The divergence of a vector field $X$ on a manifold $M$ is defined usually as the function $\text{Div}(.)$ such that $(\text{Div} X) \;\mu =L_X \mu$ for $\mu$ a volume form.
I know that there is also an alternative expression for $\text{Div} \; X$ in terms of the covariant derivative:
$\text{div} \; X= \text{trace}(\nabla X)= \nabla^aX_a.$
Can anyone explain to me how to prove this equivalence?
Best Answer
To be precise, given an oriented metric $g$ on an $n$-manifold with volume form $\mu_{b_1 \cdots b_n}$, the divergence of a vector field w.r.t. $\mu$ coincides with $\nabla^a X_a$, where $\nabla$ is the Levi-Civita connection of $g$.
One way to see this is as follows: We can write the Lie derivative of a $k$-tensor $\phi$ on $M$ w.r.t. $X$ in terms of $\nabla$ (any torsion-free connection will do) as $$(\mathcal{L}_X \phi)_{b_1 \cdots b_k} = X^c \nabla_c \phi_{b_1 \cdots b_k} + \sum_{i = 1}^k (\nabla_{b_i} X^c) \phi_{b_1 \cdots c \cdots b_k}.$$
If we take $\phi$ to be $\mu$, then the first term is zero (the volume form of a Riemannian metric is parallel with respect to the Levi-Civita connection). Now, we may write the second term as $$-(n + 1) \nabla_{[c} X^c \mu_{b_1 \cdots b_n]} + \nabla_c X^c \mu_{b_1 \cdots b_n}.$$ Here the bracket $[ \cdots ]$ indicates antisymmetrization over the enclosed indices, so the first term in this expression is the trace of a tensor of type $(n + 1, 1)$ which is skew in its $n + 1$ lower indices. But any such tensor on an $n$-manifold is zero, so only the second term survives, giving (now in classical notation, for emphasis) $$\mathcal{L}_X \mu = (\mathrm{tr} \nabla X) \mu = (\operatorname{div} X) \mu.$$