Let $X$ be an algebraic variety (say, smooth and projective) over $\mathbb{C}$, and fix $$\alpha\in H^i(X^{\text{an}}, \mathbb{Q})$$ with $i>0$.

Does there always exist a variety $Y$ and a smooth proper morphism $f: Y\to X$ such that $f^*\alpha$ is integral, i.e., it is in the image of $H^i(Y, \mathbb{Z})\to H^i(Y, \mathbb{Q})$?


  • If $i=1$, $\alpha$ gives a map $\pi_1(X)\to \mathbb{Q}\to \mathbb{Q}/\mathbb{Z}$. The kernel of this map defines a (finite) covering $Y$ of $X$. Pulling back to $Y$ makes $\alpha$ integral.

  • If $i=2$, there is a map $\mathbb{Q}\to \mathcal{O}_X\to\mathcal{O}_X^*$. Let $\gamma$ be the image of $\alpha$ under the induced map $H^2(X, \mathbb{Q})\to H^2(X, \mathcal{O}_X^*)$; this is an analytic Brauer class on $X$. I believe a result of Serre(?) tells us that it is also an etale-cohomological Brauer class. As $X$ is projective, Gabber tells us that $\gamma$ is in fact an honest Brauer class, so we may choose a Severi-Brauer variety $f: Y\to X$ with class $\gamma$. Then $f^*(\gamma)=0$. If $f^*\alpha$ is non-zero, it lives in $H^{1,1}(Y)\cap H^2(Y, \mathbb{Q})$. Let $n$ be such that $n\cdot f^*\alpha$ is integral; then by the Lefschetz $(1,1)$-theorem, there is a line bundle $\mathcal{L}$ on $Y$ with first Chern class $n\cdot f^*\alpha$. Let $\mathcal{Y}_\mathcal{L}\to Y$ be the $\mu_n$-gerbe of $n$-th roots of $\mathcal{L}$. This isn't quite a smooth projective variety over $X$, admittedly, but the pullback of $\alpha$ to $\mathcal{Y}_\mathcal{L}$ is integral. In any case, I'm fine with $Y$ being a DM stack. (There is always some construction that works involving higher stacks in the analytic category, but this is not what I'm looking for.) I guess I'd prefer $Y$ to be an actual variety, so this argument reduces the $i=2$ case to the following special case:

Let $X$ be as above, and let $\mathcal{L}$ be a line bundle on $X$. Fix a positive integer $n$. Is there always some smooth proper $f: Y\to X$ so that $f^*\mathcal{L}$ has an $n$-th root?

  • If $i>2$, I have no idea what to do.

Some more remarks:

  • For varieties so that the cup product $H^1(X, \mathbb{Q})^{\otimes n}\to H^n(X, \mathbb{Q})$ is surjective (say, Abelian varities or curves of genus at least $1$), this is certainly possible. Namely, write $\alpha$ as a linear combination of things coming from $H^1$, and then make all of those integral.

  • Qiaochu addresses the topological version. This is already difficult, I think, but it is slightly easier than the algebraic one. For example, I don't know how to make classes in $H^2(\mathbb{P}^1, \mathbb{Q})$ integral, but the Hopf map $S^3\to S^2$ makes every class in $H^2(S^2, \mathbb{Q})$ integral, since $H^2(S^3)=0$.

  • To do the case $i=2$, it suffices to consider the case when $X=\mathbb{P}^n$. To see this, observe that my argument above (under Examples, $i=2$) reduces to the case where $\alpha\in H^{1,1}(X)\cap H^2(X, \mathbb{Q})$. Then use that ample classes generate $H^{1,1}(X)\cap H^2(X, \mathbb{Q})$.

  • Jason Starr suggests that every smooth projective morphism $Y\to \mathbb{P}^1$ might have a section, which would give a negative answer for $X=\mathbb{P}^1, i=2$. Is that true? I certainly don't know a counterexample...

