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\begin{document}
\title[Atiyah's Conjecture]
{A Proof of Atiyah's Conjecture on Configurations of Four Points in Euclidean
Three-Space}
\author[Michael Eastwood]{Michael Eastwood}
\author[Paul Norbury]{Paul Norbury}
\address{\hskip-\parindent
Pure Mathematics Department\\
Adelaide University\\
South Australia 5005}
\email{meastwoo@maths.adelaide.edu.au}
\email{pnorbury@maths.adelaide.edu.au}
\thanks{Support from the Australian Research Council and the Mathematical
Sciences Research Institute is gratefully acknowledged.
Research at MSRI is supported in part by NSF grant DMS-9701755.}
\begin{abstract}
From a configuration of $n$ points in Euclidean three-space, Atiyah constructed
a determinant and conjectured that it was always non-zero.  In this article we
prove the conjecture for the case of four points.
\end{abstract}
\maketitle
\renewcommand{\thefootnote}{\sharp}

Consider $n$ points in Euclidean three-space.  Fixing attention on one of
these points, the others gives rise to $n-1$ points on its sphere of
vision.  Thinking of this as the Riemann sphere gives a polynomial of
degree $n-1$, having as its zeroes these $n-1$ points.  This
polynomial is only defined up to scale but this is not going to matter
for the simplest conjecture: just choose any scale and regard its
coefficients as a complex $n$-vector.  Repeating this exercise for
each of the $n$ points gives $n$ such vectors and hence an $n\times n$
matrix.  In~\cite{a1,a2}, Atiyah conjectured that a matrix constructed
in this way cannot be singular.  In~\cite{as}, Atiyah and Sutcliffe
amass a great deal of numerical evidence for this conjecture and
formulate a series of further conjectures based on the geometry that
their numerical studies apparently reveal.

In spite of overwhelming evidence in its favour, the basic conjecture, as
stated above, remains surprisingly resistant. The case $n=3$ is not too hard:
a geometric argument is given in \cite{a1} and an algebraic one in~\cite{a2}.
In this article we establish the case $n=4$.

\section{Normalisation}
In describing Atiyah's conjecture above we were satisfied with
the directions defined by pairs of points amongst the $n$ points.  Now we take
the scale into account. For this, we may use the Hopf mapping
$${\mathbb C}^2\ni\left\lgroup\!\begin{array}c w_1\\ w_2
\end{array}\!\right\rgroup
\stackrel{h}{\longmapsto}
\left\lgroup\!\begin{array}c
(|w_1|^2-|w_2|^2)/2\\ w_1\bar{w}_2\end{array}\!\right\rgroup
\in{\mathbb R}\times{\mathbb C}\cong{\mathbb R}^3$$
as follows.  This mapping intertwines the action of ${\mathrm{SU}}(2)$
on ${\mathbb C}^2$ with the action of ${\mathrm{SO}}(3)$ on~${\mathbb
R}^3$ and descends to an isomorphism from ${\mathbb{CP}}_1$ to the
sphere of rays through the origin in ${\mathbb R}^3$.  Therefore, for
each point in ${\mathbb R}^3\setminus\{0\}$, we may choose a
corresponding point in ${\mathbb C}^2\setminus\{0\}$ defined up to
phase.  If we make such choices in formulating Atiyah's conjecture,
then the determinant $\det M$ of the resulting $n\times n$ matrix is
well-defined up to phase.

