=\ell(m)$ and
$\mu_{b}(d,e;\textbf{m})>\sqrt{\frac{a}{2b}}$, and $b\in[1,2]$. Set
$v_{M}:=\frac{(d+be)\sqrt{2b}}{q(b+1)\sqrt{a}}$ where $q$ is the last
denominator in the weight expansion $\textbf{w}(a)$. Then

  1. (1)

If $v_{M}\leq\frac{1}{2}$, then $v_{M}>\frac{1}{3}$ and
$\sigma^{\prime}\leq\frac{1}{2}$. If $v_{M}\leq\frac{2}{3}$, then
$\sigma^{\prime}\leq\frac{7}{9}$. with $\delta=y(a)-\frac{1}{q}$ and
$y(a)=a+1-2\frac{b+1}{\sqrt{2b}}\sqrt{a}$, we have

(2.8) $\sqrt{q+\left\lfloor a\right\rfloor+1}\geq 1+\delta v_{M}q$

One can also show the bound from [2, 3], which is

(2.9) $2be+h\leq\frac{\sqrt{2ba}}{\delta}\left(\sqrt{\sigma
q}-(1-h(1-\frac{1}{b})\right)\leq\frac{\sqrt{2ba}}{\delta}\left(\frac{\sigma}{\delta
v_{M}}-\left(1-h(1-\frac{1}{b})\right)\right)$

The following lemma, [9, Lem. 2.1.8] will also be necessary for our arguments.

######  Lemma 2.9.

Assume that $(d,e;\textbf{m})$ is an exceptional class such that
$\mu(d,e;\textbf{m})>\sqrt{\frac{a}{2b}}$ for some $a$. Let
$J:=\\{k,...,k+s-1\\}$ be a block of $s-1$ consecutive integers for which
$w(a_{i}),i\in J$ is constant. Then $(m_{1},...,m_{k+1})$ is of the form

$\displaystyle(m,...,m)\newline $ $\displaystyle(m-1,m,...,m)\newline $
$\displaystyle(m,...,m+1).$

Moreover, there is at most one block of length $s\geq 2$ on which the $m_{i}$
are not all equal, and if $m_{1}\neq m_{k+1}$, then
$\sum_{i=k}^{k+s-1}\epsilon_{i}^{2}\geq\frac{s-1}{s}$.

###  2.2. Reduction at a Point

Though the first method is a necessary and sufficient condition for the
embedding of an ellipsoid into a polydisc, in principle one might need to
check more exceptional classes than is computationally feasible. The following
method straightforwardly determines whether an embedding exists but the
complexity of the process depends strongly on the value of $a$.

######  Definition 2.10.

The defect $\delta$ of an ordered vector $(d;m_{1},...,m_{n})$ is the sum
$d-m_{1}-m_{2}-m_{3}$.

The following is established in [1, 4].

######  Theorem 2.11.

An embedding $E(1,a)\hookrightarrow P(\lambda,\lambda b)$ exists if there
exists a finite sequence of Cremona moves that transforms the ordered vector

$((b+1)\lambda;b\lambda,\lambda,\textbf{w}(a))$

to an ordered vector with non-negative entries and defect $\delta\geq 0$.

We will apply this repeatedly in the proofs of both Theorem 1.1 and Theorem
1.2.

Notice the requirement that the Cremona moves be performed on ordered vectors.
The following fact [2, Prop. 2.2] allows us to avoid re-ordering.

######  Proposition 2.12.

Let $\alpha=(\mu;a_{1},...,a_{n})$ be a vector with $\mu\geq 0$ and
$\alpha^{2}\geq 0$ and assume that there is a sequence
$\alpha=\alpha_{0},...,\alpha_{m}$ of vectors such that $\alpha_{j+1}$ is
obtained from $\alpha_{j}$ by a sequence of Cremona moves. If
$\alpha_{m}=(\hat{\mu};\hat{\alpha}_{1},...,\hat{\alpha}_{n})$ is reduced and
$\hat{\alpha}_{1},...,\hat{\alpha}_{n}\geq 0$, then
$\alpha\in\overline{\mathcal{C}}_{K}(X_{n})$.

Thus to find embeddings we need only prove enough about the ordering to ensure
that the defect at a certain step is non-negative, provided that we have non-
negativity of the terms in each vector. So, only knowledge of the ordering for
the first three terms is strictly necessary. We will usually verify the
ordering at each step, except when doing so results in distinguishing too many
cases.

##  3\. Classification of Exceptional Classes at $a=8.$

In this section, we narrow the search for exceptional classes at $a=8$ and for
all $1<b\leq 2$ by considering the constraints that Equations 3.3 and 3.4 put
on candidate vectors of the form $\langle d,e;\mathbf{m}\rangle.$ For each
$n\geq 0,$ recall that we have

(3.1) $S_{k}:=\langle 2k^{2}+k,2k^{2}-k;(k^{2})^{\times 7},(k^{2}-1)\rangle,$

and let

(3.2) $T_{k}:=\langle(2k+1)(k+1),k(2k+1);(k^{2}+k+1),(k^{2}+k)^{\times
7}\rangle.$

We seek to prove the following result:

######  Lemma 3.1.

For $1<b\leq 2,$ the only exceptional classes $\langle d,e;\mathbf{m}\rangle$
with $\mu_{b}(d,e;\mathbf{m})(8)>\sqrt{8/2b}$ are $S_{k}$ and $T_{k}$ for some
$n\geq 0.$

To begin, recall that if $\langle d,e;\mathbf{m}\rangle$ is obstructive, where
$\mathbf{m}=(m_{1},\ldots,m_{k}),$ then

(3.3) $2de+1=\sum_{i=1}^{k}m_{i}^{2},$

and

(3.4) $2(d+e)-1=\sum_{i=1}^{k}m_{i}.$

We also make use of the fact that if $a$ is an integer, then the shape of
$\mathbf{m}$ is well-understood:

######  Lemma 3.2.

If $\langle d,e;\mathbf{m}\rangle$ is obstructive at $a=8,$ then
$\mathbf{m}=(m^{\times 8}),(m^{\times 7},m-1)$ or $(m+1,m^{\times 7}).$

The next number-theoretic lemma will be needed to calculate values of $d,e$
explicitly.

######  Lemma 3.3.

Let $x\in\mathbb{Z}.$ Then $8|(x^{2}-1)$ if and only if $x=\pm 1\pmod{4}.$

###  3.1. Proof of Lemma 3.1

By [10, Eqn 4.2, p.48], we may assume $d\geq e.$ Suppose $\langle
d,e;\mathbf{m}\rangle$ is obstructive at $a=8.$ If $\mathbf{m}=(m^{\times
8}),$ then $2de+1=8m^{2}$ by Equation 3.3, which is a contradiction because
$d,e$ and $m$ are all integers. Therefore, the only two cases we need to
consider are $\mathbf{m}=(m^{\times 7},m-1)$ and $\mathbf{m}=(m+1,m^{\times
7}).$

###  3.2. Case I: $\mathbf{m}=(m^{\times 7},m-1)$

In this case, we have:

(3.5) $2de+1=8m^{2}-2m+1,$

and

(3.6) $2(d+e)-1=8m-1.$

By Equation 3.6, $m=(d+e)/4.$ Substituting $m$ into Equation 3.5, we see that

(3.7) $de=\frac{(d+e)^{2}}{4}-\frac{d+e}{4}.$

Let $h=d+e.$ Then Equation 3.7 becomes

(3.8) $2h=1+4e\pm\sqrt{8e+1}.$

Since $h\in\mathbb{Z},$ it follows that $8e+1$ is a perfect square. In
particular, if $x^{2}=8e+1$ for some $x\in\mathbb{Z},$ then, by Lemma 3.3,
$x=\pm 1+4n$ for some $n\in\mathbb{Z}.$ Thus, either

(3.9) $e=2n^{2}+n,$

or

(3.10) $e=2n^{2}-n.$

If $e=2n^{2}-n,$ then $8e+1=(4n-1)^{2},$ and we have
$2h=8n^{2}-4n+1\pm(4n-1).$ So either $h=4n^{2}$ or $h=4n^{2}-4n+1.$ The latter
case cannot happen as $4|h,$ so we must have $h=4n^{2}.$ Therefore, if
$e=2n^{2}-n$ for some $n\in\mathbb{Z},$ then

$d=h-e=2n^{2}+n.$

If $e=2n^{2}+n,$ then $8e+1=(4n+1)^{2}$, and like before we have $h=4n^{2}$ so
that $d=2n^{2}-n.$ However, since we are assuming that $d\geq e,$ this
solution is excluded.

In summary, if $\langle d,e;\mathbf{m}\rangle$ is obstructive at $a=8,$ where
$\mathbf{m}=\langle m^{\times 7},m-1\rangle,$ then

$\langle d,e;\mathbf{m}\rangle=\langle 2n^{2}+n,2n^{2}-n;(n^{2})^{\times
7},(n^{2}-1)\rangle.$

Since $d\geq e,$ we have $2n^{2}+n\geq 2n^{2}-n$ which implies $n\geq 0.$

###  3.3. Case II: $\mathbf{m}=(m+1,m^{\times 7})$

By Equations 3.3 and 3.4, we have:

(3.11) $de=4m^{2}+m,$

and

(3.12) $d+e=4m+1.$

By Equation 3.11, $de=m(4m+1)=m(d+e).$ Solving for $m$ in Equation 3.12 and
substituting the result into the latter equation, we see that

(3.13) $4de=(d+e-1)(d+e).$

Let $h=d+e.$ Rewrite Equation 3.13 in terms of $h$ to obtain

(3.14) $h^{2}-(1+4e)h+4e^{2}=0.$

Therefore,

(3.15) $2h=1+4e\pm\sqrt{1+8e},$

which is exactly Equation 3.8.

Therefore, we have $e=2n^{2}+n,$ or $e=2n^{2}-n.$ We conduct our analysis in a
way completely analogous to what we did in Case I. Suppose $e=2n^{2}+n.$ Then
$h=4n^{2}$ or $h=4n^{2}+4n+1.$ In this scenario, the first case is impossible
because $d+e=1\pmod{4}.$ Therefore,

$d+e=4n^{2}+4n+1,$

and thus

(3.16) $d=4n^{2}+4n+1-e=2n^{2}+3n+1=(2n+1)(n+1).$

Just like above, the case where $e=2n^{2}-n$ is excluded because $d\geq e.$
Now, $m=n^{2}+n$ by Equation 3.12, so we see that if $\langle
d,e;\mathbf{m}\rangle$ is obstructive at $a=8$ for some $b\geq 1,$ where
$\mathbf{m}=(m+1,m^{\times 7}),$ then we must have

$\langle
d,e;\mathbf{m}\rangle=\langle(2n+1)(n+1),n(2n+1);(n^{2}+n+1),(n^{2}+n)^{\times
7}\rangle$

for some $n\in\mathbb{Z}.$ As before, since $d\geq e,$ we have
$2n^{2}+3n+1\geq 2n^{2}+n$ which implies $n\geq 0.$

##  4\. No obstructive classes for $a\in(8,9]$

In this section we outline the computer-assisted search method used to
investigate the interval $(8,9)$. Using the equation (1) $d=be+h$ from of the
same lemma, we find that for each $e$ the range of possible $d$ which might
appear in a candidate obstructive class ranges from $2e-2$ to $2e+2$. Using
the inequality (2.9) from Lemma 2.7, we find that the largest $q$ to check is
11, corresponding to $e\leq 40$, giving rise to 934 possible $(d,e)$-pairs.