  • $\begingroup$ could it be that you need higher stacks in the algebraic category as well? (the real reason I'm commenting is: what is the construction in the analytic category you are referring to?) $\endgroup$ Oct 22, 2014 at 2:42
  • $\begingroup$ @user125763: There's a "higher analytic stack" associated to any torsion cohomology class; one of these will always do the trick (I guess it works in the algebraic category as well, indeed). But I'd really prefer to have something as concrete as possible. $\endgroup$ Oct 22, 2014 at 2:45
  • 1
    $\begingroup$ Wait, what? If $X$ is $\mathbb{P}^1$ and if $\alpha$ is $1/2$ the Poincare dual of the homology class of a point, explain to me again how you are getting a Brauer class on $\mathbb{P}^1$? $\endgroup$ Oct 22, 2014 at 9:26
  • 2
    $\begingroup$ The statement you need for $i=2$ looks close to the existence of Bloch-Gieseker coverings (Thm 4.1.10 in Lazarsfeld's "Positivity"), but I guess there is some issue with ramification. Maybe the references there say something relevant? $\endgroup$
    – user47305
    Oct 22, 2014 at 19:40
  • 1
    $\begingroup$ @Mark: Awesome! This at least lets one do $H^2$ if one only requires $f$ to be flat and projective (which seems like it might be more doable). $\endgroup$ Oct 22, 2014 at 20:27

2 Answers 2


This is not an answer but some comments on the topological aspects of the problem. Probably I'm just telling you stuff you already know.

Since $H_i(X, \mathbb{Z})$ is finitely generated, any element of $H^i(X, \mathbb{Q}) \cong \text{Hom}(H_i(X, \mathbb{Z}), \mathbb{Q})$ can be lifted to an element of $\text{Hom}(H_i(X, \mathbb{Z}), \frac{1}{m} \mathbb{Z})$ for some $m$, and hence to an element of $H^i(X, \frac{1}{m} \mathbb{Z})$. So in the question we can replace $\mathbb{Q}$ by $\frac{1}{m} \mathbb{Z}$. Equivalently we can say the following. The short exact sequence

$$0 \to \mathbb{Z} \xrightarrow{m} \mathbb{Z} \to \mathbb{Z}_m \to 0$$

(permit me the sin of using $\mathbb{Z}_m$ for $\mathbb{Z}/m\mathbb{Z}$) induces a long exact sequence

$$\cdots \to H^i(X, \mathbb{Z}) \xrightarrow{m} H^i(X, \mathbb{Z}) \to H^i(X, \mathbb{Z}_m) \to \cdots$$

showing that a given $\alpha \in H^i(X, \mathbb{Z})$ is an $m^{th}$ root, or equivalently lies in the image of the multiplication-by-$m$ map above, iff it lies in the kernel of the reduction-mod-$m$ map to $H^i(X, \mathbb{Z}_m)$. So our problem reduces, more or less to the following question:

Given nice $X$ and $\alpha \in H^i(X, \mathbb{Z}_m)$, when can we find a nice $f : Y \to X$ such that $f^{\ast}(\alpha) = 0$?

(I'm being vague here so that the statement can apply both to algebraic and topological versions of the problem; I have in mind $f$ a fiber bundle and $X, Y$ compact manifolds. Strictly speaking, for this question to be equivalent to the original question $\alpha$ needs to lie in the kernel of the Bockstein but I can't imagine that condition is particularly important.)

For any $i$ there is a universal topological answer to this question: namely, thinking of $\alpha$ as a map to an Eilenberg-MacLane space $X \to B^i \mathbb{Z}_m$, we can take its homotopy fiber, which lies in a fiber sequence

$$B^{i-1} \mathbb{Z}_m \to \widetilde{X} \to X \xrightarrow{\alpha} B^i \mathbb{Z}_m.$$

$\widetilde{X}$ is, by definition, the universal space equipped with a map to $X$ such that the pullback of $\alpha$ is trivial, and can be thought of as a fiber bundle over $X$ with fiber $B^{i-1} \mathbb{Z}_m$. Unfortunately $B^{i-1} \mathbb{Z}_m$ is not really a manifold in general. Presumably a higher stacky version of this construction exists and is what you were alluding to in the comments.

When $i = 1$ the homotopy fiber $\widetilde{X}$ can be taken to be the finite cover you've mentioned already; here $B^0 \mathbb{Z}_m = Z_m$ is discrete. When $i = 2$ we can think of $B \mathbb{Z}_m$ as the infinite lens space $S^{\infty} / \mathbb{Z}_m$, and we might hope that we can replace the fiber sequence above with a fibration

$$S^n / \mathbb{Z}_m \to \widetilde{X} \to X$$

with fiber a finite lens space. That might solve the problem in the topological setting but I guess not in the algebraic setting.

When $i = 2$ and $m = 2$ this is the question of whether every class in $H^2(X, \mathbb{Z}_2)$ can be represented by a bundle of real projective spaces / real matrix algebras. This is true and is the real analogue of Serre's theorem that every torsion class in $H^3(X, \mathbb{Z})$ is representable by a bundle of complex projective spaces / complex matrix algebras; it is apparently due to E. Strickland.

