数理解析研究所講究録 1228 巻 2001 年 153-160

153
The group-quark matrix ?
Koichiro Harada / 原田耕一郎

Ohio State University / オハイオ州立大学\S 1

The group-quark matrix ?

The main purpose of this article is to propose aproblem. Let us
consider the following
matrix whose entries are finite groups.
$3\cross 3$

$A$

$=\{$

$2E_{6}(2).2U_{6}(2).2U_{4}(2).2$$Conway_{2}Fischer_{4}S_{6}(2)$$Conway_{3}MonsterO_{8}^{+}(2)]$The orders of relevant simple groups are :$|U_{4}(2)|=25920$ $=2^{6}3^{4}5$$|S_{6}(2)|=1451520$ $=2^{9}3^{4}5.7$$|O_{8}^{+}(2)|=174182400$ $=2^{12}3^{5}5^{2}7$$|U_{6}(2)|=2^{15}3^{6}5.7.11$$|Conway_{2}|=2^{18}3^{6}5^{3}7.11.23$$|Conway_{3}|=2^{21}3^{9}5^{4}7^{2}11.13.23$$|^{2}E_{6}(2)|=2^{36}3^{9}5^{2}7^{2}11.13.17.19$$|Fischer_{4}|=2^{41}3^{13}5^{6}7^{2}11.13.17.19.23.31.47$$|Monster|$ $=2^{46}3^{20}5^{9}7^{6}11^{2}13^{3}17.19.23.29.31.41.47.59.71$

Note that Conway groups are numbered according to their orders.
In particular, $|Conway_{1}|=2^{10}3^{7}5^{3}7.11.23$ . $U_{4}(2).2$ is the extension
ofby an outer automorphism of order 2, and $U_{6}(2).2$ and$2E_{6}(2).2$ are analogously
defined.
The columns of the matrix $A$ are indexed by the Dynkin diagramsof type ,and . Appearing in the first row of $A$ are the simple
components of the Weyl groups of type, E7 and. The correctindexing of the rows of the matrix $A$ is left for the future research.
We can, perhaps, index the rows of $A$ by three generations of quarks$U_{4}(2)$$E_{6}$$E_{7}$$E_{8}$$E_{6}$$ud$, $cs$ ,$tb$(up-down,charm-strange,top-bottom)$E_{8}$

154
Let us next give the ‘transpose-inverse$=\mathrm{t}\mathrm{r}\mathrm{a}- \mathrm{i}\mathrm{n}\mathrm{v}$$21+6^{\mathrm{r}_{2^{1+8}.S_{6}(2)}}.(S_{3}\cross S_{3})2^{1+22}Conway_{2}$${}^{t}A^{-1}=[2^{1+4}.(S_{3}\cross S_{3})2^{1+20}.U_{6}(2).22^{1+8}.U_{4}(2).2$’of the matrix $A$ .

$2^{1+8}.(S_{3}\cross S_{3}\cross S_{3})2^{1+24}Conway_{3}2^{1+8}O_{8}^{+}(2)]$Ifis the $(i,j)$ entry of the matrix $A$ , then the correspondis the centralizer of an involution in the centering entry of. Heredenotes theof aSylow 2-subgroup of the group, which. An exception isextral-special group of orderis almost extra-special but not exactly so. The main problem pr0posed here is: Investigate the group-quark matrix $A$ algebr0geometrically.
$A_{ij}$${}^{t}A^{-1}$$2^{1+2\mathrm{n}}$$A_{ij}$$2^{1+2n}$\S 2$\Gamma_{27}$and$2^{1+6^{*}}$$\Gamma_{28}$Let be the cubic surface defined in the projective spaceby the equations:$S$$P^{4}(\mathbb{C})$$x_{0}^{3}+x_{1}^{3}+x_{2}^{3}+x_{3}^{3}+x_{4}^{3}=0$$\{$$x_{0}+x_{1}+x_{2}+x_{3}+x_{4}=0$.

The (projective) line defined by$x_{0}=0$$\{$$x_{1}+x_{2}=0$$x_{3}+x_{4}=0$lies completley on the surface . Applying the permutations on theindex set {0, 1, 2, 3, 4}, 15 lines on can be obtained.
be azero of the quadratic equation :Next, let
$S$

$S$$\alpha(=1\pm\sqrt{5}\overline{\overline{2}})$$X^{2}-X-1=0$ ,then the line defined by:$\{\begin{array}{l}x_{0}+\alpha x_{3}+x_{4}=0x_{1}+x_{3}+\alpha x_{4}=0x_{2}-\alpha(x_{3}+x_{4})=0\end{array}$is also completely on the surface . Applying the permutations on{0, 1, 2, 3, 4} again, 12 lines can be obtained. Therefore there are
$S$

155
altogether 27 lines on S. That this is the exact number of lines oncomes from the theory of algebraic geomerty, although our special
case itself was known already in the middle of the 19th century.
$S$

Theorem. Ageneral (complex) cubic surface contains exactly
27 lines.

