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\topmatter
\title Stationary Points of the Yang-Mills Action
\endtitle
\author Lorenzo Sadun \\ and \\ Jan Segert \endauthor
\address{Courant Institute of Mathematical Sciences,
New York University,
251 Mercer Street,
New York, NY 10012. \newline Address after Sept. 1, 1991:
Department of Mathematics, University of Texas, Austin, Texas 78712}
\endaddress
\address{Department of Mathematics, University of Missouri, Columbia,
Missouri 65211}
\endaddress
\thanks {The first author was partially supported by
NSF Grant DMS-8806731. \newline The second author was
partially supported by a Bantrell Fellowship and
NSF Grant DMS-8801918.}
\endthanks
\subjclass{81E13, also 34B15, 53C05, 58E30}
\endsubjclass
\abstract{We examine the structure of a recently discovered set
of non-self-dual solutions of the Yang-Mills equations.
These solutions have a symmetry that reduces the YM equations to a
set of ODE's.
The distinct solutions are indexed by two postive odd integers.
We develop a scheme to approximate on a computer the solutions for small values of the
indexing integers, and present some numerical results.
We then analyze
the asymptotic behavior of the solutions as the indexing integers
become large. }
\endabstract
\endtopmatter
\baselineskip=15pt
\document
\heading {\bf 1. Introduction} \endheading
In this paper we consider stationary points of the $SU(2)$ Yang-Mills
action on Euclidean $\real^4$.
These are solutions to the variational equations
$$ 0={\delta S \over \delta A_\mu}\equiv \partial^\nu F_{\mu\nu}
+i [A^\nu, F_{\mu\nu}], \tag 1.1 $$
often called the {\it Yang-Mills equations}.
Local minima of the action have been
known for some time; these include the vacuum ($A_\mu =0$) and
various instanton,
anti-instanton, and multi-instanton solutions [BPST, ADHM].
All these solutions are characterized by having an
(anti)self-dual curvature tensor. That is,
%
$$ F_{\mu\nu} = \pm \tilde F_{\mu\nu}, \tag 1.2 $$
%
where the dual is
%
$$ \tilde F_{\mu\nu} \equiv
{1 \over 2} \epsilon_{\mu\nu\rho\sigma}F^{\rho\sigma}. \tag 1.3 $$
%
Recall that gauge fields are divided into topological sectors by the
instanton number $k$, which is a gauge-invariant integer,
%
$$ k = {1 \over 16 \pi^2} \int d^4x \Tr \ (F^{\mu\nu}\tilde F_{\mu\nu}).
\tag 1.4 $$
%
Defining the self-dual and anti-self-dual parts of the curvature,
%
$$ F_{\pm} = {1 \over 2} \left ( F \pm \tilde F \right ), \tag 1.5 $$
%
we can rewrite the instanton number and the action as
%
$$ \align k = & { 1 \over 16\pi^2} \int d^4x ( |F_-|^2 - |F_+|^2 ) \\
S \equiv & {1\over 2} \int d^4x |F|^2 = {1\over 2}
\int d^4x ( |F_+|^2 + |F_-|^2 ). \tag 1.6
\endalign $$
%
It follows that in each topological sector the action is at least $8\pi^2 |k|$,
and the fields that attain this minimum are precisely the
(anti)self-dual fields.
No local minima exist at higher values of the action, as was
shown by Bourguignon and Lawson [BL] and Taubes [T], so
a non-self-dual solution to (1.1) must be a saddle point of
the action.
We are interested in non-self-dual solutions on Euclidean space with
group $SU(2)$, which were only recently discovered.
Other examples of non-self-dual solutions are known, namely solutions with
larger groups [I,Ma], and solutions on manifolds that are more complicated
topologically [Ur,P2] or geometrically [P1].
For many years, the preponderance of evidence suggested that
non-self-dual solutions of the $SU(2)$ Yang-Mills equations on Euclidean
space should not exist.
However, in 1989, Sibner, Sibner, and
Uhlenbeck [SSU] demonstrated
the existence of non-self-dual solutions with zero instanton number.
We [SS2, SS3] have found a family of
non-self-dual solutions, including solutions with
nonzero instanton number.
This family of solutions is
parametrized by two odd integers $n_\pm \ge 3$.
The instanton number of the solution corresponding to the pair $(n_+,n_-)$ is
$k=(n_-^2-n_+^2)/8$, as explained in [SS1,ASSS].
Every integer $k$ except $\pm 1$ can be obtained from this formula, so
we have solutions with {\it arbitrary\/} instanton
number (except $k=\pm 1$).
For some $k$ there are several distinct solutions. For example,
$k=5$ can arise in two ways, as $(7^2-3^2)/8$ or as
$(11^2-9^2)/8$.
There are an infinite number
of examples with instanton number zero; just take $n_+=n_-$.
In Section 2 we review the general properties of solutions in this
family;
the details are in [SS2, SS3].
In Section 3 we describe a scheme for numerically approximating these solutions,
and we give results for some solutions with small $n_\pm$.
In Section 4 we develop an asymptotic description of the solutions as
the integers $n_\pm$ become large.
\medskip
\heading {\bf 2. Symmetric Gauge Fields } \endheading
The Yang-Mills action and the (anti)\,self-dual equations are
conformally invariant, so solutions on Euclidean space
correspond to
solutions on the four-sphere $S^4$, and vice-versa.
We work on $S^4$, because our solutions have a symmetry group that acts
by isometries (orthogonal transformations) on $S^4$,
while the corresponding group action on $\real^4$ is more complicated.
Let $V \simeq \real^5$ be the space of symmetric, traceless, real
$3 \times 3$ matrices $Q$, with inner product $\langle Q,Q' \rangle
= \half Tr(QQ')$.
The matrices
$$ Q_0 = {1 \over \sqrt{3}} \pmatrix -1 & 0 & 0 \cr 0 & -1 & 0 \cr
0 & 0 & 2 \endpmatrix ;
\qquad Q_1 = \pmatrix 0 & 0 & 1 \cr 0 & 0 & 0 \cr 1 & 0 & 0 \endpmatrix ;
\qquad Q_2 = \pmatrix 0 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0
\endpmatrix ; $$
$$ Q_3 = \pmatrix -1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 0 \endpmatrix;
\qquad Q_4 = \pmatrix 0 & 1 & 0 \cr 1 & 0 & 0 \cr 0 & 0 & 0
\endpmatrix \tag 2.1 $$
are an orthonormal basis.
A matrix $g$ in $SO(3)$ acts on $V$ by conjugation,
$g(Q) \to gQg^{-1}$.
$SU(2)$ is the double cover of $SO(3)$, so there is an action of $SU(2)$ on
$V$, which is just the spin-2 representation.
The unit sphere $S^4$ in $V$ inherits this $SU(2)$ action.
Since all matrices in $V$ are diagonalizable, it is not hard to
check that every $Q \in S^4$ is related by the
group action to a unique $Q_\theta = \cos(\theta)Q_0 +
\sin(\theta)Q_3$ with
$0 \le \theta \le \pi/3$. The orbits of $Q_0$ and $Q_{\pi/3}$
are two-dimensional, while all other orbits are three-dimensional.
We define a
coordinate chart $(y^0=\theta, y^1, y^2, y^3)$ by the correspondence
$$ (\theta, \vec y) \mapsto \exp(-\vec y \cdot \vec \sigma) \left (
Q_\theta \right ), \tag 2.2 $$
where $\sigma_1$, $\sigma_2$, and $\sigma_3$ are the
(antihermitian) generators of $SU(2)$,
$\sigma_i$ corresponding to the generators of the $SO(3)$ rotation about
the $i$-th axis of $\real^3$.
The $SO(3)$ matrix corresponding to
$\exp(- \vec y \cdot \vec \sigma)$ acts on $Q_\theta$ by
conjugation as above.
As $\theta$ ranges from $0$ to $\pi/3$, $Q_\theta$ moves along a
geodesic segment on the sphere.
On this segment the tangent vectors
$\partial_\theta \equiv \partial / \partial \theta$ and $\partial_i
\equiv \partial / \partial y^i$ are orthogonal but not orthonormal.
