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...

  • 1
    $\begingroup$ 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 $\mathrm{Bockstein}(v^2) \neq 0 \in H^5(M; \mathbb{Z})$. $\endgroup$ – Theo Johnson-Freyd Oct 19 '19 at 14:34
  • $\begingroup$ (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.) $\endgroup$ – 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.

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

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.