# explicit cocycle representing Stiefel-Whitney class in Milnor and Stasheff

I am trying to do Problem 7A in Characteristic Classes by Milnor and Stasheff which asks the reader to do the following:

Identify explicitly the cocycle $C^r(G_n) \cong H^r(G_n)$ which corresponds to the Stiefel-Whitney class $w_r(\gamma^n)$, where $\gamma^n$ is the universal n-plane bundle over the Grassmanian $G_n$ of n-planes in $\mathbb{R}^\infty$.

The book gives the usual cell structure for $G_n$ using Schubert varieties $\overline{e(\sigma)}$, which are n-planes $X$ that have jumps in dimension at $\sigma_1, \cdots, \sigma_n$ - that is dim($X\cap \mathbb{R}^{\sigma_i}) =i$ and dim($X\cap \mathbb{R}^{\sigma_i-1}) =i-1$. So the answer should be in terms of these cells.

Based on the material in the rest of Chapter 7 in the book, I think the approach should be something like this: Consider the n-plane bundle $\gamma^1 \times \cdots \times \gamma^1$ over $\mathbb{R}P^\infty \times \cdots \times \mathbb{R}P^\infty$. By the universal property of the universal bundle, we have a map $$f: \mathbb{R}P^\infty \times \cdots \times \mathbb{R}P^\infty \mapsto G_n$$ such that $\gamma^1 \times \cdots \times \gamma^1 = f^* \gamma^n$. Also, $H^*(\mathbb{R}P^\infty \times \cdots \times \mathbb{R}P^\infty) = \mathbb{Z}[a_1, \cdots, a_n]$ and under the map $f^*$ the $w_r(\gamma^n)$ goes to the $r$th symmetric polynomial in the $a_i$'s. Also, $f^*$ is known to be injective. So if I can find cochains in $H^r(G_n)$ that map to the symmetric polynomials I would be done. One issue is that I don't know exactly what the map $f^*$ is or even what $f$ is.

I was thinking that the map $f$ could be something like the following map for $n = 2$ and similarly for other cases: take 2 lines $l, l'$ in $\mathbb{R}P^\infty$, pick 2 vectors in $\mathbb{R}^\infty$ with $v\in l, v'\in l'$, then consider vectors in $$\overline{v'} = (v_1, 0, v_2, 0, ...), \overline{v'} = (0, v'_1, 0, v'_2, 0, ...)$$ in $\mathbb{R}^\infty$ which are independent and so span a 2-plane and we say that $(l, l')$ maps to this 2-plane.

So the point of the map is that you need to ensure that if you pick n lines that are not linearly independent, you still get an n-plane. I think such a map is well-defined and maps fibers isomorphically to fibers so that $\gamma^1 \times \cdots \times \gamma^1$ is the pullback of $\gamma^n$ by $f^*$. However, I am still not sure how to solve problem 7A using this and this approach seems a bit too complicated for the problem.

I think this is not the right direction to solve the problem. Recall that we have a map $$\mathbb{RP^{\infty}}\times \cdots \mathbb{RP^{\infty}}\rightarrow G_{n}$$ which is surjective by mapping the element $\langle l_{1},l_{2}\cdots l_{n}\rangle\rightarrow \langle v_{1},v_{2}\cdots v_{n}\rangle$, where $l_{i}=\langle v_{i}\rangle$. Here when $\langle l_{1},l_{2}\cdots l_{n}\rangle$ does not span an $n$-plane we add-up $n-m$ trivial orthogonal vectors in other coordinates to make it an $n$-plane as in problem 6-C. Then the map is surjective and the induced map on cohomology is injective: $$H^{*}(G_{n})\rightarrow H^{*}(\mathbb{RP^{\infty}}\times \cdots \mathbb{RP^{\infty}})$$ But I do not think you can identify the cells explicitly this way, because after all we know $w_{r}(\gamma_{n})\rightarrow w_{r}(\xi)=\sum_{ij\not=ik,j\not=k} a_{i1}a_{i2}\cdots a_{ir}$, we do not know in priori what is the element corresponding to the cup product of $a_{i1}a_{i2}\cdots a_{ir}$ in the space $\mathbb{RP^{\infty}}\times \cdots \mathbb{RP^{\infty}}$. We only know $a_{1},a_{2}$, etc by the pull-back of the bundle $\mathbb{RP}^{1}\rightarrow \mathbb{RP}^{\infty}$, and since the tautological line bundle over $\mathbb{RP^{\infty}}$ only give us the vector over the points in $\mathbb{RP}^{\infty}$, we in fact cannot speak of the cup product $a_{1}a_{2}\cdots a_{r}$ represented by any $r$-dimensional cell in $\mathbb{RP}^{\infty}\times\cdots$.
Here is another approach I think might work. Notice that the restriction homomorphism $$i^{*}:H^{r}(G_{n}(\mathbb{R}^{\infty})\rightarrow H^{r}(G_{n}(\mathbb{R}^{n+k}))$$ is an isomorphism for $r<k$. This is problem 6-B in the book and can be proved relatively easily using techniques in that section. So our real difficulty is to identity the cohomology class of $w_{r}(G_{n}(\mathbb{R}^{n+r+1})$. We know that the $r$-cells in this grassmanian corresponds to parititions such that $$\sum i_{s}=r, s\le n$$
So if we know how to compute the Stiefel-Whitney class of $G_{n}(\mathbb{R}^{n+r+1})$, then we are done. Using problem 6-C, we have an isomorphism which carries the $r$-cell of $G_{n}(\mathbb{R}^{m})$ to $r$-cell of $G_{n+1}(\mathbb{R}^{m+1})$. Therefore we may shift the dimension, and we are only need to identify the $r$-th dimensional cell of $$G_{1}^{r+1}\cong \mathbb{RP}^{r+1}$$ But there is only one by Corollary 6.5 in page 80. I believe this should coincide with the usual cell-decomposition of $\mathbb{RP}^{r+1}$. Reverse the isomorphism we can find an "explicit" description of the cocycle corresponding to $w_{r}(G_{n})$.
• In problem 6-C you only have an embedding of $G_n(\mathbb{R}^m)$ into $G_{n+1}(\mathbb{R}^{m+1})$, and which sends $r$-cells to $r$-cells. However, I think that not all $r$-cells have to be in the image of this embedding (as can be seen by counting $r$-cells). In particular, the induced map in cohomology $H^r(G_{n+1}(\mathbb{R}^{m+1})) \to H^r(G_{n}(\mathbb{R}^{m}))$ is not injective. Therefore, you cannot deduce the Stiefel-Whitney class in $H^r(G_{n+1}(\mathbb{R}^{m+1}))$ from the one in $H^r(G_{n}(\mathbb{R}^{m}))$. Or at least, I don't see how. – Daan Michiels Apr 10 '14 at 16:34