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We introduce in $\mathbb{R}^3$ with the usual Euclidean metric $\langle \cdot, \cdot \rangle$, the connection defined in Cartesian ccoordinates $(x^1, x^2, x^3)$ by $$\Gamma_{jk}^i = \omega \varepsilon_{ijk}$$ where $\omega: \mathbb{R}^3 \rightarrow \mathbb{R}$ is a smooth function and $$\varepsilon_{ijk} = \begin{cases} +1 \quad \text{if }(i,j,k) \text{ is an even permutation of }(1,2,3)\\ -1 \quad \text{se }(i,j,k) \text{ is an odd permutation of }(1,2,3)\\ 0 \quad \text{ otherwise} \end{cases}$$ Show that $\nabla$ is compatible with $\langle \cdot, \cdot \rangle$;

I've taken a local coordinate system $(x_1, x_2, \dots, x_n)$ and used the

We know that a connection $\nabla$ is compatible with a metric $\langle \cdot, \cdot \rangle$ if and only if for every differentiable curve $c: I \rightarrow M$ and every pair $V, \,W$ of vector fields differentiable along $c$ we have $$\frac{d}{dt} \langle V, W \rangle = \left\langle \frac{DV}{dt}, W \right\rangle +\left\langle V, \frac{DW}{dt} \right\rangle, \quad t \in I$$

Let $(x_1(t), x_2(t), x_3(t))$ be the parametrization of $c(t)$ in the local coordinate system, and let $X_i = \frac{\partial}{\partial x_i}$, and write $V$ as $V =\sum_{j=1}^3 v^j X_j$ (and similarly for $W$), where $v_j = v_j(t)$. We can express the covariant derivative as

$$ \begin{align}\frac{DV}{dt} &= \sum_{k=1}^3 \left[ \frac{dv^k}{dt} + \sum_{i,j} v^j \frac{dx_i}{dt} \Gamma_{ij}^k\right] X_k\\ &= \sum_{k=1}^3 \left[ \frac{dv^k}{dt} + \sum_{i,j} v^j \frac{dx_i}{dt} \omega \varepsilon_{ijk}\right] X_k \end{align}$$

I thought about calculating $\frac{d}{dt} \langle V, W \rangle$ and $ \left\langle \frac{DV}{dt}, W \right\rangle +\left\langle V, \frac{DW}{dt} \right\rangle$ in these terms and showing they are equal, but it would be too messy. I think there must be a smart solution that doesn't require too many calculations.

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    $\begingroup$ You're using Godinho & Natário, right? Great book. $\endgroup$
    – Ivo Terek
    Feb 14 '21 at 22:48
  • $\begingroup$ Yes, I'm using that book. $\endgroup$
    – José
    Feb 15 '21 at 0:06
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Since $\nabla\langle \cdot,\cdot\rangle$ is a tensor, it suffices to check that $(\nabla_{\partial_k}\langle\cdot,\cdot\rangle)(\partial_i,\partial_j) = 0$ for all $i,j,k \in \{1,2,3\}$. In other words, we must show that $$\partial_k \langle \partial_i,\partial_j\rangle = \langle \nabla_{\partial_k}\partial_i,\partial_j\rangle + \langle \partial_i, \nabla_{\partial_k}\partial_j\rangle$$for all $i,j,k \in \{1,2,3\}$. The left side is obviously zero, so the situation doesn't look so bad. By the definition of $\nabla$, we have that $$\begin{align} \langle \nabla_{\partial_k}\partial_i,\partial_j\rangle + \langle \partial_i, \nabla_{\partial_k}\partial_j\rangle &= \langle \omega \varepsilon^r_{~ki}\partial_r,\partial_j\rangle + \langle \partial_i, \omega \varepsilon^r_{~kj}\partial_r\rangle \\ &= \omega \varepsilon^r_{~ki} \delta_{rj} + \omega \varepsilon^r_{~kj}\delta_{ir} \\ &= \omega(\varepsilon^j_{~ki} + \varepsilon^i_{~kj}) \\ &= 0,\end{align}$$because since $(j,k,i) \mapsto (i,k,j)$ is an odd permutation, we have that $\varepsilon^j_{~ki} =- \varepsilon^i_{~kj}$.

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