research35 min studyLesson 606 in Mathematics

Poincaré series and Petersson inner products

Average a Fourier seed over a coset space and unfold an invariant integral with every measure and conjugation convention fixed.

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What you’ll understand

  • Define absolutely convergent full-level Poincaré series at even weight k>2.
  • Check invariance and convergence of the Petersson pairing on cusp forms.
  • Recover a Fourier coefficient by unfolding and explain why a Poincaré series can vanish.

Symmetrization over a finite group is a finite sum. For the infinite modular group, the analogous construction requires a quotient by the seed's stabilizer and an analytic convergence argument. Poincaré series provide that construction, while the Petersson pairing reveals exactly which Fourier data they detect. We use the full group Γ=SL₂(ℤ), even integer weight k>2, and Γ∞={±Tⁿ:n∈ℤ}. The inclusion of both signs prevents a hidden factor of two.

Average the seed over distinct translates

For an integer m≥1, start with exp(2πimτ), invariant under Γ∞ in the weight-k slash action. Summing its slash translates over left cosets of Γ∞ in Γ defines Pₖ,ₘ. Right multiplication by another group element permutes these cosets and gives the modular law once convergence is established. At even k>2, comparison with a convergent lattice sum gives absolute local uniform convergence, and the resulting function is a cusp form. Weight two is excluded from this elementary convergence statement.

A nonzero seed need not produce a nonzero averaged form. Averaging can cancel, and a cusp-form space of dimension zero forces every member Pₖ,ₘ to vanish. The case m=0 gives an Eisenstein-type average rather than the cuspidal construction used here. It is important to distinguish an individual orbit sum from a basis theorem: infinitely many Poincaré series may be linearly dependent even when their span is the whole finite-dimensional cusp space. Neither convergence nor symmetrization alone asserts that they are independent.

Pk,m(τ)=∑γ∈Γ∞\Γ(cγτ+dγ)−ke2πimγτ,Γ∞={±Tn:n∈Z},k>2 even, m≥1P_{k,m}(\tau)=\sum_{\gamma\in\Gamma_\infty\backslash\Gamma}(c_\gamma\tau+d_\gamma)^{-k}e^{2\pi i m\gamma\tau},\qquad \Gamma_\infty=\{\pm T^n:n\in\mathbb Z\},\quad k>2\text{ even},\ m\ge1
There is one term per left coset. Summing instead over translations without the central signs would double this series at even weight.↔ Scroll wide equations horizontally.

The invariant measure compensates the weight

Define the Petersson pairing as the integral of f(τ) times the conjugate of g(τ), multiplied by y^k, against hyperbolic measure dx dy/y² over one modular fundamental region. We do not divide by the domain volume or group index. This convention is linear in f and conjugate-linear in g. The factor y^k cancels the absolute-square automorphy factors, so the density descends to the quotient. Replacing it with ordinary Euclidean area without the corresponding power adjustment breaks that invariance.

For cusp forms, exponential decay at the cusp dominates the polynomial y factor, giving a finite integral. The pairing is positive definite on the cusp space because the integral of |f|² times a positive density is positive for a nonzero holomorphic function. A noncusp Eisenstein series generally has divergent norm with this unregularized integral; E₄ tends to one, leaving an integral comparable to the divergent integral of y². Pairing a cusp form with a holomorphic noncusp form can still converge, but that does not extend a finite norm to all modular forms.

⟨f,g⟩=∫Ff(τ)g(τ)‾ykdx dyy2,⟨f,Pk,m⟩=Γ(k−1)(4πm)k−1am(f)\langle f,g\rangle=\int_{\mathcal F}f(\tau)\overline{g(\tau)}y^k\frac{dx\,dy}{y^2},\qquad \langle f,P_{k,m}\rangle=\frac{\Gamma(k-1)}{(4\pi m)^{k-1}}a_m(f)
The coefficient identity uses the full-level cusp width one and the stated linear-first, unnormalized pairing. Its constant is (k−2)!/(4πm)^(k−1).↔ Scroll wide equations horizontally.

