Nonholomorphic Eisenstein series, the Kronecker limit formula, and the hyperbolic Laplacian

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Nonholomorphic Eisenstein series, the Kronecker limit formula, and the hyperbolic Laplacian Jordan Bell jordan.bell@gmail.com Department of Mathematics, University of Toronto November 7, 04 Nonholomorphic Eisenstein series Let H = {x + iy C : y > 0} For τ = x + iy H and s = σ + it, σ >, we define the nonholomorphic Eisenstein series G(τ, s = (0,0 (m,n Z ys mτ + n s. The function (τ, a, b aτ + b is continuous H S C, and for all τ H and (a, b S we have aτ + b 0. It follows that if K is a compact subset of H then there is some C K > 0 such that aτ + b C K for all τ K, (a, b S. Then, for all τ K and for all (0, 0 (m, n Z, mτ + n = m m + n τ + n (m + n C K (m + n, m + n and hence y s mτ + n s = Because σ >, y σ mτ + n σ y σ (C K (m + n σ. (m + n σ <. (0,0 (m,n Z It follows that for any s = σ + it with σ >, the function τ G(τ, s is continuous H C. It is sometimes useful to write G in another way. For τ = x + iy H and Re s >, define E(τ, s = (c,d Z,gcd(c,d= y s cτ + d s.

Theorem. For all τ H and Re s >, G(τ, s = ζ(se(τ, s. Proof. First we remark that for 0 a Z, gcd(a, 0 = a. For (0, 0 (m, n Z, with ν = gcd(m, n, ( m gcd ν, n =. ν Then G(τ, s = = = ν (m,n Z,gcd(m,n=ν ν (c,d Z,gcd(c,d= y s (c,d Z,gcd(c,d= = ζ(se(τ, s. y s mτ + n s y s νcτ + νd s cτ + d s ν 0 ν s Modular functions ( a b Theorem. For SL c d (Z, τ H, and Re s >, ( aτ + b G cτ + d, s = G(τ, s. Proof. aτ + b cτ + d = aτ + b cτ + d cτ + d cτ + d = ac τ + adτ + bcτ + bd cτ + d, so, for τ = x + iy and aτ+b cτ+d = u + iv, using that ad bc =, u = ac τ + bd + x(ad + bc y cτ + d, v = cτ + d ; we shall only use the expression for v. Also, for (m, n Z, ( aτ + b (ma + ncτ + mb + nd m + n =. cτ + d cτ + d Then, ( aτ + b G cτ + d, s = = (0,0 (m,n Z (0,0 (m,n Z ( s y (ma + ncτ + mb + nd cτ + d cτ + d ys (ma + nbτ + mb + nd s. s

( a b But SL c d (Z implies that (m, n (ma + nc, mb + nd is a bijection Z \ {(0, 0} Z \ {(0, 0}, so (0,0 (m,n Z (ma + nbτ + mb + nd s = and thus we get completing the proof. 3 Fourier expansion ( aτ + b G cτ + d, s = G(τ, s, We now derive the Fourier series of G(, s. K s Theorem 3. If τ H and Re s >, then where µτ + ν s, (0,0 (µ,ν Z denotes the Bessel function. G(τ, s = ζ(sy s Γ ( s + π ζ(s y s+ + πs n s σ s+ (nf s (πny cos(πnx, F s (w = ( w π / K s (w. Proof. Define S(z, s = n Z We can write G(τ, s using this as G(τ, s = n 0 y s, z = x + iy, y 0, Re s >. z + n s y s n s + m 0 n Z y s mτ + n s = ys ζ(s + S(mτ, s m s. m 0 The Poisson summation formula states that if f : R C is continuous and of locally bounded variation, then for all x R, f(ke πikx, n Z f(x + n = k Z Henri Cohen, Number Theory, vol. II: Analytic and Modern Tools, p., Theorem 0.4.3. Henri Cohen, Number Theory, vol. I: Tools and Diophantine Equations, p. 46, Corollary..7. 3

