Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 8 down vote accepted

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.

share|cite|improve this answer
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. – Dave Apr 27 '13 at 15:48
@DaveHartman: thanks for the catch. I've edited my answer accordingly. – Eric O. Korman Apr 27 '13 at 17:30
@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? – Bilateral Sep 17 '15 at 9:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.