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I know that $ \omega\wedge \star \eta=\langle \omega, \eta \rangle\mathrm{dVol}_g$. I want to know

Q1: why the following make sense? $$\langle \nabla \omega, \eta \rangle \mathrm{dVol}_g=\nabla \omega \wedge \star \eta,\qquad\omega,\eta\in\Omega^k(M).$$

I am asking this because $\nabla \omega$ is not a differential form in general. It is a $(0,k+1)$-tensor that has been wedge producted with a $(n-k)$-form $\star\eta$.

Q2: I know that $\langle \nabla \omega, \eta \rangle$ is a $(0,1)$-tensor, so what is the product between thatand $\mathrm{dVol}_g$ in the above displayed relation?

Also

Q3: Does $\star\nabla\omega$ make sense? i.e. Hodge star acting on tensors!! (because in general $\star\nabla=\nabla\star$ then it must make sense I think but how it works in practice?)

p.s. Motivating MO post.

C.F.G
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  • It's best to think of $\nabla \omega$ as a 1-form-valued $k$-form and therefore, $\nabla\omega\wedge \star\eta$ as a covector-valued $n$-form. In other words, given $k$ tangent vectors $v_1, \dots, v_k$, $\nabla\omega(v_1, \dots, v_k)$ is the $1$-form $v \mapsto \nabla\omega(v,v_1,\dots, v_m)$. – Deane Mar 29 '21 at 19:55
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    So $\langle \nabla \omega, \eta \rangle \mathrm{dVol}_g$ is $\langle \nabla \omega, \eta \rangle\otimes \mathrm{dVol}_g$? – C.F.G Mar 29 '21 at 20:25
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    Yes. You can work this out with respect to a local orthonormal frame – Deane Mar 29 '21 at 21:23

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