# How to push up / lift a connection one-form from the base manifold to the total 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 prinicipal 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 tanget 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 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 $$E$$.

• Do you know the book 'foundations of differential Geometry' by Kobayashi & Nomizu? It is a very good introduction to all the maths behind Riemannian manifolds, fibre, principle and frame bundles. Commented Feb 23, 2023 at 16:24
• Thank you for the recommendation, any hint to the answr for this question?
– VVM
Commented Feb 23, 2023 at 16:33
• What do you mean by "being invariant under the action of the structure group"? Note that when you pull back by different sections $s_\alpha$ and $s_\beta$, you should get something like $s_\beta^*\omega = g^{-1}(s_\alpha^*\omega) g + g^{-1}dg$, where $s_\beta = g\cdot s_\alpha$. Commented Feb 23, 2023 at 21:50
• Thank you @TedShifrin, well the question is that how to push forward a form via a section to the frame bundle!
– VVM
Commented Mar 9, 2023 at 19:09

You don't push forward forms, of course. Here's the idea: Cover $$M$$ by open sets $$U_\alpha$$ over which the frame bundle $$P$$ is trivial, and let $$s_\alpha\colon U_\alpha\to P$$ be sections for all $$\alpha$$. Define the projection $$\psi_\alpha\colon P|_{U_\alpha}\to G$$ for all $$\alpha$$. Provided that we have $$\omega_\beta = g_{\alpha\beta}^{-1}\omega_\alpha g_{\alpha\beta} + g_{\alpha\beta}^{-1}dg_{\alpha\beta} \quad\text{with } s_\beta=g_{\alpha\beta}\cdot s_\alpha \text{ on } U_\alpha\cap U_\beta,$$ then we define the $$\mathfrak g$$-valued $$1$$-form $$\omega$$ on $$P$$ by $$\omega = (\text{Ad}\,\psi_\alpha^{-1})\pi^*\omega_\alpha + \psi_\alpha^*\phi,$$ where $$\phi$$ is the left-invariant Maurer-Cartan form on $$G$$. I leave you to check well-definedness. You can find this all done carefully in Kobayashi-Nomizu (section 1 of Chapter II).
• Well, no, I meant that this formula gives a globally-defined connection form on $P$ (with the usual properties), so that when $U_\alpha$ and $U_\beta$ overlap, the two formulas agree (this is largely the purpose of the transformation law we have to assume in the first place). Commented Mar 12, 2023 at 0:15