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Theta Functions for SL(n) versus GL(n)

arXiv:alg-geom/9303004v1 28 Mar 1993Theta Functions for SL(n) versus GL(n)Ron Donagi and Loring W. TuNovember 1, 1992§1. Theta bundles§2.

A Galois covering§3. Pullbacks§4.

Proof of Theorem 1§5. A conjectural dualityOver a smooth complex projective curve C of genus g one may consider two types of modulispaces of vector bundles, M := M(n, d), the moduli space of semistable bundles of rank n anddegree d on C, and SM := SM(n, L), the moduli space of those bundles whose determinant isisomorphic to a fixed line bundle L on C. We call the former a full moduli space and the lattera fixed-determinant moduli space.

Since the spaces SM(n, L) are all isomorphic as L varies inPicd(C), we also write SM(n, d) to denote any one of them.On both moduli spaces there are well-defined theta bundles, as we recall in Section 1. Whilethe theta bundle θ on SM is uniquely defined, the theta bundles θF on M depend on the choiceof complementary vector bundles F of minimal rank over C. For any positive integer k, sectionsof θkF generalize the classical theta functions of level k on the Jacobian of a curve, and so we callsections of θk over SM and θkF over M theta functions of level k for SL(n) and GL(n) respectively.Our goal is to study the relationship between these two spaces of theta functions.

We prove asimple formula relating their dimensions, and then formulate a conjectural duality between thesetwo spaces.Theorem 1 If h = gcd(n, d) is the greatest common divisor of n and d, and L ∈Picd(C), thendim H0(SM(n, L), θk) · kg = dim H0(M(n, d), θkF) · hg.Faltings [F] has proven the Verlinde formula for semisimple groups, which gives in particularthe dimension of H0(SM, θk). The dimension of H0(M, θkF) is thus determined by Theorem 1.When k = 1 and d = 0, [BNR] computes explicitly the two spaces in Theorem 1.

Their result isa forerunner of Theorem 1.Theorem 1 is consistent with and therefore lends credence to another, so far conjectural,relationship between these two types of theta functions.To explain this, start with integers¯n, ¯d, h, k such that ¯n, h, k are positive and gcd(¯n, ¯d) = 1. Let F ∈M(¯n, ¯d) and writeSM1 = SM(h¯n, (det F)h)andM2 = M(k¯n, k(¯n(g −1) −¯d)).The tensor product map τ sends SM1 × M2 to M(hk¯n2, hk¯n2(g −1)).1

Conjecture 2 The tensor product map induces a natural duality between H0(SM1, θk) andH0(M2, θhF).For further discussion of this duality, including supporting evidence, see Section 5.Notation and Conventions.h0( ) = dim H0( )Jd = Picd(C) ={isomorphism classes of line bundles of degree d on C}J = J0 = Pic0(C)L1 ⊠L2 = π∗1L1 ⊗π∗2L2 if Li is a line bundle on Xi and πi : X1 × X2 →Xi is the i-thprojectionShC = the hth symmetric product of CTn = the group of n-torsion bundles on C1Theta bundlesWe recall here the definitions of the theta bundles on a fixed-determinant moduli space and on afull moduli space. Our definitions are slightly different from but equivalent to those in [DN].For L ∈Picd(C), the Picard group of SM := SM(n, L) is Z and the theta bundle θ on SMis the positive generator of Pic(SM).When n and d are such that χ(E) = 0 for E ∈M(n, d), i. e., when d = (g −1)n, there is anatural divisor Θ ⊂M(n, n(g −1)):Θ = closure of {E stable in M(n, n(g −1)) | h0(E) ̸= 0}.The theta bundle θ over M(n, n(g −1)) is the line bundle corresponding to this divisor.We say that a semistable bundle F is complementary to another bundle E if χ(E ⊗F) = 0.We also say that F is complementary to M(n, d) if χ(E ⊗F) = 0 for any E ∈M(n, d).

