Suppose thatA,B,C, andD are (invertible)n×n matrices. What conclusion can we draw fromφA,B◦φC,D=φC,D◦φA,B(where we take the common domain)? Ifn=1, this
implies thatφA,B=φC,Das is easy to see from the 2×2 matrix realization of fractional
linear transformations. Whenn≥2, some additional phenomena can occur.
Suppose thatλis a complex number unequal to 1, and{A,B,C,D}is a set of fourn×n
invertible matrices. We say that the quadruple (A,B,C,D) isλ-linkedif the following three equations hold:
A=λCAC−1, B=λDBD−1, AB=CD. (B.1)
Ifλ=1, thenλn=1 (take determinants); moreover, ifλis a primitivekth root of unity,
there exists a change of basis so thatA=diag(1,λ,λ2,. . .,λk−1)⊗RandC=P
k⊗S,Pk is
the (permutation) matrix of thek-cycle, andRandSaren/k×n/kcommuting invertible matrices. (Ifnis prime, thenRandSare scalars.) In particular, trA=trB=trC=trD=
0. Consequences of the definition include the following:
C=λA−1CA, D=λB−1DB, BA=DC. (B.2)
(The last follows from substitutingλBD−1=D−1BintoAB=CD.) We require an elementary lemma. The converse is trivial.
LemmaB.1. Suppose that{R,S,T,U} ⊂GL(n,C). Suppose that for allXinMnC,RXS= TXU. Then there exists a nonzero scalarλsuch thatT=λRandU=λ−1S.
Proof. We have (T−1R)·X·(SU−1)=X for all X. Let v be a right eigenvector with eigenvalueaforT−1R, and letwbe a left eigenvector forSU−1with eigenvalueb. With
X=vw, we deduce abvw=vw. Henceab=1. It follows immediately that bothT−1R andSU−1have just one eigenvalue. IfT−1Ris not a scalar multiple of the identity, then there exists a columnv1such thatT−1Rv1=av1+v. Now setX=v1w. Thenv1w=X=
T−1Rv
1wSU−1=b(av1+v)w=ab(v1w) +bvw. Asab=1, we havevw=0, which is obvi- ously impossible. SoT−1Ris a scalar multipleaof the identity, and thusR=aT. Similarly,
PropositionB.2. Suppose that{A,B,C,D} ⊂GL(n,C). The condition,φA,B◦φC,D=φC,D ◦φA,B, is equivalent to
AB=CD, A=λCAC−1, B=λDBD−1, (B.3)
for some scalarλ.
Proof. There exists an open neighbourhood of 000 in MnCsuch that for allXtherein, both
left and right sides are defined onX. For suchX, we have (I−A(I−CXD)−1B)−1=(I−
C(I−AXB)−1D)−1. Inverting and subtracting the results from I, then inverting again, we obtain
B−1(I−CXD)A−1=D−1(I−AXB)C−1 (B.4)
for all suchX. Since the relation is linear inXand is equality on an open set, we deduce that it is true forXin MnC. WhenXis set to 000, we obtain (AB)−1=(CD)−1, that is,
AB=CD. (B.5)
We deduce that for allX,
B−1C·X·DA−1=D−1A·X·BC−1. (B.6)
FromLemma B.1, there existsλsuch that
B−1C=λD−1A, (B.7)
DA−1=λ−1BC−1. (B.8)
Multiplying (B.5) by (B.7), we obtainAB·B−1C=λCD·D−1A, whenceAC=λCA, and thusA=λCAC−1.
Similarly, (B.8) multiplied by (B.5) gives usDA−1·AB=λ−1BC−1·CD, which yields
B=λDBD−1.
To show that the converse holds, it suffices to establish (B.7) and (B.8). FromAB=CD, we deduceCA−1=DB−1; fromB=λDBD−1, we haveB−1=λ−1DB−1D−1, soDB−1=
λB−1D. ThereforeC−1A=λB−1D, and thusBC−1=λDA−1, which is (B.8).
Similarly,AB=CDentailsBD−1=A−1C=λ−1CA−1, so thatλD−1A=B−1C. This situation can arise withλ=1—set A=(1 00−1)= −D and B=(0 11 0)=C. With
n=2 andλ= −1, we haveAB=CD,A=λCAC−1, andB=λ−1DBD−1. Of course,φ
A,B= φC,D.
Whenλ=1, there is another characterization.
Recall that ifQis a subset of MnC, thenQwill denote its centralizer,{A∈MnC|Aq= qAfor allqinQ},Qits double centralizer, and so forth.
LemmaB.3. LetA,B,C,Dbe a subset ofGL(n,C). Then
AB=CD, AC=CA, BD=DB, (B.9)
Proof. SupposeAB=CD. It suffices to show thatC−1A·X·BD−1=Xfor a relatively open set ofXin{A,B,C,D}. SetE=C−1A; thenEbelongs to{A,C}. SinceE=DB−1,
Ealso belongs to{B,D}. HenceEbelongs to{A,B,C,D}= {A,B,C,D}. Thus for all
Xin{A,B,C,D},EXE−1=X, so thatC−1A·X·BD−1=X.
Conversely, fromAXB=CXDfor a dense set ofXin{A,B,C,D}, we have equality
for allX in{A,B,C,D}, in particular for allXin the algebra generated by{A,B,C,D}. SettingX=I, we deduceAB=CD. SettingX=(BA)−1, we have I=C(BA)−1D, whence
BA=DC.
