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.


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!


1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .