1 Definitions and results
For n≥1, the linear group GL(n,ℂ) has a natural action on ℂn which induces an action on the set of vector spaces contained ℂn. Since the center of GL(n,ℂ) acts trivially on this set, there is a natural action of PGL(n,ℂ) on this set. A subgroup H of PGL(n,ℂ) is irreducible if the only H-invariant subspaces of ℂn are {0} and ℂn. A subgroup H of PGL(n,ℂ) is completely reducible if any H-invariant subspace of ℂn admits a H-invariant complement. Whereas we will mainly focus on PGL(n,ℂ), it is worth noticing that these definitions can be generalized to any complex reductive group (see [8] or [3]).
If H is a subgroup of a group G, then the centralizer (resp. normalizer) of H in G is denoted by ZG(H) (resp. NG(H)). In particular, the center of G, i.e. the centralizer of G in G, is denoted by Z(G).
A bad subgroup H of PGL(n,ℂ) is an irreducible subgroup whose centralizer ZPGL(n,ℂ)(H) in PGL(n,ℂ) is not trivial. In this paper, we classify centralizers of bad subgroups of PGL(n,ℂ) up to conjugation. Note that the centralizer of an irreducible subgroup of PGL(n,ℂ) is necessarily finite. More generally, a subgroup of PGL(n,ℂ) is irreducible if and only if it is completely reducible and has a finite centralizer (see [8, Corollary 17]).
Our motivation for this is to describe the singular locus of character varieties, see [2] or [3, Paragraph 7] for a definition. We recall that the representation variety is the set of representations of Γ into G. The character variety of a finitely generated group Γ into a complex reductive algebraic group G is the GIT quotient of the representation variety by the conjugation action of G. When Γ is a free group, the understanding of this singular locus admits a satisfactory answer, see [3, Theorem 7.4]: for “most” (the precise assumption can be found in the statement of the theorem) reductive groups G, the singular locus of the character variety of free groups into G is the union of the set of conjugacy classes of non-irreducible representations and the set of bad representations (i.e. whose image is a bad subgroup of G).
When Γ is Fuchsian and G is a complex semi-simple group, one can prove that any irreducible representation ρ:Γ→G is a smooth point of the representation variety (one should use Serre’s duality to show the vanishing of the second cohomology group H2(Γ,Lie(G)Ad∘ρ)). In particular, this implies (see [8]) that the conjugacy class of an irreducible representation whose centralizer in G is equal to the center of G is a smooth point of the character variety. Therefore, among conjugacy classes of irreducible representations, algebraic singularities of character varieties of Fuchsian groups are to be found in the set of conjugacy classes of bad representations.
As highlighted in a recent paper of Kapovich and Millson [6], algebraic singularities of character varieties of 3-manifold groups into SL2 can be particularly messy. There is, therefore, no hope to generalize this kind of result to 3-manifold groups.
When p is a prime number, centralizers of bad subgroups in PGL(p,ℂ) are either conjugate to a cyclic group of order p or to the direct product of two cyclic groups of order p, this is [4, Theorem 1]. This leads to a simple description of the singular locus of character varieties of free groups and surface groups in PGL(p,ℂ) (see loc. cit. for details). In the more general case at stake here, we generalize [4, Theorem 1]. In PGL(n,ℂ), centralizers of bad subgroups remain finite abelian groups. There may appear isomorphic but not conjugate centralizers of bad subgroups. In order to classify them up to conjugation, it is necessary to add an extra structure on them.
Definition 1.
Let A be a finite abelian group and C an abelian group. A C-alternate bilinear form on A is a map ϕ:A×A→C such that for all a∈A, both ϕ(a,⋅) and ϕ(⋅,a) are group morphisms and ϕ(a,a)=0. A C-alternate module is a finite abelian group A endowed with an alternate bilinear form.
We say that two C-alternate modules (A,ϕ) and (B,ψ) are isometric if there exists a group isomorphism f:A→B such that f*ψ=ϕ.
The structure of C-alternate module naturally arises when studying extensions of abelian groups by C (see Proposition 4 and Lemma 3, see also [5]). In the present paper, the classification of centralizers of irreducible subgroups reduces to the study of extensions of abelian groups by Z(SL(n,ℂ)). As a result we will mainly study C-alternate modules when C is cyclic.
Remark 1.
When C=ℤ/n and a generator c0 of C is given, one can include C inside ℚ/ℤ by sending c0k to kn. There is therefore a way to associate a ℚ/ℤ-alternate module structure to a C-structure. Even though the ℚ/ℤ-alternate module constructed depends on the generator chosen, its isometry class does not.
Following this remark, if we want to study alternate modules over cyclic groups, it suffices to understand ℚ/ℤ-alternate modules. In the sequel, ℚ/ℤ-alternate modules will simply be called alternate modules. The reader should be warned that what is true for alternate modules has no reason to be true for C-alternate modules if C is not a cyclic group nor ℚ/ℤ.
They have been studied in Tignol and Wadsworth’s book [9, Chapter 7] and in the paper of Wall [10]. These objects may be thought of as analogues of symplectic vector spaces. We are going to review some definitions and properties.
In an alternate module (A,ϕ), two elements a and b of A are orthogonal if ϕ(a,b)=0. Since ϕ is an alternate form, this relation is both reflexive and symmetric. The radicalKϕ of an alternate module (A,ϕ) is the set of elements which are orthogonal to all elements in the alternate module. A symplectic module is an alternate module whose radical is trivial.
Definition 2.
Let (A,ϕ) be an alternate module. Then a subspace of A is
isotropic if it is contained in its orthogonal,
Lagrangian if it is its own orthogonal.
It can be checked directly that being a Lagrangian is equivalent to being maximal isotropic. Applying [9, Proposition 7.5] which gives the order of Lagrangians in symplectic modules to A/Kϕ, it is easy to prove that the order of a Lagrangian of (A,ϕ) is equal to n(A,ϕ):=|A||Kϕ| and therefore only depends on (A,ϕ) and not on the specific Lagrangian. This value will be of interest to us.
If B is a finite abelian group, then we denote by B* the set of group morphisms Mor(B,ℚ/ℤ). Furthermore, ϕB:(B×B*)2→ℚ/ℤ sending ((b1,f1),(b2,f2)) to f2(b1)-f1(b2) is an alternate bilinear form and S(B):=(B×B*,ϕB) is even a symplectic module. Finally, any symplectic module is isometric to S(B) for some finite abelian group B. The result can be found in [9, Corollary 7.4], see also [10, Lemma 2].
Let A be a finite abelian group; there exist e1,…,er in A of respective order d1,…,dr such that A=〈e1〉×⋯×〈er〉 and d1∣d2∣⋯∣dr. To define an alternate bilinear form ϕ on A, one only needs to give ϕ(ei,ej) for 1≤i<j≤r with the condition that
ϕ(ei,ej)=λi,jdi
with λi,j∈ℤ i.e. that ϕ(ei,ej) is of order dividing di. This defines one and only one alternate bilinear form on A.
As stated before, the centralizer of a bad subgroup of PGL(n,ℂ) can be endowed with a natural structure of alternate module whose isometry class is invariant by conjugation in PGL(n,ℂ), see Proposition 4 for the construction. Using this, we have:
Theorem 1.
Let n≥1. Let H1¯ and H2¯ be two irreducible subgroups of PGL(n,C). Then ZPGL(n,C)(H1¯) is conjugate to ZPGL(n,C)(H2¯) if and only if the associated alternate modules are isometric.
