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Is there a proof that $BO(k)$ is not of the homotopy type of a finite dimensional complex?

The Grassmannian $BO(k) := \{ k\text{-dim subspaces of } \mathbb{R}^\infty \}$ classifies the $k$-dimensional vectorbundles on a $CW$ complex $X$ as: $Vect^k(X) \cong [X,BO(k)]$.

$BO(k)$ can be constructed as the direct limit of finite dimensional Grassmann manifolds $G_k(\mathbb{R}^n)$, but it is not finite dimensional itself. Is there a proof that there cannot be a better (=finite-dimensional) construction ?

My idea was showing that the cohomology of $BO(k)$ is not bounded in dimension. This works for $BO$ as follows: There are bundles with non-vanishing $k$th Stiefel-Whitney $\omega_k$ class for all $k$. (Take for example the $k$th product $\gamma_1 \times \dots \times \gamma_1$ of the tautological line bundle $\gamma_1$ over the real projective space $\mathbb{R}P^1$.) By naturality $\omega_k$ of the classifying bundle over $BO(k)$ cannot vanish and so $H^k(BO(k),\mathbb{Z}/2)$ is non-trivial. Then it should follow that $BO= colim_\to BO(k)$ cannot be finite-dimensional.

However, this idea does not work for $BO(k)$ because all characteristic classes I know "live" in the cohomology of dimension atmost rank of the bundle. Are there unstable classes in higher degrees, or is there some other way to proof the result?

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We have $H^*(BO(k); \mathbb{Z}_2) \cong \mathbb{Z}_2[w_1, \dots, w_k]$ where $\deg w_i = i$. In particular, $H^n(BO(k); \mathbb{Z}_2) \neq 0$ for every $n$ as $w_1^n$ is a non-zero element. Therefore $BO(k)$ cannot be homotopy equivalent to a finite-dimensional CW complex.

That is, the degrees of the usual choice of generators of the $\mathbb{Z}_2$ cohomology ring of $BO(k)$ are bounded by $k$, but they generate a graded ring which is non-zero in each degree.

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  • $\begingroup$ Thank you a lot! But I'm still confused about one point: Is there an easy example of a $k$-dimensional bundle for which $w_1^n$ does not vanish? I think Milnor and Stasheff use $\gamma_1 \times \dots \times \gamma_1$ to prove that there are no algebraic relations between $w_1, \dots w_k$ but there is no classifying map from this bundle to $G_k$, is there ? $\endgroup$
    – Jan Steinebrunner
    Commented Aug 8, 2016 at 10:21
  • $\begingroup$ Ah, I got it now: I have to take $\gamma_1$ the tautological linebundle over $G_1 = \mathbb{R}P^\infty$ and not $\mathbb{R}P^1$. Then $w_1$ is the generator of $H^*(\mathbb{R}P^\infty, \mathbb{Z}/2) \cong \mathbb{Z}/2[w_1]$ and therefore $w_1^n$ does not vanish. This is of cause just what you wrote for $k=1$, but I somehow did not want to use that we know the cohomology of $BO(k)$ for all $k$. $\endgroup$
    – Jan Steinebrunner
    Commented Aug 8, 2016 at 10:33

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