The vector $\partial_\theta$ is normalized, but the length of the vector
$\partial_i$ at $Q_\theta$ is $f_i(\theta)$, where
$$ f_1(\theta) = 2 \sin(\pi/3 + \theta); \qquad
f_2(\theta) = 2 \sin(\pi/3 - \theta); \qquad
f_3(\theta) = 2 \sin(\theta). \tag 2.3 $$
Note that $f_3$ vanishes at $\theta = 0$, as $Q_0$ is invariant
under rotations about the $3$-axis of $\real^3$. Similarly, $f_2$ vanishes at
$\pi/3$.
>From this it is easy to see how the duality operator acts on
2-forms:
$$ \tilde F_{0i} = {f_j f_k \over f_i} F_{jk}; \qquad
\tilde F_{jk} = {f_i \over f_j f_k} F_{0i}
\tag 2.4 $$
where $(i,j,k)$ are cyclic permutations of (1,\,2,\,3). To
simplify the notation we define the functions
$$G_1 = {f_2 f_3 \over f_1}; \qquad G_2={f_1 f_3 \over f_2}; \qquad
G_3 = {f_1 f_2 \over f_3}. \tag 2.5 $$
$G_1$ and $G_2$ have zeroes at $\theta=0$, while $G_3$
has a pole. Similarly, $G_1$ and $G_3$ have zeroes at $\theta=\pi/3$,
while $G_2$ has a pole.
A symmetric gauge potential is determined by its values on
the segment $ \{ Q_\theta \}$ where $ 0 \le \theta \le \pi/3 $.
This potential must also be invariant under
rotations by $\pi$ about the $1$, $2$, and $3$
axes, as these rotations take the point $Q_\theta$ to itself.
The most general potential that is invariant under
these rotations is
$$ A = - \sum_{i=1}^3 a_i(\theta) dy^i \otimes \sigma^i, \tag 2.6 $$
where $a_1$, $a_2$, and $a_3$ are real-valued functions of $\theta$.
We call such a triplet of functions $a = (a_1, a_2, a_3)$ a
{\it reduced potential} or a {\it reduced connection}.
Given a symmetric potential $A$, the curvature $F$ is easily computed:
$$ F = \big (
(a_1 + a_2 a_3) dy^2 \wedge dy^3
- a_1' d\theta \wedge dy^1
\big ) \otimes \sigma^1 + (cyclic), \tag 2.7 $$
where ${}'$ denotes $d/d\theta$, and $(cyclic)$ denotes the other
cyclic permutations of the indices (1,\,2,\,3). By equation (2.4) the
(anti)\,self-duality equations are then
$$
-a_1' = \pm {(a_1+a_2a_3) \over G_1}, \qquad
-a_2' = \pm {(a_2+a_1a_3) \over G_2}, \qquad
-a_3' = \pm {(a_3+a_1a_2) \over G_3}, \tag 2.8 $$
where $+$ denotes self-duality and $-$ denotes anti-self-duality.
>From $F$ we compute the action, with the result
$$ \split S(A) = {\pi^2} \int_0^{\pi/3} d\theta & \Big [
(a_1')^2 G_1 +
(a_2')^2 G_2 +
(a_3')^2 G_3 \\ & +
{(a_1+a_2a_3)^2 \over G_1} +
{(a_2+a_1a_3)^2 \over G_2} +
{(a_3+a_1a_2)^2 \over G_3} \Big ]. \endsplit \tag 2.9
$$
Finite action reduced potentials must have well-defined boundary values
$r=a_3(0)$ and $t=a_2(\pi/3)$. Also, since $G_1$ and $G_2$ have
zeroes at $\theta=0$, finite-action reduced potentials must have
$$a_1(0)+ a_2(0) a_3(0) = a_2(0) + a_1(0) a_3(0) = 0. \tag 2.10 $$
If $r \ne \pm 1$, these conditions imply that both $a_1(0)$
and $a_2(0)$ equal zero. Similarly, if $t \ne \pm 1$ then
$a_1(\pi/3)=a_3(\pi/3)=0$.
Not all finite-action reduced potentials correspond to
smooth potentials on all of $S^4$. For smooth potentials,
$r+1$ and $t+1$ must both be integers divisible by 4 [BoMo], in which
case
$n_+ \equiv |r|$ and $n_- \equiv |t|$ are positive odd integers.
Otherwise our reduced potential corresponds
to a singular potential with non-integer Chern
number [FHP,\,SiSi], with the singularities occurring
at the orbits of $Q_0$ and $Q_{\pi/3}$.
In [SS2,\,SS3] we showed that for every pair $(n_+,n_-)$
with $n_+\!>\!1$ and $n_-\! > \!1$, there exists a stationary point of
the action that is neither self-dual nor anti-self-dual.
This result was proven by finding minima of the action within each
symmetry
class. By the principle of symmetric criticality [Pal], these minima
are solutions of the full Yang-Mills equations.
We then proved that self-dual or anti-self-dual solutions do not exist
in these symmetry classes, so these Yang-Mills solutions must
be non-self-dual.
This existence proof is nonconstructive,
and does not give the exact form of the solutions.
The remainder of the present paper examines the form of these solutions.
\medskip
\heading {\bf 3. Fourier Analysis} \endheading
The functions $a_i(\theta)$ can be extended to functions on the entire
circle $0 \le \theta \le 2\pi$ via equation (2.6), since the points
$Q_\theta$ are well-defined for all $\theta$. The points
$Q_\theta$, $Q_{-\theta}$, and $Q_{\theta \pm 2\pi/3}$ are related by
rotations that permute the coordinates $\{y^i\}$ and also
permute the $\{\sigma^i\}$'s. As a result, the functions $a_i$ satisfy
$$a_3(-\theta)=a_3(\theta), \qquad a_2(\theta)=a_3(\theta+2\pi/3),
\qquad a_1(\theta)=a_3(\theta-2\pi/3).
\tag 3.1 $$
Thus the behavior of the three functions $a_i$ on $[0,\pi/3]$ is
expressed by the single even function $a_3$ on $[0,2\pi]$.
The potential $A$ is smooth on all of $S^4$ if and only if $a_3$ is
smooth on the entire circle [BoMo].
Since the three functions $G_i(\theta)$ also satisfy the symmetry relations
$$G_2(\theta)=G_3(\theta+2\pi/3), \qquad G_1(\theta)=G_3(\theta-2\pi/3),
\tag 3.3 $$
we can rewrite the action functional (2.9) as a
function of $a_3$ alone:
$$ \align S & = {\pi^2} \left [\int_0^{\pi/3}\!\!+ \int_{2\pi/3}^{\pi}
\!\!+ \int_{4\pi/3}^{5\pi/3} \right] d \theta \left[
(a_3')^2 G_3 + {(a_3+a_1a_2)^2 \over G_3} \right] \tag 3.3 \\
= &
{\pi^2} \left [\int_0^{\pi/3}\!\! + \int_{2\pi/3}^{\pi}
\!\! + \int_{4\pi/3}^{5\pi/3} \right] d \theta
\left [(a_3'(\theta))^2 G_3 +
{(a_3(\theta)\!+\!a_3(\theta\!+\!2\pi/3)a_3(\theta\!-\!2\pi/3))^2 \over G_3}
\right] \endalign $$
To simplify the notation, we let $\ival$ denote the region of
integration \hfil\break
$[0,\pi/3] \cup [2 \pi/3, \pi] \cup [4\pi/3, 5\pi/3]$,
and let
$$ b(\theta) = a_3(\theta) + a_3(\theta-2\pi/3) a_3(\theta+2\pi/3).
\tag 3.4 $$
This lets us write the action as
$$ S = \pi^2 \int_{\ival} (a_3')^2 G_3 + b^2/G_3, \tag 3.5 $$
and the first variation of the action as
$$ \delta S = 2\pi^2 \int_{\ival} (a_3')(\delta a_3')G_3 + (\delta b)b/G_3.
\tag 3.6 $$
The Euler-Lagrange equations $\delta S=0$ are
$$ \align (G_1 a_1')'= & {a_1+a_2a_3 \over G_1} +
{a_3(a_2+a_1a_3) \over G_2} + {a_2(a_3 + a_1 a_2) \over G_3} \tag 3.7a \\
(G_2 a_2')'= & {a_2+a_1a_3 \over G_2} +
{a_1(a_3+a_1a_2) \over G_3} + {a_3(a_1 + a_2 a_3) \over G_1} \tag 3.7b \\
(G_3 a_3')'= & {a_3+a_1a_2 \over G_3} +
{a_2(a_1+a_2a_3) \over G_1} + {a_1(a_2 + a_1 a_3) \over G_2} \tag 3.7c
\endalign $$
on the interval $[0,\pi/3]$, or equivalently any {\it one\/} of these equations
(say, 3.7c) on the entire circle.
