A NNALES DE L ’I. H. P., SECTION A
T ERRY G ANNON
The classification of SU(3) modular invariants revisited
Annales de l’I. H. P., section A, tome 65, n
o1 (1996), p. 15-55
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15
The classification of SU(3)
modular invariants revisited
Terry
GANNONDepartment of Math, Concordia University,
1455 de Maisonneuve Blvd, Montreal, Canada H3G 1 M8.
Dedicated to the memory of Claude Itzykson
Vol. 65, n° 1, 1996, Physique theorique
ABSTRACT. - The
SU(3)
modular invariantpartition
functions were firstclassified in ref.
[ 1 ].
Here weaccomplish
theSU(3)
classificationusing only
the most basic facts: modularinvariance;
EZ~
andMoo
= 1(in [ I ],
theMoore-Seiberg naturality
"theorem" forpartition
functions wasalso
used).
A number ofsignificant simplifications
to thegeneral
argumentare included here. Hidden in
[ 1 ]
were a number of smaller results which could be ofindependent
value. These areexplicity
mentioned here. We also include a survey of known tools for WZWclassifications,
and sketch thebeginnings
of a program forclassifying
all WZWpartition
functions for any affinealgebra g -
indeed theproof
here has beendesigned specifically
to
suggest
thisgeneralization.
RESUME. - Les fonctions de
partition
invariantes modulaires pourSU(3)
etaient classifiees dans la reference
[ 1 ] .
Dans cememoire,
nous etablissons la classification deSU(3)
utilisant seulement les faits lesplus fondamentaux :
1’ invariancemodulaire;
EZ2::;
etMoo
= 1(quelques
resultatsmoins elementaires obtenus par Moore et
Seiberg
etaient utilises par la .reference
[ 1 ]).
Nous incluons dans ce memoireplusieurs simplifications
ala preuve, ainsi que certains resultats additionnels de moindre
importance.
Le theme
majeur
est celui de lageneralisation
a unealgebre
arbitraire g.Annales de l’Institut Henri Poincaré - Physique theorique - 0246-0211 Vol. 65/96/01/$ 4.00/@ Gauthier-Villars
1. INTRODUCTION
This paper focuses on the classification of
A2
Wess-Zumino-Wittenpartition
functions. This was firstaccomplished
in ref.[ 1 ] .
There are twomain purposes of this paper: to do the classification
assuming only (PI), (P2),
and(P3) (defined
later thissection) 1 -
in ref.[1] some
lesselementary
results from
[2]
wererequired;
and to rewrite theproof
so that apossible generalization
to all otheralgebras begins
to takeshape.
In the process,we
get
aproof
which is both moreexplicit
and in mostplaces
farsimpler.
Most of the
underlying
statements and formulae will be made forA2,
butcan be
naturally
extended to anyalgebra.
It should be stressed however that inspite
of thesignificant
recent progress([25], [26], [29])
which this paperincorporates,
much remains to be done before the classification forgeneral
g can be realized.
(This point
is discussed in more detail in theconclusion.)
A
partition function
forA2
WZW theories at level I~(we
write this~2,A-)
looks like~03BA03BB
is thenormalized
character[3]
of therepresentation
of the affineKac-Moody algebra
with(horizontal, p-shifted) highest weight A
andlevel
1~;
it can bethrought
of as acomplex-valued
function of the Cartansubalgebra
of~4~B
i.e. a function of acomplex
2-vector z andcomplex numbers , u (we
willalways
take u =0).
The(finite)
sum inequation ( 1.1 )
is over the horizontal
p-shifted highest weights A, ~
E~~
of level &:We will
always identify
aweight A
with itsDynkin
labelsAi, ~2 -
e.g.p ==
( 1, 1 ) .
Thequantity 03BA
+ 3 will appearthroughout.
It is called theheight
and will be denoted n.The
characters x
in this paper willdepend
on acomplex
2-vector z.Many people working
on theseproblems
use "restrictedcharacters",
i.e.put z
= 0.However,
this seems to be amistake,
and the math(if
not thephysics)
appears to demand that z be included.Because
(and only because)
wehave z ~ 0,
there is a one-to-onecorrespondence
between thepartition
function Z and its coefficient matrix M. We willfreely identify
them.I The desirability of doing this was most emphatically made to the author by Claude Itzykson.
de l’Institut Henri Poiricoré - Physique theorique
The characters
Xa
for fixed k define aunitary representation [4]
of themodular group
SL (2, Z).
