Is it possible to construct a vector bundle over a given base $X$ such that the $n$th stiefel whitney class vanishes for a given $n?$ What about for some set of integers? Can we make vector bundles of arbitrary stiefel whitney classes?
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2$\begingroup$ Well, the trivial bundle always has vanishing $n$-th Stiefel-Whitney class for any $n$. If you're asking whether we can freely prescribe them, that's not always possible. (For example, they have to be compatible with the Wu formula) $\endgroup$– Achim KrauseCommented May 7, 2015 at 23:31
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1$\begingroup$ Problem $8$-$B$ from Milnor & Stasheff provides a restriction. Namely, if $w(\xi) \neq 1$, then the smallest $i$ such that $w_i(\xi) \neq 0$ is a power of $2$. $\endgroup$– Michael AlbaneseCommented May 8, 2015 at 2:18
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Let me give a more detailed answer:
A rank $n$ vector bundle on $X$ is precisely given by a homotopy class of maps $X\rightarrow BO(n)$. The Stiefel-Whitney classes of the vector bundle describe how that map acts on cohomology with $\mathbb{F}_2$ coefficients.
In general, not all maps on cohomology can be realized by an actual map of spaces. (For spectra in the place of spaces this takes very precise form in the structure of the Adams spectral sequence)
As a rule of thumb, homological information lifts back to topological information only in very special cases, and you should really consider these things just as a crude (albeit far more computable) approximation.
answered May 7, 2015 at 23:44