From Fourier analysis on finite groups to l-adic Fourier transform
Fourier analysis on finite abelian groups
Given a finite abelian group $G$, written additively, what we want to do here is to define a space $L^2(G)$ similar to the Hilbert space $L^2(X)$ of square integrable functions $X \longrightarrow \mathbb{C}$ (modulo equal almost everywhere relation) for $X$ being a measurable space. Then it is possible to develop a Fourier transform on $L^2(G)$. The finiteness seems to be a technical condition that you can see to be useful in every step. At least with this hypothesis, we do not to worry about the convergence of sums. We do not go into the "Hilbert theory" of $L^2(G)$ deeply but rather go straight to the Plancherel formula and Fourier inverse formula and see how it can be generalized to $l$-adic cohomology.
We define a character of $G$ to be a group homomorphism $\psi: G \longrightarrow (\mathbb{C}^{\times},\times)$. We call it trivial if $\psi(x) = 1$ for each $x \in G$. Since $G$ is finite, every $\psi(x)$ ($x \in G$) is a root of unity. In particular, every character takes values in the circle $S^1$.
Example 1. If $G = \mathbb{F}_{p}$, a field with $p$ elements, then $\psi(x) = e^{2\pi i x/p}$ is a character.
Example 2. If $\psi$ is a character, then $\overline{\psi}$. They are different if $\psi$ is not identical to $1$. Note that
$$\psi(-x) = \psi(x)^{-1} = \overline{\psi(x)}$$ since it lies on $S^1$.
Example 3. If $\psi,\varphi$ are characters, then $\psi\varphi$ is a character as well.
Proposition 4. If $\psi$ is a non-trivial character of $G$, then $\sum_{x \in G}\psi(x)=0$.
Proof. Since $\psi$ is non-trivial, there exists $y \in G$ with $\psi(y) \neq 1$. We have
$$\psi(y)\sum_{x \in G}\psi(x) = \sum_{x \in G}\psi(x+y) = \sum_{x \in G}\psi(x)$$ and hence the sum itself is zero because $\psi(y) \neq 1$.
Corollary 5. Let $\psi$ and $\varphi$ be characters of $G$. Then
$$\sum_{x \in G}\overline{\psi(x)}\varphi(x) = \begin{cases} \left|k \right| & \psi = \varphi \\ 0 & \psi \neq \varphi \end{cases}$$ Proof. If $\psi = \varphi$, then it is the consequence of the fact $\psi(x)^{-1} = \overline{\psi(x)}$ while if $\psi \neq \varphi$ then $\overline{\psi}\varphi$ is a non-trivial character, hence it follows from proposition 4.
Now we work in the case where $G = k$ is a finite field, then we have a, being motivated from the classical case
$$\hat{f}(x) = \int_{-\infty}^{\infty} f(y)e^{-2\pi i xy} dy$$
we define the Fourier transform
$$\begin{align*} T_{\psi}f: k & \longrightarrow \mathbb{C}^{\times} \\ x & \longmapsto \sum_{y \in k}f(y)\psi(-xy).\end{align*}$$ The Fourier transform is clearly linear, i.e. $T_{\psi}(f+g) = T_{\psi}(f) + T_{\psi}(g)$ and $T_{\psi}(af) = aT_{\psi}(f)$. The Fourier inversion formula in this case becomes almost trivial.
Proposition 6. (Fourier inverse). We have
$$T_{1/\psi}(T_{\psi}f) = \left|k \right|f$$ Proof. We compute the LHS explicitly
$$\begin{align*}T_{1/\psi}(T_{\psi}f)(x) & = \sum_{y \in k}T_{\psi}(f)(y)\overline{\psi}(-xy) \\ & = \sum_{y \in k}\left(\sum_{z \in k}f(z)\psi(-yz) \right)\overline{\psi}(-xy) \\ & = \sum_{y,z \in k} f(z)\psi(y(x-z)) \\ & = \sum_{z \in k}f(z)\left(\sum_{y \in k}\psi(y(x-z)) \right) \\ & = \left|k \right|f(x) \end{align*}$$ thanks to corollary 5.
We can endorse the vector space $\mathbb{C}^k$ of functions $k \longrightarrow \mathbb{C}$ with an inner product
$$\left< f, g \right> = \sum_{x \in k}\overline{f(x)}g(x)$$ then we have an analogue of the usual Plancherel formula.
