2.1 Double cosets

Let $H, K$ be subgroups of a finite group G. The double coset decomposition is a standard tool of elementary group theory. The original proof of Sylow’s first theorem uses double cosets, as does Mackey’s basic theorems on decomposing restrictions of induced representations. The Hecke algebra $\mathrm {End}_{G}(G/H)$ —the linear maps of the right H-invariant functions that commute with the action of G—has a basis induced by H-H double cosets. Hecke algebras are basic objects of study in modern number theory. For a detailed survey, see [Reference Diaconis and Simper24] or [Reference Curtis and Reiner14]. Double cosets can have very different sizes and [Reference Diaconis and Simper24], [Reference Paguyo48] develop a probabilistic and enumerative theory. For present applications, an explicit description of the double cosets is needed.

Example 2.2. Let M be a finite group, $G = M \times M$ , and $H = K = M$ embedded diagonally as subgroups of G (that is, $\{(m, m): m \in M \}$ ). The conjugacy classes in G are products of conjugacy classes in each coordinate of M. In the double coset equivalency classes, note that

$$\begin{align*}(s, t) \sim (id, s^{-1}t) \sim (id, k^{-1} s^{-1} t k), \end{align*}$$

and so double cosets can be indexed by conjugacy classes of M. If $Q_1$ is a conjugacy invariant probability on M, then $Q = Q_1 \times \delta _{id}$ is conjugacy invariant on G. The random walk on G induced by Q maps to the random walk on M induced by $Q_1$ . In this way, the double coset walks extend conjugacy invariant walks on M. Example 1.1 in the introduction is a special case.

Of course, the conjugacy classes in M (and so the double cosets) can be difficult to describe. Describing the conjugacy classes of $U_n(q)$ —the unit upper-triangular matrices in $GL_n(q)$ —is a well-known ‘wild’ problem. See [Reference Aguiar, André, Benedetti, Bergeron, Chen, Diaconis, Hendrickson, Hsiao, Isaacs, Jedwab, Johnson, Karaali, Lauve, Le, Lewis, Li, Magaard, Marberg, Novelli, Pang, Saliola, Tevlin, Thibon, Thiem, Venkateswaran, Vinroot, Yan and Zabrocki2] or [Reference Diaconis and Malliaris17] for background and details. Describing the double cosets for the Sylow p-subgroup in $S_n$ seems difficult.

Example 2.3. Let G be a finite group of Lie type, defined over $\mathbb {F}_q$ , with Weyl group W. Let $\mathcal {B}$ be the Borel subgroup (maximal solvable subgroup). Take $H = K = \mathcal {B}$ . The Bruhat decomposition gives

$$\begin{align*}G = \bigsqcup_{\omega \in W} \mathcal{B} \omega \mathcal{B}, \end{align*}$$

so the double cosets are indexed by W. See [Reference Carter11], Chapter 8, for a clear development in the language of groups with a $(\mathcal {B}, N)$ pair.

Conjugacy invariant walks on G have been carefully studied in a series of papers by David Gluck, Bob Guralnick, Michael Larsen, Martin Liebeck, Aner Shalev, Pham Tiep and others. These authors develop good bounds on the character ratios needed. See [Reference Guralnick, Larsen and Tiep30] for a recent paper with careful reference to earlier work. Of course, Example 2 with $G = GL_n(q), W = S_n$ is a special case. The present paper shows what additional work is needed to transfer character ratio results from G to W.

The double cosets form a basis for the algebra of H-K bi-invariant functions $L(H \backslash G /K)$ with product

$$\begin{align*}f \star g(s) = \sum_t f(t) g(st^{-1}). \end{align*}$$

This is usually developed for $H = K$ [Reference Curtis and Reiner14], [Reference Ceccherini Silberstein, Scarabotti and Tolli12], [Reference Diaconis15], but the extra flexibility is useful. We add a caveat: When $H = K$ , the algebra of bi-invariant functions (into $\mathbb {C}$ ) is semisimple and with a unit. This need not be the case for general H and K. David Craven tried many pairs of subgroups of $S_4$ and found examples which were not semisimple. For $G=S_4$ , Marty Issacs produced the example H the cyclic subgroup generated by $(1234)$ and K the cyclic subgroup generated by $(1243)$ . The algebra doesn’t have a unit and so cannot be semisimple. This occurs even for some pairs of distinct parabolic subgroups of $S_n$ . There are also distinct pairs of parabolics where the algebra is semisimple. Determining when this occurs is an open question.

Further examples are in Section 7. Since the theory is developed for general $H, K, G$ there is a large set of possibilities. What is needed are examples where the double cosets are indexed by familiar combinatorial objects and the walks induced on $H \backslash G / K$ are of independent interest.

2.2 Markov chain theory

Let $H, K$ be subgroups of a finite group G, and Q a probability on G. See [Reference Levin and Peres43] for an introduction to Markov chains; see [Reference Diaconis15] or [Reference Saloff-Coste53] for random walks on groups.

Proposition 2.4. Let Q be a probability on G which is H-conjugacy invariant ( $Q(s) = Q(h^{-1}s h) $ for $h \in H, s \in G$ ). The image of the random walk driven by Q on G maps to a Markov chain on $H \backslash G /K$ with transition kernel

$$\begin{align*}P(x, y) = Q(HyKx^{-1}). \end{align*}$$

The stationary distribution of P is $\pi (x) = |HxK|/|G|$ . If $Q(s) = Q(s^{-1})$ , then $(P, \pi )$ is reversible.

Proof. The kernel P is well defined; that is, it is independent of the choice of double coset representatives for $x, y$ . Dynkin’s criteria ([Reference Kemeny and Snell40] Chapter 6, [Reference Pang49]) says that the image of a Markov chain in a partitioning of the state space is Markov if and only if for any set in the partition and any point in a second set, the chance of the original chain moving from the point to the first set is constant for points in the second set.

Fixing $x, y$ , observe

$$\begin{align*}Q(HyK(hxk)^{-1}) = Q(HyK x^{-1}h^{-1}) = Q(HyKx^{-1}). \end{align*}$$

Since the uniform distribution on G is stationary for the walk generated by Q, the stationary distribution of the lumped chain is $\pi (x) = |HxK|/|G|$ . Finally, any function of a reversible chain is reversible and $Q(s) = Q(s^{-1})$ gives reversibility of the walk on G.

In all of our examples, the measure Q is a class function ( $Q(s) = Q(t^{-1}st)$ for all $s, t \in G$ ), which is a stronger requirement than that in Proposition 2.4. The eigenvalues of the walk on G can be given in terms of the irreducible complex characters of G. Let $\widehat {G}$ be an index set for these characters. We write $\lambda \in \widehat {G}$ and $\chi _{\lambda }(\mathcal {C})$ for the character value at the conjugacy class $\mathcal {C}$ . Let

$$\begin{align*}\beta_{\lambda} = \frac{1}{\chi_{\lambda}(1)} \sum_{s \in G} Q(s) \chi_{\lambda}(s). \end{align*}$$

If Q is simply concentrated on a single conjugacy class $\mathcal {C}$ , then $\beta _{\lambda }$ is the character ratio

$$\begin{align*}\beta_{\lambda} = \frac{\chi_{\lambda}(\mathcal{C})}{\chi_{\lambda}(1)}. \end{align*}$$

For a review of a large relevant literature on character ratios and their applications, see [Reference Guralnick, Larsen and Tiep30].

The restriction of $\chi _{\lambda }$ to H is written $\chi _{\lambda }|_H$ and $\left \langle \chi _{\lambda } |_H, 1 \right \rangle $ is the number of times the trivial representation of H appears in $\chi _{\lambda }|_H$ . By reciprocity, this is $\left \langle \chi _{\lambda }, \mathrm {Ind}_H^G(1) \right \rangle $ , where $\mathrm {Ind}_H^G$ is the induced representation from H to G.

Proposition 2.7. Let Q be a class function on G. The induced chain $P(x, y)$ of Proposition 2.4 has eigenvalues

(2.1) $$ \begin{align} \beta_{\lambda} = \frac{1}{\chi_{\lambda}(1)} \sum_{s \in G} Q(s) \chi_{\lambda}(s), \end{align} $$

with multiplicity

(2.2) $$ \begin{align} m_{\lambda} = \left\langle \chi_{\lambda}|_H, 1 \right\rangle \cdot \left\langle \chi_{\lambda}|_K, 1 \right\rangle. \end{align} $$

The average square total variation distance to stationarity satisfies

(2.3) $$ \begin{align} \sum_x \pi(x) \|P_x^{\ell} - \pi \|_{TV}^2 \le \frac{1}{4} \sum_{\lambda \in \widehat{G}, \lambda \neq 1} m_{\lambda} \beta_{\lambda}^{2 \ell}. \end{align} $$

Proof. The eigenvalues of a lumped chain are always some subset of the eigenvalues of the original chain. To determine the multiplicity of the eigenvalue $\beta _{\lambda }$ in the lumped chain, fix $\lambda : G \to GL_{d_{\lambda }}$ an irreducible representation of G. Let $M^{\lambda }$ be the $d_{\lambda } \times d_{\lambda }$ matrix representation of $\lambda $ . That is, each entry $M_{ij}^{\lambda }: G \to \mathbb {C}$ is a function of G. These functions are linearly independent and can be chosen to be orthogonal with respect to

$$\begin{align*}\left\langle f_1, f_2 \right\rangle := \frac{1}{|G|} \sum_{g \in G} f_1(g) \overline{f_2}(g) \end{align*}$$

(see Chapter 3 of [Reference Serre58]). Let $V_{\lambda }$ be the space of all linear combinations of the functions $M_{ij}^{\lambda }$ . If $f \in V_{\lambda }$ , then

$$ \begin{align*} Pf(x) = \sum_{y \in G} P(x, y)f(y) = \beta_{\lambda} f(x) \end{align*} $$

That is, $V_{\lambda }$ is the eigenspace for the eigenvalue $\beta _{\lambda }$ and it has dimension $d_{\lambda }^2 = \chi _{\lambda }(1)^2$ .

In the lumped chain on $H \backslash G / K$ , a basis for the eigenspace for eigenvalue $\beta _{\lambda }$ are the $H \times K$ invariant functions in $V_{\lambda }$ [Reference Boyd, Diaconis, Parrilo and Xiao9]. To determine the dimension of this subspace, note that $G \times G$ can act on $V_{\lambda }$ by $f^{g_1, g_2}(x) = f(g_1^{-1} x g_2)$ . This gives a representation of $G \times G$ on $V_{\lambda }$ . The matrix of this representation is isomorphic to $M \otimes M$ , since $M_{ij}(s^{-1} t u) = M_{ij}(s^{-1})M_{ij}(t)M_{ij}(u)$ .

This representation restricts to a representation $M_H \otimes M_K$ of $H \times K$ , and the dimension of the $H \times K$ invariant functions in $V_{\lambda }$ is the multiplicity of the trivial representation on $M_H \otimes M_K$ . This is

$$\begin{align*}\left\langle \chi_{\lambda}|_H \otimes \chi_{\lambda}|_K, 1 \right\rangle = \left\langle \chi_{\lambda}|_H, 1 \right\rangle \cdot \left\langle \chi_{\lambda}|_K, 1 \right\rangle. \end{align*}$$

To note the total variation inequality, let $1 = \beta _1 \ge \beta _2 \ge \dots \ge \beta _n \ge \dots \ge \beta _{|S_n|} \ge -1$ be the eigenvalues with eigenfunctions $f_j$ (chosen to be orthonormal with respect to $\pi $ ), and we have

$$\begin{align*}\chi^2_x(\ell) := \| \frac{P_x^\ell}{\pi} - 1\|_{2, \pi}^2 = \sum_{j \neq 1} f_j(x)^2 \beta_j^{2 \ell} \end{align*}$$

where $\| \cdot \|_{2, \pi }$ denotes the $\ell ^2$ norm with respect to the distribution $\pi $ . Multiplying by $\pi (x)$ and summing over all x in the state space gives

(2.4) $$ \begin{align} \sum_x \pi(x) \left\| \frac{P^{\ell}_x}{\pi} - 1 \right\|_{2, \pi}^2 = \sum_x \pi(x) \sum_{j \neq 1} f_j(x)^2 \beta_j^{2\ell} = \sum_{\lambda \neq 1} m_{\lambda} \beta_{\lambda}^{2 \ell}, \end{align} $$

using orthonormality of $f_j$ . The total variation bound arises since $4 \|P^{\ell }_x - \pi \|_{TV}^2 \le \|P^{\ell }_x/\pi - 1 \|_{2, \pi }^2$ .

2.3 Random transpositions and coagulation-fragmentation processes

Let $G = S_n$ be the symmetric group. The random transpositions Markov chain, studied in [Reference Diaconis and Shahshahani22], is generated by the measure

$$\begin{align*}Q(\omega) = \begin{cases} 1/n, & \omega = id, \\ 2/n^2, & \omega = (ij), \\ 0, & \text{otherwise}. \end{cases} \end{align*}$$

This was the first Markov chain where a sharp cutoff for convergence to stationarity was observed. A sharp, explicit rate is obtained in [Reference Saloff-Coste and Zúñiga54]. They show

$$\begin{align*}\|Q^{\ell} - u \|_{TV} \le 2 e^{-c}, \quad \text{with} \,\,\, \ell = \frac{1}{2}n(\log n + c). \end{align*}$$

The asymptotic ‘profile’ (the limit of $\|Q^{\ell } - u\|_{TV}$ as a function of c for n large) is determined in [Reference Teyssier63]. Schramm [Reference Schramm56] found a sharp parallel between random transpositions and the growth of an Erdős-Rényi random graph: Given vertices $1, 2, \dots , n$ , for each transposition $(i, j)$ chosen, add an edge from vertices i to j to generate a random graph. See [Reference Berestycki, Schramm and Zeitouni6] for extensions and a comprehensive review. The results, translated by the coagulation-fragmentation description of the cycles, give a full and useful picture for the simple mean field model described in the introduction.

