5

Let $BO(n)$ denote the classifying space of the orthogonal group $O(n)$. Then there is the well-known ring isomorphism

$$H^*(BO(n);\mathbb{Z}/2) \cong \mathbb{Z}/2[w_1,\dots,w_n] $$ where $w_i \in H^i(BO(n);\mathbb{Z}/2)$ is the $i$-th universal Steifel-Whitney class.

From the short-exact sequence $\mathbb{Z} \stackrel{\cdot 2}{\to} \mathbb{Z} \to \mathbb{Z}/2$ there is a natural Bockstein homomorphism $\beta\colon H^k(-;\mathbb{Z}/2) \to H^{k+1}(-;\mathbb{Z})$, which in particular has the property that $\beta(c) = 0$ iff $c$ is the mod-$2$ reduction of some integral class. Then we can define the integral Steifel-Whitney classes

$$ W_i = \beta(w_{i-1}) \in H^i(BO(n);\mathbb{Z}).$$

I haven't found much information about these in the usual sources other than the definition. My question is whether these are all non-zero, and if not whether there is a complete description of which ones vanish. In particular I would like to know if the universal $W_4$ is $0$.

This question is (tangentially) related to this other question about the Universal Coefficient Theorem, which led me to wonder if there is $2$-torsion in $H^4(BO(n);\mathbb{Z})$ for $n\geq 4$. I know that $H^*(BO(n);\mathbb{Z})$ consists of a free part generated by Pontryagin classes and a $2$-torsion part given by $im(\beta)$, but I can't determine whether $\beta(w_3)=0$ or not.

Note: I am aware of papers by Brown and Feshbach giving fairly explicit/complete descriptions of the ring $H^*(BO(n);\mathbb{Z})$, but I have only been able to find them on JSTOR and I don't have access.

Edit: An idea I had was to try to use the formula for $Sq^i(w_j)$ (for example here) and then I think it's true that $Sq^1 = (\text{reduction mod-}$2$)\circ\beta$. I computed $Sq^1(w_3) = w_1w_3 + w_4$, which is non-zero in $H^*(BO(n);\mathbb{Z})$ by algebraic independence of SW classes, but then this would mean $W_4 = \beta(w_3) \neq 0$. Is this argument valid?

William
  • 9,730
  • William, your edit is exactly the argument I came up with. I think it checks out. So interestingly, for spin bundles it lifts the fourth SW class. – Connor Malin May 27 '20 at 16:26
  • Ok, thank you @ConnorMalin. It looks like a similar argument can be used to show that $W_j\neq 0$ for $j > 1$. I will write it up as an answer to my own question. – William May 27 '20 at 16:32
  • I was very tempted to try to prove that as well, but the combination of 3 indices, binomial coefficients, and weird conventions for negative indices in binomial coefficients discouraged me to do the general result. – Connor Malin May 27 '20 at 16:40
  • I wonder what the compatibility is between the Chern classes and the integral Stiefel Whitney classes is. – Connor Malin May 27 '20 at 16:41
  • @ConnorMalin It turns out the formula is very simple if $i=1$, as long as I didn't make a mistake, since it is the upper bound on the sum so there are only $2$ terms. I'm not sure exactly the relation between Chern classes and integral SW classes, but I feel like I've seen something somewhere. – William May 27 '20 at 16:46
  • There is a paper by Pittie in which he computes the integral cohomlogy of $SO_n$. I don't recall his results, but I'm sure you will find many examples of your integral SW classes which survive even after applying the cohomology suspsension. – Tyrone May 27 '20 at 16:50

1 Answers1

5

Let $\rho\colon H^*(-;\mathbb{Z}) \to H^*(-;\mathbb{Z}/2)$ be the natural "reduction mod-$2$" map induced by $\mathbb{Z} \to \mathbb{Z}/2$. Then we can use the fact that the first steenrod square $Sq^1$ is actually the composition $\rho \circ \beta$, in conjuction with the formula (linked in the question)

$$Sq^i(w_j) = \sum_{t=0}^i\binom{j-i-1+t}{t}w_{i-t}w_{j+t} $$

where $i < j$ (the Steenrod square vanishes if $j > i$ for degree reasons, and $Sq^j w_j = w_j^2$). In particular, let $i = 1$ and $j> 1$: then the formula simplifies considerably to

$$Sq^1(w_j) = w_1w_j + (j-1)w_{j+1}. $$

Since the Steifel-Whitney classes are algebraically independent in $H^*(BO(n);\mathbb{Z}/2)$ it follows from this formula that $Sq^1(w_j)$ and hence $\beta(w_j)$ are non-zero for $j > 1$, therefore $W_j \neq 0$ for $j > 2$; moreover $Sq^1(w_1) = w_1^2 \neq 0$ so $W_2$ is non-zero as well.

Edit: As Connor Malin observes in a comment the class $W_1$ is actually $0$, since the natural transformation $\beta\colon H^0(-;\mathbb{Z}/2) \to H^1(-;\mathbb{Z})$ is given by a map $\mathbb{Z}/2 \to K(\mathbb{Z},1)\simeq S^1$ which cannot be homotopically non-trivial. In other words it's not just $Sq^1\colon H^0(-;\mathbb{Z}/2) \to H^1(-;\mathbb{Z}/2)$ which is trivial in this degree, but $\beta$ itself.

William
  • 9,730
  • 3
    If $B$ denotes the integral Bockstein, it’s action on $H^0(-;\mathbb{Z}/2)$ can be deduced by looking at the looped map $\mathbb{Z}/2 \rightarrow K(\mathbb{Z},1)$ which is obviously trivial. Hence, your $W_1$ is trivial. This is also how one would prove that Steenrod squares are trivial on classes below their degree. – Connor Malin May 27 '20 at 16:53
  • 3
    Alternatively, the kernel of $\beta$ is exactly the classes that can be lifted to integral cohomology (this is immediate from the definition of $\beta$ as the connecting homomorphism in the long exact sequence associated to a short exact sequence of coefficients), and every class in $H^0(-;\mathbb{Z}/2)$ obviously lifts to integral cohomology by explicitly describing $H^0$ as functions on path-components. – Eric Wofsey May 27 '20 at 16:58
  • 2
    Another way to see that $W_1$ is zero is to note that it is 2-torsion, but the first integral cohomology group is always torsion-free. – Michael Albanese Jul 01 '20 at 13:11