Unfolding turns geometric averaging into coefficient extraction

Insert the Poincaré sum into the second slot and use absolute convergence to interchange summation with integration. The translated fundamental regions tile a strip for Γ∞, so the integral unfolds to x between zero and one and y>0. Conjugating the Fourier seed produces exp(−2πimx)exp(−2πmy). The x integral annihilates every Fourier mode of f except m, and the remaining y integral is a gamma integral with exponential rate 4πm. Both factors of two in that rate have a visible origin.

If a cusp form is orthogonal to every Pₖ,ₘ, the extraction formula makes all its Fourier coefficients zero; the identity theorem then makes the form zero. This proves that the Poincaré series span the cusp space. It also tests nonvanishing in particular cases: a nonzero extracted coefficient forces the corresponding Poincaré series to be nonzero. The logic does not require positivity of every individual summand. The inner product combines the symmetry and the Fourier expansion into a useful duality between a function space and coefficient functionals.

Compare two Poincaré coefficient functionals

For normalized Δ in weight 12, compute ⟨Δ,P₁₂,₁⟩ and the ratio ⟨Δ,P₁₂,₂⟩/⟨Δ,P₁₂,₁⟩. Decide whether either Poincaré series vanishes.

  1. The extraction constant at k=12 is Γ(11)=10! in the numerator. Since a₁(Δ)=1, the first pairing is 10!/(4π)¹¹, a positive nonzero number.

  2. Since a₂(Δ)=−24, the second pairing is −24·10!/(8π)¹¹. Dividing gives −24/2¹¹=−3/256, with a negative sign allowed for a mixed pairing.

  3. Neither pairing is zero, so neither P₁₂,₁ nor P₁₂,₂ is zero. Since S₁₂ has dimension one they are proportional; in the stated conjugate-linear second slot their proportionality is the conjugate of this ratio, here the same real number −3/256.

The result

The numerical ratio depends on both the Fourier coefficient and the gamma-integral scaling. It is not merely the ratio −24 of the Fourier coefficients.

Explore the model

Interactive lab

Interactive lab · waves

A nonzero component can disappear under averaging

Reset, set phase difference=0° and amplitude ratio=1; predict a sum with twice the component amplitude. Compare a fresh Reset, phase difference=180° and amplitude ratio=1, where the two nonzero components cancel. This finite two-term cancellation illustrates why a symmetrized seed need not survive. It is not the modular coset sum or a calculation of a Petersson pairing.

A nonzero component can disappear under averagingTwo waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Current readings: Sum at x = 0: 0.866; Resultant amplitude: 1.732; Phase difference: 60°. All parameters are adjustable with the labeled controls below.−3.14−1.5701.573.14−3.4−1.701.73.4x · radiansAmplitude
SumFirst componentSecond component
Sum at x = 0
0.866
Resultant amplitude
1.732
Phase difference
60°
t = 0 s

Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forNonzero ingredients do not by themselves imply a nonzero symmetrized result.

Check your understanding

What can happen to P₄,₁ for the full modular group?

Try it yourself

For the same normalization, let f=2iΔ. Compute ⟨f,P₁₂,₁⟩ and ⟨P₁₂,₁,f⟩. Explain which one extracts a₁(f) directly.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The pairing is linear in its first slot and Hermitian, so reversing the arguments conjugates the value.

Compare with the solution

Let C=10!/(4π)¹¹, which is real and positive. Then ⟨2iΔ,P₁₂,₁⟩=2iC, directly equal to C times a₁(f)=2i. Hermitian symmetry gives ⟨P₁₂,₁,2iΔ⟩=−2iC. The coefficient formula in this lesson places f first; changing the slot without conjugating the coefficient would reverse the sign incorrectly. Neither calculation is a norm, so an imaginary value causes no conflict with positive definiteness.

Take it with you

  • Cosets and central signs fix the Poincaré-series normalization.
  • The unscaled Petersson pairing is finite and positive on cusp forms.
  • Unfolding extracts coefficients with a precise gamma factor and slot convention.

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