It must be emphasized that this mean field model is a very special case of coagulation-fragmentation models studied in the chemistry-physics-probability literature. These models study the dynamics of particles diffusing in an ambient space, and allow general collision kernels (e.g., particles close in space may be more likely to join). The books by Bertoin [Reference Bertoin7] and Pitman [Reference Pitman50] along with the survey paper of Aldous [Reference Aldous3] are recommended for a view of the richness of this subject. On the other hand, the sharp rates of convergence results available for the mean field model are not available in any generality.

There is a healthy applied mathematics literature on coagulation-fragmentation. A useful overview which treats discrete problems such as the ones treated here is [Reference Ball and Carr5]. A much more probabilistic development of the celebrated Becker–Doring version of the problem is in [Reference Hingant and Yvinec36]. This develops rates of convergence using coupling. See also [Reference Durrett, Granovsky and Gueron27] for more models with various stationary distributions on partitions.

Other lumpings of random transpositions include classical urn models—the Bernoulli–Laplace model [Reference Diaconis and Shahshahani23], [Reference Eskenazis and Nestoridi28], and random walks on phylogenetic trees [Reference Diaconis and Holmes26]. The sharp analysis of random transpositions transfers, via comparison theory, to give good rates of convergence for quite general random walks on the symmetric group [Reference Diaconis and Saloff-Coste20], [Reference Helfgott, Seress and Zuk34]. For an expository survey, see [Reference Diaconis16].

2.4 Transvections

Fix n, a prime p and $q = p^a$ for some positive integer a. A transvection is an invertible linear transformation of $\mathbb {F}_q^n$ which fixes a hyperplane, is not the identity and has all eigenvalues equal to $1$ . Transvections are convenient generators for the group $SL_n(q)$ because they generalize the basic row operations of linear algebra. These properties are carefully developed in [Reference Suzuki62] Chapter 1, 9; [Reference Artin4] Chapter 4.

Using coordinates, let $\mathbf {a}, \mathbf {v} \in \mathbb {F}_q^n$ be two nonzero vectors with $\mathbf {a}^{\top } \mathbf {v} = 0$ . A transvection, denoted $T_{\mathbf {a}, \mathbf {v}} \in GL_n(q)$ is the linear map given by

$$\begin{align*}T_{\mathbf{a}, \mathbf{v}}(\mathbf{x}) = \mathbf{x} + \mathbf{v}(\mathbf{a}^{\top} \mathbf{x}), \quad \text{for} \,\,\, \mathbf{x} \in \mathbb{F}_q^n. \end{align*}$$

It adds a multiple of $\mathbf {v}$ to $\mathbf {x}$ , the amount depending on the ‘angle’ between $\mathbf {a}$ and $\mathbf {x}$ . As a matrix, . Multiplying $\mathbf {a}$ by a nonzero constant and dividing $\mathbf {v}$ by the same constant doesn’t change $T_{\mathbf {a}, \mathbf {v}}$ . Let us agree to normalize $\mathbf {v}$ by making its last nonzero coordinate equal to $1$ . Let $\mathcal {T}_{n, q} \subset SL_n(q)$ be the set of all transvections.

An elementary count shows

(2.5) $$ \begin{align} |\mathcal{T}_{n, q}| = \frac{(q^n - 1)(q^{n-1} - 1)}{q - 1}. \end{align} $$

It is easy to generate $T \in \mathcal {T}_{n, q}$ uniformly: Pick $\mathbf {v} \in \mathbb {F}_q^n$ uniformly, discarding the zero vector. Normalize $\mathbf {v}$ so the last nonzero coordinate, say index j, is equal to $1$ . Pick $a_1, a_2, \dots , a_{j-1}, a_{j+1}, \dots a_n$ uniformly in $\mathbb {F}_q^{n-1} - \{0 \}$ , and set $a_j$ so that $\mathbf {a}^{\top } \mathbf {v} = 0$ . The transvection $T_{\mathbf {a}, \mathbf {v}}$ fixes the hyperplane $\{\mathbf {x}: \mathbf {a}^{\top } \mathbf {x} = 0 \}$ .

Example 2.8. Taking $\mathbf {v} = \mathbf {e}_1, \mathbf {a} = \mathbf {e}_2$ gives the transvection with matrix

This acts on $\mathbf {x}$ by adding the second coordinate to the first. Similarly, the basic row operation of adding $\theta $ times the ith coordinate to the jth is given by $T_{\mathbf {e}_j, \theta \mathbf {e}_i}$ .

Proof. Let $M \in GL_n(q)$ , so $M T_{\mathbf {e}_2, \mathbf {e}_1} M^{-1}$ is conjugate to $T_{\mathbf {e}_2, \mathbf {e}_1}$ . Then,

$$\begin{align*}M T_{\mathbf{e}_2, \mathbf{e}_1} M^{-1}(\mathbf{x}) = M ( M^{-1} \mathbf{x} + (M^{-1} \mathbf{x})_2 \mathbf{e}_1 ) = \mathbf{x} + (M^{-1} \mathbf{x})_2 M \mathbf{e}_1. \end{align*}$$

Let $\mathbf {a}$ be the second column of $(M^{-1})T_{\mathbf {e}_2, \mathbf {e}_1}$ and $\mathbf {v}$ the first column of M, and check this last is $T_{\mathbf {a}, \mathbf {v}}$ (and $\mathbf {a}^{\top } \mathbf {v} = 0$ ). Thus, transvections form a conjugacy class.

2.5 Gaussian elimination and the Bruhat decomposition

The reduction of a matrix $M\in GL_n(q)$ to standard form by row operations is a classical topic in introductory linear algebra courses. It gives efficient, numerically stable ways to solve linear equations, compute inverses and calculate determinants. There are many variations.

Example 2.10. Consider the sequence of row operations

$$\begin{align*}M = \begin{pmatrix} 0 & 3 & 2 \cr 1 & 2 & 0 \cr 3 & 0 & 5 \end{pmatrix} \to \begin{pmatrix} 0 & 3 & 2 \cr 1 & 2 & 0 \cr 0 & -6 & 5 \end{pmatrix} \to \begin{pmatrix} 0 & 3 & 2 \cr 1 & 2 & 0 \cr 0 & 0 & 9 \end{pmatrix} \to \begin{pmatrix} 1 &2 & 0 \cr 0 & 3 & 2 \cr 0 & 0 & 9 \end{pmatrix} = U. \end{align*}$$

The first step subtracts $3$ times row $2$ from row $3$ , multiplication by

$$\begin{align*}L_1 = \begin{pmatrix} 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & -3 & 1 \end{pmatrix}. \end{align*}$$

The second step adds $2$ times row $1$ to row $3$ , multiplication by

$$\begin{align*}L_2 = \begin{pmatrix} 1 & 0 & 0 \cr 0 & 1 & 0 \cr 2 & 0 & 1 \end{pmatrix}. \end{align*}$$

The third (pivot) step brings the matrix to upper-triangular form by switching rows $1$ and $2$ , which corresponds to multiplication by the matrix by

$$\begin{align*}\omega_1 = \begin{pmatrix} 0 & 1 & 0 \cr 1 & 0 & 0 \cr 0 & 0 & 1 \end{pmatrix}. \end{align*}$$

This gives $\omega _1 L_2 L_1 M = U \implies M = L_2^{-1} L_1^{-1} \omega _1^{-1} U = L \omega U$ with $L = L_2^{-1} L_1^{-1}, \omega = \omega _1^{-1} = \omega _1$ .

If $\mathcal {L}, \mathcal {B}$ are the subgroups of lower and upper-triangular matrices in $GL_n(q)$ , this gives

(2.6) $$ \begin{align} GL_n(q) = \bigsqcup_{\omega \in S_n} \mathcal{L} \omega \mathcal{B}. \end{align} $$

Any linear algebra book treats these topics. A particularly clear version which uses Gaussian elimination as a gateway to Lie theory is in Howe [Reference Howe37]. Articles by Lusztig [Reference Lusztig45] and Strang [Reference Strang61] have further historical, mathematical and practical discussion.

Observe that carrying out the final pivoting step costs $d_c(\omega ,\mathrm {id})$ operations, where $d_c(\omega ,\mathrm {id})$ , the Cayley distance of $\omega $ to the identity, is the minimum number of transpositions required to sort $\omega $ (with arbitrary transpositions $(i,j)$ allowed). Cayley proved $d_c(\omega ,\mathrm {id}) = n - \#\mbox {cycles in} \ \omega $ (see [Reference Diaconis16]). In the example above $n=3$ , $\omega = 213$ has two cycles and $3-2 =1$ —one transposition sorts $\omega $ .

How many pivot steps are needed ‘on average’? This becomes the question of the number of cycles in a pick from Mallows measure $\pi _q$ . Surprisingly, this is a difficult question. Following partial answers by Gladkich and Peled [Reference Gladkich and Peled29], this problem was recently solved by Jimmy He, Tobias Möller and Teun Verstraaten in [Reference He, Möller and Verstraaten33]. They show that, when $q> 1$ , the limiting behavior of the number of even cycles under $\pi _q$ has an approximate normal distribution with mean and variance proportional to n, and that the number of odd cycles has bounded mean and variance.

The Bruhat decomposition

In algebraic group theory, one uses

(2.7) $$ \begin{align} GL_n(q) = \bigsqcup_{\omega \in S_n} \mathcal{B} \omega \mathcal{B}. \end{align} $$

This holds for any semisimple group over any field with $\mathcal {B}$ replaced by the Borel group (the largest solvable subgroup) and $S_n$ replaced by the Weyl group.

Let $\omega _0 = \left (\begin {smallmatrix} 1 & 2 & \cdots & n\\ n & n-1 & \cdots & 1 \end {smallmatrix}\right )$ be the reversal permutation in $S_n$ . Since $\mathcal {L} = \omega _0 \mathcal {B} \omega _0$ , equation (2.7) is equivalent to the LU decomposition (2.6). Given $M \in GL_n(q)$ , Gaussian elimination on $\omega _0 M$ can be used to find $\omega _0 M \in L \omega ' U$ and thus $M \in B \omega B$ with $\omega = \omega _0 \omega '$ .

The subgroup $\mathcal {B}$ gives rise to the quotient $GL_n(q)/\mathcal {B}$ . This may be pictured as the space of ‘flags’. Here, a flag F consists of an increasing sequence of subspaces with $\text {dim}(F_i) = i$ . Indeed, $GL_n(q)$ operates transitively on flags and the subgroup fixing the standard flag is exactly $\mathcal {B}$ . This perspective will be further explained and used in Section 6 to study a function of the double coset Markov chain on $\mathcal {B} \backslash GL_n(q) / \mathcal {B}$ .

Remark 2.11. The double cosets of $GL_n(q)$ define equivalence classes for any subgroup of $GL_n(q)$ . For the matrices $SL_n(q)$ with determinant $1$ , these double cosets again induce the Mallows distribution on permutations. More precisely, for $x \in SL_n(q)$ , let $[x]_{SL_n(q)} = \{x' \in SL_n(q) : x' \in \mathcal {B} x \mathcal {B} \}$ be the equivalence class created by the double coset relation $\mathcal {B} \backslash GL_n(q) / \mathcal {B}$ , within $SL_n(q)$ . Note that two matrices $x, x' \in SL_n(q)$ could be in the same double coset with $x' = b_1 x b_2$ , but $b_1, b_2 \notin SL_n(q)$ (necessarily, $\mathrm {det}(b_1) = \mathrm {det}(b_2)^{-1}$ ).

Then, $|[\omega ]_{SL_n(q)}|/|SL_n(q)| = p_q(\omega )$ . This follows since $|GL_n(q)| = (q-1) \cdot |SL_n(q)|$ , and $|\mathcal {B} \omega \mathcal {B}| = (q - 1) \cdot |[\omega ]_{SL_n(q)}|$ . If $M \in GL_n(q)$ and $M \in B \omega B$ , then $M/\det (M) \in [\omega ]_{SL_n(q)}$ . Conversely, for each $M \in SL_n(q)$ there are $(q - 1)$ unique matrices in $GL_n(q)$ created by multiplying M by $1, 2, \dots , q - 1$ .

2.6 The Metropolis algorithm

The Metropolis algorithm is a basic algorithm of scientific computing which arises in describing the random walk induced by transvections on the double cosets $\mathcal {B}_n \backslash GL_n(q) / \mathcal {B}_n$ (Section 3.2). This section gives background.

Given a probability distribution $\pi $ on a space $\mathcal {X}$ , the Metropolis algorithm gives a way of changing the output of a Markov chain with transition matrix $K(x, y)$ to have stationary distribution $\pi $ on $\mathcal {X}$ . For simplicity, suppose the original chain is symmetric, $K(x, y) = K(y, x)$ (as in our examples). This implies that $K(x, y)$ has a uniform stationary distribution. Define the Metropolis Markov chain with the transition matrix:

$$\begin{align*}M(x, y) = \begin{cases} K(x, y), & \text{if} \,\,\, \pi(y) \ge \pi(x), \, x \neq y, \\ K(x, y) \cdot \frac{\pi(y)}{\pi(x)}, & \text{if} \,\,\, \pi(y) < \pi(x), \, x \neq y, \\ K(x, x) + \sum_{z: \pi(z) < \pi(x)} K(x, z) \left( 1 - \frac{\pi(z)}{\pi(x)} \right), & \text{if} \,\,\, x = y. \end{cases} \end{align*}$$

These transition probabilities have a simple implementation: From x, pick y according to $K(x, y)$ . If $\pi (y) \ge \pi (x)$ , move to y. If $\pi (y) < \pi (x)$ , flip a coin with heads probability $\pi (y)/\pi (x)$ . If the coin is heads, move to y. If the coin is tails, stay at x. Elementary calculations show that $\pi (x) M(x, y) = \pi (y) M(y, x)$ , that is, M has $\pi $ as stationary distribution. Note that the normalizing constant of $\pi $ is not needed.

