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A symplectic restriction problem
Valentin Blomer
Universitat Bonn
Automorphic Forms in Budapest
Valentin Blomer A symplectic restriction problem Automorphic Forms in Budapest 1 / 23
Periods ofautomorphic forms
����L -functions //
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Arithmetic
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Valentin Blomer A symplectic restriction problem Automorphic Forms in Budapest 2 / 23
Examples
Example 1: (Riemann)
ζ(s) =∞∑
n=1
1ns =
(2π)s/2
2Γ(s/2)
∫ ∞
0
(θ(iy) − 1
)ys/2 dy
y
θ(z) =∑n∈Z
e(n2z)
analytic properties of ζ ! analytic properties of θ.
Example 2: (Hecke) Let f ∈ Sk be a Hecke eigenform. We have
L(f , s)Γ(s +
k − 12
)(2π)−s =
∫ ∞
0f(iy)ys dy
y
Valentin Blomer A symplectic restriction problem Automorphic Forms in Budapest 3 / 23
... continued
Example 3: Let X = Γ\G/K ⊇ X0. Let Φ, φ0 be constituents (“harmonics”) inspectral decomposition of L2(X) resp. L2(X0). Is there a connection∣∣∣∣ ∫
X0
Φ(y)φ0(y)dy∣∣∣∣2 ! L − value ?
−! Gross-Prasad conjecture
Application: Apply Parseval for the restriction norm:∫X0
|Φ(y)|2dy =
∫X0
∣∣∣∣ ∫X0
Φ(y)φ0(y)dy∣∣∣∣2dµ(φ0)
Examples 1 and 2: G = SL2(R), G0 ={(∗
∗
)⊆ SL2(R)
}� R∗
Valentin Blomer A symplectic restriction problem Automorphic Forms in Budapest 4 / 23
Siegel modular forms
Let
G = Sp4(R), G0 ={(∗
∗
)⊆ Sp4(R)
}� GL2(R).
The Siegel upper half space is
H(2) ={Z = X + iY ∈ Sym2(C) | Y > 0
}.
The group G acts on H(2) by Mobius transforms
MZ = (AZ + B)(CZ + D)−1, M =(A BC D
)∈ G.
We define
S(2)k = {Siegel modular forms F for Sp4(Z) of weight k }.
Fourier coefficients of F are indexed by integral positive binary quadratic forms Q .
Valentin Blomer A symplectic restriction problem Automorphic Forms in Budapest 5 / 23
Spectral decomposition
We write X = Sp4(Z)\H(2) with dim X = 6 and
X0 := SL2(Z)\Pos2(R) � SL2(Z)\H × R>0 =: X∗0 × R>0
Y =
√r
y
( 1 −x−x x2 + y2
) ! (x + iy, r)
with dim X0 = 3. Harmonics on X0 are Maaß forms and powers.
Is ∫X0
F(iY) u(z)rs dx dyy2
drr
an L -value?
Yes, it is a Koecher-Maaß series.
Valentin Blomer A symplectic restriction problem Automorphic Forms in Budapest 6 / 23
The Koecher-Maaß series
For a Siegel modular form F with Fourier coefficients a(Q), Q = (a, b , c) anintegral positive binary quadratic form, and a Maaß form u define
L(F × u, s) =∑
Q mod SL2(Z)
a(Q)u(HQ)
(det Q)s
where
HQ =−b + i
√|b2 − 4ac |
2a∈ SL2(Z)\H
is the Heegner point associated with Q .
This L -function has
a functional equation,
but no Euler product.
Valentin Blomer A symplectic restriction problem Automorphic Forms in Budapest 7 / 23
The restriction norm
Let F be a Siegel modular form of weight k . Then
N(F) :=vol(X)
vol(X∗0)
1‖F‖22
∫X0
|F(iY)|2(det Y)k dY(det Y)3/2
=
∫X∗0
∫R
|L(F × u, 1/2 + it)|2|G(F × u, 1/2 + it)|2dt du
where G(s,F × u) are suitable gamma factors.
For large k , the gamma factors decay at t , tu � k 1/2, so this is an average of sizek 3/2, and the L -function squared has conductor k 8.
Lindelof hypothesis: N(F) � k ε.
In absence of an Euler product it is not clear if the Lindelof hypothesis is true.
But we expect more...
Valentin Blomer A symplectic restriction problem Automorphic Forms in Budapest 8 / 23
The mass equidistribution conjecture
Let f ∈ Sk be a classical Hecke eigenform of weight k , h a test function.
Theorem (Holowinsky-Soundararajan 2009)∫SL2(Z)\H
|f(z)|2yk
‖f‖22h(z)
dx dyy2
k!∞−!
1vol(SL2(Z)\H)
∫SL2(Z)\H
h(z)dx dy
y2 .
