8
$\begingroup$

It is well known that in $\mathbb{Z}_2$-valued simplicial cohomology (and other cohomologies) $$ Sq^1 = \beta\;,$$ where $Sq^1$ is the first Steenrod square and $\beta$ is the Bockstein homomorphism for the short exact sequence $$ \mathbb{Z}_2 \overset{\cdot 2}{\rightarrow} \mathbb{Z}_4 \overset{\mod 2}{\rightarrow} \mathbb{Z}_2\;.$$ I was wondering whether there is a way to also interpret $Sq^2$ or even higher Steenrod squares in a similar fashion. One could think that instead of $$\beta = g^{-1} d h^{-1}$$ for a short exact sequence of abelian groups $$A\overset{g}{\rightarrow} B\overset{h}{\rightarrow} C$$ one could try something like $$g^{-1}dh^{-1}di^{-1}$$ for an exact sequence $$A\overset{g}{\rightarrow} B\overset{h}{\rightarrow} C\overset{i}{\rightarrow} D\;.$$ However, if I'm not mistaken, the long exact sequence splits into two short exact sequences, and the above is just the product of the two corresponding Bockstein homomorphisms. E.g., when applied to the long exact sequence $$ \mathbb{Z}_2 \overset{\cdot 2}{\rightarrow} \mathbb{Z}_4 \overset{\cdot 2}{\rightarrow} \mathbb{Z}_4 \overset{\mod 2}{\rightarrow} \mathbb{Z}_2$$ we get $\beta^2$ for the short exact sequence given in the beginning, which is trivial.

Crossed module extensions are a more non-trivial analogue to ordinary group extensions which also come with an action of $C$ on $B$. It is known that crossed module extensions give rise to group 3-cocycles in $H^3(BD, A)$ just as ordinary extensions give rise to 2-cocycles. If $D$ acts trivially on $A$, the group 3-cocycle can be used to map a (simplicial) 1-cocycle to a 3-cocycle, and the "higher Bockstein" should generalize this to a map from $i$-cocycles to $i+2$-cocycles. However, I'm not sure how to use the action of $C$ on $B$ to define sich a higher Bockstein.

So I basically have two questions: 1) Is there a non-trivial analogue of a Bockstein operation for some sort of longer exact sequences/crossed module extensions? 2) If yes, is any of these operations equal to $Sq^2$ (in general, or maybe at least when applied to a specific degree)?

$\endgroup$
1
  • 1
    $\begingroup$ Note that $Sq^2$ is trivial on $H^1(X,Z/2)$ for space $X$. $\endgroup$
    – user43326
    Jan 19 at 15:53

1 Answer 1

11
$\begingroup$

$\DeclareMathOperator{\Sq}{Sq}\newcommand{\Z}{\mathbb{Z}}$The short version is that every cohomology operation can be interpreted as a Bockstein operator for an "exact sequence" (read: fiber sequence) of grouplike $E_\infty$ spaces.

Any cohomology operation $\delta:H^*(-;A)\Rightarrow H^{*+k}(-;B)$ such as $\Sq^i$ gives a morphism of Eilenberg-MacLane spectra $f_\delta:HA\to \Sigma^k HB$. You can take the fiber of $f_\delta$ to obtain a spectrum $F$ which defines a generalized cohomology theory $E_F^*:X\mapsto \pi_{-*} F^X$, and there is a resulting long exact sequence $$ \dots\to E_F^i(X)\to H^i(X;A)\xrightarrow{\delta}H^{i+k}(X;B)\to E_F^{i+1}(X)\to\dots $$ If $\delta = \Sq^1$, the fiber is again an Eilenberg-MacLane spectrum (namely $F\simeq H\Z/4$). If the degree of $\delta$ is bigger than $1$, it will have two non-zero homotopy groups, namely $A$ in degree $0$ and $B$ in degree $k-1$.

The connection to the crossed modules you mention is that applying the functor $\Omega^{\infty-1}$ to the fiber of $\Sq^2$ gives rise to a $2$-group, i.e. a homotopy type $X$ whose homotopy groups vanish outside degrees $1$ and $2$. As you mention, these are classified by the two groups $A = \pi_1(X),B = \pi_2(X)$, the action of the former on the latter, and a $k$-invariant in $H^3(A;B)$, which together can be packaged into the datum of a crossed module. However, these deloop once if and only if the action is trivial and the $k$-invariant vanishes. It is still possible to find reasonably easy algebraic models for spaces whose homotopy groups vanish outside degrees $k$ and $k+1$ (given by braided ($k=2$) and symmetric ($k\ge 3$) monoidal Picard (every object has a tensor inverse) groupoids), although I do not know a definition of $\Sq^2$ in this language.

