So I am trying to prove the following:

Let $V \rightarrow M$ be a vector bundle $\nabla$ a connection on $V$. Then there is a unique sequence of linear maps $$ \Omega^0(M;V) \xrightarrow{\nabla} \Omega^1(M;V) \xrightarrow{\nabla} \cdots $$ such that $\nabla$ coincdies with the connection for $p=0$ and such that $$\nabla (w \wedge s ) = dw \wedge s + (-1)^{|w|} w \wedge \nabla s (*) $$

Where $\Omega^k(M;V) := \Omega^k(M) \otimes_{\Omega^0(M)} \Omega^0(V)$, where $\Omega^k(M)$ are the smooth $k$-forms ($k=0$ give smooth functions), and $\Omega^0(V)$ are the smooth sections of $V \rightarrow M$.

A connection $\nabla:\Omega^0(M;V) \rightarrow \Omega^1(M;V)$ is defined to be a map that satisfies $$ \nabla (fs) = df \otimes s + f \nabla s $$

So I wanted to define $\nabla$ locally. Since given a local frame $e_i$ for $V$, we may write $s = \sum w_i \otimes e_i$ and we must have $$ \nabla s = \sum_i dw_i \otimes e_i + w _i \wedge \nabla e_i $$

The problem is I could not show this is coordinate independent.

My failed attempt:

given another local frame $f_1, \ldots, f_r$ of $V$. Suppose $e = Af$, so $e_i = \sum A_{ij} f_j $ where $A_{ij} \in C^\infty(U)$.

Then \begin{align*} \sum_i \Big(dw_i \otimes \sum A_{ij} f_j +w_i \wedge \nabla (\sum A_{ij} f_j ) \Big) &= \sum_i \Big(\sum_j A_{ij} dw_i \otimes f_j+ w_i \wedge (\sum_j dA_{ij} \otimes f_j + A_{ij} \nabla f_j ) \Big) \end{align*}


Since $ A_{ij} dw_i + w_i dA_{ij} = d(w_i A_{ij})$ for all $i$ and $j$, the expression on the right-hand side reduces to $$ \sum_{i}\sum_j \left( d(w_i A_{ij}) \otimes f_j + (w_i A_{ij}) \nabla f_j \right),$$ which is precisely the expression you want.

Edit: If the $w_i$'s are $k$-forms (rather than just smooth functions), then you would need to introduce an extra sign in your proposed definition: $$ \nabla s = \sum_i dw_i \otimes e_i + (-1)^k w_i \wedge \nabla e_i.$$ You'll end up with a corresponding factor of $(-1)^k$ on the right-hand side of your equation. Fortunately, everything works out because $A_{ij} dw_i + (-1)^k w_i \wedge dA_{ij} = d(w_iA_{ij})$.

  • $\begingroup$ Sorry, I am still unclear, I thought my desired expression is $$ \sum dw_j \otimes f_j + w_j \nabla f_j$$ I thought $A_{ij}$ should only come in play to the sections $\Omega^0(V)$ and I still use the same local coordinates for $\Omega^1(M)$. $\endgroup$ – Cy L Shih Jan 17 at 8:42
  • $\begingroup$ @CL. No, $s = \sum_{j} w'_j \otimes f_j$, where $w'_j := \sum_j w_i A_{ij} $. So you should be aiming for $s = \sum_j \left( dw'_j \otimes f_j + w'_j \nabla f_j \right) = \sum_j \left( d(w_i A_{ij}) \otimes f_j + (w_i A_{ij}) \nabla f_j \right)$ $\endgroup$ – Kenny Wong Jan 17 at 8:46
  • $\begingroup$ Yes, thanks a lot. $\endgroup$ – Cy L Shih Jan 17 at 8:54
  • $\begingroup$ thanks for the edit too . $\endgroup$ – Cy L Shih Jan 22 at 8:32

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