Inparticular:
W in
( 1.3d)
is theA2 Weyl
group. The matrices and areunitary
and
symmetric.
Our task will be to find all Z in
equation ( 1.1 ) satisfying
thefollowing
3
properties:
(PI)
modular invariance. Thisequivalent
to the matrixequations:
(P2)
the coefficients inequation ( 1.1 )
must benon-negative integers;
and
(P3)
we must haveMpP
=1,
where p =(1, 1 ) .
We will can any modular invariant function Z of the form
equation ( 1.1 ),
an invariant. Z will be called
positive
if in addition each >0,
andphysical
if it satisfies(PI), (P2),
and(P3).
Our task is to find allphysical
invariants for each level ~. There are otherproperties
aphysically
reasonable
partition
function shouldsatisfy,
but for a number of reasonsit is
preferable
to limit attention to as small a number ofproperties
aspossible.
In this paper,only (Pl )-(P3)
will be used.The
A2
classificationproblem
has had afairly long history.
Ref.[5]
tried to understand the space of all
invariants,
for anyalthough
this
approach
works for it was too messy even forA2, ~.
But thiswork was used
by [6]
to prove theA2, k
classification for k + 3prime.
It also led to the
parity rule,
which turned out to be soimportant
in theA2, ~
classification - this wasindependently
discovered in[7]
and[8].
Inwork done
simultaneously
butindependently
of[ 1 ],
ref.[9]
classified theautomorphism
invariants of~2, A- (see equation (2.2a below).
Ref.[8]
usedit,
and anamazing
coincide with the Fermat curves[ 10],
to prove theVol. 65, n ° 1-1996.
classification for k + 3
coprime
to6,
and for k + 3 =2i
and k + 3 =3B
Ref.
[11] ]
used the Knizhnik-Zamolodchikovequations
to find all localextensions of the
A2
chiralalgebra;
in ourlanguage
thisgives
thepossible p-couplings (see
section 4below).
But the first andonly
classification ofA2, k physical
invariants for all k wasgiven
in[ 1 ] .
It was doneindependent
of - in fact obivious to - all the above
work, apart
from[5] (and
the fusionrules,
calculated in[ 12]).
Within this context, the
only thing
thispresent
paperreally
adds isthat is
accomplishes
the classificationusing only
theproperties (P 1 )-(P3).
But it also
simplifies
andexpands
out moreexplicitly
thearguments
in[ 1 ], completely rewriting
most of them. This should make the wholeproof
muchmore readable. It also makes
explicit
some results hidden inside[ 1 ];
theseshould be useful in other classifications - e.g.
Proposition
1 is used in[13].
And unlike the
proof
in[ 1 ],
itsuggests generalization
to otheralgebras.
The 6 outer
automorphisms
of aregenerated by
C(order 2)
and A(order 3).
These act on the horizontalweights ~2)
E~~
in this way:Note that
A2 ~2) _ (~2, n - ~2).
The A~ are calledsimple
currents, and C is called the
(charge) conjugation. They obey
the relations03BB1-03BB2
is called thetriality of 03BB, and
where 03C9 =Write (9 for this 6 element group, and O03BB for the orbit of 03BB under 0.
Write
Oo
for the 3 elementsubgroup generated by
AOur
goal
is to prove that theonly
level kA2 physical
invariants are:Annales ’ de l’Institut Hefiri Physique theorique +
together
with theirconjugations
Z~ underC,
definedby:
Note that
D3 = D~, D~, ~~ = ~91) ~,
and£21 == £21.
The invariants
equation ( 1.7b)
were first found in[ 14],
whileequation ( 1.7c)
was found in[ 15] .
Theexceptionals equations ( 1.7d,
e,g)
were found in
[ 16],
whileequation (1.7/)
was found in[2].
The remainder of this paper is devoted toward
proving
thatequations ( 1.7)
exhaust all
A2 physical
invariants.5?c. 2: We state here the tools which we will use to
accomplish this,
many of which were not
yet
available for[ 1 ] .
Vol. 65, n ° 1-1996.
3: We find all
automorphism
invariants for each level l~. Thisargument
iscompletely
rewritten andconsiderably
shortened.4: For each
k,
we use the"parity
rule" to find allweights 03BB
EPk
which can
"couple
top".