For background, applications, and theoretical properties of the Metropolis algorithm, see the textbook of Liu [Reference Liu44] or the survey [Reference Diaconis and Saloff-Coste21]. Sharp analysis of rates of convergence of the Metropolis algorithm is still an open research problem. The special cases developed in Section 3.2 show it can lead to fascinating mathematics.

3.1 Hecke algebras and the Metropolis algorithm

The set of $\mathcal {B}$ - $\mathcal {B}$ double cosets of $GL_n(q)$ has remarkable structure. For $\omega \in S_n$ , let $T_{\omega } = \mathcal {B} \omega \mathcal {B}$ . Linear combinations of double cosets form an algebra (over $\mathbb {C}$ , for example).

Definition 3.1. The Iwahori–Hecke algebra $H_n(q)$ is spanned by the symbols $\{T_{\omega } \}_{\omega \in S_n}$ and generated by $T_i = T_{s_i}$ for $s_i = (i, i+1), 1 \le i \le n -1$ , with the relations

(3.1) $$ \begin{align} \begin{cases} T_{s_i}T_{\omega} = T_{s_i\omega} & \text{if } I(s_i\omega) = I(\omega)+1, \\ T_{s_i}T_{\omega} = qT_{s_i \omega} +(q-1)T_{\omega} & \text{if } I(s_i\omega) = I(\omega)-1, \end{cases} \end{align} $$

where $I(\omega )$ is the usual length function on $S_n$ ( $I(s_i \omega ) = I(\omega )\pm 1$ ).

Consider the flag space $\mathcal {F} = G/\mathcal {B}$ . The group $GL_n(q)$ acts on the left on $\mathcal {F}$ . One can see $H_n(q)$ acting on the right of $\mathcal {F}$ and in fact

$$\begin{align*}H_n(q) = \mathrm{End}_{GL_n(q)}(GL_n(q)/\mathcal{B}). \end{align*}$$

The Hecke algebra is the full commuting algebra of $GL_n(q)$ acting on $GL_n(q)/\mathcal {B}$ . Because transvections form a conjugacy class, the sum of transvections is in the center of the group algebra $\mathbb {C}[GL_n(q)]$ , and so it may be regarded as an element of $H_n(q)$ . This will be explicitly delineated and the character theory of $H_n(q)$ used to do computations.

3.2 The Metropolis connection

The relations (3.1) can be interpreted probabilistically. Consider what equation (3.1) says as linear algebra: Left multiplication by $T_{s_i}$ can take $\omega $ to $\omega $ or $s_i \omega $ . The matrix of this map (in the basis $\{T_{\omega }\}_{\omega \in S_n}$ ) has $\omega , \omega '$ entry

$$ \begin{align*} \begin{cases} 1, & \text{if} \quad I(s_i \omega) = I(\omega) + 1, \,\, \text{and} \,\,\, \omega' = s_i \omega, \\ q, & \text{if} \quad I(s_i \omega) = I(\omega) - 1, \,\, \text{and} \,\,\, \omega' = \omega, \\ q - 1, & \text{if} \quad I(s_i \omega) = I(\omega) -1, \,\, \text{and} \,\,\, \omega' = s_i \omega, \\ 0, & \text{otherwise}. \end{cases} \end{align*} $$

For example, on $GL_3(q)$ using the ordered basis $T_{id}, T_{s_1}, T_{s_2}, T_{s_1s_2}, T_{s_2s_1}, T_{s_1s_2s_1}$ , the matrix of left multiplication by $s_1$ is

$$\begin{align*}T_{s_1} = \begin{pmatrix} 0 & q & 0 & 0 & 0 & 0 \cr 1 & q-1 & 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & q & 0 & 0 \cr 0 & 0 & 1 & q-1 & 0 & 0 \cr 0 & 0 & 0 & 0 & 0 & q \cr 0 & 0 & 0 & 0 & 1 & q-1 \cr \end{pmatrix}. \end{align*}$$

The first column has a $1$ in row $s_1$ because $I(s_1)> I(id)$ . The second column has entries q and $q-1$ in the first two rows because $I(s_1^2) = I(id) < I(s_1)$ .

We can also write the matrices for multiplication defined by $T_{s_2}$ and $T_{s_1s_2s_1}$ as

$$ \begin{align*} &T_{s_2} = \begin{pmatrix} 0 & 0 & q & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 & q & 0 \cr 1 & 0 & q-1 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 & 0 & q\\ 0 & 1 & 0 & 0 & q-1 & 0 \cr 0 & 0 & 0 & 1 & 0 & q-1 \cr \end{pmatrix} \\ & T_{s_1s_2s_1} = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & q^3 \cr 0 & 0 & 0 & q^2 & 0 & q^2(q - 1) \cr 0 & 0 & 0 & 0 & q^2 & q^2(q - 1) \cr 0 & q & 0 & q(q - 1) & q(q-1) & q(q-1)^2 \cr 0 & 0 & q & q(q-1) & q(q-1) & q(q-1)^2 \cr 1 & q-1 & q-1 & (q-1)^2 & (q-1)^2 & (q - 1)^3 + q(q-1)\cr \end{pmatrix}. \end{align*} $$

Observe that all three matrices above have constant row sums (q, q and $q^3$ , respectively). Dividing by these row sums gives three Markov transition matrices: $M_1, M_2$ and $M_{121} = M_1M_2M_1$ .

These matrices have a simple probabilistic interpretation: Consider, for $\overline {q} = 1/q$ , the matrix defined

$$\begin{align*}M_1 = \begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 \cr \overline{q} & 1 - \overline{q} & 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 1 & 0 & 0 \cr 0 & 0 & \overline{q} & 1 - \overline{q} & 0 & 0 \cr 0 & 0 & 0 & 0 & 0 & 1 \cr 0 & 0 & 0 & 0 & \overline{q} & 1 - \overline{q} \cr \end{pmatrix}. \end{align*}$$

The description of this Markov matrix is: From $\omega $ , propose $s_1 \omega $ :

• • If $I(s_1 \omega )> I(\omega )$ , go to $s_1 \omega $ .

• • If $I(s_1 \omega ) < I(\omega )$ , go to $s_1 \omega $ with probability $1/q$ , else stay at $\omega $ .

This is exactly the Metropolis algorithm on $S_n$ for sampling from $\pi _q(\omega )$ with the proposal given by the deterministic chain ‘multiply by $s_1$ ’. The matrices $M_i, 1 \le i \le n -1$ , have a similar interpretation and satisfy

$$\begin{align*}\pi_q(\omega) M_i(\omega, \omega') = \pi_q(\omega') M_i(\omega', \omega). \end{align*}$$

The Metropolis algorithm always results in a reversible Markov chain. See Section 2.6 or [Reference Diaconis and Saloff-Coste21] for background. It follows that any product of $\{M_i \}$ and any convex combination of such products yields a $\pi _q$ reversible chain. Note also that the Markov chain on $S_n$ is automatically reversible since it is induced by a reversible chain on $GL_n$ .

Proof. Up to normalization, the matrix D in Theorem 1.6 is a positive linear combination of Markov chains corresponding to multiplication by

$$\begin{align*}T_{(i,j)} = T_{s_i}T_{s_{i+1}} \cdots T_{s_{j-2}}T_{s_{j-1}}T_{s_{j-2}}\cdots T_{s_{i+1}}T_{s_i}. \end{align*}$$

This yields a combination of the reversible chains .

Example 3.3. The transition matrix of the transvections chain on $GL_3(q)$ lumped to $S_3$ is $\frac {1}{|\mathcal {T}_{n, q}|}D$ , with

$$ \begin{align*} D &= (2q^2 - q - 1) \mathbf{I} \\ & \quad + (q-1)\begin{pmatrix} 0 &q^2 &q^2 &0 &0 &q^3 \\ q &q(q-1) &0 &q^2 &q^2 &q^2(q-1) \\ q &0 &q(q-1) &q^2 &q^2 &q^2(q-1) \\ 0 &q &q &2q(q-1) &q(q-1) &q^2+q(q-1)^2 \\ 0 &q &q &q(q-1) &2q(q-1) &q^2+q(q-1)^2 \\ 1 &q-1 &q-1 &q+(q-1)^2 &q+(q-1)^2 &(q-1)^3+3q(q-1) \end{pmatrix}. \end{align*} $$

When $q=2$ , the lumped chain has transition matrix

$$ \begin{align*}\frac{1}{21} \begin{pmatrix} 5 &4 &4 &0 &0 &8 \\ 2 &7 &0 &4 &4 &4 \\ 2 &0 &7 &4 &4 &4 \\ 0 &2 &2 &9 &2 &6 \\ 0 &2 &2 &2 &9 &6 \\ 1 &1 &1 &3 &3 &12 \end{pmatrix}. \end{align*} $$

We report that this example has been verified by several different routes including simply running the transvections chain, computing the double coset representative at each step and estimating the transition rates from a long run of the chain.

Remark 3.4. The random transvections Markov chain on $S_n$ is the ‘q-deformation’ of random transpositions on $S_n$ . That is, as q tends to $1$ , the transition matrix tends to the transition matrix of random transpositions. To see this, recall $|\mathcal {T}_{n, q}| = (q^n - 1)(q^{n-1} - 1)/(q-1)$ and use L’Hopitals rule to note, for any integer k,

$$ \begin{align*} &\lim_{q \to 1^+} \frac{(q - 1)^2 q^k(q-1)^{\ell}}{(q^n - 1)(q^{n-1} - 1)} = \begin{cases} 0, & \text{if} \,\, \ell> 0, \\ \frac{1}{n(n-1)}, & \text{if} \,\,\, \ell = 0. \end{cases} \end{align*} $$

The interpretation of multiplication on the Hecke algebra as various ‘systematic scan’ Markov chains is developed in [Reference Diaconis and Ram18], [Reference Bufetov and Nejjar10]. It works for other types in several variations. We are surprised to see it come up naturally in the present work.

The following corollary provides the connection to [Reference Ram52 Reference Ram, (3.16),(3.18),(3.20)] and [Reference Diaconis and Ram18, Proposition 4.9].

Corollary 3.5. Let $J_1 = 1$ , and let $J_k = T_{s_{k-1}}\cdots T_{s_2}T_{s_1}T_{s_1}T_{s_2}\cdots T_{s_{k-1}}$ , for $k\in \{2, \ldots , n\}$ . Then

$$ \begin{align*}D = \sum_{k=1}^n q^{n-k}(J_k-1).\end{align*} $$

Proof. Using that $J_k = T_{s_{k-1}}J_{k-1}T_{s_{k-1}}$ , check, by induction, that

$$ \begin{align*}J_k = q^{k-1} + (q-1)\sum_{j=1}^{k-1} q^{j-1}T_{(j,k)}.\end{align*} $$

Thus,

$$ \begin{align*} \sum_{k=1}^n q^{n-k}(J_k-1) &= 0+ \sum_{k=2}^n \left(q^{n-k+(k-1)} -q^{n-k} + (q-1)\sum_{j=1}^{k-1} q^{n-k+j-1}T_{(j,k)}\right) \\ &=(n-1)q^{n-1} - [n-1]_q + (q-1)D_{(21^{n-2})} = D, \end{align*} $$

where $D_{(21^{n-2})} := \sum _{i < j} q^{(n - 1) - (j-i)} T_{(i, j)}$ .

3.3 Eigenvalues and multiplicities

Hildebrand [Reference Hildebrand35] determined the eigenvalues of the random walk driven by Q on $GL_n(q)$ . His arguments use Macdonald’s version of J.A. Green’s formulas for the characters of $GL_n(q)$ along with sophisticated use of properties of Hall–Littlewood polynomials. Using the realization of the walk on the Hecke algebra, developed below in Section 5, and previous work of Ram and Halverson [Reference Halverson and Ram31], we can find cleaner formulas and proofs. Throughout, we have tried to keep track of how things depend on both q and n (the formulas are easier when $q = 2$ ).

The argument uses the representation of the Markov chain as multiplication on the Hecke algebra. This is developed further in Section 5.

Proof. (a): Let D be the sum of transvections as in Theorem 1.6. By [Reference Halverson and Ram31] (3.20), the irreducible representation $H_n^{\lambda }$ of $H_n$ indexed by $\lambda $ has a ‘seminormal basis’ $\{ v_T \mid \widehat {S}_n^{\lambda } \}$ such that $J_k v_T = q^{k-1} q^{c(T(k))} v_T$ , where $T(k)$ is the box containing k in T. Thus,

$$\begin{align*}Dv_T = \sum_{k = 1}^n q^{n - k} (q^{k - 1}q^{c(T(k))} - 1)v_T = \left( - \left( \frac{q^n - 1}{q - 1} \right) + \sum_{k = 1}^n q^{n-1} q^{c(T(k))} \right) v_T. \end{align*}$$

(b): The dimension of the irreducible representation of $H_n$ indexed by $\lambda $ is the same as the dimension of the irreducible representation of $S_n$ indexed by $\lambda $ , which is well known as the hook-length formula [Reference Macdonald46, Ch. I(7.6)(ii), §6 Ex.2(a)]:

$$\begin{align*}\text{dim}(H_n^{\lambda}) = \text{Card}(S_n^{\lambda}) = \frac{n!}{\prod_{b \in \lambda} h(b)}. \end{align*}$$

(c): With $G = GL_n(q), H = H_n$ , the result follows since [Reference Macdonald46, Ch. IV(6.7)],

$$\begin{align*}\text{dim}(G^{\lambda}) = \frac{[n]_q!}{\prod_{b \in \lambda} [h(b)]_q} \quad \text{and} \quad \mathbf{1}_{\mathcal{B}}^G = \sum_{\lambda \in \widehat{S_n}} G^{\lambda} \otimes H^{\lambda}. \end{align*}$$

Example 3.7. Equation 3.2 for some specific partitions gives

The following simple lemma will be used several times in subsequent sections. It uses the usual majorization partial order (moving up boxes) on partitions of n, [Reference Macdonald46]. For example, when $n = 4$ the ordering is $1111 \prec 211 \prec 22 \prec 31 \prec 4$ .