Let
be the graph of 27 lines with their configuration on ageneral cubic surface. Thensatisfies the following properties :$\Gamma_{27}$$\Gamma_{27}$(1). Any lineofmeets exactly ten other lines of F27.
Those ten lines split into five pairs,,, and if
andmeet and the triangle$i=1,2,3,4,5$ , thenisformed. There are 5 $27/3=45$ triangles so formed. (Note. If $i\neq j$ ,thenanddo not meet. In particular, there are no three linesthat meet at apoint. This applies to ageneral cubic surface. A
specialization of it may contain three lines that meet at apoint.)$A$$\Gamma_{27}$$(B_{1}, C_{1})$$B_{i}$$\ldots$$(B_{5}, C_{5})$$AB_{i}C_{i}$$C_{i}$$\cdot$$B_{i}$$C_{j}$(2). Let $ABC$ , $A’B’C’$ be any two triangles having no side in common. Then they determine uniquely athird triangle $A’B’C’$ suchthat each of three triples of lines $\{A, A’, A’\}$ , $\{B, B’, B’\}$ , $\{C, C’, C’\}$intersect and form three new triangles $AA’A’$ , $BB’B’$ , $CC’C^{u}$ .

Those two properties (1), (2) uniquely determines the configuration of 45 triangles formed by the elements of.
$\Gamma_{27}$Theorem (C.Jordan). Aut$(\Gamma_{27})\cong U_{4}(2).2\cong \mathrm{A}\mathrm{u}\mathrm{t}(U_{4}(2))$

This is the $(1, 1)$ entry of the matrix $A$ . The isomorphisms ofsimple groups$U_{4}(2)\cong S_{4}(3)\cong O_{5}(3)\cong O_{6}^{-}(2)$is significant in the history of group theory.
Let us next discuss the $(1, 2)$ entry of the matrix $A$ . The graphof the quartic curve$x^{4}+y^{4}+x^{2}y^{2}-8(x^{2}+y^{2})+16.25=0$

156
is drawn at the end of this article.
It is easy to see that $28=4+(12$ . 4/2 double tangents to thecurve can be drawn. If the constant 16.25 is replaced by anumber
smaller than about 15.5 then four regions merge into asingle region
and if it is replaced by anumber larger than about 17.5, then we
get four convex regions and only 24 double tangents can actually bevisible.
In general, it is known :$)$

Theorem. Anonsingular (complex) plane curve of degree 4possesses exactly 28 double tangents.
The number of double tangents to anonsingular plane curve ofdegree $m$ is given by the formula of Pliicker :Number of double tangents $= \frac{1}{2}m(m-2)(m^{2}-9)$ .
Let
be the set of 28 double tangents. The configuration satisfied by the 28 double tangents was investigated by Steiner, Aronhold
and many others.
$\Gamma_{28}$(1). (Steiner) Let
there exist five pairs
and if we put$x_{1},y_{1}$be two distinct elements of F28. Then,,,of elements in$(x_{2}, y_{2})$$(x_{3},y_{3})$$\ldots$$(x_{6}, y_{6})$$\Gamma_{28}$$6=\{(x:, y_{i})|i=1,2,3, \cdots, 6\}$

then, the eight tangent points of any pair of double tangents,$(x_{j}, y_{j})\in 6$ lie on asame conic (an irreducible plane curve of degree2). 6is called aSteiner complex.
possesses 63 Steiner complexes
in total.
$(x_{i}, y_{i})$$\Gamma_{28}$, P7 be seven points given in the complex plane. TheLet ,cubic curves passing through these seven points form avector space
T. Every pair of curvesof intersect two more points by
Bezout’s theorem. If these two points coinside then the pairpossesses acommon tangent. The totality of common tangents soobtained forms aplane curve $D’$ of class 4, or equivalently the dual
curve of aplane curve of degree 4.
$P_{1}$$\cdots$$\{C_{1}, C_{2}\}$$\mathfrak{T}$$\{C_{1}, C_{2}\}$

