Chủ đề này có 7 trả lời

[bangbang1412]

Đây là một bài mình viết sau khi đi nghe seminar do giáo sư Ngô Bảo Châu báo cáo hôm 17/8 tại viện Toán học với tựa đề Perverse sheaves and fundamental lemmas tuy nhiên giáo sư không có đủ thời gian để đi vào cả hai chủ đề mà bài nói xoay quanh việc đánh giá tổng Kloosterman bằng cách chuyển ngôn ngữ hàm số sang ngôn ngữ đối đồng điều và áp dụng giả thuyết Weil. Do đó mình để tựa đề như trên. Để thuận tiện, mình sẽ sử dụng tiếng anh.

 

Follow Katz's lectures on Weil II, let me spend some momemt to recall the motivating problem: given a prime $p$ and an integer $a$ s.t. $(a,p)=1$, the Kloosterman sum is defined as the complex number

$$\mathrm{Kl}(a,p) = \sum_{(x,y) \in \mathbb{F}_p: xy = a} \operatorname{exp}\left(\frac{2\pi i}{p}(x+y) \right).$$ By an elementary argument, one can see that this sum is a real number and in the early time when Kloosterman studied the Hardy-Littlewood circle method, he wanted to bound this sum by a function of $p$.

 

Some motivations

 

Định lý

(Kloosterman 1926) For any $\epsilon > 0$, we have $\left |\mathrm{Kl}(a,p) \right| < Cp^{3/4+\epsilon}$. 

Kloosterman's proof was quite elementary, however, the bound can be sharpen much more as follows.

Định lý

(Weil) We have $\left |\mathrm{Kl}(a,p) \right| \leq 2\sqrt{p}$.

This estimate is a consequence of Weil's proof of the "early Riemann hypothesis". The analytic version of Kloosterman sums is

$$\int_{-\infty}^{\infty}e^{i(ax+x^{-1})}dx$$ which is clearly not convergent, but we can approximate it by $\sqrt{a}K(\sqrt{a})$ where $K$ is the Bessel function. More generally, one can consider the Kloosterman sum

$$\mathrm{Kl}(a,p) = \sum_{xy=a \in \mathbb{F}_p} \psi(x+y) = \sum_{x \in \mathbb{F}_p}\psi(ax+x^{-1})$$ for any character $\psi:\mathbb{F}_p \longrightarrow \mathbb{C}^{\times}$, i.e. $\psi(x+y)=\psi(x)\psi(y)$. Here we can also prove that $\left| \mathrm{Kl}(a,p) \right| \leq 2 \sqrt{p}$ but even more, we can prove that

$$\mathrm{Kl}(a,p) = \alpha + \overline{\alpha}$$ where $\alpha$ is a complex number with $\left| \alpha \right| = \sqrt{p}$. This remains true if we replace $p$ by some of its power. This is where algebraic geometry enters the play. The first task is to transfer functions to sheaves. At the level of sheaves, we have more operations to manipulate (at least functions do not have something like duality).  But before one can see why we have to translate everything to cohomology language, one needs to have some clues about Grothendieck's formalism of six operations in $l$-adic cohomology.

 

l-adic cohomology

 

Let's fix a finite field $k=\mathbb{F}_{q}$ (where $q = p^n$ and $p$ prime) and $X/k$ be a variety. Given an integer $n$ invertible on $k$, then we can define that derived category $D^b_c(X,\mathbb{Z}/n)$ of chain complexes (modulo quasi-isomorphisms) of etale sheaves having cohomology sheaves are constructible. If $l \neq p$ is another prime, we define

$$D^b_c(X) = D^b_c(X, \overline{\mathbb{Q}}_l) = \left( \underset{\longleftarrow}{\lim} \ D^b_c(X,\mathbb{Z}/l^n\mathbb{Z}) \right) \otimes_{\mathbb{Z}_l} \overline{\mathbb{Q}}_l.$$ This definition is subtle and technical so one might follow Bhatt and Scholze's instruction to pretend that $D^b_c(X,\overline{\mathbb{Q}}_l)$ is some full subcategory of a derived category $D^b(X,\overline{\mathbb{Q}}_l)$. This is in fact does not cause any harm because almost every result for $D^b_c(X)$ is already true at the level $D^b_c(X,\mathbb{Z}/n\mathbb{Z})$. As far as I understand, the dissatisfaction with this limit-taking step is one of the reasons why Scholze introduced the pro-etale site.

Denote by $\mathfrak{TR}$ to be $2$-category of essentially small triangulated categories, then the family 

$$D^b_c: \mathrm{Var}/k \longrightarrow \mathfrak{TR} \  \ X \longmapsto D^b_c(X)$$ defines a $2$-functor admitting a formalism of six operations $(f^*,f_*,f_!,f^!,\otimes,\underline{\mathrm{Hom}})$, e.g. proper + smooth base change theorems, purity, Poincare duality,...

Objects of $D^b_c(X)$ are called $\mathbb{Q}_l$-sheaves or $l$-adic sheaves. The tensor product admits a unit denoted $\mathbb{Q}_{l,X}$ corresponding to the "constant" $l$-adic sheaf. For a $l$-adic sheaf $\mathcal{F}$, we define the $i$-th $l$-adic cohomology by setting

$$H^i(X \otimes_k \overline{k},\mathcal{F} \otimes_k \overline{k}) = \mathrm{Hom}_{D^b_c(X)}(\mathbb{Q}_{l,X},p_*\mathcal{F}[n]).$$ if $p: X \longrightarrow \mathrm{Spec}(k)$ is the structural morphism. Similarly, 

$$H^i_c(X \otimes_k \overline{k},\mathcal{F} \otimes_k \overline{k}) = \mathrm{Hom}_{D^b_c(X)}(\mathbb{Q}_{l,X},p_!\mathcal{F}[n]).$$ There is a subcategory of this category called smooth $l$-adic sheaves. Instead of treating (smooth) $l$-adic sheaf as complexes, we follow a shorter path:

Định lý

Let $X/k$ be an algebraic variety and $\overline{x} \longrightarrow X$ be a geometric point, then there is an equivalent of categories
$$\left \{\text{etale} \ \overline{\mathbb{Q}}_l-\text{sheaves} \right \} \overset{\sim}{\longrightarrow} \left \{\text{continuous rep. of} \ \pi_1(X,\overline{x}) \ \text{of} \ \overline{\mathbb{Q}}_l-\text{vector spaces} \right \}.$$ and moreover, smooth $l$-adic sheaves correspond to those representations which are of finite dimension. The equivalence is given by sending each etale $\overline{\mathbb{Q}}_l$ to its fiber over $\overline{x}$.

Frobenii

 

During the study of this subject, I found out that the definitions of the Frobenius morphism and their traces are ambiguous, precisely, there are several definitions of Frobenii, and the question is: which one is the right one that is used in our calculations and how are they related to others? I'll discuss few approaches to this definition, the explicit one and the abstract one. We still fix $k = \mathbb{F}_q$ and $k_n= \mathbb{F}_{q^n}$, the unique finite extension of degree $n$ of $k$.

 

Explicit definition

 

Although there are some different notions, they all arise from a single one, namely, the absolute Frobenius.

