# Show that an Ehresmann connection on a principal G bundle is equivalent to a Lie Algebra Valued one form.

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.

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 $y=\xi(e)$.