# Concrete example of calculation of $\ell$-adic cohomology

Let $p$ and $\ell$ be distinct prime numbers.

Consider in the affine plane $\mathbb{A}^2_{\mathbb{F}_p}$ with coordinates $(x,y)$ the union $L$ of the axes $x = 0$ and $y = 0$.

How does one compute the $\ell$-adic cohomology groups with compact support $H^i_c(L,\mathbb{Q}_\ell)$? I thought I had some idea of what $\ell$-adic cohomology is, but I don't even manage to do this example...

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Have you thought about the example of just $L$ as a single axis? Then it is smooth and you have the tools of Poincare duality etc (the compactification map is quite explicit in this example). I don't remember how these standard theorems behave for singular varieties. –  Matt Jul 16 '12 at 16:26
Yes. More generally, Poincaré duality gives us that $H^i_c(\mathbb{A}^n_{\mathbb{F}_p},\mathbb{Q}_\ell) = 0$ for $i \neq 2n$, and $H^{2n}_c(\mathbb{A}^n_{\mathbb{F}_p},\mathbb{Q}_\ell)$ is one-dimensional. But I don't know what to do with the "simplest" non-singular case, which is exactly the one I described in my post. –  Evariste Jul 16 '12 at 21:05
This may be wrong as I know very little about $\ell$-adic cohomology. Writing $A$ for the coefficient ring and $j:\mathbb{G}_m^2 \hookrightarrow \mathbb{A}^2$ and $i: L\hookrightarrow \mathbb{A}^2$. Then there should be an exact sequence, $0\to j_!A \to A \to i_*A \to 0$. Applying $H^*_c(\mathbb{A}^2,-)$ we get a long exact sequence. Use that $H^*_c(\mathbb{A}^2,j_!A) = H^*_c(\mathbb{G}_m^2,A)$ and $H^*_c(\mathbb{A}^2,i_*A) = H^*_c(L,A)$ since $i$ is a closed immersion. The same reasoning should be valid for $\ell$-adic cohomology starting with $A = \mathbb{Z}/\ell^n$. –  AFK Jul 16 '12 at 23:45
If I'm not mistaken, $\ell$-adic cohomology with compact support is not a derived functor. So I have some doubts about the fact that you'd be able to get the long exact sequence you want... –  Evariste Jul 17 '12 at 0:26
I think I understand your objection: $H^*_c$ may not be a derived functor but that's not what we're interested in. What we need is that it is a $\delta$-functor meaning that it sends short exact sequences to long exact sequences. The fact is that this is much better understood in the framework of derived categories. I'll try to write a complete answer tomorrow. –  AFK Jul 17 '12 at 1:18

(1) All we really need to know to make the computation is that cohomology with compact support $H^*_c$ (with values in $K$-vector spaces) satisfies the following properties:

• (Localization sequence) If $i: Z \hookrightarrow X$ is a closed immersion and $j:U\hookrightarrow X$ the complementary open immersion, then for any sheaf $F$ we have a long exact sequence $$\to H^{i}_c(U) \to H^{i}_c(X) \to H^{i}_c(Z) \to H^{i+1}_c(U) \to$$
• (Cohomology of the affine space) $H^{i}_c(\mathbb{A}^n) = \begin{cases} K & if~ i=2n \cr 0 & else \end{cases}$

(2) Now we can play with the long exact sequences

Writing the localization sequence for $\mathbb{A}^1 = \{0\} \coprod \mathbb{G}_m$ one finds $H^{i}_c(\mathbb{G}_m) = \begin{cases} K & if~ i=1,2 \cr 0 & else \end{cases}$ This is dual to $H^{2-i}(\mathbb{G}_m)$ as expected.

Then the localization sequence for $L = \mathbb{A}^1 \coprod \mathbb{G}_m$ reduces to
$$0\to H^{0}_c(L) \to 0 \to K \to H^{1}_c(L) \to 0 \to K \to H^{2}_c(L) \to K \to 0$$ so $H^{i}_c(L) = \begin{cases} K & if~ i=1 \cr K^2 & if~ i=2 \cr 0 & else \end{cases}$

(3) I should add a few word about the localizing sequence. For any immersion $j: U\hookrightarrow X$, we have 3 functors on sheaves

• the restriction functor $j^*$
• its right adjoint: the classical direct image $j_*$
• the extension by zero $j_!$.

Facts (see J.S. Milne's lecture notes for example):

• $j_!$ is always exact
• If $j$ is a closed immersion then $j_! = j_*$.
• If $i: Z \hookrightarrow X$ is a closed immersion and $j:U\hookrightarrow X$ the complementary open immersion, then for any sheaf $F$ we have an exact sequence $$0 \to j_!j^*F \to F \to i_*i^*F \to 0$$

Now if the variety $X$ is not proper, you can always find a dense open immersion $u:X\hookrightarrow \overline{X}$ into a proper variety. Since $u_!$ is exact, that $u_!i_* = u_!i_! = (ui)_!$ and applying $H^*(\overline{X},-)$ we obtain a long exact sequence $$\to H^{i}_c(U,j^*F) \to H^{i}_c(X,F) \to H^{i}_c(Z,i^*F) \to H^{i+1}_c(U,j^*F) \to$$

For the computation $H^*_c(\mathbb{A}^n)$ it reduces to that of $H^*_c(\mathbb{P}^n)$ by the localizing sequence. But since $\mathbb{P}^n$ is proper, we have $H^{*}_c(\mathbb{P}^n) = H^*(\mathbb{P}^n) = K[h]/(h^{n+1})$ where $h$ is the class of any hyperplane and has degree 2. Moreover, the morphism $\mathbb{P}^n \hookrightarrow \mathbb{P}^{n+1}$ induces the natural projection.

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