Định nghĩa

Let $A/k$ be an algebra, the Frobenius endomorphism $\mathrm{Frob}:A \longrightarrow A$ is simply the ring homomorphism $a \longmapsto a^p$. This construction is carried to schemes as it should be: if $X/k$ is a scheme, then the absolute Frobenius endomorphism $\mathrm{Frob}: X \longrightarrow X$ is a homeomorphism at the level of underlying topological spaces but on the structure sheaf is $f \longmapsto f^p$. Alternatively, it is defined locally by the Frobeninus endomorphism of affine pieces.

Caution. The Frobenius endomorphism is not an isomorphism in general. 

Bổ đề

The Frobeinus endomorphism $\mathrm{Frob}: X \longrightarrow X$ is finite of degree $q^{\dim(X)}$.

Proof. I strongly recommend you to prove this result with $X = \mathrm{Spec}(k[x_1,...,x_n])$ and move to the general case. Otherwise you can look at Milne's note.

Bổ đề

If $f: X \longrightarrow Y$ is a morphism of $k$-schemes, then $\mathrm{Frob}_Y \circ f = f  \circ \mathrm{Frob}_X$. In other words, the Frobenius construction is natural. 

Proof. Obvious.

 

Much much more stronger is the following.

Định lý

If $f: U \longrightarrow X$ is an etale morphism of $k$-varieties, then the diagarm \begin{xy}
\xymatrix {
 U \ar[r]^{\mathrm{Frob}} \ar[d]_{f} & U \ar[d]_f \\
                             X \ar[r]_{\mathrm{Frob}}  &  X
}
\end{xy}

is cartesian.

Proof. By the previous lemma, there exists a canonical morphism, which is called the relative Frobenius morphism $\mathrm{Frob}_{X/U}: U \longrightarrow X \times_X U$. Note that since $f$ is etale, its base change, the projection onto the first factor $pr_X: X \times_X U \longrightarrow X$ is also etale. But $pr_X \circ \mathrm{Frob}_{X/U} = f$ so that $\mathrm{Frob}_{X/U}$ is etale. The absolute Frobenii are universally bijective (as noted in the definition), this forces $\mathrm{Frob}_{X/U}$ to be universally bijective. A morphism which is universally bijective and etale must be an isomorphism due to StackProject.

We can consider others Frobenii

• The relative Frobenius $\mathrm{Frob}_r = \mathrm{Frob}_X \times \mathrm{id}_{\overline{k}}: X \otimes_k \overline{k} \longrightarrow X \otimes_k \overline{k}$. This one is a special case of the one in the proof above.
• The arithmetic Frobenius $\mathrm{Frob}_a = \mathrm{id}_X \times \mathrm{Frob}_{\overline{k}}:X \otimes_k \overline{k} \longrightarrow X \otimes_k \overline{k}$.
• The geometric Frobenius $\mathrm{Frob}_g = \mathrm{id}_X \times \mathrm{Frob}_{\overline{k}}^{-1}:X \otimes_k \overline{k} \longrightarrow X \otimes_k \overline{k}$.

The relative and arithmetic are automorphisms while the geometric and the absolute are not.

Bổ đề

Given a variety $X/k$, then we have $X(k_r)  = \overline{X}^{\mathrm{Frob}_r^n}$ where the relative Frobenius acts on $\overline{X}$ on the first factor. In other words, the set of $k_n$-points of $X$ is the set of closed points of $\overline{X}$ which is fixed under the $r$-iteration of the Frobenius.

Proof. Check on affine pieces.  

 

The next point is to formulate the Grothendieck trace formula, which (I think people may not drop this point at the first reading) is our main tool of computation. We have to find a natural way to define an endormophism, denoted $\mathrm{Frob}^*$

$$\mathrm{Frob}^*: H^i_c(X \otimes_k \overline{k}, \mathcal{F} \otimes_k \overline{k}) \longrightarrow H^i_c(X \otimes_k \overline{k}, \mathcal{F} \otimes_k \overline{k})$$ for every $l$-adic sheaf $\mathcal{F}$ and its pullback $\mathcal{F} \otimes_k \overline{k}$ to $X \otimes_k \overline{k}$.

 

Think topologically and remember how people thought about sheaves in the beginning days. Well, sheaves are actually sheaves of sections of etale spaces (by this, I really mean we have some equivalence of categories), the same thing happens here: for every $l$-adic sheaf $\mathcal{F}$ on $X$, there exists an algebraic space (which plays the role of an etale space in the topological world) $[\mathcal{F}]$ together with an etale morphism $f: [\mathcal{F}] \longrightarrow X$ such that $\mathcal{F}$ becomes the sheaf of sections of this morphism. As a consequence, we may identify $\mathcal{F}$ with $[\mathcal{F}]$. By base change, we obtain an etale morphism $f \otimes_k \overline{k}: [\mathcal{F}] \otimes_k \overline{k} \longrightarrow X \otimes_k \overline{k}$ and in a similar to the theorem above, the diagram

\begin{xy}
\xymatrix {
\overline{\mathcal{F}} = \mathcal{F} \otimes_k \overline{k} \ar[r]^{\mathrm{Frob}} \ar[d]_{f} & \overline{\mathcal{F}} \ar[d]_f \\
                             X \ar[r]_{\mathrm{Frob}}  &  X
}
\end{xy}

is cartesian. That being said, $\overline{\mathcal{F}} \simeq  \mathrm{Frob}^*\overline{\mathcal{F}}$ where by $\mathrm{Frob}^*$ I really mean pullback of a sheaf. This isomorphism yields two important facts:

• The composition $$\mathrm{Frob}^*: H_c^i(X \otimes_k \overline{k}, \overline{\mathcal{F}}) \longrightarrow H_c^i(X \otimes_k \overline{k},\mathrm{Frob}^*\overline{\mathcal{F}}) \simeq H_c^i(X \otimes_k \overline{k}, \overline{\mathcal{F}})$$ is the one that we are seeking, where the first morphism is the natural morphism. 
• If $x \in X \otimes_k \overline{k}$ is fixed by the $n$-iteration of the absolute Frobenius, then taking stalks induces an isomorphism $\mathrm{Frob}_x^{*n}: \mathcal{F}_x \overset{\sim}{\longrightarrow} \mathcal{F}_x$.

Định lý

(Grothendieck-Lefschetz trace formula). With these data, we have

$$\sum_{x \in X(k_n)}\mathrm{Trace}(\mathrm{Frob}_x^{*n},\mathcal{F}_x) = \sum_i (-1)^i\mathrm{Trace}(\mathrm{Frob}^{*n},H^i_c(X \otimes_k \overline{k},\overline{\mathcal{F}})).$$ In particular, 

$$\sum_{x \in X(k)}\mathrm{Trace}(\mathrm{Frob}_x^{*},\mathcal{F}_x) = \sum_i (-1)^i\mathrm{Trace}(\mathrm{Frob}^{*},H^i_c(X \otimes_k \overline{k},\overline{\mathcal{F}})).$$

If we set

$$\mathrm{Trace}_{\mathcal{F}}(x) =  \mathrm{Trace}(\mathrm{Frob}_x^{*},\mathcal{F}_x)$$ for each $x \in X(k)$, then this constitues a function

$$\mathrm{Trace}: X(k) \longrightarrow \overline{\mathbb{Q}}_l = \mathbb{C}$$ with the following properties