Following Atiyah and Sutcliffe~\cite{as}, we may normalize $\det M$ further.
Consider the mapping
$${\mathbb C}^2\ni\left\lgroup\!\begin{array}c w_1\\ w_2
\end{array}\!\right\rgroup
\stackrel{\sigma}{\longmapsto}\left\lgroup\!
\begin{array}c -\bar{w}_2\\ \bar{w}_1\end{array}\!\right\rgroup
\in{\mathbb C}^2,$$
observing that $h(\sigma(w))=-h(w)$ for all~$w\in{\mathbb C}^2$.
Also note that
\begin{equation}\label{phase}
\sigma(e^{i\theta}w)=e^{-i\theta}\sigma(w)\quad\mbox{and}\quad
\sigma(\sigma(w))=-w.
\end{equation}
Fix an ordering for our original $n$ points in ${\mathbb R}^3$.  Each
pair of these points contributes twice to $\det M$, once when the
later point is viewed from the earlier and once when this view is
reversed.  We mandate using $w$ and $\sigma(w)$, respectively, in
lifting to ${\mathbb C}^2$.  By virtue of~(\ref{phase}), both the
phase ambiguity $w\mapsto e^{i\theta}w$ and the ordering ambiguity
cancel from~$\det M$.  In conclusion, $\det M$ is invariant under
Euclidean motions.  It is easy to check that $\det M$ is replaced by
its complex conjugate under reflection.  In particular, if all points
lie in a plane then the determinant is real.  For further details
see~\cite{a2,as}.  We shall call~$\det M$, normalised in this way, the
{\em Atiyah determinant}.  In~\cite{as} a scale invariant
normalisation $D$ is used.  The two normalisations are related by
\[\det M=D\prod_{i>j}(2r_{ij})\] where $r_{ij}$ is the distance between
the $i^{\mathrm{th}}$ and $j^{\mathrm{th}}$ points.

We are free to use Euclidean motions to place points in convenient
locations.  Let us do this to verify the conjecture when $n=3$,
choosing the three points in ${\mathbb R}\times{\mathbb C}$ to be
$$\left\lgroup\!\begin{array}c0\\ 0\end{array}\!\right\rgroup\quad
\left\lgroup\!\begin{array}c0\\ a\end{array}\!\right\rgroup\quad
\left\lgroup\!\begin{array}c0\\ z\end{array}\!\right\rgroup$$
with $a$ real.
They form a triangle with side lengths
$a$, $b=|z|$, and $c=|a-z|$.
We may use the following
$$\begin{array}{ll}
\frac{1}{\sqrt a}\left\lgroup\!\begin{array}c a\\ a\end{array}\!\right\rgroup
\stackrel{h}{\mapsto}
\left\lgroup\!\begin{array}c0\\a\end{array}\!\right\rgroup&
\frac{1}{\sqrt b}\left\lgroup\!\begin{array}c b\\ \bar{z}\end{array}
\!\right\rgroup\stackrel{h}{\mapsto}
\left\lgroup\!\begin{array}c0\\ z\end{array}\!\right\rgroup\\ \\[-5pt]
\frac{1}{\sqrt a}\left\lgroup\!\begin{array}c -a\\ a\end{array}\!\right\rgroup
\stackrel{h}{\mapsto}
\left\lgroup\!\begin{array}c0\\-a\end{array}\!\right\rgroup&
\frac{1}{\sqrt c}\left\lgroup\!\begin{array}c c\\ \bar{z}-a
\end{array}\!\right\rgroup\stackrel{h}{\mapsto}
\left\lgroup\!\begin{array}c0\\ z-a\end{array}\!\right\rgroup\\ \\[-5pt]
\frac{1}{\sqrt b}\left\lgroup\!\begin{array}c -z\\ b\end{array}
\!\right\rgroup\stackrel{h}{\mapsto}
\left\lgroup\!\begin{array}c0\\ -z\end{array}\!\right\rgroup&
\frac{1}{\sqrt c}\left\lgroup\!\begin{array}c a-z\\
c\end{array}\!\right\rgroup
\stackrel{h}{\mapsto}
\left\lgroup\!\begin{array}c0\\a-z\end{array}\!\right\rgroup
\end{array}$$
in computing $\det M$.  We obtain
$$\begin{array}l
\displaystyle\frac{1}{abc}\left|\begin{array}{ccc}
ab&-ac&-z(a-z)\\
a\bar{z}+ab&-a(\bar{z}-a)+ac& -zc+b(a-z)\\
a\bar{z}&a(\bar{z}-a)&bc
\end{array}\right|\\ \\[-8pt]
\quad=a((z+\bar{z})(c-a-b)+2b(a+b+3c))\\[4pt]
\qquad=(a^2+b^2-c^2)(c-a-b)+2ab(a+b+3c)\\[4pt]
\quad\qquad=d_3(a,b,c)+8abc,\end{array}$$
where
\begin{equation}\label{deethree}d_3(a,b,c)=(a+b-c)(b+c-a)(c+a-b).\end{equation}
The triangle inequalities imply that $d_3(a,b,c)\geq 0$ with equality if and
only if the points lie on a line.  Therefore $\det M\geq 8abc>0$ and, in
particular, is non-zero, as required.