\medskip
\noindent{\bf Fourier Decomposition of $a_3$}
>From now on we take $n_\pm \ge 3$, and so have $a_3(\pi/3)=a_3(2\pi/3)=0$.
We express $a_3(\theta)$ as a Fourier series
$$ a_3(\theta) = \sum_{n=-\infty}^\infty A_n \alpha^n, \tag 3.8 $$
where $\alpha \equiv e^{i\theta}$.
Since $a_3$ is real and even, we have
$$ A_n = A_n^* = A_{-n} \tag 3.9$$ for all $n$.
The function $a_3$ is
described by the sequence $A_0$, $A_1, \ldots$. These coordinates
are not independent, as $a_3$ satisfies the boundary conditions
$$ a_3(0)=r, \qquad a_3(\pi/3)=a_3(2\pi/3)=0, \qquad a_3(\pi)=t.
\tag 3.10 $$
To see the effect of these conditions, we define, for $m=0,1,2,\dots,5$
$$ c_m = \sum_{l=0}^{\infty} A_{6l+m}, \tag 3.11 $$
and let
$C_0=2c_0-A_0$, $C_1=c_1+c_5$, $C_2=c_2+c_4$, and
$C_3=c_3$.
Our boundary conditions are
$$\align C_0+2C_1+2C_2+2C_3=r, & \qquad C_0+C_1-C_2-2C_3=0, \tag 3.12 \\
C_0 - C_1-C_2 + 2C_3 = 0, & \qquad C_0 - 2C_1 + 2C_2 - 2C_3 = t.
\endalign $$
Solving these equations gives
$$C_0=C_2 = (r+t)/6, \quad C_1=2C_3= (r-t)/6,
\tag 3.13 $$
fixing the values of $A_0$, $A_1$, $A_2$, and $A_3$
as a function of the remaining coordinates.
Our allowed variations are therefore as follows. Every change in
$A_{6l+m}$, $l\ge 1$, $m=0,1,2,\dots,5$, must be compensated
by the opposite change in $A_m$, thereby keeping $c_m$ fixed. Also,
any change in $A_4$ and $A_5$ must be compensated for by a change in
$A_2$ and $A_1$, respectively, keeping $C_2$ and $C_1$ fixed.
That is, the allowed variations in
$a_3$ are spanned by the functions
$$h_{l,m}(\theta) = \alpha^{6l+m} + \alpha^{-6l-m} - \alpha^m
- \alpha^{-m} \tag 3.14$$
with $l\ge 1$ and $0\le m\le 5$, or with $l=1$ and $m=-1$ or $-2$.
These last two possibilities correspond to variations of $A_5$ and $A_4$,
respectively.
\vfill\eject
\noindent{\bf Finite Mode Approximations}
We are now ready to look for solutions to the variational (Yang-Mills)
equations (3.7).
As a first try we look for solutions
that have only a finite number of nonzero Fourier coefficients.
Unfortunately, there are only three such solutions, and they have
(anti)\,self-dual curvature. They are the flat potential (gauge
equivalent to the vacuum) with the zeroth Fourier coefficient $A_0=-1$
and all other
coefficients zero, the round BPST instanton with $A_0=A_1=A_{-1}=1$ and all
other coefficients zero, and the
round BPST anti-instanton with $A_0=1$, $A_1=A_{-1}=-1$ and all other
coefficients zero. These solutions have $\npnp=$ (1,1), (3,1)
and (1,3) respectively.
To see that no other finite-term solutions exist, we look at
equation (3.7c). If $A_k$ is the highest order non-zero coefficient,
then the right-hand-side of (3.7c) has a (nonzero) term proportional
to $\alpha^{3k-1}$, while the left hand side has terms only up to
$\alpha^{k+1}$. Thus there are no solutions with $3k-1>k+1$,
i.e. $k>1$. The equations
for $k \le 1$ are easy to solve, and give the three solutions above.
All remaining solutions must have infinite Fourier expansions.
However, since solutions of the Yang-Mills equations
are known to be smooth in an appropriate gauge, we expect $a_3$ to
be smooth, and so expect the Fourier coefficients $A_n$ to
decrease exponentially with $n$. As a result, we can get very good
approximate solutions by considering only the first few
coefficients. Specifically, we pick a small integer $l_{max}$,
and consider functions $a_3$ with only the first
$6l_{max}+6$ components nonzero.
This is a $6l_{max}+2$ dimensional space ($6l_{\max}+6$ variables with
4 boundary conditions). In this space we minimize the action numerically,
using Newton's method.
The results for $n_+=5$, $n_-=3$ are listed in Table 1 for several
values of $\lmax$. In this example even taking $\lmax =1$ is
enough to give a good estimate, and taking $\lmax=4$ gives
14-digit accuracy.
% *****INSERT TABLE 1 HERE ******
For larger values of $n_\pm$ the
coefficients $A_n$ do not decrease quite so rapidly, so larger
and larger values of $\lmax$ are needed.
As a result, this method of computing approximate solutions is
only practical for $(n_++n_-)$ up to 100 or so. For larger
values of $n_\pm$ we must use the asymptotic analysis of the next
section.
The actions of 19 low-lying non-self-dual stationary points of the
Yang-Mills actions are listed
in Table 2, as computed with $\lmax=5$.
In each case, the last significant digit is the last
digit listed. We display $S/8\pi^2$ rather than $S$, as (anti)self-dual
solutions have $S/8\pi^2=|k|$, and in particular since the BPST instanton
has $S/8\pi^2=1$.
% ***** INSERT TABLE 2 HERE *******
\heading {\bf 4. Asymptotic analysis: a succession of models} \endheading
In this section we study the one-dimensional variational problem
for large $|r|$ and $|t|$, which will yield information about
Yang-Mills solutions
corresponding to large odd integers $n_+$ and $n_-$.
We work with the three functions
$a_i$ on the interval $[0, \pi/3]$ rather than the single function
$a_3$ on the circle.
Notice that the action is invariant under the change of sign of
any two of the $a_i$'s. We can therefore assume, without loss
of generality, that both $r$ and $t$ are positive.
To get the asymptotic form of the Yang-Mills solutions and their actions
as $r$ and $t$ become large,
we look at a succession of models.
We start by solving a simple toy model, and add more and more features
until we reach the reduced
Yang-Mills action functional.
Models 1--4 are symmetric models, in the sense that $r=t$. The Yang-Mills
problem is Model 4, with the key results being equations (4.4.2) and
(4.4.4). The Yang-Mills problem with $r \ne t$ is Model 6, with
the key results tabulated in Table 4.
We will make repeated use of the following useful fact, which
applies in the absence of catastrophes.
\proclaim {Useful Fact}
Let $f(p) = \{ f_1, \dots, f_k \}$ be a collection of functions on the interval
$[a,b]$, satisfying boundary conditions that depend on a parameter $p
\in \real$.
Let $\uS (p) $ be the minimum value of the functional
$S(f) = \int_a^b \sum_i [ (f_i')^2 + V(f)] dx$ subject to the boundary
conditions specified by $p$, and let $\uf =
\{\uf_1, \dots, \uf_k \}$
be the corresponding minimizing functions.
Then
$$ d \uS /dp =
2 \sum_i \big ( {\uf_i}'(b) \dot \uf_i(b) -
{\uf_i}'(a) \dot \uf_i(a)
\big ), \tag 4.1$$
where dot $(\; \dot{} \; )$ denotes $d/dp$.
\endproclaim
The proof is just integration by parts:
$$d \uS/dp = \sum_i
\int 2 {\uf_i}' \dot {\uf_i}' + \dot \uf_i \partial_i V(\uf)
= \sum_i 2 \dot \uf_i {\uf_i}' \big |_a^b
+ \sum_i \int \dot \uf_i(\partial_i V(\uf) - 2{\uf_i}''). \tag 4.2 $$
The terms in parentheses are all zero by the variational equations.
