Polar Factorization Theorem

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In the theory of optimal transport, 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 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 map. Then, $\sigma$ is said to be measure preserving if for every $\nu$-measurable subset $\Omega$ of $Y$, $\sigma ^{-1}(\Omega )$ is $\mu$-measurable and $\mu (\sigma ^{-1}(\Omega ))=\nu (\Omega )$, that is: $\textstyle \int _{X}f\circ \sigma d\mu =\int _{Y}fd\nu$ with $f$ that is $\nu$-integrable and $\sigma \circ f$ that 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. Assume $\xi \left(x\right)=Mx$ where $M$ is an invertible $d\times d$ matrix. Let $\mu$ be the ${\mathcal {N}}\left(0,I_{d}\right)$ probability measure. Then 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 allows to recover Helmholtz decomposition. Let $x\rightarrow V\left(x\right)$ be a smooth vector field. Then it can 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.

References

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