This is the most difficultpart
of the paper; it is based on sect. 4 of[ 1 ],
but thearguments
aregiven
in more detailhere,
and the mostcomplicated
case in[1] (namely n -
2(mod 4))
has beencompletely
rewritten. Thearguments
in this section areelementary
buttedious and involve
investigating
several cases.Sec. 5:
Everything
isput together
here. This sectioncompletely
rewrittenand
considerably simplified
from sect. 5 of[ 1 ] .
The main task here is tofind all
automorphisms
of the"simple
current extension" when 3 divides 1~.Sect. 6: This new section
explicitly
handles the four anomolous levels I~ =5,9,21,57.
2. THE PARITY RULE AND OTHER TOOLS
In this section we collect
together
the various tools we will beusing.
Allthese
apply
to any rational conformal fieldtheory,
but we will state andprove
only
what we need.Throughout
this paper we will often abbreviate"a - b
(mod c)"
as"a =c
b".The
weight
p is veryspecial.
For onething,
there is theimportant
property
thatfor all 1 a
2014_2014,
henceEquality
holds inequation (2.1 b)
iff A EEquation (2.1 b)
and(P3) together suggest
thepossibility
that the values may beimportant.
Indeed this is the case: our first three lemmas
given
below all tell usglobal
information about
M, given
the localknowledge
Because is
diagonal, equation ( 1.4a)
is easy to solve: M commutes with iffAnnnles de l’Institut Henri Poincaré - Physique theorique
A much harder task is to obtain useful information from
equation ( 1.4b).
This is the purpose of this section.
First some definitions. Call a
physical
invariant M aautomorphism
invariant if there exists a
permutation 03C3
ofPk
such thatFor a
given
invariantM,
letand define
PR, ,7R
ands R similarly -
e.g.s R
=L MPa S(n)03BB .
Let[03BB]
denote the orbit,7L 03BB,
and[M]’
denote the orbitJR
M.LEMMA 1. -
(a)
Let M be anypositive
invariant. For each~,
M Epk,
both s03BBL, s R ~ 0. Also s03BBL > 0 iff 03BB~PL; s R > 0 iff ~PR.
(b)
Let M be anyphysical
invariant. Then,7L
and,~R
are groups, andso
equal either {A0}
orO0 = { A0, A1, A2 }.
(i)
Let M be anyphysical.
Then =for
any Aa E~L, Ab
E~R .
(d)
Let M be aphysical
invariant.Suppose MaP ::; MPa
‘d ~ E Then= ‘d ~ E
~~
and1CL
=ICR.
(e)
Let M be aphysical invariant,
and supposeJ’CL
=~p~, 1CR
=~p~’.
Then the
cardinalities ~JL~ ==
areequal,
soJL
= andPL equals
the setof
allweights
with zerocharge
withrespect
to,7L,
i.e.Proof . - (a) Evaluating (9~
=(see equation ( 1.4b)) gives
usThe RHS of
equation (2.3a) is ~ 0,
since each >0, by equation (2.1b),
and each >
0, by (P2).
Thisgives
us the firstpart
of(a).
In fact theRHS of
equation (2.3a)
will be > 0 iff someM~,,~
>0,
i.e. iff A E Thisgives
us the second.Vol. 65, n ° 1-1996.
(b)
From M = andequation ( 1. 6b)
weget
In
deriving
this we also used(P2)
andequation (2.1&). Equality
willhappen
in
equation (2.3b)
iff~(A) =30
for all A E7~
so(b)
follows.(c)
As in(b),
weget
for anyAa E ~L, ~
The third
equal sign
appears inequation (2.3c)
because we learned in theproof
of(b)
that A" E~L iff,
for all A’ E7~ ~x t, ( ~’ ) =30.
(d) By ( 1.4b)
weget
But
by equation (2.lb),
each = > 0. Becauseequality
must holdin
equation (2.3J),
the desired conclusion holds.(e)
Eache { 0, 1}, by (c).
follows from(b), (a~.
In theproof
of(b)
we found that A E~L
iff A has zerocharge
with
respect
ot allLemma 1 was first
proved
in[ 17].
It willplay
animportant
role insection 5. Of
these, (c)
is the mostimportant.
Thehypothesis
in(d)
isoften
satisfied,
because most M havee {0, 1}.
