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This question is somewhat related to my previous one on Grassmanians. The few times I've encountered the Haar measure in the course of my mathematical education, it's always been used in a very theoretical setting: in the right setting, it exists, it is unique (if the setting is really nice), and you can integrate against it to define new objects that will have nice properties because the measure itself does.

So my question is: how practical is it to compute with? I'm talking about very concrete examples here, e.g. "G=O(4), I integrate f(M)=[some explicit function with a matrix input] and the answer I get is 42". From conversations, I got the feeling that the construction of the Haar measure allows you in principle to write such a computation explicitly. I'm concerned with the tractability of the computation itself. Examples would be great.

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It is unique only up to a positive scaling factor. If you make a construction using a choice of Haar measure, you need to make sure your construction is independent of scaling your measure (trivial example: L^2-spaces w.r.t. Haar measure) to know the construction is intrinsic. –  KConrad Aug 18 '10 at 21:13
It is very easy to come up with examples of functions defined on $\mathbb R$ which are impossibly complicated to carry out... Integrating is difficult to do explicitely, independently of Haar! –  Mariano Suárez-Alvarez Aug 18 '10 at 21:29
I think Thierry is aware of both the facts mentioned in the comments above. How about the following line of thought, related to something I've been playing with: we know that in some cases we can exploit the Plancherel formula (just as when doing integrals on the circle we could use Parseval to try and express the integral as a suitable inner product in $\ell^2$). Are there more sophisticated version to handle, say, a product of three coefficient functions? –  Yemon Choi Aug 18 '10 at 21:36
Couldn't refrain myself from saying this - isn't the answer to all our problems supposedly 42? –  Somnath Basu Aug 19 '10 at 0:05
Well... some of them, especially if we use Davidac897's example with p = 43 –  stankewicz Aug 24 '10 at 14:01
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12 Answers

You might be interested in Weyl's Integration Formula. There are versions of this for all compact Lie groups; I'll state it for the unitary group.

Let $f$ be a conjugacy invariant function on $U(n)$. A unitary matrix always has eigenvalues of the form $(e^{i \theta_1}, e^{i \theta_2}, \ldots, e^{i \theta_n})$, and $f$ is a symmetric function of $(\theta_1, \theta_2, \ldots, \theta_n)$. Then

$$\int_{U(n)} f(A) \ dA = \frac{1}{(2 \pi)^n n!} \int_{\theta_1=0}^{2 \pi} \int_{\theta_2=0}^{2 \pi} \ldots \int_{\theta_n=0}^{2 \pi} \prod_{j<k} |e^{i \theta_j} - e^{i \theta_k}|^2 \cdot f(\theta_1, \ldots, \theta_n) \ d \theta_1 \ \cdots \ d \theta_n$$

Here Haar measure is normalized so that the unitary group has volume $1$.

There are plenty of proofs in books and online; I like the write-up in Fulton and Harris's Representation Theory.

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To make computations you need to find an example of a Haar measure on your group. The first few exercises in Section 5, Chapter XII of Lang's Real and Functional Analysis give formulas for Haar measure on some groups (exercise 9 is a nonabelian group). Chapter 14 of Royden's Real Analysis gives a method of Hurwitz for computing Haar measure on Lie groups. Hurwitz himself worked this out for orthogonal groups in the late 19th century, before general invariant measures on locally compact groups were known to exist.

