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Let $X$ be a non-singular projective variety over $\mathbb{Q}$. Consider on the one hand $H^i_B(X(\mathbb{C}),\mathbb{Z}_\ell)$ the singular cohomology with value in $\mathbb{Z}_\ell$, and on the other hand $\varprojlim H^i_B(X(\mathbb{C}),\mathbb{Z}/\ell^n\mathbb{Z})$.

Are these two groups equal ? If so why.

Motivation : the comparison theorem between étale and singular cohomology states that $$H^i_{ét}(X,\mathbb{Z}/\ell^n\mathbb{Z}) \simeq H^i_B(X(\mathbb{C}),\mathbb{Z}/\ell^n\mathbb{Z}),$$ hence there is an isomorphism of the $\ell$-adic cohomology $$H^i(X,\mathbb{Z}_\ell):= \varprojlim H^i_{ét}(X,\mathbb{Z}/\ell^n\mathbb{Z}) \simeq\varprojlim H^i_B(X(\mathbb{C}),\mathbb{Z}/\ell^n\mathbb{Z}).$$ I was wondering why this implies that $H^i(X,\mathbb{Z}_\ell) \otimes_{\mathbb{Z}_\ell} \mathbb{C}$ is isomophic to $H_B^i(X(\mathbb{C}),\mathbb{C}) \cong H_B^i(X(\mathbb{C}),\mathbb{Z}) \otimes_\mathbb{Z} \mathbb{C} \cong H_{dR}^i(X(\mathbb{C}),\mathbb{C})$. I know that the $\ell$-adic cohomology is different from the étale cohomology of the constant sheaf $\mathbb{Z}_\ell$ (however I don't know why).

I will be glad if moreover someone can point out a reference about this (I have looked at Milne's notes and book and Lei Fu's book, but they don't talk about this).

Edit: I might rather ask this question instead.

Where can I find a proof of the isomorphisms $H^i(X_{\bar{\mathbb{Q}}},\mathbb{Z}_\ell) \otimes \mathbb{C} \simeq H^i_{dR}(X(\mathbb{C}),\mathbb{C})$.

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    $\begingroup$ Étale cohomology with coefficients in an infinite constant sheaf is sometimes not what you expect. For example, in $H^1$, under good conditions, for a constant abelian group $A$, $H^1_{\text{ét}} (X, A) \cong \mathrm{Hom}(\pi_1^{\text{ét}} (X), A)$, where the RHS is the group of continuous homomorphisms. Since $\pi_1^{\text{ét}} (X)$ is profinite, this will be trivial if every finite subgroup of $A$ is trivial. $\endgroup$ – Zhen Lin Jun 11 '13 at 17:27
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    $\begingroup$ Also, when people say "$\ell$-adic cohomology" they usually mean $H^i_{et}(X_{\overline{\mathbb{Q}}}, \mathbb{Z}/\ell^n\mathbb{Z})$ which is where the isomorphism works. This isn't the same thing as $H^i_{et}(X, \mathbb{Z}/\ell^n\mathbb{Z})$ which is a perfectly well-defined different thing. To see the importance of base-changing to a separable closure first, just consider $X=Spec(\mathbb{Q})$. $\endgroup$ – Matt Jun 11 '13 at 23:22
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    $\begingroup$ @ZhenLin: Dear Zhen, Regarding your comment, you may be interested in this discussion, especially James's answers. The basic point is that if $X$ has singularities, then the etale $H^1$ can be non-trivial, even with $\mathbb Z$-coefficients. Regards, $\endgroup$ – Matt E Nov 3 '13 at 11:14
  • $\begingroup$ @Matt: Dear Matt, As an aside: people do sometimes consider the $\ell$-adic cohomology without base-changing to a separable closure of the ground field. They sometimes call this the {\em absolute} $\ell$-adic cohomology. It is the $\ell$-adic analogue of Deligne cohomology, or, if you like, the $\ell$-adic avatar of the motivic cohomology of $X$. In number theory it comes in the study of Selmer groups associated to the (usual, i.e. over the algebraic closure of the base-field) $\ell$-adic cohomology groups of $X$. Regards, $\endgroup$ – Matt E Nov 3 '13 at 11:43
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$\def\ZZ{\mathbb{Z}}$I now understand the issue that user10676 is concerned about: For $X$ a reasonable topological space, why is $H^k(X, \ZZ_{\ell})$ isomorphic to $\lim_{\infty \leftarrow n} H^k(X, \ZZ/\ell^n)$? I thought that this would be most easily done from the universal coefficient theorem but, now that I try to write out a proof, I find it easiest to do directly.

