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Let $U(n)$ be the compact manifold of unitary $(n \times n)$-matrices and let $\mu_n$ denote the Haar-probability measure on $U(n)$. For $m < n$ does there exists a measurable (maybe even continuous or smooth) map $$ F: \ U(n) \rightarrow U(m) $$ with the property, that $$ \mu_m(A) = \mu_n(F^{-1}(A)) $$ for every $A \in \mathcal{B}(U(n))$?

Are there maybe certain necessary conditions on $(m,n)$?

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    $\begingroup$ Essentially all non-atomic "reasonable" probability spaces are equivalent to $[0,1]$ with Lebesgue. So there should even exist bijective probability-preserving Lebesgue bi-measurable maps between the various $U(n)$. Continuity is of course another question. $\endgroup$
    – YCor
    Oct 20 at 7:41
  • $\begingroup$ Thanks, this already helps. I would still be interested in continuity though. I suspect there to be some rather strong conditions on $(m,n)$, like that $n$ might need to be a multiple of $m$. $\endgroup$
    – Tardis
    Oct 20 at 7:55
  • $\begingroup$ I presume $\mathcal{B}$ means Borel functions? $\endgroup$ Oct 20 at 8:13
  • $\begingroup$ In this case Borel sets on $U(n)$. In other words I am looking for $\mu_m = (F)_* \mu_n$ to hold. It would however already suffice to show $\mu_m(A) = \mu_n(F^{-1}(A))$ for all open sets in $U(n)$. $\endgroup$
    – Tardis
    Oct 20 at 8:17
  • $\begingroup$ For $m=1$ it exists: the determinant. I'm not sure (in the continuous/smooth category) about the simplest case beyond, say $m=2$, $n=3$. $\endgroup$
    – YCor
    Oct 20 at 8:23
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For any $m<n$ the $n\times n$ unitary matrix $\Omega$ has the block decomposition $$\Omega=\begin{pmatrix} A&B\\ C&D\end{pmatrix},$$ where $A$ has dimensions $m\times m$, $D$ has dimensions $(n-m)\times(n-m)$, $B$ has dimensions $m\times(n-m)$ and $C$ has dimensions $(n-m)\times m$. Up to a set of measure zero, the matrix $D$ will not have a unit eigenvalue, so $I-D$ is invertible. We then define the continuous map $F$ from $U(n)$ to $U(m)$ by $$F(\Omega)=A+B(I-D)^{-1}C.$$ One readily checks that $F(\Omega)$ is unitary$^\ast$ and as David Speyer points out $F(\Omega)$ inherits$^{\ast\ast}$ the Haar measure from $\Omega$.


$^\ast$ More generally, for any $\Omega\in U(n)$ and $V\in U(n-m)$ the matrix $U=A+B(I-VD)^{-1}VC$ is unitary. One can think of $V$ as the reflection matrix of a barrier that closes off $n-m$ scattering channels. Then the remaining $m$ channels have scattering matrix $U=A+\sum_{k=0}^\infty B(VD)^kVC$, where $k+1$ counts the reflections off the barrier. As a further check for the unitarity of $U$, I can offer a Mathematica Notebook.

$^{\ast\ast}$ Since for any $g\in U(m)$, $G={{g\;\;0}\choose{0\;\;I}}\in U(n)$ one has $F(G\Omega)=gF(\Omega )$, the measure remains left-invariant.

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    $\begingroup$ To see that $F$ inherits Haar measure, note that if you replace $\Omega$ by $\text{diag}(g, \mathrm{Id}_{n-m})$ then you replace $F(\Omega)$ by $g F(\Omega)$. So the induced probability measure on $U_m$ is left invariant and must be Haar measure. $\endgroup$ Oct 20 at 15:06
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    $\begingroup$ I haven't figured out why $F(\Omega)$ is unitary, though. My suggestion was going to be to write $A$ is in polar form as $UP$ and then send $\Omega$ to $U$; this works for the same reason, on the measure 1 set where $A$ is invertible. $\endgroup$ Oct 20 at 15:08
  • $\begingroup$ well, my naive way of checking unitarity is to write $F(\Omega)=A+BC+BDC+BD^2C+BD^3C+\cdots$ and noting that this series contains all possible scattering sequences from $m$ channels in to $m$ channels out. So no probability is lost and if $\Omega$ is unitary so must $F(\Omega)$ be. $\endgroup$ Oct 20 at 15:11
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    $\begingroup$ This is the only place where "using my strong intuitive understanding of quantum mechanics" is called "naive", but I agree, that works. $\endgroup$ Oct 20 at 15:13
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    $\begingroup$ Thanks! I first made a mistake when checking if the matrix is unitary with algebraic methods, but now I'm convinced. $\endgroup$
    – Tardis
    Oct 20 at 21:10
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You can get a map which is continuous outside a set of lower dimension. Let $K\subset L$ be compact Lie groups and let $s:K\backslash L\to L$ be a section to the projection $L\to K\backslash L$. Now this section can be chosen continuous outside a set of lower dimension, since the projection is a fibre bundle. For $x\in L$ we write $s(x)$ for $s(Kx)$. We define a map $f:L\to K$ by $f(x)=xs(x)^{-1}$. Since $f(kx)=kf(x)$ for $k\in K$, the measure $f_*($Haar$)$ is invariant, hence Haar.

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