Classifying map

Question

Let $\xi=(E,p,B)$ a principal $G$-bundle and $\eta=(P,\pi,Q)$ a real vector bundle such that $\operatorname{rank}(\eta)=n$. We can consider a classificant space $BG$. What is the classifying map $f:X \rightarrow BG$? Why the name "classifying"? How can build classifying map for $\eta$?

Answer 1

Given any principal $G$-bundle $E' \to X'$, and a map $X \to X'$, the pullback $E' \times_X' X \to X$ is a map that's easily seen to be a principal $G$-bundle. The idea of the classifying space is that it's a space $BG$, with a principal $G$-bundle $EG \to BG$, such that any principal $G$-bundle over any space $X$ is a pullback of this bundle. Thus, the principal $G$-bundles over $X$ (up to isomorphism) are in bijective correspondence with the maps $X \to BG$ (up to homotopy). That is, maps to $BG$ 'classify' principal $G$-bundles.

In many cases, once you prove that such a space exists (which isn't at all obvious), you don't pay much attention to the explicit construction of classifying maps. Indeed, part of the point of the thing is to replace a somewhat opaque theory -- the study of vector bundles and principal bundles -- with a theory which has more machinery developed for it -- the study of homotopy classes of maps. This is just my opinion, though -- your mileage may vary.

One way to construct $BG$ is to let $EG$ be any contractible space with a free $G$-action, and $BG = EG/G$. Clearly $EG \to BG$ is a principal $G$-bundle, and the contractibility of and freeness of the $G$-action on $EG$ implies that there are no obstructions to defining a $G$-map from a principal $G$-bundle to $EG$, which descends to a map $X \to BG$. For more details, see e.g. these notes.

Now, this answers your questions about principal $G$-bundles. The secret to the last question is that real vector bundles really are principal bundles as well -- they're principal $O(n)$-bundles! To turn a vector bundle $E \to X$ into a principal $O(n)$-bundle, replace it by the frame bundle $\mathrm{Fr}(E) \to X$, which is a bundle whose fiber over a point $x \in X$ is the set of orthonormal bases for $E_x$. (You have to pick a metric on the vector bundle first, but this always exists if $X$ is, for example, paracompact.) Conversely, given a principal $O(n)$-bundle $P \to X$, you can replace it by the bundle whose fibers are $P_x \times \mathbb{R}^n$ mod its diagonal $O(n)$-action, which is a real vector bundle. These mappings are mutually inverse, so when we talk about real vector bundles, we might as well be talking about principal $O(n)$-bundles.

So real rank $n$ vector bundles are classified by $BO(n)$, which there actually is a nice geometric construction of. Let $EO(n)$ be the space of sequences of $n$ orthonormal vectors in an infinite-dimensional inner product space. This clearly has a free $O(n)$-action, and you can check that it's contractible. The quotient $EO(n)/O(n)$ is the space of $n$-dimensional planes (we've forgotten their bases) in this infinite-dimensional space, also called the Grassmannian $\mathrm{Gr}_n(\mathbb{R}^\infty)$. In particular, one-dimensional vector bundles are classified by $\mathrm{Gr}_1(\mathbb{R}^\infty) = \mathbb{R}P^\infty$.

Similar remarks apply to other other sorts of vector bundles. For example, complex rank $n$ vector bundles are the same as principal $U(n)$-bundles, and these are classified by the Grassmannian of copies of $\mathbb{C}^n$ in $\mathbb{C} P^\infty$.