It followseasily from the Riemann-Roch theorem that if E ∈M(n, d), h = gcd(n, d), n = h¯n, and d = h ¯d,then F has rank nF and degree dF, wherenF = k¯nanddF = k(¯n(g −1) −¯d)for some positive integer k.If F is complementary to M(n, d), letτF : M(n, d) →M(nnF, nnF(g −1))be the mapE 7→E ⊗F.Pulling back the theta bundle θ from M(nnF, nnF(g−1)) via τF gives a line bundle θF := τ ∗Fθ overM(n, d). (This bundle may or may not correspond to a divisor in M(n, d).) Let det : M(n, d) →Jd(C) be the determinant map.

When rk F is the minimal possible:rk F = ¯n = n/h, then θFis called a theta bundle over M(n, d); otherwise, it is a multiple of a theta bundle. Indeed, weextract from [DN] the formula:2

Proposition 3 Let F and F0 be two bundles complementary to M(n, d). If rk F = a rk F0, thenθF ≃θ⊗aF0 ⊗det ∗(det F ⊗(det F0)−a),where we employ the usual identification of Pic0(C) with Pic0(J0).In particular, θF depends only on rk F and det F.If θF is a theta bundle on M(n, d), then for any L ∈Picd(C), θF restricts to the theta bundleon SM(n, L).2A Galois coveringLet τ : Y →X be a covering of varieties, by which we mean a finite ´etale morphism.

A decktransformation of the covering is an automorphism φ : Y →Y that commutes with τ.Thecovering is said to be Galois if the group of deck transformations acts transitively (hence simplytransitively) on a general fiber of the covering.Denote by J = Pic0(C) the group of isomorphism classes of line bundles of degree 0 on thecurve C, and G = Tn the subgroup of torsion points of order n.Fix L ∈Picd(C) and letSM = SM(n, L), J = J0(C), and M = M(n, d). Recall that the tensor product mapτ : SM × J→M(E, M)7→E ⊗Mgives an n2g-sheeted ´etale map ([TT], Prop.

8). The group G = Tn acts on SM × J byN.

(E, M) = (E ⊗N−1, N ⊗M).It is easy to see that G is the group of deck transformations of the covering τ and that it actstransitively on every fiber. Therefore, τ : SM × J →M is a Galois covering.Proposition 4 If τ : Y →X is a Galois covering with finite abelian Galois group G, then τ∗OY isa vector bundle on X which decomposes into a direct sum of line bundles indexed by the charactersof G:τ∗OY =Xλ∈ˆGLλ,where ˆG is the character group of G.Proof.Write O = OY .The fiber of τ∗O at a point x ∈X is naturally a complex vectorspace with basis τ −1(x).Hence, τ∗O is a vector bundle over X.The action of G on τ −1(x)induces a representation of G on (τ∗O)(x) equivalent to the regular representation.

Because G isa finite abelian group, this representation of G decomposes into a direct sum of one-dimensionalrepresentations indexed by the characters of G:(τ∗O)(x) =Xλ∈ˆGLλ(x).Thus, for every λ ∈ˆG, we obtain a line bundle Lλ on X such such τ∗O =Pλ Lλ.✷3

3PullbacksWe consider the tensor product mapτ : SM(n1, L1) × M(n2, d2)→M(n1n2, n1d2 + n2d1)(E1, E2)7→E1 ⊗E2,where d1 = deg L1. For simplicity, in this section we write SM1 = SM(n1, L1), M2 = M(n2, d2),and M12 = M(n1n2, n1d2 + n2d1).Proposition 5 Let F = F12 be a bundle on C complementary to M12.

Thenτ ∗θF ≃θc ⊠θE1⊗Ffor any E1 ∈SM(n1, L1), wherec := n2 rk Frk F1=n2 rk Fn1/ gcd(n1, d1)and F1 is a minimal complementary bundle to E1.Proof. For E2 ∈M(n2, d2), letτE2 : SM1 →M12be tensoring with E2.

Then(τ ∗θF)|SM×{E2} = τ ∗E2θF = τ ∗E2τ ∗Fθ = τ ∗E2⊗Fθ = θc,where by Proposition 3c=rk (E2 ⊗F)/ rk F1=n2 rk Fn1/ gcd(n1, d1).Similarly,(τ ∗θF)|{E1}×M2=τ ∗E1θF = τ ∗E1τ ∗Fθ=τ ∗E1⊗Fθ = θE1⊗F.Note that the bundle θE1⊗F depends only on rk (E1 ⊗F) = n1 rk F and det(E1 ⊗F) = L rk F1⊗(det F)n1. Hence, both (τ ∗θF)|SM1×{E2} and (τ ∗θF)|{E1}×M2 are independent of E1 and E2.

