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Every locally compact group has left-invariants haar measures. In particular, the compact groups O(n) and U(n) have them.

I was wondering if there is a realization of such a measure on these groups, or its integral operator. Of course, right invariant ones are as good.

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  • $\begingroup$ Compact groups are unimodular (ie the left invariant measure is also right invariant) $\endgroup$ Sep 1 '13 at 2:49
  • $\begingroup$ Ok, but I still would like a concrete realization. $\endgroup$ Sep 1 '13 at 2:53
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    $\begingroup$ This thread may be interesting: mathoverflow.net/questions/36025/… $\endgroup$
    – John M
    Sep 1 '13 at 3:23
  • $\begingroup$ Wow! This thread is great John M $\endgroup$ Sep 1 '13 at 3:45
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I recently gave an explicit parameterization of $SO(n)$ and corresponding explicit formula for the Haar measure as an answer to this question. Here I will do the same thing for $U(n)$. I highly recommend reading my other answer on $SO(n)$ before reading this one.

Let $S^{2n-1}$ denote the unit sphere in $\mathbb{C}^n$, and note that $U(n)$ acts transitively on $S^{2n-1}$. I will use this action give an explicit inductive description of a parameterization of $U(n)$. The idea will be to first perform an arbitrary unitary transformation of the first $n-1$ complex coordinates, and then perform a unitary transformation that maps $\textbf{e}_n$ to any possible location on $S^{2n-1}$.

For convenience, we will use the following notation. If $\textbf{v}\in\mathbb{C}^n$ is a vector, let $\textbf{v}^a\in\mathbb{C}^{n+1}$ be the vector obtained by augmenting $\textbf{v}$ with a zero, i.e. $$ (v_1,\ldots,v_n)^a \;=\; (v_1,\ldots,v_n,0). $$ Similarly, if $M$ is an $n\times n$ complex matrix, let $M^a$ be the $(n+1)\times(n+1)$ complex matrix with the following block diagonal form: $$ M^a \;=\; \begin{bmatrix}M & \textbf{0} \\ \textbf{0}^T & 1\end{bmatrix}. $$ We will also use the notation $\textbf{e}_1,\ldots,\textbf{e}_n$ for the standard basis vectors in $\mathbb{C}^n$.


Complex Hyperspherical Coordinates

These are a coordinate system for specifying a point $\boldsymbol{\Sigma}_n(\phi_1,\ldots,\phi_n,\theta_1,\ldots,\theta_{n-1})$ on $S^{2n-1}$ given $2n+1$ angles $\phi_1,\ldots,\phi_n,\theta_1,\ldots,\theta_{n-1}$. The first few complex hyperspherical coordinate systems are given by \begin{align*} \boldsymbol{\Sigma}_1(\phi_1) &\;=\; e^{i\phi_1}, \\[3pt] \boldsymbol{\Sigma}_2(\phi_1,\phi_2,\theta_1) &\;=\; (e^{i\phi_1}\sin\theta_1,\,e^{i\phi_2}\cos\theta_1), \\[3pt] \text{and }\boldsymbol{\Sigma}_3(\phi_1,\phi_2,\phi_3,\theta_1,\theta_2) &\;=\; (e^{i\phi_1}\sin\theta_1\sin\theta_2,\,e^{i\phi_2}\cos\theta_1\sin\theta_2,\,e^{i\phi_3}\cos\theta_2), \end{align*} and in general the $k$th complex Cartesian coordinate $\Sigma_{n,k}$ of $\boldsymbol{\Sigma}_n$ is given by the formula $$ \Sigma_{n,k}(\phi_1,\ldots,\phi_n,\theta_1,\ldots,\theta_{n-1}) \;=\; \begin{cases}e^{i\phi_1}\sin \theta_1 \cdots \sin \theta_{n-1} & \text{if } k=1, \\[3pt] e^{i\phi_k}\cos \theta_{k-1} \sin \theta_k \cdots \sin \theta_{n-1} & \text{if }2\leq k \leq n.\end{cases} $$ The function $\boldsymbol{\Sigma}_n$ can also be defined inductively by the formula $$ \boldsymbol{\Sigma}_n(\phi_1,\ldots,\theta_{n-1}) \;=\; (\sin \theta_{n-1})\,\bigl(\boldsymbol{\Sigma}_{n-1}(\phi_1,\ldots,\phi_{n-1},\theta_1,\ldots,\theta_{n-2})\bigr)^a \,+\, (e^{i\phi_n}\cos \theta_{n-1})\,\textbf{e}_n. $$ with base case $\boldsymbol{\Sigma}_1$.