157
The dual of the statement above will read as follows.
(2) (Aronhold). Let ,,be seven lines on the plane. The
totality of all curves of class 3containing these seven lines forms avector space . Every pair of curves
incontains twomore lines
in common. If $L_{8}=L_{9}$ , then the pairpossesses atangent point and is on acurve $D$ of degree 4uniquelydetermined by , , . Moreover, Li,,are double tangents$D$of this curve .
Let $D$ be the curve of degree 4uniquely determined by the sevenlines. Then $D$ possesses 28 double tangents. Moreover, the following properties hold.
(i).
,,is amaximal asyzygetic set (defined below) of F28.
(ii). The remaining 21 double tangents are rationally constructible,by ,(their coefficients are rational functions of the coeffi,
cients of ,).
(iii). Every curve of degree 4without double points can be obtaind
by this construction.
(iv). Every asyzygetic set of seven double tangents ofdefines $D$ .
$L_{1}$$\cdots$$L_{7}$$\mathfrak{T}’$$\{C_{1}’, C_{2}’\}$$\mathfrak{T}’$$\{L_{8}, L_{9}\}$$\{C_{1}’, C_{2}’\}$$z$$L_{1}$$\cdots$$z$$L_{7}$$\cdots$$L_{7}$$\{L_{1}, \cdots, L_{7}\}$$\Gamma_{28}=$$\{L_{1}, L_{2}, \ldots, L_{28}\}$$L_{1}$$L_{1}$$L_{7}$$\cdots$$L_{7}$$\cdots$$L_{1}$$\cdots$$L_{7}$$\Gamma_{28}$, ,Letbe three distinct lines in. Those three lines
determine six tangent points. If those six tangent points are on asame conic, then the triple
is called syzygetic. In thecontrary case, the triple is called asyzygetic. Asubset of F28 iscalled asyzygetic if every triple of is asyzygetic.
Let us call amaximal asyzygetic seven-line set mentioned in (i)an Aronhold set Therefore, an Aronhold set is amaximal asyzygetic
subset ofconsisiting of seven elements. It is known thatcantains exactly 288 Aronhold sets.
$L_{1}$$L_{2}$$L_{3}$$\Gamma_{28}$$\{L_{1}, L_{2}, L_{3}\}$

$S$

$S$

$\Gamma_{28}$$\Gamma_{28}$Theorem(Jordan). $Aut(\Gamma_{28})\cong S_{6}(2)$ .
Note that $|S_{6}(2)|=288\cross 7!$ . In fact,transitively permutes
all Aronhold sets and the fixing subgroup of an Aronhold set acts
as the symmetric group of degree 7on .
can not be determind only by vertices and edges since Aut(F2s)
acts doubly transitively on the 28 points. Therefore,
is not a$S_{6}(2)$

$A$

$A$

$\mathrm{I}_{28}^{\backslash }$$\Gamma_{28}$158graph in an usual sense.
, then there are 10 elements
Let , be apair of elements in$X$ insuch that $\{L_{1}, L_{2}, X\}$ is asyzygetic triple. In fact, all.
such $X$ are in the Steiner complex determined by the pairpossesses 28.27.10/6 $=.1260$ syzygetic triples. IfTherefore,, then the configuration ofall syzygetic triples are given in
is completely determined. The author is not aware if any combiis known. (Note. Acombinatorial
natorial characterization ofis known as mentioned in this article before.)characterization of. Consider. For apair
Let be an element of$X$$X$, if , , is syzygetic, connect $X$ andof elements , inby an edge. Then agraph of 27 vertices and 135 edges is obtained.
Theis isomorphic withdiscussed before (Geiger, 1869).
We have thus obatined the $(1,2)$ entry of the matrix $A$ .
$L_{1}$$L_{2}$$\Gamma_{28}$$\Gamma_{28}$$\{L_{1}, L_{2}\}$$\Gamma_{28}$$\Gamma_{28}$$\Gamma_{28}$$\Gamma_{28}$$\Gamma_{27}$$L$$\Gamma_{27}’=\Gamma_{28}\backslash \{L\}$$\Gamma_{28}$$L$$\Gamma_{27}’$$\mathrm{Y}$$\Gamma_{27}’$$\mathrm{Y}$$\mathrm{Y}$$\Gamma_{27}$Problem. Define the (1,3) entry of the group-quark matrix $A$
algebr0-geometrically.

