Zeons, Permanents, the Johnson Scheme, and Generalized Derangements

Abstract

Starting with the zero-square “zeon algebra,” the connection with permanents is shown. Permanents of submatrices of a linear combination of the identity matrix and all-ones matrix lead to moment polynomials with respect to the exponential distribution. A permanent trace formula analogous to MacMahon's master theorem is presented and applied. Connections with permutation groups acting on sets and the Johnson association scheme arise. The families of numbers appearing as matrix entries turn out to be related to interesting variations on derangements. These generalized derangements are considered in detail as an illustration of the theory.

1. Introduction

Functions acting on a finite set can be conveniently expressed using matrices, whereby the composition of functions corresponds to multiplication of the matrices. Essentially, one is considering the induced action on the vector space with the elements of the set acting as a basis. This action extends to tensor powers of the vector space. One can take symmetric powers, antisymmetric powers, and so forth, that yield representations of the multiplicative semigroup of functions. An especially interesting representation occurs by taking nonreflexive, symmetric powers. Identifying the underlying set of cardinality 𝑛 with {1,2,…,𝑛}, the vector space has basis 𝑒₁,𝑒₂,…. The action we are interested in may be found by saying that the elements 𝑒_𝑖 generate a “zeon algebra,” the relations being that the 𝑒_𝑖 commute, with 𝑒^2𝑖=0, 1≤𝑖≤𝑛. To get a feeling for this, first we recall the action on Grassmann algebra where the matrix elements of the induced action arise as determinants. For the zeon case, permanents appear.

An interesting connection with the centralizer algebra of the action of the symmetric group comes up. For the defining action on the set {1,…,𝑛}, represented as 0-1 permutation matrices, the centralizer algebra of 𝑛×𝑛 matrices commuting with the entire group is generated by 𝐼, the identity matrix, and 𝐽, the all-ones matrix. The question was if they would help determine the centralizer algebra for the action on subsets of a fixed size, ℓ-sets, for ℓ>1. It is known that the basis for the centralizer algebra is given by the adjacency matrices of the Johnson scheme. Could one find this working solely with 𝐼 and 𝐽? The result is that by computing the “zeon powers”, that is, the action of 𝑠𝐼+𝑡𝐽, linear combinations of 𝐼 and 𝐽, on ℓ-sets, the Johnson scheme appears naturally. The coefficients are polynomials in 𝑠 and 𝑡 occurring as moments of the exponential distribution. And they turn out to count derangements and related generalized derangements. The occurrence of Laguerre polynomials in the combinatorics of derangements is well known. Here, the ₂𝐹₁ hypergeometric function, which is closely related to Poisson-Charlier polynomials, arises rather naturally.

Here is an outline of the paper. Section 2 introduces zeons and permanents. The trace formula is proved. Connections with the centralizer algebra of the action of the symmetric group on sets are detailed. Section 3 is a study of exponential polynomials needed for the remainder of the paper. Zeon powers of 𝑠𝐼+𝑡𝐽 are found in Section 4 where the spectra of the matrices are found via the Johnson scheme. Section 5 presents a combinatorial approach to the zeon powers of 𝑠𝐼+𝑡𝐽, including an interpretation of exponential moment polynomials by elementary subgraphs. In Section 6, generalized derangement numbers, specifically counting derangements and counting arrangements, are considered in detail. The Appendix has some derangement numbers and arrangement numbers for reference, as well as a page of exponential polynomials. An example expressing exponential polynomials in terms of elementary subgraphs is given there.

2. Representations of Functions Acting on Sets

Let 𝒱 denote the vector space ℚ^𝑛 or ℝ^𝑛. We will look at the action of a linear map on 𝒱 extended to quotients of tensor powers 𝒱^⊗ℓ. We work with coordinates rather than vectors. First, recall the Grassmann case. To find the action on 𝒱^∧ℓ consider an algebra generated by 𝑛 variables 𝑒_𝑖 satisfying 𝑒_𝑖𝑒_𝑗=−𝑒_𝑗𝑒_𝑖. In particular, 𝑒^2𝑖=0.

Notation 2. The standard 𝑛-set {1,…,𝑛} will be denoted [𝑛]. Roman caps I, J, A, and so forth denote subsets of [𝑛]. We will identify them with the corresponding ordered tuples. Generally, given an 𝑛-tuple (𝑥₁,…,𝑥_𝑛) and a subset I⊂[𝑛], we denote products 𝑥_I=_𝑗∈I𝑥_𝑗,(2.1) where the indices are in increasing order if the variables are not assumed to commute.
As an index, we will use U to denote the full set [𝑛].
Italic 𝐼 and 𝐽 will denote the identity matrix and all-ones matrix, respectively.

For a matrix 𝑋_IJ, say, where the labels are subsets of fixed size 𝑙, dictionary ordering is used. That is, convert to ordered tuples and use dictionary ordering. For example, for 𝑛=4, 𝑙=2, we have labels 12,13,14,23,24,and34 for rows one through six, respectively.

A basis for 𝒱^∧ℓ is given by products 𝑒_I=𝑒_𝑖1𝑒_𝑖2⋯𝑒_𝑖ℓ,(2.2)I⊂[𝑛], where we consider I as an ordered ℓ-tuple. Given a matrix 𝑋 acting on 𝒱, let 𝑦_𝑖=_𝑗𝑋_𝑖𝑗𝑒_𝑗,(2.3) with corresponding products 𝑦_I, then the matrix 𝑋^∧ℓ has entries given by the coefficients in the expansion 𝑦_I=_J𝑋^∧ℓ_IJ𝑒J(2.4) where the anticommutation rules are used to order the factors in 𝑒_J. Note that the coefficient of 𝑒_𝑗 in 𝑦_𝑖 is 𝑋_𝑖𝑗 itself. And for 𝑛>3, the coefficient of 𝑒₃₄ in 𝑦₁₂ is det𝑋¹³𝑋₁₄𝑋₂₃𝑋₂₄.(2.5) We see that in general the IJ entry of 𝑋^∧ℓ is the minor of 𝑋 with row labels I and column labels J. A standard term for the matrix 𝑋^∧ℓ is a compound matrix. Noting that 𝑋^∧ℓ is (^𝑛ℓ)×(^𝑛ℓ), in particular, ℓ=𝑛 yields the one-by-one matrix with entry equal to (𝑋^∧𝑛)_UU=det𝑋.

In this work, we will use the algebra of zeons, standing for “zero-ons”, or more specifically, “zero-square bosons”. That is, we assume that the variables 𝑒_𝑖 satisfy the properties 𝑒_𝑖𝑒_𝑗=𝑒_𝑗𝑒_𝑖,𝑒^2𝑖=0.(2.6) A basis for the algebra is again given by 𝑒_I, I⊂[𝑛]. At level ℓ, the induced matrix 𝑋^∨ℓ has IJ entries according to the expansion of 𝑦_I,𝑦_𝑖=_𝑗𝑋_𝑖𝑗𝑒_𝑗⟶𝑦_𝐼=_𝐽𝑋^∨ℓ_IJ𝑒J(2.7) similar to the Grassmann case. Since the variables commute, we see that the IJ entry of 𝑋^∨ℓ is the permanent of the submatrix with rows I and columns J. In particular, (𝑋^∨𝑛)_UU=per𝑋. We refer to the matrix 𝑋^∨ℓ as the “ℓth zeon power of 𝑋.”

2.1. Functions on the Power Set of [𝑛]

Note that 𝑋^∨ℓ is indexed by ℓ-sets. Suppose that 𝑋_𝑓 represents a function 𝑓∶[𝑛]→[𝑛]. So it is a zero-one matrix with (𝑋_𝑓)_𝑖𝑗=1, the single entry in row 𝑖 if 𝑓 maps 𝑖 to 𝑗. The ℓth zeon power of 𝑋 is the matrix of the induced map on ℓ-sets. If 𝑓 maps an ℓ-set I to one of lower cardinality, then the corresponding row in 𝑋^∨ℓ has all zero entries. Thus, the induced matrices in general correspond to “partial functions”.

However, if 𝑋 is a permutation matrix, then 𝑋^∨ℓ is a permutation matrix for all 0≤ℓ≤𝑛. So, given a group of permutation matrices, the map 𝑋→𝑋^∨ℓ is a representation of the group.