Every complex algebraic variety is homeomorphic to a finite simplicial complex. (If I recall correctly, you can find a proof in "Algorithms in Real Algebraic Geometry".) So I'll use simplicial rather that singular cohomology. We work with a fixed triangulation for the rest of this proof.

Let $C^k$ be the group of $k$-co-chains with $\ZZ$ coefficients. Let $Z^k$ be the co-cycles and $B^k$ the co-boundaries. So $H^k(X, \ZZ) = Z^k/B^k$. Let the finitely generated abelian group $H^k(X, \ZZ)$ be $\bigoplus_{1 \leq i \leq r} \ZZ/d_i \oplus \ZZ^{\oplus s}$ where $d_1$, $d_2$, ..., $d_r$ is a sequence of positive integers where $d_1$ divides $d_2$ divides ... divides $d_r$. The following lemma is useful in many computations about cohomology:

Lemma Let $C^{k-1} \to C^{k} \to C^{k+1}$ be a complex of free finitely generated abelian groups. Then we can write this complex as a direct sum of complexes of the following forms: $$\ZZ \longrightarrow 0 \longrightarrow 0 \quad (1)$$ $$0 \longrightarrow \ZZ \longrightarrow 0 \quad (2)$$ $$0 \longrightarrow 0 \longrightarrow \ZZ \quad (3)$$ $$\ZZ \stackrel{d}{\longrightarrow} \ZZ \longrightarrow 0 \quad (4)$$ $$0 \longrightarrow \ZZ \stackrel{d}{\longrightarrow} \ZZ \quad (5)$$ where $d$ is a positive integer. (The case $d=1$ is permitted.)

Proof First, choose arbitrary bases for $C^{k-1}$ and $Z^k$. The map $C^{k-1} \to Z^k$ is given by an integer matrix. Now change bases so that this matrix is in Smith normal form. In these bases, $C^{k-1} \to Z^k \to 0$ is a sum of complexes of types (1), (2) and (4).

Now, $B^{k+1}$ is a subgroup of the free $\ZZ$-module $C^{k+1}$, so it is free. Thus, we can choose a splitting of $C^k \to C^k/Z^k \cong B^{k+1}$. Let $A$ be the image of this splitting, so $C^k \cong Z^k \oplus A$. Put the map $A \to C^{k+1}$ into Smith normal form as before. Then $0 \to A \to C^{k+1}$ is a direct sum of complexes of the form (3) and (5). (Since $A \to C^{k+1}$ is injective, there are no zeroes on the diagonal of the Smith normal form, and (2) does not occur.) Our original complex is the direct sum of $C^{k-1} \to Z^k \to 0$ and $0 \to A \to C^{k+1}$, so it breaks up as a direct sum of complexes of the five types listed above. $\square$

If we now change coefficients to a new abelian group $G$, we get the same complexes with $\ZZ$ replaced by $G$. All our computations distribute over direct sum, so we just need to work out what each of these five complexes do for the groups $G = \ZZ_{\ell}$ and $G = \ZZ/\ell^n$.

For complexes (1) and (3), we get $0$ in both cases.

For complex (2), we get $\lim_{\infty \leftarrow n} \ZZ/\ell^n$ on one side and $\ZZ_{\ell}$ on the other (with the obvious surjections in the inverse limit).

For complex (4), let $d = \ell^i m$ with $GCD(\ell, m) =1$. For $n \geq i$, we get $\lim_{\infty \leftarrow n} \ZZ/\ell^i$ on one side and $\ZZ/\ell^i$ on the other, where the maps in the inverse limit are the identity for $n > i$.

Complex (5) is the hard one. On the $\ZZ_{\ell}$ side, we get $0$. Write $d = \ell^i m$ as above. For $n \geq i$, we again get $\lim_{\infty \leftarrow n} \ZZ/\ell^i$. However, this time the map in the inverse limit is multiplication by $\ell$ (for $n>i$). This inverse limit is $0$, so we win.

Remark It is probably worth stating the general version of this result: If $R$ is a PID and $C^{\bullet}$ is a complex of free $R$-modules, then $C^{\bullet}$ can be written as a direct sum of complexes of the forms $\cdots \to 0 \to 0 \to R \to 0 \to 0 \to \cdots$ and $\cdots \to 0 \to 0 \to R \stackrel{d}{\longrightarrow} R \to 0 \to \cdots$ for various $d \in r$, and where the nonzero terms can occur in any position. The proof is basically the same: submodules of a free module over a PID are free; surjections to free modules split; Smith normal form is valid over a PID. A lot of modern ring theory can be thought of as classifying the types of complexes which exist over different rings.

