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

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

and

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

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

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

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

and

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

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

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

and

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

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.

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 21:13isdifficult to do explicitely, independently of Haar! – Mariano Suárez-Alvarez♦ Aug 18 '10 at 21:29threecoefficient functions? – Yemon Choi Aug 18 '10 at 21:36