Furthermore, we characterize alternate modules associated to centralizers of irreducible subgroups.
Theorem 2.
Let n≥1 and let (A,ϕ) be an alternate module. Then the following assertions are equivalent:
There exists an irreducible subgroup of PGL(n,ℂ) whose centralizer has an associated alternate module which is isometric to (A,ϕ).
The order of Lagrangians in (A,ϕ) divides n.
There exists an abelian group B of order n such that (A,ϕ) is isometrically embedded in S(B).
The implication from (2) to (3) in this theorem is proved using the following result in Section 4.
Theorem 3.
Let (A,ϕ) be an alternate module. Then there exists an abelian group B of order n(A,ϕ) such that (A,ϕ) is contained in the symplectic module S(B).
The proof of this theorem is constructive and independent of the rest of the paper. It only uses the decomposition of finite abelian groups. In Section 3.2, we review some of the consequences of Theorems 1–2. In particular, we define a graph whose vertices are isometry classes of alternate modules whose Lagrangians are of order dividing n with an ordered arrow from M to N if M is (isometric to) a proper submodule of N and M maximal in N (see Examples 1–3). Using Theorem 2, this graph defines by duality a stratification of χbad(Γ,PGL(n,ℂ)) which is expected to generalize the description given in [4, Theorem 3] when n is a prime number (see Remark 9).
As we mentioned already, alternate modules share many properties with vector spaces endowed with an alternate bilinear form, e.g. complete classification of symplectic ones and explicit determination of the size of Lagrangians. On the other hand, whereas any K-vector space endowed with a K-alternate bilinear form can be decomposed as an orthogonal sum of its radical with a symplectic subspace, the analogue of this statement is false for alternate modules. In conclusion, there is no known general method to classify alternate modules up to isometry. However, Theorem 3 makes it possible to list isometry classes of alternate modules whose Lagrangians are of order dividing n by checking through submodules of a few symplectic modules of order n2.
In the rest of the introduction we give some notation that will be used throughout this paper. Let G be a group. The commutator[g,h] of g and h in a group G is defined as ghg-1h-1. The following lemma is straightforward.
Lemma 1.
Let n≥1, let A,B be two matrices in GL(n,C) and λ∈C* such that [A,B]=λIn. Let v∈Cn and μ∈C* such that Av=μv. Then ABv=λμBv. Therefore B acts on the spectrum of A by multiplication by λ.
The next lemma boils down to applying the definitions.
Lemma 2.
Let G be a group and N a subgroup of Z(G), the center of G. Define π:G→G/N the quotient map. If H¯ is a subgroup of G/N, define H:=π-1(H¯) and U:=π-1(ZG/N(H¯)). The function
ϕ:U→Mor(H,N),u↦(h↦[u,h])
is well defined. It is a group morphism whose kernel is ZG(H).
Let n be a positive integer. The natural map from SL(n,ℂ) to PGL(n,ℂ) is onto and will be denoted by π. Furthermore, ξ will always denote a fixed primitive n-th root of the unity in the complex field. The identity matrix in SL(n,ℂ) will be denoted In.
Definition 3.
For any subgroup H in SL(n,ℂ), we define
the projective centralizer of H as PZn(H):=ZPGL(n,ℂ)(π(H)),
the extended centralizer of H as Un(H):=π-1(PZn(H)).
Since a subgroup of PGL(n,ℂ) is irreducible if and only if its pull-back in SL(n,ℂ) is, it follows that classifying projective centralizers of irreducible subgroups of SL(n,ℂ) is equivalent to classifying centralizers of irreducible subgroups of PGL(n,ℂ). As we shall see later, working in SL(n,ℂ) rather than in its quotient enables us to directly make use of finite group representation theory.
2 Properties of centralizers in PGL(n,ℂ)
In the first subsection, we provide a bound for the exponent and the order of the centralizers of irreducible subgroups in PGL(n,ℂ), we also prove that they are abelian. In the second subsection, we shall see how to faithfully associate an alternate module to the centralizer of an irreducible subgroup in PGL(n,ℂ).
2.1 Basic features
Applying Lemma 2, we immediately have:
Proposition 1.
Let n≥1 and let H be an irreducible subgroup of SL(n,C). Then the projective centralizer of H is abelian of exponent dividing n.
Proof.
According to Lemma 2, define the group morphism
ϕn:Un(H)→Mor(H,〈ξIn〉),u↦(h↦[u,h]).
The kernel of ϕn is the centralizer ZSL(n,ℂ)(H) of H in SL(n,ℂ). Since H is irreducible, Schur’s lemma implies that ZSL(n,ℂ)(H)=Z(SL(n,ℂ))=〈ξIn〉. Hence Un(H)/Ker(ϕn)=Un(H)/〈ξIn〉=π-1(PZn(H))/〈ξIn〉 which is equal to PZn(H). On the other hand Un(H)/Ker(ϕn) is isomorphic to a subgroup of Mor(H,〈ξIn〉). Whence the result.∎
Here, for x∈ℂ*, x denotes an arbitrarily chosen square-root of x. In the next proposition, we compute conjugacy classes of elements in the extended centralizer of an irreducible subgroup in SL(n,ℂ).
Proposition 2.
Let n≥1 and let H be an irreducible subgroup of SL(n,C). If u∈Un(H) and π(u) is of order d in Zn(H), then there exists λ∈C* such that u is conjugate to
λ(In/dξndIn/d⋱ξnd(d-1)In/d),
where
λ∈{〈ξIn〉if d is odd or d even and nd even,ξ-n(d-1)d〈ξIn〉if d is even and nd odd.
Proof.
Since H is irreducible, it follows that π(H) is also irreducible and therefore PZn(H) is finite, whence Un(H) is finite too. Hence u is of finite order whence diagonalizable. Applying Lemma 2,
s:H→ℤ/n,h↦sh,
where sh is defined by [h,u]=ξsh, is a group morphism. Let ξt be a generator of its image s(H) with t dividing n.
Lemma 1 implies that H acts on the spectrum Sp(u) of u, if h∈H and μ∈Sp(u), then h⋅μ:=ξshμ. Let X be an orbit in Sp(u) for this action, then ⊕:μ∈XEμ(u) of ℂn is stable by H.
Since H is irreducible, this non-trivial subspace is the whole space ℂn. In particular, the action of H on Sp(u) is transitive and the eigenspaces have the same dimension v>0. Remarking that H acts through the group morphism s whose image is generated by ξt, we can say (if λ is some eigenvalue of u) that Sp(u)=λ〈ξt〉 and since u is diagonalizable, u is conjugate to
λ(IvξtIv⋱ξt(nt-1)Iv).
Using the dimensions, we have n=cardSp(u)×v=ntv, hence t=v. Since π(u) is of order d, this implies that t=ndk with 0<k<d prime to d. Hence u is conjugate to
λ(In/dξndIn/d⋱ξnd(d-1)In/d).
The condition on λ is given by the equation det(u)=1 i.e. λnξndn(d-1)2=1.∎
Remark 2.
When n is prime (see [4]), d is either 1 or n and there is therefore only one non-central element up to conjugation which commutes with irreducible subgroups of PGL(n,ℂ). It makes the whole study a lot easier since any bad subgroup can be found, up to conjugation, in the centralizer of a fixed single element.
