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I'm trying to show that if $\omega$ is a 1-form on $(\mathbb R^m,\delta)$, the action of the Laplacian is given by $$\Delta\omega=-\sum_{\mu=1}^m\frac{\partial^2\omega_\nu}{\partial x^\mu\partial x^\mu}dx^\nu$$ What I've calculated is, for a general metric $g_{\mu\nu}$, $$dd^\dagger\omega=(-1)^{m+1}\partial_\lambda\partial_\nu(\sqrt{|g|}\omega_\mu g^{\mu\nu})dx^\lambda$$ $$d^\dagger d\omega=(-1)^m\partial_\lambda(\partial_\mu\omega_\nu\sqrt{|g|}g^{\mu\lambda}g^{\nu\sigma})g_{\sigma\rho}dx^\rho$$ When I then specify the Euclidean metric, I get the desired term (up to a factor $(-1)^m$) from $d^\dagger d\omega$, but $dd^\dagger\omega$ is superfluous/of the wrong form - namely, I get $$dd^\dagger\omega=(-1)^m\frac{\partial^2\omega_\nu}{\partial x^\nu\partial x^\mu}dx^\mu $$ I'm not really sure what I'm doing wrong here, but any help would be appreciated.

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