# Volume form is parallel with respect to Levi-Civita connection

Let $(M,g)$ be a Riemannian manifold of dimension $m$. I would like to prove the following stament.

The riemannian volume form $\omega$ is parallel with respect to Levi-Civita connection.

This is my attempt:

Let $\omega$ be the riemannian volume form. Let $X$ be an arbitrary vector field such that $\gamma$ is an integral curve for $X$ starting at $p$. Now, let's choose $v_1,\ldots, v_m$ linear positively oriented independent vectors in $T_p M$ with respect to $\omega_p$. We can extend these vectors to $V_1,\ldots, V_m$, parallel vectors fields along $\gamma$, by parallel transport. So now we have: $$(\nabla_{X_p} \omega)(V_1,\ldots,V_m)={X_p} \omega(V_1,\ldots,V_m)-\sum_{i=1}^m\omega(V_1,\ldots,\nabla_{X_p} V_i, \ldots, V_m).$$ Since $V_i$ is parallel along $\gamma$, $\nabla_{X_p} V_i=0$, for each $i=1,\ldots, m$. Hence, $$(\nabla_{X_p} \omega)(V_1,\ldots,V_m)={X_p} \omega(V_1,\ldots,V_m)=X_p\bigg(\sqrt{\det(g(V_i,V_j))}\bigg)=0,$$ where the last term is zero since $g(V_i,V_j)$ is constant along $\gamma$.

Your proof is correct. You could also just work in normal coordinates centered at $$p$$ so that you don't need to extend $$V$$, since it's enough to check the equality on vector fields of the form $$(V_{1},\ldots,V_{m})=\left(\frac{\partial}{\partial x^{i_1}},\ldots,\frac{\partial}{\partial x^{i_m}}\right).$$