Bythe seesaw theorem,τ ∗θF ≃θc ⊠θE1⊗F.✷Corollary 6 Let L ∈Picd(C) andτ : SM(n, L) × J0 →M(n, d)be the tensor product map. Suppose F is a minimal complementary bundle to M(n, d).

ChooseN ∈Picg−1(C) to be a line bundle such that Nn = L ⊗(det F)h, where h = gcd(n, d). Thenτ ∗θF = θ ⊠θn2/hN.4

Proof. Apply the Proposition with rk F = n/h and n1 = n, d1 = d, n2 = 1, d2 = 0.

Then c = 1.By Proposition 3,θE1⊗F=θn2/hN⊗det ∗(det(E1 ⊗F) ⊗N−n2/h)=θn2/hN.✷4Proof of Theorem 1We apply the Leray spectral sequence to compute the cohomology of τ ∗θkF on the total space of thecovering τ : SM×J →M of Section 2. Recall that SM = SM(n, d), J = J0, and M = M(n, d).Because the fibers of τ are 0-dimensional, the spectral sequence degenerates at the E2-term andwe haveH0(SM × J, τ ∗θkF) = H0(M, τ∗τ ∗θkF).

(1)By Cor. 6 and the K¨unneth formula, the left-hand side of (1) isH0(SM × J, τ ∗θkF))=H0(SM × J, θk ⊠θkn2/hN))=H0(SM, θk) ⊗H0(J, θkn2/hN).By the Riemann-Roch theorem for an abelian variety,h0(J, θkn2/hN) = (kn2/h)g.So the left-hand side of (1) has dimensionh0(SM, θk) · (kn2/h)g.(2)Next we look at the right-hand side of (1).

By the projection formula and Prop. 4,τ∗τ ∗θkF=θkF ⊗τ∗O=θkF ⊗Xλ∈ˆGLλ=Xλ∈ˆGθkF ⊗Lλ.Our goal now is to show that for any character λ ∈ˆG,H0(M, θkF ⊗Lλ) ≃H0(M, θkF).

(3)This will follow from two lemmas.Lemma 7 The line bundle Lλ on M is the pullback under det : M →Jd of some line bundle Nλof degree 0 on Jd := Picd(C).5

Lemma 8 For F a vector bundle as above, k a positive integer, and M a line bundle of degree 0over C,H0(M, θkF ⊗M) ≃H0(M, θkF).Assuming these two lemmas, let’s prove (3). By Proposition 3,θF ⊗M = θF ⊗det ∗MnF ;hence,θkF ⊗M = θkF ⊗det ∗MnF k.If Lλ = det ∗Nλ, and we choose a root M = N1/(nF k)λ, thenθkF ⊗Lλ = θkF ⊗det ∗Nλ = θkF ⊗M.Equation (3) then follows from Lemma 8.Proof of Lemma 7.

Define α : SM × J →J to be the projection onto the second factor,β : M →J to be the composite of det : M →Jd followed by multiplication by L−1 : Jd →J, andρ : J →J to be the n-th tensor power map. Then there is a commutative diagramSM × Jτ→Mα↓↓βJρ→J.Furthermore, in the map α : SM × J →J we let G = Tn act on J byN.M = N ⊗M,M ∈J,and in the map β : M →J we let G act trivially on both M and J.

Then all the maps in thecommutative diagram above are G-morphisms.By the push-pull formula ([H], Ch. III, Prop.

9.3, p. 255),τ∗α∗OJ = β∗ρ∗OJ.By Proposition 4, ρ∗OJ is a direct sum of line bundles Vλ on J, where λ ∈ˆG. In fact, these Vλare precisely the n−torsion bundles in J; in particular, their degrees are zero.