Proof. Moving a single box (at the corner of the diagram of $\lambda $ ) in position $(i, j)$ to position $(i', j')$ necessitates $i' < i$ and $j'> j$ , and so $q^{j-i} < q^{j' - i'}$ . Since any $\lambda \prec \lambda '$ can be obtained by successively moving up boxes, the proof is complete. For example,

The partition $1^n = (1, 1, \dots , 1)$ is the minimal element in the partial order, and since $\beta _{(1^n)} = 0$ , we have the following:

Corollary 3.9. If $\lambda \vdash n$ , then

$$\begin{align*}\beta_{\lambda} \ge 0. \end{align*}$$

Corollary 3.10. If is a partition of n with $\lambda _1 = n - j$ and $j \le n/2$ , then

$$ \begin{align*} \beta_{\lambda} \le \beta_{(n-j, j)} &= \frac{q^j(q^{n - j - 1} - 1)(q^{n - j} - 1) + (q^{n - 2} - 1)(q^j - 1)}{(q^{n-1} - 1)(q^n -1)} \\ &\le q^{-j}(1 + q^{-(n - 2j + 1)})(1 + q^{-(n - 1)})(1 + q^{-(n - 2)}). \end{align*} $$

Proof. The first inequality follows from Lemma 3.8. The formula for $\beta _{(n - j, j)}$ is a simple calculation from equation (3.2). Recall the elementary inequalities:

$$\begin{align*}\frac{1}{q^r} < \frac{1}{q^r - 1} < \frac{1}{q^{r-1}}, \quad \frac{1}{1 - q^{-r}} = 1 + \frac{1}{q^r - 1} < 1 + q^{-(r -1)}. \end{align*}$$

These give the inequality for $\beta _{(n - j, j)}$ :

$$ \begin{align*} \beta_{(n - j, j)} &= \frac{q^j(q^{n - j - 1} - 1)(q^{n - j} - 1) + (q^{n - 2} - 1)(q^j - 1)}{(q^{n-1} - 1)(q^n -1)} \\ &\le \left( \frac{q^j q^{n - j - 1}q^{n - j} + q^{n-2}q^j}{q^{n-1} q^n} \right) \frac{1}{(1 - q^{-(n-1)})(1 - q^{-n})} \\ &\le (q^{-j} + q^{-(n - j + 1))})(1 + q^{-(n-2)})(1 + q^{-(n-1)}). \end{align*} $$

In the following sections, we will use a further bound from Corollary 3.10.

Corollary 3.11. Define $\kappa _{n, q, j} := (1 + q^{-(n - 2j + 1)})(1 + q^{-(n - 1)})(1 + q^{-(n - 2)})$ and

$$\begin{align*}\alpha_{n, q} := \max_{1 \le j \le n/2} \frac{\log(\kappa_{n, q, j})}{j}. \end{align*}$$

Then for all $n, q$ ,

$$\begin{align*}\alpha_{n, q} \le \frac{6}{n}. \end{align*}$$

Proof. Using $1 + x \le e^x$ , there is the initial bound

$$ \begin{align*} \kappa_{n, q, j} &\le \exp \left( q^{-(n - 2j + 1)} + q^{-(n-1)} + q^{-(n-2)} \right) \le \exp \left( 3 q^{-(n - 2j)} \right), \end{align*} $$

which uses that $j \ge 1$ and so $(n - 2) \ge (n -2j)$ . Then,

$$ \begin{align*} \frac{\log(\kappa_{n, q, j})}{j} \le \frac{3 q^{-(n - 2j)}}{j}. \end{align*} $$

With $f(x) = 3q^{-(n - 2x)}/x$ , we have

$$\begin{align*}f'(x) = \frac{3q^{-(n - 2x)}(2x \log q - 1)}{x^2}. \end{align*}$$

Since $2 \log (2)> 1$ , we see that $f(x)$ is increasing for $x \ge 1$ and any $q \ge 2$ . Thus, for $1 \le j \le n/2$ , $f(j)$ is maximized for $j = n/2$ , which gives

$$\begin{align*}\alpha_{n, q} \le \max_{1 \le j \le n/2} \frac{3 q^{-(n - 2j)}}{j} \le \frac{6}{n}. \end{align*}$$

4.1 Eigenvalue bounds

The following result from [Reference Diaconis and Ram18] will be the main tool for achieving bounds on the chi-square distance of the chain from different starting states.

In general, the right-hand side of (a) could be difficult to calculate, but it simplifies for the special cases $x = id, x = \omega _0$ . These calculations, and the analysis of the sum, are contained in the following sections.

The right-hand side of the equations in Proposition 4.1 involves the following quantities, defined for $\lambda \vdash n$ :

• • $n_{\lambda } = \sum _{i = 1}^{|\lambda |} (i - 1) \lambda _i$ ,

• • $c_{\lambda } = \sum _{b \in \lambda } ct(b)$ , where $ct(b) = j - i$ if box b is in column j and row i,

• • $t_{\lambda } = q^{n_{\lambda }} \cdot r_{\lambda }$ , where $r_{\lambda } = \frac {[n]_q!}{\prod _{b \in \lambda } [h_b]_q}$ , $[k]_q = (q^k - 1)/(q - 1)$ and ,

• • $f_{\lambda }$ , which is the number of standard Young tableau of shape $\lambda $ . The formula is

  $$\begin{align*}f_{\lambda} = \frac{n!}{\prod_{b \in \lambda} h(b)}. \end{align*}$$

Table 1 shows these values for $n = 4$ and general q. From this example, we can observe that $c_{\lambda }$ is increasing with respect to the partial order on partitions, while $n_{\lambda }$ is decreasing.

Table 1 The quantities involved in the eigenvalue and multiplicity calculations for $n = 4$ .

Let us record that

(4.1) $$ \begin{align} \beta_{(n-1,1)} = \frac{q^{n-2}-1}{q^{n-1}-1}, \qquad t_{(n-1,1)} = q\cdot \frac{q^{n-1}-1}{q-1} \qquad\mbox{and}\qquad f_{(n-1,1)} = n-1. \end{align} $$

Since $n!\le n^n = e^{n\log n}$ then

(4.2) $$ \begin{align} \frac{n!}{e^{n\log n}} \le 1. \end{align} $$

We will use the following bounds from [Reference Diaconis and Ram18, Lemma 7.2]:

(4.3) $$ \begin{align} t_{\lambda} \le q^{-\binom{\lambda_1}{2}+\binom{n}{2}}f_{\lambda}, \quad \text{and} \,\,\, \sum_{\lambda \vdash n} f_{\lambda}^2 = n!. \end{align} $$

In addition, we need the following bounds for sums of $f_{\lambda }$ . Part (b) of the proposition below is Lemma 7.2(b) in [Reference Diaconis and Ram18]; the proof there is incomplete, so we give the simple proof below.

Proof. Recall that $f_{\lambda }$ is the number of standard tableau of shape $\lambda $ , that is, the elements $1, 2, \dots , n$ are arranged in the shape $\lambda $ so that rows are increasing left to right and columns are increasing top to bottom. If $\lambda _1 = n -j$ , then there are $\binom {n}{j}$ ways to choose the elements not in the first row of the tableau. For a fixed partition $\lambda $ , the number of ways of arranging the remaining j elements is at most the number of Young tableau corresponding to the partition of j created from the remaining rows of $\lambda $ . This number is at most $\sqrt {j!}$ (Lemma 3 in [Reference Diaconis and Shahshahani22]). Thus,

$$ \begin{align*} \sum_{\lambda: \lambda_1=n-j} f_{\lambda} \le \binom{n}{j} \cdot \sqrt{j!} \cdot p(j), \end{align*} $$

where $p(j)$ is equal to the number of partitions of j. It is well known that $\log (p(n)) \sim B \cdot \sqrt {n}$ for a constant B. More precisely, from (2.11) in [Reference Hardy and Ramanujan32], there is a universal constant $K>0$ such that for all $n \ge 1$ ,

$$\begin{align*}p(n) < \frac{K}{n} e^{2 \sqrt{2n}}. \end{align*}$$

This gives (a).

For part (b), we again use the inequality $f_{\lambda } \le \binom {n}{j} f_{\lambda ^*}$ , where $\lambda ^* = (\lambda _2, \dots , \lambda _k)$ is the partition of j determined by the rest of $\lambda $ after the first row. Then,

$$ \begin{align*} \sum_{\lambda: \lambda_1 = n - j} f_{\lambda}^2 \le \binom{n}{j}^2 \sum_{\lambda^* \vdash j} f_{\lambda^*}^2 = \binom{n}{j}^2 \cdot j! = \left( \frac{n!}{(n - j)!j!} \right)^2 \cdot j! \le \frac{n^{2j}}{j!}. \end{align*} $$

Proposition 4.3. The function $s(\lambda ) := q^{c_{\lambda }} t_{\lambda }$ is monotone with respect to the partial order on partitions. For any $\lambda \vdash n$ ,

$$\begin{align*}s(\lambda) \le q^{\binom{n}{2}}. \end{align*}$$

Proof. Suppose that $\lambda \prec \widetilde {\lambda }$ and $\widetilde {\lambda }$ is obtained from $\lambda $ by ‘moving up’ one box. Suppose the box at position $(i, j)$ is moved to $(i', j')$ , with $i' < i, j'> j$ .

Let $g(\lambda ) = c_{\lambda } + n_{\lambda }$ . Then,

$$ \begin{align*} &c_{\widetilde{\lambda}} = c_{\lambda} + (j' - j) + (i - i') \\ &n_{\widetilde{\lambda}} = n_{\lambda} - (i - 1) + (i' - 1) = n_{\lambda} + (i' - i). \end{align*} $$

This implies that

$$\begin{align*}g(\widetilde{\lambda}) = g(\lambda) + (j' - j) + (i - i') - (i - 1) + (i' - 1) = g(\lambda) + (j' - j). \end{align*}$$

Now, consider the change in $r_{\lambda }$ . The hook lengths of $\widetilde {\lambda }$ are:

$$ \begin{align*} &\widetilde{h}(i, j) = h(i, j) - 1 = 0, \quad \widetilde{h}(i', j') = h(i, j) + 1 = 1 \\ &\widetilde{h}(k, j) = h(k, j) - 1, \quad k < i, k \neq i' \\ &\widetilde{h}(i, l) = h(i, l) - 1, \quad l < j \\ &\widetilde{h}(k, j') = h(k, j') + 1, \quad k < i' \\ &\widetilde{h}(i', l) = h(i', l) +1, \quad l < j', l \neq j \end{align*} $$

and $\widetilde {h}(k, l) = h(k, l)$ for all other boxes $b = (k, l)$ . Thus,

$$ \begin{align*} \sum_b \widetilde{h}(b) - \sum_b h(b) &= (j' - 1) + (i' - 1) - (i - 1) - (j - 1) \\ &= (j' - j) + (i' - i). \end{align*} $$

Using the inequalities $q^{r-1} < (q^r - 1) < q^r$ , we have

$$ \begin{align*} r(\widetilde{\lambda}) &= \frac{[n]_q!(q - 1)^n}{\prod_b (q^{h(b)} - 1)} \\ &\le \frac{[n]_q!(q - 1)^n}{q^{\sum_b \widetilde{h}(b) - n}} = \frac{1}{q^{(j' - j) + (i' - i) -n} } \frac{[n]_q!(q - 1)^n}{q^{\sum_b h(b)}} \\ &\le \frac{1}{q^{(j' - j) + (i' - i) -n} } \frac{[n]_q!(q - 1)^n}{\prod_b (q^{h(b)} - 1)} = \frac{1}{q^{(j' - j) + (i' - i) -n} } r(\lambda). \end{align*} $$

Combining this with the result for $g(\lambda )$ :

$$ \begin{align*} s(\widetilde{\lambda}) = q^{g(\widetilde{\lambda})} r_{\lambda} = q^{g(\lambda) + (j' - j)} \cdot \frac{1}{q^{(j' - j) + (i' - i) - n}} r(\lambda) = s(\lambda) q^{i - i' + n}> s(\lambda) \end{align*} $$

since $i> i'$ .