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I think the answer to your question is "the universal coefficient theorem", but I might not understand your question correctly.

To remove ambiguity, every cohomology group below will have a subscript, either $et$ to mean etale cohomology, $sing$ for singular or $DR$ for deRham.

The universal coefficient theorem says that, for a topological space $Y$ and an abelian group $A$, there is a non-canonically split short exact sequence: $$0 \to Ext(H^{sing}_{k-1}(Y, \mathbb{Z}), A) \to H_{sing}^k(Y,A) \to Hom(H^{sing}_k(Y,\mathbb{Z}), A) \to 0.$$

From this sequence, one can compute $$\lim_{\infty \leftarrow n} H_{sing}^k(X(\mathbb{C}),\mathbb{Z}/(\ell^n)) \cong H_{sing}^k(X(\mathbb{C}),\mathbb{Z}_{\ell}) \cong Hom(H^{sing}_k(X(\mathbb{C}),\mathbb{Z}_{\ell})). \quad (\ast)$$ Even if $H_{k-1}^{sing}(X(\mathbb{C}), \mathbb{Z})$ contains some $\ell$-torsion, so that $Ext(H^{sing}_{k-1}(X(\mathbb{C}), \mathbb{Z}), \mathbb{Z}/\ell^n)$ is nonzero for all $n$, the inverse limit of the $Ext$ groups is still zero.

Now, you say you already understand that $H^k_{et}(X_{\overline{\mathbb{Q}}}, \mathbb{Z}/\ell^n) \cong H^k_{sing}(X(\mathbb{C}), \mathbb{Z}/\ell^n)$, so all the terms in $(\ast)$ are also isomorphic to $\lim_{\infty \leftarrow n} H^k_{et}(X_{\overline{\mathbb{Q}}}, \mathbb{Z}/\ell^n)$.

Also from universal coefficients, $$H^k_{sing}(X(\mathbb{C}), \mathbb{C}) \cong Hom(H^{sing}_k(X(\mathbb{C}),\mathbb{C})) $$

So, chaining together the isomorphisms we know, you want to show that $$Hom(H^{sing}_k(X(\mathbb{C}),\mathbb{Z}_{\ell})) \otimes_{\mathbb{Z}_{\ell}} \mathbb{C} \cong Hom(H^{sing}_k(X(\mathbb{C}),\mathbb{C}))$$

This has nothing to do with topology; it's just that $$Hom(G, R) \otimes_R \mathbb{C} \cong Hom(G, \mathbb{C})$$ for any subring $R$ of $\mathbb{C}$ and any finitely generated abelian group $G$.

Finally, you ask about bringing in de Rham cohomology. If you just mean de Rham cohomology of smooth differential forms, this is a question of differential geometry; you can read that $H^k_{sing}(Y, \mathbb{C}) \cong H^k_{DR}(Y, \mathbb{C})$ in Bott and Tu's book, for example. If you meant algebraic de Rham cohomology, you want Grothendieck, On the de Rham cohomology of algebraic varieties.

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    $\begingroup$ How do you prove the first isomoprhism in $(*)$ (it was my first question) ? Is it a consequence of the Univ-Coef-Thm or is it an exercise of commutative algebra ? There is a map $Z_B^k(X,Z_\ell) \rightarrow \varprojlim H_B^k(X,Z/\ell^n)$. An element $x$ in the kernel is a cocycle which is a coboundary modulo $\ell^n$ (i.e $x \mod \ell^n=\partial y_n$, $\forall n$), but I can't see why we can choose $(y_n)$ such that $y_{n+1} = y_n \mod \ell^n$. $\endgroup$ – user10676 Jun 14 '13 at 15:52
  • $\begingroup$ @user10676 Thanks for this comment! I now understand the issue, see the answer I've just added. $\endgroup$ – David E Speyer Jun 20 '13 at 15:34
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Following David's advice, here's a way to use the universal coefficient theorem (UCT) to prove the statement in slightly greater generality.

