PART 2: SIEGEL MODULAR FORMS AARON POLLACK 1. Siegel modular forms We now begin to discuss Siegel modular forms. Siegel modular forms are special automorphic forms for the symplectic group Sp2n. Recall that Sp = fg 2 GL : g 0 1n gt = 0 1n g: 2n 2n −1n 0 −1n 0 Equivalently, and more canonically if you prefer, Sp2n is the group preserving a non-degenerate alternating bilinear form on a 2n-dimensional vector space. Equivalently, if k is a field and W is a 2n-dimensional k-vector space, then Sp2n is the subgroup of GL(W ) that preserves a non-degenerate alternating form on W . In more detail, a bilinear form h; i : W × W ! k is a symplectic form if • hx; yi = −hy; xi for all x; y 2 W and • hx; yi = 0 for all y 2 W implies x = 0. Then Sp2n = Sp(W ) is the algebraic group of g 2 GL(W ) such that hgx; gyi = hx; yi for all x; y 2 W . A symplectic basis of W is a basis e1; : : : en; f1; : : : ; fn such that hei; eji = hfi; fji = 0 for all 2n i; j and hei; fji = δij. Every symplectic space W has a symplectic basis. In case W = k with symplectic form hx; yi = xtJ y where J = 0 1n , the usual coordinate basis is a symplectic n n −1n 0 basis. We now describe a few subgroups of Sp2n. First, there is the Siegel parabolic subgroup P ⊆ Sp2n. a b It is the subgroup of g = c d in n × n block form with c = 0. We have the following lemma. a 0 t −1 1 x Lemma 1. One has 0 d in Sp2n if and only if d = a . The element ( 0 1 ) is in Sp2n if and only if x = xt. t Proof. This is a direct matrix calculation with the definition g Jng = Jn for g 2 Sp2n. Another subgroup of Sp2n is a unitary group of size n: Example This is example works over any field, but we explain for k = R. Consider Cn, with the hermitian form (x; y) = x∗y where x; y 2 Cn are column vectors and ∗ denotes conjugate transpose. Now, let W = Cn, but considered as a 2n-dimensional real vector space. Put on W the R-bilinear form given as hx; yi = Im((x; y)) = Im(x∗y). This a non-degenerate alternating form. As U(n) is the subgroup of GLn(C) that fixes the Hermitian pairing ( ), we see that U(n) fixes h ; i as well, so U(n) ⊆ Sp2n(R). Let us now define the Siegel upper half-space hn. The Siegel modular forms will be holomorphic functions on hn that satisfy a certain equivariance property. Set t hn = fZ 2 Mn(C): Z = Z; Im(Z) > 0g: So, if Z 2 hn then Z = X + iY with X; Y symmetric and Y positive-definite. −1 a b The group Sp2n(R) acts on hn via gZ = (aZ + b)(cZ + d) for g = c d 2 Sp2n(R). More precisely, one has the following proposition. a b Proposition 2. Suppose z 2 hn and g = c d Sp2n. Then 1 (1) The n × n matrix cz + d is invertible; −1 (2) the matrix (az + b)(cz + d) is in hn; −1 (3) the map Sp2n(R) × hn ! hn given by (g; z) 7! (az + b)(cz + d) defines an action of Sp2n(R) on hn. Proof. For Z; W 2 Mn(C), note the identity ∗ ∗ ∗ (Z; W )Jn(Z; W ) = ZW − WZ : We now prove (1). For this, note that we have ∗ ∗ 2iy = z − z = (z; 1)Jn(z; 1) t ∗ = (z; 1)gJng (z; 1) ∗ = (az + b; cz + d)Jn(az + b; cz + d) = (az + b)(cz + d)∗ − (cz + d)(az + b)∗: To prove that cz + d is invertible, we prove that ξ 2 Cn and ξ(cz + d) = 0 implies ξ = 0. To see this, suppose ξ(cz + d) = 0. Then from the above, we get 2iξyξ∗ = xi ((az + b)(cz + d)∗ − (cz + d)(az + b)∗) ξ∗ = 0: But since y is positive definite, ξyxi∗ = 0 implies ξ = 0, as desired. For part (2), we first prove that gZ := (az + b)(cz + d)−1 is symmetric and then that Im(gz) is positive definite. For the symmetry, first note the identity t t t (Z; W )Jn(Z; W ) = ZW − WZ : Thus t t 0 = z − z = (z; 1)Jn(z; 1) t t = (z; 1)gJng (z; 1) t = (az + b; cz + d)Jn(az + b; cz + d) t t = cz + d(gz; 1)Jn(gz; 1) (cz + d) = (cz + d)(gz − (gz)t)(cz + d)t: From this, one concludes that gz is symmetric. By the same argument, one also proves