Assuming the monotonicity, then if $\lambda _1 \ge n/2$ we have $s(\lambda ) \le s((\lambda _1, n - \lambda _1))$ . To calculate this quantity:

$$ \begin{align*} g(\lambda) &\le g((\lambda_1, n - \lambda_1)) = c_{(\lambda_1, n - \lambda_1)} + n_{(\lambda_1, n - \lambda_1)} \\ &= \left( \frac{\lambda(\lambda_1 - 1)}{2} + \frac{(n - \lambda_1 - 1)(n - \lambda_1 - 2)}{2} - 1 \right) + (n - \lambda_1) \\ &= \binom{\lambda_1}{2} + \binom{n - \lambda_1}{2} = \frac{n(n-1)}{2} - \lambda_1(n - \lambda_1). \end{align*} $$

For $r_{\lambda }$ , note that the hook lengths of $(\lambda _1, n - \lambda _1)$ are

If $\lambda _1 = n - j$ , we see the terms that cancel:

This uses the inequality $q^{r-1} < q^r - 1 < q^r$ . Thus, if $\lambda _1 \ge n/2$ , we have shown

$$\begin{align*}s(\lambda) = q^{g(\lambda)} r_{\lambda} \le q^{\binom{n}{2} - \lambda_1(n - \lambda_1)} \cdot q^{\lambda_1(n- \lambda_1)} = q^{\binom{n}{2}}. \end{align*}$$

Now, suppose $\lambda _1 \le n/2$ , so $s(\lambda ) \le s((n/2, n/2))$ (assume n is even). To calculate this,

$$ \begin{align*} g(\lambda) &\le g((n/2, n/2)) = c_{(n/2, n/2)} + n_{(n/2, n/2)} \\ &= \frac{(n/2 - 1)n/2}{2} + \frac{(n/2 - 2)(n/2 - 1)}{2} - 1 + \frac{n}{2} = \frac{n^2}{4} - \frac{n}{2}. \end{align*} $$

To bound $r_{\lambda }$ , use the same calculation as before to get

$$ \begin{align*} r_{(n/2, n/2)} \le q^{n^2/4}, \end{align*} $$

and so in total $s((n/2, n/2)) \le q^{n^2/2 - n/2} = q^{\binom {n}{2}}$ .

4.2 Starting from $\mathrm {id}$

Theorem 4.4 shows that restricting the random transvections walk from $GL_n(q)$ to the double coset space only speeds things up by a factor of $2$ when started from the identity. Hildebrand [Reference Hildebrand35] shows that the total variation distance on all of $GL_n(q)$ is only small after $n + c$ steps. Note this is independent of q.

Proof. (a): The inequality follows from Proposition 4.1 (b):

$$ \begin{align*} \chi^2_{id}(\ell) &= q^{-2 I(id)} \sum_{\lambda \neq (n)} t_{\lambda} \chi_H^{\lambda}(T_{id} K^{2\ell}T_{id}) \\ &= \sum_{\lambda \neq (n)} t_{\lambda} \chi_H^{\lambda}(K^{2\ell}) = \sum_{\lambda \neq (n)} t_{\lambda} f_{\lambda} \beta_{\lambda}^{2 \ell}. \end{align*} $$

(b): From Corollary 3.10 if $\lambda _1 = n - j$ , then $\beta _{\lambda } \le \kappa _{n, q, j} q^{-j}$ , where

$$ \begin{align*} \kappa_{n,q, j} &= (1 + q^{-(n - 2j + 1)})(1 + q^{-(n-1)})(1 + q^{-(n-2)}). \end{align*} $$

Using the bound on $t_{\lambda }$ from equation (4.3), for $1 \le j \le \lfloor n/2 \rfloor $ , we have

(4.5) $$ \begin{align} \sum_{\lambda: \lambda_1 = n - j} t_{\lambda} f_{\lambda} \beta_{\lambda}^{2 \ell} & \le (\kappa_{n, q, j} q^{-j} )^{2\ell} \sum_{\lambda: \lambda_1 = n -j} q^{-\binom{n-j}{2} + \binom{n}{2}} f_{\lambda}^2 \notag \\ &\le (\kappa_{n, q, j} q^{-j} )^{2\ell} q^{nj - j(j+1)/2} \cdot \frac{n^{2j}}{j!} \notag \\ &\le \frac{1}{j!} \exp \left( -2 \ell j \left(\log q - \frac{\log(\kappa_{n, q, j})}{j} \right) + \log q (nj - j(j + 1)/2) + 2j \log n \right) \notag \\ &\le \frac{1}{j!} \exp \left( -2 \ell j \left(\log q - \alpha_{n, q} \right) + \log q (nj - j(j + 1)/2) + 2j \log n \right). \end{align} $$

Recall the final inequality follows since $\alpha _{n, q} := \max _{1 \le j \le n/2} \log (\kappa _{n, q, j})/j$ . If $\ell = \frac {n \log q/2 + \log n + c}{(\log q - \alpha _{n, q})}$ , then the exponent in equation (4.5) is

$$ \begin{align*} - 2j \left( n\log q/2 + \log n + c \right) &+ \log q (nj - j(j + 1)/2) + 2j \log n \\ &= -j(2c + \log q(j+1)/2). \end{align*} $$

This gives

$$\begin{align*}\sum_{j = 1}^{\lfloor n/2 \rfloor}\sum_{\lambda: \lambda_1 = n - j} t_{\lambda} f_{\lambda} \beta_{\lambda}^{2 \ell} \le \sum_{j = 1}^{\lfloor n/2 \rfloor} \frac{e^{-2jc }}{j!} \le e^{e^{-2c}} - 1. \end{align*}$$

Next, we need to consider the partitions $\lambda $ with $\lambda _1 \le n/2$ . For these partitions,

$$\begin{align*}\beta_{\lambda} \le \beta_{(n/2, n/2)} \le \kappa_{n, q, n/2} q^{-n/2}. \end{align*}$$

Then we have

$$ \begin{align*} \sum_{j = n/2}^{n-1} \sum_{\lambda: \lambda_1 = n -j} t_{\lambda} f_{\lambda} \beta_{\lambda}^{2 \ell} &\le \beta_{(n/2, n/2)}^{2\ell} \sum_{j = n/2}^{n-1} \sum_{\lambda: \lambda_1 = n -j} t_{\lambda} f_{\lambda} \\ &\le \beta_{(n/2, n/2)}^{2\ell} \sum_{j = n/2}^{n-1} \sum_{\lambda: \lambda_1 = n -j} q^{nj - j(j+1)/2} f_{\lambda}^2 \\ &\le n! \cdot \exp \left( -2 \ell (n/2) \left( \log q + \frac{\kappa_{n, q, n/2}}{n/2} \right) + n^2/2 \right) \\ &\le \exp \left( -n (n \log q/2 + \log n + c) + n^2/2 \right) \sum_{\lambda \vdash n} f_{\lambda}^2 \\ &\le n! \exp \left( -n (n \log q/2 + \log n + c) + n^2/2 \right), \end{align*} $$

using $\sum _{\lambda } f_{\lambda }^2 = n!$ and that if $n/2 \le j \le n$ , then

$$\begin{align*}nj - j(j + 1)/2 \le n^2 - n^2/4 < n^2 - n(n + 1)/2 \le n^2/2 \end{align*}$$

since the function is increasing in j. Note also that if $q \ge 3$ , then $\log q> 1$ and $n! \le n^n = e^{\log n}$ . To finish the bound,

$$ \begin{align*} \sum_{j = n/2}^{n-1} \sum_{\lambda: \lambda_1 = n -j} t_{\lambda} f_{\lambda} \beta_{\lambda}^{2 \ell} &\le \exp\left( n \log n -n (n \log q/2 + \log n + c) + n^2/2 \right) \\ &= \exp\left( -(n^2/2) (\log q - 1) - cn \right) \le e^{-cn}. \end{align*} $$

(c): The lower bound comes from considering the $\lambda = (n - 1, 1)$ term from the sum in (a). Using the quantities (4.1), this gives

$$ \begin{align*} \chi^2_{id}(\ell) &= \sum_{\lambda \neq (n)} t_{\lambda} f_{\lambda} \beta_{\lambda}^{2\ell} \ge t_{(n-1, 1)} f_{(n-1, 1)} \beta_{(n-1, 1)}^{2 \ell} \\ &= q \cdot \frac{q^{n-1} - 1}{q -1} (n-1) \left( \frac{q^{n-2} - 1}{q^{n-1} - 1} \right)^{2 \ell} \ge (q^{n-1} - 1)(n-1) q^{-4\ell}. \end{align*} $$

This uses that $(q^{n-2} - 1)/(q^{n-1} - 1) \ge q^{-2}$ .

(d): From the alternative version of the walk on the Hecke algebra, involving $D/|\mathcal {T}_{n, q}|$ with D from 1.6, the walk proceeds by picking a transposition $(i,j), i < j$ with probability proportional to

$$\begin{align*}q^{-(j-i)} \end{align*}$$

and multiplying by $T_{ij}$ . As described in Section 3.2, multiplication by $T_{ij}$ corresponds to proposing the transposition $(i, j)$ and proceeding via the Metropolis algorithm. Thus, multiplication by $T_{ij}$ induces at most $2(j - i)$ inversions, always less than $2n$ . From [Reference Diaconis and Shahshahani22] Theorem 5.1, under $\pi _q$ , a typical permutation has $\binom {n}{2} - \frac {n - q}{q + 1} + O(\sqrt {n})$ inversions (and the fluctuations are Gaussian about this mean). If $\ell = n/8$ , the measure $P_{id}^{\ell }(\cdot )$ is concentrated on permutations with at most $n^2/4$ inversions and $\pi _q$ is concentrated on permutations with order $n^2/2 - (n - 1)/(q + 1) + O(\sqrt {n})$ inversions.

4.3 Starting from $\omega _0$

Proof of Theorem 4.5.

(a): By Proposition 4.9 in [Reference Diaconis and Ram18], if $\omega _0$ is the longest element of W, then

$$\begin{align*}\rho^{\lambda}(T_{\omega_0}^2) = q^{I(\omega_0) + c_{\lambda}}\mathrm{Id}, \end{align*}$$

where $\rho ^{\lambda }$ is the irreducible representation indexed by $\lambda $ . Using this and 4.1 (a),

$$ \begin{align*} \chi^2_{\omega_0}(\ell) &= q^{-2 I(\omega_0)}\sum_{\lambda \neq \mathbf{1}} t_{\lambda} \chi_H^{\lambda}(T_{\omega_0^{-1}} K^{2 \ell} T_{\omega_0}) \\ &= q^{-2 I(\omega_0)}\sum_{\lambda \neq \mathbf{1}} t_{\lambda} \chi_H^{\lambda}( K^{2 \ell} T_{\omega_0} T_{\omega_0^{-1}}) \\ &= q^{-2 I(\omega_0)}\sum_{\lambda \neq \mathbf{1}} t_{\lambda} \chi_H^{\lambda}( K^{2 \ell} T_{\omega_0}^2) \\ &= q^{-2 I(\omega_0)}\sum_{\lambda \neq \mathbf{1}} t_{\lambda} \chi_H^{\lambda}( K^{2 \ell} q^{c_{\lambda} + I(\omega_0)}) \\ &= q^{-I(\omega_0)} \sum_{\lambda \neq \mathbf{1}} q^{c_{\lambda}} t_{\lambda} \chi_H^{\lambda}( K^{2 \ell} ) = q^{-I(\omega_0)} \sum_{\lambda \neq \mathbf{1}} q^{c_{\lambda}} t_{\lambda} f_{\lambda} \beta_{\lambda}^{2 \ell} \end{align*} $$

since $K \in Z(H)$ , that is, K commutes with all elements of the Hecke algebra.

(b): Suppose $\lambda _1 = n -j$ for $1 \le j \le n/2$ . Recall the definition $s(\lambda ) = q^{c_{\lambda }}t_{\lambda }$ . From Proposition 4.3, $s(\lambda ) \le q^{\binom {n}{2}}$ . Then,

$$ \begin{align*} q^{-\binom{n}{2}} \sum_{\lambda: \lambda_1 = n - j} q^{c_{\lambda}} t_{\lambda} f_{\lambda} \beta_{\lambda}^{2 \ell} & = q^{-\binom{n}{2}} \sum_{\lambda: \lambda_1 = n - j} s(\lambda) f_{\lambda} \beta_{\lambda}^{2 \ell} \\ & \le (\kappa_{n, q, j} q^{-j} )^{2\ell} q^{-\binom{n}{2}} \sum_{\lambda: \lambda_1 = n -j} q^{ \binom{n }{2} } f_{\lambda} \\ & \le (\kappa_{n, q, j} q^{-j} )^{2\ell} \sum_{\lambda: \lambda_1 = n -j} f_{\lambda} \\ &\le (\kappa_{n, q, j} q^{-j} )^{2\ell} \cdot \frac{n^{j}}{\sqrt{j!}} \cdot \frac{K}{j}e^{2 \sqrt{2j}} \\ &\le \frac{K}{\sqrt{j!}} \exp \left( -2 \ell j \left(\log q - \frac{\log(\kappa_{n, q, j})}{j} \right) + (j-1)\log n + 2 \sqrt{2j} \right). \end{align*} $$

The third inequality uses Proposition 4.2 for $\sum _{\lambda : \lambda _1 = n - j} f_{\lambda }$ . Recall $\alpha _{n, q} := \max _{1 \le j \le n/2} \log (\kappa _{n, q, j})/j$ . If $\ell = (\log n/2 + c)/(\log q - \alpha _{n, q})$ , then the bound becomes

$$ \begin{align*} \sum_{j = 1}^{n/2} q^{-\binom{n}{2}} \sum_{\lambda: \lambda_1 = n - j} q^{c_{\lambda}} t_{\lambda} f_{\lambda} \beta_{\lambda}^{2 \ell} &\le \sum_{j = 1}^{n/2}\frac{K}{\sqrt{j!}} \exp \left( - j \log n - 2jc + (j-1)\log n + 2 \sqrt{2j} \right) \\ &\le 2 K \sum_{j = 1}^{n/2} \frac{e^{-2jc + 2 \sqrt{2j}}}{j}, \end{align*} $$

using the loose bound $\sqrt {j!}> j/2$ for all $j \ge 1$ . With the assumption that $c \ge 2 \sqrt {2}$ , we have $-2jc + 2 \sqrt {2j} \le -jc$ for all $j \ge 1$ . Finally,

$$ \begin{align*} 2 K \sum_{j = 1}^{n/2} \frac{e^{-jc}}{j} & \le 2 K \sum_{j = 1}^{\infty} \frac{e^{-jc}}{j} \\ &= - 2K \log(1 - e^{-c}). \end{align*} $$

