Show a connection $\nabla$ is compatible with a metric $\langle \cdot, \cdot \rangle$ of $\mathbb{R}^3$

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.

• You're using Godinho & Natário, right? Great book. Feb 14 '21 at 22:48
• Yes, I'm using that book.
– José
Feb 15 '21 at 0:06

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 $$\varepsilon^j_{~ki} =- \varepsilon^i_{~kj}$$.