Fourier Analysis and Number Theory III: The Poisson summation formula as a trace formula

Yin Yang

Yin Yang


Yin and Yang are important concept in the Chinese philosophy. They usually describe two opposite things which are not only complement of each other but also have deep interrelationship. In mathematics, we called them duality. For example (1) primes vs zeros of Riemann zeta-functions, (2) in the theory of finite group representation, irreducible characters vs conjugacy classes, (3) singular homology vs deRham cohomology, (4) length of closed geodesic vs eigenvalues of Laplacian, (5) particles vs wave. See Arthur, Harmonic Analysis and Group Representation.

Poisson summation formula can be viewed as a formula that links the values related to {f} and values related to the representations of {L^2({\mathbb Z} \backslash {\mathbb R})}. Let {\Gamma} be a discrete subgroup of a measurable, topological group {G}. Trace formulas are formulas relate the representations of {L^2(\Gamma \backslash G)} (spectral side) and a summation over {\Gamma} (geometric side). The formulas have many application to number theory, for example, it can be used to prove some special cases of Langlands functoriality.

In this note, we give another proof of the Poisson summation formula (See Lecture 2 Theorem 5 ) by using the trace formula.

1. Group representations

Let {G} be a group and {V} a linear vector space over {{\mathbb C}}. A group representation of {G} on {V} is a group homomorphism {\pi: G \rightarrow \mathrm{GL}(V)}, where {\mathrm{GL}(V)} is the set of invertible endomorphism of {V}. Equivalently, for {g, g' \in G} and {v \in V}, {\pi(g)v \in \mathrm{End}(V)} such that \pi(e)=I_V and {\pi(g) \pi(g') v = \pi(g g') v}.

2. Right regular representations and kernel functions

Recall that {{\mathbb Z} \backslash {\mathbb R}} is a topological group isomorphic to the multiplicative group {S^1 = \{ z \in {\mathbb C} | |z|=1\}} by {\theta \mapsto e^{2 \pi i \theta}}.

The right regular representation is a representation of {{\mathbb R}} on {L^2({\mathbb Z} \backslash {\mathbb R})} defined by

\displaystyle  (R(y)\varphi)(x) = \varphi(x+y).

It can be easily shown that {R} is a representation. Also

\displaystyle  \|R(y) \varphi\|_2 = \|\varphi\|_2.

Let {f} be a measurable function on {{\mathbb R}}. Formally define a linear operator on {L^2({\mathbb Z} \backslash {\mathbb R})} by

\displaystyle  R(f) = \int_{\mathbb R} f(y) R(y) dy.

More precisely, for {\varphi \in L^2({\mathbb Z} \backslash {\mathbb R})},

\displaystyle   (R(f) \varphi)(x) = \int_{\mathbb R} f(y) (R(y)\varphi)(x) dy = \int_{\mathbb R} f(y) \varphi(x+y) dy. \ \ \ \ \ (1)

{R(f)} is like a weighted sum of {R(f)} with weight {f(y)}. To justify the well-definedness, we need to check the absolutely convergence of the above integral. For example

Proposition 1 Suppose {f \in L^2({\mathbb R})}, then the integral (1) is absolutely convergent and is bounded by {\|f\|_2 \|\varphi \|_2}. Thus {R(f)} is a bounded linear operator with norm {\le \|f\|_2}.

Proof:

\displaystyle  \int_{\mathbb R} |f(y) \varphi(x+y)| dy \le \left( \int_{\mathbb R} |f(y)|^2 d y \right)^{1/2} \left( \int_{\mathbb R} |\varphi(x+y)|^2 dy \right)^{1/2} = \|f\|_2 \|\varphi\|_2.

\Box

Remark: If {f \in L^1({\mathbb R})}, then (1) is also absolutely convergent. Let {\varphi \in L^2({\mathbb Z} \backslash {\mathbb R})}. Then

\displaystyle  (R(f) \varphi)(x) = \int_{\mathbb R} f(y) \varphi(x+y) dy = \int_{\mathbb R} f(y-x) \varphi(y) dy

\displaystyle  = \int_{{\mathbb Z} \backslash {\mathbb R}} \sum_{n \in {\mathbb Z}} f(n+y-x) \varphi(n+y) dy = \int_{{\mathbb Z} \backslash {\mathbb R}} \left( \sum_{n \in {\mathbb Z}} f(n+y-x) \right) \varphi(y) dy.

Let

\displaystyle   K(x,y) = \sum_{n \in {\mathbb Z}} f(n+y-x). \ \ \ \ \ (2)

The function is called a kernel function. The above shows that

\displaystyle   (R(f) \varphi)(x) = \int_{{\mathbb Z}\backslash {\mathbb R}} K(x,y) \varphi(y) dy. \ \ \ \ \ (3)

We have the following proposition.

Proposition 2 If {f(x) \ll \frac 1{1+|x|^2}}, then {K(x,y)} is a continuous function.