\section{The case $n=4$}
\begin{thm}
For any four points in ${\mathbb R}^3$, the Atiyah determinant is
non-zero.
\end{thm}
\begin{proof}
Choose the four points in ${\mathbb R}^3={\mathbb R}\times{\mathbb C}$
to be
$$\left\lgroup\!\begin{array}c0\\ z_1\end{array}\!\right\rgroup\quad
\left\lgroup\!\begin{array}c0\\ z_2\end{array}\!\right\rgroup\quad
\left\lgroup\!\begin{array}c0\\ z_3\end{array}\!\right\rgroup\quad
\left\lgroup\!\begin{array}cr\\ 0\end{array}\!\right\rgroup.$$ Put
$z_{ij}=z_i-z_j$ for $i>j$ and label the distances between points
by~$r_{ij}$.  We define $z_4=0$ so that $z_{4j}=-z_j$.  Thus,
$r_{ij}^2=|z_{ij}|^2$ for $i<4$ and $r_{4j}^2=r^2+|z_{4j}|^2$.

For $j<i<4$ the vector running from the $j^{\mathrm{th}}$ point to the
$i^{\mathrm{th}}$ point may be lifted to
\[w=\frac{1}{\sqrt{r_{ij}}}\left\lgroup\!\begin{array}c r_{ij}\\
\bar{z}_{ij} \end{array}\!\right\rgroup\mbox{, with }\sigma(w)=
\frac{1}{\sqrt{r_{ij}}}\left\lgroup\!\begin{array}c -z_{ij}\\
r_{ij} \end{array}\!\right\rgroup.\]
Similarly, if we put $R_{4j}=r_{4j}+r$ for $j<4$, then
\[w=\frac{1}{\sqrt{R_{4j}}}\left\lgroup\!\begin{array}cR_{4j}\\
\bar{z}_{4j}\end{array}\!\right\rgroup\mbox{ and }
\sigma(w)=\frac{1}{\sqrt{R_{4j}}}\left\lgroup\!\begin{array}c -z_{4j}\\R_{4j}
\end{array}\!\right\rgroup\]
lift to ${\mathbb C}^2$ the vectors in ${\mathbb R}^3$ joining the
$j^{\mathrm{th}}$ point to the $4^{\mathrm{th}}$ point and vice versa.
For each of the four points, the coefficients of the corresponding third
degree polynomial give the following four vectors:--
\[v_1=\frac{1}{\sqrt{r_{21}r_{31}R_{41}}}\left\lgroup\!\begin{array}{c}
r_{21}r_{31}R_{41}\\
r_{21}r_{31}\bar{z}_{41}+r_{21}R_{41}\bar{z}_{31}+r_{31}R_{41}\bar{z}_{21}\\
r_{21}\bar{z}_{31}\bar{z}_{41}+r_{31}\bar{z}_{21}\bar{z}_{41}
+R_{41}\bar{z}_{21}\bar{z}_{31}\\
\bar{z}_{21}\bar{z}_{31}\bar{z}_{41}\end{array}\!\right\rgroup\]
\[v_2=\frac{1}{\sqrt{r_{32}R_{42}r_{21}}}\left\lgroup\!\begin{array}{c}
-r_{32}R_{42}z_{21}\\-r_{32}z_{21}\bar{z}_{42}+r_{32}R_{42}r_{21}
-z_{21}R_{42}\bar{z}_{32}\\
r_{32}r_{21}\bar{z}_{42}-z_{21}\bar{z}_{32}\bar{z}_{42}
+R_{42}\bar{z}_{32}r_{21}\\
\bar{z}_{32}r_{21}\bar{z}_{42}\end{array}\!\right\rgroup\]
\[v_3=\frac{1}{\sqrt{R_{43}r_{31}r_{32}}}\left\lgroup\!\begin{array}{c}
z_{31}z_{32}R_{43}\\
z_{31}z_{32}\bar{z}_{43}-z_{31}R_{43}r_{32}-z_{32}R_{43}r_{31}\\
-z_{31}r_{32}\bar{z}_{43}-z_{32}r_{31}\bar{z}_{43}+R_{43}r_{31}r_{32}\\
r_{31}r_{32}\bar{z}_{43}\end{array}\!\right\rgroup\]
\[v_4=\frac{1}{\sqrt{R_{41}R_{42}R_{43}}}\left\lgroup\!
\begin{array}{c}
-z_{41}z_{42}z_{43}\\
z_{41}z_{42}R_{43}+z_{41}z_{43}R_{42}+z_{42}z_{43}R_{41}\\
-z_{41}R_{42}R_{43}-z_{42}R_{41}R_{43}-z_{43}R_{41}R_{42}\\
R_{41}R_{42}R_{43}\end{array}\!\right\rgroup\]
and we may take $M$ to be the matrix with column vectors~$v_i$.  Hence,
\[\det M=P/(r_{21}r_{31}r_{32}R_{41}R_{42}R_{43})\]
where $P$ is a polynomial consisting of monomials each of which
contains one of $r_{ij}^2$, $r_{ij}z_{ij}$, $r_{ij}\bar{z}_{ij}$, or
$z_{ij}\bar{z}_{ij}$ for each $j<i<4$, and one of $R_{4j}^2$,
$R_{4j}z_{4j}$, $R_{4j}\bar{z}_{4j}$, or $z_{4j}\bar{z}_{4j}$.  Since
$z_{ij}\bar{z}_{ij}=r_{ij}^2$, each monomial is divisible by $r_{ij}$
and, since $|z_{4j}|^2=R_{4j}(r_{4j}-r)$, each monomial is divisible
by $R_{4j}$. Therefore, we can divide by the factor of
$r_{21}r_{31}r_{32}R_{41}R_{42}R_{43}$ leaving monomials with one of
$r_{ij}$, $z_{ij}$ or $\bar{z}_{ij}$ for each $j<i<4$, and one of
$(r_{4j}+r)$, $z_{4j}$, $\bar{z}_{4j}$ or $(r_{4j}-r)$.  It follows that
$\det M$ is now expressed as a homogeneous degree 6 polynomial in $r$,
$r_{ij}$, $z_{ij}$ and $\bar{z}_{ij}$ for $j<i\leq 4$.