Proposition 7 (Plancherel formula). For functions $f,g: k \longrightarrow \mathbb{C}^{\times}$ and a character $\psi:k \longrightarrow \mathbb{C}^{\times}$
$$\left<T_{\psi}f, T_{\psi}g\right> = \left|k \right|\left <f,g \right >$$ Proof. We expand everything $$\begin{align*}\left<T_{\psi}f, T_{\psi}g\right> & = \sum_{x \in k}\overline{T_{\psi}f(x)}T_{\psi}g(x) \\ & = \sum_{x \in k}\left( \sum_{y \in k}\overline{f(y)}\psi(xy) \right)\left( \sum_{z \in k}g(z)\psi(-xz) \right) \\ & = \sum_{x,y,z \in k}\overline{f(y)}g(z)\psi(x(y-z)) \\ & = \sum_{y,z \in k}\overline{f(y)}g(z)\left(\sum_{x \in k}\psi(x(y-z)) \right). \end{align*}$$ We analyze the sum $\sum_{x \in k}\psi(x(y-z))$.
- If $y = z$ then this sum is $\left|k \right|$.
- If $y \neq z$ then this sum is zero by proposition 4.
Hence
$$ \sum_{y,z \in k}\overline{f(y)}g(z)\left(\sum_{x \in k}\psi(x(y-z)) \right) = \sum_{y \in k }\left|k \right|\overline{f(y)}g(y) = \left|k \right| \left<f,g \right>.$$ We are now able to motivate the definitions in the $l$-adic cohomology.
l-adic Fourier transform
We restrict ourself to the definition of something called $l$-adic Fourier transform on the affine line $\mathbb{A}^1_k$ where $k$ is a finite field. More precisely, we want to define some operator
$$T_{\psi}: D_c^b(\mathbb{A}^1,\overline{\mathbb{Q}}_l) \longrightarrow D_c^b(\mathbb{A}^1,\overline{\mathbb{Q}}_l)$$ associated to any character $\psi: k\longrightarrow \mathbb{C}^{\times}$ and is forced to satisfy the Fourier inverse formula and the Plancherel formula. To do this, we have to have some sheaf-to-functions correspondence, for each variety $X/k$, naturally, we have the following: for any $K \in D_c^b(X,\overline{\mathbb{Q}}_l)$
$$\begin{align*} f^K: k & \longrightarrow \mathbb{C} \\ x & \longmapsto \mathrm{Trace}(\mathrm{Frob}^*_{\overline{x}},K_{\overline{x}}) = \sum_i (-1)^i \mathrm{Trace}(\mathrm{Frob}^*_{\overline{x}},\mathscr{H}^i(K)_{\overline{x}}) \end{align*}$$ where $\mathscr{H}^i$ denote cohomological sheaves, which are assumed to be constructible.
Given any character $\psi: k \longrightarrow
\overline{\mathbb{Q}}_l^{\times}$, one then has a local system of rank
$1$ from the composition
$$\mathcal{L}_{\psi}:
\pi_1(\mathbb{A}_k^1) \longrightarrow k \overset{\psi}{\longrightarrow}
\overline{\mathbb{Q}}_l^{\times}$$ denoted $\mathcal{L}_{\psi}$, called
the Artin-Schreier sheaf of $\psi$. We claim that
Lemma 8. $f^{\mathcal{L}_{\psi}}(x) = \psi(-x)$.
Proof.
The next thing is how can we translate operations of functions to operations of sheaves:
- (Product formula) The product of functions should correspond to tensor product of sheaves: $f^{K \otimes L}(x) = f^K(x)f^L(x)$ for every $x \in X(k)$
- (Pullback formula) Pullback of functions should correspond to pullback of functions, which is just composition $f^{f^*K}(x) = f^K(f(x))$ for every morphism $f: X \longrightarrow Y$ of $k$-varieties and $x \in X(k)$.
- (Sum formula) Proper pushforwards should correspond to taking sums or integrals: this is a consequence of Grothendieck-Lefschetz trace formula and proper base change theorem. For every morphism $f: X \longrightarrow Y$ of $k$-varieties and $K \in D^b_c(X,\overline{\mathbb{Q}}_l)$, we have $$f^{f_!K}(y) = \sum_{x \in X_y(k)}f^K(x)$$ for any $y \in Y(k)$.
Consider the diagram
where $\pi^1$ are projections and $m$ the multplication $(x,y) \longmapsto xy$. The $l$-adic Fourier transform is defined to be
$$\begin{align*} T_{\psi}: D^b_c(\mathbb{A}^1,\overline{\mathbb{Q}}_l) & \longrightarrow D^b_c(\mathbb{A}^1,\overline{\mathbb{Q}}_l) \\ K & \longmapsto \pi_!^1(\pi^{2*}K \otimes m^*\mathcal{L}_{\psi})[1] \end{align*}$$ where the shift $[1]$ is put in order to preserve the perversity, which is not of our interest here. The sheaf $m^*\mathcal{L}_{\psi}$ plays the role of the character $\psi(-xy)$ in the formula
$$T_{\psi}f(x) = \sum_{y \in k}f(y)\psi(-xy).$$ We prove that our definition is really a sheaf-theoretic version of the discrete Fourier transform (up to a sign).
