I am currently trying to make Exercise 20 of these lecture notes. It should be a very simple exercise, but I am swamped due to the amount of bookkeeping. I fear I am missing a very simple nuance that makes it impossible for me to solve this problem. I'll briefly mention the exercise below with the relevant notes from the lecture notes.
The Exercise
Exercise 20. Let $E \to M$ be a vector bundle equipped with the connection $\nabla$. Let $\gamma = \gamma(t,s) \colon I_1 \times I_2 \to M$ be a smooth map defined on the product of two intervals $I_1, I_2 \subset \mathbb{R}^2$ and $u = u(t,s) \colon I_1 \times I_2 \to E$ a smooth covering. Show that $$ \frac{\nabla^2 u}{ds \, dt} - \frac{\nabla^2 u}{ds \, dt} = k_\nabla \left( \frac{d\gamma}{ds}, \frac{d\gamma}{dt} \right)(u), $$ where $k_\nabla$ in the induced curvature.
Clarifying the notation
Let me elucidate the notation $\frac\nabla{dt}$ and $\frac\nabla{ds}$. If $\gamma \colon I \to M$ is a path, then we can use sections $s \in \Gamma(E)$ to produce smooth coverings $u = s \circ \gamma \colon I \to E$. This implies that the notation $$ \nabla_{\dot \gamma}(s)(\gamma(t)) $$ makes sense as you can view $\dot \gamma$ as a sort of vector field. We denote this as $$ \frac{\nabla(s \circ \gamma)}{dt}. $$ So more generally, we can define the operator $\frac{\nabla}{dt}$ to define a new path for smooth coverings of $\gamma$ $$ \frac{\nabla u}{dt} \colon I \to E. $$ Locally with respect to a frame $e = (e_j)_j$, we may write $u(t) = \sum_j u^je_j(\gamma(t))$. Then using formula (1.18) from these lecture notes, one may write the $i$'th component of $\frac{\nabla u}{dt}$ with respect to this frame as $$ \left( \frac{\nabla u}{dt} \right)^i = \frac{d u^i}{dt}(t) + \sum_j(t) \omega^i_j(\dot \gamma(t)). $$ Here $\omega = (\omega^i_j)$ represents the connection matrix, representing the one-forms that you obtain from the equation $\nabla_X(e_j) = \sum_i \omega_i^j(X) e_i$.
My strategy, my work and what I find hard
For this exercise I had the following strategy in mind. I was planning to use the last formula twice: twice to calculate $\frac{\nabla^2 u}{ds \, dt}$ and twice to calculate $\frac{\nabla^2 u}{dt \, ds}$. From there I hoped it would resemble the local form of the curvature.
What I find hard is the following:
- By applying that formula, you get a lot of terms, and I guess that my bookkeeping skills can't keep up at the moment. Maybe I am just going about it the wrong way. Here is my work until now.
- I also found it hard to properly express the local form of the curvature in terms of the connection: $k = d \omega + \omega \wedge \omega$, where $(\omega \wedge \omega)^i_j = \sum_k \omega^i_k \wedge \omega^k_j$ is the local connection matrix. I was never able to prove this (maybe I should open a separate thread for this, but so be it).
So am I missing an important nuance? Is this the right strategy? It this just really a matter of bookkeeping and am I just messing up?
Thanks in advance for the feedback!