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 $H^i(X_{\bar{\mathbb{Q}}},\mathbb{Z}_\ell) \otimes \mathbb{C} \simeq H^i_{dR}(X(\mathbb{C}),\mathbb{C})$.

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

$\def\ZZ{\mathbb{Z}}$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.


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.

  • 1
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.