Involutions by Descents/Ascents and Symmetric Integral Matrices. Alan Hoffman Fest - my hero Rutgers University September 2014

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1 by Descents/Ascents and Symmetric Integral Matrices Richard A. Brualdi University of Wisconsin-Madison Joint work with Shi-Mei Ma: European J. Combins. (to appear) Alan Hoffman Fest - my hero Rutgers University 9-20 September 204

2 Some Background 2

3 Some Background RSK-correspondence S n : the set of all permutations of {, 2,..., n}. Y n : the set of standard Young tableaux on n cells: a Ferrers diagram containing each of {, 2,..., n} increasing along rows from left to right and along columns from top to bottom. For example. with n = 8, (of shape 4, 3, ). The RSK-correspondence (an algorithm) provides a bijection between S n and pairs of standard Young tableaux of the same shape. σ RSK (P, Q) (P is the insertion tableau and Q is the recording tableau). Fact: σ RSK (P, Q) = σ RSK (Q, P)

4 Some Background RSK-correspondence S n : the set of all permutations of {, 2,..., n}. Y n : the set of standard Young tableaux on n cells: a Ferrers diagram containing each of {, 2,..., n} increasing along rows from left to right and along columns from top to bottom. For example. with n = 8, (of shape 4, 3, ). The RSK-correspondence (an algorithm) provides a bijection between S n and pairs of standard Young tableaux of the same shape. σ RSK (P, Q) (P is the insertion tableau and Q is the recording tableau). Fact: σ RSK (P, Q) = σ RSK (Q, P)

5 Some Background RSK-correspondence S n : the set of all permutations of {, 2,..., n}. Y n : the set of standard Young tableaux on n cells: a Ferrers diagram containing each of {, 2,..., n} increasing along rows from left to right and along columns from top to bottom. For example. with n = 8, (of shape 4, 3, ). The RSK-correspondence (an algorithm) provides a bijection between S n and pairs of standard Young tableaux of the same shape. σ RSK (P, Q) (P is the insertion tableau and Q is the recording tableau). Fact: σ RSK (P, Q) = σ RSK (Q, P)

6 Some Background Fact: σ RSK (P, Q) = σ RSK (Q, P) I n : the set of involutions in S n, that is, permutations σ S n such that σ = σ. Thus in the RSK-correspondence, Q = P, and hence σ RSK (P, P) P is a bijection between the set I n of involutions of order n and the set Y n of standard Young tableaux on n cells. Result of Schutzenberger (977) : The number of fixed points of an involution σ of order n equals the number of columns of the corresponding standard Young tableau P with odd length.

7 Some Background Fact: σ RSK (P, Q) = σ RSK (Q, P) I n : the set of involutions in S n, that is, permutations σ S n such that σ = σ. Thus in the RSK-correspondence, Q = P, and hence σ RSK (P, P) P is a bijection between the set I n of involutions of order n and the set Y n of standard Young tableaux on n cells. Result of Schutzenberger (977) : The number of fixed points of an involution σ of order n equals the number of columns of the corresponding standard Young tableau P with odd length.

8 Some Background Fact: σ RSK (P, Q) = σ RSK (Q, P) I n : the set of involutions in S n, that is, permutations σ S n such that σ = σ. Thus in the RSK-correspondence, Q = P, and hence σ RSK (P, P) P is a bijection between the set I n of involutions of order n and the set Y n of standard Young tableaux on n cells. Result of Schutzenberger (977) : The number of fixed points of an involution σ of order n equals the number of columns of the corresponding standard Young tableau P with odd length.

9 Some Background Fact: σ RSK (P, Q) = σ RSK (Q, P) I n : the set of involutions in S n, that is, permutations σ S n such that σ = σ. Thus in the RSK-correspondence, Q = P, and hence σ RSK (P, P) P is a bijection between the set I n of involutions of order n and the set Y n of standard Young tableaux on n cells. Result of Schutzenberger (977) : The number of fixed points of an involution σ of order n equals the number of columns of the corresponding standard Young tableau P with odd length.

10 Generating polynomial of by descents descent of a permutation π S n : a position i such that π(i) > π(i + ). des(π) is the number of descents of π. Generating polynomial of the set of involutions I n by descents: I n (t) = n t des(π) = I (n, k)t k, π I n where I (n, k) is the number of involutions of order n with k descents. The first few I n (t) are (see the on-line encycl. of integer sequences): I (t) =, I 2 (t) = + t, I 3 (t) = + 2t + t 2, k=0 I 4 (t) = + 4t + 4t 2 + t 3, I 5 (t) = + 6t + 2t 2 + 6t 3 + t 4.

11 Generating polynomial of by descents descent of a permutation π S n : a position i such that π(i) > π(i + ). des(π) is the number of descents of π. Generating polynomial of the set of involutions I n by descents: I n (t) = n t des(π) = I (n, k)t k, π I n where I (n, k) is the number of involutions of order n with k descents. The first few I n (t) are (see the on-line encycl. of integer sequences): I (t) =, I 2 (t) = + t, I 3 (t) = + 2t + t 2, k=0 I 4 (t) = + 4t + 4t 2 + t 3, I 5 (t) = + 6t + 2t 2 + 6t 3 + t 4.

12 Known Properties of I n (t), that is, of the sequence I (n, 0), I (n, ),..., I (n, n ) symmetric (Strehl, 98) unimodal (Dukes (2006) and Guo & Zeng (2006) not log-concave (which would have implied unimodal) (Barnabei, Bonetti & Silimbani (2009)

13 Descents in Permutation Matrices Example: Let n = 5 and let π = (2, 4,, 5, 3) with two descents: 4, and 5, 3. The corresponding permutation matrix is A descent in terms of the corresponding permutation matrix means that the in some row i + is in an earlier column than the in row i. An involution of order n corresponds to an n n symmetric permutation matrix. Thus I (n, k) counts the number of n n symmetric permutation matrices with k descents.

14 Theorem on the Involution Polynomial I n (t) Recall: n I n (t) = I (n, k)t k, k=0 where I (n, k) is the number of involutions of order n with k descents. Let T(n, i) be the of i i symmetric matrices with nonnegative integral entries with no zero rows or columns and sum of entries equal to n, and let T (n, i) = T(n, i). Theorem: I n (t) = n i= T (n, i)ti ( t) n i.

15 Theorem on the Involution Polynomial I n (t) Recall: n I n (t) = I (n, k)t k, k=0 where I (n, k) is the number of involutions of order n with k descents. Let T(n, i) be the of i i symmetric matrices with nonnegative integral entries with no zero rows or columns and sum of entries equal to n, and let T (n, i) = T(n, i). Theorem: I n (t) = n i= T (n, i)ti ( t) n i.

16 Theorem on the Involution Polynomial I n (t) Recall: n I n (t) = I (n, k)t k, k=0 where I (n, k) is the number of involutions of order n with k descents. Let T(n, i) be the of i i symmetric matrices with nonnegative integral entries with no zero rows or columns and sum of entries equal to n, and let T (n, i) = T(n, i). Theorem: I n (t) = n i= T (n, i)ti ( t) n i.

17 Theorem on the Involution Polynomial I n (t) I n (t) = n I (n, k)t k = k=0 Making the substitution, t = n T (n, i)t i ( t) n i i= x +x, this is equivalent to n I (n, k)x k+ ( + x) n k = k=0 and solving for T (n, i),... n T (n, i)x i, i=

18 Theorem on the Involution Polynomial I n (t) I n (t) = n I (n, k)t k = k=0 Making the substitution, t = n T (n, i)t i ( t) n i i= x +x, this is equivalent to n I (n, k)x k+ ( + x) n k = k=0 and solving for T (n, i),... n T (n, i)x i, i=

19 Equivalent Formulation, solving for T (n, i) i ( ) n k T (n, i) = I (n, k) i k k=0 (i =, 2,..., n). A permutation with k descents has (n k) ascents, and a little manipulation shows that this equation is equivalent to T (n, i) = n j=n i I (n, j) ( ) j n i (i =, 2,..., n), where I (n, j) = I (n, j), that is, the cardinality of the set I (n, j) of involutions of {, 2,..., n} with j ascents.

20 Equivalent Formulation, solving for T (n, i) i ( ) n k T (n, i) = I (n, k) i k k=0 (i =, 2,..., n). A permutation with k descents has (n k) ascents, and a little manipulation shows that this equation is equivalent to T (n, i) = n j=n i I (n, j) ( ) j n i (i =, 2,..., n), where I (n, j) = I (n, j), that is, the cardinality of the set I (n, j) of involutions of {, 2,..., n} with j ascents.

21 The Identity T (n, i) = n j=n i ( ) j I (n, j) n i (i =, 2,..., n), where I (n, j) is the number of n n symmetric permutation matrices with j ascents, and T (n, i) is the number of i i symmetric matrices with nonnegative integral entries with no zero rows or columns and sum of entries equal to n. This suggests that there may be a nice mapping F n i from the set j n i I (n, j) (all the n n symmetric permutation matrices P with j n i ascents) to subsets of T (n, i), such that F n i (P) = ( j n i ), and {F n i (P) : P j n i I (n, j)} is a partition of T (n, i).

22 The Identity T (n, i) = n j=n i ( ) j I (n, j) n i (i =, 2,..., n), where I (n, j) is the number of n n symmetric permutation matrices with j ascents, and T (n, i) is the number of i i symmetric matrices with nonnegative integral entries with no zero rows or columns and sum of entries equal to n. This suggests that there may be a nice mapping F n i from the set j n i I (n, j) (all the n n symmetric permutation matrices P with j n i ascents) to subsets of T (n, i), such that F n i (P) = ( j n i ), and {F n i (P) : P j n i I (n, j)} is a partition of T (n, i).

23 The Mapping F n i : j n i I (n, j) P(T (n, i)) (i) Let P be an n n symmetric perm. matrix with j n i ascents. (ii) Let the ascents occur in the pairs of row indices of P given by {p, p +}, {p 2, p 2 +},..., {p j, p j +} ( p < p 2 < < p j < n). (iii) Choose a set X of (n i) of these pairs. (iv) The chosen pairs determine an ordered partition U, U 2,..., U i of [n] into maximal sets of consecutive integers: if U k = {m, m +, m + 2,..., q, q} (m and q depend on k), then with Uk = {{m, m + }, {m +, m + 2},..., {q, q}} X we have U U 2 U i = X and Uk U l = for k l. (v) The partition in (iv) determines a partition of the rows and columns of P into an i i block matrix [P rs ]. Let A = [a rs ] be the i i matrix where a rs equals the sum of the entries of P rs. (vi) Then A is an i i symmetric, nonnegative integral matrices without any zero rows and columns, the sum of whose entries equals n. Since P has j n i ascents, P gives ( j n i) such matrices A. Alan Hoffman Fest, 9-20 September 204

24 Example of F n i : j n i I (n, j) P(T (n, i)) Example: Let n = 8 and consider the involution 5, 7, 8, 6,, 4, 2, 3 with corresponding symmetric permutation matrix P. The ascents occur in the pairs of positions (row indices) {, 2}, {2, 3}, {5, 6}, and {7, 8}. Choosing the pairs {, 2}, {2, 3}, and {7, 8}, we obtain the partition U = {, 2, 3}, U 2 = {4}, U 3 = {5}, U 4 = {6}, U 5 = {7, 8} of {, 2,..., 8} and corresponding partition of P given by

25 Inverting F n i A: an i i symmetric, nonnegative integral matrix with no zero rows and columns where the sum of the entries equals n. r k : the sum of the entries in row (and column) k of A. If A is to result from an n n symmetric permutation matrix P as described, then, since P has exactly one in each row and column, it must use the partition of the row and column indices of P into the sets: U = {,..., r }, U 2 = {r +,..., r +r 2 },..., U i = {r +r 2 + +r i +, n}. There must be a string of (r k ) consecutive ascents corresponding to the positions in each U k, where there are ascents or descents in the position pairs: (r, r +), (r +r 2, r +r 2 +),..., (r +r 2 + +r i, r +r 2 + +r i +). Then it can be shown that there is exactly one involution (symmetric permutation matrix) with these restrictions.

26 Inverting F n i, an Example Example: A = Then n =, and r = 2, r 2 = 6, r 3 = 3. We seek an symmetric permutation matrix of the form with one ascent in rows and 2, five ascents in rows 3 to 8, and two ascents in rows 9,0,. It is easy to see that the only possibility is:

27 Inverting F n i, an Example (n =, r = 2, r 2 = 6, r 3 = 3) equivalently, the involution , 4;, 2, 5, 9, 0, ; 6, 7, 8. Notice that the pairs of positions which could be either ascents or descents, namely {2, 3} and {8, 9}, are both descents in this case.,

28 Reference Enumeration of by Descents and Symmetric Matrices. RAB and Shi-Mei Ma, European Journal of Combinatorics, to appear. Belated Happy 90th Birthday, May 30

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