Lemma 9. $f^{T_{\psi}(K)}(x) = - \sum_{y \in k} f^K(y)\psi(-xy)$ for any $x \in k$ and $K \in D^b_c(\mathbb{A}^1,\overline{\mathbb{Q}}_l)$.
Proof. By the last part of our remark above, we have (we delete the shift since it is irrelevant here)
$$\begin{align*} f^{T_{\psi}(K)}(x) & = \sum_i (-1)^i \mathrm{Trace}\bigg(\mathrm{Frob}^*_{\overline{x}}, \mathscr{H}^i\big( \pi_!^1(\pi^{2*}K \otimes m^*\mathcal{L}_{\psi})_{\overline{x}} \big) \bigg) \\ & = - \sum_i (-1)^i \sum_{y \in k} \mathrm{Trace}(\mathrm{Frob}^*_{(\overline{x},\overline{y})}, \mathscr{H}^i(\pi^{2*}K \otimes m^*\mathcal{L}_{\psi})) \\ & =- \sum_i (-1)^i \sum_{y \in k} \mathrm{Trace}(\mathrm{Frob}^*_{(\overline{x},\overline{y})}, \mathscr{H}^i(\pi^{2*}K) \otimes \mathscr{H}^i( m^*\mathcal{L}_{\psi})) \\ & = - \sum_i (-1)^i \sum_{y \in k} \mathrm{Trace}(\mathrm{Frob}^*_{(\overline{x},\overline{y})}, \mathscr{H}^i(\pi^{2*}K)) \mathrm{Trace}(\mathrm{Frob}^*_{(\overline{x},\overline{y})}, \mathscr{H}^i( m^*\mathcal{L}_{\psi})) \\& = - \sum_i (-1)^i \sum_{y \in k} \mathrm{Trace}(\mathrm{Frob}^*_{(\overline{x},\overline{y})}, \mathscr{H}^i(\pi^{2*}K)) \psi(-xy) \\ & = - \sum_{y \in k} \sum_i (-1)^i \mathrm{Trace}(\mathrm{Frob}^*_{(\overline{x},\overline{y})}, \mathscr{H}^i(\pi^{2*}K)) \psi(-xy) \\ & = - \sum_{y \in k} f^K(y)\psi(-xy) \end{align*}$$ where we have used:
- The sum formula for $\pi^1: \mathbb{A}^1 \times \mathbb{A}^1 \longrightarrow \mathbb{A}^1$ in the first equality.
- In the second one, homology commutes with tensor product.
- In the third one, we applied the product formula.
- In the rest, we applied the pullback formula and lemma 8.
Let us now verify the Plancherel formula before the Fourier inverse formula (which is more complicated).
Theorem 10 (Plancherel). We have
$$\left<f^{T_{\psi}(K)},f^{T_{\psi}(L)}\right> = \left|k \right|\left<f^K,f^L\right> \ \forall \ K,L \in D^b_c(\mathbb{A}^1,\overline{\mathbb{Q}}_l)$$ Proof. This is just a formal manipulation based on lemma 9 and proposition 7.
Theorem 11 (Fourier inverse). We have
$$T_{\psi^{-1}}T_{\psi}K = K(-1)$$ where $(-1)$ denotes the Tate twist.
For this result, we need an auxiliary lemma, omitted proof, but can be understood heuristically as the Fourier transform of the canonical character equals the dirac delta function.
