Suppose I have a covector field $\omega$ and a covariant derivative $\nabla_{X}$ for some vector field $X$ on a Riemannian manifold $(M, g)$.

Define $X^{\flat} \in \mathfrak{X}^{*}(M)$ as $X^{\flat}(Y) = g(X, Y)$, and $\omega^{\sharp}$ as the unique vector for which $\omega(X) = g(\omega^{\sharp}, X)$; it's easily checked that these musical maps are mutually inverse, so they're isomorphisms between $\mathfrak{X}(M)$ and $\mathfrak{X}^{*}(M)$.

Anyway, I want to prove that, for the Levi-Civita connection $\nabla$ on $(M, g)$, we have $\nabla_{X}(\omega^{\sharp}) = (\nabla_{X}\omega)^{\sharp}.$ In several books I've been reading, it says that this follows immediately from $\nabla g = 0$.

However, I keep getting that this is wrong, for example if $(E_{1},...,E_{n})$ is a local orthonormal frame on $M$, I get that $\nabla_{E_{i}}(E_{j}^{*}{^{\sharp}})$ does not equal $(\nabla_{E_{i}}E_{j}^{*})^{\sharp}$. Here's what I've been trying:

1. Finding $\nabla_{E_{i}}(E_{j}^{*}{^{\sharp}})$: I start by finding the components of $E_{j}^{*}{^{\sharp}}$

$$(E_{j}^{*}{^{\sharp}})^{k} = \sum_{l} g^{kl} (E_{j}^{*})_{l} = \delta_{kj}.$$

Therefore, $E_{j}^{*}{^{\sharp}} = E_{j}$. Now, I want to find $\nabla_{E_{i}} E_{j}^{*}{^{\sharp}} = \nabla_{E_{i}} E_{j}$.

$$\nabla_{E_{i}} E_{j} = \sum_{k} \Gamma_{ij}^{k}E_{k},$$

where $\Gamma_{ij}^{k}$ are the Christoffel symbols with respect to this frame.

2. Finding $(\nabla_{E_{i}}E_{j}^{*})^{\sharp}$: again, I begin by finding the components, this time of $(\nabla_{E_{i}}E_{j}^{*})^{\sharp}$:

$$ ((\nabla_{E_{i}}E_{j}^{*})^{\sharp})^{k} = \sum_{l} g^{kl} (\nabla_{E_{i}}E_{j}^{*})_{l},$$ so I want to find $\nabla_{E_{i}}E_{j}^{*}(E_{l})$. Since $\nabla_{E_{i}}$ is a tensor derivative, we have:

$$ 0 = \nabla_{E_{i}}(E_{j}^{*}(E_{l})) = \nabla_{E_{i}}E_{j}^{*}(E_{l}) + E_{j}^{*}(\nabla_{E_{i}}E_{l}),$$

so $\nabla_{E_{i}}E_{j}^{*}(E_{l}) = - E_{j}^{*}(\sum_{k}\Gamma_{il}^{k}E_{k}) = - \Gamma_{il}^{j}$. Finally,

$$ ((\nabla_{E_{i}}E_{j}^{*})^{\sharp})^{k} = \sum_{l} g^{kl} (\nabla_{E_{i}}E_{j}^{*})_{l} = -\Gamma_{ik}^{j},$$

so $(\nabla_{E_{i}}E_{j}^{*})^{\sharp} = \sum_{k} (-\Gamma_{ik}^{j}E_{k}) = -\sum_{k}\Gamma_{ik}^{j}E_{k}$.

Since I don't know that $\Gamma_{ik}^{j} + \Gamma_{ij}^{k} = 0$, I don't see why these two are equal. Did I make a mistake somewhere; is there an easier proof of this fact?


We have that $(\nabla_X \omega)^\sharp$ is such that $$g((\nabla_X \omega)^\sharp, Y) = \nabla_X \omega(Y)$$ but, $$\nabla_X \omega(Y) = X\omega(Y) - \omega(\nabla_X Y) = Xg(\omega^\sharp,Y) - g(\omega^\sharp,\nabla_X Y) $$ $$\nabla_X \omega(Y) = g(\nabla_X(\omega^\sharp),Y) + g(\omega^\sharp,\nabla_X Y) - g(\omega^\sharp,\nabla_X Y)$$ Thus $g((\nabla_X \omega)^\sharp, Y) = g(\nabla_X(\omega^\sharp),Y)$ for all $Y$vector fields in $M$. Therefore $(\nabla_X \omega)^\sharp = \nabla_X(\omega^\sharp)$.


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