It then remains to verify that no classes with these values of $(d,e)$ are
obstructive. To accomplish this task, we perform a search along the lines of
what appears in [9, Appendix B]. Our search is significantly less economical
(to an extreme degree) but it does not depend at all of the value of $b$.

For fixed $(d,e)$ the self-intersection and Chern number conditions in
equations (2.4) and (2.5) constrain the possible m which can appear. For
example, by (2.5), the first term $m_{1}$ in m can be no larger than
$\lfloor\sqrt{2de+1}\rfloor$. If $m_{1}$ takes this maximal value, the
subsequent term can be no larger than

$\left\lfloor\sqrt{2de+1-\lfloor\sqrt{2de+1}\rfloor^{2}}\right\rfloor.$

This recursive structure suggests a straightforward method for enumerating
candidate obstructive classes. Pseudocode for an algorithm Extend() carrying
this out appears in (1), along with an example recursion tree. Computer code
(written in Python) used to perform this search is available on GitHub. While
this method will clearly produce all possible classes eventually, it has the
significant downside of a high time complexity. Indeed, we may estimate an
upper bound for this runtime in terms of $e$. Using $d\leq 2e+2$ and the
right-hand sides of (2.4) and (2.5), the upper bound for the $\ell^{2}$ norm
as an input of Extend() is $2e^{2}+1$. As the loop in line 15 of (1) may call
the function $\lfloor\sqrt{2e^{2}+1}\rfloor$ times, this worst-case time
complexity is $\mathcal{O}(e^{2e^{2}+1})$.

Fortunately, there are improvements possible on the version described above
and shown in Algorithm (1). Lemma 2.6 aids in computation by obviating the
need to calculate past the bound on $q$ from Lemma 2.7. (1) of this lemma
bounds $q$ by 11, and the weight expansion of any number of the form
$8\frac{n}{11}$ has length at most 19, so there is no need to to continue a
search past this point. Even so, given the bound $e\leq 40$, the time and
memory demands of the method necessitate high-performance computing resources
both to generate possible classes and check them for possible obstructions.
The additional memory available allows for the use of memoization, caching
previously computed values rather than re-computing them.

After generating all possibilities, we use Lemma 2.9 to eliminate those with
more than one block on which the entries differ by more than 1. As this
depends on the weight sequence of $a$, we verify this for each of the possible
$a$-values.

For an example computation, consider the class $(42,22;16,15^{\times 7},3,3)$.
While this satisfies the conditions of Lemma 2.9 and Lemma 2.6 at
$a=8\frac{1}{2}$, we can easily calculate that for any $b\in(1,2)$,

$\frac{\langle(16,15^{\times 7},3,3),(1^{\times
8},\frac{1}{2},\frac{1}{2})\rangle}{42+b\cdot 22}<\frac{\langle(16,15^{\times
7},3,3),(1^{\times 8},\frac{1}{2},\frac{1}{2})\rangle}{42+2\cdot
22}<\sqrt{\frac{8\frac{1}{2}}{2\cdot 1}}<\sqrt{\frac{8\frac{1}{2}}{2\cdot b}}$

Data: $\ell^{2}$ = remaining $\ell^{2}$-norm, $\ell^{1}$ = remaining
$\ell^{1}$-norm, $p$ = previous term

Result: List of tuples Extensions={$\textbf{m}_{k}$}

$\textbf{Extensions}\leftarrow\\{\\}$;

if  _$\ell^{2} <\ell^{1}$ OR $\ell^{2}<0$ OR $\ell^{1}<0$_ then

return Extensions

end if

if  _$\ell^{2}$ =$\ell^{1}$_ then

return $\\{(1^{\times\ell^{2}})\\}$

end if

if  _$p\neq 0$_ then

start = min($p$, $\sqrt{\ell^{2}})$

end if

else

start $=\sqrt{\ell^{2}}$ ;

end if

for  _$i\leftarrow$ start to $1$ by _-1__ do

foreach  _ mk in Extend($\ell^{2}-i^{2}$, $\ell^{1}-i$)_ do

Append {(i,mk)} to Extensions

end foreach

end for

Extend($\ell^{2}=9$, $\ell^{1}=7$, $p=0$)Extend($\ell^{2}=0$, $\ell^{1}=4$,
$p=3$)$(3)$$i=3$Extend($\ell^{2}=5$, $\ell^{1}=5$, $p=2$)$(2,1^{\times
5})$$i=2$

Example recursion tree, producing the class $(2,2;2,1^{\times 5})$.

Algorithm 1 Extend: a recursive algorithm for solutions to Equations (2.4) and
(2.5)

##  5\. Reduction Calculations

We follow the notational convention in [12]. In this sectoin we use Cremona
reductions to show that the volume constraint is sufficient to guarantee
embeddings if $a\geq 9$ and $1\leq b\leq 2.$ Throughout, we write

$C_{k}=(h||t_{1},t_{2},t_{3},\ldots)$

if $C_{k}$ is an unordered weight vector, where the ordering of
$t_{1},t_{2},\ldots$ is not necessarily known in full. If $C_{k}$ is an
ordered weight vector, we write