Now, for the $\lambda $ with $\lambda _1 \ge n/2$ , we have

$$ \begin{align*} \sum_{j = n/2}^{n-1} \sum_{\lambda: \lambda_1 = n -j} q^{-\binom{n}{2}} q^{c_{\lambda}} t_{\lambda} f_{\lambda} \beta_{\lambda}^{2 \ell} &\le \beta_{(n/2, n/2)}^{2\ell} \sum_{j = n/2}^{n-1} \sum_{\lambda: \lambda_1 = n -j} f_{\lambda} \\ &\le \beta_{(n/2, n/2)}^{2\ell} \sum_{\lambda \vdash n} f_{\lambda} \\ &\le \beta_{(n/2, n/2)}^{2\ell} \left(\sum_{\lambda \vdash n} f_{\lambda}^2\right)^{1/2} \cdot p(n)^{1/2}, \end{align*} $$

where $p(n)$ is equal to the number of partitions of n (the inequality is Cauchy–Schwarz). Since $p(n) \le \frac {K}{n} e^{2 \sqrt {2n}}$ for a constant $K> 0$ ([Reference Hardy and Ramanujan32]) and $\sum _{\lambda \vdash n} f_{\lambda }^2 = n!$ , this gives

$$ \begin{align*} &\le \sqrt{n!} \cdot \sqrt{\frac{K}{n}} \exp \left( \sqrt{2 n} -2 \ell (n/2) \left( \log q - \frac{\kappa_{n, q, n/2}}{n/2} \right) \right) \\ &\le \sqrt{n!} \cdot \sqrt{\frac{K}{n}} \exp \left(\sqrt{2 n} -n \log n/2 -2n c \right) \\ &\le \sqrt{K} \exp \left( -nc - n(c - \sqrt{2}n^{-1/2}) \right) \end{align*} $$

since $\sqrt {n!} \le e^{n \log n/2}$ . Since $c> 2 \sqrt {2}$ , then the bound is $\le \sqrt {K} e^{-cn}$ for any $n \ge 1$ .

(c): A lower bound comes from using equation (4.1) for the lead term on the right-hand side of (a):

$$ \begin{align*} \chi_{\omega_0}^2(\ell) &\ge q^{-\binom{n}{2}} q^{c_{(n-1, 1)}} t_{(n-1, 1)} f_{(n-1, 1)} \beta_{(n-1, 1)}^{2 \ell} \\ &= q^{-\binom{n}{2}} q^{\binom{n-1}{2} + 1} \cdot q \frac{q^{n-1} - 1}{q - 1} \cdot (n-1) \left( \frac{q^{n-2} - 1}{q^{n-1} - 1} \right)^{2 \ell} \\ &\ge q^{-(n-2)}(q^{n-1} -1)(n-1) q^{-4 \ell}. \end{align*} $$

4.4 Starting from a typical site

In analyzing algorithms used repeatedly for simulations, as the algorithm is used, it approaches stationarity. This means the quantity

$$\begin{align*}\sum_{x \in S_n} \pi_q(x) \|P_x^{\ell} - \pi_q \|_{TV} \end{align*}$$

is of interest. For the problem under study, $\pi _q$ is concentrated near $\omega _0$ so we expect rates similar to those in Theorem 4.5.

Proof. (a): This is simply a restatement of 4.1 part (b).

(b): We will divide the sum depending on the first entry of the partition. By Proposition 4.2, we have the bound (true for any $1 \le j \le n$ )

$$\begin{align*}\sum_{\lambda: \lambda_1 = n - j} f_{\lambda}^2 \le \frac{n^{2j}}{j!}. \end{align*}$$

Combining this with Corollary 3.10, for $j \le \lfloor n/2 \rfloor $ ,

$$ \begin{align*} \sum_{\lambda: \lambda_1 = n - j} \beta_{\lambda}^{2 \ell} f_{\lambda}^2 &\le \sum_{\lambda: \lambda_1 = n - j} \beta_{(n - j, j)}^{2 \ell} f_{\lambda}^2 \le \beta_{(n - j, j)}^{2 \ell} \cdot \frac{n^{2j}}{j!} \\ &\le \kappa_{n, q, j}^{2\ell} q^{-2 \ell j} \cdot \frac{n^{2j}}{j!} = \frac{1}{j!} \exp \left( 2 \ell( \log(\kappa_{n, q, j}) - j\log q) + 2 j \log n \right) \\ &= \frac{1}{j!} \exp \left( 2 \ell j \left( \frac{ \log(\kappa_{n, q, j})}{j} - \log q\right) + 2 j \log n \right). \end{align*} $$

Define

$$\begin{align*}\alpha_{n, q} := \max_{1 \le j \le n/2} \frac{\log(\kappa_{n, q, j})}{j}, \end{align*}$$

so then if $\ell = (\log n + c)/(\log q - \alpha _{n, q})$ , the bound is

$$ \begin{align*} &\le \frac{1}{j!} \exp \left( 2 \ell j( \alpha_{n, q} - \log q) + 2 j \log n \right) \\ & = \frac{1}{j!} \exp \left( -2 j \log n - 2j c + 2j \log n \right) = \frac{e^{-2jc}}{j!}. \end{align*} $$

Then summing over all possible j gives

$$\begin{align*}\sum_{j = 1}^{\lfloor n/2 \rfloor} \sum_{\lambda: \lambda_1 = n - j} f_{\lambda}^2 \beta_{\lambda}^{2 \ell} \le \sum_{j = 1}^{\lfloor n/2 \rfloor} \frac{e^{-2jc}}{j!} \le \left( e^{e^{-c}} - 1 \right). \end{align*}$$

Now, we have to bound the contribution from partitions $\lambda $ with $\lambda _1 \le n/2$ . Because $\beta _{\lambda }$ is monotone with respect to the order on partitions, we have for all $\lambda $ such that $\lambda _1 \le n/2$ ,

$$\begin{align*}\beta_{\lambda} \le \beta_{(n/2, n/2)} \le \kappa_{n, q, n/2} q^{-n/2} \end{align*}$$

since $\lambda \preceq (n/2, n/2)$ (assuming without essential loss that n is even). Then,

$$ \begin{align*} \sum_{j = n/2}^{n-1} \sum_{\lambda: \lambda_1 = n -j}f_{\lambda}^2 \beta_{\lambda}^{2 \ell} &\le \beta_{(n/2, n/2)}^{2\ell} \sum_{\lambda} f_{\lambda}^2 \\ &\le n! \cdot \exp \left( 2 \ell (n/2)( \log q + \frac{\kappa_{n, q, n/2}}{n/2} ) \right) \\ &\le n! \cdot \exp \left( -n (\log n + c) \right) \\ &\le e^{-cn}, \end{align*} $$

using $\sum _{\lambda } f_{\lambda }^2 = n! \le n^n$ .

(c): The sum is bounded below by the term for $\lambda = (n - 1, 1)$ . This is

$$\begin{align*}f_{(n-1, 1)}^2 \cdot \beta_{(n-1, 1)}^{2 \ell} = (n-1)^2 \left( \frac{q^{n-2} - 1}{q^{n-1} - 1} \right)^{2 \ell} \ge (n-1)^2 q^{-4 \ell}. \end{align*}$$

5.1 Overview

Let $\mathbb {C}[G]$ denote the group algebra for $G = GL_n(q)$ . This is the space of functions $f: G \to \mathbb {C}$ , with addition defined $(f + g)(s) = f(s) + g(s)$ and multiplication defined by

$$\begin{align*}f_1 \ast f_2(s) = \sum_{t\in G} f_1(t) f_2(st^{-1}). \end{align*}$$

Equivalently, $\mathbb {C}[G] = \mathrm {span} \{g \mid g \in G \}$ and we can write an element $f = \sum _{g} c_g g$ for $c_g \in \mathbb {C}$ , so $f(g) = c_g$ .

Define elements in $\mathbb {C}[G]$ :

Note that if $b \in \mathcal {B}$ , then

If $g \in \mathcal {B} \omega \mathcal {B}$ , so $g = b_1 \omega b_2$ with $b_1, b_2 \in \mathcal {B}$ , then

The Hecke algebra is and has basis $\{T_{\omega } \mid \omega \in S_n \}$ . Note that $H_n(q)$ are all functions in $\mathbb {C}[G]$ which are $\mathcal {B}$ - $\mathcal {B}$ invariant, that is, $f(b_1 g b_2) = f(g)$ .

Now, let P be the transition matrix for G defined by multiplying by a random transvection. We can also write this

$$\begin{align*}P = \frac{1}{|\mathcal{T}_{n, q}|} \sum_{T \in \mathcal{T}_{n, q}} M_T, \end{align*}$$

where $M_T$ is the transition matrix ‘multiply by T’. In other words, $M_T(x, y) = \textbf {1}(y = Tx)$ .

Then P defines a linear transform on the space $\mathbb {C}[G]$ , with respect to the basis $\{g \mid g \in G \}$ . The matrix $M_T$ is just the function: Multiply by T in the group algebra. This means P is equivalent to multiplication by $\frac {1}{|\mathcal {T}_{n, q}|} D$ , with $D = \sum _{T \in \mathcal {T}_{n, q}} T$ as an element in $\mathbb {C}[G]$ . The Markov chain lumped to $S_n = \mathcal {B} \backslash GL_n(q) / \mathcal {B}$ is then equivalent to multiplication by D on $H_n(q)$ . Since D is the sum of all elements in a conjugacy class, it is in the center $Z(\mathbb {C}[G])$ . This means if $g \in \mathcal {B} \omega \mathcal {B}$ , then

In conclusion, to determine how D acts in $H_n(q)$ , we can compute

. The remainder of the section proves the following.

Theorem 5.1. Let $D = \sum _{T \in \mathcal {T}_{n, q}} T \in \mathbb {C}[G]$ . Then,

where $s_{ij}$ is the transposition switching i and j.

5.2 Row reduction

Let $G = GL_n(q)$ and $\mathcal {B}$ be the upper-triangular matrices. For $1 \le i \le n -1$ and $c \in \mathbb {F}_q$ , define $y_i(c) \in G$ by

That is, multiplication on the left by $y_i(c)$ acts by adding c times the ith row to the $(i+1)$ th row, then permuting the i and $i + 1$ rows.

Let $\omega \in S_n$ . We can write the reduced word

, for $s_i = (i, i+1)$ the simple reflections. Then,

and $|\mathcal {B} \omega \mathcal {B}|/|\mathcal {B}| = q^{I(\omega )}$ and $G = \bigsqcup _{\omega \in S_n} \mathcal {B} \omega \mathcal {B}$ . This provides a very useful way for determining the double coset that a matrix M belongs to, which just amounts to performing row reduction by multiplying by different matrices $y_i(c)$ .

5.3 Transvections

For every transvection $T \in \mathcal {T}_{n, q}$ , we can perform row operations, multiplying by $y_i(c)$ , to determine which double coset T belongs to.

A transvection is defined by two vectors $\mathbf {a}, \mathbf {v} \in \mathbb {F}_q^n$ with $\mathbf {v}^{\top } \mathbf {a} = 0$ , with the last nonzero entry of $\mathbf {v}$ normalized to be $1$ . Assume this entry is at position j. Assume the first nonzero entry of $\mathbf {a}$ is at position i. That is, the vectors look like:

$$ \begin{align*} &\mathbf{a} = (0, \dots, 0, a_i, a_{i+1}, \dots, a_n)^{\top}, \quad a_i \neq 0, \\ &\mathbf{v} = (v_1, \dots, v_{j-1}, 1, 0, \dots, 0)^{\top}. \end{align*} $$

Let $T_{\mathbf {a}, \mathbf {v}} = I + \mathbf {v} \mathbf {a}^{\top }$ . We consider the possible cases for i and j to prove the following result.

Proposition 5.2. Let $T_{\mathbf {a}, \mathbf {v}} = I + \mathbf {v} \mathbf {a}^{\top }$ be the transvection defined by nonzero vectors $\mathbf {a}, \mathbf {v}$ with $\mathbf {v}^{\top } \mathbf {a} = 0$ and the last nonzero entry of $\mathbf {v}$ equal to $1$ . Let j be the index of the last nonzero entry of $\mathbf {v}$ and i the index of the first nonzero entry of $\mathbf {a}$ . If $i> j$ , then $T_{\mathbf {a}, \mathbf {v}} \in \mathcal {B}$ . If $i < j$ , then

$$\begin{align*}T_{\mathbf{a}, \mathbf{v}} \in \mathcal{B} s_{j-1}\cdots s_{i+1}s_is_{i+1}\cdots s_{j-1} \mathcal{B}. \end{align*}$$

The case $i = j$ does not occur.

Proof. Case 1 $i> j$ . If $i> j$ , then $T_{\mathbf {a}, \mathbf {v}} \in \mathcal {B}$ . To see this, suppose $k> l$ . Then

$$\begin{align*}T_{\mathbf{a}, \mathbf{v}}(k, l) = (\mathbf{v} \mathbf{a}^{\top})_{k, l} = v_k a_l = 0 \end{align*}$$

because $v_k a_l$ can only be nonzero if $k \le j < i \le l$ . Now, we can count how many transvections satisfy this. There are $(q - 1)q^{n - i}$ choices for $\mathbf {a}$ (because $a_i$ must be nonzero) and $q^{j-1}$ choices for $\mathbf {v}$ (because $v_j$ is fixed at $1$ ). In total for this case, there are

Case 2 $i = j$ . In this case, $\mathbf {v}^{\top } \mathbf {a} = a_i \neq 0$ . This does not satisfy the condition $\mathbf {v}^{\top } \mathbf {a} = 0$ , so this case cannot occur.

