The Dimension of the Hitchin Component for Triangle Groups

D.D.LONG M. B.THISTLETHWAITE

Let p, q, r be positive integers satisfying 1/p + 1/q + 1/r < 1, and let ∆(p, q, r) be a geodesic triangle in the hyperbolic plane with angles π/p , π/q , π/r. Then there exists a tiling of thehyperbolicplaneby triangles congruentto ∆(p, q, r),and we define the triangle group T(p, q, r) to be the group of orientation preserving isometries of this tiling. Representation varieties of closed surface groups into SL(n, R) have been studied extensively by Hitchin and Labourie, and the dimension of a certain distinguished component of the variety was obtained by Hitchin using Higgs bundles. Here we determine the corresponding dimension for representations of triangle groups into SL(n, R), generalising some earlier work of Choi and Goldman in the case n = 3.

22E40; 20H10

D.D. Long Department of Mathematics, University of California, Santa Barbara, CA 93106 [email protected]

M.B. Thistlethwaite Department of Mathematics, University of Tennessee, Knoxville, TN 37996 [email protected] 865–974–4268 2 D. D. Long and M. B. Thistlethwaite

1 Introduction

It is well known that the angle sum of a triangle is greater than π in the round 2– sphere, equal to π in the Euclidean plane and less than π in the hyperbolic plane. Let p, q, r be integers greater than or equal to 2, and let ∆(p, q, r) be a triangle in the appropriate geometry with angles π/p , π/q , π/r at vertices P, Q, R respectively. We assume that the ordering (P, Q, R) of vertices is consistent with a chosen orientation of the ambient space. The triangle group associated with ∆(p, q, r) is the group of isometries T(p, q, r) generated by rotations rP , rQ , rR about P, Q, R through angles +2π/p , +2π/q , +2π/r respectively. It is easily checked that the orbit of ∆(p, q, r) under the action of T(p, q, r) constitutes a tiling of the homogeneous space (S2 , R2 or 2 H ) by triangles congruent to ∆(p, q, r), and that the product rP ◦ rQ ◦ rR (composition being as usual from right to left) is the identity. The quotient under the action of T(p, q, r) is a 2–sphere with three cone points of orders p, q, r, whose (orbifold) fundamental group is easily shown to be ha , b , c | ap = bq = cr = abc = 1i, the generators corresponding to the rotations rP , rQ , rR . From elementary covering space theory in the orbifold setting it follows that the group T(p, q, r) also admits this presentation, i.e. T(p, q, r) = ha , b , c | ap = bq = cr = abc = 1i . For triangles ∆(p, q, r) in the 2–sphere, the group T(p, q, r) is finite; for each of the cases (p, q, r) = (2, 2, n) it is a dihedral group, and for each of the remaining cases (p, q, r) = (2, 3, 3) , (2, 3, 4) , (2, 3, 5) it is the group of rotational symmetries of a platonic solid. The three Euclidean triangle groups T(2, 3, 6) , T(2, 4, 4) , T(3, 3, 3) all generate familiar tilings of the Euclidean plane. In this paper we consider exclusively hyperbolic triangle groups, namely those for which 1/p + 1/q + 1/r < 1. Let us fix a triangle ∆(p, q, r) in the hyperbolic plane. The holonomy representation φ embeds the resulting triangle group T(p, q, r) as a discrete subgroup of Isom+(H2) ≈ PSL(2, R). Since a triangle in the hyperbolic plane is determined up to congruence by its angles, φ is rigid, in the sense that its only deformations are those afforded by conjugation. However, if we compose φ with the (unique) irreducible representation ρn of PSL(2, R) into PSL(n, R), for all sufficiently large n we obtain a representation of degree n of T(p, q, r) that admits essential deformations, i.e. deformations that do not arise from post•composition with an inner automorphism of PSL(n, R). The purpose of this article is to establish a general formula for the dimension of the deformation space of such a representation. The interest in representation varieties of hyperbolic triangle groups lies partly in the fact that these groups contain surface groups as subgroups of finite index, allowing one The Dimension of the Hitchin Component for Triangle Groups 3 to draw on deep work of N. Hitchin [2] and F. Labourie [3]. Borrowing terminology of [3], we call the component of the representation variety containing ρn ◦ φ the Hitchin component. The Hitchin component is homeomorphic to Rk for some k, and all representations therein are discrete and faithful [3]. In the case where there are no essential deformations, we have k = n2 − 1. The statement of the main theorem involves a certain arithmetic function of two variables σ(n, p) (n, p ≥ 2). Let Q , R be the quotient and remainder on dividing n by p, i.e. n = Q p + R (0 ≤ R ≤ p − 1) . Then σ(n, p) = (n + R)Q + R . For example, for n = 7 , p = 4, we have Q = 1 , R = 3; thus σ(7, 4) = 13. It will be seen later that the function σ(n, p) is related to the dimension of the centralizer in SL(n, R) of a certain diagonalizable matrix whose eigenvalues are roots of unity.