$C_{k}=(h;t_{1},t_{2},t_{3},\ldots),$

and we let

(5.1) $\delta_{k}=h-t_{1}-t_{2}-t_{3}$

be the defect of $C_{k}.$

Let $a\geq 9,$ and let

$\mathcal{W}(a)=(1^{\times\left\lfloor a\right\rfloor},d_{1}^{\times
l_{1}},\ldots,d_{k}^{\times l_{k}})$

be the weight expansion of $a.$ Let $1\leq b\leq 2.$

Note that

(5.2) $d_{1}=a-\left\lfloor a\right\rfloor.$

Recall that the initial unordered vector is of the form

(5.3)
$C_{0}=((b+1)\lambda||b\lambda,\lambda,\mathcal{W}(a))=((b+1)\lambda||b\lambda,\lambda,1^{\times\left\lfloor
a\right\rfloor},d_{1}^{\times l_{1}},\ldots,d_{t}^{\times l_{t}}).$

Now $b\lambda\geq\lambda$ because $b\geq 1.$ Since $a\geq 9,$ we have
$b\lambda>2,$ so it follows that

(5.4) $b\lambda-1>1.$

Either $\lambda\geq 2$ or not. If $\lambda\geq 2,$ then of course

$C_{0}=((b+1)\lambda;b\lambda,\lambda,1^{\times\left\lfloor
a\right\rfloor},\ldots),$

with defect $\delta_{0}=-1,$ and so after performing a Cremona transformation,
we obtain

$C_{1}=((b+1)\lambda-1;b\lambda-1,\lambda-1,1^{\times\left\lfloor
a\right\rfloor-1},\ldots),$

with defect $\delta_{1}=0.$ Therefore, we assume

(5.5) $b\lambda-1>1>\lambda-1,$

and

$C_{1}=((b+1)\lambda-1||b\lambda-1,1^{\times\left\lfloor
a\right\rfloor-1},\lambda-1,d_{1}^{\times l_{1}},\ldots),$

where the only remaining ambiguity in the ordering of the first several terms
of $C_{1}$ is whether $\lambda-1\geq d_{1}$ or $\lambda-1<d_{1}.$ Both cases
can occur for suitable $a$ and $b,$ and we address these cases below.

###  5.1. Case I: $\lambda-1\geq d_{1}$

In this case, we have

(5.6) $b\lambda-1>1>\lambda-1\geq d_{1}$

and thus

$C_{1}=((b+1)\lambda-1;b\lambda-1,1^{\times\left\lfloor
a\right\rfloor-1},\lambda-1,d_{1}^{\times l_{1}},\ldots),$

with defect $\delta_{1}=\lambda-2.$

For each $k\in\mathbb{Z},$ define

(5.7) $h(k)=b\lambda-1+k(\lambda-2).$

######  Lemma 5.1.

For each $k\in\mathbb{Z},$ we have:

  1. (1)

$h(k+1)=h(k)+\lambda-2;$

  2. (2)

$h(k)\geq 1\implies h(k+1)\geq\lambda-1;$ and

  3. (3)

$h(k)+3-2\lambda=h(k-2)-1.$

###### Proof.

(1) is clear; (2) follows by adding $\lambda-2$ to both sides; and for (3) we
have

$\displaystyle h(k)+3-2\lambda$ $\displaystyle=$ $\displaystyle
h(k)+(4-2\lambda)-1$ $\displaystyle=$
$\displaystyle(h(k)+(2-\lambda))+(2-\lambda)-1$ $\displaystyle=$
$\displaystyle(h(k-1)+2-\lambda)-1$ $\displaystyle=$ $\displaystyle h(k-2)-1.$

∎

As an immediate and important consequence of Lemma 5.1, we have

$h(0)>h(1)>h(2)>\ldots$

######  Lemma 5.2.

Let $k\geq 0$ be an integer. If $h(k)\geq 1$ and $\left\lfloor
a\right\rfloor-(2k+1)\geq 0,$ then

$C_{k+1}=(h(k)+\lambda;h(k),1^{\times\left\lfloor
a\right\rfloor-(2k+1)},(\lambda-1)^{\times 2k+1},d_{1}^{\times
l_{1}},\ldots).$

###### Proof.

$C_{1}$ has the stated form by 5.6 with defect $\delta_{1}=\lambda-2.$ The
claim then follows by induction on $k$ combined with the identities of Lemma
5.1.∎

Throughout this section, it will be useful to know when, for positive real
numbers $p,q,$ we have $(b+p)\lambda-q\geq 0$ for all $b\in[1,2].$ Now,
$(b+p)\lambda-q\geq 0\iff a\geq 2q^{2}b/(b+p)^{2}.$ A straightforward analysis
of the function $f(x)=2q^{2}x/(x+p)^{2}$ on the interval $[1,2]$ shows that it
has an absolute maximum at $x=2$ if $p\geq 2.$

So

$(b+p)\lambda-q\geq 0\iff a\geq 4q^{2}/(2+p)^{2}.$

In light of the above inequalities, if $p$ and $q$ are positive real numbers
with $p\geq 2,$ define

(5.8) $m(p,q):=\frac{4q^{2}}{(2+p)^{2}}.$

In the following several subsections, we conduct the reduction calculations on
a useful partition of $[9,16).$

####  5.1.1. $a\in[9,10)$

We have

(5.9) $(b+2)\lambda-6\geq 0$

because $m(2,6)=9\leq a.$ In particular, $h(2)\geq 1.$ Therefore,

$h(0)>h(1)>h(2)\geq 1$

by 5.9, so

$C_{2}=(h(1)+\lambda;h(1),1^{\times 6},(\lambda-1)^{\times 3},d_{1}^{\times
l_{1}},\ldots),$

with $\delta_{2}=\lambda-2,$ which implies

$C_{3}=(h(2)+\lambda;h(2),1^{\times 4},(\lambda-1)^{\times 5},d_{1}^{\times
l_{1}},\ldots),$

with $\delta_{3}=\lambda-2.$

So

$C_{4}=(h(3)+\lambda||h(3),1^{\times 2},(\lambda-1)^{\times 7},d_{1}^{\times
l_{1}},\ldots).$

