# Does every real vector bundle admit a metric?

Let $E \longrightarrow B$ be a vector bundle.

I know that if B is paracompact, the bundle admits a metric.
My question is the following: is this true for any B?

I also know that a reduction of the structure group from $GL(n,R)$ to $O(n)$ is possible but I don't know about the conditions on B.
It seems to me that if it were true for any B, then any vector bundle would admit a metric. This seems strange

Here are two links that have caused confusion:
http://en.wikipedia.org/wiki/Reduction_of_the_structure_group
http://www.math.washington.edu/~julia/Quillen/prin.pdf
p.19 "In other words, every real vector bundle admits a Euclidean metric. Similarly, every GLnC-bundle is induced from a U(n)-bundle; equivalently, every complex vector bundle admits a Hermitian metric".

Thank you

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Your second reference says p. 18: "We assume for the rest of this section that the base space B is a CW-complex" (hence paracompact, as noted on p. 12). –  Francois Ziegler Jun 28 '13 at 21:58
In the special case that $E$ is the tangent space on $B$, then paracompactness of $B$ is equivalent to the existence of a metric: ams.org/journals/proc/1969-020-02/S0002-9939-1969-0236876-3/… –  Otis Chodosh Jun 28 '13 at 22:00
The standard definition of vector bundles (or more general fibre bundles) is not really the right definition if $B$ is not paracompact. You should instead take as part of the definition that there is a trivialising cover that is numerable; otherwise, very little of the theory will work properly. With that additional assumption, every vector bundle admits a metric. –  Neil Strickland Jun 28 '13 at 22:05
Hello, thank you for answers. @NeilStrickland can you please elaborate or give me references ? –  Lepanais Jun 29 '13 at 12:51