# Connection of dual bundle

Let $E \to M$ be a vector bundle on $M$.Then It was written in a book that the connection on the dual bundle $E^* \to M$ is given as $(\nabla^*_X{\theta})\sigma = X(\theta(\sigma)) - \theta(\nabla_X\sigma)$ where $\theta \in \Gamma(E^*)$,$\sigma \in \Gamma(E)$ and $\nabla$and $\nabla^*$ are connections on the vector bundles $E$ and $E^*$ respectively . Generally, a connection on a vector bundle $E$ is defined as a map from $\Gamma(TM) \times \Gamma(E) \to \Gamma(E)$ as $(X, Y) \mapsto \nabla_X Y$ so that the connection on the dual bundle will be a map $\Gamma(TM) \times \Gamma(E^*) \to \Gamma(E^*)$ as $(\theta, \sigma) \mapsto \nabla^*_{\theta}{\sigma}$. I want to ask two questions :

1. How did we get the expression of dual connection given above in second line?I am not getting its proof.

2. As $\nabla^*_X \theta \in \Gamma(E^*)$ $\therefore$ it is a map from $M \to E^*$ then how $(\nabla^*_X{\theta})\sigma$ is meaningful where $\sigma \in \Gamma(E)$? or where does $(\nabla^*_X{\theta})\sigma$ belongs to?

The definition comes from wanting a "product rule" (think of $\theta(\sigma)$ as a contraction of $\theta\otimes\sigma$). But the book is giving a definition here; you must check that it satisfies the properties of a connection. For your second question, perhaps you'd prefer another set of parentheses: You're pairing once again a section of $E^*$ with a section of $E$, obtaining a function.
• what is meant by contraction of $\theta \otimes \sigma$? I am a beginner to differential geometry so can you please explain me more preisely. For second question as you said pairing a section of $E^*$ and a section of $E$ is giving a function but what function? where would it be defined? Commented Feb 18, 2014 at 4:18
• This is a matter of understanding what a dual vector space is. Make sure you understand that, and then you know what's going on (locally) with vector bundles. The contraction to which I referred is, similarly, the map $V^*\otimes V\to\Bbb R$ given by $\phi\otimes v \mapsto \phi(v)$. Commented Feb 18, 2014 at 4:24