Domain and Volume Form

If we let $D_n$ be the subset of the parameter space defined by $$ 0\leq\phi_k\leq 2\pi, \qquad\text{and}\qquad 0\leq \theta_k\leq \pi/2 $$ for all $k$, then $\boldsymbol{\Sigma}_n$ maps $D_n$ onto $S^{2n-1}$, and is one-to-one on the interior of $D_n$. The $(2n-1)$-dimensional volume form with respect to $\boldsymbol{\Sigma}_n$ is $$ dV \;=\; \left(\prod_{k=1}^{n-1} \cos \theta_k \sin^{2k-1} \theta_k\right)\,d\phi_1\cdots d\phi_n\,d\theta_1 \cdots d\theta_{n-1}. $$


An Orthonormal Basis

Before writing down the parameterization of $U(n)$, we need to extend $\{\boldsymbol{\Sigma}_n(\phi_1,\ldots,\phi_n,\theta_1,\ldots,\theta_{n-1})\}$ to an orthonormal basis of $\mathbb{C}^n$. The basis is $$ \bigl\{\textbf{U}_{n,1},\ldots,\textbf{U}_{n,n-1},\boldsymbol{\Sigma}_n\bigr\} $$ where $$ \textbf{U}_{n,k}(\phi_1,\ldots,\phi_n,\theta_1,\ldots,\theta_{n-1}) \;=\; \frac{1}{(\sin \theta_{k+1}) \cdots (\sin \theta_{n-1})}\frac{\partial \boldsymbol{\Sigma}(\phi_1,\ldots,\phi_n,\theta_1,\ldots,\theta_{n-1})}{\partial \theta_k}. $$ That is, $\textbf{U}_{n,k}$ is the unit vector tangent to $S^{2n-1}$ in the direction of increasing $\theta_k$.

The vectors $\textbf{U}_{n,k}(\phi_1,\ldots,\theta_{n-1})$ can also be defined inductively by the formula $$ \textbf{U}_{n,n-1}(\phi_1,\ldots,\theta_{n-1}) \;=\; (\cos \theta_{n-1})\,\bigl(\boldsymbol{\Sigma}_{n-1}(\phi_1,\ldots,\phi_{n-1},\theta_1,\ldots\theta_{n-2})\bigr)^a \,-\, (e^{i\phi_n}\sin \theta_{n-1})\,\textbf{e}_{n} $$ and $\textbf{U}_{n,k}(\phi_1,\ldots,\phi_n,\theta_1,\ldots,\theta_{n-1}) = \bigl(\textbf{U}_{n-1,k}(\phi_1,\ldots,\phi_{n-1},\theta_1,\ldots,\theta_{n-2})\bigr)^a$ for $k<n-1$.

Let $M_n(\phi_1,\ldots,\phi_n,\theta_1,\ldots,\theta_{n-1})$ denote the $n\times n$ unitary matrix whose columns are the vectors of this orthonormal basis: $$ M_n \;=\; \begin{bmatrix}\textbf{U}_{n,1} & \cdots & \textbf{U}_{n,n-1} & \boldsymbol{\Sigma}_n\end{bmatrix}. $$ So $M_n(\phi_1,\ldots,\phi_n,\theta_1,\ldots,\theta_{n-1})$ maps $\textbf{e}_n$ to an arbitrary point $\boldsymbol{\Sigma}_{n}(\phi_1,\ldots,\phi_n,\theta_1,\ldots,\theta_{n-1})$ on the unit $(2n-1)$-sphere.


Parameterization of $U(n)$

Our parameterization for $U(n)$ will be an inductively defined function $\Phi_n$, which will take the $n^2$ angles $\{\phi_{jk}\}_{1\leq j\leq k\leq n}\cup\{\theta_{jk}\}_{1\leq j < k\leq n}$ as input, and output an $n\times n$ matrix in $U(n)$. It is defined inductively by the rule $$ \Phi_1(\phi_{11}) \;=\; e^{i\phi_{11}}. $$ and $$ \Phi_n\bigl(\phi_{jk}\text{'s},\theta_{jk}\text{'s}\bigr) \;=\; M_n(\phi_{1,n},\ldots,\phi_{n,n},\theta_{1,n},\ldots,\theta_{n-1,n})\, \bigl(\Phi_{n-1}(\phi_{jk}\text{'s},\theta_{jk}\text{'s})\bigr)^a $$ where the product is a matrix product, and the inputs to $\Phi_{n-1}$ include all variables for which $k\leq n-1$. Conceptually, the second factor performs an arbitrary unitary transformation on the first $n-1$ coordinates, and then the first factor performs a specific unitary transformation that maps $\textbf{e}_n$ to an arbitrary point on $S^{2n-1}$.