Questions in the comments

  1. You already discuss the resulting cohomology operation in the case that the action of $A$ on $D$ is trivial. For the general case, one first construts a natural transformation from $H^1(-;A)$ to local systems: $H^1(-;A)$ are isomorphism classes of $A$-principal bundles, and this natural transformation sends a principal bundle $P$ to $P\times_A D$. A cohomology class in $H^k(A;D)$ then gives a natural transformation from $H^1(-;A)$ to $H^k$ of this local system, by the same construction as in the trivial case.
  2. Yes, for a $(2,3)$-type $X$ the $2$-group $\Omega X$ is split, i.e. the action of $\pi_1(\Omega X)\cong \pi_2(X)$ on $\pi_2(\Omega X)\cong \pi_3(X)$ is trivial (this can be shown by the Eckmann-Hilton argument) and the $k$-invariant vanishes.
  3. For degree $2$ cohomology operations, there is in fact a complete classification (compare (Co)homology of the Eilenberg-MacLane spaces K(G,n) and the cited references): operations $H^2(-;A)\to H^4(-;B)$ are given by quadratic functions $q:A\to B$ (i.e. such that $q(x+y) - q(x) - q(y)$ is bilinear and $q(kx) = k^2q(x)$), and the resulting cohomology operation is given by a suitable version of the Pontryagin square. For $k\ge 3$, operations are in bijection with linear maps from $A\otimes \Z/2$ to $B$ (observe that such a linear map is also quadratic by the "freshman's dream"), and the resulting cohomology operation is the composition

$$ H^*(-;A)\to H^*(-;A\otimes Z/2)\xrightarrow{\Sq^2} H^{*+2}(-;A\otimes\Z/2)\to H^{*+2}(-;B) $$

The relation to Picard groupoids is a consequence of the Homotopy hypothesis, and given a braided Picard groupoid $C$, you can associate to it its abelian group $\pi_0 C$ of isomorphism classes and the (abelian!) group $\pi_1 C$ of automorphisms of the unit $1$, together with the map $q: \pi_0 C\to \pi_1 C$ which sends $x$ to the composition

$$ 1\cong x\otimes x^{-1}\xrightarrow{\beta_{x,x^{-1}}} x^{-1}\otimes x\cong 1 $$

It's a fun exercise to show that $q$ is quadratic in the above sense. It is not straightforward to give an inverse to this construction, i.e. construct the braided Picard groupoid from the quadratic map; for a reference in the symmetric setting, see Cegarra, A. M.; Khmaladze, E., Homotopy classification of graded Picard categories, Adv. Math. 213 (2007).

A chain level representative of $\Sq^2$ (and higher Steenrod squares) can be found in Ralph M. Kaufmann, Anibal M. Medina-Mardones. Cochain level May-Steenrod operations.

$\endgroup$
7
  • $\begingroup$ Thanks a lot for your great answer! As a physicist who doesn't know much about spectra etc. I have some very basic questions: $\endgroup$
    – Andi Bauer
    Jan 20 at 10:21
  • $\begingroup$ 1) Given a 2-group/crossed module, is there a way to construct a cohomology operation from that? 2) You say $Sq^2$ gives rise to a crossed module, but do I understand it correctly that later you say that this crossed module is trivial (i.e. equivalent to the trivial crossed module), since it has to "deloop once"? 3) Are you saying that even though the 2-group is trivial for $Sq^2$, there is another algebraic object (the "braided Picard groupoid" you mention) which describes $Sq^2$? Is there a way to explicitly construct a cohomology operation from such a Picard groupoid? $\endgroup$
    – Andi Bauer
    Jan 20 at 10:21
  • $\begingroup$ If we apply the non-trivial group 3-cocycle in $H^3(B\mathbb{Z}_2, \mathbb{Z}_2)$ to a 1-cocycle $C$, we get $C\cup C\cup C$ which is a non-trivial cohomology operation. However, $Sq^2$ is trivial on 1-cocycles (as also remarked in a comment to the question). Is that another way to see that the crossed module of $Sq^1$ has to be trivial? (I was hoping that maybe $Sq^2$ might correspond to a crossed module only when applied to certain degrees, but from what you're writing it doesn't sound like this would make sense.) $\endgroup$
    – Andi Bauer
    Jan 20 at 10:31
  • 1
    $\begingroup$ I've tried to answer your questions in the comments. Let me just also mention that for general (unstable!) cohomology operations, one can investigate its "delooping" (operations one degree higher which give the operation after conjugation with the suspension isomorphism). It's easy to see that the operation mist be additive for a delooping to exist. For ordinary cohomology, it turns out that this is sufficient, and that there is a unique delooping in this case. $\endgroup$ Jan 21 at 8:29
  • 1
    $\begingroup$ The action of $A$ on $D$ defines a principal $D$-bundle $EA\times_A D\to BA$, and since degree $1$ cohomology classes in $H^1(X;A)$ are homotopy classes from $X$ to $BA$, you can pull it back to get a principal $D$-bundle over $X$. Sections of this bundle are then a sheaf which is locally isomorphic to locally constant functions (aka a local system) to $D$, and you can define its sheaf cohomology as usual. A cocycle in $H^k(BA;D)$ then defines a degree $k$ cohomology classpulled back from the universal example $X = BA$, whose sheaf cohomology gives group cohomology. $\endgroup$ Jan 24 at 8:22

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.