2.2. Zeon Powers of 𝑠𝐼+𝑡𝑋

Our main theorem computes the ℓth zeon power of 𝑠𝐼+𝑡𝑋 for an 𝑛×𝑛 matrix 𝑋, where 𝑠 and 𝑡 are scalar variables. Figure 1 illustrates the proof.

Figure 1 

Configuration of sets: B=I⧵A, C=J⧵A.

Theorem 2.1. For a given matrix 𝑋, for 0≤ℓ≤𝑛, and indices |I|=|J|=ℓ, (𝑠𝐼+𝑡𝑋)^∨ℓ_IJ=_{0≤𝑗≤ℓ}𝑠^ℓ−𝑗𝑡^𝑗_{𝐴⊂I∩J||A||=ℓ−𝑗}𝑋^∨𝑗_I⧵A,J⧵A.(2.8)

Proof. Start with 𝑦_𝑖=𝑠𝑒_𝑖+𝑡𝜉_𝑖, where 𝜉_𝑖=∑_𝑗𝑋_𝑖𝑗𝑒_𝑗. Given I=(𝑖₁,…,𝑖_ℓ), we want the coefficient of 𝑒_J in the expansion of the product 𝑦_𝐼=𝑦_𝑖1⋯𝑦_𝑖ℓ. Now, 𝑦_𝐼=𝑠𝑒_𝑖1+𝑡𝜉_𝑖1⋯𝑠𝑒_𝑖ℓ+𝑡𝜉_𝑖ℓ.(2.9) Choose A⊂I with |A|=ℓ−𝑗, 0≤𝑗≤ℓ. A typical term of the product has the form 𝑠^ℓ−𝑗𝑡^𝑗𝑒_A𝜉_B(2.10) where A∩B=∅, B=I⧵A. 𝜉_B denotes the product of terms 𝜉_i with indices in B. Expanding, we have e_𝐴𝜉_B=_𝐶𝑋^∨𝑗_BC𝑒_C,𝜉_B=_𝐶𝑋^∨𝑗_BC𝑒_A𝑒_C.(2.11) Thus, for a contribution to the coefficient of 𝑒_J, we have A∪C=J, where A∩C=∅. That is, C=J⧵A and A⊂I∩J. So, the coefficient of 𝑠^ℓ−𝑗𝑡^𝑗 is as stated.

2.3. Trace Formula

Another main feature is the trace formula which shows the permanent of 𝐼+𝑡𝑋 as the generating function for the traces of the zeon powers of 𝑋. This is the zeon analog of the theorem of MacMahon for representations on symmetric tensors.

Theorem 2.2. One has the formula per(𝑠𝐼+𝑡𝑋)=_𝑗𝑠^𝑛−𝑗𝑡^𝑗tr𝑋^∨𝑗.(2.12)

Proof. The permanent of 𝑠𝐼+𝑡𝑋 is the UU entry of (𝑠𝐼+𝑡𝑋)^∨𝑛. Specialize I=J=U in Theorem 2.1. So A is any (𝑛−𝑗)-set with I⧵𝐴=J⧵𝐴=𝐴^, its complement in [𝑛]. Thus, per(𝑠𝐼+𝑡𝑋)=(𝑠𝐼+𝑡𝑋)^∨𝑛_UU=_{0≤𝑗≤𝑛}𝑠^𝑛−𝑗𝑡^𝑗_||A_{||=𝑛−𝑗}𝑋^∨𝑗_A^′_,A^′=_{0≤𝑗≤𝑛}𝑠^𝑛−𝑗𝑡^𝑗tr𝑋^∨𝑗,(2.13) as required.

2.4. Permutation Groups

Let 𝑋 be an 𝑛×𝑛 permutation matrix. We can express per(𝐼+𝑡𝑋) in terms of the cycle decomposition of the associated permutation.

Proposition 2.3. For a permutation matrix X, per(𝐼+𝑡𝑋)=_{0≤ℓ≤𝑛}1+𝑡^ℓ^{𝑛𝑋(ℓ)},(2.14) where 𝑛_𝑋(ℓ) is the number of cycles of length ℓ in the cycle decomposition of the corresponding permutation.

Proof. Decomposing the permutation associated to 𝑋 yields a decomposition into invariant subspaces of the underlying vector space 𝒱. So per(𝐼+𝑡𝑋) will be the product of per(𝐼+𝑡𝑋_𝑐) as 𝑐 runs through the corresponding cycles with 𝑋_𝑐 the restriction of 𝑋 to the invariant subspace for each 𝑐. So we have to check that if 𝑋 acts on 𝒱^ℓ as a cycle of length ℓ, then per(𝐼+𝑡𝑋)=1+𝑡^ℓ. For this, apply Theorem 2.2. Apart from level zero, there is only one set fixed by any 𝑋^∨𝑗, namely when 𝑗=ℓ. So the trace of 𝑋^∨𝑗 is zero unless 𝑗=ℓ and then it is one. The result follows.

2.4.1. Cycle Index: Orbits on ℓ-sets

Now, consider a group, 𝐺, of permutation matrices. We have the cycle index 𝑍_𝐺𝑧₁,𝑧₂,…,𝑧_𝑛=1||𝐺||_𝑋∈𝐺𝑧^𝑛𝑋1(1)𝑧^𝑛𝑋2(2)⋯𝑧^{𝑛𝑋𝑛(𝑛)},(2.15) each 𝑧_ℓ corresponding to ℓ-cycles in the cycle decomposition associated to the 𝑋s. From Proposition 2.3, we have an expression in terms of permanents. Combining with the trace formula, we get the following.

Theorem 2.4. Let 𝐺 be a permutation group of matrices, then one has 1||𝐺||_𝑋∈𝐺per(𝐼+𝑡𝑋)==𝑍_𝐺ℓ1+𝑡,1+𝑡²,…,1+𝑡^𝑛𝑡^ℓ#(orbitson𝑙-sets).(2.16)

Remark 2.5. This result refers to three essential theorems in group theory acting on sets. Equality of the first and last expressions is the “permanent” analog of Molien's theorem, which is the case for a group acting on the symmetric tensor algebra, that the cycle index counts orbits on subsets is an instance of Polya Counting, with two colors. The last expression is followed by the Cauchy-Burnside lemma applied to the groups 𝐺^∨ℓ={𝑋^∨ℓ}_𝑋∈𝐺.

2.4.2. Centralizer Algebra and Johnson Scheme

Given a group, 𝐺, of permutation matrices, an important question is to determine the set (among all matrices) of matrices commuting with all of the matrices in 𝐺. This is the centralizer algebra of the group. For the symmetric group, the only matrices are 𝐼 and 𝐽. For the action of the symmetric group on ℓ-sets, a basis for the centralizer algebra is given by the incidence matrices for the Johnson distance. These are the same as the adjacency matrices for the Johnson (association) scheme. Recall that the Johnson distance between two ℓ-sets I and J is dist_JS(I,J)=12||IΔJ||=||I⧵J||=||J⧵I||.(2.17) The corresponding matrices JS^𝑘𝑛ℓ are defined by JS^𝑘𝑛ℓ_IJ=1ifdist^JS0,otherwis(eI.,J)=𝑘,(2.18) As it is known, [1, page 36], that a basis for the centralizer algebra is given by the orbits of the group 𝐺², acting on pairs, the Johnson basis is a basis for the centralizer algebra. Since the Johnson distance is symmetric, it suffices to look at 𝐺^∨2.

Now, we come to the question that is a starting point for this work. If 𝐼 and 𝐽 are the only matrices commuting with all elements (as matrices) of the symmetric group, then since the map 𝐺→𝐺^∨ℓ is a homomorphism, we know that 𝐼^∨ℓ and 𝐽^∨ℓ are in the centralizer algebra of 𝐺^∨ℓ. The question is how to obtain the rest? The, perhaps surprising, answer is that in fact one can obtain the complete Johnson basis from 𝐼 and 𝐽 alone. This will be one of the main results, Theorem 4.1.

2.4.3. Permanent of 𝑠𝐼+𝑡𝐽

First, let us consider 𝑠𝐼+𝑡𝐽.

Proposition 2.6. One has the formula per(𝑠𝐼+𝑡𝐽)=𝑛!_{0≤ℓ≤𝑛}𝑠^ℓ𝑡^𝑛−ℓℓ!.(2.19)

Proof. For 𝑋=𝐽, we see directly, since all entries equal one in all submatrices, that 𝐽^∨ℓ_IJ=ℓ!(2.20) for all I and J. Taking traces, tr𝐽^∨ℓ=ℓ𝑛ℓ!,(2.21) and by the trace formula, Theorem 2.2, per(𝑠𝐼+𝑡𝐽)=_ℓℓ𝑛ℓ!𝑠^𝑛−ℓ𝑡^ℓ=_ℓ𝑛!(𝑛−ℓ)!𝑠^𝑛−ℓ𝑡^ℓ.(2.22) Reversing the order of summation yields the result stated.

