# $H^i\left(BO_m ; \mathbb{Z}_2\right) \cong H^i\left(B O_n ; \mathbb{Z}_2\right)$ for $i \leq m$ and $i \leq n.$

The notes I am reading say that groups in the title are isomorphic. Could someone explain to me why it is the case?

Here by $$BO_n$$ I mean the infinite Grassmann manifold of $$n$$-dimensional subspaces. It seems that it should somehow follow from the general theorem that $$H^*(B O_r ; \mathbb{Z}_2) \cong \mathbb{Z}_2\left[w_1, \ldots, w_r\right],$$ where $$w_1,\ldots, w_r$$ are the Stiefel-Whitney classes, however I don't really get it.

Thanks.

• There is a fiber sequence $S^n\rightarrow BO(n)\rightarrow BO(n+1)$, hence the inclusion $BO(n)\rightarrow BO(n+1)$ is $n$-connected, which almost implies the result without doing any specific computation, but I'm not sure how to get injectivity in degree $n$ without a computation. Apr 13 at 13:28

As you state we have $$H^*(BO_r; \mathbb{F}_2) \cong \mathbb{F}_2[w_1, \ldots, w_r]$$, but you neglect to mention that $$|w_i| = i$$; with this the result is immediate since $$\operatorname{rank}((\mathbb{F}_2[w_1, \ldots, w_r])_i) = \operatorname{rank}((\mathbb{F}_2[w_1, \ldots, w_i])_i)$$ for all $$i \leq r$$.
• Thank you! I was not sure about the grading on $\mathbb{F}_2\left[w_1, \ldots, w_r\right].$ By $(\mathbb{F}_2[w_1, \ldots, w_r])_i$ you mean the $i$th graded component of $\mathbb{F}_2[w_1, \ldots, w_r],$ right? In that case, isn't it true that $(\mathbb{F}_2[w_1, \ldots, w_r])_i=(\mathbb{F}_2[w_1, \ldots, w_i])_i$ for $i \leq r$? Apr 13 at 14:16
• @Haldot You're welcome! Yes, I mean the $i$th graded component. For the last point, also yes. Not sure why I only wrote $\operatorname{rank}$ there, but either way the claim follows :) Apr 13 at 14:43