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Excellent! I'll be sure to check out these references--if the new semester will give me a fighting chance! Of course, 19th century mathematicians were so much more focused on explicit computations too. So primary sources might be very helpful to get a handle on this. –  Thierry Zell Aug 18 '10 at 21:43
Thierry, ever notice that, despite apparently "analytic" nature of construction, for all matrix gps you've seen, left Haar is integration against top-degree difftl form that's "algebraic" (wrt matrix entries)? (NB: variety product does not have product topology!) Reason is left-invariant top-degree diff'tl forms (used to make the volume form on a Lie gp) works algebraically. That is, on any smooth group variety over any field, vector space of left-inv't top-deg difftl forms is 1-dim'l. See sec. 4.2 of "Neron models", a functorial/scheme version of Lie gp proof, constructive in affine case. –  BCnrd Aug 18 '10 at 22:04
Thierry, the same kind of algebraic methods also give a conceptual proof that the modulus character is likewise "algebraic" with respect to the matrix entries (i.e., an algebraic homomorphism $G \rightarrow {\rm{GL}}_1$). The content is that the 1-dimensional vector group of left-invariant top-degree difftl forms is an algebraic repn space for the group (best seen by thinking functorially via Yoneda's Lemma, hence allowing the ring to be general and not just a field); see Prop. 4 in sec. 4.2 of the book "Neron Models". –  BCnrd Aug 18 '10 at 22:07
Last point (from me) on this theme. One consequence of algebraicity of the modulus character is that as long as $G(\overline{k})$ is its own derived group, the modulus character must be trivial (since can compute that on $k$-points). This applies even when the group $G(k)$ of rational points over the ground field (like reals, complex, p-adic numbers, etc.) is not its own derived group. Good example for that is ${\rm{PGL}}_ n$ (for $n$ such that $k^{\times}$ is not its own $n$th-power subgroup), and various other connected semisimple groups that aren't simply connected. –  BCnrd Aug 19 '10 at 2:05
Dear Florian: Oops, indeed I forgot to put in the absolute value (or idelic norm in the case of group of adelic points). I was thinking of the algebraic character as the more basic object (since it interacts well with ground field extension, and passage to adelic points). I should have also relaxed the condition $G(\overline{k})$ is its own derived group'' to the slightly more ubiquitous condition $G(\overline{k})$ is generated by its derived group and center'' (to explain unimodularity in connected reductive cases). –  BCnrd Aug 19 '10 at 14:06
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Couldn't resist to point this out: actually, $$ \lim_{N \rightarrow \infty} \frac{1}{N^{k^2}} \mathbb{E}_{U\in U(N)} |Z_U|^{2k} = \frac{G(k+1)^2}{G(2k+1)}, $$ with $G$ the Barnes $G$ function, $Z_U$ the characteristic polynomial of the matrix $U$ in $U(N)$, taken according to Haar measure and evaluated somewhere along the unit circle in $\mathbb{C}$, say 1 (where on the circle is irrelevant as Haar measure is rotationally invariant).

When $k=3$, this actually evaluates to $$ \frac{\mathbf{42}}{9!}, $$ and accounts for the 3rd (and first new) case in the Keating-Snaith discovery of the interest of random matrices for quantitative formulations of analytic number theory conjectures, as explained in their paper "Random Matrix Theory and $\zeta (1/2 + i t )$" or less formally in http://seedmagazine.com/content/article/prime_numbers_get_hitched/ .

Incidentally, the connections between random matrix theory and number theory indeed lead to many practical computations for Haar-random matrices in classical compact groups. Some are easier to understand (for instance via the Weyl integration formula or reformulations in terms of Selberg integrals), while some are much less clear (for instance, subsitute in the above statement $|Z_U'|$ instead of $|Z_U|$ and study again the behaviour for large k, or analytic continuation in k of the RHS. The renormalization in that case, however, is know, and would be $\frac{1}{N^{k^2+2k}}$).

Since this is a reference request, look also at papers by Conrey or Hughes for examples of such explicit computations.

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The right or left Haar measures for a matrix group can be obtained in a completely straightforward manner with the aid of the right or left Maurer-Cartan form, respectively.

I will show the procedure for the stochastic group of invertible stochastic matrices (i.e., invertible matrices in $GL(B)$ whose rows sum to unity), though much of it generalizes in an obvious way. (The motivation I had for figuring this out is a gauge theory of random walks on the root lattice $A_n$ which I'll finish up one of these days.)

Let $R, R' \in STO(B)$, and let $R$ be parametrized by (say) $\{R_{jk}\} \equiv \{R_{(j,k)}\}$ for $1 \le j \le B, k \ne j$. Now if

$\left(\mathcal{R}^{-1}\right)_{(j,k)}^{(l,m)} := \frac{\partial(RR')_{(j,k)}}{\partial R'_{(l,m)}} \Bigg|_{R'=I}$

then the right Maurer-Cartan form on $STO(B)$ is $\omega_{(j,k)}^{(\mathcal{R})} = \mathcal{R}_{(j,k)}^{(l,m)}dR_{(l,m)}$.

