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

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

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

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

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

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

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

Dimension 1 edit

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

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

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

Polar decomposition of matrices edit

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

${\displaystyle M=SO}$ 

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

Helmholtz decomposition edit

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

${\displaystyle V=w+\nabla p}$ 

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

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

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

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

${\displaystyle \int _{\Omega }f(x+\epsilon V(x))dx=\int _{\Omega }f((\nabla \varphi _{\epsilon })\circ \sigma _{\epsilon }\left(x\right))dx=\int _{\Omega }f(\nabla \varphi _{\epsilon }\left(x\right))dx}$ 

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

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

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