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?
$BO$ 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.
$BO$ 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.
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.