Let n≥1 and let H be an irreducible subgroup of SL(n,ℂ). We define the standard representation of Un(H) as the natural inclusion ιH of Un(H) in SL(n,ℂ). Its character Tr∘ιH will be denoted by χH. The computation of its character is straightforward from the preceding proposition.
Proposition 3.
Let n≥1, let H be an irreducible subgroup in SL(n,C) and u in Un(H). Then
χH(u)={0if u∉〈ξIn〉,nξkif u=ξkIn, where k∈Z/n.
Since Un(H) is a finite group, we may use the theory of finite group representations to get some additional properties of PZn(H). For instance:
Corollary 1.
Let n≥1 and let H be an irreducible subgroup of SL(n,C). Then the order of PZn(H) divides n2. Furthermore, PZn(H) is irreducible if and only if its order is n2.
Proof.
First, we sum up classical results of the theory of finite group representations (see, for instance, [7]). Say we are given a representation ρ:G→GL(n,ℂ) of a finite group G. Then its character χ:=Tr∘ρ has a norm defined by
∥χ∥2:=1|G|∑g∈Gχ(g)χ(g-1).
We know that ∥χ∥2 is an integer and ρ is irreducible if and only if ∥χ∥2=1. Applying this to the standard representation ιH of Un(H), we have
∥χH∥2=1|Un(H)|∑u∈Un(H)χH(u)χH(u-1)
=n3|Un(H)| (by Proposition 3).
Since ∥χH∥2 must be an integer, |Un(H)| divides n3 and therefore |PZn(H)| divides n2. Furthermore, PZn(H) is irreducible if and only if ∥χH∥2=1 if and only if |PZn(H)|=n2.∎
Remark 3.
We shall see later in Proposition 6 that the bound for the order of projective centralizers is always reached.
The second corollary deals with conjugacy classes of centralizers of irreducible subgroups in PGL(n,ℂ).
Corollary 2.
Let H1 and H2 be two irreducible subgroups in SL(n,C). Then PZn(H1) is conjugate to PZn(H2) if and only if there exists an abstract group isomorphism f from Un(H1) to Un(H2) which fixes Z(SL(n,C)) pointwise.
Proof.
We begin with the assumption that PZn(H1) is conjugate to PZn(H2) in PGL(n,ℂ). Then there exists g∈SL(n,ℂ) such that
π(g)PZn(H1)π(g)-1=PZn(H2)
and therefore, the element g conjugates Un(H1) to Un(H2). As a result, the conjugation morphism by g restricted to Un(H1) will be such an isomorphism.
Conversely, say such an f:Un(H1)→Un(H2) exists. Recall that for i=1,2, ιHi defines the injection of Un(Hi) in SL(n,ℂ). We define the representation ρ:=ιH2∘f from Un(H1) into SL(n,ℂ) and χ:=Tr∘ρ its character. Observe that for u∈Un(H1),
f(u){∉Z(SL(n,ℂ))if u∉Z(SL(n,ℂ)),=uif u∈Z(SL(n,ℂ)).
As a result, Proposition 3 leads to χH1=χH. In particular, ιH1 and ρ share the same character and are therefore conjugate. Since the respective images of ιH1 and ρ are Un(H1) and Un(H2), PZn(H1) and PZn(H2) are conjugate, too. ∎
A last consequence of Proposition 3 is that any subgroup of PZn(H) is the centralizer of an irreducible subgroup of PGL(n,ℂ). More precisely:
Corollary 3.
Let n≥1 and let H be an irreducible subgroup of SL(n,C). For any subgroup A of PZn(H), ZPGL(n,C)(ZPGL(n,C)(A))=A.
Proof.
We denote B:=ZPGL(n,ℂ)(ZPGL(n,ℂ)(A)). Clearly A≤B≤PZn(H).
Let A0 (resp. B0) be π-1(A) (resp. π-1(B)). Then A0≤B0 and ZSL(n,ℂ)(A0) is contained in ZSL(n,ℂ)(B0). Straight from the definitions,
ZPGL(n,ℂ)(B)=ZPGL(n,ℂ)(A)
and therefore Un(A0)=Un(B0). Hence
[Un(A0):ZSL(n,ℂ)(A0)]
=[Un(B0):ZSL(n,ℂ)(A0)]
=[Un(B0):ZSL(n,ℂ)(B0)][ZSL(n,ℂ)(B0):ZSL(n,ℂ)(A0)].
Applying Lemma 2 and using the finiteness of A0, [Un(A0):ZSL(n,ℂ)(A0)] is finite. Hence, [ZSL(n,ℂ)(B0):ZSL(n,ℂ)(A0)] is finite.
Assume now that A≠B; then A0≠B0. We denote ρA0 (resp. ρB0) the inclusion of A0 (resp. B0) in SL(n,ℂ), its character will be denoted by χA0 (resp. χB0).
Since A0<B0≤π-1(PZn(H))=Un(H), it follows that these representations are restrictions of the standard representation ιH of Un(H). Using the computation of this character in Proposition 3, this leads to
∥χA0∥2=n3|A0| and |χB0∥2=n3|B0|.
Since |A0|<|B0|, we have ∥χA0∥>∥χB0∥. In terms of finite group representations (cf. [7]), this means that there exists an irreducible B0-module of ℂn which is decomposed as a non-trivial sum of sub A0-modules. In particular, the centralizer of the representation ρB0 is of infinite index in the centralizer of the representation ρA0 which is a contradiction. As a result, A=ZPGL(n,ℂ)(ZPGL(n,ℂ)(A)). ∎
In the next subsection, we introduce a correspondence between projective centralizers of irreducible subgroups of SL(n,ℂ) and alternate modules.
2.2 Projective centralizers as alternate modules
In this subsection, we study projective centralizers as extensions of abelian groups by Z(SL(n,ℂ)). More generally, extensions of groups are classified using group cohomology (see [1, Chapter IV], for instance). In the sequel, we are only considering cohomology of abelian groups with trivial action on the coefficients since we are studying central extensions, but it is worth noticing that it may also be defined with non-trivial action on the coefficients and a non-commutative group acting as well. The group laws on abelian groups will be denoted additively whereas the group laws on non-necessarily abelian will be denoted multiplicatively. Let A and C be two abelian groups, a 2-cocyclez is a map from A×A→C verifying for all a1,a2,a3∈A,
z(a1,a2)+z(a1+a2,a3)=z(a2,a3)+z(a1,a2+a3).
A 2-coboundaryb is a map A×A→C such that there exists a map f:A→C verifying for all a1,a2∈A, b(a1,a2)=f(a1)+f(a2)-f(a1+a2). One directly checks that a 2-coboundary is a 2-cocycle. Denote by Z2(A,C) (resp. B2(A,C)) the group of 2-cocycles (resp. 2-coboundaries). The second cohomology group of A with coefficients in C is H2(A,C):=Z2(A,C)/B2(A,C). Elements of this group are classes of 2-cocycles and are denoted by [z] where z is a 2-cocycle.
If we are given a central extension 1→C→G→A→1, and a set-theoretic section x:A→G, we can construct a 2-cocycle for A with coefficients in C by defining
zx(a1,a2):=x(a1)x(a2)x(a1a2)-1.
It is easy to show that the cohomology class of zx in H2(A,C) does not depend on the section x. Conversely, it is possible to show that any second cohomology class is realized by an extension. Indeed, one can introduce the following group law on G=C×A given by
(c1,a1)*z(c2,a2):=(c1+c2+z(a1,a2),a1+a2).
Remark 4.