In thegeneral
Annales de l’Institut Henri Poincaré - Physique theorique
case
(i. e.
notA2 ),
we stillhave ~JL~ = ~JR~
in(e),
but this nolonger
will
necessarily
mean,~L
== Infact,
weget
a moregeneral
result.Let M be any
physical invariant,
then forintegers
a, b thefollowing
threepropositions
areequivalent:
7~ 0~
a t ( ~ ) - 3 b t ( tc )
whenever0;
= tc.
This follows from the same calculation used in
(b).
It becomesparticularly
valuable when the
algebra
issemi-simple (since
then there are so manysimple currents).
The next lemma was proven in
[7].
Theonly
additional result here is that= ~ p ~
iff= ~ p };
this followsimmediately
from(d).
LEMMA 2. - Let M be
a physical
invariant. Then K03C1L= {03C1} iff K03C1R = { 03C1 } iff
M is a~automorphism
invariant(see equation 2.2a)).
Note from
equation ( 1.4)
that the matrixproduct
MM’ of two invariantsM and M’ is
again
an invariant(at
the samelevel).
This is animportant fact,
andquite probably
has not beenexploited enough.
The next lemma isthe main
place
thisproperty
is used. It isproved using
the Perron-Frobeniustheory
ofnon-negative
matrices[ 18],
and can bethought
ofloosely
as ageneralization
of Lemma 2. It will be used in section 5 tosignificantly
restrict the
possibilities
for the sets oncegiven
~, and also to bound the values ofAny
matrix M can be written as a direct sumof
indecomposable
submatricesB.~ (i. e.
we cannot writeB~
=B~
(BB~ ).
the indices i of M "contained" in the submatrix
B~;
every index i of M(in
our case, every weitht A EPk)
lies in one andonly
one I(Bl).
We will
always put
p E Z(Blj. By
anon-negative
martix we mean a square matrix B withnon-negative
real entries.Any
such matrix has anon-negative
real
eigenvalue
r ==r (B) (called
the Perron-Frobeniuseigenvalue [18])
with the
property
thatr > Isl
for all other(possibly complex) eigenvalues
s of B. The number r
(B)
has many niceproperties,
forexample :
Vol. 65, n° 1-1996.
provided
B isindecomposable
andsymmetric,
eitherequality
holds inequation (2.4b)
iff each row sum~ Bi~
isequal,
andequality
holds inj
equation (2.4c)
iff B is the 1 x 1 matrix B =(r). Also,
there is aneigenvector
v witheigenvalue
r withcomponents vi ~
0.For
example,
consider the m x m matrixIts
eigenvalues
are 0(multiplicity
m-1 )
and m .~(multiplicity 1 ).
Thereforer
(B(~, ~.,.t~ )
= m .~. Itseigenvector
v is v =(1, ... , 1).
These matriceshave the
important property
thatthey
areproportional
to their square.They
occur
frequently
in modular invariants.LEMMA 3. -
(a)
Let M be apositive
invariant with non-zeroindecomposable
blocksB~,
where p E Z~(B1).
Then r::;
r(B1) for
all .~.
(b) Suppose
now thatBi
=r B1, for
some saclar r. Then r = rfor
all .~.
If
in additionBi
=B1,
then each::;
r.(c)
Now let M be aphysical invariant,
andICL
=~p~,
=~p~’. Suppose M03BB ~ 0,
where 03BB is not afixed point of JL (i.e.
J EJL
and J 03BB = 03BBimplies
J =A°)
and M is not afixed point of
ThenM03BBv ~
0iff
v E
[M]’
and 0iff
v E~~~ .
Proof. - (a)
FromMPP
= andequation (2.1b)
weget
the very crude boundfor any
positive
invariant M.Let B be an
arbitrary non-negative
matrix.Looking
at its Jordan blockform,
we seeimmediately
thatMoroever, r((J3/~p)
= so(2.4b)
tells usChoosing r (B1)
s andconsidering (M/s)~
forlarge j,
wefind from
(2.5&, c)
that we violateequation (2.Sa).
l’Institut Henri Poincaré - Physique theorique
(b)
First note that r =r (B1), by (2.5c). Suppose r(~)
r, for/1 V
some l. Look at the sequence of matrices
(
-M , for j
~ ~2014 thisV
7
sequence will not in
general
converge.However, by equation (2.Sa)
the/1 B ~
entries
of - M j for any j,
will be bounded aboveby 2 ) ,
/1 V
so
by
Bolzano-Weierstrass the sequence -M
will haveconvergent
Br /
subsequences.
