Integral of a trace and divergence theorem

Question

I have found the following equality in a paper I am reading and I got stuck because I am not able to check it.

We have a divergence-free, smooth vector field $V \colon \mathbb T^N \to \mathbb T^N$ defined on the torus. It is claimed that $$ \int_{\mathbb T^N} \text{Tr}[(V \otimes V) \cdot \nabla V] \, dx = 0 $$ where $dx$ is the standard Lebesgue measure on the torus. My only idea to check this is to resort to some use of integration by parts and the divergence theorem: the "trace" appearing in the integral should be reduced to the divergence of some quantity (using the fact that $\text{div } V = 0$) and then the conclusion would follow by divergence theorem indeed (since we are on the torus).

However, something breaks: in 2D an explicit computation tells me that the integrand is $$ v_1^2 \partial_1 v_1 + v_2^2 \partial_2v_2 + v_1v_2 (\partial_1 v_2 + \partial_2 v_1) $$ (with obvious notation for derivatives and $V=(v_1,v_2)$) and I fail to write this as divergence of something, not even using integration by parts or the fact that $\partial_1 v_1 = - \partial_2 v_2$.

I feel there should be some simple (general?) trick behind, but after a night of computations I’m giving up. Thanks for your help.

Answer 1

Consider the following vector fields: $$X = |V|^2 V = \sum_j V_j^2 V$$ on $\mathbb T^N$. Then by divergence theorem, $$\int_{\mathbb T^N} \operatorname{div} (X) = 0.$$ Since

\begin{align} \operatorname{div} (X) & = \operatorname{div} (|V|^2 V) \\ &= \sum_i \nabla_i (|V|^2 V_i) \\ &= 2 \sum _{i,j} (\nabla_i V_j) V_j V_i + |V|^2 \sum_i \nabla_iV_i \\ &= 2 \sum_{i,j} V_i V_j \nabla_i V_j \\ &= 2\operatorname{tr} ( V\otimes V, \nabla V), \end{align}

one obtains the result.

Answer 2

So, integrating by parts and using the fact that $\nabla\cdot v=0$ you have $$ \int_{\mathbb{T}^N} \mathrm{Tr}((v\otimes v)\cdot\nabla v) = -\int_{\mathbb{T}^N} \mathrm{Tr}(\nabla\cdot(v\otimes v) \otimes v) = - 0 -\int_{\mathbb{T}^N} \mathrm{Tr}((v\cdot\nabla v) \otimes v) = -\int_{\mathbb{T}^N} \mathrm{Tr}((v\otimes v)\cdot\nabla v) $$ so $$ 2\int_{\mathbb{T}^N} \mathrm{Tr}((v\otimes v)\cdot\nabla v) = 0. $$