If we are given two central extensions 1→C→G→A→1 and 1→C→H→A→1 such that the associated cohomology classes are equal, then there exists an isomorphism f making the following diagram commute
If z is in Z2(A,C) and B is a subgroup of A, then ResBz∈Z2(B,C) is defined as the restriction of z to B×B. Furthermore, it is straightforward to see that ResB descends to a map (still denoted by ResB) from H2(A,C) to H2(B,C).
The following proposition whose proof is left to the reader allows us to construct a C-alternate module out of an extension by an abelian group C.
Proposition 4.
Let G be a group and let C be contained in the center of G. Let A:=G/C, let π be the natural projection of G onto A, and assume that A is abelian. For a given set-theoretic section x:A→G of π, we define
fG:A×A→C,(a,b)↦[x(a),x(b)].
Then fG is well defined, does not depend on x and is a C-alternate bilinear form on A.
Remark 5.
Let H be an irreducible subgroup in SL(n,ℂ). Applying Proposition 4 to the groups G:=Un(H), C:=Z(SL(n,ℂ)), and A:=PZn(H), we define a Z(SL(n,ℂ))-alternate module (Un(H),fUn(H)).
As mentioned in the introduction, it is more convenient to see this Z(SL(n,ℂ))-alternate module as an alternate module (i.e. with coefficients in ℚ/ℤ) by choosing an injection of Z(SL(n,ℂ)) into ℚ/ℤ. The alternate module constructed this way is called the associated alternate module to PZn(H) and is denoted by (PZn(H),ϕH). Note that the isometry class of this alternate module does not depend on the chosen injection of Z(SL(n,ℂ)) into ℚ/ℤ. It is also invariant by conjugation.
Remark 6.
One can easily express
[x(a),x(b)]=x(a)x(b)x(ab)-1x(ab)x(a)-1x(b)-1,
so that [x(a),x(b)] written additively as an element of C is equal to the difference zx(a,b)-zx(b,a). Since its cohomology class does not depend on x, this actually defines a map from H2(A,C) to the group of alternate bilinear forms on A with coefficients in C sending [z] to f[z]:=fGz, where Gz is the extension constructed out of the 2-cocycle z. This map happens to be surjective, see [5] for instance or [1, Chapter V, Section 6] for a more general treatment of this question.
What we are concerned about here is the lack of injectivity of this map. The following lemma is well known (see the aforementioned references), we include its proof here for completeness.
Lemma 3.
Let A=〈e1〉⊕⋯⊕〈er〉 be a finite abelian group and let C be an abelian group. Let z1 and z2 be two 2-cocycles satisfying
the C-alternate bilinear forms f[z1] and f[z2] are equal,
for all 1≤i≤r,
Res〈ei〉[z1]=Res〈ei〉[z2].
Then [z1]=[z2].
Proof.
By considering the 2-cocycle z=z1-z2, the proof boils down to proving that an extension 1→C→K→A→1 given by a 2-cocycle z of A with coefficients in C satisfying
z(a,b)=z(b,a) for all a,b∈A,
Res〈ei〉[z]=[0] for all i
is a trivial extension, i.e. [z]=[0]. The first condition is equivalent to K being abelian. The projection from K to A is denoted by π, the projection from A to 〈ei〉 is denoted by pi and we denote by di the order of ei. For any 1≤i≤r, we take one element ki∈π-1(ei) and define, if a=a1e1+⋯+arer∈A with 0≤ai<di,
x(a):=∑i=1raiki.
We remark that x(a)=x(p1(a))+⋯+x(pr(a)). Since K is commutative, one easily computes
zx(a,b)=x(a)+x(b)-x(a+b)
=∑i=1rx(pi(a))+∑i=1rx(pi(b))-∑i=1rx(pi(a+b))
=∑i=1rx(pi(a))+x(pi(b))-x(pi(a+b))
=∑i=1rRes〈ei〉[z](pi(a),pi(b)).
It follows that zx is the sum of pi*Res〈ei〉[z] which happen to be 2-coboundaries by hypothesis, whence zx belongs to B2(A,C). ∎
Theorem 1.
Let n≥1. Let H1¯ and H2¯ be two irreducible subgroups of PGL(n,C). Then ZPGL(n,C)(H1¯) is conjugate to ZPGL(n,C)(H2¯) if and only if the associated alternate modules are isometric.
Proof.
For i=1,2 define Hi:=π-1(Hi¯). Note that
ZPGL(n,ℂ)(Hi¯)=PZn(Hi)
for i=1,2. If PZn(H1) and PZn(H2) are conjugate, Proposition 4 and Corollary 2 clearly imply that (PZn(H1),ϕH1) and (PZn(H2),ϕH2) are isometric.
Conversely, assume that (PZn(H1),ϕH1) and (PZn(H2),ϕH2) are isometric and let ψ:PZn(H1)→PZn(H2) be the isometry between them. Let zi be a 2-cocycle associated to the extension
1→Z(SL(n,ℂ))→Un(Hi)→PZn(Hi)→1.
Let E be the group defined as the extension of PZn(H2) by Z(SL(n,ℂ)) with respect to the following 2-cocycle z1(ψ-1⋅,ψ-1⋅). Then there exists an isomorphism gψ making the first two rows of this diagram commute:
To prove the theorem, it suffices (by Corollary 2) to construct g making the last two rows of the diagram commute. This will be a consequence of Lemma 3. We first remark that, by construction, the Z(SL(n,ℂ))-alternate modules (PZn(H2),fE) and (PZn(H2),fUn(H2)) are equal.
On the other hand, write
PZn(H1)=〈e1,1〉⊕⋯⊕〈e1,r〉
and define e2,j:=ψ(e1,j)∈PZn(H2). We denote by dj the order of e1,j.
For i=1,2 and j=1,…,r, let ui,j be an element in Un(Hi) such that its projection onto PZn(Hi) is ei,j. Then, by Proposition 2, ui,j is conjugate to
λ(In/djξndjIn/dj⋱ξndj(dj-1)In/dj)
with
λ∈{〈ξIn〉if dj is odd or dj even and ndj even,ξ-n(dj-1)/dj〈ξIn〉if dj is even and ndj odd.
Up to multiplying ui,j by ξk, we may take
λ=1 if d is odd or n/d is even, so that the lift ui,j of ei,j is of order dj, or
λ=ξ-n(dj-1)/dj otherwise, so that the lift ui,j of ei,j is of order 2dj.
Let us define set-theoretic sections
x:PZn(H2)→E and y:PZn(H2)→Un(H1).
If a=a1e2,1+⋯+are2,r with 0≤ai<di, then let
x(a):=gψ(u1,1)a1⋯gψ(u1,r)ar and y(a):=u2,1a1⋯u2,rar.
If a=a1e2,1+⋯+are2,r and b=b1e2,1+⋯+bre2,r, then for all 1≤j≤r,
Res〈ej〉zx(a,b)=gψ(u1,j)aj+bj-(aj+bjmoddj)
and
Res〈ej〉zy(a,b)=u2,jaj+bj-(aj+bjmoddj).