Corollary 2.7. For varying 𝑛, one will explicitly denote 𝑝_𝑛(𝑠,𝑡)=per(𝑠𝐼_𝑛+𝑡𝐽_𝑛), then, with 𝑝₀(𝑠,𝑡)=1, ^∞_𝑛=0𝑧^𝑛𝑛!𝑝^𝑛(𝑠,𝑡)=𝑒^𝑠𝑧1−𝑡𝑧.(2.23)

The Corollary exhibits the operational formula 𝑝_𝑛(𝑠,𝑡)=11−𝑡𝐷_𝑠𝑠^𝑛,(2.24) where 𝐷_𝑠=𝑑/𝑑𝑠. By inspection, this agrees with (2.19) as well.

Observe that (2.19) can be rewritten as per(𝑠𝐼+𝑡𝐽)=^∞0(𝑠+𝑡𝑦)^𝑛𝑒^−𝑦𝑑𝑦,(2.25) that is, these are “moment polynomials” for the exponential distribution with an additional scale parameter.

We proceed to examine these moment polynomials in detail.

3. Exponential Polynomials

For the exponential distribution, with density 𝑒^−𝑦 on (0,∞), the moment polynomials are defined as ℎ_𝑛(𝑥)=^∞0(𝑥+𝑦)^𝑛𝑒^−𝑦𝑑𝑦.(3.1) The exponential embeds naturally into the family of weights of the form 𝑥^𝑚𝑒^−𝑥 on (0,∞) as for generalized Laguerre polynomials. We define correspondinglyℎ_𝑛,𝑚(𝑥,𝑡)=^∞0(𝑥+𝑡𝑦)^𝑛(𝑡𝑦)^𝑚𝑒^−𝑦𝑑𝑦,(3.2) for nonnegative integers 𝑛,𝑚, introducing a factor of 𝑦^𝑚 and a scale factor of 𝑡. We refer to these as exponential moment polynomials.

Proposition 3.1. Observe the following properties of the exponential moment polynomials. (1)The generating function 1𝑡^𝑚𝑚!^∞_𝑛=0𝑧^𝑛𝑛!ℎ^𝑛,𝑚(𝑥,𝑡)=𝑒^𝑧𝑥(1−𝑡𝑧)^1+𝑚,(3.3) for |𝑡𝑧|<1. (2)The operational formula 1𝑡^𝑚𝑚!ℎ^𝑛,𝑚(𝑥,𝑡)=(𝐼−𝑡𝐷)^−(𝑚+1)𝑥^𝑛,(3.4) where 𝐼 is the identity operator and 𝐷=𝑑/𝑑𝑥. (3)The explicit form ℎ_𝑛,𝑚(𝑥,𝑡)=^𝑛_𝑗=0𝑛𝑗(𝑚+𝑗)!𝑥^𝑛−𝑗𝑡^𝑚+𝑗.(3.5)

Proof. For the first formula, multiply the integral by 𝑧^𝑛/𝑛! and sum to get ^∞0𝑦^𝑚𝑒^{𝑧𝑥+𝑧𝑡𝑦−𝑦}𝑑𝑦=𝑒^𝑧𝑥^∞0𝑦^𝑚𝑒^{−𝑦(1−𝑡𝑧)}𝑑𝑦,(3.6) which yields the stated result.
For the second, write 𝑡^𝑚𝑚!(𝐼−𝑡𝐷)^−(𝑚+1)𝑥^𝑛=𝑡^𝑚^∞0𝑦^𝑚𝑒^{−(𝐼−𝑡𝐷)𝑦}𝑥^𝑛𝑑𝑦=^∞0(𝑡𝑦)^𝑚𝑒^−𝑦(𝑥+𝑡𝑦)^𝑛𝑑𝑦,(3.7) using the shift formula 𝑒^𝑎𝐷𝑓(𝑥)=𝑓(𝑥+𝑎).
For the third, expand (𝑥+𝑡𝑦)^𝑛 by the binomial theorem and integrate.

A variation we will encounter in the following isℎ_{𝑛−𝑚,𝑚}(𝑥,𝑡)=^𝑛−𝑚_𝑗=0𝑛−𝑗𝑚(𝑚+𝑗)!𝑥^{𝑛−𝑚−𝑗}=𝑡^𝑚+𝑗(3.8)_𝑛_𝑗=𝑚𝑛𝑗−−𝑚𝑚𝑗!𝑥^𝑛−𝑗=𝑡^𝑗(3.9)_𝑛_𝑗=𝑚𝑛𝑛−−𝑚𝑗𝑗!𝑥^𝑛−𝑗=𝑡^𝑗(3.10)_𝑛−𝑚_𝑗=0𝑛−𝑗𝑚(𝑛−𝑗)!𝑥^𝑗𝑡^𝑛−𝑗,(3.11) replacing the index 𝑗←𝑗−𝑚 for (3.9) and reversing the order of summation for the last line. And for future reference, consider the integral formulaℎ_{𝑛−𝑚,𝑚}(𝑥,𝑡)=^∞0(𝑥+𝑡𝑦)^𝑛−𝑚(𝑡𝑦)^𝑚𝑒^−𝑦𝑑𝑦.(3.12)

3.1. Hypergeometric Form

Generalized hypergeometric functions provide expressions for the exponential moment polynomials that are often convenient. In the present context, we will use ₂𝐹₀ functions, defined by ₂𝐹₀𝑎-,--𝑏||||𝑥=^∞_𝑗=0(𝑎)_𝑗(𝑏)_𝑗𝑗!𝑥^𝑗,(3.13) where (𝑎)_𝑗=Γ(𝑎+𝑗)/Γ(𝑎) is the usual Pochhammer symbol. In particular, if 𝑎, for example, is a negative integer, the series reduces to a polynomial. Rearranging factors in the expressions for ℎ_n,m, via (3) in Proposition 3.1, and ℎ_n−m,m, (3.8), we can formulate these as ₂𝐹₀ hypergeometric functions.

Proposition 3.2. One has the following expressions for exponential moment polynomials: ℎ_𝑛,𝑚(𝑥,𝑡)=𝑥^𝑛𝑡^𝑚𝑚!₂𝐹₀ℎ_{𝑛−𝑚,𝑚}−𝑛,-1--+𝑚||||−𝑥𝑡,(𝑥,𝑡)=𝑥^𝑛−𝑚𝑡^𝑚𝑚!₂𝐹₀𝑚−𝑛-,--1+𝑚||||−𝑥𝑡.(3.14)

4. Zeon Powers of 𝑠𝐼+𝑡𝐽

We want to calculate (𝑠𝐼+𝑡𝐽)^∨ℓ, that is, the (^𝑛ℓ)×(^𝑛ℓ) matrix with rows and columns labelled by ℓ-subsets I,J⊂{1,…,𝑛} with the IJ entry equal to the permanent of the corresponding submatrix of 𝑠𝐼+𝑡𝐽. This is equivalent to the induced action of the original matrix 𝑠𝐼+𝑡𝐽 on the ℓth zeon space 𝒱^∨ℓ.

Theorem 4.1. The ℓth zeon power of 𝑠𝐼+𝑡𝐽 is given by (𝑠𝐼+𝑡𝐽)^∨ℓ=_𝑘ℓ_𝑗=𝑘ℓℓ−−𝑘𝑗𝑗!𝑠^ℓ−𝑗𝑡^𝑗JS^𝑘𝑛ℓ=_𝑘ℎ_{ℓ−𝑘,𝑘}(𝑠,𝑡)JS^𝑘𝑛ℓ,(4.1) where the ℎ’s are exponential moment polynomials.