• For any $x \in X(k)$ and $\mathcal{F},\mathcal{G} \in D^b_c(X)$ $$\mathrm{Trace}_{\mathcal{F}}(x)\mathrm{Trace}_{\mathcal{G}}(x) = \mathrm{Trace}_{\mathcal{F} \otimes \mathcal{G}}(x).$$
• For any morphism of $k$-varieties $f: X \longrightarrow Y$ $$\mathrm{Trace}_{f^*\mathcal{F}}(x)  = \mathrm{Trace}_{\mathcal{F}}(f(x)).$$
• For any $y \in Y(k)$ then $$\sum_{x \in X_y(k)} \mathrm{Trace}_{\mathcal{F}}(x) = \mathrm{Trace}_{f_!\mathcal{F}}(y).$$

Katz's point of view

 

Given a connected variety $X/k$ and for any point $x: k_r \longrightarrow X$, we get an induced group homomorphism

$$x_*: \pi_1(k_r,\overline{k}) \longrightarrow \pi_1(X,\overline{k})$$ by the functoriality of the etale fundamental group functor. Since $\pi_1(k_r)$ contains the Frobenius automorphism $\mathrm{Frob}_{k_r}: \overline{k} \longrightarrow \overline{k}, a \mapsto a^{\left| k_r \right|}$, we can consider its image via $x_*$ and set

$$x_*(\mathrm{Frob}_{k_r}) = \mathrm{Frob}_{k_r,x}.$$ Now given a smooth $l$-adic sheaf, i.e. a finitely dimensional representation 

$$\mathcal{F}: \pi_1(X) \longrightarrow \mathrm{GL}(r,\overline{\mathbb{Q}}_l),$$ and a $k$-point $x: k \longrightarrow X$ then it makes sense to consider the trace of the automorphism $\mathrm{Trace}(\mathcal{F}(\mathrm{Frob}_{k,x}))$ which is nothing but $\mathrm{Trace}_{\mathcal{F}}(x)$ considered before. However, I do not have any reference for this.

 

Artin-Schreier theory

 

Now with the formalism of $l$-adic cohomology in hands, we are ready to translate functions to cohomology. We introduce things called Artin-Schreier sheaf on $\mathbb{A}^1$. Here again, $k = \mathbb{F}_q, q = p^m$.

 

The Artin-Schreier sheaf is the morphism 

$$\begin{align*} L: \mathbb{A}^1_k &  \longrightarrow \mathbb{A}^1_k \\ t & \longmapsto t - t^q  \end{align*}$$ (here $t$ denotes the canonical coordinate on $\mathbb{A}^1$) is an etale covering whose whose Galois group is $\mathbb{F}_q$, i.e. $\mathrm{Aut}_{\mathbb{A}^1}(\mathbb{A}^1) = k$ and generated by $x \longmapsto x+1$.  Note that the fundamental group $\pi_1(\mathbb{A}^1)$ contains $\mathrm{Aut}_{\mathbb{A}^1}(\mathbb{A}^1)$ as an element of the projective system, so there is a canonical projection 

$$\pi_1(\mathbb{A}_k^1) \longrightarrow k.$$ Given any additive 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 $\psi$. The important fact is that

Bổ đề

$\mathrm{Trace}_{\mathcal{L}_{\psi}}(x)  = \psi(x)$ for any $x \in k = \mathbb{A}^1_k(k)$.

Proof. Since $\mathrm{Trace}_{\mathcal{L}_{\psi}}(x) = \mathrm{Trace}(\psi(\mathcal{L}_{\psi}(\mathrm{Frob}_{k,x})))$, we need to know what is $\mathcal{L}_{\psi}(\mathrm{Frob}_{k,x})$; in other words, where the Frobenius goes. We are done if we can prove that $ \mathcal{L}_{\psi}(\mathrm{Frob}_{k,x})=x$. To be continued.

 

Now we come to the main point, namely, the cohomological expression of Kloosterman sums. For any value $a$, we consider the hyperbol

$$X_a = \left \{(x,y) \in \mathbb{A}^2_k \mid xy = a \right \}$$

and consider the morphism $h_a: X_a \longrightarrow \mathbb{A}^1, (x,y) \mapsto x+y$. By theorem 5 and lemma 4, we have

$$\mathrm{Kl}(a,\psi) = \sum_{i=0}^2 (-1)^i \mathrm{Trace}(\mathrm{Frob}^*, H^i_c(X_a \otimes_k \overline{k}, h_a^*\mathcal{L}_{\psi} \otimes_k \overline{k})).$$ Note that $X_a$ is non-compact curve, so $H^0(X_a) = 0$ and by Poincare duality $H^2(X_a)=0$, therefore 

$$\mathrm{Kl}(a,\psi) = - \mathrm{Trace}(\mathrm{Frob}^*, H^1_c(X_a \otimes_k \overline{k}, h_a^*\mathcal{L}_{\psi} \otimes_k \overline{k})).$$ Note that, $$\dim \ H^1_c(X_a \otimes_k \overline{k}, h_a^*\mathcal{L}_{\psi} \otimes_k \overline{k}) = -\chi_c(X_a, h^*\mathcal{L}_{\psi})$$ the Euler characteristic with compact support. We'd like to compute this dimension first. Thanks to the Grothendieck-Ogg-Shafarevich theorem, we can compute this characteristic as follows.

Định lý

(Grothendieck-Ogg-Shafarevich). Let $\overline{X}$ be a proper smooth curve over $k$ and $X$ an open subset of $\overline{X}$ and $\mathcal{F}$ a local system on $X$. Then

$$\chi_c(X \otimes_k \overline{k},\mathcal{F}) = \chi_c(X \otimes_k \overline{k})\mathrm{rank}(\mathcal{F})  - \sum_{x \in \overline{X}\setminus X} \mathrm{Sw}_x(\mathcal{F})$$ where $\mathrm{Sw}$ are Swan conductors.

The Swan conductors are hard to be defined but in practice, one just needs to know its formal properties:

• $\mathrm{Sw}_x(\mathcal{F})$ depends only on its restriction to the punctured formal disc $\hat{X}_x^{\bullet}$. 
• $\mathrm{Sw}_x(\mathcal{F})=0$ when the restriction of $\mathcal{F}$ to $\hat{X}_x^{\bullet}$ is tame.
• If $\mathcal{G}$ is a tame local system at $\hat{X}_x^{\times}$, then $\mathrm{Sw}_x(\mathcal{F} \otimes \mathcal{G}) = \mathrm{Sw}_x(\mathcal{F})\mathrm{rank}(\mathcal{G}).$

Here are some computations.

Ví dụ

If $X = \mathbb{A}^1$ and $\mathcal{F} = \mathcal{L}_{\psi}$, then by an elementary argument, we see that

$$\mathrm{Trace}(\mathrm{Frob}^*,H_c^1(\mathbb{A}^1,\mathcal{L}_{\psi})) = \sum_{x \in k} \psi(x) = 0$$ and hence $\chi_c(\mathbb{A}^1,\mathcal{L}_{\psi})=0$. By Grothendieck-Ogg-Shafarevich formula, we see that

$$0 = \chi_c(\mathbb{A}^1)\mathrm{rank}(\mathcal{L}_{\psi}) - \mathrm{Sw}_{\infty}(\mathcal{L}_{\psi})$$ and from this we deduce that $\mathrm{Sw}_{\infty}(\mathcal{L}_{\psi}) = 1$.