The integer aj+bj-(aj+bjmoddj) is either 0 or dj. As a result, if the common order of gψ(u1,j) and u2,j is dj, both Res〈ej〉zx(a,b) and Res〈ej〉zy(a,b) are trivial, and if the common order of gψ(u1,j) and u2,j is 2dj, then Res〈ej〉zx(a,b) and Res〈ej〉zy(a,b) are both equal to the unique element of order 2 in Z(SL(n,ℂ)) (it exists because n needs to be even for this case to arise). In any case we have Res〈ej〉zx(a,b)=Res〈ej〉zy(a,b). Thus we can apply Lemma 3 to zx and zy and we end up with [zx]=[zy]. Using Remark 4, we can construct the required isomorphism g between E and Un(H2), and conclude that PZn(H1) and PZn(H2) are conjugate using Corollary 2.∎
In the next section, we shall see under which necessary and sufficient condition an alternate module (A,ϕ) is associated to the centralizer of some irreducible subgroup of PGL(n,ℂ).
3 Classification of centralizers in PGL(n,ℂ)
Corollary 1 implies that the centralizer of an irreducible subgroup of PGL(n,ℂ) is of order dividing n2. We say that it is full if its order is exactly n2.
In the first subsection, we give a necessary and sufficient condition for an alternate module to be associated to such a centralizer in terms of the order of its Lagrangians. Full centralizers play a key role in the study. Whereas it is obvious from the preceding remark that full centralizers are maximal elements of the set of centralizers of irreducible subgroups of PGL(n,ℂ), it appears that all maximal elements of this set are full centralizers. This last result is proved using Theorem 3 whose proof is postponed to Section 4. In the second subsection, we focus on the consequences of this result.
3.1 Lagrangians in projective centralizers
In order to prove our classification result, we need a correspondence between properties of the standard representation of Un(H) and geometric features of (PZn(H),ϕH).
Once again, we begin with a result from representation theory. Let G be a finite group and ρ:G→GL(n,ℂ) a representation of G acting linearly on ℂn through ρ. Then ℂn may be decomposed as a sum of maximal isotypic sub-G-modules V1, …, Vk (i.e. which are linearly equivalent to λi⋅(Wi,ρi), where (Wi,ρi) is an irreducible representation of G and λi>0). Furthermore, up to the order of (Vi), the decomposition happens to be unique. This leads to a technical lemma:
Lemma 4.
Let n≥1, and let G be a finite subgroup of GL(n,C). Define ιG to be the inclusion of G into GL(n,C). Let V1⊕⋯⊕Vk be the maximal isotypic decomposition of (Cn,ιG). Then the normalizer of G in GL(n,C) acts on the set {V1,…,Vk} of subspaces occurring in the maximal isotypic decomposition of (Cn,ιG).
Proof.
Let n be in NGL(n,ℂ)(G) and 1≤i≤k. For all g∈G and v∈Vi,
g⋅(n(vi))=nn-1gn⏟∈G⋅(vi)⏞∈Vi.
As a result, nVi is stable by G and nVi is a submodule of V. Furthermore, if (Vi,ρi′) is the induced representation on Vi by ιG, then the representation on nVi induced by ιG is isomorphic to (Vi,ρi′(n-1⋅n)). It follows that (nVi,ιG) is isotypic and is therefore included in a maximal isotypic submodule. This implies that there exists 1≤j≤r such that nVi⊆Vj if and only if Vi⊆n-1Vj.
Using the same argument, n-1Vj is also isotypic, but it contains Vi which is maximal isotypic. It follows that Vi=n-1Vj, whence nVi=Vj. ∎
With this lemma, we can bound the order of Lagrangians of alternate modules associated to centralizers of irreducible subgroups in PGL(n,ℂ).
Proposition 5.
Let n≥1 and let H be an irreducible subgroup of SL(n,C). Then the order n(PZn(H),ϕH) of Lagrangians in (PZn(H),ϕH) divides n.
Proof.
Let K be an isotropic subspace of (PZn(H),ϕH) and let K0 be its pull-back in Un(H). If k1,k2∈K0, then, by definition, we have [k1,k2]=In. Therefore, π-1(K) is an abelian subgroup in Un(H). In particular, if L is a Lagrangian in (PZn(H),ϕH), then L0:=π-1(L) is an abelian subgroup of Un(H). Denote by ι0:L0→SL(n,ℂ) the inclusion of L0 in SL(n,ℂ). Let h∈H and x∈L0. Then hxh-1∈〈ξIn〉x since x∈Un(H). Because 〈ξIn〉 is included in L0, it follows that hxh-1 belongs to L0 as well. Whence, H normalizes L0.
We denote by V1, …, Vk the maximal isotypic subspaces occurring in (ℂn,ι0). Using Lemma 4, H acts on {V1,…,Vk}. Since H is irreducible, the action is transitive. In particular, V1,…,Vk share the same dimension d>0.
Since L0 is an abelian group, each Vi can be decomposed as d copies of a 1-dimensional representation. We denote by χ0 the character of ι0 and by χi the character of (Vi,ι0,|Vi). By considerations of finite group representation theory (see [7]), ∥χi∥2=d2. Using the orthogonality of the characters χi and χj for i≠j, we have ∥χ0∥2=kd2=ddim(V)=dn. In the meantime, χ0=χH|L0, where χH is the character of the standard representation ιH of Un(H) whose character has been computed in Proposition 3. Hence
∥χ0∥2=n3|L0|.
As a result, we have |L0|=nnd. Since L0=π-1(L) and the kernel of π is of order n, we end up with |L|=nd. In particular, the order n(PZn(H),ϕH) of L divides n. ∎
Remark 7.
From the proof of Proposition 5, we see that the order of Lagrangians in (PZn(H),ϕH) is equal to the number of isotypic components of the standard representation of Un(H).
We have just given a necessary condition for an alternate module to be associated to the centralizer of an irreducible subgroup of PGL(n,ℂ). It will be proved to be a sufficient condition as well in Theorem 2. In order to prove this, we need to construct some particular examples of such centralizers.
Proposition 6.
For any n≥1 and any abelian group B of order n, there exists a finite irreducible subgroup H of SL(n,C) such that PZn(H) is isomorphic to B×B.
Proof.
We prove this by induction on n≥1. If n=1, then the property is obviously true. Assume n>1. Let B be an abelian group of order n and let d be the exponent of B. Denote by B:=B1×〈e〉, where e is of order d. The induction hypothesis implies the existence of a finite irreducible subgroup K of SL(nd,ℂ) such that PZn/d(K) is isomorphic to B1. We define a subgroup K0 of SL(n,ℂ) by blocks of size nd:
K0:={(k⋱k):k∈K}.
We also write
M:=λ(0In/dIn/d⋱⋱⋱In/d0)
and
u:=λ(In/dξndIn/d⋱ξnd(d-1)In/d),
where λ is defined as in Proposition 2 (i.e. so that det(M)=det(u)=1). Finally, let H be the subgroup of SL(n,ℂ) generated by K0, M, u and ξIn.
Since u is scalar by blocks of size nd with distinct values, it follows that
ZSL(n,ℂ)(u)={(g0⋱gd-1):g0,…,gd-1∈GL(nd,ℂ),det(g0)⋯det(gd-1)=1}.
The commutator [M,u] is of order d in Z(SL(n,ℂ)) which is the order of π(u). As a result,
Un(H)⊆Un(〈u〉)=〈ZSL(n,ℂ)(u),M〉.
Let g0,…,gd-1 be in GL(nd,ℂ) and b≥0 such that
x:=(g0⋱gd-1)Mb∈Un(H).
Then [M,x] is in Z(SL(n,ℂ)) and since π(M) is of order d, it follows that [M,x] is of order dividing d. Since [M,u] is of order d, up to multiplying x by some power of u, we may assume that [M,x]=In. But this implies (by a straightforward calculation) that g0=⋯=gd-1. Hence, we have showed that
Un(H)⊆{(g⋱g)ulMb:g∈GL(nd,ℂ), det(g)d=1,b,l≥0}.
