I'm working on problem 7B of Milnor/Stasheff:

Show that the cohomology algebra $H^*\left (G_n\left (\mathbb{R}^{n+k}\right ) \right )$ over $\mathbb{Z}/2$ is generated by the Stiefel-Whitney classes $w_1, \ldots, w_n$ of $\gamma^n$ and the dual classes $\overline{w}_1, \ldots, \overline{w}_k$ subject only to the $n+k$ relations $$ \left (1 + w_1 + \ldots + w_n\right )\left (1 + \overline{w}_1 + \ldots + \overline{w}_k \right ) = 1. $$

I tried looking at this similar question but couldn't really grasp the comment there. Milnor hinted at the beginning that the proof of this is based on a previous problem in which we can show that there is a natural embedding $G_n\left (\mathbb{R}^m\right ) \hookrightarrow G_{n+1}\left (\mathbb{R}^{m+1}\right )$ preserving the Schubert symbol of each cell and also admitting a bundle map $\varepsilon^1 \oplus \gamma^n\left (\mathbb{R}^m\right ) \to \gamma^{n+1}\left (\mathbb{R}^{m+1}\right )$. I kinda gathered that a natural way to prove this would be to use induction, but I'm getting lost in my inductive step. Here's what I have so far:

Base Case

When $n=1$ we have that $G_1\left (\mathbb{R}^{k+1}\right ) = \mathbb{P}^k$ and we already have that the cohomology ring $H^*\left (\mathbb{P}^k\right )$ is the ring of polynomials over generator $a = w_1\left (\gamma_1^1\right )$ with $\mathbb{Z}/2$ coefficients and the one relation $a^{k+1} = 0$. However, computing the dual elements we can easily find that $\overline{w}_i = a^i$ and we can just as easily verify that $(1 + w_1 )\left (1 + \overline{w}_1 + \ldots + \overline{w}_k\right ) = 1$.

Inductive Step

Assume the statement is true for $n$, hence $H^*\left (G_n\left (\mathbb{R}^{n+k}\right )\right )$ is generated by the SW classes $w_1, \ldots, w_n$ and the dual classes $\overline{w}_1, \ldots, \overline{w}_k$ and the one relation given above. We know that by the embedding indicated above that for $i \leq n$ we have $w_i\left (G_n\mathbb{R}^{n+k}\right ) = w_i\left (G_{n+1}\mathbb{R}^{n+1+k}\right )$ since each SW class is given explicitly by the dual of the cell represented by the Schubert symbol of all 1's, and the embedding preserves this symbol.

Beyond this is where I'm starting to get lost. Based on the CW structure of the Grassmannian, I'm gathering that $H^i\left (G_{n+1}\mathbb{R}^{n+1+k}\right ) = H^i\left (G_n\mathbb{R}^{n+k}\right )$ for all $i \leq n$ and the one issue is proving that $H^{n+1}\left (G_{n+1}\mathbb{R}^{n+1+k}\right )$ can be generated by a combination of $w_1, \ldots, w_{n+1}, \overline{w}_1, \ldots, \overline{w}_k$. The number of $(n+1)$-cells in $G_{n+1}\mathbb{R}^{n+1+k}$ is the partition of $n+1$ into at most $n+1$ integers each of which is $\leq k$, hence if we look at cup products of the form

$$ w_1^{r_1}w_2^{r_2}\ldots w_{n+1}^{r_{n+1}} $$

we would require that each $r_i \leq k$. I'm not sure exactly where to take this counting argument, but I also know that in computing the desired relation, we have that

\begin{eqnarray*} (1+w_1 + \ldots + w_{n+1}) \left (1 + \overline{w}_1 + \ldots + \overline{w}_k\right ) & = & (1+w_1 + \ldots + w_n)\left (1 + \overline{w}_1 + \ldots + \overline{w}_k\right ) + w_{n+1}\left (1 + \overline{w}_1 + \ldots + \overline{w}_k\right ) \\ & = & 1 + w_{n+1}\left (1 + \overline{w}_1 + \ldots + \overline{w}_k\right ) \end{eqnarray*}

hence we would like to show that $\sum_{i=0}^k w_{n+1}\overline{w}_i = 0$ somehow.

I'm very new to cohomology theory in general and I'm struggling quite a bit wrap my mind around this problem. I'd really like whatever feedback I can get on my proof so far, whether I'm close or not, or if there's something fundamentally wrong in my understanding/approach. Additionally, if I'm on the right track and fairly close to a solution, I'd like some suggestions and gentle nudges; not necessarily a full complete solution.


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