Claim: If $X$ is a topological space with finitely generated homology groups, then for all $n$, $$H^n(X,\mathbb{Z}_p) \cong \varprojlim_{m} H^n(X,\mathbb{Z}/p^m \mathbb{Z}).$$

Proof: Applying the UCT with coefficients in $\mathbb{Z}/p^m \mathbb{Z}$ yields a SES $$0 \to \text{Ext}_\mathbb{Z}^1(H_{n-1}(X),\mathbb{Z}/p^m \mathbb{Z}) \to H^n(X,\mathbb{Z}/p^m \mathbb{Z}) \to \text{Hom}_\mathbb{Z}(H_n(X),\mathbb{Z}/p^m \mathbb{Z}) \to 0. \quad (*)$$ Letting $m$ vary, we obtain vertical inverse systems over each of the terms in $(*)$. Since $\text{Ext}_\mathbb{Z}^2(A,B) = 0$ for all abelian groups $A, B$, the functor $\text{Ext}_\mathbb{Z}^1(H_{n-1}(X), \text{__})$ is right exact, so $$\text{Ext}_\mathbb{Z}^1(H_{n-1}(X),\mathbb{Z}/p^{m+1} \mathbb{Z}) \twoheadrightarrow \text{Ext}_\mathbb{Z}^1(H_{n-1}(X),\mathbb{Z}/p^m \mathbb{Z})$$ for all $m$. It follows that $\varprojlim^1 \text{Ext}_\mathbb{Z}^1(H_{n-1}(X),\mathbb{Z}/p^m \mathbb{Z}) = 0$, whence a SES by applying $\varprojlim$ to $(*)$: $$0 \to \varprojlim \text{Ext}_\mathbb{Z}^1(H_{n-1}(X),\mathbb{Z}/p^m \mathbb{Z}) \to \varprojlim H^n(X,\mathbb{Z}/p^m \mathbb{Z}) \to \varprojlim \text{Hom}_\mathbb{Z}(H_n(X),\mathbb{Z}/p^m \mathbb{Z}) \to 0.$$ Using the projections $\mathbb{Z}_p \to \mathbb{Z}/p^m \mathbb{Z}$ and the UCT for $\mathbb{Z}_p$, we now have a commutative diagram $$\require{AMScd} \begin{CD} \text{Ext}_\mathbb{Z}^1(H_{n-1}(X),\mathbb{Z}_p) @>>> H^n(X,\mathbb{Z}_p) @>>> \text{Hom}_\mathbb{Z}(H_n(X),\mathbb{Z}_p) \\ @VVV @VVV @VVV \\ \varprojlim \text{Ext}_\mathbb{Z}^1(H_{n-1}(X),\mathbb{Z}/p^m \mathbb{Z}) @>>> \varprojlim H^n(X,\mathbb{Z}/p^m \mathbb{Z}) @>>> \varprojlim \text{Hom}_\mathbb{Z}(H_n(X),\mathbb{Z}/p^m \mathbb{Z}). \\ \end{CD} $$ These are short exact sequences; I've ommitted the ending zeroes to save space. The third vertical arrow is an isomorphism because $\text{Hom}_\mathbb{Z}(H_n(X),\text{__})$ is left exact. Hence, by the five lemma, it's enough to prove the first vertical arrow is an isomorphism.

By the additivity of $\text{Ext}_\mathbb{Z}^1$ and the fundamental theorem of finitely generated abelian groups, it suffices to verify this isomorphism in the case $H_{n-1}(X)$ is cyclic (of finite or infinite order). Recall that $\text{Ext}_\mathbb{Z}^1(\mathbb{Z},B) = 0$ and $\text{Ext}_\mathbb{Z}^1(\mathbb{Z}/d \mathbb{Z},B) = B/dB$ for $d > 1$. So one gets zero in the infinite cyclic case, while $$\varprojlim \text{Ext}_\mathbb{Z}^1(\mathbb{Z}/d \mathbb{Z},\mathbb{Z}/p^m \mathbb{Z}) = \varprojlim (\mathbb{Z}/p^m \mathbb{Z})/d(\mathbb{Z}/p^m \mathbb{Z}) = \varprojlim \mathbb{Z}/\text{gcd}(p^m, d) \mathbb{Z} = \mathbb{Z}/p^{v_p(d)} \mathbb{Z};$$ here $p^{v_p(d)}$ is the highest power of $p$ dividing $d$. Since $\mathbb{Z}/p^{v_p(d)} \mathbb{Z} = \mathbb{Z}_p/d\mathbb{Z}_p = \text{Ext}_\mathbb{Z}^1(\mathbb{Z}/d \mathbb{Z},\mathbb{Z}_p)$, we're done.

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