Is this true on thin subsets?
Is this true in higher rank?
Valentin Blomer A symplectic restriction problem Automorphic Forms in Budapest 9 / 23
Thin symplectic mass equidistribution conjecture
N(F) = 4 log k + O(1) , k ! ∞
Why? Recall
X0 := SL2(Z)\Pos2(R) � SL2(Z)\H × R>0 =: X∗0 × R>0
Y =
√r
y
( 1 −x−x x2 + y2
) ! (x + iy, r)
The space X0 has infinite volume, but a Siegel cusp form of weight k decaysquickly as soon as λmin(Y) � 1/k , λmax(Y) � k . We obtain roughly an effectivevolume ∫ k 2
1/k 2
drr
= 4 log k .
Valentin Blomer A symplectic restriction problem Automorphic Forms in Budapest 10 / 23
Saito-Kurokawa lifts
M+k−1/2 � S2k−2
SK↪−! S(2)
k
dim � k dim � k 3
a(det Q) a(Q) depends only on det Q
Recall
L(F × u, s) =∑
Q mod SL2(Z)
a(Q)u(HQ)
(det Q)s =∑D<0
a(|D |)P(D, u)
|4D |s
where P(D, u) is the Heegner period
P(D, u) =∑z∈HD
u(z),
HD ={−b + i
√|D |
2a| ax2 + bxy + cy2 of disc D
}/SL2(Z)
Valentin Blomer A symplectic restriction problem Automorphic Forms in Budapest 11 / 23
The main result
Theorem 1: (B-Corbett 2019)Let W be a smooth weight function with support in [1, 2], ω =
∫W(x)x dx. Then
Nav(K) =1ω
12K2
∑k∈2N
W( kK
) ∑f∈B2k−2
N(SK(f)) = 4 log K + O(1).
This is an averaged version of
the thin symplectic mass equidistribution conjecture and
(a strong form of) the Lindelof hypothesis for L(F × u, s).
Valentin Blomer A symplectic restriction problem Automorphic Forms in Budapest 12 / 23
Steps of the proof...
Apply the Parseval period formula and an approximate functional equation:∑k
∑f∈B2k−2
∑j
∫R
|L(SK(f) × uj , 1/2 + it)|2|G(· · · )|2dt .
separate treatment of the constant function u0 where P(D, u0) = H(D) is theHurwitz class number
Valentin Blomer A symplectic restriction problem Automorphic Forms in Budapest 13 / 23
Interlude I: Voronoi formula for Hurwitz class numbers
Let c ∈ N, 4 | c, (a, c) = 1. Let φ be a smooth function with compact support in(0,∞). Then∑
D<0
H(D)
|D |1/4e(a |D |
c
)φ(|D |)
=
√2
c1/2
(−ca
)εae
(38
)∑D<0
H(D)
|D |1/2e(−
a |D |c
) ∫ ∞
0sin
(4π√|D |t
c
)φ(t)
dtt1/4
+12
∑n=�
e( an
c
) ∫ ∞
0exp
(−
4π√
ntc
)φ(t)
dtt1/4
+
∫ ∞
0φ(x)
( 14x1/4 −
π
3cx1/4
)dx
.
Valentin Blomer A symplectic restriction problem Automorphic Forms in Budapest 14 / 23
Steps of the proof... continued
Apply the Parseval period formula and an approximate functional equation.∑k
∑f∈B2k−2
∑j
∫R
|L(SK(f) × uj , 1/2 + it)|2|G(· · · )|2dt
separate treatment of the constant function u0 where P(D, u0) = H(D) is theHurwitz class number
sum over f ∈ B2k−2 by a half-integral weight Kohnen-Petersson formula.
the diagonal term: by Waldspurger/Zhang we have
|P(D, uj)|2 = |D |1/2
L(uj , 1/2)L(uj × χD , 1/2)
4L(sym2uj , 1)
Valentin Blomer A symplectic restriction problem Automorphic Forms in Budapest 15 / 23
Interlude II: a commutative diagram
P(D, u)Katok-Sarnak
Biro
Duke-Imamoglu-Toth//
Zhang
''
aSh−1(u)(|D |)
Baruch-Mao
��L(u × χD , 1/2)
Valentin Blomer A symplectic restriction problem Automorphic Forms in Budapest 16 / 23
Steps of the proof... continued
Apply the Parseval period formula and an approximate functional equation.∑k
∑f∈B2k−2
∑j
∫R
|L(SK(f) × uj , 1/2 + it)|2|G(· · · )|2dt
separate treatment of the constant function u0 where P(D, u0) = H(D) is theHurwitz class number
sum over f ∈ B2k−2 by a half-integral weight Kohnen-Petersson formula.