If $h(3)\geq 1,$ then $\delta_{4}=\lambda-2,$ and

$C_{5}=(h(4)+\lambda;h(4),(\lambda-1)^{\times 9},d_{1}^{\times l_{1}},\ldots)$

because $h(3)\geq 1\implies h(4)\geq\lambda-1.$ Therefore,
$\delta_{5}=2-\lambda>0.$

Now suppose $h(3)<1.$ Now $h(2)\geq 1\implies h(3)\geq\lambda-1,$ and

$1>h(3)\implies\lambda-1>h(4).$

Putting everything together, we get $1>h(3)\geq\lambda-1>h(4),$ and

$C_{4}=(h(3)+\lambda;1^{\times 2},h(3),(\lambda-1)^{\times 7},d_{1}^{\times
l_{1}},\ldots),$

with $\delta_{4}=\lambda-2.$ Thus

$C_{5}=(h(4)+\lambda||(\lambda-1)^{\times 9},h(4),d_{1}^{\times
l_{1}},\ldots).$

Since $\lambda-1$ exceeds $h(4)$ and $d_{1},\ldots,d_{t},$ it follows that

$\delta_{5}=h(4)+3-2\lambda=h(2)-1\geq 0,$

where the last equality follows from the last part of Lemma 5.1.

####  5.1.2. $a\in[10,10.24)$

In this situation, $h(2)>1,$ and by Lemma 5.2 we get

$C_{3}=(h(2)+\lambda;h(2),1^{\times 5},(\lambda-1)^{\times 5},d_{1}^{\times
l_{1}},\ldots)$

with $\delta_{3}=\lambda-2,$ so

$C_{4}=(h(3)+\lambda||h(3),1^{\times 3},(\lambda-1)^{\times 7},d_{1}^{\times
l_{1}},\ldots).$

If $h(3)\geq 1,$ then $\delta_{3}=\lambda-2$ so

$C_{5}=(h(4)+\lambda||h(4),1^{\times 1},(\lambda-1)^{\times 9},d_{1}^{\times
l_{1}},\ldots),$

and since $h(4)\geq\lambda-1,$ we have $\delta_{5}=0.$

Otherwise, $1>h(3),$ so

$C_{4}=(h(3)+\lambda;1^{\times 3},h(3),(\lambda-1)^{\times 7},d_{1}^{\times
l_{1}},\ldots),$

and $\delta_{4}=h(3)+\lambda-3=h(4)-1.$ Since $\lambda-1>h(4),$ we have
$h(4)-1<0.$ Therefore,

$C_{5}=(h(3)+h(4)+\lambda-1||h(3),(\lambda-1)^{\times 7},h(4)^{\times
3},d_{1}^{\times l_{1}},\ldots).$

Since $h(3)>\lambda-1,$ we have $\delta_{5}=h(4)+1-\lambda<0.$ Therefore,

$\displaystyle C_{6}$ $\displaystyle=$
$\displaystyle(h(3)+2h(4)||h(3)+h(4)+1-\lambda,(\lambda-1)^{\times
5},h(4)^{\times 5},d_{1}^{\times l_{1}},\ldots)$ $\displaystyle=$
$\displaystyle(3h(3)+2\lambda-4||2h(3)-1,(\lambda-1)^{\times 5},h(4)^{\times
5},d_{1}^{\times l_{1}},\ldots).$

Now

$2h(3)-1\geq\lambda-1\iff(b+5/2)\lambda-7\geq 0\iff a\geq m(5/2,7)\approx
9.68.$

Therefore, $\delta_{6}=h(3)-1<0,$ so

$C_{7}=(4h(3)+2\lambda-5||3h(3)-2,(\lambda-1)^{\times 3},h(4)^{\times
7},d_{1}^{\times l_{1}},\ldots).$

Similar to above reasoning, we have

$3h(3)-2\geq\lambda-1\iff(b+8/3)\lambda-22/3\geq 0\iff a\geq
m(8/3,22/3)\approx 9.88.$

Again, we have $\delta_{7}=h(3)-1,$ so

$C_{8}=(5h(3)+2\lambda-6||4h(3)-3,(\lambda-1)^{\times 1},h(4)^{\times
9},d_{1}^{\times l_{1}},\ldots).$

Now

$4h(3)-3\geq\lambda-1\iff(b+11/4)\lambda-15/2\geq 0\iff a\geq
m(11/4,15/2)\approx 9.97.$

Moreover, $d_{1}=a-\left\lfloor a\right\rfloor\leq.24,$ and we have

$h(4)\geq.24\iff(b+4)\lambda-9.24\geq 0\iff a\geq m(4,9.24)\approx 9.49.$

Putting these estimates together, we conclude that the ordering to the right
of the double bars in $C_{8}$ above is accurate and

$\displaystyle\delta_{8}$ $\displaystyle=$ $\displaystyle
5h(3)+2\lambda-6-4h(3)+3-\lambda+1-h(4)$ $\displaystyle=$ $\displaystyle
h(3)+\lambda-2-h(4)$ $\displaystyle=$ $\displaystyle h(4)-h(4)=0.$

####  5.1.3. $a\in[10.24,12)$

First, we note that

$h(3)=(b+3)\lambda-7>1$

for all $1\leq b\leq 2$ because $a\geq m(3,8)=10.24.$

Assume $10.24\leq a<11.$ By Lemma 5.2, we have

$C_{4}=(h(3)+\lambda;h(3),1^{\times 3},(\lambda-1)^{\times 7},d_{1}^{\times
l_{1}},\ldots)$

with $\delta_{4}=\lambda-2.$ Therefore,

$C_{5}=(h(4)+\lambda||h(4),1^{\times 1},(\lambda-1)^{\times 9},d_{1}^{\times
l_{1}},\ldots).$