Recall that $\det M$ is invariant under Euclidean motions.  Moreover, the six
distances $r_{ij}$ determine our configuration of four points.  Also, notice
that these distances are constrained only by triangle inequalities.  Hence, the
Atiyah determinant $\det M$ may be regarded as a function of the independent
variables~$r_{ij}$.  We claim that
\begin{itemize}
\item $\Re(\det M)$ is a polynomial in~$r_{ij}$ (homogeneous of degree 6).
\item $|\det M|^2$ is a polynomial in~$r_{ij}$ (homogeneous of degree 12).
\end{itemize}
Having done this we shall use the triangle inequalities on the four
faces of our configuration to show that, in fact, $\Re(\det M)>0$.
This is more than sufficient to finish the proof.

It is convenient to set $z_{ij}=-z_{ji}$ and $r_{ij}=r_{ji}$ when
$j>i$.  The monomials in our expression for $\det M$ contain an equal
number of $z_{ij}$ and $\bar{z}_{kl}$.  Consider the product
$z_{ij}\bar{z}_{kl}$.  There are two cases:-- \newline
\indent
(i) $\{i,j\}$ and $\{k,l\}$ have an element in common (suppose $l=j$):
\begin{equation}  \label{eq:adef}
    z_{ij}\bar{z}_{kj}=(1/2)\left(r_{ij}^2+r_{kj}^2-r_{ki}^2\right)+
    2A_{ijk}\sqrt{-1}
\end{equation}
where $A_{ijk}$, defined by (\ref{eq:adef}), equals plus or minus 
the area of the $ijk^{\mathrm{th}}$ triangle under the projection
${\mathbb R}^3={\mathbb R}\times{\mathbb C}\to{\mathbb C}$ onto the
complex plane; \newline
\indent(ii) $\{ i,j,k,l\}=\{1,2,3,4\}$:
\[z_{ij}\bar{z}_{kl}=z_{ij}\bar{z}_{kj}z_{kj}\bar{z}_{kl}/r_{kj}^2\]
and we can rewrite the numerator as in (i).\newline
We claim that all quadratic expressions in the $A_{ijk}$ may be
written as polynomials in the $r_{ij}$ and $r^2$.  Specifically, when
all four points lie in the complex plane, then one may verify that
\begin{equation}\label{quad}
16A_{ijk}A_{ijl}=2r_{ij}^2(r_{ik}^2+r_{il}^2-r_{kl}^2)
-(r_{ij}^2+r_{ik}^2-r_{jk}^2)(r_{ij}^2+r_{il}^2-r_{jl}^2)
\end{equation}
and when the fourth point lies off the plane ($r>0$), we replace
$r_{4j}^2$ by $r_{4j}^2-r^2$.