Ví dụ

For each $a \neq 0$, we see that $X_a \simeq \mathbb{G}_m$ so by Grothendieck-Ogg-Shafarevich formula,

$$\chi_c(X_a,h_a^*\mathcal{L}_{\psi}) = \chi_c(X_a) - \mathrm{Sw}_0(h_a^*\mathcal{L}_{\psi}) - \mathrm{Sw}_{\infty}(h_a^*\mathcal{L}_{\psi}) = - \mathrm{Sw}_0(h_a^*\mathcal{L}_{\psi}) - \mathrm{Sw}_{\infty}(h_a^*\mathcal{L}_{\psi}).$$ By the properties of Swan conductors

$$\begin{align*} \mathrm{Sw}_0(h_a^*\mathcal{L}_{\psi}) & = \mathrm{Sw}_0(x^*\mathcal{L}_{\psi} \otimes y^*\mathcal{L}_{\psi}) \\ & = \mathrm{Sw}_0(x^*\mathcal{L}_{\psi})\mathrm{rank}(y^*\mathcal{L}_{\psi})  = 1 \end{align*}$$ since $y^*\mathcal{L}_{\psi}$ is even unramified (not just tame) and by the previous example. By symmetry, $\mathrm{Sw}_{\infty}(h_a^*\mathcal{L}_{\psi}) = 1$ and finally this all implies that $\chi_c(X_a,h_a^*\mathcal{L}_{\psi})=2$.

 

Weight theory of Deligne

 

We fix once for all an identification $\iota: \overline{\mathbb{Q}}_l  \overset{\sim}{\longrightarrow} \mathbb{C}$ so that we can speak of an absolute on $\overline{\mathbb{Q}}_l$. Given a smooth $\mathbb{Q}_l$-sheaf $\mathcal{F}$ on an algebraic variety $X/k$, $k_n/k$ a finite extension of $k$.

$$\mathcal{F}: \pi_1(X) \longrightarrow \mathrm{GL}(r,\mathbb{C})$$ and a point $x \in X(k_n)$, then we say that

• $\mathcal{F}$ is pure of weight $w$ if for each $n$, every eigenvalue of $\mathrm{Frob}^{*n}_x$ has eigenvalues with absolute values $\left|k \right|^{w/2}$.
• $\mathcal{F}$ is mixed of weight $\geq w$ if if for each $n$, every eigenvalue of $\mathrm{Frob}^{*n}_x$ has eigenvalues with absolute values $\geq \left|k \right|^{w/2}$.
• $\mathcal{F}$ is mixed of weight $\leq w$ if if for each $n$, every eigenvalue of $\mathrm{Frob}^{*n}_x$ has eigenvalues with absolute values $\leq \left|k \right|^{w/2}$.

We call the celebrated theorem due to Deligne, originally known as Weil conjectures.

Định lý

(Deligne) Let $X/k$ be a variety and $\mathcal{F}$ is a $l$-adic sheaf mixed of weight $\leq 0$, then every eigenvalue of 

$$\mathrm{Frob}^*:H_c^i(X \otimes_k \overline{k},  \mathcal{F} \otimes_k \overline{k}) \longrightarrow H_c^i(X \otimes_k \overline{k},  \mathcal{F} \otimes_k \overline{k})$$ has absolute values $\leq \left |k \right|^{i/2}$. 

In Weil II, Deligne proved something much stronger where one replaces $U \longrightarrow \mathrm{Spec}(k)$ by a morphism $f: X \longrightarrow Y$, then $R^if_!\mathcal{F}$ is mixed of weight $\leq w +i$  whenever $\mathcal{F}$ is mixed of weight $\leq w$. However, the Target theorem is enough to deduce the last part of the Weil conjectures and estimates of Kloosterman sums. 

 

From Deligne's weight theorems, the computation $\dim \ H^1_c(X_a \otimes_k \overline{k} ,h_a^*\mathcal{L}_{\psi} \otimes_k \overline{k}) = 2$, and 

$$\mathrm{Kl}(a,\psi) = - \mathrm{Trace}(\mathrm{Frob}^*, H^1_c(X_a \otimes_k \overline{k}, h_a^*\mathcal{L}_{\psi} \otimes_k \overline{k})).$$ we see that

$$\left | \mathrm{Kl}(a,\psi) \right| \leq 2p^{1/2}.$$

More generally, if we define a generalized Kloosterman sum as 

$$\mathrm{Kl}_m(a,\psi) = \sum_{x_1\cdots x_m = a, x_i \in k}\psi(x_1 + \cdots + x_m)$$ then we have an estimate $\left |\mathrm{Kl}_m(a,\psi) \right | \leq mp^{(m-1)/2}$.  

Bài viết đã được chỉnh sửa nội dung bởi bangbang1412: 22-08-2023 - 00:00

[bangbang1412]

In this post, I'd like to give a rapid introduction to the theory of $l$-adic Fourier transform developed by Laumon-Deligne-... My goal is not to how can we apply $l$-adic Fourier transform to prove the Weil conjectures but rather to see why their definitions are natural in comparison with the classical theory. My feeling is that it is easier to present Fourier transforms on finite fields than on measurable spaces (which require a lot of work and details) and the $l$-adic one is formally adapted from the one for finite fields.

 

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 $S^1$.

 

Ví dụ

If $G = \mathbb{F}_{p}$, a field with $p$ elements, then $\psi(x) = e^{2\pi i x/p}$ is a character.

Ví dụ

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 $S^1$.

Ví dụ

If $\psi,\varphi$ are characters, then $\psi\varphi$ is a character as well.

Mệnh đề

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 $\psi(y) \neq 1$.

Hệ quả

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.

 

Mệnh đề

(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.

 

Mệnh đề

(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

 

Bổ đề

$f^{\mathcal{L}_{\psi}}(x) = \psi(-x)$. 

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

\begin{xy}

\xymatrix{

& \mathbb{A}^1 \times \mathbb{A}^1 \ar[rr]^m \ar[dr]^{\pi^2} \ar[dl]_{\pi^1} & & \mathbb{A}^1 \\

 \mathbb{A}^1 & & \mathbb{A}^1 &

}

\end{xy}

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).

 

Bổ đề

$f^{T_{\psi}(K)}(x) = - \sum_{y \in k} f^K(y)\psi(-xy)$ 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). 

 

 

Định lý

(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 6 and proposition 3.

 

Định lý

(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.

 

Bổ đề

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

$$T_{\psi}(\overline{\mathbb{Q}}_l[1]) = \delta(-1).$$

Nhận xét

. 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

 

\begin{xy}
\xymatrix {
&  & \mathbb{A}^1 \times \mathbb{A}^1 \times \mathbb{A}^1 \ar[dr]^{\pi^{23}} \ar[dl]_{\pi^{13}} &  & \\
                       & \mathbb{A}^1 \times \mathbb{A}^1  \ar[dl]_{\pi^1} \ar[dr]^{\pi^2} & & \mathbb{A}^1 \times \mathbb{A}^1 \ar[dl]_{\pi^1} \ar[dr]^{\pi^2}& \\