Lemma 12. Let $i: 0 \hookrightarrow \mathbb{A}^1$ be the canonical closed immersion, then $i_*\overline{\mathbb{Q}}_l = \delta$ is the skyscraper sheaf at the origin, we then have
$$T_{\psi}(\overline{\mathbb{Q}}_l[1]) = \delta(-1).$$ Remark. The occurence of Tate twist here is understandable because it corresponds to multiplying $1/2\pi$ in the formula
$$\delta(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{ixy}dy$$ Proof of Fourier inverse. Let's consider the diagram, in which the square is cartesian
and define $\alpha: \mathbb{A}^3 \longrightarrow \mathbb{A}^2$ by $(x,y,z) \longmapsto (x,y-z)$, then
$$\begin{align*} T_{\psi^{-1}}T_{\psi}K & = \pi^1_!\bigg(\pi^{2*}\big(\pi_!^1(\pi^{2*}K \otimes m^*\mathcal{L}_{\psi}\big) \otimes m^*\mathcal{L}_{\psi^{-1}} \bigg)[2] \\ & = \pi^1_! \bigg( \pi_!^{12}\pi^{23*}\big( \pi^{2*}K \otimes m^*\mathcal{L}_{\psi} \big) \otimes m^*\mathcal{L}_{\psi^{-1}} \bigg)[2] & \text{proper base change} \\ & =\pi^1_! \pi_!^{12}\big( \pi^{23*}\pi^{2*}K \otimes \pi^{23*}m^*\mathcal{L}_{\psi} \otimes \pi^{12*}m^*\mathcal{L}_{\psi^{-1}} \big)[2] & \text{projection formula} \\ & = \pi^1_! \pi_!^{12}(\pi^{23*}\pi^{2*}K \otimes \alpha^*m^*\mathcal{L}_{\psi})[2] & \\ & = \pi^1_! \pi_!^{13}(\pi^{13*}\pi^{2*}K \otimes \alpha^*m^*\mathcal{L}_{\psi})[2] & \pi^1 \pi^{12} = \pi^1\pi^{13} \ \text{and} \ \pi^2 \pi^{23} = \pi^2\pi^{13} \\ & = \pi^1_! (\pi^{2*}K \otimes \pi_!^{13}\alpha^*m^*\mathcal{L}_{\psi})[2] & \text{projection formula}\end{align*}$$ Consider the cartesian diagram
where $\beta(x,z)=z-x$, then by base change we get
$$\begin{align*} T_{\psi^{-1}}T_{\psi}K & = \pi^1_!(\pi^{2*}K \otimes \beta^*\pi^2_!m^*\mathcal{L}_{\psi})[2] & \\ & = \pi^1_!(\pi^{2*}K \otimes \beta^*T_{\psi}\overline{\mathbb{Q}}_l[-1])[2] & \\ & = \pi^1_!(\pi^{2*}K \otimes \beta^*i_*\overline{\mathbb{Q}}_l(-1)[-2])[2] & \text{previous lemma} \\ & = \pi^1_!(\pi^{2*}K \otimes \beta^*i_*\overline{\mathbb{Q}}_l)(-1) \end{align*}$$ But the square
is cartesian, where $\Delta$ is the diagonal, note that $i$ is proper so by base change again, we have
$$\begin{align*} T_{\psi^{-1}}T_{\psi}K & = \pi^1_!(\pi^{2*}K\otimes \Delta_!\overline{\mathbb{Q}}_l(-1)) & \\ & = \pi_!^!\Delta_!(\Delta^*\pi^{2*}K\otimes \overline{\mathbb{Q}}_l(-1)) & \text{projection formula} \\ & = K(-1) & \pi^1\Delta = \mathrm{id} \ \text{and} \ \pi^2 \Delta = \mathrm{id} \end{align*}$$ as desired.
We finish the proof by proving the following.
Lemma. The following holds
$$\pi^{12*}(m^*\mathcal{L}_{\psi^{-1}}) \otimes \pi^{23*}(m^*\mathcal{L}_{\psi}) = \alpha^*m^*\mathcal{L}_{\psi}.$$ Proof. Set $\mathbb{A}^1 = \mathrm{Spec}(k[t])$ and $\mathbb{A}^3 = \mathrm{Spec}(k[x,y,z])$ and considre
$$X
= \mathrm{Spec}(k[x,y,z,u,v]/(u^{\left|k \right|}-u-xy,v^{\left|k
\right|}-v-yz)) \longrightarrow \mathbb{A}^3$$ which is a Galois
covering whose group of Deck transformations is isomorphic to $k \times
k$. There are three projections of $X$ onto $\mathbb{A}^1$ given by
$$t \longmapsto u, t \longmapsto v-u, t \longmapsto v.$$ We view $\mathbb{A}^1$ as a scheme over itself by Artin-Schreier morphism, then the three morphisms above induce homomorphisms of groups of deck transformations
$$k \times k \longrightarrow k$$ given respectively by
$$(a,b) \longmapsto a, (a,b) \longmapsto b - a, (a,b) \longmapsto b.$$ We compose the original character $\psi$ with these three morphisms to get three new characters
$$\pi_1(X) \longrightarrow k \times k \longrightarrow k \overset{\psi}{\longrightarrow} \mathbb{C}^{\times}.$$ The three new characters correspond to the sheaves involved in the equation
$$\begin{align*} \pi^{12*}m^*\mathcal{L}_{\psi^{-1}} &\longrightarrow \psi_1(a,b) = \psi(a) \\ \alpha^*m^*\mathcal{L}_{\psi} & \longrightarrow \psi_2(a,b) = \psi(a)/\psi(b) \\ \pi^{23*}m^*\mathcal{L}_{\psi} & \longrightarrow \psi_3(a,b) = \psi(b) \end{align*}$$ then the question boils down to the trivial fact that $\psi_2\psi_3=\psi_1$.
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