Im(gz) = (cz + d)−1Im(z)((cz + d)∗)−1, from which one concludes that Im(gz) is positive definite. −1 We leave the proof that z 7! (az + b)(cz + d) is an action to the reader. We also have Lemma 3. The Siegel parabolic subgroup P ⊆ Sp2n, and thus Sp2n, acts transitively on hn, and the stabilizer of i 2 hn is the unitary group U(n). Proof. For the first part, suppose that x + iy 2 hn. Because y is positive definite, there exists t 1 x m m 2 GLn(R) such that mm = y. Then note that ( 1 )( tm−1 )·i = x+iy, so P acts transitively. AB For the second part, if A + iB 2 U(n), with A; B 2 Mn(R), one can check that −BA 2 Sp2n directly from the definition. This is how U(n) is embedded in Sp2n via matrices. One computes immediately that such elements fix i, and moreover that they are the full stabilizer. As a corollary of the proof of the lemma, one has that Sp2n(R) = P (R)U(n). We now come to the definition of Siegel modular forms. 2 Definition 4. Suppose w ≥ 1 is an integer. A holomorphic function f : hn ! C is said to be a Siegel modular form of weight w if f(γz) = j(γ; z)wf(z) for all γ in some congruence subgroup Γ of Sp2n(Z). If n = 1 one also imposes a slow growth condition on f(z), which turns out to be automatic for n > 1. Siegel modular forms have a very nice Fourier expansion, as follows: Denote by Sn(Z) the n × n _ _ symmetric matrices, and Sn (Z) the dual space under trace pairing, so that Sn (Z) consists of the half-integral symmetric matrices. Then 1 V Lemma 5. Suppose f is a Siegel modular form of level Γ, and that Γ contains the elements 0 1 P 2πi(T;Z) for every V 2 S (Z). Then f(Z) = _ a (T )e for some complex numbers a (T ). n T 2Sn(Z) f f Proof. We have f(Z) = f(Z + V ) for every V 2 Sn(Z). This plus holomorphy immediately implies the Fourier expansion. We will shortly prove that af (T ) 6= 0 implies T ≥ 0, i.e. that Siegel modular forms can only have nonzero Fourier coefficients corresponding to T that are positive semidefinite. We begin with a lemma, which is useful to know regardless. For simplicity, from now on we assume (for simplicity) that if f is a Siegel modular form of level Γ, where Γ contains P (Q) \ Sp2n(Z), with P the Siegel parabolic. t Lemma 6. Suppose f is a Siegel modular form of weight w and m 2 GLn(Z). Then af ( mT m) = w det(m) af (T ). Proof. One has X 2πi(T;Z) f(Z) = af (T )e _ T 2Sn(Z) X t 2πi(mtT m;Z) = af (m T m)e _ T 2Sn(Z) and w t X w 2πi(T;mZmt) det(m) f(mZm ) = det(m) af (T )e _ T 2Sn(Z) X w 2πi(mtT m;Z) = det(m) af (T )e : _ T 2Sn(Z) By the equivariance of f, one has det(m)wf(mZmt) = f(Z). Comparing the two displayed equa- tions now gives the lemma. We now prove one of the first very interesting properties of Siegel modular forms. Denote by Sn the n × n symmetric matrices. Proposition 7. Suppose f is a Siegel modular form of genus n > 1. Then af (T ) 6= 0 implies that T is positive semi-definite. Proof. We have Z −2π tr(T;Y0) −2πi tr(T;X0) af (T )e = e f(Y0 + X) dX Sn(Z)nSn(R) 2π tr(T ) from which one obtains the bound jaf (T )j ≤ Cf e for a positive constant Cf that is indepen- dent of T . Thus for m 2 GLn(Z), t 2π tr(mtT m) jaf (T )j = jaf (m T m)j ≤ e : 3 But now if T is not positive semi-definite, then tr(mtT m) can be made arbitrarily negative as m varies over GLn(Z). Thus af (T ) = 0 for such T . We now give some examples of Siegel modular forms. Example 1 Siegel Eisenstein series. For ` sufficiently large (in terms of n), set E`(Z) = P j(γ; Z)−`. When ` is so large that the series converges absolutely, it is clear γ2Pn(Z)n Sp2n(Z) ` that E`(δZ) = j(δ; Z) E`(Z). Example 2 Pullbacks. Denote by W2m a symplectic space of dimension 2m. Then W2n1 ⊕W2n2 = W2(n1+n2) defines a symplectic space of dimension 2n1 +2n2 by taking a direct sum of the symplectic pairings.