I need to know all relations between StiefelWhitney 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.
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 $x\in H^{np}(X)$, $n=\dim X$.) 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 StiefelWhitney 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, $w_1w_3=w_1^2w_2=0$.

$\begingroup$ Why is u_3=w_1 w_2? I thought there is also w_1^3 and w_3 there. $\endgroup$ – Anton Kapustin Feb 27 '14 at 16:17

$\begingroup$ 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. $\endgroup$ – Alex Degtyarev Feb 27 '14 at 18:16