If τL−1 : Jd →J ismultiplication by the line bundle L−1, we set Nλ := τ ∗L−1Vλ. Thenτ∗OSM×J=β∗Xλ∈ˆGVλ=det ∗τ ∗L−1XVλ=Xdet ∗Nλ.By Prop.

4, τ∗OSM×J = P Lλ. Since both Lλ and det ∗Nλ are eigenbundles of τ∗OSM×Jcorresponding to the character λ ∈ˆG,Lλ = det ∗Nλ.6

✷Proof of Lemma 8. Tensoring with M ∈J0(C) gives an automorphismτM : M→ME7→E ⊗M,under whichθF ⊗M = τ ∗MθF .Hence,θkF ⊗M = τ ∗M(θkF)and the lemma follows.✷Returning now to Eq.

(1), its right-hand side isH0(M, τ∗τ ∗θkF)=Xλ∈ˆGH0(M, θkF ⊗Lλ)≃Xλ∈ˆGH0(M, θkF),(by (3))which has dimensionh0(M, θkF) · n2g.By (2) the left-hand side of Eq. (1) has dimensionh0(SM, θk) · (kn2/h)g.Equating these two expressions givesh0(M, θkF) = h0(SM, θk) · (kh)g.This completes the proof of Theorem 1.5A conjectural dualityAs in the Introduction we start with integers ¯n, ¯d, h, k such that ¯n, h, k are positive andgcd(¯n, ¯d) = 1.

Taken1 = h¯n, d1 = h ¯d, n2 = k¯n, d2 = k(¯n(g −1) −¯d), and L1 ∈Picd1(C).The tensor product induces a mapτ : SM(n1, L1) × M(n2, d2) →M(n1n2, n1n2(g −1)).As before, write SM1 = SM(n1, L1), M2 = M(n2, d2), and M12 = M(n1n2, n1n2(g −1)). LetF2 = F and F12 = O be complementary to M2 and M12 respectively.By the pullback formula (Proposition 5)τ ∗θO = θn2/¯n ⊠θE1.7

But by Proposition 3,θE1 = θhF ⊗det ∗(L ⊗(det F)−h).If L = (det F)h, then θE1 = θhF andτ ∗θO = θk ⊠θhF.By the K¨unneth formula,H0(SM1 × M2, τ ∗θO) = H0(SM1, θk) ⊗H0(M2, θhF).In [BNR] it is shown that up to a constant, θO has a unique section s over M12. Then τ ∗s is asection of H0(SM1 × M2, τ ∗θO) and therefore induces a natural mapH0(SM1, θk)∨→H0(M2, θhF).

(4)We conjecture that this natural map is an isomorphism.Among the evidence for the duality (4), we cite the following.i) (Rank 1 bundles) The results of [BNR] thatH0(SM(n, O), θ)∨≃H0(M(1, g −1), θnO)andH0(M(n, n(g −1)), θO) = C,are special cases of (4), for (n2, d2) = (1, g −1) and (n1, d1) = (1, 0) respectively.ii) (Consistency with Theorem 1) Given a triple of integers (n1, d1, k), we define h, ¯n, ¯d byh = gcd(n1, d1), n1 = h¯n, d1 = h ¯dand let n2, d2 be as before:n2 = k¯n, d2 = k(¯n(g −1) −¯d).Assuming n1 and k to be positive, it is easy to check that the function(n1, d1, k) 7→(n2, d2, h)is an involution. Write v(n, d, k) = h0(M(n, d), θkF) and s(n, d, k) = h0(SM(n, d), θk).

ThenTheorem 1 assumes the formv(n, d, k) · hg = s(n, d, k) · kg. (5)The duality (4) implies that there is an equality of dimensionss(n1, d1, k) = v(n2, d2, h).

(6)Because (n1, d1, k) 7→(n2, d2, h) is an involution, it follows thats(n2, d2, h) = v(n1, d1, k). (7)Putting (5), (6), and (7) together, we getv(n2, d2, h)kg = s(n2, d2, h)hg,which is Theorem 1 again.8

iii) (Elliptic curves) We keep the notation above, specialized to the case of a curve C of genusg = 1:n1 = h¯n, d1 = h ¯d, n2 = k¯n, d2 = −k ¯d.Set C′ := Pic¯d(C). The map sending a line bundle to its dual gives an isomorphism C′ ≃Pic−¯d(C).