Again, if we let $E_n$ be the subset of parameter space defined by $0\leq \phi_{jk}\leq 2\pi$ and $0\leq \theta_{jk} \leq \pi/2$, then $\Phi_n$ maps $E_n$ onto $U(n)$ and $\Phi_n$ is one-to-one on the interior of $E_n$.

The volume form on $U(n)$ corresponding to Haar measure is $$ dV \;=\; \left(\prod_{1\leq j < k \leq n} \cos \theta_{jk} \sin^{2j-1} \theta_{jk} \right) d\phi_{11} \cdots d\theta_{n-1,n}. $$ Note that this measure isn't normalized. Instead, the total volume of $U(n)$ is the product $$ \prod_{k=1}^{n} \mathrm{Vol}(S^{2k-1}), $$ where $\mathrm{Vol}(S^{2k-1})$ denotes the $(2k-1)$-dimensional volume (i.e. surface area) of the unit sphere in $\mathbb{C}^k$.


Some Examples

For $n=2$, we are parameterizing $U(2)$ using $4$ variables $\phi_{11},\phi_{12},\phi_{22},\theta_{12}$, where $0\leq\phi_{jk}\leq 2\pi$ and $0\leq \theta_{12} \leq \pi/2$. The parameteriztion $\Phi_2(\phi_{11},\phi_{12},\phi_{22},\theta_{12})$ is the following matrix product $$ \begin{bmatrix} e^{i\phi_{12}} \cos\theta_{12} & e^{i\phi_{12}}\sin \theta_{12} \\ -e^{i\phi_{22}}\sin\theta_{12} & e^{i\phi_{22}}\cos \theta_{12} \end{bmatrix} \begin{bmatrix}e^{i\phi_{11}} & 0 \\ 0 & 1\end{bmatrix}. $$ The volume form is $$ dV \;=\; \cos \theta_{12} \sin \theta_{12} \,d\phi_{11}\,d\phi_{12}\,d\phi_{22}\,d\theta_{12}, $$ and the total volume of $U(2)$ is $(2\pi)(2\pi^2) = 4\pi^3$.

For $n=3$, we are parameterizing $U(3)$ with nine parameters $\phi_{11},\phi_{12},\phi_{22},\phi_{13},\phi_{23},\phi_{33},\theta_{12},\theta_{13},\theta_{23}$, where $0\leq\phi_{jk}\leq 2\pi$ and $0\leq \theta_{jk} \leq \pi/2$. The parameterization $\Phi_3(\phi_{11},\phi_{12},\phi_{22},\phi_{13},\phi_{23},\phi_{33},\theta_{12},\theta_{13},\theta_{23})$ is the product of the matrix $$ \begin{bmatrix} e^{i\phi_{13}} \cos \theta_{13} & e^{i\phi_{13}} \sin \theta_{13} \cos \theta_{23} & e^{i\phi_{13}} \sin \theta_{13} \sin \theta_{23} \\ -e^{i\phi_{23}} \sin \theta_{13} & e^{i\phi_{23}} \cos \theta_{13} \cos \theta_{23} & e^{i\phi_{23}} \cos \theta_{13} \sin \theta_{23} \\ 0 & -e^{i\phi_{33}} \sin \theta_{23} & e^{i\phi_{33}} \cos \theta_{23} \end{bmatrix} $$ with $\begin{bmatrix}\Phi_2(\phi_{11},\phi_{12},\phi_{22},\theta_{12}) & \textbf{0} \\ \textbf{0}^T & 1\end{bmatrix}$. The volume form is $$ dV \;=\; \bigl(\cos\theta_{12}\sin \theta_{12}\bigr) \bigl(\cos\theta_{13}\sin \theta_{13}\bigr) \bigl(\cos \theta_{23}\sin^3 \theta_{23}\bigr)\,d\phi_{11}\cdots d\theta_{23}, $$ and the total volume of $U(3)$ is $(2\pi)(2\pi^2)(\pi^3) = 4\pi^6$.

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  • $\begingroup$ can one divide this parametrization of a random matrix from U(N) by its determinant to get a random matrix from SU(N)? $\endgroup$ Mar 5 '20 at 0:58

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