For $X$ a topological space, from the short exact sequence

$$ 0 \rightarrow \mathbb{Z}/2 \rightarrow \mathbb{Z}/4 \rightarrow \mathbb{Z}/2 \rightarrow 0 $$

we get a Bockstein homomorphism

$$H^i(X, \mathbb{Z}/2) \rightarrow H^{i+1}(X, \mathbb{Z}/2)$$

This is also known as the Steenrod square $Sq^1$.

Now suppose instead that $X$ is a variety over a (not algebraically closed) field. We still get a sequence

$$ 0 \rightarrow \mathbb{Z}/2 \rightarrow \mathbb{Z}/4 \rightarrow \mathbb{Z}/2 \rightarrow 0 $$

inducing a Bockstein homomorphism in etale cohomology

$$H^i_{et}(X, \mathbb{Z}/2) \rightarrow H^{i+1}_{et}(X, \mathbb{Z}/2).$$

However, there is also a short exact sequence

$$ 0 \rightarrow \mathbb{Z}/2 \rightarrow \mathbb{Z}/4(1) \rightarrow \mathbb{Z}/2 \rightarrow 0 $$

where $\mathbb{Z}/4(1)$ denotes the Tate twist, probably less confusingly written as $\mu_4$. This also induces a [presumably different] Bockstein map in etale cohomology

$$H^i_{et}(X, \mathbb{Z}/2) \rightarrow H^{i+1}_{et}(X, \mathbb{Z}/2).$$

Question: Which of these is the "right" Bockstein homomorphism?

This question is a little open-ended. My main criterion for "right" is that the map be "$Sq^1$ on etale cohomology", meaning it fits into an action of the Steenrod algebra.

There are other possible criteria. For instance, a literature search revealed that people have defined notions of "Bockstein homomorphism" and "Steenrod operations" on Chow rings, motivic cohomology, ... so "right" could also mean "compatible with these other things". (Hopefully the answer is the same.)

Relevant literature:

Unfortunately, I can't really parse what's happening in these papers.

Some motivation/example: For $X \subset Y$ a closed embedding of smooth varieties of codimension $r$, I have a cycle class $[X] \in H^{2r}_X(Y; \mathbb{Z}/2)$. I would like to show that $Sq^1 [X]=0$ (it may not be true). Since the cycle class lifts to $H^{2r}_X(Y;\mathbb{Z}_{2}(r))$, this depends on which sequence is related to $Sq^1$.


You maybe want to have a look at

  • P. Brosnan and R. Joshua. Comparison of motivic and simplicial operations in mod-$\ell$ motivic and étale cohomology. In: Feynman amplitudes, periods and motives, Contemporary Math. 648, 2015, 29-55.

There are two sequences of cohomology operations in motivic and étale cohomology which can rightfully be called Steenrod operations. One comes from a "geometric" model of the classifying space of the symmetric groups, the other one from a "simplicial" model. The first one is related to Voevodsky's motivic Steenrod algebra, the second one to the more classical Steenrod algebra (as in Denis Nardin's answer).

The paper mentioned above provides a comparison between these sequences of cohomology operations (see Theorem 1.1 part iii for the étale cohomology) in terms of cup products with powers of a motivic Bott element in $H^0(\operatorname{Spec} k,\mathbb{Z}/\ell\mathbb{Z}(1))$.

Addendum: Actually, the Bockstein operations are the same in both sequences of cohomology operations, and they are the ones defined by the $\mathbb{Z}/4\mathbb{Z}$ extension. (So probably the first one should be the "right".) This follows from the paper of Brosnan and Joshua as well as the paper by Guillou and Weibel in the question. Note that $\mu_2^{\otimes i}\cong\mu_2^{\otimes 2 i}\cong\mathbb{Z}/2\mathbb{Z}$ and the $Sq^1$ in the paper of Guillou-Weibel actually fits the Bockstein story.

Another note: If the motivation/example is defined over a field with $4$-th root of unity then $\mathbb{Z}/4\cong\mu_4$ (compatible with the extensions) and so both operations you defined agree.

