Suppose that u is suitably regular (e.g. $C^2(\mathbb{R}^N,\mathbb{R}^N)$ or $W^{1,2}(\mathbb{R}^N)^N$) and we write $$\det (\nabla u)=\nabla u^1 \cdot\sigma$$
for some $\sigma$ (obtained via the Laplace expansion), where $\nabla u$ the $N\times N$ matrix of first derivatives of $u$ and $\nabla u^1$ is the first row of $\nabla u$.
I believe one can show that $$\mathrm{div } \sigma =0.$$ It is possible to show it in 2 and 3 dimensions by expanding but I am trying to show it for general N. I feel I am probably missing a simple trick with determinants but I am struggling to calculate it. Please can you tell me if there is a nice way to calculate the divergence? I have tried induction but it got a bit messy and didn't seem to help.
Thanks!
Best Answer
OK, so I think i worked it out now. If we use the Physics/ Einstein notation we can write: $$\det(\nabla u)= \varepsilon_{i_1,\cdots,i_n}\partial_{i_1} u^1\cdots\partial_{i_n} u^n$$ where $\varepsilon_{i_m,\cdots,i_m}=0 $ and $\varepsilon$ is anti symmetric considered as an n-linear form. Writing this equation more explicitly in the first summed component we see that in the notation of the question, $$\nabla u^1\cdot\sigma=\Sigma_k(\partial_{k} u^1) (\varepsilon_{k,\cdots,i_n}\partial_{i_2} u^2 \cdots\partial_{i_n} u^n).$$ Then $\sigma_k=\varepsilon_{k,\cdots,i_n}\partial_{i_2} u^2 \cdots\partial_{i_n} u^n.$ Therefore, $$\text{div}\sigma=\varepsilon_{k,\cdots,i_n}\partial_{k}(\partial_{i_2} u^2 \cdots\partial_{i_n} u^n).$$ Finally we can use the product rule, the commutativity of the partial derivatives on our function $u$ and the anti-symmetry of $\varepsilon$ to see that $\text{div}\sigma=0.$ To see this note that we have a sum with terms of the form $$\sum_{k,i_j}\varepsilon_{k,i_2\cdots,i_j\cdots i_n}\partial_{k,i_j}u^j\prod_{l\neq k,i_j} \partial_lu^l=0.$$