# Is there a geometric reason for why every map has a divergence-free Jacobian cofactor matrix?

$\newcommand{\Cof}{\text{Cof}}$

Consider the following claim:

Let $f:\mathbb{R}^n \to \mathbb{R}^n$ be a $C^2$ map. Then $$\text{div} (\Cof df)=0,$$ where $\Cof df$ is the cofactor matrix of $df$, and the divergence is taken row-by-row.

In other words $$\sum_{j=1}^n \frac{\partial(Cof(Du))_{kj}}{\partial x_j} = 0,$$ for every $1 \le k \le n$.

This is proved in Evan's PDE book, in section 8.1 of the Calculus of Variation.

This identity is "universal", that is, it is satisfied by any smooth (or $C^2$) map $\mathbb{R}^n \to \mathbb{R}^n$.

I follow his proof, but is there an intuitive/geometric way to see why this should be true? I seems like an arbitrary identity - how would someone know this?

$$\newcommand{\Cof}{\operatorname{Cof}}$$ $$\newcommand{\R}{\mathbb{R}}$$

My previous answer was insufficient, and did not really prove the required identity. The following is an updated, completely new answer.

Here is a geometric way to think about this result. A convenient way for describing it is to use differential forms:

First, a lemma:

Lemma 1: Let $$f_0, f_1:M\rightarrow N$$ smooth homotopic maps between smooth manifolds $$M$$ and $$N$$. Suppose $$M$$ is compact with no boundary. Then, for every closed form $$\omega\in \Omega^m(N)$$ (where $$m=\dim M$$), $$\int_{M}f_0^*\omega=\int_Mf_1^*\omega.$$

The proof is essentially Stokes theorem.

This lemma implies that the functional $$E:C^{\infty}(M,N) \to \mathbb{R}$$ defined by $$E(f)=\int_{M}f^*\omega$$ is a null-Lagrangian, i.e. every smooth map is a critical point of it.

In, particular we have the following corollary:

Let $$\Omega \subseteq \R^n$$. Then $$E:C^{\infty}(\Omega,\R^n) \to \mathbb{R}$$ defined by $$E(f)=\int_{\Omega} f^*\text{Vol}_{\R^n}=\int_{\Omega} \det df \text{Vol}_{\R^n}$$ is a null-Lagrangian.

Now, it is easy to prove that the Euler-Lagrange equation of $$E$$ is exactly $$\text{div} \Cof (df)=0,$$ when the divergence is taken row-wise. Indeed,

Let $$f_t=f+tV$$ for some smooth variation field $$V:\Omega \to \R^n$$. Then $$\left. \frac{d}{dt} \right|_{t=0} E(f_t)=\int_{\Omega} \left. \frac{d}{dt} \right|_{t=0} \det df_t=\int_{\Omega} \left. \frac{d}{dt} \right|_{t=0} \det (df+tdV)=\int_{\Omega} \langle \Cof df,dV \rangle=\int_{\Omega} \langle \text{div} \Cof df,V \rangle,$$

where we used two facts:

1. The cofactor is the gradient of the determinant.
2. The divergence is the adjoint of the gradient $$d$$. (This is essentially Stokes theorem or integration by parts).

Thus, a map is critical if and only if $$\text{div} \Cof df=0. \tag{1}$$

Now, since any map $$f:\Omega \to \R^n$$ is critical (by lemma 1) this proves that identity $$(1)$$ holds universally.

Further discussion:

It turns out that there is nothing special in the Euclidean case; everything here can be generalized to arbitrary Riemannian manifolds:

Let $$M,N$$ be oriented $$d$$-dimensional Riemannian manifolds. Then $$E(f)=\int_{M}f^*\text{Vol}_N$$ is a null-Lagrangian. Its Euler-Lagranges equation is $$\delta (\Cof df)=0$$ which is satisfied by any smooth map $$f:M \to N$$.

Here $$\delta$$ and $$\Cof df$$ are natural generalizations of the cofactor matrix and the divergence operator to Riemannian settings:

1. $$\Cof df:TM \to f^*(TN)$$ is the cofactor map of $$df$$ defined by $$\Cof df= (-1)^{d-1} \star_{f^*TN}^{d-1} (\wedge^{d-1} df) \star_{TM}^1.$$

$$\Cof df$$ represents the action of $$df$$ on $$d-1$$ dimensional cubes. ($$\star$$ denotes the Hodge-dual operator, and the subscripts indicate w.r.t which metric it is taken).

1. Note that $$\Cof df \in \Omega^1(M,f^*TN)$$. $$\delta:\Omega^1(M,f^*TN) \to \Omega^0(M,f^*TN)=\Gamma(f^*TN)$$ is the adjoint of the pullback connection $$f^*\nabla^{TN}$$ of the Levi-Civita connection of $$N$$.

The full details of the derivation of the E-L equation in the Riemannian setting can be found in my paper, in lemma 2.9.

A non-variational approach for deriving this identity (as well as more information on the variational approach) can be found in this paper of mine.