In the sequel, D(g) denotes the matrix
D(g)=(g⋱g),g∈GL(nd,ℂ).
Let g∈GL(nd,ℂ) with det(g)d=1. Then det(g) is a d-th root of the unity. Up to multiplying D(g) by an nd-th root of det(g) (which is an n-th root of the unity, i.e. ξs for some integer s), we may assume that det(g)=1. Whence
(3.1)Un(H)⊆{ξsD(g)ulMb:g∈SL(nd,ℂ),s,b,l≥0}.
Let us define a function
ψ0:Un(H)→〈ξn/d〉×〈ξn/d〉,v↦([v,M],[v,u]).
The function ψ0 is a group morphism (a consequence of Lemma 2), it is surjective since ψ0(u) and ψ0(M) generate 〈ξn/d〉×〈ξn/d〉 and it factors through PZn(H). It follows that we have a surjective isomorphism ψ from PZn(H) to ℤ/d×ℤ/d.
From inclusion (3.1), it follows that Ker(ψ)⊆{π(D(g)):g∈SL(n,ℂ)}. In particular, 〈π(M),π(u)〉 is of trivial intersection with Ker(ψ). Since
ψ(〈π(M),π(u)〉)=〈ξn/d〉×〈ξn/d〉,
it follows that
PZn(H) is isomorphic to Ker(ψ)×ℤ/d×ℤ/d.
Let k∈K and g∈SL(nd,ℂ). Then [D(k),D(g)]=D([k,g]). As a result, D(g)∈Un(H) if and only if g∈Un/d(K). In particular, this leads to a well-defined group morphism modulo 〈ξIn〉:
φ:Ker(ψ)→PZn/d(K),π(D(g))↦π(g).
Because φ is injective, Ker(ψ) is isomorphic to PZn/d(K) which is isomorphic to B1×B1. As a result, PZn(H) is isomorphic to B×B. Note that H is finite and therefore completely reducible, and since its centralizer is of finite order, it is irreducible. Whence the conclusion by induction.∎
In the proof, we did not only find an irreducible subgroup H¯ of PGL(n,ℂ) whose centralizer is isomorphic to B×B but, as is, the induction also shows that H¯ is its own centralizer. We did not emphasize it there because we shall see, in the next proposition, that full centralizers have many properties including this one. We recall that for a given abelian group, S(B) denotes the symplectic module B×B* endowed with its natural alternate bilinear form
ϕ((b1,f1),(b2,f2))=f2(b1)-f1(b2).
Proposition 7.
Let n≥1 and let H be an irreducible subgroup of SL(n,C) containing the center of SL(n,C). Then the following assertions are equivalent:
PZn(H) is a full centralizer.
π(H) is its own centralizer.
(PZn(H),ϕH) is isometric to S(B), where B is some abelian group of order n.
Proof.
The equivalence (1) ⇔ (2)
is mutatis mutandis the second assertion in Corollary 1. The implication
(5) ⇒ (1) is obvious.
(2) ⇒ (3) Say PZn(H) is irreducible; then its centralizer is abelian by Proposition 1, therefore, π(H) is abelian as well.
(3) ⇒ (4) Let ρ be the inclusion of H into SL(n,ℂ) and χ its character. On one hand, since H is included in Un(H), ρ is the standard representation ιH of Un(H) restricted to H whose character has been computed in Proposition 3. It follows that ∥χ∥2=n3|H|. On the other hand, since H is irreducible, ρ is irreducible and then ∥χ2∥=1. Whence |H|=n3 and |π(H)|=n2. Since |PZn(H)|≤n2 and π(H) is included in PZn(H), it follows that π(H)=PZn(H) and π(H) is its own centralizer.
(4) ⇒ (2) If π(H) is its own centralizer, then PZn(H)=π(H) is irreducible.
(1) ⇒ (5) The alternate module (PZn(H),ϕH) is of order n2. By Proposition 5, n(PZn(H),ϕH):=|PZn(H)||KϕH| divides n. It follows that |KϕH| divides 1. Whence KϕH is trivial and (PZn(H),ϕH) is a symplectic module of order n2. By [9, Corollary 7.4], there exists an abelian group B of order n such that (PZn(H),ϕH) is isometric to S(B).
∎
We end this subsection with a characterization theorem:
Theorem 2.
Let n≥1 and let (A,ϕ) be an alternate module. Then the following assertions are equivalent:
There exists an irreducible subgroup of PGL(n,ℂ) whose centralizer has an associated alternate module which is isometric to (A,ϕ).
The order of Lagrangians in (A,ϕ) divides n.
There exists an abelian group B of order n such that (A,ϕ) is isometrically embedded in S(B).
Proof.
The implication
(1) ⇒ (2) is the statement of Proposition 5, and
(2) ⇒ (3) is a straightforward consequence of Theorem 3.
(3) ⇒ (1) Assume that (A,ϕ) is isometrically embedded in S(B). Let H be an irreducible subgroup such that PZn(H) is isomorphic (as a group) to B×B (by Proposition 6). Since |PZn(H)|=|B|2=n2, it follows from Proposition 7 that (PZn(H),ϕH) is isometric to S(B). Let f be an isometry from PZn(H) to B×B*.
Let (A0,ϕ0) the submodule of (PZn(H),ϕH) which is sent to (A,ϕ) by f. By Corollary 3, A0 is the projective centralizer of an irreducible subgroup K of SL(n,ℂ), i.e. PZn(K)=A0. Furthermore, from the definition of (A0,ϕK) and (PZn(H),ϕH), ϕ0:=ϕH|A0×A0=ϕK. Hence, (A0,ϕ0)=(PZn(K),ϕK) and (A,ϕ) is therefore isometric to (PZn(K),ϕK). ∎
3.2 Consequences of the classification
Something that may be the most obvious consequence is the following:
Corollary 4.
Let n≥1. Then the number of conjugacy classes of centralizers of irreducible subgroups in PGL(n,C) is finite.
In the next corollary, we characterize the isomorphism classes of centralizers:
Corollary 5.
Let n≥1 and let A be an abelian group. Then there exists an irreducible subgroup H of SL(n,C) such that PZn(H) is isomorphic to A if and only if there exists an abelian group B of order n such that A is isomorphic to a subgroup of B×B.
In particular, for any abelian group A, there exists an integer n≥1 and an irreducible subgroup H of SL(n,C) such that PZn(H) is isomorphic to A.
Proof.
It suffices to use the equivalence between the first and the third assertion in Theorem 2 and forget about the alternate bilinear form. ∎
Let e′, e and d be three integers such that e′ divides e and e divides d. Furthermore, let A:=ℤ/e×ℤ/d with b a generator of ℤ/e×{0} and a a generator of {0}×ℤ/d. Then ϕ(b,a):=1e′ defines an alternate bilinear form on A. In the sequel we shall write Me′,e,d for this alternate module.
We recall that an integer n≥1 is square-free if it cannot be divided by a non-trivial square.
Lemma 5.
Let n be a square-free integer and let (A,ϕ) be an alternate module of rank 2, of exponent dividing n and such that n(A,ϕ) divides n. Let A be isomorphic to 〈b〉×〈a〉 with e (the order of b) dividing d (the order of a) dividing n. Then (A,ϕ) is isometric to Me,e,d.
Proof.
