# Proving $d\varphi^{\mu_1} \wedge d\varphi^{\mu_2}... \wedge d\varphi^{\mu_k}$ is basis for space of $k$-forms

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 $$$$\omega = \omega_{\mu_1...\mu_k}d\varphi^{\mu_1} \wedge ... \wedge d\varphi^{\mu_k}$$$$ To see why this holds, note that since $$\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})$$: $$$$(d\varphi^{\mu_1} \wedge ... \wedge d\varphi^{\mu_k})(\textbf{e}_{\nu_1},...,\textbf{e}_{\nu_k})$$$$ Due to the duality of $$\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?

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