Intuition for Haar measure of random matrix

Question

What is an intuitive way to understand Haar measure as defined for random matrices, say, $N\times N$ orthogonal or unitary matrices?

My understanding for what Haar measure means for $U(1)$ is that it can be thought of as a measure over a uniform distribution of phases on a circle, i.e. a matrix representing $M \in U(1)$ can be parameterized with an angle $\theta$ so that $$d\mu(M) = \frac {d\theta}{2\pi} $$

What is a correct generalization of this intuition to $N>1$? In particular, are there any explicit parameterizations of Haar measure that resemble writing down angles that are uniformly distributed?

One possibility that came to mind is that eigenvectors, rows and/or columns have (generalized) phases that can be thought of as direction angles that are in some sense uniform over $SO(N)$ or $U(N)$? (Edit: seems like this would not be the case for rows and columns.)

Another possibility I have thought of is using Givens rotations to parameterize an orthogonal matrix using the resulting rotation angles that bring it to the identity matrix. Are there any known results about the distribution of the Givens angles? It would seem plausible that they could be uniformly distributed, but given that the Givens rotations are usually applied in a particular fashion to achieve diagonalization, that could introduce correlations that would result in nonuniformity.

(Caveat: I'm new to all this so I could very well be wrong about even trying to conceptualize such a question.)

Answer 1

You want to think of the Haar measure $d\mu(U)$ as a way of measuring uniformity in the group $U(N)$ of unitary $N\times N$ matrices.

To form your intuition, consider $N=1$. You then have $U=e^{i\phi}$, with $0<\phi\leq 2\pi$ and $d\mu(U)=d\phi$ measures the perimeter of the unit circle. This is a uniform measure, because $d(\phi+\phi_0)=d\phi$ for any fixed phase shift $\phi_0$. You could write the requirement of uniformity in the form $d\mu(UU_0)=d\mu(U)$, with $U_0=e^{i\phi_0}$ the unitary matrix corresponding to the phase shift $\phi_0$.

Once your intuition is formed for $N=1$, you simply generalize to $N>1$ using the same definition of uniformity, $d\mu(UU_0)=d\mu(U)$ for any fixed $U_0\in U(N)$. For orthogonal (or symplectic) matrices you use the same definition of uniformity, with $U_0$ now restricted to the orthogonal or symplectic subgroup of $U(N).$

To explicitly write down the Haar measure $d\mu(U)$ in terms of the matrix elements of $U$ is only easily done for a few small values of $N$. (In particular, there is no relationship to random directions of rows or columns, as Yemon Choi pointed out.) You typically do not need such explicit expressions, since integrals with the Haar measure can be evaluated by using only the definition of uniformity.

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In response to the follow-up question: If you wish to evaluate Haar-measure integrals of polynomials of matrix elements of $U$, you can use the socalled Weingarten functions.

Puchała and Miszczak - Symbolic integration with respect to the Haar measure on the unitary group gives a Mathematica program to generate these.

If you need an explicit expression for the Haar measure, the steps to take are the following:

1. parameterize your matrix $U$ in terms of a set of real parameters $\{x_i\}$.

2. calculate the metric tensor $m_{ij}$, defined by $\sum_{ij}|dU_{ij}|^2 = \sum_{ij}m_{ij}dx_i dx_j$.

3. obtain the Haar measure by equating $d\mu(U) = (\operatorname{Det} m)^{1/2}\prod_i dx_i$.

This is the general recipe. In practice, for many parameterizations the answer is in the literature. In particular, for the Haar measure in Euler angle parameterizations see:

Tilma and Sudarshan - Generalized Euler Angle Parameterization for SU(N)

Życzkowski and Kuś - Random unitary matrices

Answer 2

Dear Jiahao Chen, if you're interested in explicit expressions of the Haar measure you might want to look at the paper http://arxiv.org/abs/1103.3408 which contains simple parametrizations of U(N) and SU(N), as well as a formula of the normalized Haar measure for arbitrary N. This paper could be interesting for you as the provided framework can directly be applied to compute group integrals. Knowledge of the Weingarten functions is not required.

Answer 3

I like the following characterisation. Consider a continuous function $f:G\to\mathbb{R}$ (for any compact Lie group $G$) and the set $T=\{t_g(f): g\in G\}$ of translates of $f$, where $t_g(f)(x)=f(gx)$. Let $C$ be the closure of the convex hull of $T$. It can be shown that $C$ contains a unique constant function, and the value of that function is the inegral of $f$ with respect to Haar measure.

Answer 4

see the appendix of this paper for understanding Haar measure: Determinantal point processes in the plane from products of random matrices

intuition for Haar random orthogonal matrix: choose a vector randomly from the unit sphere in ${\mathbb R}^n$ (uniform distribution on the unit sphere). That's the first column. Now for the second column, choose a vector randomly from the unit sphere in the $n-1$ dimensional subspace orthogonal to the first column. Similarly for the third column, choose a vector randomly from the unit sphere in the $n-2$ dimensional subspace orthogonal to the first two columns...and so on....

Answer 5

Carlos Beenacker's answer is to the point, but there are more concrete constructions for $O(N)$ and related groups.

Haar on $O(N)$

Take a $N\times N$ matrix of i.i.d. standard Gaussian random variables. Think of the columns of this matrix as vector $X_1,\dots,X_N$. These vectors a.s. form a basis of $R^N$, and one can apply Gram-Schmidt to them to obtain an orthonormal basis $Y_1,\dots,Y_N $. I claim that the matrix U with columns given by the $Y$'s is distributed according to Haar measure over $O (N) $.

Proof sketch: One can check that the law of $U $ is left-invariant, meaning that, for any $R\in O (N) $, $U $ and $RU $ have the same distribution. Now let $V $ be an independent copy of $U^T $. One can see that the law of $V $ is right invariant and that $VU $ has the same law as both $V $ and $U $. Thus the laws of $U $ and $U^T $ coincide, and the law of $U $ is both left and right invariant. A similar argument shows that there can be onlyour one left-invariant (or right-invariant) distribution over $O (N) $.

Haar over $SO (N) $

Condition on the determinant of $U $ being one.

Haar over $U (N) $

Going back to the $O (N)$ we see that the same proof works if the $X_i $ are i.i.d., spherically symmetric vectors in $R^N $ with $X_1\neq 0$ a.s. . If one replaces $R^N $ with $C^N $ in the preceding sentence, and then does the same Gram-Schmidt procedure, one obtains a Haar distributed element of $U (N) $.

Remark

The basic idea in each case is that the the law of $Y_i $ should be uniform over the unit sphere of the orthogonal complement of the subspace generated by the $i-1$ previous vectors. But this is harder to implement directly.

Added in edit: this remark is basically the preceding answer.

Answer 6

In MatLab I generate Haar distributed matrices like this:

m = randn(m,m)% all elements are normally distributed 

u= qr(m) % make qr decomposition and what you get is Haar measure on "u"

So the mathematical statement is that if m is normally distributed, then u is Haar.

The reason is quite trivial — normal distribution is preserved by unitary transformations.

However writing this I begin to doubt myself about tiny details - depending how they implement qr algorithm the matrix u is not unique, it can be multiplied by $\operatorname{diag}(\pm1)$. Neverthelss most probably everything should be correct.