If L ∈Pic¯d(C), viewed as a line bundle on C, we let ℓbe the corresponding pointin C′, and OC′(ℓ) the associated line bundle of degree 1 on the curve C′. There is a naturalmapγ : Pich ¯d(C) →Pich(C′)which sends L := L1 ⊗· · · ⊗Lh ∈Pich ¯d(C) to L′ := OC′(ℓ1 + · · · + ℓh), where Li ∈Pic¯d(C)corresponds to the point ℓi ∈C′.From [A] and [T] we see that there are natural identificationsM(h¯n, h ¯d) ≃ShM(¯n, ¯d) ≃ShPic¯d(C) = ShC′andM(k¯n, −k ¯d) ≃SkM(¯n, −¯d) ≃SkPic−¯d(C) ≃SkC′.Furthermore, there is a commutative diagramM(h¯n, h ¯d)∼→ShC′det↓↓αPich ¯d(C)γ→Pich(C′).Since the fiber of the Abel-Jacobi map α : ShC′ →Pich(C′) above L′ is the projective spacePH0(C′, L′), it follows that there is a natural identificationSM(h¯n, L) ≃PH0(C′, L′).Since the theta bundle is the positive generator of SM(h¯n, L), it is the hyperplane bundle.

ForF ∈M(¯n, ¯d), let q ∈C′ be the point corresponding to the line bundle Q := det F ∈Pic¯d(C).ThenH0(SM(h¯n, (det F)h), θk)≃H0(PH0(C′, OC′(hq)), O(k))=SkH0(C′, OC′(hq))∨.Recall that each point q ∈C′ determines a divisor Xq on the symmetric product SkC′:Xq := {q + D | D ∈Sk−1C′}.The proof of Theorem 6 in [T] actually shows that if F ∈M(¯n, −¯d), then under the identificationM(k¯n, −k ¯d) ≃SkC′, the theta bundle θF corresponds to the bundle associated to the divisor Xqon SkC′, where q is the point corresponding to det F ∈Pic¯d. Therefore, by the calculation of thecohomology of a symmetric product in [T]H0(M(k¯n, −k ¯d), θhF)=H0(SkC′, O(hXq))=SkH0(C′, O(hq)).So the two spaces H0(SM(h¯n, (det F)h), θk) and H0(M(k¯n, −k ¯d), θhF) are naturally dual to eachother.9

iv) (Degree 0 bundles) Consider the moduli space SM(n, 0) of rank n and degree 0 bundles. Inthis case,n1 = n, d1 = 0, h = gcd(n, 0) = n, n2 = k, d2 = k(g −1).So the conjectural duality isH0(SM(n, O), θk)∨≃H0(M(k, k(g −1)), θnO).Because M(k, k(g −1)) is isomorphic to M(k, 0) (though noncanonically), it follows that inthe notation of ii)s(n, 0, k) = v(k, 0, n).According to R. Bott and A. Szenes, this equality follows from Verlinde’s formula.References[A]M. Atiyah, Vector bundles over an elliptic curve, Proc.

London Math. Soc.

7 (1957), 414-452. [BNR] A. Beauville, M. S. Narasimhan, and S. Ramanan, Spectral curves and the generalizedtheta divisor, J. reine angew.

Math. 398 (1989), 169-179.[DN]J.-M.

Drezet and M. S. Narasimhan, Groupe de Picard des vari´et´es de modules de fibr´essemi-stables sur les courbes alg´ebriques, Invent. Math.

97 (1989), 53-94.[F]G. Faltings, A proof of the Verlinde formula, preprint.[H]R.

Hartshorne, Algebraic Geometry, Springer-Verlag, New York, 1977.[TT]M. Teixidor and L. W. Tu, Theta divisors for vector bundles, in Curves, Jacobians, andAbelian Varieties, Contemporary Mathematics 136 (1992), 327-342.[T]L.

W. Tu, Semistable bundles over an elliptic curve, to appear in Advances in Mathematics.10

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