\mathbb{A}^1 & & \mathbb{A}^1 & & \mathbb{A}^1
}
\end{xy}

 

and define $\alpha: \mathbb{A}^3 \longrightarrow \mathbb{A}^2$ by $(x,y,z) \longmapsto (x,y-z)$, then

$$\begin{align*} T_{\psi^{-1}}T_{\psi}K & = \pi^1_!\bigg(\pi^{2*}\big(\pi_!^1(\pi^{2*}K \otimes m^*\mathcal{L}_{\psi}\big) \otimes m^*\mathcal{L}_{\psi^{-1}} \bigg)[2] \\ & = \pi^1_! \bigg( \pi_!^{12}\pi^{23*}\big( \pi^{2*}K \otimes m^*\mathcal{L}_{\psi} \big) \otimes m^*\mathcal{L}_{\psi^{-1}} \bigg)[2] &  \text{proper base change} \\ & =\pi^1_! \pi_!^{12}\big( \pi^{23*}\pi^{2*}K \otimes \pi^{23*}m^*\mathcal{L}_{\psi} \otimes \pi^{12*}m^*\mathcal{L}_{\psi^{-1}} \big)[2] &  \text{projection formula} \\ & =  \pi^1_! \pi_!^{12}(\pi^{23*}\pi^{2*}K \otimes \alpha^*m^*\mathcal{L}_{\psi})[2] & \text{by the last lemma below} \\ & = \pi^1_! \pi_!^{13}(\pi^{13*}\pi^{2*}K \otimes \alpha^*m^*\mathcal{L}_{\psi})[2] &  \pi^1 \pi^{12} = \pi^1\pi^{13} \ \text{and} \ \pi^2 \pi^{23} = \pi^2\pi^{13} \\ & = \pi^1_! (\pi^{2*}K \otimes \pi_!^{13}\alpha^*m^*\mathcal{L}_{\psi})[2] & \text{projection formula}\end{align*}$$  Consider the cartesian diagram

\begin{xy}
\xymatrix {
\mathbb{A}^3 \ar[r] \ar[d]_{\pi^{13}} \ar[r]^{\alpha} & \mathbb{A}^2 \ar[d]_{\pi^2} \\
                             \mathbb{A}^2 \ar[r]_{\beta}  &  \mathbb{A}^1
}
\end{xy}

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

 

\begin{xy}
\xymatrix {
\mathbb{A}^1 \ar[r] \ar[d]_{\Delta} & 0 \ar[d]_i \\
                             \mathbb{A}^2 \ar[r]_{\beta}  &  \mathbb{A}^1
}
\end{xy}

 

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.

 

Bổ đề

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

$$\begin{align*} \pi^{12*}m^*\mathcal{L}_{\psi^{-1}} &\longrightarrow   \psi_1(a,b) = 1/\psi(a) \\ \alpha^*m^*\mathcal{L}_{\psi} & \longrightarrow \psi_2(a,b) = \psi(b)/\psi(a) \\ \pi^{23*}m^*\mathcal{L}_{\psi} & \longrightarrow \psi_3(a,b) = \psi(b) \end{align*}$$ then the question boils down to the trivial fact that $\psi_1\psi_3=\psi_2$.

Bài viết đã được chỉnh sửa nội dung bởi bangbang1412: 22-08-2023 - 18:50

[Zaraki]

Objects of $D^b_c(X)$ are called $\mathbb{Q}_l$-sheaves or $l$-adic sheaves. The tensor product admits a unit denoted $\mathbb{Q}_{l,X}$ corresponding to the "constant" $l$-adic sheaf. For a $l$-adic sheaf $\mathcal{F}$, we define the $i$-th $l$-adic cohomology by setting

$$H^i(X \otimes_k \overline{k},\mathcal{F} \otimes_k \overline{k}) = \mathrm{Hom}_{D^b_c(X)}(\mathbb{Q}_{l,X},p_*\mathcal{F}[n]).$$ if $p: X \longrightarrow \mathrm{Spec}(k)$ is the structural morphism. Similarly, 

$$H^i_c(X \otimes_k \overline{k},\mathcal{F} \otimes_k \overline{k}) = \mathrm{Hom}_{D^b_c(X)}(\mathbb{Q}_{l,X},p_!\mathcal{F}[n]).$$ There is a subcategory of this category called smooth $l$-adic sheaves. Instead of treating (smooth) $l$-adic sheaf as complexes, we follow a shorter path:

Định lý

Let $X/k$ be an algebraic variety and $\overline{x} \longrightarrow X$ be a geometric point, then there is an equivalent of categories
$$\left \{\text{etale} \ \overline{\mathbb{Q}}_l-\text{sheaves} \right \} \overset{\sim}{\longrightarrow} \left \{\text{continuous rep. of} \ \pi_1(X,\overline{x}) \ \text{of} \ \overline{\mathbb{Q}}_l-\text{vector spaces} \right \}.$$ and moreover, smooth $l$-adic sheaves correspond to those representations which are of finite dimension. The equivalence is given by sending each etale $\overline{\mathbb{Q}}_l$ to its fiber over $\overline{x}$.

 

Bằng có biết làm thế nào để định nghĩa smooth $\ell$-adic sheaves as complexes không? Có cảm giác cái này sẽ tương đương với smooth functions $X(k)\to \overline{\mathbb{Q}_{\ell}}$, nhưng như thế nào mới được coi là smooth từ $X(k)$?

 

Như vậy là cái tên "etale-$\mathbb{Q}_{\ell}$-sheaves" tương đương với local system/locally constant sheaves?

 

The next point is to formulate the Grothendieck trace formula, which (I think people may not drop this point at the first reading) is our main tool of computation. We have to find a natural way to define an endormophism, denoted $\mathrm{Frob}^*$

$$\mathrm{Frob}^*: H^i_c(X \otimes_k \overline{k}, \mathcal{F} \otimes_k \overline{k}) \longrightarrow H^i_c(X \otimes_k \overline{k}, \mathcal{F} \otimes_k \overline{k})$$ for every $l$-adic sheaf $\mathcal{F}$ and its pullback $\mathcal{F} \otimes_k \overline{k}$ to $X \otimes_k \overline{k}$.

 

Think topologically and remember how people thought about sheaves in the beginning days. Well, sheaves are actually sheaves of sections of etale spaces (by this, I really mean we have some equivalence of categories), the same thing happens here: for every $l$-adic sheaf $\mathcal{F}$ on $X$, there exists an algebraic space (which plays the role of an etale space in the topological world) $[\mathcal{F}]$ together with an etale morphism $f: [\mathcal{F}] \longrightarrow X$ such that $\mathcal{F}$ becomes the sheaf of sections of this morphism. As a consequence, we may identify $\mathcal{F}$ with $[\mathcal{F}]$. By base change, we obtain an etale morphism $f \otimes_k \overline{k}: [\mathcal{F}] \otimes_k \overline{k} \longrightarrow X \otimes_k \overline{k}$ and in a similar to the theorem above, the diagram

\begin{xy}
\xymatrix {
\overline{\mathcal{F}} = \mathcal{F} \otimes_k \overline{k} \ar[r]^{\mathrm{Frob}} \ar[d]_{f} & \overline{\mathcal{F}} \ar[d]_f \\
                             X \ar[r]_{\mathrm{Frob}}  &  X
}
\end{xy}

is cartesian. That being said, $\overline{\mathcal{F}} \simeq  \mathrm{Frob}^*\overline{\mathcal{F}}$ where by $\mathrm{Frob}^*$ I really mean pullback of a sheaf. This isomorphism yields two important facts:

• The composition $$\mathrm{Frob}^*: H_c^i(X \otimes_k \overline{k}, \overline{\mathcal{F}}) \longrightarrow H_c^i(X \otimes_k \overline{k},\mathrm{Frob}^*\overline{\mathcal{F}}) \simeq H_c^i(X \otimes_k \overline{k}, \overline{\mathcal{F}})$$ is the one that we are seeking, where the first morphism is the natural morphism. 
• If $x \in X \otimes_k \overline{k}$ is fixed by the $n$-iteration of the absolute Frobenius, then taking stalks induces an isomorphism $\mathrm{Frob}_x^{*n}: \mathcal{F}_x \overset{\sim}{\longrightarrow} \mathcal{F}_x$.