Now $h(4)>\lambda-1$ by Lemma 5.1. In particular,

$\lambda-1=\min\\{1,\lambda-1,h(4)\\}.$

Therefore, $\delta_{5}=0.$

Now assume $11\leq a<12.$ Lemma 5.2 shows

$C_{4}=(h(3)+\lambda;h(3),1^{\times 4},(\lambda-1)^{\times 7},d_{1}^{\times
l_{1}},\ldots),$

with defect $\delta_{4}=\lambda-2.$ So

$C_{5}=(h(4)+\lambda||h(4),1^{\times 2},(\lambda-1)^{\times 9},d_{1}^{\times
l_{1}},\ldots).$

Assume $h(4)\geq 1.$ Recall that $h(4)\geq 1\implies h(5)\geq\lambda-1.$
Therefore,

$C_{5}=(h(4)+\lambda;h(4),1^{\times 2},(\lambda-1)^{\times 9},d_{1}^{\times
l_{1}},\ldots),$

with $\delta_{5}=\lambda-2.$ Applying another Cremona transformation, we see
that

$C_{6}=(h(5)+\lambda;h(5),(\lambda-1)^{\times 11},d_{1}^{\times
l_{1}},\ldots),$

with defect $\delta_{6}=2-\lambda>0.$

Now suppose $h(4)<1.$ Similar to previous reasoning, we have

(5.10) $1>h(4)>\lambda-1>h(5).$

So

$C_{5}=(h(4)+\lambda;1^{\times 2},h(4),(\lambda-1)^{\times 9},d_{1}^{\times
l_{1}},\ldots),$

with $\delta_{5}=\lambda-2.$ By 5.10, we get

$C_{6}=(h(5)+\lambda||(\lambda-1)^{\times 11},h(5),d_{1}^{\times
l_{1}},\ldots),$

with $\delta_{6}=h(5)+3-2\lambda=h(3)-1>0,$ where the latter equality follows
from the last part of Lemma 5.1.

####  5.1.4. $a\in[12,16)$

We start by making the following observations:

$\displaystyle h(5)\geq 1$ $\displaystyle\iff$ $\displaystyle a\geq
m(5,12)=11.755.$ $\displaystyle h(6)\geq 1$ $\displaystyle\iff$ $\displaystyle
a\geq m(6,14)=12.25.$ $\displaystyle h(7)\geq 1$ $\displaystyle\iff$
$\displaystyle a\geq m(7,16)\approx 12.642.$

If $12\leq a<13,$ then $h(5)\geq 1,$ so Lemma 5.2 implies

$C_{6}=(h(5)+\lambda;h(5),1^{\times 1},(\lambda-1)^{\times 11},d_{1}^{\times
l_{1}},\ldots)$

with $\delta_{6}=0.$

If $13\leq a<14,$ then $h(6)\geq 1,$ so

$C_{7}=(h(6)+\lambda;h(6),(\lambda-1)^{\times 13},d_{1}^{\times
l_{1}},\ldots),$

with $\delta_{7}=2-\lambda>0.$

For $14\leq a<15,$ we still use $h(6)\geq 1$ and Lemma 5.2 to get

$C_{7}=(h(6)+\lambda;h(6),1^{\times 1},(\lambda-1)^{\times 13},d_{1}^{\times
l_{1}},\ldots),$

with $\delta_{7}=0.$

Finally, if $15\leq a<16,$ Lemma 5.1 gives

$C_{8}=(h(7)+\lambda;h(7),1^{\times 1},(\lambda-1)^{\times 14},d_{1}^{\times
l_{1}},\ldots),$

with $\delta_{8}=0.$

###  5.2. Case II: $d_{1}>\lambda-1$

We begin this case with the following lemma.

######  Lemma 5.3.

We have $\lambda-1>1/2.$ In particular, if $a$ is such that $d_{1}>\lambda-1,$
and $l_{1}$ is the multiplicity of $d_{1}$ in the weight expansion of $a,$
then $l_{1}\leq 1.$

###### Proof.

We have $\lambda>3/2\iff 2a>9b.$ Since $a\geq 9$ and $b\leq 2,$ we have
$2a\geq 18\geq 9b.$ ∎

A suitable modification of Lemma 5.2 states that under the conditions of Lemma
5.2 we have

(5.11) $C_{k+1}=(h(k)+\lambda||h(k),1^{\times\left\lfloor
a\right\rfloor-(2k+1)},d_{1}^{\times l_{1}},(\lambda-1)^{\times 2k+1},\ldots)$

We handle this case on intervals of the form $[k,k+1)$ where $k$ is an integer
as we did above.

####  5.2.1. $a\in[9,10)$

Since $h(2)\geq 1,$ Equation 5.11 gives us

$C_{3}=(h(2)+\lambda||h(2),1^{\times 4},d_{1},(\lambda-1)^{\times 5},\ldots)$

with defect $\delta_{3}=\lambda-2<0.$ So

$C_{4}=(h(3)+\lambda||h(3),1^{\times 2},d_{1},(\lambda-1)^{\times 7},\ldots).$

From this point, there are several cases to consider, each of which yields
different calculations. The cases stem from how $h(3),1,$ and $d_{1}$ relate
to one another. Each case below is viewed as a subcase of the assumption that
$a\in[9,10).$ Such will be the case in future subsections where similar
computations are demonstrated. Computations of subsequent $\delta_{k}$ will
make liberal use of the identities in Lemma 5.1.

