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Let $X$ be a smooth $n$-manifold with a local chart $(U,\varphi)$. Taking the wedge product of the chart induced covariant basis $\{d\varphi^\mu\}$: \begin{align} \{d\varphi^{\mu_1} \wedge d\varphi^{\mu_2}... \wedge d\varphi^{\mu_k}:1\leq \mu_1 \leq ... \leq \mu_k \leq n\} \end{align} gives an basis for the space of $k$-forms $\Omega^k(X)$.

I have already proven linear independence and now have to prove that these $k$-forms span $\Omega^k(X)$. Let $\omega \in \Omega^k(X)$ and define: \begin{align} \omega_{\mu_1...\mu_n} = \omega(\textbf{e}_{\mu_1},...,\textbf{e}_{\mu_k}) \end{align} where $\{\textbf{e}_{\mu}\}$ is the chart-induced contravariant basis. We claim that \begin{equation} \omega = \omega_{\mu_1...\mu_k}d\varphi^{\mu_1} \wedge ... \wedge d\varphi^{\mu_k} \end{equation} To see why this holds, note that since $\omega$ is multi-linear we just need to look at the action of our candidate basis on $(\textbf{e}_{\nu_1},...,\textbf{e}_{\nu_k})$: \begin{equation} (d\varphi^{\mu_1} \wedge ... \wedge d\varphi^{\mu_k})(\textbf{e}_{\nu_1},...,\textbf{e}_{\nu_k}) \end{equation} Due to the duality of $\textbf{e}_{\nu_j}$ and $d\varphi^{\mu_i}$, the contributing terms will be delta functions $\delta^{\mu_i}_{\nu_j}$. We then have that: \begin{align} (d\varphi^{\mu_1} \wedge ... \wedge d\varphi^{\mu_k})(\textbf{e}_{\nu_1},...,\textbf{e}_{\nu_k})&=\sum_{\pi \in \text{S}_{k}} \text{sgn}(\pi) (d\varphi^{\mu_1} \otimes...\otimes d\varphi^{\mu_k})(\textbf{e}_{\pi(\nu_1)},...,\textbf{e}_{\pi(\nu_k)})\\ &=\sum_{\pi \in \text{S}_{k}} \text{sgn}(\pi) \delta^{\mu_1}_{\pi(\nu_1)}... \delta^{\mu_k}_{\pi(\nu_k)} \end{align} implying that \begin{align} \omega_{\mu_1...\mu_k} (d\varphi^{\mu_1} \wedge ... \wedge d\varphi^{\mu_k})(\textbf{e}_{\nu_1},...,\textbf{e}_{\nu_k})&=\sum_{\pi \in \text{S}_{k}} \text{sgn}(\pi) \omega_{\mu_1...\mu_k} \delta^{\mu_1}_{\pi(\nu_1)}... \delta^{\mu_k}_{\pi(\nu_k)}\\ &=\sum_{\pi \in \text{S}_{k}} \text{sgn}(\pi) \omega_{\pi(\nu_1)...\pi(\nu_k)} = k!\omega(\textbf{e}_{\nu_1},...,\textbf{e}_{\nu_k}) \end{align} where $\text{sgn}(\pi) \omega_{\pi(\nu_1)...\pi(\nu_k)}=\omega(\textbf{e}_{\nu_1},...,\textbf{e}_{\nu_k})$ by anti-symmetry. However, I should not have this $k!$ factor in my result, where did I go wrong?

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    $\begingroup$ Fix the order of indices, that would clear it up. $\endgroup$
    – Berci
    Commented Oct 12, 2021 at 8:00
  • $\begingroup$ @Berci The order of which indices? $\endgroup$
    – Kouta Dagnino
    Commented Oct 12, 2021 at 8:06
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    $\begingroup$ We can assume $\mu_1<\mu_2<\dots$ and $\eta_1<\eta_2<\dots$. $\endgroup$
    – Berci
    Commented Oct 12, 2021 at 8:12
  • $\begingroup$ So the terms in the sum with $\mu$ restricted gives $1/k!$ of the terms in the sum with $\mu$ unrestricted? $\endgroup$
    – Kouta Dagnino
    Commented Oct 12, 2021 at 8:16
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    $\begingroup$ You still have an error. You need $\mu_1<\mu_2$, etc. $\endgroup$
    – Ted Shifrin
    Commented Oct 12, 2021 at 17:29

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