Let e′ be the order of ϕ(b,a). Then Kϕ contains 〈be′〉 of order ee′ and 〈ae′〉 of order de′. Since those two groups are in trivial intersection, Kϕ contains their direct product and |Kϕ| is divided by de(e′)2. Remark that |A|=de. It follows that dee′ divides |A||Kϕ|=:n(A,ϕ).
Since n(A,ϕ) divides n by assumption, dee′ divides n. Assume that e′ is a strict divisor of e and take p to be a prime number dividing ee′; then p divides d and, hence, p2 divides n which contradicts the fact that n is square-free. Hence the order of ϕ(b,a) is e. Define ϕ(b,a)=λe∈ℚ/ℤ. Since ϕ(b,a) is of order e, it follows that λ is prime to e. Let μ be an integer such that λμ=1 mod e. Then the automorphism f:A→A sending b to bμ and a to a is an isometry between (A,ϕ) and Me,e,d. ∎
Remark 8.
If p is a prime number, conjugacy classes of centralizers of irreducible subgroups in PGL(p,ℂ) are classified by their isomorphism classes. Indeed, there are only two of them which are ℤ/p and ℤ/p×ℤ/p (this is [4, Theorem 1]). It appears that this property still holds provided n is square-free.
Proposition 8.
Let n≥1. Then conjugacy classes of centralizers of bad subgroups of PGL(n,C) are classified by their isomorphism classes if and only if n is square-free.
Proof.
Assume that n is square-free. Let H1 and H2 be two irreducible subgroups in SL(n,ℂ) such that PZn(H1) and PZn(H2) are isomorphic. First note that, by Corollary 5, PZn(Hi) is necessarily isomorphic to a subgroup of B×B, where B is abelian of order n. Since n is square-free, B is actually cyclic. It follows that the common rank of PZn(H1) and PZn(H2) is less than or equal to 2.
If the common rank of PZn(H1) and PZn(H2) is either 0 or 1, then for i=1,2, (PZn(Hi),ϕHi) is necessarily a trivial module (i.e. ϕHi=0). Since PZn(H1) and PZn(H2) are isomorphic, it is clear that (PZn(H1),ϕH1) and (PZn(H2),ϕH2) are isometric. Otherwise, the common rank of PZn(H1) and PZn(H2) is 2. Let ℤ/e×ℤ/d be the isomorphism class of both PZn(H1) and PZn(H2). Since n(PZn(H1),ϕH1) and n(PZn(H2),ϕH2) both divide n, we are in the situation where Lemma 5 applies.
In any case, the modules (PZn(H1),ϕH1) and (PZn(H2),ϕH2) are isometric which implies that PZn(H1) and PZn(H2) are conjugate using Theorem 1.
Assume that n is not square-free and let p be a prime number such that p2∣n. We define two structures of alternate module on ℤ/p×ℤ/p: (ℤ/p×ℤ/p,0) the trivial structure and M2:=S(ℤ/p) the symplectic structure. The second criterion in Theorem 2 justifies that both alternate modules are associated to centralizers of irreducible subgroups of PGL(n,ℂ). However, these centralizers are not conjugate because of Theorem 1. Thus, in this case, we have two isomorphic centralizers of irreducible subgroups of PGL(n,ℂ) which are not conjugate to each other. ∎
As a result, when n is square-free, we can compute the number of conjugacy classes of centralizers of irreducible subgroups in PGL(n,ℂ).
Corollary 6.
Let n be a square-free integer and let r be the number of distinct prime numbers dividing n. There are exactly 3r conjugacy classes of centralizers of irreducible subgroups in PGL(n,C).
Now we would like to highlight a last consequence that might be the most fruitful. Let ℳn be the set of isometry classes of alternate modules which are n-subsymplectic. We see (ℳn,≤) as an ordered set, where ≤ is the usual relation of inclusion (up to isometry). For any n≥1, we define a graph structure Gn, by taking for the set of vertices, the set ℳn and having an edge from M to N if M is (isometric to) a proper submodule of N and M maximal in N, or equivalently, if |M|/|N| is a prime number.
Before giving some examples we give notation for some alternate modules. In the sequel, a cyclic group ℤ/k endowed with a trivial alternate form is denoted by Ck. Furthermore, ⊕ denotes the orthogonal sum. The next examples are easily computed. We only draw the associated graphs.
Example 1.
As seen in Figure 1, there are exactly three conjugacy classes of centralizers of irreducible subgroups in PGL(p,ℂ), where p is a prime number.
Example 2.
As seen in Figure 2, there are exactly nine conjugacy classes of centralizers of irreducible subgroups in PGL(p2,ℂ) where p is a prime number.
Example 3.
As seen in Figure 3, there are exactly 24 conjugacy classes of centralizers of irreducible subgroups in PGL(p3,ℂ), where p is a prime number.
Let us focus briefly on character varieties. Let Γ be a finitely generated group. We denote by χbad(Γ,PGL(n,ℂ)) the set of conjugacy classes of bad representations. Our goal is to decompose χbad(Γ,PGL(n,ℂ)).
Let M be the isometry class of an alternate module in ℳn. Because of Theorem 2, there exists ZM, the centralizer of some irreducible subgroup in PGL(n,ℂ) whose associated alternate module is M. For any M∈ℳn, we define a subset χM of the character variety as
χM:={PGL(n,ℂ)⋅ρ:ρ is irreducible and ZPGL(n,ℂ)(ρ)≤ZM}.
Since χM only depends on the conjugacy class of ZM, it only depends on M. We end up with the decomposition
(3.2)χbad(Γ,PGL(n,ℂ))=⋃M∈ℳn|M|>1χM.
Considering ℒ𝒮n to be the set of all χM for M being an element of ℳn partially ordered by the inclusion, we have:
Proposition 9.
For n≥2 and Γ a finitely generated group, the function from Mn to LSn sending M to χM is decreasing. Furthermore, if M∈Mn is of order n2, then χM is finite.
Proof.
Assume that M≤N with M,N∈ℳn. Then we may choose up to conjugation ZM≤ZN. Hence, we have ZPGL(n,ℂ)(ZN)≤ZPGL(n,ℂ)(ZM) which leads to χN⊆χM.
For the second assertion, note that if M is of order n2, then ZM is a full centralizer. By Proposition 7, ZM is irreducible and self-centralizing:
ZPGL(n,ℂ)(ZM)=ZM.
It follows that for any representation ρ of Γ such that its conjugacy class belongs to χM, the image of ρ must be equal to ZM (taking H:=π-1(ρ(Γ)) in Proposition 7).
Consider, in Γ, the set X of normal subgroups N such that Γ/N is isomorphic to ZM. Since ZM is abelian, X is in bijection with the set X′ of subgroups A in ΓAb such that ΓAb/A is isomorphic to ZM. Since Γ is finitely generated, ΓAb is finitely generated. This implies that the set X′ is necessarily finite. For any representation ρ whose conjugacy class belongs to χM, Ker(ρ)∈X.
If we denote χM,N the set of conjugacy classes of representations whose kernel is equal to N, then χM,N is exactly the set of conjugacy classes of representations from Γ/N onto ZM. Since we are dealing with finite group representations, χM,N is necessarily finite. Since χM is a union of χM,N for N∈X and X is finite, it follows that χM is finite as well. ∎
With this proposition, the decomposition in equation (3.2) of the bad locus of character varieties is governed by the graph Gn we defined earlier. Each vertex M of Gn leads to a stratum of the bad locus, and the minimal strata of the bad locus, namely the χM’s where M is a symplectic module of order n2, are simply finite unions of points.
