Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I need to know all relations between Stiefel-Whitney classes for closed manifolds of dimensions 3 and 4. Unfortunately, I found the literature on the subject quite confusing. The answer for all dimensions appears to be contained in E. H. Brown and F. P. Peterson, Bull. AMS 69 (1963), p. 228, but I found it rather cryptic.

share|improve this question
add comment

1 Answer

up vote 12 down vote accepted

What is cryptic? All relations follow from $u_i=0$ for $2i>\dim X$, where $$ u=\operatorname{Sq}^{-1}w=1+w_1+(w_2+w_1^2)+w_1w_2+(w_4+w_1w_3+w_2^2+w_1^4)+\ldots $$ is the total Wu class. (The reason is the fact that $(\operatorname{Sq}^px)[X]=(u_p\smile x)[X]$ for any class $x\in H^{n-p}(X)$, $n=\dim X$.) Explicitly, in small dimensions we have:

$w_1^2+w_2=w_1w_2=0$ for $\dim X=3$,

$w_1w_2=w_1^4+w_2^2+w_1w_3+w_4=0$ for $\dim X=4$.

Of course, these relations generate an ideal in the algebra of Stiefel--Whitney classes invariant under the Steenrod operations. Hence, in dimension $3$ we also have $w_3=w_1^3=0$, i.e., all classes of dimension $3$ are trivial (any $3$-manifold is null cobordant, Rokhlin's theorem). In dimension $4$, we get, in addition, $w_1w_3=w_1^2w_2=0$.

share|improve this answer
    
Why is u_3=w_1 w_2? I thought there is also w_1^3 and w_3 there. –  Anton Kapustin Feb 27 at 16:17
    
Well, that's just the way it is. You need to compute $\def\Sq{\operatorname{Sq}}\Sq^{-1}=1+\Sq^1+\Sq^2+\Sq^2\Sq^1+\ldots$ and use Wu formulas. –  Alex Degtyarev Feb 27 at 18:16
    
Thanks, I got it now. –  Anton Kapustin Mar 4 at 21:21
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.