Am I right in thinking that the structure group of a fibre bundle is any group $G$ of homeomorphisms of the fibre $F$ such that all transition functions map into $G$? Or is $G$ somehow the minimal such group, for all possible trivialisations?

Another way of phrasing the question: am I correct in thinking that there are potentially many $G$-bundles which are the same as fibre bundles?


Yes, the structure group is not unique. For example, a vector bundle $E$ has, by definition, a structure group $GL(n)$. Some additional structures on $E$ are equivalent to reductions of the structure group to a subgroup of $GL(n)$ and (even if they exist) you may not want to specify these additional structures and thus may not care about getting a smaller structure group.

For example $E$ will always admit a metric (at least if the base is nice like a manifold) and specifying a metric is equivalent to giving a reduction of the structure group to $O(n) < GL(n)$ (given a metric you consider only orthonormal local frames, which shows that the transition functions take values in $O(n)$ and conversely given a reduction to $O(n)$ you take an $O(n)$ trivialization of the bundle and define a metric by making those frames orthonormal). If you don't care about using a metric then you may not want to reduce the structure group (which requires a choice and so is not natural).

Another example is if the bundle is trivializable then its structure group is the trivial group. Again, there may not be a natural trivialization so you may not want to think of the structure group as trivial.

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    $\begingroup$ Since any real vector bundle $E$ of rank $n$ always admits a metric its structure group can always be reduced to $O(n)$. The structure group of $E$ can be reduced to $SO(n)$ iff the vector bundle $E$ is orientable. $\endgroup$ – Dave Apr 27 '13 at 15:48
  • $\begingroup$ @DaveHartman: thanks for the catch. I've edited my answer accordingly. $\endgroup$ – Eric O. Korman Apr 27 '13 at 17:30
  • $\begingroup$ @EricO.Korman: Let $P\to M$ be a principal $G$-bundle over $M$. Suppose that $P$ admits a reduction to a subgroup $H\subset G$. Then $P$ is a $H$ bundle over $P/H$. How can this bundle be equivalent to the first one if the dimension of the bases is not the same? $\endgroup$ – Bilateral Sep 17 '15 at 9:01

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