Since$[O_{8}^{+}(2) :S_{6}(2)]=120$,the algebr0-geometric model on whichacts should contain
120 elements in it. Let us denote the object by Fi20. The fixing
subgroup of apoint ofshould be.
Therefore,is, as an-set, equivalent to the quotientspace. The action of
on $O_{8}^{+}(2)/S_{6}(2)$ is wellknown and it induces arank 3-permutation representation. Equivalently one point stabilizer
has exactly two orbits on the re. The suborbit lengths are 56 and 63,
maining 119 points
and the stabilizer of apoint inisorrespectively.
Let us write$O_{8}^{+}(2)$$\alpha$$S_{6}(2)$$\Gamma_{120}$$O_{8}^{+}(2)$$\Gamma_{120}$$O_{8}^{+}(2)/S_{6}(2)$$O_{8}^{+}(2)$$S_{6}(2)$$\Gamma_{120}\backslash \{\alpha\}$$S_{6}(2)$$U_{4}(2)$$E_{32}.S_{5}$$\Gamma_{120}=\{\alpha\}+\Delta+\Omega$where, $|\Delta|=56$ , $|\Omega|=63$ .

We are assuming that the configuration graph ofcontains
as asubgraph. Therefore, we should be able to identify and. is of length 63 and so it is natural to assume
in terms ofthat is the totality of all Steiner complexes$\Gamma_{120}$$\triangle$$\Gamma_{28}$$\Omega$$\Gamma_{28}$$\Omega$$\Omega$

159
There are 28 double tangents and so obviously there are 56 tangent points. Therefore, it is natural again to choose to be the setof all (double) tangent points of the plane curve of degree 4that we
initially began with.
In Heinrich Weber’s Lehrbuch der Algebra,, thereis a50 page chapter entirely devoted to the structure of F28. In
it, it is proved also thatis the automorphism group of theconfiguration.
There are other 120 mathematical objects.
(1). Anonsingular plane curve of degree 5posseses 120 doubletangents (easy by Pliicker’s formula).
(2). There is acurve (called del Pizzo surface) of degree 6and of
genus 4possessing 120 tritangents planes.
(3). The root system of typepossesses 240 roots. If the sign ofeach root is ignored then aset of 120 objects and its graph areobtained.
It must be an interesting problem to investigate the configuration
purely group theoretically also.
$\Delta$$\mathrm{V}\mathrm{o}\mathrm{l}11(1899)$$S_{6}(2)$$E_{8}$$\Gamma$$\Gamma_{120}$

Q3

The second and third rows of $A$ .

The second and third rows of the group-quark matrix are up
in the air at this moment. McKay [Finite Groups, Proceedingsof Symposia in Pure Mathematics, Vol. 37, Amer. Math. Soc.
1980] observed that if and are involutions of the Monster both ofwhich are conjugate to the involutions of $2A$ type, then its productbelongs to the conjugacy classes of the Monster of type
$s$

$t$

$st$

1A, $2A$ , $3A$ , $4A$ , $5A$ , $6A$ , $3C$ , $4B$ , $2B$ .
Recall that
then$\mathrm{i}\mathrm{f}-\alpha_{0}$is the highest root of the Lie algebra of type

$E_{8}$,

1 $\alpha_{0}+2\alpha_{1}+3\alpha_{2}+4\alpha_{3}+5\alpha_{4}+6\alpha_{5}+3\alpha_{6}+4\alpha_{7}+2\alpha_{8}=0$ .

The numbers {1, 2, 3, 4, 5, 6, 3, 4, 2} are called the weights of.
McKay lists $Fischer_{3}$ and $Fischer_{4}$ as groups having similar property with respect toand E7, respectively. $Fischer_{3}$ is replacedby $2E_{6}(2)$ in this article, since it fits better if we consider the $(2,1)$
entry of the-invof the group-quark matrix$E_{8}$$E_{6}$$\mathrm{t}\mathrm{r}\mathrm{a}$${}^{t}A^{-1}$160
Similar coincidences between weights of Dynkin diagrams andorders of groups elements have been observed by Glauberman andNorton [to appear in the Proceedings of Monster Workshop at Montreal, 1999]. At Kyoto symposium, the $(2,1)$ and $(3,1)$ entries ofthe matrix $A$ were the sporadic simple groups Suzuki and Fischers
and $2E_{6}(2)$ , however, appear torespectively. The new entries
better, although leaving the main realm-inv matrix
fit itsof the 3-transposition groups may be aproblem.
$U_{6}(2)$$\mathrm{t}\mathrm{r}\mathrm{a}$${}^{t}A^{-1}$

Ithank M.Kitazume, J.Matsuzawa, T.Asai, and J.McKay forvaluable discussions.

$\mathrm{x}^{4}+\mathrm{y}^{4}+\mathrm{x}^{2}\mathrm{y}^{2}- 8(\mathrm{x}^{2}+\mathrm{y}^{2})+1$$6.25<0$