Proof. Choose I and J with |I|=|J|=ℓ. By Theorem 2.1, we have, using the fact that all of the entries of 𝐽^∨𝑗 are equal to 𝑗!, (𝑠𝐼+𝑡𝐽)^∨ℓ_IJ=_{0≤𝑗≤ℓ}𝑠^ℓ−𝑗𝑡^𝑗_{𝐴⊂I∩J||A||=ℓ−𝑗}𝐽^∨𝑗_I⧵A,J⧵A=_{0≤𝑗≤ℓ}𝑠^ℓ−𝑗𝑡^𝑗_A⊂I∩_{J||A||=ℓ−j}𝑗!.(4.2) Now, if dist_JS(I,J)=𝑘, then |I∩J|=ℓ−𝑘, and there are ^{ℓ−𝑘ℓ−𝑗} subsets A of I∩J satisfying the conditions of the sum. Hence the result.

Note that the specialization ℓ=𝑛, 𝑘=0, recovers (2.19).

We can write the above expansion using the hypergeometric form of the exponential moment polynomials, Proposition 3.2, (𝑠𝐼+𝑡𝐽)^∨ℓ=_𝑘𝑠^ℓ−𝑘𝑡^𝑘𝑘!₂𝐹₀𝑘−ℓ-,--1+𝑘||||−𝑠𝑡JS^𝑘𝑛ℓ.(4.3)

4.1. Spectrum of the Johnson Matrices

Recall, for example, [2, page 220], that the spectrum of the Johnson matrices for given 𝑛 and ℓ is the set of numbersΛ^𝑘𝑛ℓ(𝛼)=_𝑖ℓ−𝑖𝛼𝑛−ℓ−𝑖𝛼+𝑖ℓ𝑘−−𝑖𝑖(−1)^𝑘−𝑖,(4.4) where the eigenvalue for given 𝛼 has multiplicity (^𝑛𝛼)−(^𝑛𝛼−1).

For ℓ-sets, the Johnson distance takes values from 0 to min(ℓ,𝑛−ℓ), with 𝛼 taking values from that same range.

4.2. The Spectrum of (𝑠𝐼+𝑡𝐽)^∨ℓ

Recall that as the Johnson matrices are symmetric and generate a commutative algebra, they are simultaneously diagonalizable by an orthogonal transformation of the underlying vector space. Diagonalizing the equation in Theorem 4.1, we see that the spectrum of (𝑠𝐼+𝑡𝐽)^∨ℓ is given by _𝑘ℎ_{ℓ−𝑘,𝑘}(𝑠,𝑡)Λ^𝑘𝑛ℓ(𝛼).(4.5)

Proposition 4.2. The spectrum of (𝑠𝐼+𝑡𝐽)^∨ℓ is given by 𝑠^𝛼𝑡^{𝑛−ℓ−𝛼}(𝑛−ℓ−𝛼)!ℎ^{ℓ−𝛼,𝑛−ℓ−𝛼}(𝑠,𝑡)=_𝑖𝑠^ℓ−𝑖𝑡^𝑖ℓ−𝑖𝛼𝑛−ℓ−𝑖𝛼+𝑖𝑖!,(4.6) for 0≤𝛼≤min(ℓ,𝑛−ℓ), with respective multiplicities (^𝑛𝛼)−(^𝑛𝛼−1).

Proof. In the sum over 𝑖 in (4.4), only the last two factors involve 𝑘. We have _𝑘ℎ_{ℓ−𝑘,𝑘}(𝑠,𝑡)(−1)^𝑘−𝑖=ℓ𝑘−−𝑖_𝑘𝑖^∞0(𝑠+𝑡𝑦)^ℓ−𝑘(𝑡𝑦)^𝑘(−1)^𝑘−𝑖ℓ𝑘−−𝑖𝑖𝑒^−𝑦=𝑑𝑦setting𝑘=𝑖+𝑚_𝑚^∞0(𝑠+𝑡𝑦)^{ℓ−𝑖−𝑚}(𝑡𝑦)^𝑖+𝑚(−1)^𝑚ℓ𝑚−𝑖𝑒^−𝑦=𝑑𝑦_∞0=𝑠_ℓ−𝑖(𝑠+𝑡𝑦−𝑡𝑦)^ℓ−𝑖(𝑡𝑦)^𝑖𝑒^−𝑦𝑑𝑦𝑡^𝑖𝑖!,(4.7) using the binomial theorem to sum out 𝑚. Filling in the additional factors yields _𝑘ℎ_{ℓ−𝑘,𝑘}(𝑠,𝑡)Λ^𝑘𝑛ℓ(𝛼)=_𝑖𝑠^ℓ−𝑖𝑡^𝑖𝑖!ℓ−𝑖𝛼𝑛−ℓ−𝑖𝛼+𝑖,(4.8) Taking out a denominator factor of (𝑛−ℓ−𝛼)! and multiplying by 𝑠^−𝛼𝑡^{𝑛−ℓ−𝛼} gives _𝑖𝑠^{ℓ−𝛼−𝑖}𝑡^{𝑛−ℓ−𝛼+𝑖}ℓ−𝑖𝛼(𝑛−ℓ−𝛼+𝑖)!,(4.9) which is precisely ℎ_{ℓ−𝛼,n−ℓ−𝛼} as in the third statement of Proposition 3.1.

As in Proposition 3.2, we can express the eigenvalues as follows.

Corollary 4.3. The spectrum of (𝑠𝐼+𝑡𝐽)^∨ℓ consists of the eigenvalues 𝑠^ℓ2𝐹₀𝛼−ℓ,1+---𝑛−ℓ−𝛼||||−𝑠𝑡,(4.10) for 0≤𝛼≤min(ℓ,𝑛−ℓ), with corresponding multiplicities as indicated above.

4.3. Row Sums and Trace Identity

For the row sums, we know that the all-ones vector is a common eigenvector of the Johnson basis corresponding to 𝛼=0. These are the valencies Λ_𝑘(0). For the Johnson scheme, we have Λ^𝑘𝑛ℓ(0)=ℓ𝑘𝑛−𝑘ℓ.(4.11) for example, see [2, page 219], which can be checked directly from the formula for Λ^𝑘𝑛ℓ(𝛼), (4.4), with 𝛼 set to zero. Setting 𝛼=0 in Proposition 4.2 gives1𝑡^𝑛−ℓ(𝑛−ℓ)!ℎ^{ℓ,𝑛−ℓ}(𝑠,𝑡)=_𝑖ℓ𝑖𝑛−ℓ𝑖+𝑖𝑖!𝑠^ℓ−𝑖𝑡^𝑖,(4.12) for the row sums of (𝑠𝐼+𝑡𝐽)^∨ℓ.

4.3.1. Trace Identity

Terms on the diagonal are the coefficient of JS^0𝑛ℓ, which is the identity matrix. So, the trace is tr(𝑠𝐼+𝑡𝐽)^∨ℓ=ℓ𝑛ℎ_ℓ,0(𝑠,𝑡)=ℓ𝑛_𝑘ℓ𝑘𝑘!𝑠^ℓ−𝑘𝑡^𝑘.(4.13) Cancelling factorials and reversing the order of summation on 𝑘 yields the following formula.tr(𝑠𝐼+𝑡𝐽)^∨ℓ=𝑛!(𝑛−ℓ)!_{0≤𝑘≤ℓ}𝑠^𝑘𝑡^ℓ−𝑘𝑘!.(4.14)

Now, Proposition 4.2 gives the trace tr(𝑠𝐼+𝑡𝐽)^∨ℓ=_{0≤𝛼≤min(ℓ,𝑛−ℓ)}𝛼𝑛−𝛼−𝑛1_𝑖𝑠^ℓ−𝑖𝑡^𝑖ℓ−𝑖𝛼𝑛−ℓ−𝑖𝛼+𝑖𝑖!.(4.15)

Proposition 4.4. Equating the above expressions for the trace yields the identity _0≤𝛼≤m_{in(ℓ,𝑛−ℓ)}𝛼𝑛−𝛼−𝑛1_𝑖𝑠^ℓ−𝑖𝑡^𝑖ℓ−𝑖𝛼𝑛−ℓ−𝑖𝛼+𝑖𝑖!=(𝑛−𝑛!ℓ)!_{0≤𝑗≤ℓ}𝑠^𝑗𝑡^ℓ−𝑗𝑗!.(4.16)