For higher $i$ it's not clear to me that even the topological version of the problem is solvable; I guess a suitable negative answer to that problem would imply a negative answer to the original problem?

Edit: When $i = 2$ we can do a bit more than the question asks for, at least in the topological setting: instead of causing any element of $H^2(X, \mathbb{Z})$ to have an $m^{th}$ root we can in fact cause any of $H^2(X, \mathbb{Z})$ to vanish. If $\alpha$ is such an element, thought of as a map $X \to B^2 \mathbb{Z}$, its homotopy fiber fits into a fiber sequence

$$B \mathbb{Z} \to \widetilde{X} \to X \to B^2 \mathbb{Z}$$

where here our saving grace is that $B \mathbb{Z} \cong S^1$ can be modeled by a compact manifold. Said another way, $\alpha$ is the first Chern class of a unique principal $\text{U}(1)$-bundle, and it vanishes when pulled back to the total space of this bundle.

For example, when we do this to a generator of $H^2(S^2, \mathbb{Z})$ we get the Hopf fibration $S^1 \to S^3 \to S^2$. More generally, when we do this to a generator of $H^2(\mathbb{CP}^n, \mathbb{Z})$ we get the familiar fibration $S^1 \to S^{2n+1} \to \mathbb{CP}^n$. The construction in this case can be interpreted as computing the $2$-connected cover, and if we could take $i$-connected covers of compact manifolds that were still compact manifolds for all $i$ then we would be done, but I don't think that's possible; e.g. the $3$-connected cover of $\text{Spin}(n), n \ge 3$ is $\text{String}(n)$, which can't be modeled by a finite-dimensional Lie group.

  • 1
    $\begingroup$ Indeed, this is exactly the topological version of the "higher stacks" construction I was alluding to. I don't know how to solve this in the setting of manifolds, but as you say, if you allow infinite CW complexes it is relatively easy. $\endgroup$ Oct 22, 2014 at 13:30
  • $\begingroup$ One possible sense in which the topological (by which I mean, I am thinking about smooth submersions of manifolds) is easier than the algebraic one: suppose we are thinking about $\mathbb{P}^1$. Then this question is really about one of the differentials in the Leray spectral sequence for $f: X\to \mathbb{P}^1$. For smooth submersions, this can be pretty much anything (consider e.g. the Hopf fibration), but for smooth projective morphisms of varieties, it is very restricted since the Leray spectral sequence for $\mathbb{Q}$-cohomology degenerates at $E_2$. $\endgroup$ Oct 22, 2014 at 20:29
  • $\begingroup$ If I haven't fumbled a spectral sequence computation then it looks like the $3$-connected cover of $S^3$ has nontrivial cohomology in arbitrarily high dimensions. So I would be pessimistic about prospects for solving this problem for $i = 3$ and $X = S^3$ topologically. I guess the closest algebraic example is something like $i = 4$ and $X = \mathbb{CP}^2$? $\endgroup$ Oct 23, 2014 at 5:27
  • $\begingroup$ Hmm...I'm a bit confused by your heuristic. Can't whatever space you're thinking about just be truncated above degree $1000$ or something (which won't change cohomology) to obtain a finite CW complex with the desired property? $\endgroup$ Oct 23, 2014 at 15:17
  • $\begingroup$ @Daniel: I have in mind the new space $Y$ being a compact manifold rather than just a finite CW complex; that seemed closer to the algebraic version of the problem, but I could be mistaken. $\endgroup$ Oct 23, 2014 at 19:21

As Jason Starr remarks in the comments, this answer to a question of his implies the answer to both of my questions is "no." For the latter, one may take:

$X=\mathbb{P}^1, \mathcal{L}=\mathcal{O}(1)$

Then there is no smooth proper morphism $f: Y\to X$ such that $f^*\mathcal{L}$ has a square root, for example. (Jason's question is about projective morphisms, but I think it's fairly easy to deduce the proper case.)

For the first question, $X=\mathbb{P}^1, i=2$ is a counterexample.

I would be curious to see a simpler proof of this result. I would also be curious to know the answer with the smoothness condition relaxed to flatness.

  • 3
    $\begingroup$ As already indicated by Mark in the comments, the Bloch–Gieseker cover provides a positive answer for the flat version of your second question, and therefore by your reduction to the $i = 2$ case of the first question. Using the splitting principle, this also gives a positive answer for classes coming from $\operatorname{CH}^*(X)_{\mathbb Q} \cong K^0(X)_{\mathbb Q}$, which under the Hodge conjecture answers the question for rational $(i,i)$-classes. $\endgroup$ Sep 23, 2017 at 16:47

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.