3. Spectral expansion of the kernel functions

Let {\varphi_n(x) = e^{2 \pi i n x}}. Then {\{\varphi_n|n \in {\mathbb Z}\}} is an orthonormal basis of {L^2({\mathbb Z} \backslash {\mathbb R})}. Formally we define

\displaystyle   \Phi(x,y) = \sum_{n \in {\mathbb Z}} (R(f)\varphi_n)(x) \overline{\varphi_n(y)}. \ \ \ \ \ (4)

Note that

\displaystyle  (R(f) \varphi_n) (x) = \int_{\mathbb R} f(y) e^{2\pi i n (x+y)}dy

\displaystyle  = e^{2\pi i n x} \int_{\mathbb R} f(y) e^{2\pi n y} dy = \widehat{f}(-n) \varphi_n(x).

Proposition 3 Suppose {f} has a continuous second derivative satisfying

\displaystyle  f(x) \ll \frac 1{1+|x|^2} \text{ and } f'(x) \ll \frac 1{1+|x|^2}

and {f'' \in L^1({\mathbb R})}. Then

\displaystyle  \sum_{n \in {\mathbb Z}} |(R(f)\varphi_n)(x) \overline{\varphi_n(y)}| \ll 1

and {\Phi(x,y)} is a continuous function.

Proof: Because {f'' \in L^1({\mathbb R})} and is continuous, {f''(\pm \infty)=0}. By Lecture 1, lemma 2 (i) for {t\neq 0}, {\widehat{f}(t) = (4\pi i t)^{-2} \widehat{f''}(t) \ll t^{-2}}. Thus for {n \neq 0}, {(R(f)\varphi_n)(x) \overline{\varphi_n(y)} \ll \frac{1}{n^2}}. The proposition follows easily. \Box

Proposition 4 Suppose {f} satisfies the conditions of Proposition 3. Suppose {\varphi \in L^2({\mathbb Z} \backslash {\mathbb R})}. Then

\displaystyle  \int_{{\mathbb Z} \backslash {\mathbb R}} \Phi(x,y) \varphi(y) dy

is absolutely convergent and is a bounded linear operator. Also

\displaystyle   (R(f)\varphi)(x) = \int_{{\mathbb Z} \backslash {\mathbb R}} \Phi(x,y) \varphi(y) dy. \ \ \ \ \ (5)

Proof: By the previous proposition, {\Phi(x,y)} is bounded, say by {C}.

\displaystyle  \int_{{\mathbb Z} \backslash {\mathbb R}} |\Phi(x,y) \varphi(y)| dy \le \left( \int_{{\mathbb Z} \backslash {\mathbb R}} |\Phi(x,y)|^2 dy \right)^{1/2} \left(\int_{{\mathbb Z} \backslash {\mathbb R}} |\varphi(y)|^2 dy \right)^{1/2} \le C \|\varphi\|_2.

This proves the absolutely convergence and the boundedness.

Let {\varphi = \varphi_m}. Then by the previous proposition,

\displaystyle  \int_{{\mathbb Z} \backslash {\mathbb R}} \Phi(x,y) \varphi(y) dy = \int_{{\mathbb Z} \backslash {\mathbb R}} \sum_{n \in {\mathbb Z}} (R(f)\varphi_n)(x) \overline{\varphi_n(y)} \varphi_m(y) dy

\displaystyle  = \sum_{n \in {\mathbb Z}} (R(f)\varphi_n)(x)\int_{{\mathbb Z} \backslash {\mathbb R}} \overline{\varphi_n(y)} \varphi_m(y) dy = (R(f) \varphi_m)(x).

Thus (5) is valid for {\varphi = \varphi_m}. Because {\{\varphi_m\}} is an orthonormal basis of {L^2({\mathbb Z}\backslash{\mathbb R})} and both sides of (5) are bounded linear operator, (5) is valid for any {\varphi \in L^2({\mathbb Z}\backslash {\mathbb R})}. \Box

Suppose {f} satisfies the conditions of Proposition 3. Comparing (3) and (5), for any {\varphi \in L^2({\mathbb Z} \backslash {\mathbb R})},

\displaystyle  \int_{{\mathbb Z}\backslash {\mathbb R}} K(x,y) \varphi(y) dy = \int_{{\mathbb Z}\backslash {\mathbb R}} \Phi(x,y) \varphi(y) dy.

Hence

\displaystyle  \int_{{\mathbb Z}\backslash {\mathbb R}} (K(x,y)-\Phi(x,y)) \varphi(y) dy = 0.

Because both {K} and {\Phi} are continuous function,

\displaystyle  K(x,y) = \Phi(x,y).

Thus

\displaystyle   \sum_{n\in {\mathbb Z}} f(n-x+y) = \sum_{n \in {\mathbb Z}} \widehat{f}(-n) \varphi_n(x) \overline{\varphi_n(y)} \ \ \ \ \ (6)

Letting {x=y}, we obtain

\displaystyle  \sum_{n\in {\mathbb Z}} f(n) = \sum_{n \in {\mathbb Z}} \widehat{f}(-n) = \sum_{n \in {\mathbb Z}} \widehat{f}(n).

This is the Poisson summation formula.

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