Subcase 1: $h(3)\geq 1.$ This case is the most straightforward. In this case,

$h(3)\geq 1\geq d_{1}>\lambda-1$

so $\delta_{4}=\lambda-2<0.$ Therefore,

$C_{5}=(h(4)+\lambda||h(4),d_{1},(\lambda-1)^{\times 9},\ldots).$

If $h(4)\geq d_{1},$ then $\delta_{5}=1-d_{1}\geq 0.$ If
$d_{1}>h(4)>\lambda-1,$ then still $\delta_{5}=1-d_{1}\geq 0.$ Finally, if
$d_{1}>\lambda-1>h(4),$ then

$\displaystyle\delta_{5}$ $\displaystyle=$ $\displaystyle h(4)+\lambda-
d_{1}-2(\lambda-1)$ $\displaystyle=$ $\displaystyle h(4)-d_{1}+2-\lambda$
$\displaystyle\geq$ $\displaystyle h(4)-1+2-\lambda$ $\displaystyle=$
$\displaystyle h(3)-1\geq 0.$

Subcase 2: $1>h(3)>d_{1}.$ This case is very similar to the previous since, as
before, we have $\delta_{4}=\lambda-2,$ and

$C_{5}=(h(4)+\lambda||h(4),d_{1},(\lambda-1)^{\times 9},\ldots).$

The analysis directly above shows that $\delta_{5}\geq 0$ in this case.

Subcase 3: $1>d_{1}>h(3).$ To begin, we need a key inequality.

######  Lemma 5.4.

If $d_{1}>h(3)$ and $9\leq a<10,$ we have

(5.12) $4h(3)-3d_{1}>\lambda-1>h(4)+1-d_{1}.$

###### Proof.

On $[9,10),$ $d_{1}=a-9,$ so the first inequality in 5.12 holds if and only if

(5.13) $(4b+11)\lambda-3a>0.$

Fix $b,$ and view the left-hand side of 5.13 as a function of $a.$ The
derivative of the left-hand side is $(4b+11)\lambda^{\prime}-3,$ which is
negative if and only if
$\lambda^{\prime}=\frac{1}{4b}\sqrt{\frac{2b}{a}}<3/(4b+11),$ or equivalently,

$\sqrt{\frac{2b}{a}}<\frac{12b}{4b+11}.$

Now $12b/(4b+11)\geq 12/15$ for $b\in[1,2],$ and $\sqrt{2b/a}\leq\sqrt{2b}/3$
on $[9,10).$ The above inequality will hold, then, if $\sqrt{2b}<36/15,$ which
is certainly true. Therefore, $(4b+11)\lambda-3a$ is decreasing as a function
of $a$ on $[9,10).$ Now

$\lim_{a\to 10^{-}}(4b+11)\lambda-3a=(4b+11)\sqrt{5/b}-30>0,$

where the latter inequality follows from familiar considerations using
differential calculus. In summary, the function in question is positive on
$[9,10),$ and the first inequality in 5.12 is hence established.

For the second inequality in 5.12, note that $\lambda-1>h(4)+1-d_{1}$ if and
only if $d_{1}>h(4)+2-\lambda=h(3),$ which is given. ∎

As an immediate consequence of Lemma 5.4 and our key assumption that
$d_{1}>h(3),$ we have

(5.14) $h(3)>2h(3)-d_{1}>3h(3)-2d_{1}>4h(3)-2d_{1}>\lambda-1>h(4)+1-d_{1},$

where the last inequality follows directly from the assumption that
$d_{1}>h(3)$ and the familiar identity $h(3)+\lambda-2=h(4).$

These inequalities in mind, we are ready to begin reduction calculations in
this case. Throughout, let $d_{2}=1-d_{1}.$ We have

$C_{4}=(h(3)+\lambda;1^{\times 2},d_{1},h(3),(\lambda-1)^{\times 7}||\ldots),$

so that $\delta_{4}=h(3)+\lambda-2-d_{1}=h(4)-d_{1}.$

If $\delta_{4}<0,$ then

$C_{5}=(h(3)+h(4)+\lambda-d_{1}||(h(4)+d_{2})^{\times
2},h(4),h(3),(\lambda-1)^{\times 7},\ldots).$

Now $1>h(3)\implies\lambda-1>h(4),$ so

$C_{5}=(h(3)+h(4)+\lambda-d_{1};h(3),(\lambda-1)^{\times
7},(h(4)+d_{2})^{\times 2},h(4)||\ldots),$

and $\delta_{5}=h(3)-d_{1}<0.$ Now $\lambda-1+h(3)-d_{1}=h(4)+d_{2}$ which,
combined with 5.14, implies

$C_{6}=(2h(3)+h(4)+\lambda-2d_{1};2h(3)-d_{1},(\lambda-1)^{\times
5},(h(4)+d_{2})^{\times 4},h(4)||\ldots),$

so that again $\delta_{6}=h(3)-d_{1}<0.$ Continuing, we see

$C_{7}=(3h(3)+h(4)+\lambda-3d_{1};3h(3)-2d_{1},(\lambda-1)^{\times
3},(h(4)+d_{2})^{\times 6},h(4)||\ldots),$

so that $\delta_{7}=h(3)-d_{1}<0.$ Finally, we see that

$C_{8}=(4h(3)+h(4)+\lambda-4d_{1};4h(3)-3d_{1},(\lambda-1)^{\times
1},(h(4)+d_{2})^{\times 8},h(4),\ldots),$