Let M~ be the limit of any suchsubsequence. Clearly,
M~will be a
positive
invariant.Suppose
r(B~)
r for someB~ 7~ 0,
and letA e
I(Bl).
Then A ePL (M) (since Bl ~ 0)), but 03BB~ PL (M’) because
( - Bl)j
~ 0by (2.5b)).
Butby hypothesis
= rM’03C1
for all , soby
Lemma 1(a)
we must have~ ( M) _ ~L ( M’ ) - a
contradiction.That each
M~ ~
r follows nowby looking
at rr
(Bi
=r2, M~ by equation (2.4c).
(c)
Let m= ~y~!! (which by
Lemma 1(e)).
Write out thedecomposition (B~B~
for as inequation (2.4a).
Thenby hypothesis, Bi
=B(m, ~) equation (2.4a~),
sor(~)
=?7~
for each(by
Lemma 3
(b)).
Now,
letBi
be the block with A eT(B,).
Then for allJ, J’
e~, by
Lemma 1(c)
weget
Let
Bi
denote the matrixThen
element-wise, Bi by equation (2.5d),
soby
p. 57 of[18]
weget
that r( Bi ) >
r(J3~),
withequality
iffBi =
ButBi
=so
m2
= r( Bi ) >
r =m2
Therefore = 1 andBi
=B,.
Using MT
M inplace
ofM MT,
weget
thecorresponding
result forLemma 2 is a
corollary
of Lemma 3(c).
In section5,
Lemma 3(b, c)
will be
applied
to the case = =Oo
p. There is aunique
fixedpoint
there:
f
=(n/3, n/3).
So Lemma 3(c)
tells us about most of theweights;
the main value of Lemma 3
(b)
for us will be inanalysing
thepossible
value of but it will also be useful in section 6.
Vol. 65, n° 1-1996.
The final observation we will use is the
parity
rule. Its shortest derivationis in
[ 19];
there is no need torepeat
it here. We willonly
state theresult,
as it
applies
toA2, ~.
For any real numbers define
by {.r ~~
theunique
numbercongruent
to x
satisfying 0 ~ { x }y
?/. Considerany 03BB=(03BB1, 03BB2), 03BBi
EZ,
A notnecessarily
in Define theparity ~ (A)
of 03BB to beThen it can be shown that e
(A) 7~
0 iff there exists aunique
root latticevector v = ~’
(2, -1)~-rrz (1,1), ~
m EZ,
and aunique Weyl
transformationúJ E
W (A2),
such thatin this = det w.
LEMMA 4. -
(a)
Let M be any invariant. Choose any~, ~c
EThen, for
eaeh f ’coprime to
, 3(.~ (.~ /
0 and(b)
Now let M be anypositive
invariant. Thenfor coprime to
3 n,Ma~, ~
0The more useful one for our purposes is Lemma 4
(b).
Infact,
we willbe
mostly
interested inapplying
itto
= p. This willgive
us an upper bound on the sets R, and from there Lemmas1, 2,
3 can be used.Equation (2.7)
will be used in section 6.The
parity
rule isextremely powerful.
Forexample,
for thespecial
casen
coprime
to6,
4(b)
alone isenough
toimply
thatC 0 A,
for anypositive
invariant M. We will not use that here(the complicated proof
isgiven
in[10]). Incidently,
a similar result holds for it is natural to ask how this extends tohigher
rankAe.
The "catch" is that
proving anything using
theparity
rule seems to meanimmersing
oneself in mazes of minute details andspecial
cases. Below isthe
parity
rule forp-couplings
for theAl
modular invariantclassification;
itwill be used in section 4
below,
where we will find that theAl
classification is embedded in some way in theA2
one. Theproof
of Lemma5,
and thoseAnnales de l’Institut Henri Poincaré - Physique theorique
in section
4,
look solong
andcomplicated
because we havedeliberately
included all details to the
arguments.
