# A 'projective' property of the Haar U(n) measure

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 $$(m,n)$$?

• 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.
– YCor
Oct 20 at 7:41
• 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 $m$. Oct 20 at 7:55
• I presume $\mathcal{B}$ means Borel functions? Oct 20 at 8:13
• 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 $U(n)$. Oct 20 at 8:17
• For $m=1$ it exists: the determinant. I'm not sure (in the continuous/smooth category) about the simplest case beyond, say $m=2$, $n=3$.
– YCor
Oct 20 at 8:23

For any $$m the $$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.

• 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. Oct 20 at 15:06
• 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. Oct 20 at 15:08
• 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. Oct 20 at 15:11
• This is the only place where "using my strong intuitive understanding of quantum mechanics" is called "naive", but I agree, that works. Oct 20 at 15:13
• Thanks! I first made a mistake when checking if the matrix is unitary with algebraic methods, but now I'm convinced. Oct 20 at 21:10

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.