Theorem 1.1 Let H be the Hitchin component for the representation ρn ◦ φ of the triangle group T(p, q, r). Then dim H = (2n2 + 1) − (σ(n, p) + σ(n, q) + σ(n, r)) .

Remarks (i) For p = n, q = 1 and r = 0, and for p > n, q = 0 and r = n. It follows that for p ≥ n, σ(n, p) = n; therefore if none of p, q, r is less than n, dim H = 2n2 + 1 − 3n = (2n − 1)(n − 1). (ii) If one wishes to “factor out” conjugation and consider the moduli space H∗ of essential deformations, one simply subtracts dim PSL(n, R) = n2 − 1 from the expression of Theorem 1.1, obtaining ∗ dim H = (n2 + 2) − (σ(n, p) + σ(n, q) + σ(n, r)) . From the definition of σ(n, p) a quick computation shows that σ(n, p) ≡ n (mod 2), from which we see that dim H∗ is always even. (iii) The case n = 3 is covered in S. Choi’s and W.M. Goldman’s paper [1]. The special case of their result that pertains to hyperbolic triangle groups can be summarized as follows: if one of p, q, r is 2, T(p, q, r) is rigid, and otherwise dim H∗ = 2. From Theorem 1.1 one finds that triangle groups of type (2, 3, r) (r ≥ 7) cease to be rigid at 4 D. D. Long and M. B. Thistlethwaite n = 6, and those of type (2, q, r) (4 ≤ q < r) cease to be rigid at n = 4 (see [4], §4 for examples).

(iv) Using Higgs bundle techniques, N. Hitchin [2] proved that for a compact surface S, the Hitchin component of the representation variety of π1(S) into PSL(n, R) has dimension χ(S)(1 − n2).

In outline, the proof of Theorem 1.1 proceeds as follows: in §3 the deformation spaces of the finite cyclic groups hai , hbi , hci are studied, and then in §4 the role of the relation abc = 1 is determined by means of a transversality argument in conjunction with Schur’s Lemma.

2 From PSL(n, R) to SL(n, R)

Our primary aim is to examine deformations of ρn ◦ φ : T(p, q, r) → PSL(n, R), where φ : T(p, q, r) → PSL(2, R) is the holonomy representation and ρn : PSL(2, R) → PSL(n, R) is the irreducible representation, namely the projective version of the familiar representation σn : SL(2, R) → SL(n, R) induced by the action of SL(2, R) on two• variable homogeneous polynomials of degree n − 1.

Naturally, in discussions we prefer to work with matrix representatives of elements of projective linear groups, being mindful that each matrix is deemed equivalent to its negative. Therefore we would like to translate the task of proving Theorem 1.1 to a problem involving matrix groups. This involves a certain technical issue for the case where n is even and at least one of p, q, r is even.

Let U(p, q, r) be the pullback of the pair consisting of the natural projection ̟2 : SL(2, R) → PSL(2, R) and the injection φ : T(p, q, r) → PSL(2, R). Thus U(p, q, r) comes with an injection ψ : U(p, q, r) → SL(2, R) and a two•to•one epimor• phism ̟0 : U(p, q, r) → T(p, q, r), as part of the following commutative diagram:

ψ σn U(p, q, r) SL(2, R) SL(n, R)

̟0 ̟2 ̟n

φ ρn T(p, q, r) PSL(2, R) PSL(n, R) The Dimension of the Hitchin Component for Triangle Groups 5

We must address the mildly inconvenient fact that if at least one of p, q, r is even, φ does not admit a lift to SL(2, R). To see this, note that a rotation through 2π/k about a point of H2 is represented by a matrix conjugate to

cos π − sin π τ = k k , k sin π cos π  k k  of order 2k rather than k; moreover, if k is even, this order doubling feature cannot be corrected by negating the matrix τk . The issue disappears in SL(n, R) for odd n, as then SL(n, R) = PSL(n, R); however the issue does persist in SL(n, R) with n even.

