Polar factorization theorem

In optimal transport, a branch of mathematics, polar factorization of vector fields is a basic result due to Brenier (1987),^[1] with antecedents of Knott-Smith (1984)^[2] and Rachev (1985),^[3] that generalizes many existing results among which are the polar decomposition of real matrices, and the rearrangement of real-valued functions.

Notation. Denote ${\displaystyle \xi _{\#}\mu }${\displaystyle \xi _{\#}\mu } the image measure of ${\displaystyle \mu }${\displaystyle \mu } through the map ${\displaystyle \xi }${\displaystyle \xi }.

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

${\displaystyle \int _{X}(f\circ \sigma )(x)\mu (dx)=\int _{X}(\sigma ^{*}f)(x)\mu (dx)=\int _{Y}f(y)(\sigma _{\#}\mu )(dy)=\int _{Y}f(y)\nu (dy)}${\displaystyle \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 ${\displaystyle f}${\displaystyle f} is ${\displaystyle \nu }${\displaystyle \nu }-integrable and ${\displaystyle f\circ \sigma }${\displaystyle f\circ \sigma } is ${\displaystyle \mu }${\displaystyle \mu }-integrable.

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

${\displaystyle \xi =\left(\nabla \varphi \right)\circ \sigma }${\displaystyle \xi =\left(\nabla \varphi \right)\circ \sigma }

In addition, ${\displaystyle \nabla \varphi }${\displaystyle \nabla \varphi } and ${\displaystyle \sigma }${\displaystyle \sigma } are uniquely defined almost everywhere.^{[1] [4]}

In dimension 1, and when ${\displaystyle \mu }${\displaystyle \mu } is the Lebesgue measure over the unit interval, the result specializes to Ryff's theorem.^[5] When ${\displaystyle d=1}${\displaystyle d=1} and ${\displaystyle \mu }${\displaystyle \mu } is the uniform distribution over ${\displaystyle \left[0,1\right]}${\displaystyle \left[0,1\right]}, the polar decomposition boils down to

${\displaystyle \xi \left(t\right)=F_{X}^{-1}\left(\sigma \left(t\right)\right)}${\displaystyle \xi \left(t\right)=F_{X}^{-1}\left(\sigma \left(t\right)\right)}

where ${\displaystyle F_{X}}${\displaystyle F_{X}} is cumulative distribution function of the random variable ${\displaystyle \xi \left(U\right)}${\displaystyle \xi \left(U\right)} and ${\displaystyle U}${\displaystyle U} has a uniform distribution over ${\displaystyle \left[0,1\right]}${\displaystyle \left[0,1\right]}. ${\displaystyle F_{X}}${\displaystyle F_{X}} is assumed to be continuous, and ${\displaystyle \sigma \left(t\right)=F_{X}\left(\xi \left(t\right)\right)}${\displaystyle \sigma \left(t\right)=F_{X}\left(\xi \left(t\right)\right)} preserves the Lebesgue measure on ${\displaystyle \left[0,1\right]}${\displaystyle \left[0,1\right]}.

When ${\displaystyle \xi }${\displaystyle \xi } is a linear map and ${\displaystyle \mu }${\displaystyle \mu } is the Gaussian normal distribution, the result coincides with the polar decomposition of matrices. Assuming ${\displaystyle \xi \left(x\right)=Mx}${\displaystyle \xi \left(x\right)=Mx} where ${\displaystyle M}${\displaystyle M} is an invertible ${\displaystyle d\times d}${\displaystyle d\times d} matrix and considering ${\displaystyle \mu }${\displaystyle \mu } the ${\displaystyle {\mathcal {N}}\left(0,I_{d}\right)}${\displaystyle {\mathcal {N}}\left(0,I_{d}\right)} probability measure, the polar decomposition boils down to

${\displaystyle M=SO}${\displaystyle M=SO}

where ${\displaystyle S}${\displaystyle S} is a symmetric positive definite matrix, and ${\displaystyle O}${\displaystyle O} an orthogonal matrix. The connection with the polar factorization is ${\displaystyle \varphi \left(x\right)=x^{\top }Sx/2}${\displaystyle \varphi \left(x\right)=x^{\top }Sx/2} which is convex, and ${\displaystyle \sigma \left(x\right)=Ox}${\displaystyle \sigma \left(x\right)=Ox} which preserves the ${\displaystyle {\mathcal {N}}\left(0,I_{d}\right)}${\displaystyle {\mathcal {N}}\left(0,I_{d}\right)} measure.

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

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

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

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

${\displaystyle \xi _{\epsilon }(x)=x+\epsilon V(x)}${\displaystyle \xi _{\epsilon }(x)=x+\epsilon V(x)}

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

${\displaystyle \int _{\Omega }f(x+\epsilon V(x))dx=\int _{\Omega }f((\nabla \varphi _{\epsilon })\circ \sigma _{\epsilon }\left(x\right))dx=\int _{\Omega }f(\nabla \varphi _{\epsilon }\left(x\right))dx}${\displaystyle \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 ${\displaystyle \sigma _{\epsilon }}${\displaystyle \sigma _{\epsilon }} was preserving the Lebesgue measure was used in the second equality.

In fact, as ${\displaystyle \textstyle \varphi _{0}(x)={\frac {1}{2}}\Vert x\Vert ^{2}}${\displaystyle \textstyle \varphi _{0}(x)={\frac {1}{2}}\Vert x\Vert ^{2}}, one can expand ${\displaystyle \textstyle \varphi _{\epsilon }(x)={\frac {1}{2}}\Vert x\Vert ^{2}+\epsilon p(x)+O(\epsilon ^{2})}${\displaystyle \textstyle \varphi _{\epsilon }(x)={\frac {1}{2}}\Vert x\Vert ^{2}+\epsilon p(x)+O(\epsilon ^{2})}, and therefore ${\displaystyle \textstyle \nabla \varphi _{\epsilon }\left(x\right)=x+\epsilon \nabla p(x)+O(\epsilon ^{2})}${\displaystyle \textstyle \nabla \varphi _{\epsilon }\left(x\right)=x+\epsilon \nabla p(x)+O(\epsilon ^{2})}. As a result, ${\displaystyle \textstyle \int _{\Omega }\left(V(x)-\nabla p(x)\right)\nabla f(x))dx}${\displaystyle \textstyle \int _{\Omega }\left(V(x)-\nabla p(x)\right)\nabla f(x))dx} for any smooth function ${\displaystyle f}${\displaystyle f}, which implies that ${\displaystyle w\left(x\right)=V(x)-\nabla p(x)}${\displaystyle w\left(x\right)=V(x)-\nabla p(x)} is divergence-free.^{[1] [6]}

• polar decomposition – Representation of invertible matrices as unitary operator multiplying a Hermitian operator

• Measures (measure theory)
• Theorems involving convexity

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