Example 4.5. For 𝑛=4, ℓ=2, we have ⎡⎢⎢⎢⎢⎢⎢⎢⎣𝑠²+2𝑠𝑡+2𝑡²𝑠𝑡+2𝑡²𝑠𝑡+2𝑡²𝑠𝑡+2𝑡²𝑠𝑡+2𝑡²2𝑡²𝑠𝑡+2𝑡²𝑠²+2𝑠𝑡+2𝑡²𝑠𝑡+2𝑡²𝑠𝑡+2𝑡²2𝑡²𝑠𝑡+2𝑡²𝑠𝑡+2𝑡²𝑠𝑡+2𝑡²𝑠²+2𝑠𝑡+2𝑡²2𝑡²𝑠𝑡+2𝑡²𝑠𝑡+2𝑡²𝑠𝑡+2𝑡²𝑠𝑡+2𝑡²2𝑡²𝑠²+2𝑠𝑡+2𝑡²𝑠𝑡+2𝑡²𝑠𝑡+2𝑡²𝑠𝑡+2𝑡²2𝑡²𝑠𝑡+2𝑡²𝑠𝑡+2𝑡²𝑠²+2𝑠𝑡+2𝑡²𝑠𝑡+2𝑡²2𝑡²𝑠𝑡+2𝑡²𝑠𝑡+2𝑡²𝑠𝑡+2𝑡²𝑠𝑡+2𝑡²𝑠²+2𝑠𝑡+2𝑡²⎤⎥⎥⎥⎥⎥⎥⎥⎦.(4.17) One can check that the entries are in agreement with Theorem 4.1. The trace is 6𝑠²+12𝑠𝑡+12𝑡². The spectrum is eigeenivgaelnuveal𝑠u²e+𝑠₂6𝑠𝑡+12𝑡²,withmultiplicity1,eigenvalue+𝑠₂2𝑠𝑡,withmultiplicity3,,withmultiplicity2,(4.18) and the trace can be verified from these as well.

Remark 4.6. What is interesting is that these matrices have polynomial entries with all eigenvalues polynomials as well, and furthermore, the exact same set of polynomials produces the eigenvalues as well as the entries. Specializing 𝑠 and 𝑡 to integers, a similar statement holds. All of these matrices will have integer entries with integer eigenvalues, all of which belong to closely related families of numbers. We will examine interesting cases of this phenomenon later on in this paper.

5. Permanents from 𝑠𝐼+𝑡𝐽

Here, we present a proof via recursion of the subpermanents of 𝑠𝐼+𝑡𝐽, thereby recovering Theorem 4.1 from a different perspective.

Remark 5.1. For the remainder of this paper, we will work with an 𝑛×𝑛 matrix corresponding to an ℓ×ℓ submatrix of the above discussion. Here, we have blown up the submatrix to full size as the object of consideration.

Let 𝑀_𝑛,ℓ denote the 𝑛×𝑛 matrix with 𝑛−ℓ entries equal to 𝑠+𝑡 on the main diagonal, and 𝑡’s elsewhere. Note that 𝑀_𝑛,0=𝑠𝐼+𝑡𝐽 and 𝑀_𝑛,𝑛=𝑡𝐽, where 𝐼 and 𝐽 are 𝑛×𝑛. Define𝑃_𝑛,ℓ=per𝑀_𝑛,ℓ(5.1) to be the permanent of 𝑀_𝑛,ℓ.

For ℓ=0, define 𝑃_0,0=1, and, recalling (2.19),𝑃_𝑛,0=per(𝑠𝐼+𝑡𝐽)=^𝑛_𝑗=0𝑛!𝑗!𝑠^𝑗𝑡^𝑛−𝑗=^𝑛_𝑗=0𝑛!(𝑛−𝑗)!𝑠^𝑛−𝑗𝑡^𝑗.(5.2) We have also 𝑃_𝑛,𝑛=per(𝑡𝐽)=𝑛!𝑡^𝑛 for 𝐽 of order 𝑛×𝑛. These agree at 𝑃_0,0=1.

Theorem 5.2. For 𝑛≥1, 1≤ℓ≤𝑛, one has the recurrence 𝑃_𝑛,ℓ=𝑃_𝑛,ℓ−1−𝑠𝑃_{𝑛−1,ℓ−1}.(5.3)

Proof. We have 0≤ℓ≤𝑛 so 𝑛−(ℓ−1)=𝑛−ℓ+1≥1, that is, the matrix 𝑀_𝑛,ℓ−1 contains at least 1 entry on its main diagonal equal to 𝑠+𝑡. Write the block form 𝑀_𝑛,ℓ−1=𝑠𝐴+𝑡𝐴_𝑇𝑀_{𝑛−1,ℓ−1},(5.4) with 𝐴=[𝑡,𝑡,…,𝑡] the 1×(𝑛−1) row vector of all 𝑡s, and 𝐴^𝑇 is its transpose. Now, compute the permanent of 𝑀_𝑛,ℓ−1 expanding along the first row. We get 𝑃_𝑛,ℓ−1=per𝑀_𝑛,ℓ−1=(𝑠+𝑡)per𝑀_{𝑛−1,ℓ−1}+𝐹𝐴,𝐴^𝑇,𝑀_{𝑛−1,ℓ−1},(5.5) where 𝐹(𝐴,𝐴^𝑇,𝑀_{𝑛−1,ℓ−1}) is the contribution to 𝑃_𝑛,ℓ−1 involving 𝐴. Now, 𝑡per𝑀_{𝑛−1,ℓ−1}+𝐹𝐴,𝐴^𝑇,𝑀_{𝑛−1,ℓ−1}=per𝐴𝑡𝐴_𝑇=per𝐴_𝑇𝑀_{𝑛−1,ℓ−1}𝑀_{𝑛−1,ℓ−1}=per𝑀𝑡𝐴_{𝑛−1,ℓ−1}𝐴^𝑇=𝑃_𝑛,ℓ𝐴𝑡.(5.6) Thus, from (5.5), 𝑃_𝑛,ℓ−1==𝑠𝑠𝑃per𝑀_{𝑛−1,ℓ−1𝑛−1,ℓ−1}+𝑡per𝑀_{𝑛−1,ℓ−1}+𝐹𝐴,𝐴^𝑇,𝑀_{𝑛−1,ℓ−1}+𝑃_𝑛,ℓ,(5.7) and hence the result.

We arrange the polynomials 𝑃_𝑛,ℓ in a triangle, with the columns labelled by ℓ≥0 and rows by 𝑛≥0, starting with 𝑃_0,0=1 at the top vertex𝑃_0,0𝑃_1,0𝑃_1,1𝑃_2,0𝑃_2,1𝑃_2,2𝑃⋮⋱⋱_𝑛−1,0⋯𝑃_{𝑛−1,𝑛−2}𝑃_{𝑛−1,𝑛−1}𝑃_𝑛,0𝑃_𝑛,1⋯𝑃_{𝑛,𝑛−1}𝑃_𝑛,𝑛(5.8) The recurrence says that to get the 𝑛,ℓ entry, you combine elements in column ℓ−1 in rows 𝑛 and 𝑛−1, forming an L-shape. Thus, given the first column {𝑃_𝑛,0}_𝑛≥0, the table can be generated in full.

Now, we check that these are indeed our exponential moment polynomials. Additionally, we derive an expression for 𝑃_𝑛,ℓ in terms of the initial sequence 𝑃_𝑛,0. For clarity, we will explicitly denote the dependence of 𝑃_𝑛,ℓ on (𝑠,𝑡).

Theorem 5.3. For ℓ≥0, one has (1)the permanent of the 𝑛×𝑛 matrix with 𝑛−ℓ entries on the diagonal equal to 𝑠+𝑡 and all other entries equal to 𝑡 is𝑃_𝑛,ℓ(𝑠,𝑡)=ℎ_{𝑛−ℓ,ℓ}(𝑠,𝑡)=^𝑛_𝑗=ℓ𝑛𝑛−−ℓ𝑗𝑗!𝑠^𝑛−𝑗𝑡^𝑗,(5.9)(2)𝑃_𝑛,ℓ(𝑠,𝑡)=^ℓ_𝑗=0ℓ𝑗(−1)^𝑗𝑠^𝑗𝑃_{𝑛−𝑗,0}(𝑠,𝑡),(5.10)(3) the complementary sum is 𝑠^𝑛=^𝑛_𝑗=0ℓ𝑛(−1)^ℓ𝑃_𝑛,ℓ(𝑠,𝑡).(5.11)