Under the circumstances, a reasonable option is simply to work with the group U(p, q, r). This group has generators α , β , γ with respective orders 2p, 2q, 2r; their images under ψ are conjugates of τp , τq , τr , and their images under ̟2 ◦ ψ are φ(a) , φ(b) , φ(c), respectively. The central element z = αp = βq = γr is mapped by ψ to −I, and αβγ ∈ {1 , z}. We may identify T(p, q, r) with the quotient group U(p, q, r)/hzi. The homomorphism σn ◦ ψ is an absolutely irreducible representation of U(p, q, r) into SL(n, R).

Although it is not needed for the sequel, the apparent ambiguity regarding the product αβγ can be resolved as follows.

Proposition 2.1 Let α,β,γ, z ∈ U(p, q, r) be as above. If none of p, q, r is 2, then αβγ = z, and if p = 2, then α can be chosen so that αβγ = z. It follows that for appropriately chosen generators α , β , γ , U(p, q, r) admits the presentation

U(p, q, r) = hα , β , γ | αp = βq = γr = αβγ = z , z2 = 1i .

Proof The reason why the case p = 2 is considered separately is that it is the only case where τp is conjugate to its negative. If p > 2, both eigenvalues of τp have strictly positive real part, and the condition that α is conjugate to τp excludes ambiguity in the choice between α and its negative.

If ∆ is an arbitrary triangle in the hyperbolic plane, or indeed in the Euclidean plane, with angles θ1 , θ2 , θ3 , then it is an easy geometric fact that the product of rotations through 2θ1 , 2θ2 , 2θ3 about the respective vertices, taken in the correct order, is the identity. The reason is simply that a rotation through 2θi about a vertex with angle θi is the product of reflections in the two sides incident to the vertex. Therefore to prove the Proposition it is sufficient to check the conclusion for a single case, the general case following by continuity. 6 D. D. Long and M. B. Thistlethwaite

Here are matrices for α , β , γ in the case p = q = r = 4, obtained from a specific triangle in the hyperbolic plane; it is easily checked that each of α , β , γ is conjugate to τ4 and that αβγ = −I.

1 √2 1 α = 2 " 2 √2 # − 1 √2 1 + √2 2(1 + √2) β = − 2  2 1 + √2 + 2(1 + √2) √q2  −      q  1 √2 1 1 + √2 1 + √2 1 + √2 γ = − − 2  2 1 + √2p+ 1 + √ 2 √2 1 + p1 + √2  −   p   p  

Let us define ψn = σn ◦ ψ , φn = ρn ◦ φ, i.e. ψn , φn are the representations given by the top and bottom rows of the above commutative diagram.

Proposition 2.2 Let H , K be the deformation spaces of φn , ψn respectively. Then H , K are diffeomorphic.

Proof We proceed to define a smooth map F : H → K and a smooth inverse to F, G : K→H.

′ ′ Let φn ∈ H, and let Φt (0 ≤ t ≤ 1) be a path in H from φn to φn . For each g ∈ T(p, q, r), Φt(g) can be regarded as a pair of paths {mt(g) , −mt(g)} in SL(n, R), this pair being well defined by continuity. We define a path Ψt in K by

mt(g) (ψn(g) = m0(g)) Ψt(g) = (−mt(g) (ψn(g) = −m0(g))

′ ′ and define F(φn) by F(φn)(g) =Ψ1(g).

′ ′ Conversely, let ψn ∈ K, and let Ψt (0 ≤ t ≤ 1) be a path in K from ψn to ψn . Since ψn(z) = ±In is central in SL(n, R), Ψt(z) is kept constant at ±In . It follows that Ψt induces a path Φt in H with ̟n ◦ Ψt =Φt ◦ ̟0 for each t ∈ [0, 1], and we define ′ G(ψn) =Φ1 . The Dimension of the Hitchin Component for Triangle Groups 7

3 Deformations of finite cyclic groups

Any deformation of ψn restricts to deformations of the cyclic subgroups hαi , hβi , hγi of orders 2p, 2q, 2r respectively. Let g be one of the generators α , β , γ , and let us consider the subgroup hgi. By the deformation space of hgi we mean the component of the representation variety of hgi in SL(n, R) containing the restriction of ψn : U(p, q, r) → SL(n, R) to the ′ subgroup hgi. Let us denote this component Dn(hgi). As representations ψn of hgi ′ travel through Dn(hgi), ψn(g) must remain within the conjugacy class of ψn(g), as the ′ R order of the element ψn(g) remains constant; also, by connectedness of SL(n, ) any conjugate of ψn(g) is so attainable. Since any homomorphism of hgi is determined ′ ′ by its effect on g, the correspondence ψn 7→ ψn(g) defines a diffeomorphism from Dn(hgi) to the conjugacy class of ψn(g) in SL(n, R), or equivalently the space of right cosets of the centralizer of ψn(g) in SL(n, R). Denoting this centralizer C(ψn(g)), we have dim Dn(hgi) = dim SL(n, R) − dim C(ψn(g)) .

