# How to define a connection on the dual bundle?

I'm reading Terence Tao's blog and he says

Once one has a connection on a bundle $V$, one automatically can define a connection on the dual bundle $V^*$.

Can anyone tell me how?

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Assume that $V$ is a vector bundle over $M$. The dual connection is uniquely defined by requiring that it satisfies the product rule with respect to the natural pairing between $V$ and $V^*$. Let $\xi \in \Gamma(V)$ and $\varphi \in \Gamma(V^*)$, and denote by $$\left< \varphi, \xi \right> = \varphi(\xi ) \in C^\infty(M)$$ the natural pairing between the bundles. That is, at each point $p \in M$, you apply the functional $\varphi_p$ to the vector $\xi_p$ to get a number $\varphi_p(\xi_p)=(\varphi(\xi))(p)$. Then, the dual connection satisfies the "product" rule $$\nabla_X \left< \varphi, \xi \right> = X \left< \varphi, \xi \right> = \left< \nabla_X \varphi, \xi \right> + \left< \varphi, \nabla_X \xi \right>$$ for all $X \in \Gamma(TM)$.
When define a connection on a vector bundle, we are given a function $f\in\Gamma(V)$ and a vector field $X\in\Gamma(TM)$. But according to your answer, we must be given TWO functions, one in $\Gamma(V)$ and one in $\Gamma(V^*)$, to define a dual connection?.. –  hxhxhx88 Nov 13 '12 at 7:41
No. A connection on $V$ eats a section $\xi \in \Gamma(V)$ and a vector field $X \in \Gamma(TM)$ and returns a section in $\Gamma(V)$. Here, $\varphi$ a section of the dual vector bundle $\Gamma(V^*)$ and so in $\nabla_X \varphi$. Instead of defining it directly, I gave the property that characterizes it. Note, that if you move the second term of the equality on the right to the other side, you understand how the functional $\nabla_X \varphi$ acts on a vector $\xi$ in terms of the connection on $V$, for any $\xi$. This defines $\nabla_X \varphi$ uniquely. –  levap Nov 13 '12 at 9:35
If $V$ is a vector space, how do you define how a functional $\varphi \in V^*$? You say what it does to any vector $v \in V$. That is, you give $\varphi(v) = \left< \varphi, v \right>$. This is the same. –  levap Nov 13 '12 at 9:37