When clear from context, we will not distinguish between the
functional $S$ and its minimum value $S^*$, or between $f_i$ and $\uf_i$.
\medskip
\noindent{\bf Toy Model Number 1}
\medskip
In this model we have two functions, $f$ and $g$, on the interval
$[-1,1]$, and we want to minimize the functional
$$S_1(f,g)= \int_{-1}^1 (f')^2 + (g')^2 + b^2f^2g^2 , \tag 4.1.1$$
with the boundary conditions $g(1)= f(-1)=0$, $f(1)=g(-1)=n$. Here
$b$ is a parameter independent of $n$. By the
Useful Fact this means that $dS_1/dn = 2(f'(1)-g'(-1))=4f'(1)$, which we
can integrate to get $S_1$.
We first consider $b=1$.
The variational equations for this system are
$$f'' = fg^2; \qquad g'' = gf^2. \tag 4.1.2 $$
These equations are invariant under translation $x \to x + a$,
and under the scale change $f(x),g(x) \to \lambda f(\lambda x), \lambda
g(\lambda x)$.
It is clear that $f$ and $g$ cannot be large at the same places, or
else the $f^2 g^2$ term would get huge. Therefore, for large $n$
we expect $f$ to be small for $x$ significantly negative, and expect $g$ to be
small for $x$ significantly positive, and for there to be a small overlap
region near zero. By the variational equations,
for $x$ much bigger than
$0$, $f$ will approach a straight line, and $g$ will decay rapidly.
By the WKB approximation, this decay goes as $\exp(-\sqrt{n}x^2/2)$.
We now look for solutions to (4.1.2) with $g$
going to zero as $x \to \infty$, with $f$ going to zero as $x \to -\infty$,
and symmetric about the origin ($f(x)=g(-x)$). These will not be
exact solutions to our variational problem, since $g(1)$ will not be
exactly zero, but the error introduced is exponentially small in $\sqrt{n}$.
By rescaling we can make the asymptotic slope of $f$ equal to 1, so
that, for $x$ much bigger than 1,
$$ f(x) \approx x + c, \tag 4.1.3$$
for some yet-to-be determined constant $c$.
To get the boundary condition $f(1)=g(-1)=n$, we must rescale again,
this time by a factor $\lambda$. For $x$ much
bigger than $1/\lambda$, our solution is now
$$f(x) = \lambda^2 x + \lambda c, \tag 4.1.4 $$
with the same constant $c$ as before. We identify $n$ with
$f(1)=\lambda^2 + \lambda c$, and get
$$\lambda = \sqrt{n + c^2/4} - c/2, \tag 4.1.5 $$
and so
$$ dS_1/dn = 4f'(1)=4\lambda^2 = 4n + 2c^2 -4c \sqrt{n+c^2/4}. \tag 4.1.6 $$
Integrating we get
$$S_1(n) = 2n^2 + 2c^2n -{8c \over 3} (n + c^2/4)^{3/2} + K, \tag 4.1.7$$
plus exponentially small terms. Here $K$ is a constant of integration.
(Note, by the way, that the overlap region is of size
$O(1/\lambda)=O(n^{-1/2})$).
All that remains is to find $c$ (and $K$). We put this model on the
computer, dividing the interval [-1,1] into 50, 100, and 200 slices,
for $n$ = 10, 20, 30, 40, and 50. $c$ is best estimated not from the
limiting actions, but from the limiting solutions, since
$c= (n- f'(1))/\sqrt{f'(1)}$. From these studies we get $c=0.64847 \pm
0.00002$. This fits the numerical actions as well, to within 0.2 or so,
with $K=0$, from which we deduce that $K$ is not larger than 0.5 or so,
and may in fact be zero.
We next consider $b \ne 1$. The variational equations are then
$$ f''=b^2 fg^2, \qquad g'' = b^2 gf^2. \tag 4.1.8$$
Under the (vertical) rescaling $f(x) \to b f(x)$, $g(x)\to b g(x)$
these equations become the same as (4.1.2). As a result we have
$$S_1(b,n) = b^{-2} S_1(1,bn) = 2n^2 -
{8c \over 3 \sqrt{b}}n^{3/2} +O(n). \tag 4.1.9$$
\vfill\eject
\noindent{\bf Toy Model Number 2}
\medskip
Next we add a mass term, and do our calculation on the interval
$[-T, T]$, for some constant $T$, rather than on $[-1,1]$.
$$S_2 = \int_{-T}^T (f')^2 + (g')^2 + f^2 + g^2 + b^2f^2g^2. \tag 4.2.1$$
For now we take $b=1$; other values of $b$ are treated as in Model 1.
Note that the mass term $f^2+g^2$, restricted to the overlap region,
contributes only $O(\sqrt{n})$ to the action, and so can be
ignored in looking for the $O(n^{3/2})$ behavior of the system. This
is therefore Model 1 in the overlap region. Outside the overlap
region, either $f$ or $g$ is near zero, and the system evolves into
hyperbolic functions. So instead of $f(x)$ asymptotically approaching
$\lambda^2 x + \lambda c$, $f(x)$ is asymptotically $\lambda^2 \sinh(x)
+\lambda c \cosh(x)$. We identify $n$ with $f(T)$, and so
$$ n = \lambda^2 \sinh(T) + \lambda c \cosh(T). \tag 4.2.2$$
It is convenient to define $\N=n/\sinh(T)$, and have
$$ \lambda = \sqrt{\N+c^2 \coth^2(T)/4} - c \coth(T)/2, \tag 4.2.3$$
so
$$ \align
f'(T) = & \lambda^2 \cosh(T) +\lambda c \sinh(T) \\
= & \N \cosh(T) -\lambda c /\sinh(T) \tag 4.2.4\\
= & \N \cosh(T) - c \sqrt{\N}/ \sinh(T) + O(1)
\endalign $$
Now we use our Useful Fact and integrate to get $S_2(n)$.
(If $b \ne 1$ the subleading term must then be divided by $\sqrt{b}$, as in
Model 1.) Since
$${dS_2 \over d\N} = \sinh(T) {dS_2 \over dn}= 4 \sinh(T) f'(T), \tag 4.2.5$$
we get
$$ S_2 = 2\N^2 \cosh(T)\sinh(T) - {8 c \over 3\sqrt{b}} \N^{3/2} + O(\N).
\tag 4.2.6 $$
In terms of $n$ this is
$$ S_2 = 2n^2 \coth(T) - {8c n^{3/2}\over 3 \sqrt{b} \sinh(T)^{3/2}} +O(n).
\tag 4.2.7$$
%\smallskip
\noindent{\bf Toy Model Number 3}
\medskip
Next we consider a functional involving our Yang-Mills fields
$a_i$:
$$ S_3(a_2,a_3) = {\pi^2} \int_0^{\pi/3} d\theta \;
\left[
(a_3')^2 G_3 + (a_2')^2 G_2
+ a_3^2/G_3 + a_2^2/G_2
+ a_2^2 a_3^2 / G_1 \right] . \tag 4.3.1 $$
This is our true Yang-Mills action functional, only with $a_1$
identically zero. It is also just a small perturbation of Model 2.
We define the new coordinate
$$ x(\theta) = - \int_\theta^{\pi/6} dy/G_3(y), \tag 4.3.2 $$
for $\theta < \pi/6$, and
$$ x(\theta) = \int^\theta_{\pi/6} dy/G_2(y), \tag 4.3.3 $$
for $\theta > \pi/6$. We also let $f(x)$=$a_2(\theta)$,
$g(x)=a_3(\theta)$, and $T=\int_0^{\pi/6} dy/G_3(y)$. Our action
functional is then
$$ \align S_3 = & \pi^2\int_{-T}^0 dx [(g')^2 + g^2 + f^2 G_3/G_2
+ (f')^2 G_2/G_3 + f^2g^2 G_3/G_1] \\
& + \pi^2 \int^T_0 dx [(f')^2 + f^2 + g^2 G_2/G_3 + (g')^2 G_3/G_2
+ f^2g^2 G_2/G_1], \tag 4.3.4 \endalign $$
which we approximate as
$$\tilde S_3 = \pi^2 \int_{-T}^T (f')^2 + (g')^2 + f^2 + g^2 +
4 f^2g^2. \tag 4.3.5$$
To the left of the overlap region, $f$ is exponentially small, so the
difference between $G_3/G_2$ and 1 contributes only $o(1)$ to the action,
and can be ignored. Similarly to the right of the overlap region,
where $g$ is small. In the overlap region $G_3/G_2=1+O(x)$, and
so the difference between $G_3/G_2$ and 1 contributes $O(1)$ to the
action (since $f^2$ and $g^2$ are $O(n)$ and the width of the region
is $O(n^{-1/2})$). Finally, in the overlap region the coefficient of
$f^2g^2$ is $4+O(x)$. The difference between this and 4 contributes
$O(n)$ to the action.
