Which elements of $H^2(M,\mathbb{Z}/2)$ are the $w_2(E)$ for a real bundle $E$?

Any element of $$H^1(M,\mathbb{Z}/2)$$ is the $$w_1(E)$$ of a real line bundle $$E$$ over $$M$$. I wonder how to characterize (probably using the Steenrod squares) which elements of $$H^2(M,\mathbb{Z}/2)$$ are the $$w_2(E)$$ of a real vector bundle $$E$$ over $$M$$.

Considering tensor products and tensoring by line bundles, it is clear that such elements form a subgroup of $$H^2(M,\mathbb{Z}/2)$$. I know that any element killed by $$Sq^1$$ is realizable this way, since if $$Sq^1 v=0$$ for $$v\in H^2(M,\mathbb{Z}/2)$$ then $$v$$ is a reduction of an element $$c$$ in $$H^2(M,\mathbb{C})$$. Then one can pick a complex line bundle whose $$c_1$$ is $$c$$, which always exists...

• I don't know the answer, but I can tell you one obstruction. Recall that if $w_2(E) = 0$, then $E$ carries a fractional Pontryagin class $\frac{p_1}2(E)$. This is part of a stronger statement even when $w_2(E) \neq 0$, namely that there is a degree-$4$ integral cochain "$\frac{p_1}2(E)$" solving $d\frac{p_1}2(E) = \mathrm{Bockstein}(w_2(E)^2)$, where $\mathrm{Bockstein} : C^4(M;\mathbb{Z}_2) \to C^5(M;\mathbb{Z})$ is (a cochain model for) the integral Bockstein. Taking cohomology, we find that $v$ is not of the form $w_2(E)$ if $\mathrm{Bockstein}(v^2) \neq 0 \in H^5(M; \mathbb{Z})$. – Theo Johnson-Freyd Oct 19 '19 at 14:34
• (I am assuming you are looking for an oriented bundle. If $w_1(E) \neq 0$, then there is a more complicated formula that I never worked out.) – Theo Johnson-Freyd Oct 19 '19 at 14:36

This is an obstruction theory problem; you regard $$v\in H^2(M;\mathbb{Z}/2)$$ as a homotopy class of maps $$v: M\to K(\mathbb{Z}/2,2)$$, then ask if $$v$$ lifts through the universal Stiefel-Whitney class $$w_2:BSO\to K(\mathbb{Z}/2,2)$$. The first obstruction is $$\beta(v^2)\in H^5(M;\mathbb{Z})$$, as mentioned by Theo Johnson-Freyd in the comments, coming from the fact that $$w_2^2(E)$$ is the mod $$2$$ reduction of $$p_1(E)$$ for any real vector bundle $$E$$. The secondary obstruction is more subtle (it will have indeterminacy given by a choice of integral lift of $$v^2$$, for example) but I would be surprised if there is no information on it in the literature.

I think the state of the art on this question might still be Teichner's paper

Teichner, Peter, 6-dimensional manifolds without totally algebraic homology, Proc. Am. Math. Soc. 123, No. 9, 2909-2914 (1995). ZBL0858.57033.

Teichner shows that if $$\operatorname{dim}(M)\le 5$$, then $$\beta(v^2)$$ is the only obstruction to $$v$$ being $$w_2(E)$$ for some real vector bundle $$E$$. On the other hand, he constructs examples of manifolds in all dimensions $$n\ge 6$$ with classes $$v\in H^2(M;\mathbb{Z}/2)$$ such that $$\beta(v^2)\neq 0$$. He also shows that any $$v\in H^2(M;\mathbb{Z}/2)$$ which is Poincaré dual to a codimension $$2$$ submanifold $$N\subseteq M$$ is $$w_2(E)$$ for some $$E$$.

An older paper which also seems relevant is

Suzuki, H., On the realization of the Stiefel-Whitney characteristic classes by submanifolds, Tohoku Math. J., II. Ser. 10, 91-115 (1958). ZBL0107.17001.

To elaborate on Mark Grant's answer, and sticking to the oriented case, there is a fibration sequence (all of infinite loop maps) $$BSpin \rightarrow BSO \xrightarrow{w_2} K(\mathbb Z/2,2) \xrightarrow{d} BBSpin$$ The homotopy groups of $$BBSpin$$ are known from Bott periodicity: $$\pi_5(BBSpin) = \mathbb Z$$, $$\pi_9(BBSpin) = \mathbb Z$$, $$\pi_{10}(BBSpin) = \mathbb Z/2$$, etc.

So a first obstruction to lifting through $$w_2$$ will be the composite $$K(\mathbb Z/2,2) \xrightarrow{d} BBSpin \rightarrow K(\mathbb Z,5)$$ which identifies as $$\beta Sq^2$$ as noted above. The next obstruction would live in 9 dimensional cohomology, etc., and so these obstructions would vanish on spaces of dimension 8 or less.

Finally, I believe the answer doesn't change if one considers nonoriented bundles, because $$BO \xrightarrow{w_2} K(\mathbb Z/2,2)$$ factors through $$BSO \xrightarrow{w_2} K(\mathbb Z/2,2)$$, if I am not mistaken.

• Thanks. I would have accepted this as an answer too, if MO allows one to select more than one answers. – Yuji Tachikawa Oct 21 '19 at 8:21