If we set

$$\mathrm{Trace}_{\mathcal{F}}(x) =  \mathrm{Trace}(\mathrm{Frob}_x^{*},\mathcal{F}_x)$$ for each $x \in X(k)$, then this constitues a function

$$\mathrm{Trace}: X(k) \longrightarrow \overline{\mathbb{Q}}_l = \mathbb{C}$$

 

Với cái định nghĩa trace như thế này thì ở chỗ nào ta dùng điều kiện constructible của $\mathcal{F}$ nhỉ? Vì tớ tưởng ta chỉ có thể dùng function-sheaf dictionary cho complexes of constructible sheaves. 

Bài viết đã được chỉnh sửa nội dung bởi Zaraki: 23-08-2023 - 19:32

[bangbang1412]

Bằng có biết làm thế nào để định nghĩa smooth $\ell$-adic sheaves as complexes không? Có cảm giác cái này sẽ tương đương với smooth functions $X(k)\to \overline{\mathbb{Q}_{\ell}}$, nhưng như thế nào mới được coi là smooth từ $X(k)$?

 

Như vậy là cái tên "etale-$\mathbb{Q}_{\ell}$-sheaves" tương đương với local system/locally constant sheaves?

Hi Toàn, cái chữ smooth này thực ra rất nguy hiểm. Nó không phải smooth theo nghĩa function $X(k) \longrightarrow \overline{\mathbb{Q}}_l$, mà tớ cũng không rõ lý do thật sự họ dùng chữ smooth.

 

Về định nghĩa, ta phải quay lại định nghĩa bó $l$-adic trên một lược đồ noether $X$, được định nghĩa là một họ $(F_n)$ sao cho mỗi bó $F_n$ là bó các $\mathbb{Z}/l^n$-module sao cho $F_{n+1} \otimes_{\mathbb{Z}/l^{n+1}} \mathbb{Z}/l^{n} \simeq F_n$. Gọi một bó $l$-adic là constructible nếu mỗi bó $F_n$ đều là constructible và gọi là smooth nếu nó locally constant.

 

Còn nếu cậu muốn xem nó như các complex thì thực chất cậu đang muốn định nghĩa derived category $D^b_c(X,\mathbb{Z}_l)$. Nó được định nghĩa là "giới hạn" (theo nghĩa nào đó)

$$D^b_c(X,\mathbb{Z}_l) = \underset{\longleftarrow}{\lim} D^b_{ctf}(X,\mathbb{Z}/l^n)$$

của các derived categories theo nghĩa thông thường, ở đây $D^b_{ctf}(X,\mathbb{Z}/l^n)$ là derived cat của các phức có bó đối đồng điều là constructible và ta yêu cầu nó đẳng cấu với một $\mathbb{Z}/l^n$-flat complex. Nếu muốn chuyển qua $\mathbb{Q}_l$-sheaves $D^b_c(X,\mathbb{Q}_l)$ thì ta chỉ "tensor hình thức" mọi thứ với $\mathbb{Q}_l$. Cụ thể hơn, định nghĩa projective limit ở trên phải hiểu theo nghĩa sau (tớ chỉ nói về objects)

 

Objects: Một object của $D^b_c(X,\mathbb{Z}_l)$ sẽ là một projective system $(K_n^{\bullet})$ sao cho mỗi $K_n^{\bullet} \in D^b_{ctf}(X,\mathbb{Z}/l^n)$ sao cho có các quasi-isomorphism $K^{\bullet}_{n+1} \otimes_{\mathbb{Z}/l^{n+1}} \mathbb{Z}/l^n \simeq K^{\bullet}_n$ bên trong $D^b_c(X,\mathbb{Z}/l^n)$.

 

Mọi thứ phức tạp hơn một chút nữa nếu muốn $\overline{\mathbb{Q}}_l$, nhớ rằng bao đóng đại số thì là giới hạn của các mở rộng hữu hạn $E/\mathbb{Q}_l$. Nên một cách tự nhiên ta sẽ định nghĩa

$$D^b_c(X,\overline{\mathbb{Q}}_l) = \underset{\longrightarrow}{\lim} D^b_{c}(X,E)$$

trong đó $D^b_c(X,E)$ được định nghĩa giống hệt với $D^b_c(X,\mathbb{Q}_l)$ (nhớ rằng $E$ cũng là local field) nên thay vì $l$-adic ta sẽ nói tới $\pi$-adic với $\pi$ là uniformizer nào đó.

 

Cuối cùng giả sử $(F_n)$ là một bó $l$-adic (technical condition: torision free in some Artin-Rees category) thì $(F_n)$ sẽ định nghĩa một vật trong $D^b(X,\mathbb{Z}_l)$ bằng cách xem các $F_n$ là các complex concentrated at degree zero. Như vậy chuyển qua giới hạn thì mỗi $l$-adic sheaf cho ta một $E$-sheaf ($E/\mathbb{Q}_l$: finite extension) và một $\overline{\mathbb{Q}}_l$-sheaf.

Bài viết đã được chỉnh sửa nội dung bởi bangbang1412: 23-08-2023 - 21:47

[bangbang1412]

Với cái định nghĩa trace như thế này thì ở chỗ nào ta dùng điều kiện constructible của $\mathcal{F}$ nhỉ? Vì tớ tưởng ta chỉ có thể dùng function-sheaf dictionary cho complexes of constructible sheaves. 

Chỗ này theo tớ không thật sự dùng constructible, constructible là về mặt cohomology.

 

Edit: tớ hiểu ý cậu rồi, tớ đoán là cậu đang hiểu function-sheaf dictionary cho constructible sheaves rồi extend cho complex, từ sheaves lên complexes of sheaves thì mình dùng tổng đan dấu của các cohomology (giống kiểu Euler-characteristic).

Bài viết đã được chỉnh sửa nội dung bởi bangbang1412: 23-08-2023 - 20:10

[Zaraki]

Còn nếu cậu muốn xem nó như các complex thì thực chất cậu đang muốn định nghĩa derived category $D^b_c(X,\mathbb{Z}_l)$. Nó được định nghĩa là "giới hạn" (theo nghĩa nào đó)

$$D^b_c(X,\mathbb{Z}_l) = \underset{\longleftarrow}{\lim} D^b_{ctf}(X,\mathbb{Z}/l^n)$$

của các derived categories theo nghĩa thông thường, ở đây $D^b_{ctf}(X,\mathbb{Z}/l^n)$ là derived cat của các phức có bó đối đồng điều là constructible và ta yêu cầu nó đẳng cấu với một $\mathbb{Z}/l^n$-flat complex. 

Cảm ơn Bằng nhiều. Định nghĩa này nhìn có vẻ khá phức tạp ... Bằng giải thích nôm na vì sao ta cần đẳng cấu với $\mathbb{Z}/\ell^n$-flat complex không?

 

Chỗ này theo tớ không thật sự dùng constructible, constructible là về mặt cohomology.

 

Edit: tớ hiểu ý cậu rồi, tớ đoán là cậu đang hiểu function-sheaf dictionary cho constructible sheaves rồi extend cho complex, từ sheaves lên complexes of sheaves thì mình dùng tổng đan dấu của các cohomology (giống kiểu Euler-characteristic).