SettingX=B−1, we haveA=CB−1D, whence (right multiplying byC)AC=CB−1DC; thusAC=CB−1·BA=CA. Finally, withX=D−1, we haveAD−1B=C, and right mul- tiplying byDyieldsAD−1BD=CD=AB, soD−1BD=Band thusBD=DB. In particular, ifλ=1 and{A,B,C,D}generates all of MnC, thenφA,B◦φC,D=φC,D◦ φA,Bentails thatφA,B=φC,D. This corresponds to an irreducible representation of a group
we are going to define.
Ifλ is a primitive nth root of unity (wheren is the matrix size, as usual), there is a complete description of theλ-linking quadruples, up to conjugacy. LetP denote the cyclic permutation matrix (ones in the (i,i+ 1) and (n, 1) positions). Letpbe any complex polynomial which does not have anynth roots of unity (including 1) as a root, and letα
andγbe nonzero complex numbers. Up to change of basis ofCn, all possible quadruples
are obtained from the formula below:
A=diag1,λ,λ2,. . .,λn−1α,
C=Pγ,
B=C·pA−1C,
D=A−1·pA−1Cλ.
(†)
This follows from the simple observation thatB=A−1CD; on plugging this intoB=
λDBD−1, we obtainA−1CD=λDA−1CDD−1, so thatD·A−1C·D−1=λ−1A−1C. The last two remarks are part of a more general theory, namely, the irreducible finite dimensional representations of groups with appropriate generators and relations. LetG
be the group with generators{a,b,c,d,Λ}and relations
[Λ,a]=[Λ,b]=[Λ,c]=[Λ,d]=1, (B.10)
ab=cd, (B.11)
a=Λcac−1, (B.12)
b=Λdbd−1. (B.13)
The first line merely says thatΛis in the centre ofG. We want to record the finite dimen- sional irreducible representations ofG. Letπ :G→GL(n,C) be an irreducible represen- tation, and denoteπ(Λ)=λI,π(a)=A,π(b)=B, and so forth.
We require a few properties ofG. (a)ba=dc.
Proof. Equation (B.12) implies thatac−1=Λ−1c−1a, soc−1a=Λac−1. From (B.11),c−1a
=db−1; thusΛac−1=db−1. From (B.13),bd−1=Λ−1d−1b, and thusΛac−1=Λb−1d, that
is,ba=dc.
(b) The groupH:= c−1a,Λis a normal subgroup ofG.
Proof. Sete=c−1a=db−1. Thenaea−1=ac−1=Λ−1c−1a=Λ−1e. Sinceeandanormal- izeH, so doesc=a−1e−1. Similarly,b−1eb=b−1d=Λ−1db−1=Λ−1e, and obviouslyd
also normalizesH.
Now we observe thatGhas generators{Λ,e=c−1a,a,b}with relations,Λis central, and [a,e]=[b−1,e]=Λ−1. There is an obvious normal form for elements ofG, namely,
(word inaandb)ekΛl, (B.14)
and sinceHis a normal subgroup, it follows thatG/His the free group on two generators (the images ofaandb).
Suppose thatλ=1 (whereπ(Λ)=λI—sinceΛis central andπis irreducible, the image ofΛis a scalar matrix). ThenE=C−1A=DB−1is the image ofπ(e) and is thus in the centre ofπ(G). HenceE=αI for some nonzero scalarα. This entails thatC=αAand
B=α−1D, or what amounts to the same thing,φ
A,B=φC,D. In order to be irreducible,
necessary and sufficient is that{A,B}generate MnCas an algebra—this is equivalent to AandBhaving no nontrivial invariant subspace in common.
If insteadλ=1, then it is necessarily that annth root of unity and (A,B,C,D) isλ- linked; moreoverEnis a scalar matrix. We have already described the irreducible repre-
sentationsπ :G→GL(n,C) such thatπ(Λ)=λI whereλis aprimitiventh root of unity. They are given bya→A,b→B, and so forth as in (†). We see that this is irreducible, since{A,C}already generates MnC. The parameters are given by theφ(n) possible values
ofλ, the primitiventh root of unity (sincenλis the value of the trace ofπ(Λ), different choices forλyield inequivalent representations),α,γ, and the coefficients ofp.
To decide when two such representations are equivalent, it is desirable to simplify the form of π, by using the alternative generators and relations,{a,b,e,Λ} forG. Then π
restricted to the subgroup generated by{a,e,Λ}is irreducible (and induces a projective representation modulo the parametersαandγ ofZn×Zn), and we can write it in the
form
a−→A:=diag1,λ,λ2,. . .,λn−1α,
e−→E:=Pγ. (B.15)
Nowbacommutes withe, soπ(b)Amust commute withE. AsEhas distinct eigenvalues (Eis sizen, with eigenvalues{γexp(2πk/n)}),π(b)Ais a polynomial inE, so we have to setπ(b)=p(E)A−1for some polynomialp. In order to guarantee thatp(E) is invertible, we are required to have thatp(γexp(2πik/n))=0 for allk—but it is easy to see that this is the only condition required ofpforπto be a representation (of course, if we insist on special types of representations, such as unitary or of determinant one, more conditions will be imposed).
Let tr be the usual trace on MnC, and let Tr≡Trπbe the character associated to the
representation, that is, Tr(g)=trπ(g).
Onceλis fixed, we see thatαn=Tr(an) andγn=Tr(en) are invariants ofπ. Now write p(t)=k−1
j=0pjtj (we obviously can assume phas degree less thank). Then p(E)A−1=
pjEjA−1; we calculate the Tr(e−jba)—we simply obtainnpj. Hence p itself is an in-
variant of the representation. Thusλ,αn,γn,pare invariants ofπ, and it is now easy to
see that these are complete (i.e., matching invariants implies equivalent representations). What about irreducible representations ofGfor whichλk=I butk=1,n? If we do not
require that, for example,π(ak) be scalar, then there are plenty.