the diagonal term: by Waldspurger/Zhang we have
|P(D, uj)|2 = |D |1/2
L(uj , 1/2)L(uj × χD , 1/2)
4L(sym2uj , 1)
sum over uj with the Kuznetsov formula
the diagonal-diagonal term: explicit evaluation
Valentin Blomer A symplectic restriction problem Automorphic Forms in Budapest 17 / 23
Interlude III: an Euler product
The various integral and half-integral weight Hecke relations lead to the followingEuler product
∏p
(1 −
1p2 −
1p3 +
1p4
)×
∑(dδν,m)=1
∑(f ,δ)=1
∑d1 |fdd2 |fν
∑d3 |(
fdd1, fν
d2)
∑r1r2= f2dν
d1d2d23
µ(dδν)µ2(m)µ(d1)µ(d2)d3
(dν)4δ2f3m3rad(δd1d2r2)
×∏
p|δd1d2r2
(1 +
1p−
1p3
)−1
where rad(n) is the squarefree kernel. It equals
ζ(4)−1
Valentin Blomer A symplectic restriction problem Automorphic Forms in Budapest 18 / 23
Steps of the proof... continued
Apply the Parseval period formula and an approximate functional equation.∑k
∑f∈B2k−2
∑j
∫R
|L(SK(f) × uj , 1/2 + it)|2|G(· · · )|2dt
separate treatment of the constant function u0 where P(D, u0) = H(D) is theHurwitz class number
sum over f ∈ B2k−2 by a half-integral weight Kohnen-Petersson formula.
the diagonal term: by Waldspurger/Zhang we have
|P(D, uj)|2 = |D |1/2
L(uj , 1/2)L(uj × χD , 1/2)
4L(sym2uj , 1)
sum over uj with the Kuznetsov formula
the diagonal-diagonal term: explicit evaluation
the diagonal-offdiagonal term: multiple Poisson summation andHeath-Brown’s large sieve for quadratic characters.
Valentin Blomer A symplectic restriction problem Automorphic Forms in Budapest 19 / 23
The off-diagonal term
sum over k � K
interpret P(D, uj) as metaplectic Fourier coefficients (by Katok-Sarnak) andapply half-integral Voronoi summation
heart of the proof: need cancellation in∑u
P(D1, u)P(D2, u)︸ ︷︷ ︸product of 4 half-integral weight Fourier coefficients
h(tu)
“Kuznetsov formula for toric Fourier coefficients”
Valentin Blomer A symplectic restriction problem Automorphic Forms in Budapest 20 / 23
A trace formula for pairs of Heegner periodsTheorem 2: Let D1,D2 be two negative fundamental discriminants. Let
F(x, t) = Jit (x) cos(π/4 − πit/2) − J−it (x) cos(π/4 + πit/2),
Wt (n) =1
2πi
∫(2)
Γ( 12 ( 1
2 + s + 2it))Γ( 12 ( 1
2 + s − 2it))
Γ( 14 + it)Γ( 1
4 − it)πses2
n−s dss.
Then
1|D1D2 |1/4
∫X∗0
P(D1; u)P(D2; u)h(tu)du
=3π
H(D1)H(D2)
|D1D2 |1/4h(i/2) − constant function
+
∫ ∞
−∞
∣∣∣∣D1D2
4
∣∣∣∣it/2 Γ(− 14 + it
2 )e(1/2−it)2
√8πΓ( 1
4 + it2 )
L(χD1 , 1/2 + it)L(χD2 , 1/2 + it)ζ(1 + 2it)
h(t)dt4π
polar term
+ δD1=D2
∑m
χD1 (m)
m
∫ ∞
−∞
Wt (m)h(t)t tanh(πt)dt
4π2 − diagonal term
+ e(3/8)∑
n,c,m
K+3/2(|D1 |n2, |D2 |, c)χD1 (m)
n1/2cm
∫ ∞
−∞
F(4πn√|D1D2 |/c, t)
cosh(πt)h(t)Wt (nm)t
dtπ.
Valentin Blomer A symplectic restriction problem Automorphic Forms in Budapest 21 / 23
Idea of proof
We transform P(D1, u)P(D2, u) as follows:
Katok-Sarnak: av(|D1|)av(1)av(|D2|)av(1), v = Sh−1(u)
Waldspurger: av(|D1|)av(|D2|)L(u, 1/2)
approximate functional equation av(|D1|)av(|D2|)∑
n λ(n)n−1/2
Shimura:∑
n n−1/2av(|D1|n2)av(|D2|)
metaplectic Kuznetsov formula:∑
n,c K(|D1|n2, |D2|, c)
Valentin Blomer A symplectic restriction problem Automorphic Forms in Budapest 22 / 23
Periods ofautomorphic forms
����L -functions //
??
Arithmetic
__
oo
Valentin Blomer A symplectic restriction problem Automorphic Forms in Budapest 23 / 23