(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 $s^*(s)\in C^\infty(M)\cong \Gamma(M\times \mathbb R)$).


1 Answer 1


I'm not sure either, but let me share my thoughts!

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(?).

  • $\begingroup$ 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.. $\endgroup$
    – soap
    Commented Oct 23, 2018 at 9:29
  • $\begingroup$ $\nabla s(x) \in (T^*M \otimes E)_x$ (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. $\endgroup$
    – Creo
    Commented Oct 23, 2018 at 10:05

You must log in to answer this question.

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