Polar factorization theorem

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 ${\displaystyle \left[0,1\right]}$.

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.

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]