where f(ξ = R e πiξt f(tdt, ξ R. Let z = x + iy, y 0, let Re s >, and define f y : R C by f y (t = t + iy s = ((t + iy(t iy s = (t + y s, t R. Applying the Poisson summation formula we get x + n + iy s = f y (ke πikx, n Z k Z i.e., with S(z, s = y s k Z f y (k = R f y (ke πikx, ( e πikt (t + y s dt. As y 0, doing the change of variable t = yu we get f y (k = e πikyu (y u + y s y du R = y s+ e πikyu (u + s du 0 R = y s+ cos(πkyu(u + s du; the final equality is because the function u (u + s is even. We use the following identity: 3 for a > 0 and Re s >, ( a s cos(au(u + s du = π / K s (a. For k Z \ {0}, using this with a = π ky > 0 gives 0 f y (k = y s+ π / (π ky s K s (π ky = y s+ π s k s K s (π ky. Therefore ( becomes S(z, s = y s fy (0 + y s k 0 y s+ π s k s = y s fy (0 + y πs = y s fy (0 + 4 y πs k 0 k= K s (π ky eπikx k s Ks (π ky eπikx k s Ks (πk y cos(πkx. 3 Henri Cohen, Number Theory, vol. II: Analytic and Modern Tools, p. 7, Theorem 9.8.9. 4

We use the following identity for the beta function: 4 For Re b > Re a > 0, u a (u + b du = ( a 0 B, b a = Γ ( ( a Γ b a. Γ(b Using this with a = and b = s, and since Γ ( = π, f y (0 = y s+ (u + s du = y s+ π Γ ( s. Therefore 0 S(z, s = π y s+ Γ ( s + 4 y πs k s Ks (πk y cos(πkx k= = π y s+ Γ ( s + πs k s F s (πk y cos(πkx. k= We now express G(τ, s using this formula for S(z, s. For τ H and Re s >, since S(z, s = S( z, s, As km=n G(τ, s = y s ζ(s + = y s ζ(s + m 0 S(mτ, s m s S(mτ, s m= m s = y s Γ ( s ζ(s + π (my s+ m s m= + πs m s k s F s (πkmy cos(πkmx m= k= = y s Γ ( s ζ(s + π y s+ ζ(s + πs k s m s F s (πkmy cos(πkmx. k s m s = km=n k,m (km s m s = N s km=n m s+ = N s σ s+ (N, 4 Henri Cohen, Number Theory, vol. II: Analytic and Modern Tools, p. 93, Corollary 9.6.40. 5

this can be written as G(τ, s = y s Γ ( s ζ(s + π y s+ ζ(s + πs N s σ s+ (NF s (πny cos(πnx, completing the proof. N= We use the above Fourier expansion to establish that for all t H, G(τ, s has a meromorphic continuation to C and satisfies a certain functional equation. 5 The meromorphic continuation and functional equation of G(τ, s can also be obtained without using its Fourier expansion. 6 Theorem 4. For any τ H, G(τ, s has a meromorphic continuation to C whose only pole is at s =, which is a simple pole with residue π. The function satisfies the functional equation G(τ, s = π s G(τ, s G(τ, s = G(τ, s. Proof. For ν C and for w > 0 we have 7 and 8 K ν (w = Kν(w 0 e w cosh t cosh(νtdt ( w e w, w +. π Using the above identity, one checks that for w > 0, the function s K s (w is entire, and that for any s C, the function w K s (w belongs to C (R >0. We have a fortiori that for any s C, K s (w = O(e w, w +. 5 Henri Cohen, Number Theory, vol. II: Analytic and Modern Tools, p., Corollary 0.4.4. 6 Paul Garrett, The simplest Eisenstein series, http://www.math.umn.edu/~garrett/m/ mfms/notes_c/simplest_eis.pdf 7 Henri Cohen, Number Theory, vol. II: Analytic and Modern Tools, p. 3, Proposition 9.8.6. 8 Henri Cohen, Number Theory, vol. II: Analytic and Modern Tools, p. 5, Proposition 9.8.7. 6

Let Λ(s = π s Γ ( s ζ(s. The functional equation for the Riemann zeta function states that Λ has a meromorphic continuation to C whose only poles are at s = 0 and s =, which are simple poles, and satisfies, for all s 0,, Λ( s = Λ(s. Using the Fourier series for G(, s, we have that for τ H and Re s >, G(τ, s = π s G(τ, s = Λ(sy s + Λ(s y s+ + n s σ s+ (nf s (πny cos(πnx. The residue of Λ(s at s = 0 is ; the residue of Λ(s at s = is ; the residue of Λ(s at s = is ; and the residue of Λ(s at s = is. It follows that the residue of G(τ, s at s = 0 is ; the residue of G(τ, s at s = is y/ y/ = 0; the residue of G(τ, s at s = is ; and these are no other poles of G(τ, s. Because has a simple pole at s = 0, G(τ, s = πs G(τ, s does not have a pole at s = 0. The residue of G(τ, s at s = is and this is the only pole of G(τ, s. For s C, n ( s σ ( s+ (n = n s σ s (n = (ef s e s ef=n = e s f s ef=n = (ef s f s+ ef=n = n s σ s+ (n. π Γ( = π, Generally, K ν = K ν, so F s = F ( s. Thus each term in the series in the above formula for G(τ, s is unchanged if s is replaced with s, and together 7