We now put this into a more practical form. Let g ∈ U(p, q, r) be as in the previous paragraph, and suppose that g has order k. Recall that ψ(g) ∈ SL(2, R) is a matrix conjugate to cos π − sin π τ = k k k sin π cos π  k k  with eigenvalues ζ , 1/ζ , where ζ is the primitive (2k)th root of unity eπi/k . It is x 0 easily determined that the image under σ of a diagonal matrix is an n×n n 0 1/x diagonal matrix, with diagonal entries   − − − − x n+1 , x n+3 ,..., xn 3 , xn 1 .

Therefore the eigenvalues of ψn(g) = σn(τk) are − − − − ζ n+1 , ζ n+3 ,...,ζn 3 , ζn 1 .

The origin of the slightly arcane function σ(n, p) of Theorem 1.1 should now be apparent: taking the multiset − − − − E = {ζ n+1 , ζ n+3 ,...,ζn 3 , ζn 1} , and writing n = Q p + R (0 ≤ R ≤ p − 1), we see that there are R eigenvalues each occurring with multiplicity Q + 1, and that each of the remaining eigenvalues in E 8 D. D. Long and M. B. Thistlethwaite occurs with multiplicity Q. The dimension of the centralizer C(g) is one less than the sum of the squares of the multiplicities, i.e. n − R(Q + 1) dim C(g) = R(Q + 1)2 + Q2 − 1 = (n + R)Q + R − 1 , Q the reduction by 1 corresponding to the constraint that matrices have determinant 1.

Proposition 3.1 Let g be any one of the three generators α , β , γ of U(p, q, r), and let k be the order of g. Then 2 2 dim Dn(hgi) = (n − 1) − (σ(n , k) − 1) = n − σ(n , k) .

We are grateful to E. Weir for providing the simplified expression for σ(n, k).

4 Proof of Theorem 1.1

For g ∈ SL(n, R), we adopt the notation − [g] = {xgx 1 : x ∈ SL(n, R)} .

Also we define αn , βn , γn , zn to be the images of α,β,γ, z ∈ U(p, q, r) under the representation ψn : U(p, q, r) → SL(n, R). Let ′ ′ ′ ′ ′ ′ S = {αn βn γn : αn ∈ [αn] , βn ∈ [βn] , γn ∈ [γn]} . The natural map ′ ′ ′ ′ ′ ′ Π : [αn] × [βn] × [γn] → S , Π (αn , βn , γn) = αn βn γn −1 is a submersion. Since our deformation space K is naturally diffeomorphic to Π (zn), we deduce that

dim K = dim [αn] + dim [βn] + dim [γn] − dim S . In light of Proposition 3.1 and the fact that H , K have the same dimension, in order to complete the proof of Theorem 1.1 it is sufficient to show that S contains a neigh• bourhood in SL(n, R) of αn βn γn = zn . The following proposition shows that this ′ ′ objective is achieved even if we hold γn fixed, allowing only αn , βn to vary within their conjugacy classes:

Proposition 4.1 Let us define ′ ′ ′ ′ ′ S = {αn βn : αn ∈ [αn] , βn ∈ [βn]} . ′ Then S contains a neighbourhood in SL(n, R) of αn βn . The Dimension of the Hitchin Component for Triangle Groups 9

Since our discussion takes place entirely within SL(n, R), we may lighten notation by suppressing the subscripts of αn , βn without fear of confusion. The proof of Proposition 4.1 makes use of the well•known non•degenerate, indefinite bilinear form on the Lie algebra sl(n, R): hx , yi := tr(x y) .

We note immediately that h , i is respected by Adg for any g ∈ SL(n, C), since −1 −1 −1 hAdg(x) , Adg(y)i = tr((gxg )(gyg )) = tr(g(xy)g ) = tr(xy) = hx , yi .

Lemma 4.1.1 Let g ∈ SL(n, R) be any element of the image of U(p, q, r) under the representation ψn . Then sl(n, R) admits a direct sum decomposition

sl(n, R) = Ker(Adg −1) ⊕ Im(Adg −1) , orthogonal with respect to h , i.

