# Polar factorization theorem

In optimal transport, a branch of mathematics, polar factorization of vector fields is a basic result due to Brenier (1987), with antecedents of Knott-Smith (1984) and Rachev (1985), that generalizes many existing results among which are the polar decomposition of real matrices, and the rearrangement of real-valued functions.

## The theorem

Notation. Denote $\xi _{\#}\mu$ the image measure of $\mu$ through the map $\xi$.
Definition: Measure preserving map. Let $(X,\mu )$ and $(Y,\nu )$ be some probability spaces and $\sigma :X\rightarrow Y$ a measurable map. Then, $\sigma$ is said to be measure preserving iff $\sigma _{\#}\mu =\nu$, where $\#$ is the pushforward measure. Spelled out: for every $\nu$-measurable subset $\Omega$ of $Y$, $\sigma ^{-1}(\Omega )$ is $\mu$-measurable, and $\mu (\sigma ^{-1}(\Omega ))=\nu (\Omega )$. The latter is equivalent to:
    $\int _{X}(f\circ \sigma )(x)\mu (dx)=\int _{X}(\sigma ^{*}f)(x)\mu (dx)=\int _{Y}f(y)(\sigma _{\#}\mu )(dy)=\int _{Y}f(y)\nu (dy)$
where $f$ is $\nu$-integrable and $f\circ \sigma$ is $\mu$-integrable.
Theorem. Consider a map $\xi :\Omega \rightarrow R^{d}$ where $\Omega$ is a convex subset of $R^{d}$, and $\mu$ a measure on $\Omega$ which is absolutely continuous. Assume that $\xi _{\#}\mu$ is absolutely continuous. Then there is a convex function $\varphi :\Omega \rightarrow R$ and a map $\sigma :\Omega \rightarrow \Omega$ preserving $\mu$ such that
$\xi =\left(\nabla \varphi \right)\circ \sigma$
In addition, $\nabla \varphi$ and $\sigma$ are uniquely defined almost everywhere.

## Applications and connections
### Dimension 1

In dimension 1, and when $\mu$ is the Lebesgue measure over the unit interval, the result specializes to Ryff's theorem. When $d=1$ and $\mu$ is the uniform distribution over $\left[0,1\right]$, the polar decomposition boils down to
$\xi \left(t\right)=F_{X}^{-1}\left(\sigma \left(t\right)\right)$
where $F_{X}$ is cumulative distribution function of the random variable $\xi \left(U\right)$ and $U$ has a uniform distribution over $\left[0,1\right]$. $F_{X}$ is assumed to be continuous, and $\sigma \left(t\right)=F_{X}\left(\xi \left(t\right)\right)$ preserves the Lebesgue measure on $\left[0,1\right]$.

### Polar decomposition of matrices

When $\xi$ is a linear map and $\mu$ is the Gaussian normal distribution, the result coincides with the polar decomposition of matrices. Assuming $\xi \left(x\right)=Mx$ where $M$ is an invertible $d\times d$ matrix and considering $\mu$ the ${\mathcal {N}}\left(0,I_{d}\right)$ probability measure, the polar decomposition boils down to
$M=SO$
where $S$ is a symmetric positive definite matrix, and $O$ an orthogonal matrix. The connection with the polar factorization is $\varphi \left(x\right)=x^{\top }Sx/2$ which is convex, and $\sigma \left(x\right)=Ox$ which preserves the ${\mathcal {N}}\left(0,I_{d}\right)$ measure.

### Helmholtz decomposition

The results also allow to recover Helmholtz decomposition. Letting $x\rightarrow V\left(x\right)$ be a smooth vector field it can then be written in a unique way as
$V=w+\nabla p$
where $p$ is a smooth real function defined on $\Omega$, unique up to an additive constant, and $w$ is a smooth divergence free vector field, parallel to the boundary of $\Omega$.
The connection can be seen by assuming $\mu$ is the Lebesgue measure on a compact set $\Omega \subset R^{n}$ and by writing $\xi$ as a perturbation of the identity map
$\xi _{\epsilon }(x)=x+\epsilon V(x)$
where $\epsilon$ is small. The polar decomposition of $\xi _{\epsilon }$ is given by $\xi _{\epsilon }=(\nabla \varphi _{\epsilon })\circ \sigma _{\epsilon }$. Then, for any test function $f:R^{n}\rightarrow R$ the following holds:
$\int _{\Omega }f(x+\epsilon V(x))dx=\int _{\Omega }f((\nabla \varphi _{\epsilon })\circ \sigma _{\epsilon }\left(x\right))dx=\int _{\Omega }f(\nabla \varphi _{\epsilon }\left(x\right))dx$
where the fact that $\sigma _{\epsilon }$ was preserving the Lebesgue measure was used in the second equality.
In fact, as $\textstyle \varphi _{0}(x)={\frac {1}{2}}\Vert x\Vert ^{2}$, one can expand $\textstyle \varphi _{\epsilon }(x)={\frac {1}{2}}\Vert x\Vert ^{2}+\epsilon p(x)+O(\epsilon ^{2})$, and therefore $\textstyle \nabla \varphi _{\epsilon }\left(x\right)=x+\epsilon \nabla p(x)+O(\epsilon ^{2})$. As a result, $\textstyle \int _{\Omega }\left(V(x)-\nabla p(x)\right)\nabla f(x))dx$ for any smooth function $f$, which implies that $w\left(x\right)=V(x)-\nabla p(x)$ is divergence-free.