and $\delta_{8}=0.$ This concludes the case where $9\leq a<10.$

####  5.2.2. $a\in[10,11)$

Since $h(2)\geq 1,$ we start with

$C_{3}=(h(2)+\lambda;h(2),1^{\times 5},d_{1},(\lambda-1)^{\times 5}||\ldots),$

with $\delta_{3}=\lambda-2.$ So

$C_{4}=(h(3)+\lambda||h(3),1^{\times 3},d_{1},(\lambda-1)^{\times 7},\ldots).$

We claim $h(3)\geq 1$ when $d_{1}>\lambda-1.$ Indeed, $d_{1}=a-10$ on this
interval, and recall [Lemma] asserts that $\lambda-1>1/2.$ So
$d_{1}>\lambda-1\implies a-10>1/2,$ or equivalently, $a>10.5$ Now $h(3)\geq 1$
if and only if $(b+3)^{2}a/2b\geq 64,$ and since
$(b+3)^{2}a/(2b)\geq(b+3)^{2}\cdot 10.5/(2b),$ we need only hope that the
latter expression exceeds 64. If $f(x)=10.5(x+3)^{2}/2x$ on $[1,2],$ then $f$
is easily seen to be decreasing, and its least value is thus $f(2)=65.525.$
Thus $h(3)\geq 1$ as desired. As an immediate consequence,
$h(4)\geq\lambda-1.$

Since $h(3)\geq 1,$ we have $\delta_{4}=\lambda-2,$ and after re-arranging the
first three terms past the head, we get

$C_{5}=(h(4)+\lambda||1^{\times 1},d_{1},h(4),(\lambda-1)^{\times 9},\ldots).$

The term $h(4)$ may be greater than or equal to $d_{1}$ or less than $d_{1}$
on this interval. However, $h(4)\geq\lambda-1$ when $d_{1}>\lambda-1.$ Put
another way, the least of $h(4),1,$ and $d_{1}$ must be at least as great as
$\lambda-1$ on $[10,11),$ provided that $d_{1}>\lambda-1.$ In particular,
$\delta_{5}=\lambda-1-d_{1}<0,$ and we get

$C_{6}=(h(5)+\lambda+1-d_{1}||h(5)+1-d_{1},\lambda-d_{1},(\lambda-1)^{\times
10},\ldots).$

If $h(5)+1-d_{1}\geq\lambda-1,$ then $\delta_{6}=d_{1}-\lambda+1>0.$
Otherwise, $h(5)+1-d_{1}<\lambda-1,$ and we have

$\delta_{6}=h(5)+3-2\lambda=h(4)+1-\lambda\geq 0.$

####  5.2.3. $a\in[11,12)$

Before we get started here, we claim $h(5)>d_{1}.$ First, note that
$d_{1}=a-\left\lfloor a\right\rfloor=a-11.$ The desired inequality is true if
and only if $(b+5)^{2}/2b\geq a,$ which holds if the former expression is at
least as great as 12. The function $f(x)=(x+5)^{2}/2x$ on $[1,2]$ is
decreasing, and $f(2)>12,$ so the assertion is, in fact, accurate. In
particular, since $h(4)>h(5),$ it follows that $h(4)>d_{1}.$

In this case, we have

$C_{4}=(h(3)+\lambda||h(3),1^{\times 4},(\lambda-1)^{\times 7},d_{1},\ldots),$

with $\delta_{4}=\lambda-2.$ So,

$C_{5}=(h(4)+\lambda||h(4),1^{\times 2},d_{1},\ldots).$

If $h(4)\geq 1,$ we have $\delta_{5}=\lambda-2$ and

$C_{6}=(h(5)+\lambda;h(5),d_{1},(\lambda-1)^{\times 11}||\ldots)$

with $\delta_{6}=1-d_{1}\geq 0.$ If $h(4)<1,$ then

$C_{5}=(h(4)+\lambda;1^{\times 2},h(4),d_{1},(\lambda-1)^{\times 9}||\ldots),$

and $\delta_{5}=\lambda-2<0.$ Continuing, we see that

$C_{6}=(h(5)+\lambda h(5),d_{1},(\lambda-1)^{\times 11}||\ldots),$

so that $\delta_{6}=1-d_{1}\geq 0.$

####  5.2.4. $a\in[12,13)$

Here, we have

$C_{6}=(h(5)+\lambda||h(5),1^{\times 1},d_{1},(\lambda-1)^{\times 11},\ldots)$

so $\delta_{6}=\lambda-1-d<0.$ Then

$C_{7}=(h(6)+\lambda+1-d_{1}||h(6)+1-d_{1},\lambda-d_{1},(\lambda-1)^{\times
12},\ldots),$

with $\delta_{7}=d-\lambda+1>0.$

####  5.2.5. $a\in[13,14)$

Since $h(6)\geq 1,$ we get

$C_{7}=(h(6)+\lambda||h(6),d_{1},(\lambda-1)^{\times 13},\ldots),$

so that $\delta_{7}=1-d\geq 0.$

####  5.2.6. $a\in[14,15)$

We have $h(6)\geq 1$ so

$C_{7}=(h(6)+\lambda;h(6),1^{\times 1},d_{1},(\lambda-1)^{\times
13}||\ldots).$

Then $\delta_{7}=\lambda-1-d<0$ so that

$C_{8}=(h(7)+\lambda+1-d_{1};h(7)+1-d_{1},\lambda-d_{1},(\lambda-1)^{\times
14}||\ldots).$

In particular, $\delta_{8}=d_{1}-\lambda+1>0.$

####  5.2.7. $a\in[15,16)$

In this case, we have $h(7)\geq 1$ so that

$C_{8}=(h(7)+\lambda||h(7),1^{\times 1},d_{1},(\lambda-1)^{14},\ldots).$

From here, we see $\delta_{8}=\lambda-1-d_{1}<0,$ and thus

$C_{9}=(h(8)+\lambda+1-d_{1}||h(8)+1-d_{1},\lambda-d_{1},(\lambda-1)^{\times
15},...),$

and thus $\delta_{9}=d_{1}-\lambda+1>0.$

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