Since the right Maurer-Cartan form is right-invariant, the right Haar measure is given (up to an irrelevant constant multiple) by

$d\mu^{(\mathcal{R})} = \underset{(j,k)}{\bigwedge} \omega_{(j,k)}^{(\mathcal{R})}.$

A similar construction yields the left Haar measure.

For a concrete example, let $B=2$. A straightforward calculation yields

$\omega_{(1,2)}^{(\mathcal{R})} = \frac{(1-R_{21}) \cdot dR_{12} + R_{12} \cdot dR_{21}}{1-R_{12}-R_{21}}$


$\omega_{(2,1)}^{(\mathcal{R})} = \frac{R_{21} \cdot dR_{12} + (1-R_{12}) \cdot dR_{21}}{1-R_{12}-R_{21}}$.

It follows that

$d\mu^{(\mathcal{R})} = \omega_{(1,2)}^{(\mathcal{R})} \land \omega_{(2,1)}^{(\mathcal{R})} = \frac{dR_{12} \land dR_{21}}{\lvert 1-R_{12}-R_{21} \rvert}$.

(The modulus is taken in the denominator to ensure a positive rather than a signed measure.) Similarly, the left Haar measure is

$d\mu^{(\mathcal{L})} = \frac{dR_{12} \land dR_{21}}{\lvert 1-R_{12}-R_{21}\rvert^2}$.

Notice that both the right and left Haar measures assign infinite volume to the set of nonnegative stochastic matrices (i.e., the unit square in the $R_{12}$-$R_{21}$ plane). However, the singular behavior of the measures occurs precisely on the set of singular stochastic matrices. Indeed, for $0 \le \epsilon < 1$ consider the sets

$X_I(\epsilon) := \{(R_{12}, R_{21}) : 0 \le R_{12} \le 1-\epsilon, \ 0 \le R_{21} \le 1 - \epsilon - R_{12} \}$

$X_{II}(\epsilon) := \{(R_{12}, R_{21}) : \epsilon \le R_{12} \le 1, \ 1 + \epsilon - R_{12} \le R_{21} \le 1 \}$


$X(\epsilon) := X_I(\epsilon) \cup X_{II}(\epsilon)$,

i.e., $X(\epsilon)$ is the unit square minus a strip of width $\epsilon \sqrt{2}$ centered on the line $1 - R_{12} - R_{21} \equiv \det R = 0$. Then

$\int_{X(\epsilon)} d\mu^{(\mathcal{R})} = 2(\log \epsilon^{-1} - 1 + \epsilon)$


$\int_{X(\epsilon)} d\mu^{(\mathcal{L})} = 2(\epsilon^{-1} - 1)$.

It is not hard to show that for $B$ arbitrary

$d\mu^{(\mathcal{R})} = \lvert \det \mathcal{R} \rvert \underset{(j,k)}{\bigwedge} dR_{jk}$,

and similarly for the left Haar measure. The general end result is

$d\mu^{(\mathcal{R})} = \lvert \det R \rvert^{1-B} \underset{(j,k)}{\bigwedge} dR_{jk}, \quad d\mu^{(\mathcal{L})} = \lvert \det R \rvert^{-B} \underset{(j,k)}{\bigwedge} dR_{jk}$.

To see this, consider the isomorphism between the stochastic and affine groups and see, e.g. (N. Bourbaki. Elements of Mathematics: Integration II. Chapters 7-9. Springer (2004)).

Finally, a Fubini-type theorem (see, e.g., L. Loomis. An Introduction to Abstract Harmonic Analysis. Van Nostrand (1953)) applies to the special stochastic group $SSTO(B)$ (i.e., the subgroup of unit-determinant stochastic matrices). If for example elements of $SSTO(2)$ are parametrized by $R_{12}$, then $d\mu = \omega_{(1,2)} = dR_{12}$ is the (right and left) Haar measure. More generally, taking $\{R_{jk}\}$ for an appropriate choice of pairs $(j,k)$ as parameters for $SSTO(B)$, we have that $\mathcal{R} = I = \mathcal{L}$ and the Haar measure for $SSTO(B)$ is (up to normalization)

$d\mu = \underset{(j,k)}{\bigwedge} dR_{jk}$.

