# Definition of Reduction of a Structure Group Implies Triviality?

Let $$\pi:E\rightarrow B$$ be a (smooth) principle bundle with structure group $$G$$.

By definition, there exists a reduction of the structure group to a subgroup $$H, if there exists a global section $$s:B\rightarrow E/H$$.

Now, clearly, if $$G$$ acts freely and fiber-preserving on $$E$$, then so does any subgroup $$H. And if $$G$$ acts transitively on $$E$$, so does $$H$$ on $$E/H$$... so $$E/H\rightarrow B$$ is a principle $$H$$-bundle, right?

But then, there is the following fact:

If a principle bundle admits a global section, then it is already trivial.

Question So doesn't the definition of a reduction imply that $$E/H\rightarrow B$$ is a trivial bundle?

• A good example to work out is the frame bundle to $S^2$ and it's reduction to the orthogonal frame bundle (aka the unit tangent bundle to the sphere). Commented May 22 at 23:18

$$E/H \to B$$ is a fiber bundle with fiber $$G/H$$, it is not a principal $$H$$-bundle (note, $$H$$ acts trivially on $$E/H$$). Therefore, the existence of a section does not imply that the bundle is trivial.