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?

  • 1
    $\begingroup$ (Re: The only problem I see is with the Steenrod square operation which may be undefined for integral coefficients) Well, yes there are no non-trivial stable operations from H(-;Q) to H(-;Q), after all... $\endgroup$
    – Grigory M
    Apr 8, 2015 at 8:22
  • $\begingroup$ @GrigoryM But do we get something nontrivial or useful if we define the "top" class to be the Euler class and then inductively $w_i(\xi)=(\pi^*)^{-1}w_i(\xi_0)$ where $\xi_0$ is an $n-1$ bundle over $E_0$ ($E_0$ are nonzero elements of the original bundle with fiber over $e_0\in E_0$ consisting of the orthogonal complements of $e_0$) and $\pi^*$ is the isomorphism $H^*(B)\to H^*(E_0)$ in dimension $<n$? $\endgroup$ Apr 8, 2015 at 12:32

2 Answers 2


You might be interested in the definition of both Stiefel-Whitney and Chern Classes in terms of the outer tensor product with $\gamma^1 \to RP^\infty$ and $\gamma^1 \to CP^\infty$ resp. This is a refreshing different approach to show the existence of those classes. You can find it for example briefly in Ebert's script about cobordism.

Note that you use the nice cohomology rings of those base spaces for the definition. Also note that this definition uses line bundles, which are boring in the orientable case, so the canonical idea might not work.

  • $\begingroup$ Thank you, it will take me a while but I will try to study it. $\endgroup$ Apr 6, 2015 at 21:43

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.


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