Proof. The initial sequence 𝑃_𝑛,0=ℎ_𝑛,0 as noted in (5.2). We check that ℎ_{𝑛−ℓ,ℓ} satisfies recurrence (5.3). Starting from the integral representation for ℎ_{𝑛−ℓ+1,ℓ−1}, (3.2), we have ℎ_{𝑛−ℓ+1,ℓ−1}==^∞0(𝑠+𝑡𝑦)^𝑛−ℓ+1(𝑡𝑦)^ℓ−1𝑒^−𝑦𝑑𝑦_∞0=𝑠ℎ(𝑠+𝑡𝑦_{𝑛−ℓ,ℓ−1})(𝑠+𝑡𝑦)^𝑛−ℓ(𝑡𝑦)^ℓ−1𝑒^−𝑦𝑑𝑦+ℎ_{𝑛−ℓ,ℓ},(5.12) as required, where we now identify ℎ_{𝑛−ℓ+1,ℓ−1}=𝑃_𝑛,ℓ−1, ℎ_{𝑛−ℓ,ℓ−1}=𝑃_{𝑛−1,ℓ−1}, and ℎ_{𝑛−ℓ,ℓ}=𝑃_𝑛,ℓ. And (3.10) gives an explicit form for 𝑃_𝑛,ℓ.
For (2), starting with the integral representation for 𝑃_𝑛,0=ℎ_𝑛,0, we get ^ℓ_𝑗=0ℓ𝑗(−1)^𝑗𝑠^𝑗^∞0(𝑠+𝑡𝑦)^𝑛−𝑗𝑒^−𝑦𝑑𝑦=^ℓ_𝑗=0ℓ𝑗(−1)^𝑗𝑠^𝑗^∞0=(𝑠+𝑡𝑦)^𝑛−ℓ(𝑠+𝑡𝑦)^ℓ−𝑗𝑒^−𝑦𝑑𝑦_∞0=ℎ(𝑠_{𝑛−ℓ,ℓ}+𝑡𝑦)^𝑛−ℓ(𝑠+𝑡𝑦−𝑠)^ℓ𝑒^−𝑦𝑑𝑦,(5.13) as required. The proof for (3) is similar, using (3.12), 𝑃_𝑛,ℓ=ℎ_{𝑛−ℓ,ℓ}=^∞0(𝑠+𝑡𝑦)^𝑛−ℓ(𝑡𝑦)^ℓ𝑒^−𝑦𝑑𝑦,(5.14) and the binomial theorem for the sum.

5.1. (𝑠𝐼+𝑡𝐽)^∨ℓ Revisited

Now, we have an alternative proof of Theorem 4.1.

Lemma 5.4. Let I and J be ℓ-subsets of [𝑛] with dist_JS(I,J)=𝑘, then per(𝑠𝐼+𝑡𝐽)_𝐼𝐽=𝑃_ℓ,𝑘(𝑠,𝑡).(5.15)

Proof. Now, |I∩J|=ℓ−𝑘, so the submatrix (𝑠𝐼+𝑡𝐽)_IJ is permutationally equivalent to the ℓ×ℓ matrix with ℓ−𝑘 entries 𝑠+𝑡 on its main diagonal and 𝑡s elsewhere, that is, to the matrix 𝑀_ℓ,𝑘. Hence, by definition of 𝑃_ℓ,𝑘(𝑠,𝑡), (5.1), we have the result.

Thus, the expansion in the Johnson basis is(𝑠𝐼+𝑡𝐽)^∨ℓ=_𝑘ℎ_{ℓ−𝑘,𝑘}(𝑠,𝑡)JS^𝑘𝑛ℓ.(5.16)

Proof. Let I and J be ℓ-subsets of [𝑛] with Johnson distance 𝑘. By definition, the IJ entry of the LHS of (5.16) equals the permanent of the submatrix from rows I and columns J, per(𝑠𝐼+𝑡𝐽)_IJ=𝑃_ℓ,𝑘(𝑠,𝑡)=ℎ_{ℓ−𝑘,𝑘}(𝑠,𝑡), by Lemma 5.4 and Theorem 5.3(1). Now, on the RHS of (5.16), if dist_JS(I,J)=𝑘, the only nonzero contribution comes from the JS^𝑘𝑛ℓ term. This yields ℎ_{ℓ−𝑘,𝑘}(𝑠,𝑡)×1=ℎ_{ℓ−𝑘,𝑘}(𝑠,𝑡) as required.

5.2. Elementary Subgraphs and Permanents

There is an approach to permanents of 𝑠𝐼+𝑡𝐽 via elementary subgraphs, based on that of Biggs [3] for determinants.

An elementary subgraph (see [3, page 44]) of a graph 𝐺 is a spanning subgraph of 𝐺 all of whose components are 0, 1, or 2 regular, that is, all of whose components are isolated vertices, isolated edges, or cycles of length 𝑗≥3.

Let 𝐾^𝑛(ℓ) be a copy of the complete graph 𝐾_𝑛 with vertex set [𝑛] in which the first 𝑛−ℓ vertices [𝑛−ℓ]={1,2,…,𝑛−ℓ} are distinguished. We may now consider the matrix 𝑀_𝑛,ℓ as the weighted adjacency matrix of 𝐾^𝑛(ℓ) in which the weights of the distinguished vertices are 𝑠+𝑡, with all undistinguished vertices and all edges assigned a weight of 𝑡.

Let 𝐸 be an elementary subgraph of 𝐾^𝑛(ℓ), then we describe 𝐸 as having 𝑑(𝐸) distinguished isolated vertices and 𝑐(𝐸) cycles. The weight of 𝐸, wt(𝐸), is defined aswt(𝐸)=(𝑠+𝑡)^𝑑(𝐸)𝑡^{𝑛−𝑑(𝐸)},(5.17) a homogeneous polynomial of degree 𝑛.

This leads to an interpretation/derivation of 𝑃_𝑛,ℓ(𝑠,𝑡) as the permanent per(𝑀_𝑛,ℓ).

Theorem 5.5. One has the expansion in elementary subgraphs 𝑃_𝑛,ℓ(𝑠,𝑡)=_𝐸2^𝑐(𝐸)wt(𝐸).(5.18)

Proof. Assign weights to the components of 𝐸 as follows:
each distinguished isolated vertex will have weight 𝑠+𝑡;
each undistinguished isolated vertex will have weight 𝑡;
each isolated edge will have weight 𝑡²;
and each 𝑗-cycle, 𝑗≥3, will have weight 𝑡^𝑗.
To obtain wt(𝐸) in agreement with (5.17), we form the product of these weights over all components in 𝐸. The proof then follows along the lines of Proposition 7.2 of [3, page 44], slightly modified to incorporate isolated vertices and with determinant, “det,” replaced by permanent, “per,” ignoring the minus signs. Effectively, each term in the permanent expansion thus corresponds to a weighted elementary subgraph 𝐸 of the weighted 𝐾^𝑛(ℓ).

See Figure 2 for an example with 𝑛=3.

Figure 2 

𝑀3,ℓ, 𝐾3(ℓ), 𝑃3,ℓ, and the 5 weighted elementary subgraphs of 𝐾3(ℓ) for ℓ=0,1,2,3. Distinguished vertices are shown in bold.

5.3. Associated Polynomials and Some Asymptotics

Thinking of 𝑠 and 𝑡 as parameters, we define the associated polynomials 𝑄_𝑛(𝑥)=^𝑛_ℓ=0ℓ𝑛𝑥^ℓ𝑃_𝑛,ℓ.(5.19) As in the proof of (3) above, using the integral formula (3.12), we have𝑄_𝑛(𝑥)==^∞0(𝑠+𝑡𝑦+𝑥𝑡𝑦)^𝑛𝑒^−𝑦𝑑𝑦_𝑗𝑛𝑗𝑠^𝑗=𝑛!_𝑗(1+𝑥)^𝑛−𝑗𝑡^𝑛−𝑗(𝑛−𝑗)!𝑠^𝑗(1+𝑥)^𝑛−𝑗𝑡^𝑛−𝑗𝑗!.(5.20) Comparing with (5.2), we have the following.

Proposition 5.6. Consider 𝑄_𝑛(𝑥)=^𝑛_ℓ=0ℓ𝑛𝑥^ℓ𝑃_𝑛,ℓ(𝑠,𝑡)=𝑃_𝑛,0(𝑠,𝑡+𝑥𝑡).(5.21)

And one has the following.