  • $\begingroup$ Wow, this is exactly the kind of thing I need! But forgive my ignorance (I'm far from this motivic stuff), I may need more help in seeing why their theorem implies the desired equality. Are you referring to the version people.math.osu.edu/joshua.1/coh-ops.new.pdf? I see a Theorem 1.1 but no part (iii). I guess I should be taking r=0, but I don't quite see how the two things can be identified with my two Bockstein maps. $\endgroup$
    – user84144
    Mar 17 '16 at 15:40
  • $\begingroup$ @user84144: there is a difference between the preprint and the published version. The published version includes the discussion of étale cohomology. $\endgroup$ Mar 17 '16 at 21:05
  • $\begingroup$ In response to your addendum: ah, that's too bad. The equality would have made life much easier, but oh well. (This does beg the question of whether the other boundary map fits into a larger story...) I had noticed that things were okay over field containing $\mu_4$, but was really hoping to get the result for all finite fields. $\endgroup$
    – user84144
    Mar 18 '16 at 0:44
  • 1
    $\begingroup$ @user84144: Looking at it, in your motivation/example, you have a cycle class. Probably this should be an element in $H^{2r}(X,\mathbb{Z}/2(r))$ rather than $H^{2r}(X,\mathbb{Z}/2(0))$. Then $Sq^1$ should be the Bockstein associated to the $\mathbb{Z}/4$-sequence (twisted by r), i.e., the cohomology operation appearing in all the abovementioned papers. But that would be trivial if the class lifts to $\mathbb{Z}/4(r)$. $\endgroup$ Mar 18 '16 at 8:36
  • $\begingroup$ Minor point, but I think Bert spells his last name Guillou. $\endgroup$ Mar 18 '16 at 16:54

Your first map fits in an action of the Steenrod algebra.

In fact $H^*_{ét}(X;\mathbb{F}_2)$ is the homology of $C^*_{ét}(X;\mathbb{F}_2)=R\Gamma(X;\mathbb{F}_2)$, an element of the derived category of $\mathbb{F}_2$ which has the property of being an $E_\infty$-algebra.[1]

This means that not only it is a commutative algebra for the derived tensor product (thus generating a cup product on homology) but the commutative and associative relations can be upgraded with some explicit homotopies that satisfy additional relations (the precise definition is a bit of a pain to write down without the appropriate formalism, but it boils down to be "as commutative as you can hope to get").

In particular its homology $H^*_{ét}(X;\mathbb{F}_2)$ has more structure than just the cup product: it has an action of an algebra called the Dyer-Lashof algebra. The negative degree part of the Dyer-Lashof algebra is exactly the Steenrod algebra (negative degree because, as every good homotopy theorist, I always use homological grading). So, $H^*_{ét}(X;\mathbb{F}_2)$ inherits an action of the Steenrod algebra from the $E_\infty$-structure on the chain level.

Incidentally this is exactly the same mechanism underlying the action of the Steenrod algebra on the cohomology of spaces, and in general every time you have a well behaved notion of "derived global sections" you can get a Steenrod algebra action out of the deal.

Ideally we would want $H^*_{ét}(X;\mathbb{F}_2)$ to be an unstable module over the Steenrod algebra (that is $Sq^ix=0$ for $|x|>i$) but I don't know if this is true.

NOTE: The existence of two "Bocksteins" means that the algebra acting naturally on $\bigoplus_n H^*_{ét}(X;\mathbb{F}_2(n))$ is presumably bigger than just the Steenrod algebra. I think it could be interesting to figure out exactly what algebra this is. Unfortunately I'm unaware of any work in this direction.

[1] The $E_\infty$-structure is due to the fact that $R\Gamma$ is a lax monoidal functor in the derived sense, being the right adjoint of the pullback functor which is symmetric monoidal, and so it preserves algebras for every operad.

  • $\begingroup$ Thanks for this answer (though not being an algebraic topologist myself, the first paragraph was gibberish to me). $\endgroup$
    – user84144
    Mar 18 '16 at 4:00
  • $\begingroup$ @user84144 I've expanded the answer. Hopefully now it's clearer to non-experts. $\endgroup$ Mar 18 '16 at 12:46

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.