This can easily be verified explicitly for small values of $B$ with a computer algebra package.

One simplifying feature of the simple stochastic group is that it is unimodular, so the left and right Haar measures coincide. Moreover, the Haar measure of the set of nonnegative special stochastic matrices is (finite, and w/l/o/g equals) unity. (For $SSTO(B)$, the constant multiplying the RHS of the equation above and that provides this normalization can be shown to be $((B-1)!)^{B-1}(B-2)!$.) Although this set is not invariant, it is a semigroup and it is obviously privileged in probabilistic contexts.

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Very interesting question. As a prominent harmonic analyst told me recently, when I asked him where I could learn to make explicit computations on hyperbolic spaces: "not easy to find references, and it's all Sigurdur Helgason's fault". He was joking, of course, but basically he meant: there is now an implicit understanding that for each one of your questions there's a formula somewhere in some book of SH, so why are you asking? read the books. But on the contrary, those elegant and general formulas are of no help if you really want to compute something: basically you still need a lot of work, choose proper coordinates, write down explicit formulas for every Harish-Chandra thing and so on. A slower development of the subject would have been more helpful, by now we'd have available books on special cases with explicit formulas and so on.

More to the point: a beautiful example of an explicit computation using the Haar measure on $SO(3)$ is this paper on endpoint Strichartz estimates for the cubic Dirac equation. The computation is quite elementary, so you will not have troubles in reading it in case you're interested. I find it a compelling example of how useful it would be to develop some more machinery to work with Haar measures.

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The following method can be used to integrate (by hand) polynomials in the matrix elements of the group element over the special orthogonal, special unitary and symplectic groups (with respect to the Haar measure).

The description will be given for special orthogonal groups, but this method is valid for the special unitary and symplectic groups by passing to the complex numbers and the quaternions.

This method reduces the integration to a series of integrations on spheres.

The method relies on the following facts:

  • Let $V_{n,k}$ is the Stiefel manifold of orthogonal k-frames in $R^n$, then $SO(n) \cong V_{n,n-1}$, (a torsor)

This is because one can view a (special) orthogonal matrix as a collection of n-1 orthonormal unit column vectors $v^{(1)}, . . ., v^{(n-1)}$.

the method is valid for integrands which are polynomials in the group element which can always be written in the form $tr(c^t v^{(i)} + v^{(i)^t} c)$.

Let us denote the one dimensional orthogonal projectors on $v^{(i)}$ by $\theta^{(i)} = v^{(i)} v^{(i)^t}$

  • There exists a series of fibrations:

$S^{n-k} \cong V_{n-k+1,1} \rightarrow V_{n,k} \rightarrow V_{n,k-1}$

The integration method is based on sequential integrations on the spherical fibers starting from

$SO(n) \cong V_{n,n-1}$ and ending with $S^{n-1} \cong V_{n,1}$, such that each time the integration is performed on one of the unit vectors.

  • The most important part is that one must remember that the unit vectors are not independent because they are orthonormal. Thus each integration over a unit vector must be performed on the intersection of the sphere defined by it and the hyperplane orthogonal to the other unit vectors. For example if we start the integration from $v^{(1)}$, this amount replacing $v^{(1)}$ before the integration on it by: $(1 - \theta^{(2) }- . . . \theta^{(n-1)}) v^{(1)}$. This replacement unties the orthogonality constraints.

  • Integration of homogeneous polynomials over spheres can be performed by the replacement of the integration measure by a Gaussian measure and scaling the result (according to the ratio of a spherically invariant integrand of the same degree).

