6
$\begingroup$

Let $O(n)$ be the $n$-th orthogonal group and $O$ be the direct limit of $O(n)$ with respect to $n$. Let $BO(n)$ and $BO$ be the classifying spaces.

Question: Why $BO$ is an $H$-space? My supervisor said "$BO=\Omega^\infty\mathbb{E}$ where $\mathbb{E}$ is a spectrum." What does this mean?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ there's a lot to mention here. Do you know what a spectrum is? If not, it might be good to look up "infinite loop spaces," "extraordinary cohomology theories" or "complex K-theory" to get moving in that direction. $\endgroup$
    – Jonathan Beardsley
    Commented Nov 18, 2015 at 3:20
  • $\begingroup$ Ah, @Qiaochu has given a much more helpful answer. $\endgroup$
    – Jonathan Beardsley
    Commented Nov 18, 2015 at 3:20

3 Answers 3

Reset to default
12
$\begingroup$

$BO$ is the connected component of the zeroth space of a spectrum called the real K-theory spectrum. This spectrum represents a cohomology theory, namely real K-theory, and this means that $BO$ has much more structure than an H-space: it is in fact an infinite loop space, which is loosely a homotopy-theoretic version of an abelian group (as opposed to merely a monoid).

If you believe that homotopy classes of maps $X \to BO$ classify stable real vector bundles (ignoring their dimension) on $X$, then the H-space structure on $BO$ comes from direct sum of vector bundles.

Improve this answer
$\endgroup$
3
  • 8
    $\begingroup$ Details: a spectrum does not have an ``underlying space'', but it does have a zeroth space, which is an infinite loop space, and the zeroth space of the spectrum representing real K-theory is BO \times Z, not BO; otherwise the answer is on the mark. $\endgroup$
    – Peter May
    Commented Nov 18, 2015 at 4:30
  • 2
    $\begingroup$ May I add that this is related to (real) Bott periodicity, which describes the `additive' delooping of $\mathbb{Z}\times BO$, and hence its base point component $BO$. In fact $\mathbb{Z}\times BO$ is a ring space, as studies by May (together with Quinn, Ray and Tornehave) and the other (infinite) lopp space structure comes from tensor product of vector bundles. Please correct me if I am wrong, but I think, Bott's work has appeared before the machinery for stable homotopy and spectra, as we know today, was established, and later on it was interpreted in the language of stable homtopy theory. $\endgroup$
    – user51223
    Commented Nov 18, 2015 at 5:49
  • $\begingroup$ @Peter: yes, I guess "zeroth space" is a better term. I'll edit. $\endgroup$
    – Qiaochu Yuan
    Commented Nov 18, 2015 at 6:10
10
$\begingroup$

$BO$ can be defined as the colimit over $(k,n)$ of Grassmanians $G_k(\Bbb R^n)$ of $k$-dimensional linear subspaces of $\Bbb R^n$ (the limit over $n$ is defined by standard inclusions $\Bbb R^n \subset \Bbb R^{n+1}$, and the limit over $k$ is defined by the operation $X \mapsto X\oplus \Bbb R$ which takes a $k$-plane in $\Bbb R^n$ to the $(k+1)$-plane $X\oplus \Bbb R$ inside $\Bbb R^n \oplus \Bbb R = \Bbb R^{n+1}$.

The direct sum operation defines pairings $$ G_k(\Bbb R^n) \times G_{\ell}(\Bbb R^m) \to G_{k+\ell}(\Bbb R^{n+m}) $$ that are compatible with respect to taking colimits. Taking colimits induces a map $BO \times BO \to BO$ defining the $H$-space structure.

Improve this answer
$\endgroup$
6
$\begingroup$

If you do not know about spectra and just want to know that $BO$ is a monoid (and hence a H-space), instead of the stronger statement that it is an infinite loop space; you can use that $\Omega_0 U/O \simeq BO$. This is part of the Bott periodicity theorem, $U/O$ is known as the Lagrangian Grassmannian.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .