total space of the bundle of orthonormal frames over a compact manifold is compact -- how to see it?

Question

I have seen this statement on several occasions but could not seem to be able to figure it out by myself. Surely the fibers are all compact but how does the compactness of the base (riemannian) manifold imply the compactness of the total space of the bundle? Could someone please enlighten me with greater detail? Thanks!

Answer 1

Let $M$ be a compact manifold, and $U_i$ a trivialisation of $O(M)$ the bundle of ortogonal frames and $F$ its typical fibre. For every $x\in M$, there exists $i_x$ such that $x\in U_{i_x}$, there exists an open neighborhood $V_x\subset U_{i_x}$ whose adherence is compact and contained in $U_x$, since $M$ is locally compact, you can extract a finite family $V_{x_1},...,V_{x_n}$ which covers $M$. Let $W_{x_i}$ the adherence of $V_{x_i}$, the trivialisation induces a continuous map $f_i:W_{x_i}\times F\rightarrow U_{x_i}\times F\rightarrow O(M)$,

Let $N$ be the disjoint union of $W_{x_i}\times F$, it is compact since it is a finite union of compact sets. Let $f:N\rightarrow O(M)$ whose restriction to $W_{x_i}\times F$ is $f_i$, $f$ is continuous and surjective, this implies that $O(M)$, the image of $f$ is compact, since the image of a compact set by a continuous map is compact.

Answer 2

Tsemo Aristide has given a correct proof, but here's another, using a different view of compactness, namely that every ultrafilter converges. Suppose we have a fiber bundle $E\to B$ with compact base $B$ and compact fiber $F$. To show that the total space $E$ is compact, consider an arbitrary ultrafilter $U$ on $E$. Its projection to $B$ is an ultrafilter on $B$, so it converges to some $b\in B$. By definition of fiber bundle, $b$ has a neighborhood $N$ in $B$ whose pre-image in $E$ is homeomorphic to $N\times F$, with the projection of $E$ and the product projection $N\times F\to N$ agreeing on $N\times F$. Since the projection of $U$ concentrates on $N$, $U$ itself concentrates on the pre-image of $N$, which I can identify with $N\times F$. So $U$ has a projection to $F$, which converges to some $f\in F$ because $F$ is compact. Then $U$ converges to $(b,f)\in N\times F$ (by definition of the product topology) and therefore converges to the corresponding point in (the pre-image of $N$ in) $E$.