Remark 9.
This begins the generalization of [4, Theorem 3]. In [4, Theorem 3], when n is a prime number, we have two strata of bad representations. Using the alternate module construction defined in the present paper, we have a stratum associated to the module (ℤ/n,0) which is described in [4] as a finite union of closed sub-varieties (called pseudo-components) and another smaller stratum associated to the module S(ℤ/n) which consists of a finite union of points. These points appear to be the intersection points of any pairs of so-called pseudo-components. To end this generalization, one should understand how to introduce the pseudo-components when n is not necessarily prime.
To define these pseudo-components on the stratum associated to an alternate module M, it seems that one should consider an inclusion of M inside a symplectic module of order n2. Unlike the case where n is prime, it is not uniquely defined. However, it always exists because of Theorem 3.
4 Appendix on alternate modules
In this section, we prove:
Theorem 3.
Let (A,ϕ) be an alternate module. Then there exists an abelian group B of order n(A,ϕ) such that (A,ϕ) is contained in the symplectic module S(B).
The proof of this theorem follows from a straightforward recursion on the order of alternate modules provided that the following lemma is true.
Lemma 6.
Let (A,ϕ) be an alternate module whose radical is not trivial. Then there exists an alternate module (A^,ϕ^) such that A^ strictly contains A, ϕ^|A×A=ϕ and n(A,ϕ)=n(A^,ϕ^).
The proof of Lemma 6 will be elementary, although non-trivial, and constructive. We first gather a few facts whose proofs will be shortened if not totally omitted when they can be left to the reader.
Using [9, Proposition 7.1], which states that any alternate module can be decomposed as an orthogonal sum of alternate p-modules where p is prime (i.e. alternate modules where the underlying abelian group is a p-group), we have:
Fact 1.
If Lemma 6 works for alternate p-modules for all prime numbers p, then it works for any alternate module.
Another fact, deduced from the corresponding result in symplectic modules:
Fact 2.
Let (A,ϕ) be an alternate module and B a subgroup of A. If (B,ϕ|B×B) is symplectic, then the alternate module (A,ϕ) is the sum of B with its orthogonal.
In Lemma 6, the goal is to construct an extension of alternate modules without changing the order of Lagrangians. The next fact gives a simple criterion for this.
Fact 3.
Let p be a prime number, let (A,ϕ) and (A^,ϕ^) be two alternate p-modules such that (A,ϕ) is contained in (A^,ϕ^) and [A^:A]=p. If Kϕ is not contained in Kϕ^, then Kϕ^ is contained in Kϕ; furthermore, in this case n(A,ϕ)=n(A^,ϕ^).
Proof.
Remark that if e^ belongs to A^ and not to A,
then A^ is generated by A and e^. Therefore, f:Kϕ→ℚ/ℤ sending k to ϕ^(e^,k) is a group morphism whose image cannot be trivial. A straightforward consequence of this is that Ker(f) is exactly Kϕ^ which is thus contained in Kϕ. The equality n(A,ϕ)=n(A^,ϕ^) follows easily.∎
Proof of Lemma 6.
Using Fact 1, we only need to prove Lemma 6 when (A,ϕ) is an alternate p-module. The proof is itself an induction on the order of A with three distinct cases (the induction hypothesis will only be necessary for the second case). We fixe some notation: A is equal to 〈e1〉×⋯×〈er〉, the order of ei divides the order of ei+1 for 1≤i≤r-1 and Kϕ is the radical (A,ϕ). We recall that we want to find an alternate module (A^,ϕ^) strictly containing (A,ϕ) such that n(A^,ϕ^)=n(A,ϕ).
First case: Kϕ is not contained in p⋅A. Let x be in Kϕ such that x is not in p⋅A. Then x=γ1⋅e1+⋯+γr⋅er and there exists 1≤k≤r such that γk is not divisible by the prime number p. Define A^:=〈e〉×A, where e is of order p, and define the alternate form ϕ^ on A^ such that
ϕ^|A×A=ϕ and ϕ^(e,ei):={0if i≠k,1pif i=k.
Clearly, (A^,ϕ^) contains (A,ϕ) and x does not belong to Kϕ^ by construction. Applying Fact 3, n(A^,ϕ^)=n(A,ϕ) and the statement is true in this case.
Second case: There exists 2≤j≤r such that the order of ϕ(e1,ej) is equal to the order of ej. Since the order of e1 divides the order of ej and the order of ϕ(e1,ej) divides the order e1, we see that the sub-module B generated by e1 and ej is symplectic. Therefore, using Fact 2, (A,ϕ) is the orthogonal sum of B and C, the orthogonal of B. By induction hypothesis, there exists an extension C^ of C like in the lemma and therefore the alternate module (A^,ϕ^) defined as the orthogonal sum of B and C^ fulfills the needed conditions.
Third case: For 2≤j≤r, the order of ϕ(e1,ej) is smaller than the order of ej and Kϕ is contained in p⋅A. Let A^ be the group 〈e1^〉×〈e2〉×⋯×〈er〉, where the order of e1^ is p times the order of e1. The function
ιA:A→A^,e1↦p⋅e1^,e2↦ei for 2≤i≤r
is a group inclusion of A into A^. We will define an alternate form ϕ^ on A^ such that the inclusion ιA is an inclusion of alternate modules.
For 2≤j≤r, define λj to be a p-th root of ϕ(e1,ej) (i.e. such that p⋅λj=ϕ(e1,ej)). It is always possible to do so because ℚ/ℤ is a divisible group. Because of the assumption on the order of ϕ(e1,ej), the equations
{ϕ^(ei,ej):=ϕ(ei,ej)for 2≤i<j≤r,ϕ^(e1^,ej):=λjfor 2≤j≤r
lead to a well-defined alternate form ϕ^ on A^ which clearly satisfies ιA*ϕ^=ϕ. Let us show that Kϕ is not contained in Kϕ^. Let κ be the standard projection of A onto the symplectic module (A/Kϕ,ϕ¯). By [9, Corollary 7.4], we define g1,g1*,…,gs,gs* to be a symplectic base for A/Kϕ and then write
κ(e1)=∑i=1rγigi+γi*gi*,where γi,γi*∈ℤ.
Assume that κ(e1) belongs to p⋅(A/Kϕ). Since Kϕ is included in p⋅A by assumption, it follows that e1 belongs to p⋅A which is impossible. Therefore we have κ(e1)∉p⋅(A/Kϕ). As a result, there exists i0 such that (swapping gi0 and gi0* if necessary) γi0* is not divisible by p. Take a0∈A such that κ(a0)=gi0 and define pk to be the order of gi0. By definition, pk⋅a0 belongs to Kϕ but
ϕ^(e1^,pk⋅a0)=ϕ(e,pk-1⋅a0)=ϕ¯(κ(e),pk-1⋅gi0)=pk-1ϕ¯(κ(e1),gi0)
=pk-1∑i=1rγiϕ¯(gi,gi0)+γi*ϕ¯(gi*,gi0)
=pk-1γi0*ϕ¯(gi0*,gi0).
The very definition of a symplectic base ensures that ϕ¯(gi0*,gi0) and gi0 have the same order pk and pk-1γi0* is not divisible by pk. Hence pk⋅a0 is not orthogonal to e1^. As a result, Kϕ is not included in Kϕ^ and applying Fact 3, the statement is true.∎