Let
CL
denote the set of all numberscoprime
to L. In order toapply
theparity rule,
we need asystematic
way ofproducting
lots of numbers .~ inC L. Fortunately,
this is not difficult: forexample,
consider£ will lie in
C2L
iff eitherL/2i
is evenand j
=0,
orL/2i
is oddand j
> 0.The reason is that these choices of
i, j guarantee .~
isodd;
any otherprime
p
dividing
L will not divide:f:2j
so cannot divide .~. These series of £’ s arethe reason for
introducing
thebinary expansions
inequation (2.9a)
below.Choose any
integer
m > 2. LetJ~~
denote the set of allintegers
a,0 a m,
satisfying:
LEMMA 5. 2 -
Define
the set as above. Then:(a) for m ::J 6, 10, 12, 30,
we havel’Cm = ~ 1,
m -1 ~;
Proof . -
Write m =2L m’,
wherem’
is odd. Define theinteger
Mby m/2 2Al
m.First,
note that a EKm
iff m - a E The reason isthat £ E
C2~
must beodd, so {£ (m - a) ~ 2~ equals m - ~ .~ a ~ 2m
m if{ ~ ~ }2~ ?7~or3m-{~a }2~
> m if{ ~ ~ }2~
> m.Let a E Define b = so 0 b
1,
and write out itsbinary (=
base2) expansion:
So
bi
= 0means {2i b}2 ::; l,
whilebi
= 1means {2i b}2 2:
1.1
For
example, 2
hasbinary expansion
0.100 ... =0.011 ...,
while2
- = 0.0101 ....
3
2 Note that this is slightly more general than Claim 1 of [ 1 ]. The proof is mostly unchanged, apart from many more details included. The proof of Lemma 5 is fortunately as bad as it gets in this paper !
Vol. 65, nO 1-1996.
Case 1 . - m is odd
(i. e.
L =0).
This is thesimplest
case. We mayassume that a is odd
(otherwise replace
a with m -a). Choose j
so thatPut f = m -
2~ ;
it iscoprime
to 2 m.Then ~ .~ a }2m == { m - 2j a ~ 2m
=3 m - 2~ a
> m, because a is odd and because ofequation (2.9b). Therefore, by equation (2.8), { ~ }2~
> m, which canonly happen
if m -2j
0.Equation (2.9b)
now tells us a2 m/2~ 2,
a == 1.Now,
what if m is even?Putting .~
= m - 1 forces a E1C~
to be odd:for if it was even, = 2 m - a > m. We may also assume
a m/2, i. e. b 1 otherwise replace
a with m -a).
Let c
= {~/2~ }2.
There two different cases: either 0 c 1(to
becalled case
2),
or 2 > c > 1(to
be called case3). c ~ 0, 1,
because a is odd and L > 0.Case 2 . -
Define ~
== m’ +2i.
Then for i =1,
..., M -1,
each and 0.~i
m. Thendividing equation (2.8) by
m tells us thatfor each
1
iM,
either c+ { 2i b ~ 2
1 or 2 c+ { 2’& }2:
moreprecisely,
for each i =1,
..., M -1,
Choose j
so thatequation (2.9b)
holds.Suppose
for contradiction thatj
M. Thenby equation (2.10), 2~ -1 b = {2~~6}2
1 - c, but== {2~6}2
> 2 - c.Hence,
2 - c 2 -2 c,
i.e. c0,
which is false. ,
Therefore b
21-M,
i.e. am / 2 M -1 4,
so a oddimplies
eithera = 1 or 3. All that remains for case 2 is to show
3 fj.
We will prove thisby
contradiction. Note that because c =3/2L 1,
we must haveL ~ 2; a m/2
then meansm 2:
8.If m - 3 2 use £ ==
(m
+1)/3,
while if m ==31
use £ ==(m
+2)/3
+m’.
If m == 9
0, 3
take £ =m/3
+1,
while if m == 9 6 use £ =m/3
+ 3. In all cases £ E 0£
m, = 3 £ > m, soequation (2.8)
isviolated. Thus
3 fj.
and we are done case 2.Case 3 . -
Take li
=m’
+2i
=m’ - 2j .
Then fori, j
>0,
both£i, £j
EC2m.
For i =1,
..., M -1,
0£i
m;1,
..., M -L,
0
~
m; andfor j
= M - L +1,
...,M, - m ~
0. Therefore :de l’Institut Henri Poincaré - Physique theorique
for either c
{2~}2orl+{2~}2
e.(2.11c)
Now suppose for some M - L
j M,
that e{ 2~ }2
but1 +
{2j+1b}2
c.Then bj
==1,
so{2j+1 b}2 =
2{2j b} - 2,
andwe
get
c20142014, contradicting
c > 1.Similarly,
c 2 iscontradicted if 1 +
{ 2j b }2 2
c but c{2j+1
&}2.