Proof Let g ∈ ψn(U(p, q, r)). Since a triangle group has no parabolic elements, all elements of U(p, q, r) (considered as a subgroup of SL(2, R)) are diagonalizable (over C). This property is preserved by the representation ψn , and it is a simple matter to check that it holds also for the endomorphism Adg . Therefore the endomorphism Adg −1 of sl(n, R) has a full eigenspace for the eigenvalue 0. This eigenspace constitutes precisely the kernel of Adg −1, and the direct sum decomposition follows.

On the matter of orthogonality, let ξ ∈ Ker(Adg −1) , η ∈ Im(Adg −1). Let us write η = (Adg −1)(ζ). Then

hξ , ηi = hξ , (Adg −1)(ζ)

= hξ , Adg(ζ)i − hξ,ζi

= hAdg(ξ) , Adg(ζ)i − hξ,ζi = hξ,ζi − hξ,ζi = 0 .

Lemma 4.1.2 Let W be a linear subspace of sl(n, R), and let W⊥ be the orthogonal complement of W with respect to h , i. Then ⊥ dim W + dim W = dim sl(n, R) .

Proof The non•degenerate bilinear form h , i induces an isomorphism from W⊥ to ∗ the annihilator of W in the dual space sl(n, R) . 10 D. D. Long and M. B. Thistlethwaite

We recall the following facts from basic Lie theory: (i) There exists a neighbourhood of 0 ∈ sl(n, R) on which the exponential map is a diffeomorphism to a neighbourhood of In ∈ SL(n, R); −1 (ii) For ξ ∈ sl(n, R) and g ∈ SL(n, R), exp(Adg ξ) = g exp(ξ) g , from which it follows that Adg(ξ) − ξ = 0 ⇐⇒ exp(ξ) ∈ C(g).

′ Proof of Proposition 4.1 We define subsets S1 , S2 of S , each containing αβ, as follows: ′ ′ ′ ′ S1 = {α β : α ∈ [α]} , S2 = {α β : β ∈ [β]} .

It is sufficient to show that the tangent spaces of S1 , S2 at αβ generate the tangent space of SL(n, R) at αβ. Let g = exp(ξ) , h = exp(η) be elements of SL(n, R) close to the identity. Then −1 −1 the typical element gαg β of S1 can be written as exp(ξ)(exp(Adα ξ)) (αβ), and −1 to first order this is equal to exp((1 − Adα) ξ) (αβ). Similarly, αhβh ∈ S2 has the first order approximation exp((Adα − Adαβ) η) (αβ).

Our task is therefore reduced to checking that the images of 1 − Adα , Adα − Adαβ satisfy Im(1 − Adα) + Im(Adα − Adαβ) = sl(n, R) . Suppose that the sum of these images is a proper subspace of sl(n, R). Then from Lemma 4.1.2 there exists a non•zero element ζ in the orthogonal complement of Im(1 − Adα) + Im(Adα − Adαβ). In particular, ζ is orthogonal to Im(1 − Adα), hence from Lemma 4.1.1 is in Ker(1 − Adα). Note that we have just established that ζ = Adα(ζ). Similarly, ζ is in the orthogonal complement of Im(Adα − Adαβ) = Adα(Im(1 − Adβ)). Since orthogonality is preserved by Adα , this last orthogonal complement is precisely Adα(Ker(1 − Adβ)). Therefore, using ζ = Adα(ζ), we have Adα(ζ) ∈ Adα(Ker(1 − Adβ)), implying ζ ∈ Ker(1 − Adβ). Now Schur’s Lemma applied to the absolutely irreducible representation ψn gives us

Ker(1 − Adα) ∩ Ker(1 − Adβ) = {0} , from which we deduce ζ = 0, a contradiction.

5 Acknowledgments

The first author is partially supported by the National Science Foundation, and the second author is partially supported by the Simons Foundation. The Dimension of the Hitchin Component for Triangle Groups 11

References

[1] S. Choi and W.M. Goldman, The deformation spaces of convex RP2 –structures on 2–orbifolds, Amer. J. Math. 127 (2005) no.5, 1019–1102. [2] N.J. Hitchin, Lie groups and Teichmuller¨ space, Topology 31 (1992), 449 – 473. [3] F. Labourie, Anosov flows, surface groups and curves in projective space, Inventiones Math. 165 (2006), 51 – 114. [4] D.D. Long and M.B. Thistlethwaite, Zariski dense surface subgroups in SL(4, Z), Experimental Math. 15 (2016), 291–305.