Now, observe that our formulae so far for $\Re(\det M)$ and $(\Im(\det M))^2$
involve only quadratics expressions in $A_{ijk}$.  If we substitute according
to (\ref{quad}) and its non-planar version, we obtain rational expressions for
$\Re(\det M)$ and $(\Im(\det M))^2$ in the seven quantities in $r_{ij}$
and $r$, the denominator being a polynomial in the six
variables~$r_{ij}$.  Recall that a reflection such as $r\mapsto -r$ conjugates
$\det M$.  Hence, we may drop all odd powers of $r$ in the numerators, to
obtain polynomials in $r_{ij}$ and~$r^2$.

Finally, we eliminate~$r^2$ from these expressions.  This is possible
by writing the volume $V$ of the tetrahedron with vertices
our four points in two different ways.  On the one hand
\begin{eqnarray*}
    144V^2&=&-r_{21}^4r_{43}^2-r_{21}^2r_{43}^4-r_{32}^2r_{41}^4
    -r_{32}^4r_{41}^2-r_{31}^4r_{42}^2-r_{31}^2r_{42}^4\\
    &&+r_{21}^2r_{43}^2r_{31}^2+r_{21}^2r_{43}^2r_{41}^2
   -r_{21}^2r_{42}^2r_{41}^2+r_{21}^2r_{42}^2r_{43}^2\\
   &&+r_{21}^2r_{42}^2r_{31}^2+r_{21}^2r_{32}^2r_{43}^2
   -r_{21}^2r_{32}^2r_{31}^2+r_{32}^2r_{42}^2r_{41}^2\\
   &&+r_{31}^2r_{42}^2r_{41}^2+r_{32}^2r_{43}^2r_{41}^2
   -r_{32}^2r_{42}^2r_{43}^2+r_{32}^2r_{42}^2r_{31}^2\\
   &&+r_{31}^2r_{42}^2r_{43}^2+r_{32}^2r_{31}^2r_{41}^2
   -r_{31}^2r_{43}^2r_{41}^2+r_{21}^2r_{32}^2r_{41}^2.
\end{eqnarray*}
On the other hand, let $A$ denote the area of the
triangular base in~${\mathbb C}$. Then
\[ 16A^2=2r_{21}^2r_{31}^2+2r_{21}^2r_{32}^2+2r_{31}^2r_{32}^2-r_{21}^4
-r_{31}^4-r_{32}^4\]
and $V=rA/3$.  We may therefore replace $r^2$ by
$9V^2/A^2$, as required.

Thus, we may conclude that $\Re(\det M)$ and $(\Im(\det M))^2$ are rational
functions of the variables $r_{ij}$ but, if we now clear any common factors, we
claim they are, in fact, polynomials.  To see this, notice that in (ii) we had
a choice when we introduced $\bar{z}_{kj}z_{kj}/r_{kj}^2$.  We could have
insisted that $\{k,j\}\subset\{1,2,3\}$.  Then the denominators of
$\Re(\det M)$ and $(\Im(\det M))^2$ would not involve~$r_{4j}$.  However,
$\det M$ does not see the ordering of our four points.  Hence, if $r_{4j}$ are
omitted from the denominators, then so are all variables~$r_{ij}$, as required.

