# Proving that the induced connection on the dual bundle is indeed a connection

I have a hard time proving that the Leibniz property holds of the induced connection on the dual bundle. Let me first outline some definitions.

I will first give the definition of a connection on a vector bundle $$E \to M$$, after which I'll show the definition of the induced connection on the dual bundle $$E^{*} \to M$$. I will then show my work and at which part I am stuck.

## Definitions

Definition 1. A connection on the vector bundle $$E \to M$$ is a bilinear map $$\nabla$$ $$\mathfrak{X}(M) \times \Gamma(E) \to \Gamma(E),$$ satisfying $$\nabla_{fX}(s) = f \nabla_X(s), \qquad \nabla_X(fs) = f \nabla_{X}(s) + X(f)s.$$ for all $$f \in C^{\infty}(M), X \in \mathfrak{X}(M), s \in \Gamma(E)$$.

Definition 2. Let $$E^{*} \to M$$ be the dual vector bundle of $$E \to M$$. The induced connection on the dual $$E^*$$ is given by $$\nabla^{*}_X(\xi)(s) = X(\xi(s)) - \xi(\nabla_X(s)),$$ for all $$s \in \Gamma(E), \xi \in \Gamma(E^*)$$ and $$X \in \mathfrak{X}(M)$$.

A little note about the notation. Although I am not 100% sure, I suppose the notation $$(\xi)(s)$$ just means that for $$x \in M$$, $$(\xi)(s)(x) := \xi(x)(s(x))$$. In this sense $$(\xi) \in \Gamma(E^*)$$ must be linear when it comes to the sections $$s \in \Gamma(E)$$. This means that for $$f \in C^{\infty}(M)$$ and $$s \in \Gamma(E)$$, we have (where $$x \in M$$) $$(\xi)(fs)(x) = \xi(x)(f(x) s(x)) = f(x) \xi(x)(s(x))$$ for all $$x \in M$$, which can be succinctly written as $$\xi(fs) = f \cdot \xi(s)$$.

## Proving the induced connection is a connection

Bilinearity + First Condition (I managed to prove this)

1. Consider $$f_1, f_2 \in C^{\infty}(M)$$ and $$X_1, X_2 \in \mathfrak{X}(M)$$. We write \begin{align} \nabla_{f_1 X_1 + f_2 X_2}^*(\xi)(s) &= (f_1 X_1 + f_2 X_2)(\xi(s)) - \xi(\nabla_{f_1 X_1 + f_2 X_2}(s)) \\ &= f_1 X_1(\xi(s)) + f_2 X_2(\xi(s)) - \xi(f_1\nabla_{X_1}(s) + f_2\nabla_{X_2}(s)) \\ &= f_1 X_1(\xi(s)) + f_2 X_2(\xi(s)) - f_1 \xi(\nabla_{X_1}(s)) + f_2\xi(\nabla_{X_2}(s)) \\ &= f_1 \nabla^*_{X_1}(\xi)(s) + f_2 \nabla^*_{X_2}(\xi)(s) \end{align}
2. Take $$a_1, a_2 \in \mathbb{R}$$ and $$\xi_1, \xi_2 \in \Gamma(E^*)$$ (and $$s \in \Gamma(E)$$). We then write \begin{align} \nabla^*_X(a_1 \xi_1 + a_2 \xi_2)(s) &= X((a_1 \xi_1 + a_2 \xi_2)(s)) - (a_1 \xi_1 + a_2 \xi_2)(\nabla_X(s)) \\ &= X(a_1 \xi_1(s) + a_2 \xi_2(s)) - a_1 \xi_1(\nabla_X(s)) - a_2 \xi_2(\nabla_X(s)) \\ &= a_1 (X(\xi_1(s)) - \xi_1(\nabla_X(s))) + a_2 (X(\xi_2(s)) - \xi_2(\nabla_X(s))) \\ &= a_1 \nabla^*_X(\xi_1)(s) + a_2 \nabla^*_X(\xi_2)(s) \end{align}

Leibniz Condition (Here I get stuck)

From here I am a bit confused about multiple things. Namely, we now want to prove that for $$X \in \mathfrak{X}(M), \xi \in \Gamma(E^*)$$ and $$s \in \Gamma(E)$$, we have $$\nabla^*_X(f \xi)(s) = f \nabla^*_X(\xi)(s) + X(f)\xi(s).$$ I hope this is the case at least. From here if we write out from the LHS, we obtain \begin{align} \nabla^*_X(f \xi)(s) &= X(f \xi(s)) - (f \xi)(\nabla_X(s)) \\ &= f X(\xi(s)) - (f \xi)(\nabla_X(s)) \end{align} From here I don't know how to continue. What mistakes am I making. Thanks in advance!

While writing out this question, I realised the following: the equality $$X(f \xi(s)) = f X(\xi(s))$$ is not true. What is true though is, $$X(f \xi(s)) = f X(\xi(s)) + \xi(s) X(f)$$ due to the product rule. Writing out the Leibniz condition is thus: \begin{align} \nabla^*_X(f \xi)(s) &= X(f \xi(s)) - (f \xi)(\nabla_X(s)) \\ &= f X(\xi(s)) + \xi(s) X(f) - f \xi(\nabla_X(s)) \\ &= f(X(\xi(s)) - \xi(\nabla_X(s))) + \xi(s) X(f) \\ &= f \nabla^*_X(\xi)(s) + X(f)\xi(s), \end{align} which is exactly the property that I wanted to prove.