I have a question about an argument used in the proof the red tagged theorem below from A Concise Course in Algebraic Topology by P. May at page 220. Here the excerpt:
We write $\mathcal{N}_n$ for the set of cobordism classes of smooth closed $n$-manifolds where two classes $[N], [M] \in \mathcal{N}_n$ are equivalent if there exist a $n+1$ manifold $W$ with boundary such that $\partial W = N \coprod M$. We give $\mathcal{N}_*$ a ringstructure with "disjoint union" as addition and "product" $- \times -$ as multiplication.
Since $∂(M × I) = M ∐M$, the ring $\mathcal{N}_*$ becomes a $\mathbb{Z}/2$-graded ring.
Below Theorem May introduced explicitely the generators $u_n$ of $\mathcal{N}_n$ in each degree explicitely.
Using the fact that a manifold is a boundary if and only if its normal Stiefel-Whitney numbers are zero we deduce that $u_{2i}= [\mathbb{RP}^{2i}]$ are not boundaries and therefore non trivial classes and therefore generators of $\mathcal{N}_{2i}$.
On the other hand for odd numbers $\mathbb{RP}^{2i-1}$ represent zero classes so can't be generators since it's all Stiefel Whitney classes and therefore also numbers vanish.
For odd numbers not of the shape $2^s-1$ May suggests to take as generators of $\mathcal{N}_{2i-1}$ the hyperplanes as representants $H_{2^{p+1}q,2^p} \subset \mathbb{RP}^{2^{p+1}q} \times \mathbb{RP}^{2^p}$ as described in text.
My QUESTION is why these hypersurfaces are good choises? So why the $H_{2^{p+1}q,2^p} $ aren't boundaries / respectively why their Stiefel Whitney classes aren't trivial?
My ideas:
I think that the key lies in composition of following maps:
$$H_{2^{p+1}q,2^p} \hookrightarrow \mathbb{RP}^{2^{p+1}q} \times \mathbb{RP}^{2^p} \xrightarrow{pr_1} \mathbb{RP}^{2^{p+1}q}$$
Does it provide a relevant statement about Stiefel Whitney classes of $H_{2^{p+1}q,2^p}$?
Consider the map $(BO(1))^{\times n} \to BO(n)$ classifying the sum of $n$ line bundles. In mod 2 cohomology, this identifies $$H^* BO(n) \xrightarrow{\cong} (H^*(BO(1))^{\otimes n})^{\Sigma_n} \cong \mathbb{F}_2[t_1, \ldots, t_n]^{\Sigma_n}.$$ Let $s_n = \sum_i t_i^n$. This is a symmetric polynomial and so determines a characteristic class $s_n = s_n(w_1, \ldots, w_n)$ in $H^n BO$.
Thom proved the following:
Let $H = H_{a,b}$ and $n = a + b - 1$. Our strategy is to show that $\langle s_n(\nu_H), [H] \rangle$ is nonzero for particular values of $a$ and $b$. This will prove that $[H]$ is a generator of the cobordism ring in dimension $n$. I believe that this example is due to Milnor.
Since $H$ is the zero locus in $\mathbb{R}P^a \times \mathbb{R}P^b$ of a bilinear form, $\nu_H \cong i^*(\gamma_1 \otimes \gamma_2)$, where $\gamma_1$ (resp. $\gamma_2$) is the tautological line bundle on $\mathbb{R}P^a$ (resp. $\mathbb{R}P^b$) and $i: H \hookrightarrow \mathbb{R}P^a \times \mathbb{R}P^b$ is the inclusion. Write $t_1 = w_1(\gamma_1)$ and $t_2 = w_1(\gamma_2)$. Then, \begin{align} \langle s_n(\nu_H), [H] \rangle &= \langle i^*(t_1 + t_2)^n, [H] \rangle \\ &= \langle (t_1 + t_2)^{n+1}, [\mathbb{R}P^a \times \mathbb{R}P^b] \rangle \\ &= \binom{n+1}{b}. \end{align}
In the case $a = 2^{p+1} q$, $b = 2^p$, and $n = 2^p(2q + 1) - 1$, we are reduced to computing the binomial coefficient $\binom{2^p(2q + 1)}{2^p}$. It is not hard to see that this is odd, so we are done.