# Connection on the dual vector bundle

(Note: I looked at the other questions about defining a connection on the dual bundle, but the answers do not apply to my case since I use a slightly different definition of connection).

I am following some lecture notes on differential geometry.

A connection of $$E\rightarrow M$$ is defined as a linear map $$\nabla:\Gamma(E)\rightarrow \Gamma(T^*M\otimes E)$$ satisfying the Leibniz rule.

Given such a connection, the induced dual connection is defined by imposing $$d(s^*(s))=(\nabla^*s^*)(s) + s^*(\nabla s)$$

However, I do not see how to make sense of this. I understand how to make sense of $$s^*(s)$$ for $$s\in \Gamma(E)$$ and $$s^*\in \Gamma(E^*)$$. But $$\nabla^*s^*\in\Gamma(T^*M\otimes E^*)\ne \Gamma(E^*)$$ and $$\nabla s \in \Gamma(T^*M\otimes E)\ne \Gamma(E)$$, so I cannot make sense of $$(\nabla^*s^*)(s)$$ and $$s^*(\nabla s)$$.

(I think that $$d$$ above is the trivial connection, and the author of the notes is taking $$s^*(s)\in C^\infty(M)\cong \Gamma(M\times \mathbb R)$$).

As you pointed out correctly, $$s^*(s) \in C^{\infty}(M)$$, so I suppose $$d(s^*(s))$$ refers to the exterior differential, hence $$d(s^*(s)) \in \Omega^1(M)$$. Indeed, we would want to have $$\nabla^* s^* \in \Gamma(T^*M \otimes E^*)$$ and thus I'd interpret $$(\nabla^* s^*)$$ as follows: Let $$x \in M$$ be an arbitrary point. Then $$(\nabla^* s^*)(x) \in (T^*M)_x \otimes (E^*)_x$$, and so $$(\nabla^* s^*)(x)(s(x)) \in (T^*M)_x = T^*_xM$$, for if we write $$\nabla^*s^* = \omega \otimes e,$$ where $$\omega \in \Gamma(T^*M)$$ and $$e \in \Gamma(E^*)$$, we might set $$(\nabla^*s^*)s := \omega \cdot (e(s))$$, where the dot is just the scalar multiplikation. Thus, $$(\nabla^* s^*)(s)$$ is indeed a 1 form, as should be. In the same manner, I'd interpret the latter expression $$s^*(\nabla s)$$, but maybe it's a good exercise for you to write that down in detail by yourself(?).
• That seems convincing, but for the later expression it seems to not work: $\nabla s (x) \in Hom(E,T^* M)$ but $s^*(x)\in E^*$, so $\nabla s (x) s^*(x)$ does not make sense..
• $\nabla s(x) \in (T^*M \otimes E)_x$ (I just noted that I forgot the fibres in my answer, I updated it). By definition $(T^*M \otimes E)_x = (T^*M)_x \otimes E_x$ thus you can, as before, let $s^*$ act on it.