So $S_3(n)$ and $\tilde S_3(n)$ differ only by $O(n)$, and we have
$$S_3(n) = \tilde S_3(n) + O(n) = 2\pi^2 n^2 \coth(T)-
{8\pi^2 c n^{3/2}\over 3 \sqrt{2}
\sinh(T)^{3/2}} +O(n). \tag 4.3.6$$
If we do the integral for $T$ we find that
$$ \align
T & = \log(\sqrt{3}+1)-\log \sqrt{6}; \qquad \coth(T) = 4 + 3\sqrt{3}\\
& 1/\sinh(T) = \sqrt{\coth^2(T)-1} = \sqrt{42+24\sqrt{2}}.
\tag 4.3.7 \endalign $$
Plugging this back into (4.3.6) we get
$$S_3(n) = 8\pi^2(2.2990381 n^2 - 4.2246 n^{3/2}) + O(n). \tag 4.3.8$$
\medskip
\noindent{\bf Model 4. The Full System}
\medskip
Finally we consider the full Yang-Mills action:
$$ \align S_4 = \pi^2 \int_0^{\pi/3} d\theta \;&
\left [ (a_1')^2 G_1 + (a_2')^2 G_2 + (a_3')^2 G_3
\right . \\ & +
\left . (a_1+a_2a_3)^2 / G_1 + (a_2+a_1a_3)^2 / G_2 + (a_3+a_1a_2)^2/G_3
\right ]. \tag 4.4.1 \endalign $$
We claim that this differs from $S_3$ only by $O(\sqrt{n})$, and
thus that eq.(4.3.8) applies to the full action. To see this we
consider the effect of a nonzero $a_1$.
First we look at the potential energy (nonderivative) terms. To
the left of the overlap region $a_2$ is small, as is $a_2a_3$, so
a nonzero $a_1$ can only add to the potential energy. In effect,
the potential energy is changed from $a_3^2/G_3$ to
$a_3^2 (G_3^{-1} + a_1^2/G_2) + a_1^2/G_1$. Clearly, then, $a_1$
will be small in this region, and will similarly be small to the
right of the overlap region. In the overlap region, however,
a nonzero $a_1$ can help reduce the large $(a_1+a_2a_3)^2$ potential
energy term. If $a_1$ is of size $O(A)$ in this region,
$(a_1+a_2a_3)^2$ is reduced by $O(nA)$ (since $a_2$ and $a_3$ are
$O(\sqrt{n})$), which, integrated over the region of size
$O(1/\sqrt{n})$, gives a savings of $O(A\sqrt{n})$.
This must be balanced against the increase in kinetic energy,
which goes as $O(A^2\sqrt{n})$, since the change from 0 to $A$
occurs in a region of size $O(1/\sqrt{n})$. The difference between
the savings and cost is maximized when $A$ is $O(1)$, making the
net savings $O(\sqrt{n})$. In short,
$$\align S_4(n) = S_3(n) + O(\sqrt{n}) = & 2\pi^2 n^2 \coth(T)
- {8\pi^2 c n^{3/2} \over 3 \sqrt{2} \sinh(T)^{3/2}} +O(n) \\
= & 8\pi^2(2.2990381 \; n^2 - 4.2246 \; n^{3/2}) + O(n). \tag 4.4.2$$
\endalign $$
We remark that $S_4(n)$ is bounded by the quadratic term,
$2\pi^2n^2\coth(T)$, for all values of $n$, not just for $n$
sufficiently large. This is because $2\pi^2n^2\coth(T)$ is precisely
the action of the trial wavefunction
$$ a_1=0, \quad a_2=\cases 0,&x\le 0 \\ n \sinh(x)/\sinh(T), &x >0.
\endcases
\quad a_3=\cases -n \sinh(x)/\sinh(T), & x < 0 \\ 0, & x\ge 0. \endcases
\tag 4.4.3 $$
All that remains is a numerical fit of the data for
$S_4(n) - 8\pi^2(2.2990381n^2 + 4.2246 n^{3/2})$. The leading behavior
should be $O(n)$, and there should also be a $\sqrt{n}$ term, a
constant term, and an $o(1)$ remainder. There might also be a
$\log(n)$ term, from the integral of $n^{-1}$ terms in the
expansion of $f'$. (There were no such terms in the basic
Model 1, but we can't rule them out for the more complicated models).
Fitting data for $n=1,3,5, \ldots,19$ (gotten by the Fourier
expansion of Section 3), we find that
$$ S(n) \approx 8\pi^2(2.2990381 n^2 - 4.2246 n^{3/2} + 2.76 n
-1.0 \sqrt{n} + 0.16) \tag 4.4.4$$
Although we cannot rule it out, there is no evidence of a $\log(n)$
term.
Table 3 shows the $S(n)$ vs. the right hand side of (4.4.4) for
various values of $n$. Note that the estimated action (4.4.4)
is almost as accurate for $n=29$, which was not used in making the fit to
the data, as for $n$ up to 19.
% ***** INSERT TABLE 3 HERE ****
\medskip
\noindent{\bf Toy Model Number 5: $r \ne t$}
We now look at asymmetric models, with $r \ne t$.
We define the new variables
$$N=r + t; \qquad \alpha = (r-t)/N. $$
We are interested in the $O(N^2)$ and $O(N^{3/2})$ behavior of
the action for large $N$, with $\alpha$ fixed.
We consider a toy model on the interval $[-T_2, T_1]$, with action
$$S_5 = \int_{-T_2}^{T_1} (f')^2 + a^2f^2 + (g')^2 + b^2 g^2 +f^2g^2,
\tag 4.5.1$$
where $a$ and $b$ are constants, with the boundary conditions
$f(T_1)=t$, $g(-T_2)=r$. We also require that
$$ \frac{r}{t} = \frac{\sinh(bT_2)/b}{\sinh(aT_1)/a}. \tag 4.5.2$$
This insures that asymptotically the center of the overlap region,
which we call the {\it crossover point\/}, will be at
the origin. Eq.~(4.5.2) is not really an extra condition, since given
a value of $\alpha$ and an interval of size $T_1+T_2$ we can always choose
the origin in such a way that (4.5.2) holds. Note that the coefficient
of $f^2g^2$ is 1; other values of that coefficient can be treated as
in Models 1 and 2.
Unfortunately, for finite $N$ the crossover point is not exactly at
the origin, but at a point $\delta(N)$, to be computed below. Although
in the end $\delta$ does not contribute to our result, it does greatly
complicate the algebra. We
shall see that $\delta=O(N^{-1/2})$.
We proceed as in Model 2. Away from the
overlap region there is a constant $\lambda$ such that
$$ f(x) \approx \frac{\lambda^2}{a} \sinh(a(x-\delta))
+ \lambda c \cosh(a(x-\delta)), \tag 4.5.3 $$
$$ f'(x) \approx \lambda^2 \cosh(a(x-\delta))
+ \lambda ca \sinh(a(x-\delta)), \tag 4.5.4 $$
for $x$ much bigger than $\delta$, and
$$ g(x) \approx \frac{-\lambda^2}{b} \sinh(b(x-\delta))
+ \lambda c \cosh(b(x-\delta)), \tag 4.5.5 $$
$$ g'(x) \approx -\lambda^2 \cosh(b(x-\delta))
+ \lambda cb \sinh(b(x-\delta)), \tag 4.5.6 $$
for $x$ much less than $\delta$.
