Help with Kervaire paper

Question

I was trying to read Kervaire's 1960 paper where he first shows the existence of a manifold that does not admit differentiable structure and I got stuck. On the second page of the paper where he starts to define his invariant he writes: where $u_2 \in H^{10}(\Omega;\mathbf Z_2)$ is the reduction modulo $2$ of $e_2 \in H^{10}(\Omega)$

As I understand the construction cannot be done with coefficients other than $\mathbf Z_2$. But I don't understand why this is the case. On the same page further down he again uses reduction modulo $2$. Why is he forced to consider homology over $\mathbf Z_2$? It is not clear to me why integer coefficients would not work and why the construction is only well defined in $\mathbf Z_2$. I'd be very grateful if you could help me understand his paper by explaining to me why one needs $\mathbf Z_2$.

It is clear to me that homology over $\mathbf Z_2$ is oblivious to orientation of the manifold. However it does not explain why in this particular case other coefficients would make the whole constructions not well defined.

Answer 1

If you read the proof of Lemma 1.2 where he proves his invariant is well-defined, you can get a sense of why $\Bbb Z/2$ coefficients are useful. Using obstruction theory, Kervaire shows that if $f$ and $g$ are two maps such that $f^\ast(e_1) = g^\ast(e_1)$, then $$(f^\ast(u_2) - g^\ast(u_2))[s_{10}] = u_2[h\omega_{10}(f,g)[s_{10}]]$$ for any $10$-simplex $s_{10}$, where $\omega_{10}(f,g) \in C^{10}(K; \pi_{10}(\Omega S^6))$ is the obstruction cocycle, $K$ is a triangulation of $X$, and $h: \pi_{10}(\Omega S^6) \longrightarrow H_{10}(\Omega S^6)$ is the Hurewicz homomorphism.

A result of Serre implies that $u_2 [h \alpha]$ is the mod $2$ Hopf invariant of the element of $\pi_{11}(S^6)$ represented by $\alpha \in \pi_{10}(\Omega S^6)$ (recall that we have an isomorphism $\pi_{k+1}(X) \cong \pi_k(\Omega X)$). Since there are no elements in $\pi_{11}(S^6)$ with odd Hopf invariant, it follows that Kervaire's invariant is well-defined.

If instead we used $\Bbb Z$-coefficients, we would have $$(f^\ast(e_2) - g^\ast(e_2))[s_{10}] = e_2[h\omega_{10}(f,g)[s_{10}]],$$ and again by Serre we have that $e_2[h\alpha]$ is the non-reduced (integer) Hopf invariant of the class in $\pi_{11}(S^6)$ corresponding to $\alpha \in \pi_{10}(\Omega S^6)$. But in this case, for any even integer $m$, one can construct an element of $\pi_{11}(S^6)$ with Hopf invariant $m$. Hence we cannot conclude that the $\Bbb Z$-valued Kervaire invariant is well-defined. Reducing mod $2$ gets rid of this problem.