Thus either c{2jb }2
for all M - L i
::; M,
or 1 +{ 2’& }2
e for all M - L i::;
M.Moreover,
if Z > 1 thenadding equation (2.11~)
andequation (2.11c)
withi
= j
= M - 1 tells us that if L > 1 then c{ 2~ h
iff c~. 2
Subtracting equation (2.11~)
fromequation (2.11~) produces
theinequality - { 2’& h ~
for all1 ~
i~
M - L.Thus,
for thesei,
&,
== 0 iffbi+l
== 1.Summarizing,
we see that the first Mbinary digits &,
of & are fixedby
the demand that
bl
== 0(since ~ ~ ?r~/2),
andequations (2.11~-c):
and for L >
1, b
= 1 iff c~. 2
This fixes the value of b up to acorrection of
2-M,
a up tom/2M,
so a is thencompletely
fixedby
thecondition that it be odd. To eliminate this value of a
(except
for 7special
values of
m),
we will consider 5 subcases.(i)
Consider first M - L ==1,
m = 3 ’2L,
and L > 26,
12. Thenby equation (2.11J)
the first Mbinary digits
of b areb = 0.011 "’
1-,
where "-" denotes theremaining
unknowndigits bi,
i > M. Therefore
1/2 - 1/2M ~ 1/2,
a =m/2 - ~
for some0 ~ c ~ m/2M
2. Therefore a =m/2 -
1. Since m >14, .~
= 7 lies inC2m
and satisfies{ hm = ~ 2 m -
7 > m,violating equation (2.8).
(ii)
Consider next M - L =2,
i.e. m = 5 .2L
or 7 .2L,
and L > 2(i. e. 10, 20, 14, 28).
Then b = 0.0100 .’.0-,
i.e. a =m/4
+ e, for(if
m = 7 ’2L)
will violateequation (2.8).
(iii)
Now consider M - L >2,
L > 2. The first fourdigits
of b willbe b = 0 . 0101- . If c
"2 3
thenputting
= 2 inequation (2.11&) gives
c >
1. 01,
i.e. c = 1. 01- . Nowby equation (2.11J),
= -" === 1.
Putting
i = M - L + 1gives {2ib}2
+ c = 1.11 -+1.01- ~ 3,
contradicting equation
Vol. 65, n ° 1-1996.
If instead c >
3
thenputting j
= 1 inequation (2.11 b) gives
c = 1.10 - .Equation (2.11 a~
saysb~T _ L+~ _ ~ ~ ~
=b~~
=0; putting
i = M - L + 1gives {2i b}2
+ c = 0.00 - +1.10-::; 2, contradicting equation (2.11 a).
(iv)
Consider next M - L >2,
L = 2.Here,
either c = 1.01= - 5
4
or c = 1.11 =
7.
4 Consider first c =1. Ol,
then _0,
--bnl_1
=bM
= 1by equation (2.11d). Putting j
= M - L -l,
weget
c - ~ 2~ b ~ 2
= 1. O l - 1.011-0, contradicting equation (2.11 b) .
Ifinstead c =
1.11, then j
= M - L - 1gives
1.11 - 0.100- >1,
alsocontradicting equation (2.llb).
(v) Finally ( ! ),
look at M -L >2,
L = 1. Then c= 3
and m > 10. From. 1 1 2
1 2
equation (2.11 a~
we find that for Meven - - -- :S
b::; -
+. nI ,
1 2 1
3
3.2 3 3.2and for M
odd 1 3 - 3.2M
b +1 3.2M.
Thatis,
a =m / 3
+ ~, where2 4
33.2 -
3 3 2 4 2- - 2
~- 4
if M is even, andwere - - h 4
~- 2
if M is odd. So3- -3 3 3
for m -3
0,
a~ =m/3 ~ 1,
and if m - 3 :!: 1 we have a =m/3 ~ 1/3 respectively.