Proposition 5.7. As 𝑛→∞, for 𝑥≠−1, 𝑄_𝑛(𝑥)∼𝑡^𝑛(1+𝑥)^𝑛𝑛!𝑒^{𝑠/(𝑡+𝑡𝑥)},(5.22) with the special cases 𝑄_𝑛𝑄_𝑛(−1)=𝑠^𝑛,𝑄_𝑛(0)=𝑃_𝑛,0∼𝑡^𝑛𝑛!𝑒^𝑠/𝑡,(1)=_ℓℓ𝑛𝑃^𝑛,ℓ∼(2𝑡)^𝑛𝑛!𝑒^𝑠/(2𝑡).(5.23)

Proof. From (5.20), 𝑄_𝑛(𝑥)=𝑛!_𝑗𝑠^𝑗(1+𝑥)^𝑛−𝑗𝑡^𝑛−𝑗=𝑡_𝑛𝑗!(1+𝑥)^𝑛𝑛!^𝑛_𝑗=01𝑗!1𝑠+/𝑡𝑥^𝑗,(5.24) from which the result follows.

6. Generalized Derangement Numbers

The formula (2.19) is suggestive of the derangement numbers (see, e.g., [4, page 180]), 𝑑_𝑛=𝑛!^𝑛_𝑗=0(−1)^𝑗𝑗!.(6.1) This leads to the following.

Definition 6.1. A family of numbers, depending on 𝑛 and ℓ, arising as the values of 𝑃_𝑛,ℓ(𝑠,𝑡) when 𝑠 and 𝑡 are assigned fixed integer values, are called generalized derangement numbers.

We have seen that the assignment 𝑠=−1,𝑡=1 produces the usual derangement numbers when ℓ=0. In this section, we will examine in detail the cases 𝑠=−1,𝑡=1, generalized derangements, and 𝑠=𝑡=1, generalized arrangements.

Remark 6.2. Topics related to this material are discussed in Riordan [5]. The paper [6] is of related interest as well.

6.1. Generalized Derangements of [𝑛]

To start, define 𝐷_𝑛,ℓ=𝑃_𝑛,ℓ(−1,1).(6.2) Equation (5.9) and Proposition 3.2 give𝐷_𝑛,ℓ=^𝑛_𝑗=ℓ(−1)^𝑛−𝑗𝑛𝑛−−ℓ𝑗𝑗!=(−1)^𝑛−ℓℓ!₂𝐹₀ℓ−𝑛-,--1+ℓ||||1.(6.3) Equation (5.2) reads as per(𝐽−𝐼)=𝐷_𝑛,0=𝑑_𝑛,(6.4) the number derangements of [𝑛]. So we have a combinatorial interpretation of 𝐷_𝑛,0.

6.1.1. Combinatorial Interpretation of 𝐷_𝑛,ℓ

We now give a combinatorial interpretation of 𝐷_𝑛,ℓ for ℓ≥1.

When ℓ≥1, recurrence (5.3) for 𝑃_𝑛,ℓ(−1,1) gives𝐷_𝑛,ℓ=𝐷_𝑛,ℓ−1+𝐷_{𝑛−1,ℓ−1}.(6.5) We say that a subset I of [𝑛] is deranged by a permutation if no point of I is fixed by the permutation.

Proposition 6.3. 𝐷_𝑛,0=𝑑_𝑛, the number of derangements of [𝑛]. In general, for ℓ≥0, 𝐷_𝑛,ℓ is the number of permutations of [𝑛] in which the set {1,2,…,𝑛−ℓ} is deranged, with no restrictions on the ℓ-set {𝑛−ℓ+1,…,𝑛}.

Proof. For ℓ≥0, let 𝐷^∗𝑛,ℓ denote the set of permutations in the statement of the proposition. Let 𝐸_𝑛,ℓ=|𝐷^∗𝑛,ℓ|. We claim that 𝐸_𝑛,ℓ=𝐷_𝑛,ℓ.
The case ℓ=0 is immediate. We show that 𝐸_𝑛,ℓ satisfies recurrence (6.5).
Now, let ℓ>0. Consider a permutation in 𝐷^∗𝑛,ℓ. The point 𝑛 is either (1) deranged, or (2) not deranged (i.e., fixed). (1)If 𝑛 is deranged, then the (𝑛−ℓ+1)-set {1,2,…,𝑛−ℓ,𝑛} is deranged. By switching 𝑛↔𝑛−ℓ+1 in all permutations of 𝐷^∗𝑛,ℓ, we obtain a permutation in 𝐷^{∗𝑛,ℓ−1}. Conversely, given any permutation of 𝐷^{∗𝑛,ℓ−1}, we switch 𝑛↔𝑛−ℓ+1 to obtain a permutation in 𝐷^∗𝑛,ℓ where 𝑛 is deranged. Hence, the number of permutations in 𝐷^∗𝑛,ℓ with 𝑛 deranged equals 𝐸_𝑛,ℓ−1. (2)Here, 𝑛 is fixed, so if we remove 𝑛 from any permutation in 𝐷^∗𝑛,ℓ we obtain a permutation in 𝐷^{∗𝑛−1,ℓ−1}. Conversely, given a permutation in 𝐷^{∗𝑛−1,ℓ−1}, we may include 𝑛 as a fixed point to obtain a permutation in 𝐷^∗𝑛,ℓ with 𝑛 fixed. Hence, the number of permutations in 𝐷^∗𝑛,ℓ with 𝑛 fixed equals 𝐸_{𝑛−1,ℓ−1}. Combining the above two paragraphs shows that 𝐸_𝑛,ℓ satisfies recurrence (6.5).

And a quick check, 𝐷_𝑛,𝑛=𝑛!,(6.6) there being no restrictions at all in the combinatorial interpretation, in agreement with (6.3) for ℓ=𝑛.

Example 6.4. When 𝑛=3, we have 𝑑₃=𝐷_3,0=2 corresponding to the 2 permutations of [1] in which {1,2,3} is moved: 231,312.
Then, 𝐷_3,1=3 corresponding to the 3 permutations of [1] in which {1,2} is moved: 213,231,312.
Then, 𝐷_3,2=4 corresponding to the 4 permutations of [1] in which {1} is moved: 213,231,312,321.
Finally, 𝐷_3,3=3!=6 corresponding to the 3 permutations of [1] in which ∅ is moved: 123,132,213,231,312,321.

Reversing the order of summation in (6.3) gives an alternative expression𝐷_𝑛,ℓ=^𝑛−ℓ_𝑗=0(−1)^𝑗𝑛−𝑗ℓ(𝑛−𝑗)!.(6.7)

Remark 6.5. Formulation (6.7) may be proved directly by inclusion-exclusion on permutations fixing given points.

Example 6.6. Consider 𝐷_5,2=³_𝑗=0(−1)^𝑗=120−72+3𝑗18−(52−=𝑗)6!4=.305!−314!+323!−332!(6.8)

Now, from (2) of Theorem 5.3, 𝑠=−1, and 𝑡=1, we have𝐷_𝑛,ℓ=^ℓ_𝑗=0ℓ𝑗𝑑^𝑛−𝑗.(6.9) Here is a combinatorial explanation. To obtain a permutation in 𝐷^∗𝑛,ℓ, we first choose 𝑗 points from {𝑛−ℓ+1,…,𝑛} to be fixed. Then, every derangement of the remaining (𝑛−𝑗) points will produce a permutation in 𝐷^∗𝑛,ℓ, and there are 𝑑_𝑛−𝑗 of such derangements.

Example 6.7. Consider 𝐷_5,2=²_𝑗=02𝑗𝑑^5−𝑗=20𝑑₅+21𝑑₄+22𝑑₃=1×44+2×9+1×2=44+18+2=64.(6.10)

6.1.2. Permanents from 𝐽−𝐼

Theorem 4.1 specializes to (𝐽−𝐼)^∨ℓ=^{min(ℓ,𝑛−ℓ)}_𝑘=0𝐷_ℓ,𝑘JS^𝑘𝑛ℓ.(6.11) This can be written using the hypergeometric form (𝐽−𝐼)^∨ℓ=^{min(ℓ,𝑛−ℓ)}_𝑘=0(−1)^ℓ−𝑘𝑘!₂𝐹₀𝑘−ℓ-,--1+𝑘||||1JS^𝑘𝑛ℓ,(6.12) with spectrum eigenvalue(−1)^ℓ2𝐹₀occurringwithmultip𝛼li−citℓy,−𝛼𝑛𝛼-+--−𝑛−𝛼ℓ−𝑛+11||||,1,(6.13) by Corollary 4.3 and Proposition 4.2.