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it might be useful to add that these integrals of products of matrix coefficients over classical groups, with respect to the Haar measure, are given by the socalled Weingarten function en.wikipedia.org/wiki/Weingarten_function –  Carlo Beenakker May 14 '11 at 14:36
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I just stumbled upon this old thread while searching for something else, and couldn't resist saying two things: 1. if you like p-adics, the expository article http://arxiv.org/pdf/math/0205207v2 by T. C. Hales asks pretty much the same question, and gives some very interesting examples, an explanation why in general this question is very hard, and a general approach via motivic integration (a lot of progress happened in motivic integration since this article was written, but this is still a great introduction to the main ideas). 2. For a split connected reductive algebraic group over a local field, one can write down an explicit formula for the Haar measure in convenient coordinates (more precisely, one can just write down the invariant differential form that Brian Conrad mentioned): for an explicit formula, see e.g. section 2.4 in http://arxiv.org/pdf/math/0203106 (I am sure this is a classical formula, but I have never seen a reference for it -- would be grateful if someone pointed it out).

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Thanks for the links! It's always nice to see that this old question of mine still has some life in it... –  Thierry Zell May 14 '11 at 4:53
Thank you, Thierry! I thought it was a very nice question... –  JGordon May 14 '11 at 5:46
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If we consider the independent bond percolation model in the two dimensions the Haar Measure is obtained when you choose the parameter $p=\frac{1}{2}$.

We can look for this model as a collection of measures and the only Haar measure for this model it is also characterized for a critical behavior with respect to some geometric properties of the model.

The group structure of this model it seems not too much explored yet. But there are many explicit calculations using the Haar measure. The book Percolation, by Grimmett have some of this explicit calculations.

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A fairly simple example is the Haar measure on $\mathbb{Q}_p$. If we scale the measure so that $\mathbb{Z}_p$ has measure $1$, and the measure is translation invariant, it follows that $a+p\mathbb{Z}_p$ has measure $\frac{1}{p}$. We can do similarly for cosets of $p^n\mathbb{Z}_p$. See Chapter 2 of Cassels-Frohlich for details on this.

In this vein, one defines Haar measures on other number-theoretic objects, like adeles and ideles. Integration over these spaces can then be used to prove basic facts about more concrete objects, like zeta functions. For details, consult Koch's Number Theory, which gives many explicit examples of integration over $p$-adics and spaces of adeles and then uses them to prove Hecke's functional equation for the zeta function. (You can also find similar material in Cassels-Frohlich, though I find Koch to be much more readable.)

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The word you want is "vein," as in Definition III.9.b in the OED: dictionary.oed.com/cgi/entry/… –  Qiaochu Yuan Aug 24 '10 at 1:04
Davidac, we don't "know" that Z_p has measure 1. We only know that for a Haar measure on Q_p the measure of Z_p is finite since it's compact and positive since it's open (or, more broadly, since it contains an open set). Therefore we can scale Haar measure on Q_p in a unique way to give Z_p measure 1. –  KConrad Aug 24 '10 at 2:44
Right...I more meant that once we scale the Haar measure for $\mathbb{Z}_p$ to have measure $1$, we can derive using that information and the definition of a Haar measure what the measure of certain other sets is. –  David Corwin Aug 24 '10 at 3:18
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There are some explicit calclulations in Hewitt & Ross, Abstract Harmonic Analysis

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There are cases in which computation can be made easier. Consider for example a locally compact abelian group $G$ and denote by $\mu$ a Haar measure.

One hypothesis on $G$ that can make life easier is to suppose $G$ totally disconnected. In such case, $G$ has a base of neighborhoods of $0$ consisting of clopen compact subgroups. Notice also that, if we have two such neighborhoods $V_1\subseteq V_2$, then $V_2/V_1$ is finite and $\mu(V_2)=|V_2/V_1|$, provided $\mu(V_1)=1$ (we can always suppose this up to a renormalization of the measure).

This seems quite abstract but there are examples in which this could be very useful in concrete computation (see the answer of Davidac897 for a very explicit occurrence of what I'm saying).

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Integration with respect to Haar measure is used a lot in multivariate statistical analysis, see for example Muirhead: "Aspects of Multivariate Statistical Theory". There is also lots of examples in the developing thneory of special function with matrix argument, search for books/papers by Mathai ...

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