with Λ( w = Λ(w this yields G(τ, s = Λ( sy s + Λ( s y ( s+ + n s σ s+ (nf s (πny cos(πnx = Λ( (s y s + Λ( sy s + n s σ s+ (nf s (πny cos(πnx = Λ(s y s + Λ(sy s + n s σ s+ (nf s (πny cos(πnx = G(τ, s. 4 Kronecker limit formula For τ H, Theorem 4 shows that G(τ, s is meromorphic and that its only pole is at s =, which is a simple pole with residue π. It follows that G(τ, s has the Laurent expansion about s =, G(τ, s = π s + a 0(τ + a (τ (s +, and so defining π C(τ = a 0(τ, G(τ, s = π ( + C(τ + O( s, s. s We define the Dedekind eta function η : H C by η(τ = e πiτ ( q n, τ H, where q = e πiτ = e πy e πix, for τ = x + iy. We now prove the Kronecker limit formula, 9 which expresses C(τ in terms of the Dedekind eta function. Theorem 5. For τ = x + iy H, G(τ, s = π ( + C(τ + O( s, s, s with C(τ = γ log log y 4 log η(τ. 9 Henri Cohen, Number Theory, vol. II: Analytic and Modern Tools, p. 3, Theorem 0.4.6. 8

Proof. Define Γ ( s G(s = π ζ(s y s+. Then ( log G(s = ( Γ s log π + log ζ(s + ( s + log y + log. ( We use the asymptotic formula and with this gives, as s, and hence log ζ(s = log ζ(s = + γ + O( s, s, s log( + w = w + O( w, w, ( + γ + O( s s = log(s + log( + γ(s + O( s = log(s + γ(s + O( s, log ζ(s = log(s + γ(s + O( s, s. The Taylor series for log Γ(z about z = is log Γ(z = ( log π ( log + γ z + ( k ( k ζ(k k for z <, and the Taylor series of log Γ( + z about z = 0 is log Γ( + z = γz + k= ( k ζ(k k zk, k= ( z k, for z <. Using these we have ( log Γ s = log π ( log + γ(s + O( s, s and log = γ(s + O( s, s. 9

Applying these approximations with ( we get, as s, log G(s = log π log(s + γ(s + O( s + ( s + log y + log π ( log + γ(s + O( s + γ(s + O( s = log π log log(s + γ(s + ( s + log y ( log + γ(s + γ(s + O( s = log π log(s + (γ log log y(s + O( s. Taking the exponential and using e w = + w + O( w, w 0, as s we have G(s = π s ( + (γ log log y(s + O( s = π s + π (γ log log y + O( s. Using this and the fact that Theorem 3 thus yields that as s, G(τ, s = π 6 y + π + πs We have As well, and σ s+ (n = d n ζ(sy s = π y + O( s, s, 6 s + π (γ log log y + O( s n s σ s+ (nf s (πny cos(πnx. π s = π + O( s, s. n s = + O( s, s, d s+ = d n (d +O( s = σ (n+o( s, s. Finally, we use the fact that that for all t > 0, 0 K (t = π t e t, 0 Henri Cohen, Number Theory, vol. II: Analytic and Modern Tools, p., Theorem 9.8.5. 0

giving ( t F (t = K π (t = e t, and hence F s (πny = e πny + O( s, s. Therefore, as s, G(τ, s = π 6 y + π s + π (γ log log y + π σ (ne πny cos(πnx + O( s. This implies that the constant term in the Laurent expansion of G(τ, s about s = is a 0 (τ = π 6 y + π (γ log log y + π σ (ne πny cos(πnx. But, with q = e πiτ, ( Re σ (nq n = σ (nre (e πinτ so where = = σ (nre (e πny e πinx σ (ne πny cos(πnx, a 0 (τ = π 6 y + π (γ log log y + πs(τ, ( S(τ = Re σ (nq n. Using the power series for log( + z about z = 0, log ( q n = log( q n = = = m= q nm m q N d N= d N σ (Nq N, N=