Taking
l = 3 inequation (2.8)
eliminates m -32: {
la}2m
= m+ 1 > m.For m - 3
1,
take l =m 2
+ 6 :{l
a}2m
=10 m - 1
m - 2 =366 -m - 2 2
> m .All that remains is m = 3
0,
i.e. m - 366, 18,
30. For m = 366, put
.~ = 4 +m/6..~
E since .~ -65,
and .~m. ~ .~
a:i::}2m
==4 1 1
= 11 3 - .3
m :!:4 + -
m :!:6
m ==6
m + 4or 2
m -4,
In both casesviolating
equation (2.8).
For m - 36
18, put
.~ = 2 +m/6. Again
.~ m and .~ - 65,
so .~ EC2~.
{l
ahm = 11
m -E- 2or 3
m -2
, soequation (2.8)
is violated.0 2
For m - 36
30, put
l = 6 +m/6. Again
l m, and l - 65,
sol E
. {l
a= 11
m + 6or 3 2
m -6
, so for m > 36’ (t. e. m ~ 30), equation ( 2.8 )
isviolated.
2There were some
special
values of m thatslipped through
thesearguments: namely
m =6, 10, 12, 14, 20, 28,
30. These can be worked outexplicitly. []
Annales de l’Institut Henri Poincaré - Physique theorique
3. THE
AUTOMORPHISM INVARIANTS
Recall the definition of
automorphism
invariantgiven
inequation (2.2a).
Let M~ denote its coefficient matrix. In this section we will find all
automorphism
invariants.THEOREM 1. - The
only
level kautomorphism invariants for A2
arefoY
1~ =30, ~3
0.That the matrix M03C3 must commute with
(see equation ( 1.4b))
isequivalent
toOf course,
(P3)
tells us 7p= p.The first
step
in theproof
is to constrain the "fundamentalweight" (2, 1 ).
CLAIM 1. -
o- (2, 1)
E0(2, 1).
Proof of Claim
1. - Fromequation (3.1 )
it suffices to show thatis most
conveniently
evaluatedusing equation (2.1a).
Since the value of is constantalong
the orbit0A,
we may suppose~2 ~
n - a~ - ~2.
Wemay
also assume~>3-~=1,2
can be verifiedby
hand.
Obviously,
theonly possible products
for
integers
1a~ & ~ c ~ ~
are(a, b, c) = (1, 2, 3), (1, 1, c)
or(2, 2, 2) -
for any othertriple
the LHS ofequation (3.2b)
willclearly
belarger
or smaller than the RHS. In order for a =Ai,
b== A2,
c= ~ 1-I- h 2
orn -
A2,
these threetriples require respectively
a E C(2, 1), A
Eor
both 03BB = (2, 2)
and k = 3. But a Eclearly
hasS(n)03C1,(2,1).
A =
( 2, 2 )
at Jb = 3 can be evaluatedexplicitly;
we find,S’P6 ~
>~~i)
has been
established..
The
claim, together
withequation (2.1c),
tells us that theonly possibilities
for 03C3
(2, 1)
are:Vol. 65, n° 1-1996.
Note that the
possibilities
for k ~ 3 0 are realizedby
andrespectively,
and for k =3 ± 1 andDk
andrespectively.
Since the
(matrix) product M~~
of twoautomorphism
invariants M~M03C3’ is
anotherautomorphism
invariant to prove Theorem 1 for each 1~ it suffices to show that theonly automorphism
invariantsatisfying
~ ( 2, 1 )
==( 2, 1 )
isCLAIM 2. -
Suppose for
anyA, .V
EP~,
bothA’.
Proof of
Claim 2. - Therepresentation ring
of asemi-simple complex
Lie
algebra
is apolynomial ring freely generated by
theWeyl
characters of the fundamentalrepresentations (see
Ch.VI, § 3.4,
Thm. 1 of[20]).
Thusthe
Weyl
character for anyhighest weight ,~
EP+
ofA2
can bewritten as a
polynomial
Moreover,
we know from[4]
thatThen,
for anyIn
particular,
fromequation (3.4)
we find thatUnitarity
of now forces 03BB =03BB’. []
Now choose
any A
E andput ~’ _
The firstequation
inequation (3.4)
follows fromequation (3 .1 )
and ~p= ~ r(2, 1 )
==( 2, 1 ) .
The second
equation
isjust
thecomplex conjugation
of the first - seeequation (1.6&).
Thus Claim 2 tells us A ==A~
i.e. cr is theidentity.
l’Institut Henri Poincaré - Physique theorique