Evaluating at the endpoints we get
$$ \align
t=f(T_1) = & \frac{\lambda^2}{a} \sinh(a(T_1-\delta))
+ \lambda c \cosh(a(T_1-\delta)) \tag 4.5.7 \\
= & \frac{\lambda^2}{a} \sinh(aT_1)
+ \lambda c \cosh(aT_1) - \delta[\lambda^2\cosh(aT_1)+\lambda ac \sinh(aT_1)]
\endalign $$
plus $O(\delta^2)$ corrections. Similarly
$$ \align
r=g(-T_2) = & \frac{\lambda^2}{b} \sinh(b(T_2+\delta))
+ \lambda c \cosh(b(T_2+\delta)) \tag 4.5.8 \\
= & \frac{\lambda^2}{b} \sinh(bT_2)
+ \lambda c \cosh(bT_2) + \delta[\lambda^2\cosh(bT_2)+\lambda bc \sinh(bT_2)]
\endalign $$
plus $O(\delta^2)$ corrections. Looking at the derivatives we find
$$ \align f'(T_1) = &
\lambda^2\cosh(a(T_1-\delta)) + \lambda ac \sinh(a(T_1-\delta)) \\
= & \lambda^2\cosh(aT_1) + \lambda ac \sinh(aT_1) - a^2 \delta t,
\tag 4.5.9 \endalign $$
$$ \align g'(-T_2) = &
-\lambda^2\cosh(b(T_2+\delta)) - \lambda bc \sinh(b(T_2+\delta)) \\
= & -\lambda^2\cosh(bT_2) - \lambda bc \sinh(bT_2) - b^2 \delta r,
\tag 4.5.10 \endalign $$
where we have used the fact that $f'(x+\epsilon)\approx f'(x)
+ \epsilon f''(x) = f'(x) + \epsilon a^2f(x)$.
We now define the rescaled boundary value
$$M = \frac{at}{\sinh(aT_1)} = \frac{br}{\sinh(bT_2)}. \tag 4.5.11 $$
Equations (4.5.7) and (4.5.8) then become
$$ \align M = &
\lambda^2(1-a\delta \coth(aT_1)) + \lambda ac(\coth(aT_1)-a\delta)
\tag 4.5.12a \\
= & \lambda^2(1+b\delta \coth(bT_2)) + \lambda bc(\coth(bT_2)+b\delta).
\tag 4.5.12b \endalign $$
Setting these two expressions equal we can compute $\delta$:
$$ \delta = \frac {c(a\coth(aT_1)-b\coth(bT_2))}{\lambda
(a\coth(aT_1)+b\coth(bT_2)) + c(a^2+b^2)} = O(\lambda^{-1})=O(M^{-1/2}).
\tag 4.5.13 $$
Solving eq.~(4.5.12a) for $\lambda$ and expanding we find
$$ \align \lambda = & \sqrt{M} + \sqrt{M}a\delta\coth(aT_1)/2 - ca\coth(aT_1)/2
+ o(1) \tag 4.5.14a \\
\lambda^2 = & M + a\sqrt{M}\coth(aT_1)(\delta \sqrt{M}-c) + O(1),
\tag 4.5.14b \endalign $$
and so
$$ \align f'(T_1) = & (M- ac\sqrt{M}\coth(aT_1))\cosh(aT_1) +
a \delta M \coth(aT_1)\cosh(aT_1) \\ & \qquad + ac\sqrt{M} \sinh(aT_1)
- a^2\delta n_- +O(1) \\
= & a t \coth(aT_1) - c\sqrt{n_-} \left(\frac{a}{\sinh(aT_1)}\right)
^{3/2} + \frac {a^2\delta t}{\sinh^2(aT_1)} + O(1) \tag 4.5.15
\endalign $$
Similarly solving eq.~(4.5.13b) gives
$$ \align \lambda = & \sqrt{M} - \sqrt{M}b\delta\coth(bT_2)/2 - cb\coth(bT_2)/2
+ o(1) \tag 4.5.16a \\
\lambda^2 = & M - b\sqrt{M}\coth(bT_2)(\delta \sqrt{M}+c) + O(1),
\tag 4.5.16b \endalign $$
$$ \align -g'(-T_2) = & (M- bc\sqrt{M}\coth(bT_2))\cosh(bT_2) -
b \delta M \coth(bT_2)\cosh(bT_2) \\ & \qquad + bc\sqrt{M} \sinh(bT_2)
- b^2\delta r +O(1) \\
= & b r \coth(bT_2) - c\sqrt{r} \left(\frac{b}{\sinh(bT_2)}\right)
^{3/2} - \frac {b^2\delta r}{\sinh^2(bT_2)} + O(1) \tag 4.5.17
\endalign $$
We're finally ready to compute the action. By the Useful Fact we know
$$ \align dS_5 \!= & 2f'(T_1) dt - 2g'(-T_2)dr \tag 4.5.18 \\ = &
2 \left [ a t \coth(aT_1) - c\sqrt{t} \left(\frac{a}{\sinh(aT_1)}\right)
^{3\over 2}
+\frac {\delta a^2t}{\sinh^2(aT_1)} \right ] dt \\ & +
2\left [ b r \coth(bT_2) \!-\! c\sqrt{r} \left(\frac{b}{\sinh(bT_2)}\right)
^{3\over 2} -
\frac{\delta b^2r} {\sinh^2(bT_2)} \right ] dr .
\endalign $$
By eq.~(4.5.2) the terms involving $\delta$ cancel. Integrating (4.5.18)
we get at last our action
$$ S_5 = a t^2 \coth(aT_1) + br^2\coth(bT_2)
-\frac{4c}{3} \left (\frac{at}{\sinh(aT_1)} \right )^{3\over 2}
-\frac{4c}{3} \left (\frac{br}{\sinh(bT_2)} \right )^{3\over 2} +O(n).
\tag 4.5.19 $$
Note that by (4.5.2) the last two terms are equal. Also note that for
$a=b=1$, $r=t=n$, $T_1=T_2=T$, this result reduces to (4.2.7),
the result of Model 2.
\medskip
\noindent{\bf Model 6: The Full System with $r\ne t$}
\medskip
Here we look at the full Yang-Mills functional $S_4$, with $r \ne t$.
As we have seen, setting $a_1=0$ only introduces an $O(\sqrt{N})$
error in the action, which makes no difference to the $O(N^2)$ and
$O(N^{3/2})$ terms. We therefore take $a_1=0$, reducing the full action
$S_4$ to $S_3$.
Now for large $N$, the system will have a crossover point $\theta_0$,
which will be calculated below. We define the constants
$$ T = \int_0^{\theta_0} \frac{dy}{G_3(y)}; \qquad T'=\int_{\theta_0}^{\pi/3}
\frac{dy}{G_2(y)}; \qquad k=\frac{G_2(\theta_0)}{G_3(\theta_0)};
\qquad K=\frac{G_1(\theta_0)}{G_3(\theta_0)}, \tag 4.6.1 $$
and make the change of coordinates
$$ x(\theta) = - \int_\theta^{\theta_0} dy/G_3(y), \tag 4.6.2 $$
for $\theta < \theta_0$, and
$$ x(\theta) = \int^\theta_{\theta_0} k dy/G_2(y), \tag 4.6.3 $$
for $\theta \ge \theta_0$. The factor of $k$ in (4.6.3) is to make
$dx/d\theta$ continuous at $\theta_0$.
In terms of $x$ the action is
$$\align S = & \pi^2 \int_{-T}^0 a_3^2 + (a_3')^2 + a_2^2G_3/G_2
+ (a_2')^2G_2/G_3 + (a_2a_3)^2G_3/G_1 dx \tag 4.6.4 \\ &
+\pi^2 \int^{kT'}_0 \! a_2^2/k + k(a_2')^2 + a_3^2G_2/(kG_3)
+ (a_3')^2kG_3/G_2 + (a_2a_3)^2G_2/(kG_1 )dx \endalign $$
Letting $g=a_3$, $f=\sqrt{k} a_2$, we can approximate this as
$$ S \approx \pi^2 \int_{-T}^{T'} \frac{g^2}{k^2} + (g')^2
+ f^2 + (f')^2 + \frac{f^2g^2}{kK} dx. \tag 4.6.5$$
This is precisely Model 5, with $T_1=kT'$, $a=1/k$, so $aT_1=T'$,
and with $T_2=T$, $b=1$, and so $bT_2=T$, and with the factor
$1/(kK)$ multiplying the $f^2g^2$ term. Also, the boundary value at
$T'$ is $r\sqrt{k}$, not $r$.
