I am trying to make Exercise 19 of these lecture notes.
I'll briefly summarize the question below.
Exercise Let $V$ be a dinite dimensional vector space and $A \in \mathrm{End}(V)$. We consider
$$
A^* \in \mathrm{End}(V^*), \quad \text{given by } A^*(\xi)(v) = \xi(A(v)).
$$
We keep the same notation for endomorphisms of vector bundles.
Show that the curvature of $\nabla^*$ is given by
$$
k_{\nabla^*}(X,Y) = k_{\nabla}(X,Y)^*.
$$
The lecture notes use the following definition of curvature:
Definition For any connection $\nabla$ on a vector bundle $E \to M$, the curvature is given by
$$
k_{\nabla}(X,Y)s = \nabla_X \nabla_Y(s) – \nabla_Y \nabla_X(s) – \nabla_{[X,Y]},
$$
where $X,Y \in \mathfrak{X}(M)$, $s \in \Gamma(E)$ and $[\cdot, \cdot]$ denotes the Lie bracket.
So it is viewed as a map $\mathfrak{X}(M) \times \mathfrak{X}(M) \times \Gamma(E) \to \Gamma(E)$ (or $k_{\nabla} \in \Gamma(\Lambda^2 T^*M \otimes \mathrm{End}(E)) = \Omega^2(M; \mathrm{End}(E)).)$.
Also, the induced connection $\nabla^*$ on $E^*$ is characterized by the equation
$$
L_X(\xi(s)) = \nabla_X^*(\xi)(s) + \xi(\nabla_X(s)).
$$
I have attempted this exercise by writing both sides out, but it gets very messy very quickly. Is there a smarter way to tackle this question, or is it a matter of really diligently writing everything out and hoping for the best? If that's the case, I am afraid I will have more questions, but that's a worry for later 😉
Thanks in advance for the advice or help! 🙂
Best Answer
Here's the core step of the computation:
$$ \nabla_X^\ast \nabla_Y^\ast (\xi)(s) = \nabla_X^\ast (\nabla_Y^\ast (\xi))(s) = L_X (( \nabla_y^\ast \xi)(s)) - (\nabla_Y^\ast)(\nabla_Xs) \\ = L_X(L_Y\xi(s) - \xi(\nabla_Ys))-L_Y\xi(\nabla_X s)+\xi(\nabla_Y\nabla_Xs) \\= L_X L_Y\xi(s) - L_X \xi(\nabla_Ys)-L_Y\xi(\nabla_X s)+\xi(\nabla_Y\nabla_Xs) .$$
Can you take it from there?