# Vector bundles with exactly one nonzero SW-class

I am interested in seeing examples of a space $X$ (preferably a closed smooth manifold, but any finite-dimensional CW-complex would also be of interest) with a vector bundle $\xi\colon E \to X$ on it, so that there is exactly one index $i$ with $w_i(\xi) \neq 0$, and $i$ is bigger than $8$. Here are some remarks:

(1) First, this could only happen if $i$ is a power of 2: As a module over the Steenrod algebra, $H^{\ast}(BO;\mathbb F_2) = \mathbb F_2[w_1,w_2,w_3,\dots]$ is generated by $w_{2^k}$, so the first nonzero SW-class is always of degree $2^k$ (this is also an exercixe in Milnor-Stasheff).

(2) For $i=1,2,4,8$ one could take $S^1 = \mathbb RP^1, S^2 = \mathbb CP^1, S^4 = \mathbb HP^1$ and $S^8 = \mathbb OP^1$ with the canoncial bundles on these spaces.

(3) Beyond dimension 8, a sphere (or even a connected sum of products of spheres) does not give rise to any such bundle. This can be seen from analyzing $$w\colon \tilde{KO}(S^{d_1} \times \dots \times S^{d_r}) \to H^{\ast}(S^{d_1} \times \dots \times S^{d_r};\mathbb F_2),$$ the main input for understanding this map is Adams' Hopf invariant one theorem.

For $i = 2^k$ you can get an example with $X = \Bbb{RP}^{m}$ for any $m \geq 2^k$. If $L$ is the canonical line bundle on $\Bbb{RP}^{m}$ then let $E = \bigoplus_{i=1}^{2^k} L$ be the a sum of several copies of it. Then it has total Stiefel-Whitney class $$w(E) = \prod_{i=1}^{2^k} w(L) = (1+x)^{2^k} = 1 + x^{2^k},$$ where $x$ is the generator of $H^1(\Bbb{RP}^{2^k};\Bbb F_2)$.
• @Jens Oh. That's possible for $i=16$, but I only know a "universal" construction that's not simple. Stong showed that $w_{16}$ has nontrivial image in the cohomology of the 8-connected cover $Y = BO(9,\dots,\infty)$, and so you can take the $16$-skeleton $Y$ of $BO(9,\dots,\infty)$ to get a 16-dimensional finite complex with a vector bundle classified by $Y \to BO$ such that $w_{16}$ is nontrivial but all lower $w_i$ are. – Tyler Lawson May 22 '16 at 13:42
• However, if I'm reading right, Stong's calculation also shows this can't happen if you ask for it to be 16-connected and $w_{32}$ to be nontrivial, because $w_{32}$ has trivial image in the cohomology of $BO(17,\dots,\infty)$. ("Determination of $H^*(BO(k,\dots,\infty),\Bbb Z_2)$ and $H^*(BU(k,\dots,\infty),\Bbb Z_2)$", Trans. AMS, 1963) – Tyler Lawson May 22 '16 at 13:45