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 ${\xi }_{\mathrm{\#}}\mu$ the image measure of $\mu$ through the map $\xi$.

Definition: Measure preserving map. Let $(X,\mu )$ and $(Y,\nu )$ be some probability spaces and $\sigma :X\to Y$ a measurable map. Then, $\sigma$ is said to be measure preserving iff ${\sigma }_{\mathrm{\#}}\mu =\nu$, where $\mathrm{\#}$ is the pushforward measure. Spelled out: for every $\nu$-measurable subset $\mathrm{\Omega }$ of $Y$, ${\sigma }^{-1}(\mathrm{\Omega })$ is $\mu$-measurable, and $\mu ({\sigma }^{-1}(\mathrm{\Omega }))=\nu (\mathrm{\Omega })$. The latter is equivalent to:

$\underset{X}{\int }(f\circ \sigma )(x)\mu (dx)=\underset{X}{\int }({\sigma }^{*}f)(x)\mu (dx)=\underset{Y}{\int }f(y)({\sigma }_{\mathrm{\#}}\mu )(dy)=\underset{Y}{\int }f(y)\nu (dy)$

where $f$ is $\nu$-integrable and $f\circ \sigma$ is $\mu$-integrable.

Theorem. Consider a map $\xi :\mathrm{\Omega }\to R^d$ where $\mathrm{\Omega }$ is a convex subset of $R^d$, and $\mu$ a measure on $\mathrm{\Omega }$ which is absolutely continuous. Assume that ${\xi }_{\mathrm{\#}}\mu$ is absolutely continuous. Then there is a convex function $\phi :\mathrm{\Omega }\to R$ and a map $\sigma :\mathrm{\Omega }\to \mathrm{\Omega }$ preserving $\mu$ such that

$\xi =\left(\mathrm{\nabla }\phi \right)\circ \sigma$

In addition, $\mathrm{\nabla }\phi$ 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 $[0,1]$, 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 $[0,1]$. $F_X$ is assumed to be continuous, and $\sigma \left(t\right)=F_X\left(\xi \left(t\right)\right)$ preserves the Lebesgue measure on $[0,1]$.

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 $𝒩(0,I_d)$ 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 $\phi \left(x\right)=x^{\mathrm{\top }}Sx/2$ which is convex, and $\sigma \left(x\right)=Ox$ which preserves the $𝒩(0,I_d)$ measure.

Helmholtz decomposition

The results also allow to recover Helmholtz decomposition. Letting $x\to V\left(x\right)$ be a smooth vector field it can then be written in a unique way as

$V=w+\mathrm{\nabla }p$

where $p$ is a smooth real function defined on $\mathrm{\Omega }$, unique up to an additive constant, and $w$ is a smooth divergence free vector field, parallel to the boundary of $\mathrm{\Omega }$.

The connection can be seen by assuming $\mu$ is the Lebesgue measure on a compact set $\mathrm{\Omega }\subset R^n$ and by writing $\xi$ as a perturbation of the identity map

${\xi }_{ϵ}(x)=x+ϵV(x)$

where $ϵ$ is small. The polar decomposition of ${\xi }_{ϵ}$ is given by ${\xi }_{ϵ}=(\mathrm{\nabla }{\phi }_{ϵ})\circ {\sigma }_{ϵ}$. Then, for any test function $f:R^n\to R$ the following holds:

$\underset{\mathrm{\Omega }}{\int }f(x+ϵV(x))dx=\underset{\mathrm{\Omega }}{\int }f((\mathrm{\nabla }{\phi }_{ϵ})\circ {\sigma }_{ϵ}\left(x\right))dx=\underset{\mathrm{\Omega }}{\int }f(\mathrm{\nabla }{\phi }_{ϵ}\left(x\right))dx$

where the fact that ${\sigma }_{ϵ}$ was preserving the Lebesgue measure was used in the second equality.

In fact, as ${\phi }_0(x)=\frac{1}{2}\Vert x{\Vert }^2$, one can expand ${\phi }_{ϵ}(x)=\frac{1}{2}\Vert x{\Vert }^2+ϵp(x)+O({ϵ}^2)$, and therefore $\mathrm{\nabla }{\phi }_{ϵ}\left(x\right)=x+ϵ\mathrm{\nabla }p(x)+O({ϵ}^2)$. As a result, $\underset{\mathrm{\Omega }}{\int }(V(x)-\mathrm{\nabla }p(x))\mathrm{\nabla }f(x))dx$ for any smooth function $f$, which implies that $w\left(x\right)=V(x)-\mathrm{\nabla }p(x)$ is divergence-free.

See also

• Template:Annotated link