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I want to show that for an $n$-form $w$ and an $m$-form $v$, we have $d(w \wedge v)$ = $dw \wedge v + (-1)^n w \wedge dv$. What I have so far is that for $w = fdx^{i_1} \wedge ...\wedge dx^{i_n}$ and $v = gdx^{j_1} \wedge ...\wedge dx^{j_m}$,

\begin{align*} d(w \wedge v) &= d(fg\, dx^{i_1}\wedge\dots\wedge dx^{i_n}\wedge dx^{j_1}\wedge \dots\wedge dx^{j_m}) \\ &= (g\,df + f\,dg) \wedge (dx^{i_1}\wedge\dots\wedge dx^{i_n}\wedge dx^{j_1}\wedge \dots\wedge dx^{j_m}) \\ &= (g\,df \wedge (dx^{i_1}\wedge\dots\wedge dx^{i_n}\wedge dx^{j_1}\wedge \dots\wedge dx^{j_m})) + (f\,dg \wedge (dx^{i_1}\wedge\dots\wedge dx^{i_n}\wedge dx^{j_1}\wedge \dots\wedge dx^{j_m})) \\ \end{align*} But then when computing the above, I get $d(w \wedge v)$ = $(-1)^ndw \wedge v + (-1)^n w \wedge dv$, because it seems like we are moving both the $g$ in the first summand and the $dg$ in the second summand across $p$ wedge products. For example, in the second summand we evaluate it to be $(-1)^n (f\,dx^{i_1}\wedge\dots\wedge dx^{i_n})\wedge (dg\wedge dx^{j_1}\wedge \dots\wedge dx^{j_m}) = (-1)^n w\wedge dv$. Where is my math or logic wrong?

noamchomsky
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1 Answers1

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For the first term, we have \begin{align} &(g\,df \wedge (dx^{i_1}\wedge\dots\wedge dx^{i_n}\wedge dx^{j_1}\wedge \dots\wedge dx^{j_m}))\\ &=[df\wedge dx^{i_1}\wedge\cdots \wedge dx^{i_n}]\wedge [g\,dx^{j_1}\wedge \cdots \wedge dx^{j_m}]\\ &=(dw)\wedge v \end{align} Note that we were able to move $g$ over because it is simply a function (a $0$-form). If you want to be even more explicit about this, then if you evaluate everything at a point $p\in M$, then $g(p)$ is a real number, and simply based on the bilinearity of the wedge product, you can move the scalar across.

For the second term, $dg$ is a $1$-form, so in order to make it "jump across" the $dx^{i_1}\wedge\cdots \wedge dx^{i_n}$, there are $n$ 1-forms it has to jump across, so that's why there's a $(-1)^n$ for this term but not in the previous paragraph.

peek-a-boo
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