How to derive connection Lie algebra valued one-form on the frame bundle if given the pulled back of it on the physical space?

I am following this YouTube lecture by Schuller where he finds the appropriate formalism for the quantum mechanics in the physical curved space.

Everything makes sense to me but at the very end I see that we find the pull backed connection one-form on the base manifold.

He says to the end of the lecture that later on we will see how we can find the connection one-form on the total space, the principal bundle, i.e., the frame bundle, level.

This is not shown, however, in the next video.

I wonder how can we calculate the connection one-form on the total space when we only have information about the pulled back of this form on the base manfiold through a section map?

If we have a tangent vector on the base manifold clearly we can push it forward via the section map. But how about when we have defined a form, as we cannot push it forward from the base manifold to the total space via the section map.

Is the following a good way to illustrate in the case of a flat base space how to derive the connection one-form on the total space from the given pulled-back connection one-form on the base manifold?

Consider the trivial bundle $$\mathbb{R}^2 \times G \to \mathbb{R}^2$$, where $$G$$ is a Lie group, and let $$E$$ denote the principal G-bundle with fiber $$G$$. We can think of $$E$$ as the bundle whose fiber over each point in $$\mathbb{R}^2$$ is just $$G$$.

Suppose we have a connection one-form $$\widetilde \omega$$ on the base manifold $$\mathbb{R}^2$$ that takes values in the Lie algebra of $$G$$. This connection one-form is given by a one-form on $$\mathbb{R}^2$$ whose values are elements of the Lie algebra of $$G$$ at each point.

Now suppose we have a section $$s:\mathbb{R}^2 \to E$$ of the principal $$G$$-bundle $$E$$. We can think of $$s$$ as a map that assigns to each point in $$\mathbb{R}^2$$ an element of $$G$$. We can also think of $$s$$ as a map that takes a point in $$\mathbb{R}^2$$ and "lifts" it to a point in $$E$$ by taking the point in $$\mathbb{R}^2$$ and mapping it to the point in the fiber over that point determined by $$s$$.

To derive the connection one-form on the total space, we want to "lift" the pulled-back connection one-form on the base manifold to $$E$$. We can do this as follows:

1. Given a point $$p \in E$$, we can use the section map $$s$$ to identify the point $$q=s^{-1}(p)$$ in $$\mathbb{R}^2$$.

2. We can then evaluate the pulled-back connection one-form $$\widetilde{\omega} = s^* \omega$$ on $$\mathbb{R}^2$$ at the point $$q$$. Since $$\widetilde{\omega}$$ takes values in the Lie algebra of $$G$$, this evaluation gives us an element of the Lie algebra of $$G$$.

3. ** We can then "lift" this element to a Lie algebra-valued one-form on $$E$$ by extending it trivially in the direction transverse to the fiber. Specifically, for any vector $$v$$ tangent to $$E$$ at $$p$$, we can define the value of the lifted connection one-form at $$p$$ in the direction of $$v$$ to be the element of the Lie algebra of $$G$$ we obtained in step 2.**

By repeating this process for all points in $$E$$, we obtain a connection one-form on $$E$$ that reduces to $$\omega$$ when restricted to any fiber of $$E$$. This connection one-form satisfies all the conditions required of a connection and can be used to define parallel transport and curvature on $$E$$.

• Which minutes? Where does Schuller introduce a group $G$? Commented Mar 4, 2023 at 17:39
• See: en.wikipedia.org/wiki/Cartan_connection and Contemporary Geometry: Method and Applications Part 2 Doubrovin, Novikov, Fomenko, Chapter 25 Commented Mar 4, 2023 at 21:34
• @Qmechanic at minute 34, the group on the framde bundle is GL(d, R).
– VVM
Commented Mar 4, 2023 at 21:42
• You should probably change the title; as the question stands (v3) it is fundamentally about diff geo and the "quantum mechanics in curved physical space" bit only really distracts from that Commented Mar 5, 2023 at 22:18
• @NiharKarve good suggestion now updated
– VVM
Commented Mar 5, 2023 at 22:29