Case 3 $i < j$ . The transvection is of the form

where

$$\begin{align*}N_{\mathbf{a}, \mathbf{v}} = \begin{pmatrix} 1+a_iv_i &a_{i+1}v_i &a_{i+2} v_i &\cdots &a_jv_i \\ a_iv_{i+1} &1+a_{i+1}v_{i+1} & a_{i+2} v_{i+1} &\cdots &a_j v_{i+1}\\ \vdots &&&&\vdots \\ a_i &a_{i+1} &&\cdots &1+ a_j \end{pmatrix}. \end{align*}$$

Then,

where A is the upper-triangular matrix

See Appendix A for details of this row reduction calculation.

To count how many transvections fit Case 3: There are

• • $q^{n-i}$ choices for $a_{i+1}, \dots , a_n$ .

• • $q - 1$ choices of $a_i$ .

• • $q^{i-1}$ choices of $v_1, \dots , v_{i-1}$ .

• • $q^{j - 1 - i}$ choices of $v_i, \dots , v_{j-1}$ to satisfy .

The total is

$$ \begin{align*} q^{n - i}(q - 1)q^{i - 1}q^{j - 1 - i} &= (q - 1)q^{n - i + j - 2} = (q - 1)q^{n - 1 - (j - i)}q^{2(j - 1 - i) + 1} \\ &= (q - 1)q^{n - 1 - (j - i)}q^{I(s_{ij})}. \end{align*} $$

Proposition 5.2 now enables the proof of Theorem 5.1.

Proof of Theorem 5.1.

Let $\mathbb {C}[G]$ be the group algebra of $G = GL_n(q)$ , and $Z(\mathbb {C}[G])$ the center. Then since transvections are a conjugacy class, the sum of transvections is in the center,

$$\begin{align*}D = \sum_{\mathbf{a}, \mathbf{v}} T_{\mathbf{a}, \mathbf{v}} \in Z(\mathbb{C}[G]). \end{align*}$$

Since D commutes with every element of $\mathbb {C}[G]$ , we can compute

6.1 Flag representation

The subgroup $\mathcal {B}$ gives rise to the quotient $GL_n(q)/\mathcal {B}$ . This may be pictured as the space of ‘flags’.

Definition 6.2. Here, a flag F consists of an increasing sequence of subspaces

with $\text {dim}(F_i) = i$ . The standard flag is

Indeed, $GL_n(q)$ operates transitively on flags, and the subgroup fixing the standard flag is exactly $\mathcal {B}$ . There is a useful notion of ‘distance’ between two flags $F, F'$ which defines a permutation.

Definition 6.3. Let $F, F'$ be two flags. The Jordan–Holder permutation $\omega (F, F')$ is a permutation $\omega = \omega (F, F') \in S_n$ defined by $\omega (i) = j$ if j is the smallest index such that

$$\begin{align*}F_{i - 1} + F_j' \supseteq F_i. \end{align*}$$

The Jordan–Holder permutation distance satisfies

(6.1) $$ \begin{align} \omega(F, F') = \omega(F', F)^{-1} \quad \text{and} \,\,\, \omega(F, F') = \omega(MF, MF') \,\,\,\,\, \text{for} \,\,\, M \in GL_n(q). \end{align} $$

A thorough development with full proofs is in [Reference Abels1].

Lemma 6.4. For $M \in GL_n(q)$ and E the standard flag,

$$\begin{align*}M \in \mathcal{B} \omega \mathcal{B} \iff \omega = \omega(E, ME). \end{align*}$$

6.2 Top label chain

This representation is useful for analyzing a further projection of the chain on $S_n$ : Let $\{\omega _t \}_{t \ge 0}$ denote the Markov chain on $S_n$ . Let $P_1(\cdot , \cdot )$ denote the marginal transition probabilities of the first position. That is,

$$\begin{align*}P_1(j, k) = \operatorname{\mathbb{P}}(\omega_{t+1}(1) = k \mid \omega_t(1) = j ), \quad j, k \in \{1, \dots, n \}. \end{align*}$$

If $\omega $ is distributed as $\pi _q$ , then the marginal distribution on $\{1, \dots , n \}$ of the first card $\omega (1)$ is

Lemma 6.5. For $1 \le i, j \le n$ ,

$$ \begin{align*} P_1(i, j) &= \begin{cases} \dfrac{(q-1)^2q^{n + j - 3}}{(q^n - 1)(q^{n-1} - 1)}, & i \neq j, \\ \dfrac{(q^{n-1} - 1)^2 + (q - 1)^2(q^{i - 1} - 1)q^{n - 2}}{(q^n - 1)(q^{n-1} - 1)}, & i = j. \end{cases} \end{align*} $$

Remark 6.6. Observe that for $i \neq j$ the transition probabilities can be written

$$ \begin{align*} P_1(i, j) &= \frac{q^{j-1}}{(q^n - 1)/(q-1)} \left( \frac{(q - 1) q^{n-2}}{q^{n-1} - 1} \right) = \pi_{q, 1}(j) \cdot \left( \frac{(q - 1) q^{n-2}}{q^{n-1} - 1} \right) \\ P_1(j, j) &= \pi_{q, 1}(j) \cdot \left( \frac{(q - 1) q^{n-2}}{q^{n-1} - 1} \right) + \left(1 - \frac{(q - 1) q^{n-2}}{q^{n-1} - 1} \right). \end{align*} $$

Write $p:= ((q - 1) q^{n-2})/(q^{n-1} - 1)$ . This provides another description of the Markov chain: At each step, flip a coin which gives heads with probability p, tails with probability $1-p$ . If heads, move to a random sample from $\pi _{q, 1}$ . Otherwise, don’t move.

Proof. If $\omega (1) = i$ , then a flag representing the double coset has $F_1 = \mathbf {e}_i$ . Recall that a transvection T defined by vectors $\mathbf {v}, \mathbf {a}$ has $T(\mathbf {e}_i) = \mathbf {e}_i + a_i \mathbf {v}$ . The first coordinate in the new permutation is the smallest j such that

$$\begin{align*}\mathbf{e}_i + a_i \mathbf{v} \subset \left\langle e_1, \dots, e_j \right\rangle. \end{align*}$$

There are two cases when this smallest j is equal to i:

1. 1. $a_i = 0$ , and $\mathbf {v}$ can be anything: There are $(q^{n - 1} - 1)$ possibilities for $\mathbf {a}$ such that $a_i = 0$ . Then there are $(q^{n-1} - 1)/(q - 1)$ possibilities for $\mathbf {v}$ .

2. 2. $a_i \neq 0$ and $\mathbf {v}$ is such that $a_i v_i \neq - 1$ and $v_k = 0$ for all $k> i$ . Note that if $i = 1$ , then this is not possible. Since the nonzero entry at the largest index in $\mathbf {v}$ is normalized to be $1$ , there are two further possibilities:

   • • If $v_i = 1$ , then there are $(q - 2)$ possibilities for $a_i$ , $(q^{i - 1} -1)$ possibilities for the rest of $\mathbf {v}$ (note that it’s not allowed for the rest of $\mathbf {v}$ to be $0$ because then we could not get $\mathbf {a}^{\top } \mathbf {v} = 0$ ), and then $q^{n-2}$ possibilities for the rest of $\mathbf {a}$ .

   • • If $v_i = 0$ , then there are $(q^{i - 1} - 1)/(q - 1)$ possibilities for $\mathbf {v}$ and then $(q - 1)q^{n-2}$ possibilities for $\mathbf {a}$ .

In summary, if $i \neq 1$ , then

$$ \begin{align*} P_1(i, i) &= \frac{1}{|\mathcal{T}_{n, q}|} \left( \frac{(q^{n-1} - 1)(q^{n - 1} -1)}{q - 1} + (q^{i - 1} -1)(q - 2)q^{n-2} + \frac{(q^{i - 1} -1)(q - 1)q^{n-2}}{q - 1} \right) \\ &= \frac{(q^{n-1} - 1)^2 + (q - 1)^2(q^{i - 1} - 1)q^{n - 2}}{(q^n - 1)(q^{n-1} - 1)}. \end{align*} $$

Now, if $i \neq j$ we consider the two possibilities.

1. 1. $i> j$ : This transition will occur if $a_i\cdot v_i = -1$ and $\mathbf {v}$ is such that $v_j> 0$ and $v_k = 0$ for all $j < k < i$ . Since the $\mathbf {v}$ is normalized so that the entry at the largest index is $1$ , this requires $v_i = 1$ and $a_i = -1$ . There are then $(q - 1)q^{j - 1}$ such $\mathbf {v}$ , and for each $\mathbf {v}$ there are $q^{n-2}$ possibilities for $\mathbf {a}$ . This gives in total

   $$\begin{align*}(q - 1) q^{n + j - 3}. \end{align*}$$

2. 2. $i < j$ : This transition will occur if $a_i \neq 0$ , $v_j = 1$ and $v_k = 0$ for $k> j$ . There are $q^{j - 1}$ possibilities for $\mathbf {v}$ and then $(q - 1)q^{n-2}$ possibilities for $\mathbf {a}$ , so again in total

   $$\begin{align*}(q - 1)q^{n+ j - 3}. \end{align*}$$

In summary, for any $i \neq j$ ,

$$ \begin{align*} P_1(i, j) = \frac{(q-1)q^{n + j - 3}}{\vert \mathcal{T}_{n, q} \vert} = \frac{(q-1)^2q^{n + j - 3}}{(q^n - 1)(q^{n-1} - 1)}. \end{align*} $$

Lemma 6.8. Let $P_1$ be the Markov chain from Lemma 6.5 with stationary distribution $\pi _{q, 1}$ . Then $P_1$ has eigenvalues $\{1, \beta \}$ with

$$\begin{align*}\beta = \frac{q^{n -2} -1}{q^{n-1} - 1}, \end{align*}$$

which is equal to $\beta _{(n-1, 1)}$ from Section 3.3. The eigenvalue $\beta $ has multiplicity $n -1$ . If $q> 1$ is fixed and n large, then $\beta \sim 1/q$ .

Proof. Call $P_1(i, j) = c_j$ , and remember this is constant across all $i \neq j$ . Then $P_1(i, i) = p_i = 1 - \sum _{j \neq i} c_j$ . Let M be the matrix with column j constant $c_j$ , so the rows are constant $[c_1, \dots , c_n]$ and let $r = \sum c_j$ be the constant row sum. Note we can write our transition matrix as

$$\begin{align*}T = (1 - r)I + M. \end{align*}$$

Note that the diagonal entries of this matrix match what we want since $T(i, i) = (1 - r) + c_i = 1 - \sum _{j \neq i} c_j = p_i$ .

Since M has constant columns/rows it has null space of dimension $n-1$ . That is, we can find $n-1$ vectors $v_2, v_3, \dots v_n$ such that $vM = 0$ . Then

$$\begin{align*}vT = v((1 - r)I + M) = (1-r) v \end{align*}$$

(and remember of course the other eigenvector is the stationary distribution, with eigenvalue $1$ ). This says that there is only one eigenvalue $\beta = 1-r$ , with multiplicity $n -1$ —eigenfunctions are basis of null space of M. The eigenvalue $\beta $ is

$$ \begin{align*} \beta = 1 - \sum_j \frac{(q - 1)^2q^{n + j -3}}{(q^n - 1)(q^{n-1} - 1)} = 1 - \frac{q^{n-2}(q - 1)}{q^{n-1} - 1} = \frac{q^{n-2} - 1}{q^{n-1} - 1}. \end{align*} $$

Note that this is the largest eigenvalue of the full chain on $S_n$ .

6.3 Mixing time

The mixing time of the Markov chain $P_1$ is very fast; the chain reaches stationarity within a constant number of steps (the constant depends on q but not on n). This can be proven using a strong stationary time.

Strong stationary time

An strong stationary time for a Markov chain is a random stopping time $\tau $ at which the chain is distributed according to the stationary distribution. That is, if $(X_t)_{t \ge 0}$ is a Markov chain on $\Omega $ with stationary distribution $\pi $ , then $\tau $ satisfies

(6.2) $$ \begin{align} \operatorname{\mathbb{P}} \left( \tau = t, X_{\tau} = y \mid X_0 = x\right) = \operatorname{\mathbb{P}} \left( \tau = t \mid X_0 = x \right) \pi(y), \quad x,y \in \Omega. \end{align} $$

In words, $X_{\tau }$ has distribution $\pi $ and is independent from $\tau $ . See Section 6.4 of [Reference Levin and Peres43].

A strong stationary time is very powerful for bounding convergence time. Intuitively, once the strong stationary time is reached the chain has mixed, so the mixing time is bounded by the random time $\tau $ . This idea is formalized in the following.

Proposition 6.10 (Proposition 6.10 from [Reference Levin and Peres43]).

If $\tau $ is a strong stationary time for a Markov chain P on state space $\Omega $ , then for any time t,

$$\begin{align*}\max_{x \in \Omega} \|P^t(x, \cdot) - \pi \|_{TV} \le \max_{x \in \Omega} \operatorname{\mathbb{P}} \left( \tau> t \mid X_0 = x \right). \end{align*}$$

For most Markov chains it is very difficult to find an obvious strong stationary time. A simple one exists for the chain $P_1$ , using the alternate description from Remark 6.6.

