# Two definitions of divergence

Let $$M$$ be an oriented Riemannian manifold with volume form $$dV$$, and let $$X$$ be a smooth vector field on $$M$$. Recall that the divergence of $$X$$ is characterized by the formula $$d(i_X dV)=(\text{div}X) dV$$ (where $$i_X$$ is the interior multiplication by $$X$$.)

In p.115 of the book Spin Geometry (Lawson, Michelson), it is written that, at $$x\in M$$, $$\text{div}(X)(x)=\sum_j \langle \nabla_{e_j} X,e_j\rangle_x$$, where $$e_1,\dots,e_n$$ is an orthonormal frame of $$TM$$. Are these two definitions equivalent? I can't see why..

The definition on L&M is $${\rm div}(X)={\rm tr}(\nabla X)$$. So it suffices to prove that $${\rm d}(\iota_X{\rm d}V) = {\rm tr}(\nabla X){\rm d}V$$. Set $$n = \dim M$$. Here's a clean way (which doesn't really rely on coordinates):

1. For any $$\alpha \in \Omega^k(M)$$, verify that $$\mathcal{L}_X\alpha = \nabla_X\alpha + \sum_{i=1}^k \alpha(\cdot,\cdots, [\nabla X]\cdot, \cdots, \cdot),$$where $$\nabla X$$ enters in the $$i$$-th slot. Here, $$\mathcal{L}$$ stands for the Lie derivative, and this is true for any torsionfree connection. Sanity-check: $$\mathcal{L}_X$$ and $$\nabla_X$$ are derivations, so their difference is a tensor.

2. For $$k=n$$, $$\alpha$$ is closed by dimensional reasons, so $$\mathcal{L}_X\alpha = \iota_X({\rm d}\alpha) +{\rm d}(\iota_X\alpha) = {\rm d}(\iota_X\alpha).$$This is Cartan's homotopy formula.

3. The Riemannian volume form $$\alpha = {\rm d}V$$ is parallel (because the Ricci tensor of the Levi-Civita connection of the metric is symmetric).

4. The sum on the right side of (1) when $$k=n$$ equals $${\rm tr}(\nabla X)\alpha$$. This is linear algebra, and you can prove it when $$\alpha$$ is any linear $$n$$-form on a $$n$$-dimensional vector space $$V$$ and $$\nabla X$$ is any linear operator $$T:V \to V$$.

5. Put everything together to obtain $${\rm d}(\iota_X{\rm d}V) = {\rm tr}(\nabla X){\rm d}V$$.

The bottom line is that $${\rm div}(X)$$ makes sense if you have either a connection, or a volume form. For the Levi-Civita connection of a metric and the corresponding Riemannian volume form, both divergences coincide. More generally, if you choose any torsionfree connection and a parallel volume form (whose existence forces the Ricci tensor of the connection to be symmetric), the results will coincide.

• Does 1 mean that $(L_X\alpha)(v_1,\dots,v_n)=(\nabla_X \alpha)(e_1,\dots,e_n)+\sum_i \alpha(e_1,\dots,\nabla_{e_i}V,\dots,e_n)$? Aug 1 at 7:29
• Yes, exactly. ${}{}{}$ Aug 1 at 7:31