Use la notación de índice abstracto para probar la regla de Leibniz para derivada exterior

Question

Quiero usar la notación de índice abstracto para probar la regla de Leibniz para el derivado exterior del producto de cuña:

Para $\omega\in \Omega^k(U),\eta\in\Omega^l(U)$ , d $(\omega\wedge\eta)=\text{d}\omega\wedge\eta +(-1)^k\omega\wedge\text{d}\eta$ .

Answer 1

Requisito previo:

Para $\alpha \en \Omega^s(U) \text{ y } \beta\en \Omega^t(U), (\alpha\wedge\beta)_{i_1\cdots i_{s+t}} :=\frac{(s+t)!}{s!\ t!}\alpha_{[i_1\cdots i_s}\beta_{i_{s+1}\cdots i_{s+t}]}$

\begin{align} \alpha &= \Sigma_I\ \alpha_{i_1\cdots i_s} dx^{i_1} \wedge \cdots \wedge dx^{i_s} \\ &\notag\text{(the summation is over increasing } s-\text{tuples}, I = \{1\leq i_1 < \cdots < i_s \leq n\},n\text{ is dimension of } U)\\ &= \frac {1}{s!}\alpha_{i_1\cdots i_s} dx^{i_1} \wedge \cdots \wedge dx^{i_s}\text{(Einstein summation convention)}\\ \text{d}\alpha&:= \Sigma_{I,i}\ \partial_i(\alpha_{i_i\cdots i_s}) dx^i \wedge dx^{i_1} \wedge \cdots \wedge dx^{i_s}\\ &=\frac {1}{s!}\partial_i\alpha_{i_1\cdots i_s} dx^i \wedge dx^{i_1} \wedge \cdots \wedge dx^{i_s} =\frac {1}{(s+1)!}(\text{d}\alpha)_{ii_1\cdots i_s} dx^i \wedge dx^{i_1} \wedge \cdots \wedge dx^{i_s} \end{align}

Por lo tanto el coeficiente de $(\text{d}\alpha)_{ii_1\cdots i_s} = (s+1)\partial_i\alpha_{i_1\cdots i_s}$

Escribir en resumen índice de notación, \begin{align} (\text{d}\alpha)_{a_1\cdots a_{s+1}} &=\frac{1}{(s+1)!}(\text{d}\alpha)_{i_1\cdots i_{s+1}} (e^{i_1})_{a_1}\cdots(e^{i_{s+1}})_{a_{k+1}}\\ &=\frac{1}{s!}\partial_{i_1}\alpha_{i_2\cdots i_{s+1}} (e^{i_1})_{a_1}\cdots(e^{i_{s+1}})_{a_{k+1}}\\ (\text{d}\alpha)_{a_1\cdots a_{s+1}} &=(\text{d}\alpha)_{[a_1\cdots a_{s+1}]} =\frac{1}{s!}\partial_{i_1}\alpha_{i_2\cdots i_{s+1}} (e^{i_1})_{[a_1}\cdots(e^{i_{s+1}})_{a_{k+1}]} \\ &=\frac{1}{s!}\partial_{i_1}\alpha_{i_2\cdots i_{s+1}} (e^{[i_1})_{a_1}\cdots(e^{i_{s+1}]})_{a_{k+1}} =\frac{1}{s!}\partial_{[i_1}\alpha_{i_2\cdots i_{s+1}]} (e^{i_1})_{a_1}\cdots(e^{i_{s+1}})_{a_{k+1}}\\ &=(s+1)\partial_{[a_1}\alpha_{a_2\cdots a_{s+1}]} \end{align}

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Para $\omega\in \Omega^k(U) \text{ and } \eta\in\Omega^l(U), \text{d}(\omega\wedge\eta) = \text{d}\omega\wedge\eta +(-1)^k\omega\wedge\text{d}\eta$.

Prueba:

\begin{align} (\text{d}(\omega\wedge\eta))_{i_1\cdots i_{k+l+1}} &=\frac{(k+l)!}{k!\ l!}\text{d}(\omega_{[i_2\cdots i_{k+1}} \eta_{i_{k+2}\cdots i_{k+l+1}]}) \\ &=\frac{(k+l)!}{k!\ l!}(k+l+1)\partial_{[i_1}(\omega_{[i_2\cdots i_{k+1}} \eta_{i_{k+2}\cdots i_{k+l+1}]})_] \\ &=\frac{(k+l+1)!}{k!\ l!}\partial_{[i_1}(\omega_{i_2\cdots i_{k+1}} \eta_{i_{k+2}\cdots i_{k+l+1}})_] \\ &=\frac{(k+l+1)!}{k!\ l!}(\partial_{[i_1}\omega_{i_2\cdots i_{k+1}} \eta_{i_{k+2}\cdots i_{k+l+1}]} + \omega_{[i_2\cdots i_{k+1}} \partial_{i_1}\eta_{i_{k+2}\cdots i_{k+l+1}]}) \\ &=\frac{(k+l+1)!}{k!\ l!}(\partial_{[i_1}\omega_{i_2\cdots i_{k+1}} \eta_{i_{k+2}\cdots i_{k+l+1}]} + (-1)^k\omega_{[i_1\cdots i_{k}} \partial_{i_{k+1}}\eta_{i_{k+2}\cdots i_{k+l+1}]}).\end{align}

\begin{align} (\text{d}\omega\wedge\eta)_{i_1\cdots i_{k+l+1}} &=(k+1)\frac{(k+l+1)!}{(k+1)!\ l!}\partial_{[[i_1}\omega_{i_2\cdots\ i_{k+1}]}\beta_{i_{k+2}\cdots i_{k+l+1}]} \\ &=\frac{(k+l+1)!}{k!\ l!}\partial_{[i_1}\omega_{i_2\cdots\ i_{k+1}}\beta_{i_{k+2}\cdots i_{k+l+1}]}.\end{align}

\begin{align} (-1)^k (\omega\wedge\text{d}\eta)_{i_1\cdots i_{k+l+1}} &=(-1)^k \frac{(k+l+1)!}{k!\ (l+1)!}(l+1)\omega_{[i_1\cdots \ i_k} \partial_{[i_{k+1}}\eta_{i_{k+2}\cdots i_{k+l+1}]]} \\ &=(-1)^k \frac{(k+l+1)!}{k!\ l!}\omega_{[i_1\cdots \ i_k} \partial_{i_{k+1}}\eta_{i_{k+2}\cdots i_{k+l+1}]}. \end{align}

Por lo tanto d$(\omega\wedge\eta)=$d$\omega\wedge\eta +(-1)^k\omega\wedge\text{d}\eta$.