TITLE: Stiefel-Whitney classes with Z-coefficients
QUESTION [6 upvotes]: Stiefel-Whitney classes are defined (for example, in Milnor's Characteristic classes) as elements of the cohomology groups with $\mathbb{Z}_2$-coefficients as follows.
$$w_i(E)=\phi^{-1} Sq^i(u)$$ where $E$ is an $n$-plane bundle over $B$, $E_0$ is the complement of the zero section, $Sq^i: H^*(E,E_0)\to H^{*+i}(E,E_0)$, $u\in H^n(E,E_0)$ is the Thom class and $\phi: H^*(B)\to H^{*+n}(E,E_0)$ is the Thom isomorphism (using $\mathbb{Z}_2$ coefficients).
My question is: if $E$ is oriented, does a similar construction work with $\mathbb{Z}$-coefficients? If yes, does it have a name (except of the "top" Euler class)? The only problem I see is with the Steenrod square operation which may be undefined for integral coefficients. If there are problems with the Steenrod squares, could such classes be defined inductively similarly as Chern classes for complex bundles?

REPLY [3 votes]: A bit long for a comment, but these classes might not be what you want.  Note that for line bundles, the sw classes satisfy $w_1(L_1\otimes L_2)=w_1(L_1)+w_1(L_2)$. Lets call the proposed classes $f$, and that they satisfy similar axioms. For a real line bundle (over some (para)compact base say, so that we have metrics) a line bundle $L$ is isomorphic to its dual $L^*$. Then
$$
f_1(L\otimes L^*)=f_1(L)+f_1(L^*)=2 f_1(L)
$$
But $L\otimes L^*\cong Hom(L,L)$ is trivial as it has a section (the identity). So $2 f_1(L)=0.$. This does not necessarily show that the class can't exist, but that the first class lives in the torsion part of the cohomology.