Prove that the covariant derivative commutes with musical isomorphisms

Question

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?

Answer 1

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 $(\nabla_X \omega)^\sharp = \nabla_X(\omega^\sharp)$.

Answer 2

I am recently learning Riemannian Geometry and have similar questions as yours. For your calculation, I think it should be right, with $\Gamma^j_{ik} + \Gamma^k_{ij} = 0$ for the following reason.

I would use Lee's notation on P124 of his book Introduction to Riemannian Manifolds, and Prop 5.11 (c) states "In a local orthonormal frame: if $g$ is Riemannian, $(E_i)$ is a smooth local orthonormal frame on an open subset $U\subseteq M$, and let $c^k_{ij}: U\to\mathbb{R}$ be the $n^3$ smooth functions defined by \begin{equation} [E_i,E_j] = c^k_{ij}E_k. \end{equation} Then, the coefficients of the Levi-Civita connection in this frame are \begin{equation} \Gamma^k_{ij}=\frac{1}{2}\left(c^k_{ij}-c^j_{ik}-c^i_{jk}\right). \end{equation} "

Following his notation, we also know \begin{equation} \Gamma^j_{ik}=\frac{1}{2}\left(c^j_{ik}-c^k_{ij}-c^i_{kj}\right). \end{equation} By anticommunitivity of Lie brackets, \begin{equation} c^i_{jk} = -c^i_{kj}, \end{equation} which amounts to give that to be proven, $\Gamma^j_{ik} + \Gamma^k_{ij} = 0$.

By the way, the way Lee used to prove this is to use the fact that sharp and flat operators are inverses to each other and proving the result for the flat operation, to prove which he used $F^{\flat} = \text{tr}(F\otimes g)$ and $g$ is parallel (since Levi-Civita connection is metric). The proof was neat and worth taking a look IMO.