3
$\begingroup$

Let $E$ be a smooth principal $G$-bundle on M.

The vertical bundle $V$ is defined as $V=\ker(d\pi:TE\to \pi^*TM)$. An Ehresmann connection on $E$ is a smooth subbundle $H$ of $TE$ (also called the horizontal bundle), such that $TE=H\oplus V$.

Now let $H$ be invariant with respect to the $G$ action on $E$, so $H_{eg}=d(R_g)_e(H_e)$ for all $e\in E$ and all $f \in G$, where $d(R_g)_e$ is the differential of the right action of $g$ at $e$.

The Ehresmann connection, $H$ should be equal to a 1-form $\omega$ on $E$ with values in the Lie algebra $\mathfrak{g}$ of $G$. Can anyone provide me any insight as to why this is true and how one constructs this 1-form from the Ehresmann connection? I have poor knowledge of differential geometry.

$\endgroup$
4
$\begingroup$

Not quite right. The horizontal subspace, $H_e$, will be given by $\ker\omega(e)$.

For the direction you didn't ask for, the transformation properties of $\omega$ under $R_g$ (which you didn't state) will give you the invariance of $H$. For the direction you asked for, note that for any $\xi\in\mathfrak g$ you get a vector field tangent to $V$ by taking $\xi(e)=\frac{d}{dt}\big|_{t=0}R_{\exp t\xi}(e)$. Now define $\omega$ by specifying its values on $T_e E = H_e \oplus V_e$ by $\omega(e)(y) = 0$ when $y\in H_e$ and $\omega(e)(y) = \xi\in\mathfrak g$ when $y=\xi(e)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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