 

Thật ra ý tớ cậu trả lời trước khi edit rồi, tức là ta có để xây dựng function $\text{Trace}_{\mathcal{F}}: X(k)\to \overline{\mathbb{Q}_{\ell}}$ cho mọi $\mathcal{F}\in D^b(X,\overline{\mathbb{Q}_{\ell}})$, không nhất thiết phải $D^b_c(X,\overline{\mathbb{Q}_{\ell}})$? 

 

Thật ra tớ có đọc được một cách khác để định nghĩa $\text{Trace}_{\mathcal{F}}$ nhưng phải cần điều kiện $\mathcal{F}$ là constructible complex, ví dụ trong trang 3 của https://math.uchicag...u/~ngo/PCMI.pdf: Với một điểm $x: \text{Spec }x\to X$, thì $\overline{x}^*\mathcal{F}$ cũng là $\ell$-adic, cụ thể là constructible, tức constructible $H^i(\overline{x}^*\mathcal{F})$ tương ứng với continuous representation of $\text{Gal}(\overline{k}/k) \to GL_n(\overline{\mathbb{Q}_{\ell}})$. Khi đó ta có thể định nghĩa trace của Frobenius của $H^i(\overline{x}^*\mathcal{F})$. 

 

Chắc là hai định nghĩa này giống nhau? 

Bài viết đã được chỉnh sửa nội dung bởi Zaraki: 24-08-2023 - 03:50

[bangbang1412]

Cảm ơn Bằng nhiều. Định nghĩa này nhìn có vẻ khá phức tạp ... Bằng giải thích nôm na vì sao ta cần đẳng cấu với $\mathbb{Z}/\ell^n$-flat complex không?

 

 

Thật ra ý tớ cậu trả lời trước khi edit rồi, tức là ta có để xây dựng function $\text{Trace}_{\mathcal{F}}: X(k)\to \overline{\mathbb{Q}_{\ell}}$ cho mọi $\mathcal{F}\in D^b(X,\overline{\mathbb{Q}_{\ell}})$, không nhất thiết phải $D^b_c(X,\overline{\mathbb{Q}_{\ell}})$? 

 

Thật ra tớ có đọc được một cách khác để định nghĩa $\text{Trace}_{\mathcal{F}}$ nhưng phải cần điều kiện $\mathcal{F}$ là constructible complex, ví dụ trong trang 3 của https://math.uchicag...u/~ngo/PCMI.pdf: Với một điểm $x: \text{Spec }x\to X$, thì $\overline{x}^*\mathcal{F}$ cũng là $\ell$-adic, cụ thể là constructible, tức constructible $H^i(\overline{x}^*\mathcal{F})$ tương ứng với continuous representation of $\text{Gal}(\overline{k}/k) \to GL_n(\overline{\mathbb{Q}_{\ell}})$. Khi đó ta có thể định nghĩa trace của Frobenius của $H^i(\overline{x}^*\mathcal{F})$. 

 

Chắc là hai định nghĩa này giống nhau? 

Về câu hỏi thứ hai thì tớ trả lời là trong note của giáo sư Châu ban đầu ta làm với constructible sheaves (theo nghĩa thông thường) rồi extend lên complexes bằng cách xét tổng đan dấu như tớ nói. Cậu phải cần $\mathcal{F} \in D^b_c$ vì nếu constructible thì nó chính là một complex with constructible cohomology.

 

Quay lại câu hỏi thứ nhất, thực chất ta có thể viết

$$D^b_c(X,\mathbb{Z}_l) = \underset{\longleftarrow}{\lim} D^b_{c}(X,\mathbb{Z}/l^n)$$

nhưng cái này không thật sự đúng về mặt technical một cách hình thức như sau: giả sử cậu có một họ $(D_n)_{n \geq 0}$ các triangulated category và các hàm tử khớp $T_{n+1}: D_{n+1} \longrightarrow D_n$ thì cậu sẽ định nghĩa giới hạn

$$\underset{\longleftarrow}{\lim} D_n$$ như thế nào? Một dự đoán đầu tiên là ta xét phạm trù gồm các vật $(A_n,\phi_n)_{n \geq 0}$ trong đó $\phi_{n+1}:F_{n+1}(A_{n+1}) \simeq A_n$. Morphism được định nghĩa một cách tự nhiên (compatible with $\phi_n)$. Một distinguished triangle 

 

$$X \longrightarrow Y \longrightarrow Z \overset{+1}{\longrightarrow}$$ thì chỉ là một họ các

$$X_n \longrightarrow Y_n \longrightarrow Z_n \overset{+1}{\longrightarrow}$$ compatible với những gì có thể.

 

Lemma. The limit $\underset{\longleftarrow}{\lim} D_n$ is a triangulated provided that all $\mathrm{Hom}_{D_n}(K,L)$ ($K, L\in D_n$) are finite.

 

Proof. For instance, let us verify TR3 in the axioms defining a triangulated category. Given two triplets $(K,L,M) \longrightarrow (K',L',M')$ and we'd like to find some $Z \longrightarrow Z'$ making everything commutative.

 

Define $E_n$ to be the set of morphisms $M_n \longrightarrow M_n'$ such that there exist morphisms of distinguished triangles

$$(K_n,L_n,M_n) \longrightarrow (K_n',L_n',M_n')$$ (note that we fix $K_n \longrightarrow K_n',L_n \longrightarrow L_n'$). Each set $E_n$ is nonempty and finite by the TR3 axiom for each $D_n$ and the assumption on hom sets. Moreover, $(E_n)$ defines a projective system whose limit

$$E = \underset{\longleftarrow}{\lim} E_n$$

is nonempty. Any choice of a morphism in $E$ is a morphism that we are seeking.

 

Giờ ta quay lại với $D^b_c(X,\mathbb{Z}_l)$, ta cần xem nó như giới hạn $D^b_c(X,\mathbb{Z}_l) = \underset{\longleftarrow}{\lim} D^b_{c}(X,\mathbb{Z}/l^n)$ trong đó các transition là

$$\begin{align*} D^b_c(X,\mathbb{Z}/l^{n+1}) & \longrightarrow D^b_c(X,\mathbb{Z}/l^n) \\ K^{\bullet} & \longmapsto K^{\bullet} \otimes_{\mathbb{Z}/l^{n+1}} \mathbb{Z}/l^n \end{align*}$$ thì theo bổ đề trên ta cần điều kiện

$$\mathrm{Hom}_{D^b_c(X,\mathbb{Z}/l^n)}(K,L)$$ là finite. Tuy nhiên điều kiện này không đúng với $D^b_c$ mà đúng với $D^b_{ctf}$. Nói cách khác ta có

 

Proposition (SGA 4 1/2). Under some nice assumption on the base field,

$$\mathrm{Hom}_{D^b_{ctf}(X,\mathbb{Z}/l^n)}(K,L)$$

are finite.

Bài viết đã được chỉnh sửa nội dung bởi bangbang1412: 24-08-2023 - 06:50

[bangbang1412]

Here I'd like to talk more about $D^b_{ctf}(X)$ (coefficient in some $\mathbb{Z}/n$) and this should partially explain why Grothendieck pointed out to Illusie as "the right notion". Every object in $D^b_{ctf}(X)$ is in fact quasi-isomorphic to a bounded flat complex whose components are constructible! (note that my definition in the previous answer did not include the constructibility). Any such (bounded, flat, constructible components) is called a perfect complex.