so Then, because and because this becomes S(τ = Re η(τ = e πiτ S(τ = log ( log ( q n. Re log z = log z ( q n = e πy q n, q n = πy log η(τ. Thus so a 0 (τ = π 6 y + π completing the proof. π (γ log log y y π log η(τ 6 = π (γ log log y π log η(τ, C(τ = γ log log y 4 log η(τ, 5 Hyperbolic Laplacian For f C (H, we define H f : H C by ( H f(τ = y ( xf + yf(τ, τ = x + iy H. For more on H see the below references. Let (0, 0 (m, n Z and Re s >, and define f : H C by f(x, y = y s mx + n + imy s = y s (mx + n + imy s (mx + n imy s. Write g(x, y = (mx + n + imy s (mx + n imy s. Daniel Bump, Spectral Theory and the Trace Formula, http://sporadic.stanford.edu/ bump/match/trace.pdf; Fredrik Strömberg, Spectral theory and Maass waveforms for modular groups from a computational point of view, http://www.cams.aub.edu.lb/events/confs/ modular0/files/lecture_notes_spectral_theory.pdf; cf. Anton Deitmar, Automorphic Forms, p. 54, Lemma.7.3.

We calculate ( x g(x, y = s(mx + n + imy s m(mx + n imy s s(mx + n + imy s (mx + n imy s m, and ( xg(x, y = s(s + (mx + n + imy s m (mx + n imy s + s (mx + n + imy s (mx + n imy s m + s (mx + n + imy s m (mx + n imy s + s(s + (mx + n + imy s (mx + n imy s m = s(s + m (mx + n + imy g(x, y + s m (mx + n + imy (mx + n imy g(x, y + s(s + m (mx + n imy g(x, y, from which we have ( xf(x, y = s(s + m (mx + n + imy f(x, y + s m mτ + n f(x, y + s(s + m (mx + n imy f(x, y. We also calculate ( y g(x, y = s(mx + n + imy s im(mx + n imy s s(mx + n + imy s (mx + n imy s ( im, ( yg(x, y = s(s + (mx + n + imy s ( m (mx + n imy s + s (mx + n + imy s (mx + n imy s m + s (mx + n + imy s (im(mx + n imy s ( im + s(s + (mx + n + imy s (mx + n imy s ( im = s(s + m (mx + n + imy g(x, y + s m (mx + n + imy (mx + n imy g(x, y s(s + m (mx + n imy g(x, y. Now, and ( y f(x, y = sy s g(x, y + y s ( y g(x, y ( yf(x, y = s(s y s g(x, y + sy s ( y g(x, y + y s ( yg(x, y, 3

from which we have ( yf(x, y = s(s y s g(x, y s imy s (mx + n + imy g(x, y + s imy s (mx + n imy g(x, y s(s + m y s (mx + n + imy g(x, y + s m y s (mx + n + imy (mx + n imy g(x, y s(s + m y s (mx + n imy g(x, y = s(s y f(x, y s imy (mx + n + imy f(x, y + s imy (mx + n imy f(x, y s(s + m (mx + n + imy f(x, y + s m mτ + n f(x, y s(s + m (mx + n imy f(x, y. Combining the above expressions we get ( xf + y(x, y = m f(x, y (s mτ + n + s mτ + n Thus i.e., + s(s y f(x, y s imy (mx + n + imy f(x, y + s imy (mx + n imy f(x, y = m f(x, y (s mτ + n + s mτ + n + s(s y f(x, y 4s m mτ + n f(x, y = s(s y f(x, y. ( H f(x, y = s(s f(x, y, H f = s(s f. Thus we immediately get that for Re s >, H G(, s = s(s G(, s. Because the coefficients of the differential operator L = H s(s are real analytic, a function f : H C satisfying Lf = 0 is real analytic. Therefore, for Re s >, G(, s is real analytic. Lipman Bers and Martin Schechter, Elliptic Equations, in Lipman Bers, Fritz John, and Martin Schechter, eds., Partial Diferential Equations, p. 07, Chapter 4, Appendix. 4

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