It remains to calculate these polynomials.  We did this using
Maple\footnote{The program is at
ftp://ftp.maths.adelaide.edu.au/meastwood/maple/points.} and found
that $\Re(\det M)$ is a homogeneous polynomial of degree 6 with 226
terms and $|\det M|^2$ is a homogeneous degree 12 polynomial with 4500
terms.  We claim that $\Re(\det M)>0$.  To see this, we can rewrite
the output of the Maple calculation as follows:--
\begin{eqnarray*}
\Re(\det M)&\!=\!&64r_{21}r_{31}r_{32}r_{41}r_{42}r_{43}
    -4d_3(r_{21}r_{43},r_{31}r_{42},r_{32}r_{41})\\
    &&\;+12\av\left(r_{41}((r_{42}+r_{43})^2-r_{32}^2)
    d_3(r_{21},r_{31},r_{32})\right)+288V^2.
\end{eqnarray*}
Here, $d_3(a,b,c)$ is the polynomial~(\ref{deethree}) and $\av$
denotes the operation of averaging a polynomial in $r_{ij}$ under the action
of ${\mathcal S}_4$ on the vertices of our tetrahedron: for example,
$$\begin{array}{rcl}
\av(r_{21})&=&(r_{21}+r_{31}+r_{32}+r_{41}+r_{42}+r_{43})/6\\[4pt]
\av(r_{21}r_{43})&=&(r_{21}r_{43}+r_{31}r_{42}+r_{41}r_{32})/3.\end{array}$$
The final two terms are non-negative since the triangle inequality gives
$(r_{42}+r_{43})^2\geq r_{32}^2$ and $d_3(r_{21},r_{31},r_{32})\geq 0$,
and the square of the volume is non-negative.  To estimate the other terms we
may use the easily verified inequality
\[ abc\geq d_3(a,b,c),\quad \forall\ a,b,c\geq 0.\]
In conclusion,
\[\Re(\det M)\geq 60r_{21}r_{31}r_{32}r_{41}r_{42}r_{43}>0,\]
as required.  This is nearly enough for a stronger conjecture of
Atiyah and Sutcliffe \cite[Conjecture 2]{as} that
$|\det M|\geq 64r_{21}r_{31}r_{32}r_{41}r_{42}r_{43}$.
\end{proof}
A third conjecture of Atiyah and Sutcliffe \cite[Conjecture 3]{as} can
be expressed in the four point case in terms of polynomials in the
edge lengths as:
\[|\det M|^2\geq \prod_1^4 (d_3(r_{ij},r_{ik},r_{jk})+8r_{ij}r_{ik}r_{jk})\]
where the product runs over the four faces of the tetrahedron and the
left hand side is known explicitly.  We have been unable to prove this
conjecture even in the case that the four points lie in a plane (in which
case $|\det M|$ can be replaced by the simpler expression $\Re(\det M)$
given above).

\section{The planar case.}
Atiyah's basic conjecture is unresolved in general, 
even when the $n$ points lie in a
plane (in which case recall that $\det M$ is real).  Reasoning analogous to the
case of four points gives the following.
\begin{thm}
    The Atiyah determinant of $n$ points in a plane can be expressed
    as a rational function in the distances between the points.
\end{thm}
\begin{proof}
    Again, we can express $z_{ij}\bar{z}_{kl}$ as a rational function in
    the $r_{ij}$ and $A_{ijk}$.  It is no longer true that quadratic
    expressions in the $A_{ijk}$ are polynomials in the $r_{ij}$.
    Instead they are rational functions in the $r_{ij}$.  This uses the
    same trick of introducing new points in common between two
    triangles in order to apply (\ref{quad}):--
    \[ A_{ijk}A_{lmn}=\frac{(A_{ijk}A_{ijn})(A_{imn}A_{lmn})}
    {A_{ijn}A_{imn}}.\]
\end{proof}
In the general four point case, the distances $r_{ij}$ acted as variables.  The
denominator was too small to be appropriately symmetrical and therefore had to
divide the numerator, leaving a polynomial rather than a rational function.  In
the planar case (and also in the general case with more than four points), the
distances between points satisfy a set of polynomial constraints.
Symmetry arguments are no longer valid and expressions for the determinant are
no longer unique.  We suspect, however, that there is a polynomial expression.
\begin{thebibliography}X
\bibitem{a1}M.F. Atiyah,
{\em The geometry of classical particles},
Surveys in Differential Geometry vol.~7, International Press 2001, to appear.
\bibitem{a2}M.F. Atiyah,
{\em Configurations of points},
Phil. Trans. Roy. Soc. Lond. {\bf A359} (2001), to appear.
\bibitem{as}M.F. Atiyah and P.M. Sutcliffe,
{\em The geometry of point particles},
hep-th/ 0105179.

\end{thebibliography}
\end{document}