 

In some sense, the flatness is equivalent to projectiveness, and hence when we deal with derived categories of modules, we shoud replace everywhere flatness with projectiveness. Let $R$ be a ring, a complex in $D(R)$ is called perfect if it is quasi-isomorphic to a bounded complex of finite projective module. We denote by $D_{perf}(R)$ the triangulated subcategory of $D(R)$ formed by perfect complexes. This definition is slightly different with the ones in $D^b_{ctf}(X)$ but in $D^b(R)$ they are almost equivalent, namely, a complex in $D^b(R)$ is necessarily perfect provided that its cohomology are perfect (cohomology are perfect $\simeq$ cohomology are constructible) (at this point, I haven't checked why we can remove projectiveness, this is possibly due to the boudedness that we have imposed). But more importantly,

 

Proposition. For a ring, the set of compact objects of $D(R)$ are precisely perfect complexes.

 

Remind that a compact object $K$ in a category with small direct sums is a object such that $\operatorname{Hom}(K,-)$ commutes with small direct sums.

 

Proposition. The smallest strictly full triangulated subcategory stable under direct factors of $D(R)$ generated by a single object $R$ (regarded as a complex concentrated at degree $0$) is precisely the full subcategory of $D(R)$ consisting of perfect complexes.

 

Remind that a subcategory $\mathcal{C}$ of a category $\mathcal{A}$ with finite direct sums is stable under direct factors if whenever $X \oplus Y \in \mathcal{C}$ then both $X,Y \in \mathcal{C}$. These two results are subtle, I have to say that. Let me formulate in a more formal way: suppose that $\mathcal{T}$ is a triangulated having all direct sums and $\Lambda \subset \mathcal{T}$ is a set (not a proper class) of objects

• There exists a smallest triangulated subcategory $ \left < \left< \Lambda \right>\right >$ containing $\Lambda$ and stable under direct sums. If all objects of $\Lambda$ are compact and $ \left < \left< \Lambda \right>\right > = \mathcal{T}$ then we say that $\mathcal{T}$ is compactly generated. 
• There exists a smallest triangulated subcategory $ \left< \Lambda \right>^{ct}$ containing $\Lambda$ and stable under direct factors. If $\mathcal{T}$ is compactly generated by $\Lambda$ then (by a consequence of abstract Brown representability theorem) we have $\left< \Lambda \right>^{ct}$ is exactly the triangulated category of compact objects of $\mathcal{T}$.

Moreover, the fact that $\mathcal{T}$ is compactly generated by $\Lambda$ is equivalent to

$$\operatorname{Hom}(A[n],B) = 0 \ \forall \ A \in \Lambda \Rightarrow B = 0.$$ In terms of the formulation above, we can write $D(R) = \left < \left< R \right>\right >$ and $D_{perf}(R) = \left<R \right>^{ct}$. Their proofs can be ignored at first but you should think in comparison with the topological world. You replace $D(R)$ with $\mathbf{SH}$, the stable homotopy category, whose objects are sequences $X=(X_n)$ of simplicial sets together with morphisms $S^1 \wedge X_n \longrightarrow X_{n+1}$ so that you can define stable homotopy groups $\pi_n^{st}(X)$ and say that morphism is a stable weak equivalence if it induces isomorphisms on stable homotopy groups. Then $\mathbf{SH}$ is obtained by inverting all stable weak equivalence just like you invert all quasi-isomorphisms for complexes, it is a triangulated category whose distinguished triangles are those isomorphic to a cone sequence. There is a very special spectrum called the sphere spectrum $\mathbb{S} = (S^n)$, with transition $S^1 \wedge S^n \overset{\sim}{\longrightarrow} S^{n+1}$ and stable homotopy groups are stable homotpy groups of spheres. You can show that

$$\mathbf{SH} = \left < \left< \mathbb{S} \right>\right >.$$

Hence you can view every spectrum as a module over the sphere spectrum (and this is indeed the right way in the sense that: the category $\mathbb{SH}$ is not à priori a tensor category, to get a tensor structure you have to work with symmetric spectra, i.e. spectra with action of symmetric groups, and prove that two stable categories are equivalent and symmetra have $\otimes$ and $\underline{\operatorname{Hom}}$ moviated from the way of thinking every spectrum is a module over $\mathbb{S}$) and the triangulated subcategory formed by compact objects is $\left< \mathbb{S} \right>^{ct}$. The condition

$$\operatorname{Hom}(\mathbb{S}[n],B) = 0 \Rightarrow B = 0.$$ is in some sense equivalent to saying that for a CW-complex or simplicial set $X$, if $\pi_n(X)=0$ then $X=\bullet$ (Whitehead's theorem). Here comes to another subtle point; why these two worlds are so similar? The answer lies in the Dold-Kan correspondence theorem, basically it says that the category of chain complexes of $\mathbb{Z}$-modules is equivalent to the category of simplicial abelian groups. This equivalence maps homotopy groups to homology groups so that you can say homology groups are homotopy groups. The homotopy groups of $R[n]$ behaves like the homology of $S^n$ (concentrated at degree $0$ and $n$).

 

In summary,

• Both $D(R)$ and $\mathbf{SH}$ are stable model categories, and hence triangulated ones. Moreover, they are all tensor triangulated category.
• Both $D(R)$ and $\mathbf{SH}$ are compactly generated with a single generator.
• Another example that I can provide is that derived category of $\mathcal{O}_X$-module, the single generator is $\mathcal{O}_X$ itself.

It is likely to define

$$D^b(X) = \left < \left< \text{a single generator} \right> \right>,$$

where we have to specify the generator. A naive guess is the constant sheaf on $X$, but this isn't enough, as you'd like to have Poincaré duality, you have to add Tate twist into the play. Therefore, the definition should be: $D^b_{ctf}(X)$ as the smalles full triangulated subcategory stable under direct factors of $D^b(X)$ generated by objects of the form

$$f_!(\mathbb{Z}/n)(-d)[-2d]$$ where $f: Y \longrightarrow X$ is smooth of relative dimension $d$, i.e. 

$$D^b(X) = \left < \left< f_!(\mathbb{Z}/n)(-d)[-2d] \right> \right> \ \ \ \ \text{and} \ \ \ \ D^b_{ctf}(X) =  \left< f_!(\mathbb{Z}/n)(-d)[-2d] \right>^{ct}$$

Actually, this should be true though I couldn't find a reference but by this tag on StackProject, a "weaker" result holds, where we change triangulated categories to abelian categories, namely, the category of constructible sheaves of abelian groups is the smallest full subcategory of the category of sheaves of abelian groups contaning objects of the form $j_!\mathbb{Z}/n$ (with $j: U \longrightarrow X$ being étale, aka smooth of relative dimension $0$ $\Rightarrow$ no need to consider Tate twist) and closed under finite limits and colimits.

 

But this is the way people nowadays define constructible motives. For instance, in motivic homotopy theory where we offen work with the stable homotopy category $\mathbf{SH}(X)$ of Voevodsky (has small direct sums like $D^b(X)$) in which such a nice representation like a perfect complex is not available then this way of definining constuctible seems to be an appropriate way (and much easier to work with).

Bài viết đã được chỉnh sửa nội dung bởi bangbang1412: 24-08-2023 - 09:47