Plugging this all into (4.5.19) we get
$$ \align
S = & \pi^2 \Big ( r^2 b\coth(bT_2) \!+\! t^2ka
\coth(aT_1) \! -\! \frac{4c(kK)^{1\over 4}}{3} \left [ \left (
\frac {at\sqrt{k}}{\sinh(aT_1)} \right )^{3\over 2} \!+\!
\left ( \frac {br}{\sinh(bT_2)} \right )^{3\over 2} \right ] \Big ) \\
= & \pi^2 (r^2\coth(T)+t^2\coth(T'))
\tag 4.6.6 \\ & -\frac{4c\pi^2}{3} \left (
\left (\frac{t}{\sinh(T')}\right)^{3\over 2} \left ( \frac{G_1(\theta_0)
G_3(\theta_0)}{G_2(\theta_0)^2}\right)^{1\over 4}
+ \left (\frac{r}{\sinh(T)}\right)^{3\over 2} \left ( \frac{G_1(\theta_0)
G_2(\theta_0)}{G_3(\theta_0)^2}\right)^{1\over 4} \right ).
\endalign $$
So all that remains to compute $S(r,t)$ through order $N^{3/2}$
is to compute $\theta_0$ (and hence $T$, $T'$, and the hyperbolic
functions of $T$ and $T'$). $\theta_0$ is determined by setting the
derivative of the $O(N^2)$ term with respect to $\theta_0$ equal
to zero, or equivalently setting the two $O(N^{3/2})$ terms equal to
each other. As with Model 4, the quadratic term gives an upper
bound to the action for all values of $N$.
The integrals (4.6.1) are not difficult, and give
$$\coth(T)=\frac{4\mu-1}{2\mu-2}=\frac{3}{2(1-\mu)}-2, \qquad
(\hbox{where }\mu\equiv\cos(\theta_0)), \tag 4.6.7 $$
with a similar result for $\coth(T')$. The derivative of the $O(N^2)$
term is then easily taken, and setting it equal to zero gives the condition
$$\frac{r^2 \sin(\theta_0)}{(1-\cos(\theta_0))^2} =
\frac{t^2 \sin(\pi/3-\theta_0)}{(1-\cos(\pi/3-\theta_0))^2}. \tag 4.6.8 $$
Given a value of $\alpha$ this equation can be solved numerically,
and plugged back into (3.8.6) to give the $N^2$ and $N^{3/2}$
coefficients. This data is compiled in Table 4.
% *****INSERT TABLE 4 HERE****************
\heading{\bf 5. Conclusions} \endheading
We have approximated certain non-self-dual solutions of the
Yang-Mills equations on the sphere. For small $(n_+,n_-)$, the
approximate solutions were obtained by truncating the Fourier expansion
of the reduced connection $a_i (\theta)$ and minimizing on the computer.
For large values of $(n_+,n_+)$, we derived asymptotic formulas, and
fitted the constants using the computer-generated solutions for small
$(n_+,n_-)$.
No solutions on Euclidean space (equivalently the sphere) are known
explicitly,
either of this symmetry type
or of any other type.
In the case of self-dual Yang-Mills solutions, t'Hooft
constructed a class of `multi-instanton'
solutions (see [JNR], [AHS] sec. 7).
In a certain limit,
these solutions have curvature localized near a finite set of points,
with approximately one unit of `topological charge' concentrated near
each point.
Although the Yang-Mills equations are nonlinear, these solutions can
be interpreted as some sort of superposition of single instanton
solutions. This procedure does not produce all self-dual
solutions, but only a subset.
It is a possibility that some non-self-dual solutions admit a similar
interpretation, as a superposition of instantons and anti-instantons.
The idea is that in one region of $S^4$, the solution would be almost
self-dual, and in another region, the solution would be almost anti-self-dual.
These regions would be connected by an overlap region.
Sibner, Sibner, and Uhlenbeck [SSU] suggest that their
non-self-dual solutions may
correspond to a superposition of an $m$-instanton and an
$m$-anti-instanton, with $m \ge 2$.
Parker's solution on $S^4$ with a modified metric [P1,P2] can be
interpreted as a superposition of an instanton and an anti-instanton at
antipodal points.
However, our non-self-dual solutions do not admit such an interpretation.
They do not approximately
satisfy the (anti)self-dual equations (2.8) anywhere on the segment
$0 \le \theta \le \pi/3$.
To the left of the overlap region, we have $a_1 \approx a_2 \approx 0$,
so the (anti)self-duality equations reduce to $-da_3/dx = \pm a_3$,
where $x$ is given by (4.6.2) and (4.6.3). Similarly, to the right
the equations are $-da_2/dx = \pm a_2$. In either region,
$e^{-x}$ is self-dual and $e^x$ is anti-self-dual. As we have
seen, our solutions go as
$\sinh(x)$, which means that near the overlap region we have an
equal mixture of self-dual and anti-self-dual. As we move away from
the overlap region towards $\theta=0$, the amplitude of the self-dual
component ($e^{-x}$) increases somewhat, while the anti-self-dual
amplitude decreases, but the ratio of the amplitudes, namely $e^{-2x}$,
is finite even at $\theta=0$ ($x=-T$).
When the asymmetry parameter $\alpha = (r-t)/(r+t)$ is equal to zero,
this ratio at $\theta=0$ is only $(2+\sqrt{3})/3\approx 1.243$.
In the asymmetric case, when $\alpha$ is positive, $T$ will be
somewhat larger, so on the left side of the overlap region, the solution
will be more self-dual and less
anti-self-dual, while on the right side of
the overlap, the proportions will differ less.
However, as long as $\alpha$ is less than 1, the ratio of self-dual
to anti-self-dual amplitudes at $\theta=0$ will remain finite.
The solutions become approximately (anti)self-dual locally
only as $\alpha$ approaches $1$.
In that limit, the crossover point moves to
$\theta=\pi/3$, $T$ becomes infinite, and the solution (which is
supported entirely on the left side of the crossover point)
becomes entirely self-dual, taking the form
$$ a_1(\theta)=a_2(\theta)=0,
\qquad a_3(\theta)=r \exp(-\int_0^\theta dy/G_3(y)). \tag 5.1 $$
In the case $(r,t)=(n,1)$ with $n$ large, our formula
(4.6.6) gives an action $S=\pi^2n^2+O(n)$, which asymptotically matches the
topological lower bound $\pi^2(n^2-1)$.
We thank Jalal Shatah for helpful discussions. Some of the numerical
computations were done at the New York University Academic Computing
Facility. The first author was partially supported by NSF Grant
DMS-8806731. The second author was partially supported by a Bantrell
Fellowship and NSF Grant DMS-8801918.
