Choice of order in the Leibniz rule is arbitrary?

Question

One of the rules which characterizes the exterior derivative is that, for $\varphi$ a real-valued function and $\omega$ a $k$-form, we have $$d(\varphi \cdot \omega) = d\varphi \wedge \omega + \varphi \cdot d\omega.$$

I note that this is not the same thing as $\omega \wedge d\varphi + (d \omega)\cdot \varphi$ when $k$ is odd, even though in the expression $(\varphi \cdot \omega)$ it is essentially arbitrary which factor we write first. Is choosing $d(\varphi \cdot \omega) = \omega \wedge d\varphi + \varphi \cdot d\omega$ equivalent to making some other arbitrary choice (like an orientation choice?)

Answer 1

Comments to the question (v2):

1. One can define left and right exterior derivatives $$ d_L(\omega\wedge\eta)~=~(d_L\omega)\wedge\eta + (-1)^{|\omega|}\omega\wedge d_L\eta \tag{L}, $$ $$ d_R(\omega\wedge\eta)~=~(-1)^{|\eta|} (d_R\omega)\wedge\eta + \omega\wedge d_R\eta \tag{R}, $$ $$ d_R\omega~=~(-1)^{|\omega|}d_L \omega, \tag{C}$$ where $\omega, \eta\in\Omega(M)$ are differential forms of definite form degrees $|\omega|$, $|\eta|\in \mathbb{N}_0$.

2. Here $d_L\equiv d$ is the usual exterior derivative, while $d_R$ is the alternative sign convention mentioned by OP.

3. Formula (L) and (R) are examples of left and right graded derivations.

4. The left exterior derivative $d_L$ follows the Koszul sign rule:
   $$\text{Introduce a sign factor }(-1)^{|a||b|}\text{ when permuting objects }a \text{ and }b.\tag{K}$$ Here the exterior derivative $d_L$ has form-degree $|d_L|=1$.

5. It is possible to interpret the sign in (R) via the Koszul sign rule (K) if we rewrite (R) in Polish notation $$ (\omega\wedge\eta)\stackrel{\leftarrow}{d_R}~=~(-1)^{|\eta|} \omega\stackrel{\leftarrow}{d_R}\wedge\eta + \omega\wedge (\eta\stackrel{\leftarrow}{d_R}) \tag{R'}. $$ In this sense, one may view the right exterior derivative $d_R$ as acting from right.