interior product equality

Question

im having a hard time proving that for 2 differential forms $\omega\in \Omega^m(V)$ and $\eta \in \Omega^n(V)$ and a smooth vector field $v$ on $V$, that we have $j_v(\omega \wedge \eta)= j_v\omega\wedge \eta +(-1)^m\omega \wedge j_v\eta$. This looks extremely similar to this equation $d(\omega \wedge \eta)=d\omega \wedge \eta +(-1)^m\omega \wedge d\eta$, which i already proved. Where exactly is the connection between those two, and what exactly do i need to modify in my prove for the second equality to obtain the first one? ( Also im sorry for not making it more readable, because i dont know how to make new lines). Thanks in advance !

Answer 1

Here is one way to prove it: Let $\alpha^1,\dots,\alpha^k\in\Omega^1(V)$ be one-forms. Then we have $$\iota_v(\alpha^1\wedge\dots\wedge\alpha^k)=\sum_{i=1}^k(-1)^{i+1}\alpha^i(v)\alpha^1\wedge\dots\wedge\widehat{\alpha^i}\wedge\dots\wedge\alpha^k$$ where the hat over $\alpha^i$ means that this term is omitted. This is just the Leibniz expansion of determinants in the following formula $$(\alpha^1\wedge\dots\wedge\alpha^k)(v_1,\dots,v_k)=\det(\alpha^i(v_j))_{ij}$$ By linearity of $\iota_v$ we are reduced to show that the claim holds in the case where $\omega,\eta$ are of the form \begin{align*} \omega&=\alpha^1\wedge\dots\wedge\alpha^m\\ \eta&=\beta^1\wedge\dots\wedge\beta^n \end{align*} for one-forms $\alpha^i,\beta^j$. So we get: \begin{align*} \iota_v(\omega\wedge\eta)&=\iota_v(\alpha^1\wedge\dots\wedge\alpha^m\wedge\beta^1\wedge\dots\wedge\beta^n)\\ &=\sum_{i=1}^m(-1)^{i+1}\alpha^i(v)\alpha^1\wedge\dots\wedge\widehat{\alpha^i}\wedge\dots\wedge\alpha^m\wedge\beta^1\wedge\dots\wedge\beta^n\\&\hspace{0.7cm}+\sum_{j=1}^{n}(-1)^{j+1+m}\beta^i(v)\alpha^1\wedge\dots\alpha^n\wedge\beta ^1\wedge\dots\wedge\widehat{\beta ^j}\wedge\dots\wedge\beta ^m\\ &=\iota_v(\omega)\wedge\eta+(-1)^m\omega\wedge\iota_v(\eta) \end{align*}

Answer 2

The connection lies in the Lie derivative. For all vector fields $X$, the Lie derivative, as the name suggests, is an operator on forms $\mathcal L_X : \Omega^k(M)\to\Omega^k(M)$ defined as $$\mathcal L_X(\alpha):=\frac{d}{dt}\Bigg|_{t=0}(\varphi^{(X)*}_t \alpha),$$ where $\varphi^{(X)}_t$ is the flux associated to $X$. Here, the definition is unimportant: what matters is that $\mathcal L_X$ is a differential operator, and as such it satisfies the product rule $$ \mathcal{L}_{X}(\omega \wedge \eta)=\left(\mathcal{L}_{X} \omega\right) \wedge \eta+\omega \wedge\left(\mathcal{L}_{X} \eta\right); \tag1 $$ also, Cartan's "magic" formula connects it with $j_X$ and $d$ in an interesting way: $$\mathcal L_X\omega = j_X(d\omega)+d(j_X\omega). \tag2$$ This entails $$\begin{split} \mathcal L_X(\omega \wedge \eta)&=j_Xd(\omega\wedge\eta)+dj_X(\omega\wedge \eta) \\ (\mathcal L_X \omega)\wedge \eta + \omega\wedge(\mathcal L_X\eta)&=j_X((d\omega)\wedge \eta)+ (-1)^mj_X(\omega\wedge(d\eta)) \\ &\quad+d((j_X\omega) \wedge \eta + (-1)^m \omega \wedge (j_X\eta)) \\ 0 &= (-1)^{m+1}(d\omega)\wedge(j_X\eta) + (-1)^m(j_X\omega)\wedge(d\eta)\\ &\quad+(-1)^{m}(d\omega)\wedge(j_X\eta) + (-1)^{m-1}(j_X\omega)\wedge(d\eta) \\ 0&=0, \end{split}$$ by applying the magic formula on the LHS as well, and the graded product rules you stated on the RHS (twice), thus proving their consistency (at least if you trust $(1)$ and $(2)$!).