# functions with orthogonal Jacobian

I'm working on a model that would require to use vectorial functions of $\mathbb{R}^n \rightarrow \mathbb{R}^n$, such that $\forall x, y \in \mathbb{R}^n$, $\lVert \frac{df(x)}{dx}(y) \lVert_2 = \lVert y \lVert_2$, ie with an orthogonal Jacobian.

I can only think of trivial functions (like $f(x) = Ox + c$ for $O$ orthogonal and $c \in \mathbb{R}^n$).

Are there other functions that verify this property? What would it be if we add the constraint $\forall x \in \mathbb{R}^n$, $\lVert f(x) \lVert_2 = \lVert x \lVert_2$ ?

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Such maps are conformal. A theorem of Liouville says that if $n\geq 3$, the only conformal maps (defined in some region in $R^n$) are Mobius. A Mobius map is a composition of inversions in spheres. For example $x\mapsto x/|x^2|$ is the inversion in the unit sphere. Inversions in all spheres generate the Mobius group.

Derivative of a conformal map is a constant times orthogonal. So if you require it to be orthogonal, you obtain only affine maps.

Liouville's theorem does not hold in dimension $2$. Conformal maps in dimension $2$ are complex analytic function whose derivative is not equal to zero. Your condition implies that the complex derivative has constant absolute value, so it is constant, and again you obtain an affine map.

Usually they include orientation-reversing maps to the Mobius group, so conformal maps can be preserving or reversing orientation.

EDIT. Of course Liouville proved his theorem for sufficiently smooth functions, and the proof is essentially the same as in the answer of David Speyer. However this theorem holds under much les restrictive assumptions (for some maps differentiable almost everywhere), and this is one of the subjects discussed in the book of Reshetnyak, Stability theorems in geometry and Analysis.

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You have answered a more interesting question, but I don't think you answered the question which was asked, since the OP wants the Jacobian to actually be orthogonal, not to be a scalar multiple of an orthogonal matrix. – David Speyer Jan 21 at 20:03
I already edited my answer to address the original question as well. – Alexandre Eremenko Jan 21 at 20:10

$\def\RR{\mathbb{R}}$A brute force approach shows that there are no other $C^2$ solutions. Let $F: \RR^n \to \RR^n$ have orthogonal Jacobian everywhere. We will show that the Hessian of $F$ vanishes everywhere, so $F$ is linear.

It is enough to show that Hessian vanishes at $0$, since there is nothing special about $0$. Translating and rotating our coordinates, we may assume that $F(0) =0$ and the Jacobian at $0$ is the identity. So, writing the components of $F$ as $(F_1, \ldots, F_n)$, we have $$F_j(x_1, \ldots, x_n) = x_j + \sum_{a,b} Q^j_{ab} x_a x_b + (\mbox{higher order terms}).$$ Here $Q^1$, $Q^2$, ..., $Q^n$ are each symmetric $n \times n$ matrices. Our goal is to show $Q^j=0$.

Up to linear terms, the $(i,j)$ entry in the Jacobian is $\delta_i^j + 2 \sum_k Q^j_{ik} x_k$. Writing down the condition that the $j$-th column has length $1$, up to linear terms, gives $1+2 \sum Q^j_{jk} x_k = 1$. So $Q^j_{jk}=0$ and, by the symmetry of $Q^j$, we also have $Q^j_{kj}=0$.

Let $i \neq j$. Writing down the condition that the $i$-th and $j$-th column are orthogonal, up to linear order, gives $2 \sum_k Q^i_{jk} x_k + 2 \sum_k Q^j_{ik} x_k=0$, so $Q^j_{ik} = - Q^i_{jk}$ whenever $i \neq j$.

If $(i,j,k)$ are all distinct, we have $Q^i_{jk} = - Q^j_{ik} = Q^k_{ij} = - Q^i_{kj} = - Q^i_{jk}$. So $Q^i_{jk}=0$.

If $j$ and $k$ are distinct, we have $Q^j_{kk} = Q^k_{jk}=0$.

In all cases, we have shown the entries of $Q$ are $0$.

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