\vfill\eject
\centerline{\bf Table 1. Finite-Mode Approximations for $(n_+,n_-)=(5,3)$.}
\medskip
$$ \vbox{\offinterlineskip \hrule
\halign{\vrule#&\strut\hfil#\hfil&&\vrule#&\hfil#\cr
&&& \multispan{5}{\hfil $A_n$ for\hfil} & \cr
& n &&$l=1$\hfil&\omit&$l=2$\hfil&\omit&$l=3$\hfil& \cr \hline
&0 && -0.3245945605 && -0.3245713584 && -0.3245713855 & \cr \hline
&1 && -1.4142100298 && -1.4142096286 && -1.4142096235 & \cr \hline
&2 && -0.5362985122 && -0.5363137348 && -0.5361317255 & \cr \hline
&3 && -0.6687388537 && -0.6687367686 && -0.6687367756 & \cr \hline
&4 && 0.2040423555 && 0.2040428056 && 0.2040428225 & \cr \hline
&5 && 0.0885205177 && 0.0885274511 && 0.0885274545 & \cr \hline
&6 && -0.0043693864 && -0.0043523523 && -0.0043523755 & \cr \hline
&7 && -0.0074408414 && -0.0074473554 && -0.0074473502 & \cr \hline
&8 && -0.0016187954 && -0.0016410762 && -0.0016410774 & \cr \hline
&9 && 0.0020721870 && 0.0020742051 && 0.0020741999 & \cr \hline
&10 && 0.0005416187 && 0.0005755070 && 0.0005755227 & \cr \hline
&11 &&-0.0002029798 && -0.0001648856 && -0.0001648778 & \cr \hline
&12 && 0 && -0.0000286352 && -0.0000286438 & \cr \hline
&13 && 0 && -0.0000394540 && -0.0000394588 & \cr \hline
&14 && 0 && 0.0000014885 && 0.0000014772 & \cr \hline
&15 && 0 && -0.0000041032 && -0.0000041040 & \cr \hline
&16 && 0 && 0.0000016766 && 0.0000016566 & \cr \hline
&17 && 0 && 0.0000005393 && 0.0000005864 & \cr \hline
&18 && 0 && 0 && 0.0000000454 & \cr \hline
&19 && 0 && 0 && -0.0000000624 & \cr \hline
&20 && 0 && 0 && -0.0000000139 & \cr \hline
&21 && 0 && 0 && 0.0000000130 & \cr \hline
&22 && 0 && 0 && 0.0000000045 & \cr \hline
&23 && 0 && 0 && -0.0000000014 & \cr \hline \hline
&$S/8\pi^2$&& 11.4882182679 && 11.4882177866 && 11.4882177866 & \cr
}
\hrule} $$
\vfill\eject
\centerline{\bf Table 2. Actions of Some Low-Lying Solutions}
\smallskip
$$ \vbox{\offinterlineskip\hrule
\halign{\vrule#&\strut\quad\hfil#\hfil\quad&\vrule#&\qquad#\hfil\quad&
\vrule# \cr
& $(n_+,n_-)$ & & \hfil$S/8\pi^2$ & \cr \hline
&(3,3) & & \phantom{0}5.43281507547525 & \cr \hline
&(5,3) & & 11.4882177865800 & \cr \hline
&(7,3) & & 18.9925969277396 & \cr \hline
&(5,5) & & 21.9519857528157 & \cr \hline
&(9,3) & & 27.8614519776604 & \cr \hline
&(7,5) & & 34.1521844389385 & \cr \hline
&(11,3) & & 38.04236576029 & \cr \hline
&(9,5) & & 47.96544358790 & \cr \hline
&(13,3) & & 49.498844315 & \cr \hline
&(7,7) & & 51.23396533677 & \cr \hline
&(15,3) & & 62.20367167 & \cr \hline
&(11,5) & & 63.3120814979 & \cr \hline
&(9,7) & & 70.1032953685 & \cr \hline
&(17,3) & & 76.1355939 & \cr \hline
&(13,5) & & 80.135023031 & \cr \hline
&(11,7) & & 90.668949991 & \cr \hline
&(19,3) & & 91.277452 & \cr \hline
&(9,9) & & 94.14892253 & \cr \hline
&(15,5) & & 98.3908046 & \cr} \hrule} $$
\bigskip
\bigskip
\centerline{\bf Table 3. Numerical vs. Asymptotic Actions}
$$ \vbox{\offinterlineskip \hrule
\halign{&\vrule#&\strut\quad\hfil#\cr
& && Numerical \hfil\quad && Asymptotic \hfil\quad && & \cr
&n && $S(n,n)/8\pi^2$\hfil\quad && Formula (4.4.4) \hfil\quad &
& Difference\hfil\quad & \cr \hline
& 1 & & 0 & & -0.00059 & & 0.00059 & \cr \hline
& 3 & & 5.43282 & & 5.45249 & & -0.01967 & \cr \hline
& 5 & & 21.95199 && 21.97212 & & -0.02013 & \cr \hline
& 7 & & 51.23397 && 51.25092 & & -0.01695 & \cr \hline
& 9 & & 94.14892 && 94.16215 & & -0.01323 & \cr \hline
& 11 & & 151.25581 && 151.26545 & & -0.00964 & \cr \hline
& 13 & & 222.95467 && 222.95946 & & -0.00479 & \cr \hline
& 15 & & 309.54623 && 309.54693 & & -0.00070 & \cr \hline
& 17 & & 411.26764 && 411.26797 & & -0.00033 & \cr \hline
& 19 & & 528.32139 && 528.31912 & & 0.00227 & \cr \hline
& 29 & & 1348.57662 && 1348.55178 & & 0.02484 & \cr}
\hrule} $$
\vfill \eject
\centerline{\bf Table 4. Coefficients for $S/8\pi^2$}
$$ \vbox{\offinterlineskip \hrule
\halign{&\vrule#&\strut\quad\hfil#\hfil\quad \cr
& $\alpha$ && $N^2$ coefficient && $N^{3/2}$ coefficient && $\theta_0$ & \cr
\hline
& 0 && 0.5747595264 && -1.4936 && 0.5235987756 & \cr \hline
& 0.1 && 0.5713916984 && -1.4878 && 0.5585551849 & \cr \hline
& 0.2 && 0.5612469955 && -1.4703 && 0.5939076369 & \cr \hline
& 0.3 && 0.5441954125 && -1.4405 && 0.6300877011 & \cr \hline
& 0.4 && 0.5199966585 && -1.3976 && 0.6676074578 & \cr \hline
& 0.5 && 0.4882542493 && -1.3401 && 0.7071288703 & \cr \hline
& 0.6 && 0.4483208416 && -1.2657 && 0.7495894773 & \cr \hline
& 0.7 && 0.3990863929 && -1.1701 && 0.7964690692 & \cr \hline
& 0.8 && 0.3384135611 && -1.0444 && 0.8504824968 & \cr \hline
& 0.9 && 0.2610245328 && -0.8648 && 0.9181185734 & \cr \hline
& 0.91 && 0.2518690752 && -0.8414 && 0.9262302578 & \cr \hline
& 0.92 && 0.2423379341 && -0.8163 && 0.9347270475 & \cr \hline
& 0.93 && 0.2323739184 && -0.7892 && 0.9436758704 & \cr \hline
& 0.94 && 0.2218998821 && -0.7597 && 0.9531664337 & \cr \hline
& 0.95 && 0.2108070566 && -0.7268 && 0.9633242998 & \cr \hline
& 0.96 && 0.1989326069 && -0.6896 && 0.9743358721 & \cr \hline
& 0.97 && 0.1860107315 && -0.6459 && 0.9865024134 & \cr \hline
& 0.98 && 0.1715437937 && -0.5912 && 1.0003811278 & \cr \hline
& 0.99 && 0.1543234661 && -0.5129 && 1.0173034421 & \cr \hline
%& 0.991 && 0.1523371308 && -0.5024 && 1.0192858444 & \cr \hline
& 0.992 && 0.1502760714 && -0.4910 && 1.0213497358 & \cr \hline
%& 0.993 && 0.1481274280 && -0.4785 && 1.0235088791 & \cr \hline
& 0.994 && 0.1458738872 && -0.4646 && 1.0257817480 & \cr \hline
%& 0.995 && 0.1434910933 && -0.4489 && 1.0281942478 & \cr \hline
& 0.996 && 0.1409426774 && -0.4306 && 1.0307849275 & \cr \hline
%& 0.997 && 0.1381694543 && -0.4085 && 1.0336162816 & \cr \hline
& 0.998 && 0.1350610333 && -0.3797 && 1.0368043702 & \cr \hline
& 0.999 && 0.1313507206 && -0.3360 && 1.0406281236 & \cr \hline
%& 0.9991 && 0.1309217765 && -0.3299 && 1.0410713063 & \cr \hline
%& 0.9992 && 0.1304764592 && -0.3232 && 1.0415316160 & \cr \hline
%& 0.9993 && 0.1300119567 && -0.3158 && 1.0420119684 & \cr \hline
%& 0.9994 && 0.1295244877 && -0.3075 && 1.0425162820 & \cr \hline
& 0.9995 && 0.1290087378 && -0.2980 && 1.0430500586 & \cr \hline
%& 0.9996 && 0.1284567810 && -0.2868 && 1.0436214967 & \cr \hline
%& 0.9997 && 0.1278557357 && -0.2731 && 1.0442439079 & \cr \hline
%& 0.9998 && 0.1271816109 && -0.2549 && 1.0449420563 & \cr \hline
& 0.9999 && 0.1263765530 && -0.2268 && 1.0457755838 & \cr}
\hrule} $$
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\enddocument
\bye