# Is the top Stiefel-Whitney number of a topological manifold the Euler characteristic mod two?

Recall that the Stiefel-Whitney classes of a smooth manifold are defined to be those of its tangent bundle - this definition doesn't extend to topological manifolds as they don't have a tangent bundle. Wu's theorem states that for a closed smooth manifold, $w = \operatorname{Sq}(\nu)$. The expression $\operatorname{Sq}(\nu)$ makes sense for a closed topological manifold and therefore serves as a definition for the Stiefel-Whitney classes on such a manifold.

Recall that if $M$ is a closed smooth $n$-dimensional manifold, then $w_n(M)$ is equal to the mod $2$ reduction of $e(M)$, see Corollary 11.12 of Milnor and Stasheff's Characteristic Classes. In particular, the Stiefel-Whitney number $\langle w_n(M), [M]\rangle$ is the mod $2$ reduction of the Euler characteristic. Is this still true for closed topological manifolds?

Let $M$ be a closed topological $n$-dimensional manifold. If $w_n(M)$ is the top Stiefel-Whitney class of $M$, as defined above, is the Stiefel-Whitney number $\langle w_n(M), [M]\rangle$ the mod $2$ reduction of $\chi(M)$?

As you say, we define $w_n = \sum \text{Sq}^i \nu_{n - i}$, where $\nu_{n-i}$ is the Wu class, the class such that $\nu_{n-i} \cup c = \text{Sq}^{n-i} c$ for $c \in H^{i}$. So as a corollary we have $\text{Sq}^i \nu_{n - i} = \nu_i \cup \nu_{n-i}$.

Because $\nu_j$ vanishes for $j > n/2$, the sum over $i$ is only the term $\nu_{n/2}^2$ when $n$ is even, and $0$ when $n$ is odd. As the Euler characteristic of an odd-dimensional closed manifold vanishes, this gives the odd-dimensional case.

Now when $n = 2k$, using Poincare duality mod 2 we see that $\chi(M) = \text{rk } H^k(M;\Bbb Z/2) \pmod 2.$ So the claim is that $\langle \nu_k^2, [M] \rangle = \text{rk } H^k(M;\Bbb Z/2) \pmod 2.$

This is because $\nu_k$ is a characteristic vector for the symmetric bilinear cup-product form on $H^k(M;\Bbb Z/2)$; in fact, for any 'characteristic vector' $y$ for a nondegenerate symmetric bilinear form over a $\Bbb Z/2$-vector space $V$, meaning that $y \cdot x = x^2$ for all $x$, we have $\text{rk } V = y^2 \pmod 2$.

The most obvious way for me to see this is to classify nondegenerate symmetric bilinear forms over $\Bbb Z/2$ vector spaces: they are all sums of copies of $\begin{pmatrix}1\end{pmatrix}$ and $\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}$, for which the respective characteristic vectors are $(1)$ and $(0,0)$.

• The proof of the classification is a reverse-induction on dimension; if there is some vector so that $B(x,x) = 1$, then split off $x$ as a summand. If not, pick an arbitrary nonzero vector $x_1$ and use nondegeneracy to find some vector $x_2$ with $B(x_1, x_2) = 1$. Split off this subspace as a summand (this uses nondegeneracy to see that the 'complement' is actually 2 dimensions less); the assumption that $B(x, x) = 0$ for all $x$ implies the bilinear form is given by the stated matrix on this summand. – Mike Miller Sep 10 '18 at 20:27

It has been proved in the preprint (page 6) by Renee Hoekzema that the vanishing of the $w_{n}(M)$ implies $\chi(M)$ is even. The proof uses the fact that a symplectic vector space over $\mathbb{F}_{2}$ has even dimension. It is quite similar to the one Mike Miller given here without the induction procedure.

The author suggests there is a more direct proof generalizing the one from Milnor-Stasheff without using the Euler class. I am not sure it might be. The paper actually proved much more and I found it really interesting.

• It seems this is the same argument, with the clever reduction that $v_n = 0$ implies all vectors square to zero, and hence $H^k$ is a symplectic vector space. The proof I give above is basically the usual proof that a symplectic vector space is equivalent to a standard one. In any case very nice preprint. – Mike Miller Sep 10 '18 at 20:43
• @MikeMiller: I actually have something very basic to ask: If I recall correctly, Milnor-Stasheff defined the Euler class via the Thom class. Is this still doable for topological manifolds? If it is, what is the difficulty to extend the classical proof to this case? – Bombyx mori Sep 10 '18 at 20:47
• It sounds like even basic questions are too hard for me :) The Thom class comes from the tangent bundle, which at first blush sounds like we're out of luck. But topological manifolds are Poincare Duality spaces and so they have a Spivak normal fibration as a weak replacement. Maybe one can define Euler classes using that, but it is beyond my pay grade. – Mike Miller Sep 10 '18 at 20:56
• Topological manifolds have tangent microbundles, and Thom classes which live in $H^*(M\times M, M\times M-\Delta)$. See this note, for example: ams.org/journals/bull/1966-72-03/S0002-9904-1966-11537-9/… or Ch 14 of Switzer's book. – Mark Grant Sep 11 '18 at 10:49