# Coordinate independence of connections

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 $$A_{ij} dw_i + (-1)^k w_i \wedge dA_{ij} = d(w_iA_{ij})$$.
• 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 $\Omega^1(M)$. – Cy L Shih Jan 17 at 8:42
• @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 $s = \sum_j \left( dw'_j \otimes f_j + w'_j \nabla f_j \right) = \sum_j \left( d(w_i A_{ij}) \otimes f_j + (w_i A_{ij}) \nabla f_j \right)$ – Kenny Wong Jan 17 at 8:46