To restate the alternate description, let $p := ((q - 1) q^{n-2})/(q^{n-1} - 1)$ . Let $\{R_t \}_{t \ge 1}$ be a sequence Bernoulli $(p)$ random variables. The Markov chain $\{X_t \}_{t \ge 0}$ defined by $P_1$ can be coupled with the random variables $R_t$ by: Given $X_t$ ,

1. 1. If $R_{t+1} = 1$ , sample $Z \sim \pi _{q, 1}$ and set $X_{t+1} = Z$ .

2. 2. If $R_{t+1} = 0$ , set $X_{t+1} = X_t$ .

Lemma 6.11. With $\{R_t \}_{t \ge 1}$ defined above, the random time $ \tau = \inf \{ t> 0 : R_t = 1 \} $ is a strong stationary time for $\{X_t \}_{t \ge 0}$ . If

$$\begin{align*}t = c \cdot \frac{q^{n-1} - 1}{(q-1)q^{n-2}}, \quad \text{with} \,\,\, c> 0, \end{align*}$$

then

$$\begin{align*}\operatorname{\mathbb{P}}(\tau> t) < e^{-c}. \end{align*}$$

Proof. By the alternate description of the Markov chain $P_1$ , whenever $R_t = 1$ , the next state $X_t$ is a sample from $\pi _{q, 1}$ . That is,

$$\begin{align*}X_t \mid \{R_t = 1 \} \quad \text{is distributed as} \,\,\, \pi_{q, 1}. \end{align*}$$

Then, by equation (6.2),

$$\begin{align*}\operatorname{\mathbb{P}}(\tau = t, X_{\tau} = y \mid X_0 = x) = \operatorname{\mathbb{P}}(\tau =t \mid X_0 = x) \pi_{q, 1}(y), \end{align*}$$

and $\tau $ is a strong stationary time. The time $\tau $ is a geometric random variable with parameter p. Note also that $\tau $ is independent of the starting state $X_0$ . Then,

$$ \begin{align*} \operatorname{\mathbb{P}}( \tau> t) &= (1-p)^{t} \le e^{-pt} = \exp \left( -t \frac{(q - 1) q^{n-2}}{(q^{n-1} - 1)} \right) \le e^{-c}, \end{align*} $$

for $ t = c \cdot \frac {q^{n-1} - 1}{(q-1)q^{n-2}}$ .

7.1 Parabolic subgroups of $GL_n$

In [Reference Karp and Thomas39], the authors enumerate the double cosets of $GL_n(q)$ generated by parabolic subgroups.

Section 4 of [Reference Karp and Thomas39] shows that if $\alpha , \beta $ are two partitions of n, the double cosets $P_{\alpha } \backslash GL_n(q) / P_{\beta }$ are indexed by contingency tables with row sums $\alpha $ and column sums $\beta $ . Proposition 4.37 of [Reference Karp and Thomas39] contains a formula for the size of a double coset which corresponds to the table X, with $\theta = 1/q$ ,

(7.1) $$ \begin{align} \theta^{- n^2 + \sum_{1 \le i < i' \le I, 1 \le j < j' \le J} X_{ij} X_{i'j'}}(1 - \theta)^n \cdot \frac{\prod_{i = 1}^I[\alpha_i]_{\theta}! \prod_{j = 1}^J [\beta_j]_{\theta}!}{\prod_{i, j} [X_{ij}]_{\theta}!}, \quad I = |\alpha|, J = |\beta|. \end{align} $$

Dividing equation (7.1) by $(1-\theta )^n$ and sending $\theta \to 1$ recovers the usual Fisher–Yates distribution for partitions $\alpha , \beta $ .

For $\alpha = (n-1, 1), \beta = (1^n)$ , the contingency tables $P_{\alpha } \backslash GL_n(q) / P_{\beta }$ are uniquely determined by the position of the single $1$ in the second row. That is, the space is in bijection with the set $\{1, 2, \dots , n \}$ . If $X^j$ denotes a table with the $1$ in the second row in column j, that is, $X_{2j}^j = 1, X_{2k}^j = 0$ for all $k \neq j$ , then equation (7.1) becomes

$$ \begin{align*} q^{-n + n^2 - (j-1)}(q -1)^n \cdot [n-1]_{1/q}! = q^{\binom{n}{2} - (j-1)} \frac{q^n - q^{n-1}}{q^n - 1} \cdot \prod_{k = 1}^n (1 - q^k), \end{align*} $$

which uses

Dividing by $|GL_n(q)| = (q - 1)^n q^{\binom {n}{2}} \cdot [n]_q$ gives

$$ \begin{align*} q^{-(j-1)} \cdot \frac{(q^n - q^{n-1})(q - 1)}{(q^n - 1)^2(q - 1)^n} \prod_{k = 1}^{n} (q^k - 1). \end{align*} $$

Note that this distribution is equal to $\pi _{q, 1}$ from Section 6 if we map $j \to n - j$ , that is, index the double cosets by $n, n-1, \dots , 1$ instead of $1, 2, \dots , n$ . The ‘follow the top card’ chain in Section 6 is then equivalent to the induced chain on double cosets $P_{\alpha } \backslash GL_n(q) / P_{\beta }$ from random transvections on $GL_n(q)$ .

It remains an open problem to further investigate probability distributions and Markov chains on the double cosets of $GL_n(q)$ from parabolic subgroups. In a reasonable sense, for finite groups of Lie type, parabolic subgroups or close cousins are the only systematic families which can occur; see [Reference Seitz57].

7.2 Gelfand pairs

A group G with subgroup H is a Gelfand pair if the convolution of H-bi-invariant functions is commutative. Probability theory for Gelfand pairs was initiated by Letac [Reference Letac42] and Sawyer [Reference Sawyer55] who studied the induced chain on

$$\begin{align*}GL_n(\mathbb{Z}_p) \backslash GL_n(\mathbb{Q}_p) / GL_n(\mathbb{Z}_p) \end{align*}$$

as simple random walk on a p-ary tree. Many further examples of finite Gelfand pairs appear in [Reference Diaconis and Shahshahani23], [Reference Diaconis15], [Reference Ceccherini Silberstein, Scarabotti and Tolli12]. These allow analysis of classical problems such as the Bernoulli–Laplace model of diffusion and natural walks on phylogenetic trees. The literature cited above contains a large number of concrete examples waiting to be interpreted. For surprising examples relating Gelfand pairs, conjugacy class walks on a ‘dual group’ and ‘folding’, see [Reference Lawther41].

We also note that Gelfand pairs occur more generally for compact and noncompact groups. For example, $\mathcal {O}_n/\mathcal {O}_{n-1}$ is Gelfand and the spherical functions become the spherical harmonics of classical physics (this example is further discussed in Section 7.5 below). Gelfand pairs are even useful for large groups such as $S_{\infty }$ and $U_{\infty }$ , which are not locally compact; see [Reference Borodin and Olshanski8] and [Reference Neretin47] for research in this direction.

7.3 Contingency tables

Simper [Reference Simper59] considers the symmetric group $S_n$ with parabolic subgroups $S_{\lambda }$ and $S_{\mu }$ , for $\lambda = (\lambda _1, \dots , \lambda _I), \mu = (\mu _1, \dots , \mu _J)$ partitions of n. Then $S_{\lambda } \backslash S_n / S_{\mu }$ can be interpreted as $I \times J$ contingency tables $\{T_{ij} \}_{1 \le i \le I, 1 \le j \le J}$ with row sums $\lambda $ and column sums $\mu $ . Such tables appear in every kind of applied statistical work. See [Reference Diaconis and Simper24] Section 5 for references. The stationary distribution,

$$\begin{align*}\pi(T) = \frac{1}{n!} \prod_{i, j} \frac{\lambda_i! \mu_j!}{T_{ij}!}, \end{align*}$$

is the familiar Fisher–Yates distribution underlying ‘Fisher’s exact test for independence’. Markov chains with the Fisher–Yates distribution as stationary were studied in [Reference Diaconis and Sturmfels25]. If Q is the random transposition measure on $S_n$ , [Reference Simper59] uses knowledge of the Q chain to give an eigen-analysis of the chain induced by Q on contingency tables.

7.4 Chevalley groups

Let G be a finite Chevalley group (a finite simple group of Lie type). These come equipped with natural notions of Borel subgroups B and Weyl groups W. The Bruhat decomposition

$$\begin{align*}G = \bigsqcup_{\omega \in W} \mathcal{B} \omega \mathcal{B} \end{align*}$$

is in force, and conjugacy invariant probabilities Q on G induce Markov chains on W. Let U be a minimal unipotent conjugacy class in G ([Reference Carter11], Chapter 5) and Q the uniform probability on U. Conjugacy invariance implies that convolving by Q induces an element of $\mathrm {End}_G(G/\mathcal {B})$ . This may be identified with the Hecke algebra of B-bi-invariant functions. James Parkinson has shown us that the argument of Section 2.4 (for the computation of D in the Hecke algebra) goes through for a general Chevalley groups G over a finite field $\mathbb {F}_q$ . Although a similar formula holds in full generality, for simplicity we will state it in the equal parameter case (nontwisted) and when G is not of type $C_n$ with character lattice equal to the root lattice. Let $\theta $ be the highest root, and let $\mathcal {X}_{\theta } = \{ x_{\theta }(c)\ |\ c\in \mathbb {F}_q\}$ be the corresponding root subgroup. The conjugacy class of $x_{\theta }(1)$ is an analogue, for general Chevalley groups, of the conjugacy class of transvections in $GL_n(q)$ . The sum of the elements in the conjugacy class of $x_{\theta }(1)$ provides a Markov chain on $\mathcal {B}\backslash G/\mathcal {B}$ which acts the same way as

$$ \begin{align*} D &= (q-1)\sum_{\alpha\in \Phi^+_l} q^{\frac12(I(s_{\theta})-I(s_{\alpha}))}(1+T_{s_{\alpha}}), \end{align*} $$

where $\theta $ is the highest root, $\Phi ^+_l$ is the set of positive long roots, and $s_{\beta }$ denotes the reflection in the root $\beta $ and $T_w$ is the element of the Iwahori–Hecke algebra for $\mathcal {B}\backslash G/\mathcal {B}$ corresponding to the element w in the Weyl group.

Following the ideas in [Reference Diaconis and Ram18], the $T_{s_{\alpha }}$ can be interpreted via the Metropolis algorithm applied to the problem of sampling from the stationary distribution $\pi (x) = Z^{-1}q^{I(x)}$ on W by choosing random generators. We have not worked out any further examples but would be pleased if someone would.

7.5 A continuous example

Most of the generalities above extend to compact groups G and closed subgroups $H, K \neq G$ . Then, $X = H \backslash G /K$ is a compact space and an H-conjugacy invariant probability Q on G induces a Markov chain on X.

To consider a simple example, let $G = \mathcal {O}_n$ , the usual orthogonal group over $\mathbb {R}$ and $H = K = \mathcal {O}_{n-1}$ embedded as all $m \in \mathcal {O}_n$ fixing the ‘north pole’ $e_1 = (1, 0, 0 \dots , 0)^{\top }$ . Then, $\mathcal {O}_n/\mathcal {O}_{n-1}$ can be thought of as the $(n-1)$ -sphere $\mathcal {S}_{n-1}$ . The double coset space $\mathcal {O}_{n-1} \backslash \mathcal {O}_n / \mathcal {O}_{n-1}$ codes up the ‘latitude’. Consider the sphere $\mathcal {O}_n / \mathcal {O}_{n-1}$ defined by ‘circles’ orthogonal to $e_1$ (see Figure 1).

Figure 1 The space $\mathcal {O}_n/\mathcal {O}_{n-1}$ is defined by the circles on the sphere orthogonal to $e_1$ .

Then $\mathcal {O}_{n-1} \backslash \mathcal {O}_n / \mathcal {O}_{n-1}$ simply codes which circle contains a given point on the sphere. Thus, $\mathcal {O}_{n-1} \backslash \mathcal {O}_n / \mathcal {O}_{n-1}$ may be identified with $[-1, 1]$ .

Represent a uniformly chosen point on the sphere as $x = z/\|z \|$ with $z = (z_1, z_2, \dots , z_n)$ independent standard normals. The latitude is $z_1/\|z\|$ and so $\pi (x)$ is the distribution of the square root of a $\beta (1/2, (n-1)/2)$ distribution on $[-1, 1]$ . When $n = 3$ , $\pi (x)$ is uniform on $[-1, 1]$ (theorem of Archimedes).

One simple choice for a driving measure Q on $\mathcal {O}_n$ is ‘random reflections’. In probabilistic language, this is the distribution of $I - 2 U U^{\top }$ , with U uniform on $\mathcal {S}_{n-1}$ . There is a nice probabilistic description of the induced walk on the sphere.

Figure 2 Illustration of procedure from Lemma 7.2.

The induced chain on $[-1, 1] \cong \mathcal {O}_{n-1} \backslash \mathcal {O}_n / \mathcal {O}_{n-1}$ is obtained by simply reporting the latitude of y. Thanks to Sourav Chatterjee for the following probabilistic description. For simplicity, it is given here for $n = 3$ (so $\pi (x)$ is uniform on $[-1, 1]$ ).

Lemma 7.4. For $n = 3$ , the Markov chain on $[-1, 1]$ described above is; from $x_0 \in [-1, 1]$ , the chain jumps to

$$\begin{align*}X_1 := (1 - 2U_1^2)x_0 + 2(\cos(\pi U_2)) |U_1| \sqrt{ 1 - U_1^2} \sqrt{1 - x_0^2}, \end{align*}$$

where $U_1, U_2$ are i.i.d. uniform on $[-1, 1]$ random variables. (We have checked that the uniform distribution is stationary using Monte Carlo.)