The entries of (𝐽−𝐼)^∨ℓ are from the set of numbers 𝐷_𝑛,ℓ. For the spectrum, start with 𝛼=0. From (6.3), we have (−1)^ℓ2𝐹₀−ℓ,𝑛-−--ℓ+1||||1=(𝑛−1ℓ)!𝐷^{𝑛,𝑛−ℓ}.(6.14) As 𝛼 increases, we see that the spectrum consists of the numbers (−1)^𝛼(𝑛−ℓ−𝛼)!𝐷^{𝑛−2𝛼,𝑛−ℓ−𝛼}.(6.15) Think of moving in the derangement triangle, as in the appendix, starting from position 𝑛,𝑛−ℓ, rescaling the values by the factorial of the column at each step, then the eigenvalues are found by successive knight's moves, up 2 rows and one column to the left, with alternating signs.

Example 6.8. For 𝑛=5, ℓ=3, we have (𝐽−𝐼)^∨3=⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣2333343344323343343433243343343342333443343323434343333244333343442333343434323343344333234443333332⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦,(6.16) with characteristic polynomial 𝜆⁵(𝜆−32)(𝜆+3)⁴.(6.17)

Remark 6.9. Except for ℓ=2, the coefficients in the expansion of (𝐽−𝐼)^∨ℓ in the Johnson basis will be distinct. Thus, the Johnson basis itself can be read off directly from (𝐽−𝐼)^∨ℓ. In this sense, the centralizer algebra of the action of the symmetric group on ℓ-sets is determined by knowledge of the action of just 𝐽−𝐼 on ℓ-sets.

6.2. Generalized Arrangements of [𝑛]

Given [𝑛], 0≤𝑗≤𝑛, a 𝑗-arrangement of [𝑛] is a permutation of a 𝑗-subset of [𝑛]. The number of 𝑗-arrangements of [𝑛] is 𝐴(𝑛,𝑗)=𝑛!(𝑛−𝑗)!.(6.18) Note that there is a single 0-arrangement of [𝑛], from the empty set.

Define 𝐴_𝑛,ℓ=𝑃_𝑛,ℓ(1,1). So, similar to the case for derangements, (5.9) gives𝐴_𝑛,ℓ=^𝑛_𝑗=ℓ𝑛𝑛−−ℓ𝑗𝑗!=ℓ!²𝐹₀ℓ−𝑛-,--1+ℓ||||−1.(6.19) Now, define 𝑎_𝑛=𝐴_𝑛,0, so 𝑎_𝑛=per(𝐼+𝐽)=^𝑛_𝑗=0𝑛!(𝑛−𝑗)!=^𝑛_𝑗=0𝐴(𝑛,𝑗)(6.20) is the total number of j-arrangements of [𝑛] for 𝑗=0,1,…,𝑛. Thus, we have a combinatorial interpretation of 𝐴_𝑛,0.

6.2.1. Combinatorial Interpretation of 𝐴_𝑛,ℓ

We now give a combinatorial interpretation of 𝐴_𝑛,ℓ for ℓ≥1.

When ℓ≥1, recurrence (5.3) for 𝑃_𝑛,ℓ(1,1) gives𝐴_𝑛,ℓ=𝐴_𝑛,ℓ−1−𝐴_{𝑛−1,ℓ−1}.(6.21)

Proposition 6.10. 𝐴_𝑛,0=𝑎_𝑛, the total number of arrangements of [𝑛]. In general, for ℓ≥0, 𝐴_𝑛,ℓ is the number of arrangements of [𝑛] which contain {1,2,…,ℓ}.

Proof. For ℓ≥0, let 𝐴^∗𝑛,ℓ denote the set of arrangements of [𝑛] which contain [ℓ]. With [0]=∅, we note that 𝐴^∗𝑛,0 is the set of all arrangements. Let 𝐵_𝑛,ℓ=|𝐴^∗𝑛,ℓ|. We claim that 𝐵_𝑛,ℓ=𝐴_𝑛,ℓ.
The initial values with ℓ=0 are immediate. We show that 𝐵_𝑛,ℓ satisfies recurrence (6.21).
Consider 𝐴^{∗𝑛,ℓ−1}. Let 𝐴∈𝐴^{∗𝑛,ℓ−1}, so 𝐴 is an arrangement of [𝑛] containing [ℓ−1]. If ℓ=1, then 𝐴∈𝐴^∗𝑛,0 is any arrangement. Now, either ℓ∈𝐴 or ℓ∉A.
If ℓ∈𝐴, then 𝐴∈𝐴^∗𝑛,ℓ, and so the number of arrangements in 𝐴^{∗𝑛,ℓ−1} which contain ℓ equals 𝐵_𝑛,ℓ.
If ℓ∉A, then by subtracting 1 from all parts of 𝐴 which are ≥ℓ+1, we obtain an arrangement of [𝑛−1] which contains [ℓ−1], that is, an arrangement in 𝐴^{∗𝑛−1,ℓ−1}. Conversely, given an arrangement in 𝐴^{∗𝑛−1,ℓ−1}, adding 1 to all parts ≥ℓ yields an arrangement in 𝐴^{∗𝑛,ℓ−1} which does not contain ℓ. Hence, the number of arrangements in 𝐴^{∗𝑛,ℓ−1} which do not contain ℓ equals 𝐵_{𝑛−1,ℓ−1}.
We conclude that 𝐵_𝑛,ℓ−1=𝐵_𝑛,ℓ+𝐵_{𝑛−1,ℓ−1}; hence, This is the result.

Example 6.11. When 𝑛=3, we have 𝑎₃=𝐴_3,0=16 corresponding to the 16 arrangements of [1]: [],1,2,3,12,21,13,31,23,32,123,132,213,231,312,321.
Then, 𝐴_3,1=11 corresponding to the 11 arrangements of [1] which contain {1}: 1,12,21,13,31,123,132,213,231,312,321.
Then, 𝐴_3,2=8 corresponding to the 8 arrangements of [1] which contain {1,2}: 12,21,123,132,213,231,312,321.
Finally, 𝐴_3,3=3!=6 corresponding to the 6 arrangements of [1] which contain {1,2,3}: 123,132,213,231,312,321.

Rearranging the factors in (5.9), we have 𝑃_𝑛,ℓ(𝑠,𝑡)=^𝑛_𝑗=ℓ𝐴(𝑗,ℓ)𝐴(𝑛−ℓ,𝑗−ℓ)𝑠^𝑛−𝑗𝑡^𝑗,(6.22) With 𝑠=𝑡=1, this gives𝐴_𝑛,ℓ=^𝑛_𝑗=ℓ𝐴(𝑗,ℓ)𝐴(𝑛−ℓ,𝑗−ℓ).(6.23) Here is a combinatorial explanation of (6.23).

For any 𝑗≥ℓ, to obtain a 𝑗-arrangement 𝐴 of [𝑛] containing [ℓ], we may place the ℓ points of {1,2,…,ℓ} into these j positions in 𝐴(𝑗,ℓ) ways. Then, the remaining (𝑗−ℓ) positions in 𝐴 can be filled in by a (𝑗−ℓ)-arrangement of the unused (𝑛−ℓ) points in 𝐴(𝑛−ℓ,𝑗−ℓ) ways.

Example 6.12. Consider 𝐴_5,2=⁵_𝑗=2===𝐴22×+(2𝐴11,82(+𝑗)+,62𝐴7)×2(𝐴33+,+(031),21+𝑗02−𝐴×=2(623)1+,22.2)0𝐴×(63,1)+𝐴(4,2)𝐴(3,2)+𝐴(5,2)𝐴(3,3)(6.24) Finally, from (2) of Theorem 5.3, 𝑠=1, and 𝑡=1, we have 𝐴_𝑛,ℓ=^ℓ_𝑗=0(−1)^𝑗ℓ𝑗𝑎^𝑛−𝑗.(6.25)

Example 6.13. Consider 𝐴_5,2=²_𝑗=0(−1)^𝑗2𝑗𝑎^5−𝑗=20𝑎₅−21𝑎₄+22𝑎₃=1×326−2×65+1×16=326−130+16=212.(6.26)

6.2.2. Permanents from 𝐼+𝐽

Theorem 4.1 specializes to (𝐼+𝐽)^∨ℓ=^{min(ℓ,𝑛−ℓ)}_𝑘=0𝐴_ℓ,𝑘JS^𝑘𝑛ℓ.(6.27) This can be written using the hypergeometric form (𝐼+𝐽)^∨ℓ=^{min(ℓ,𝑛−ℓ)}_𝑘=0𝑘!₂𝐹₀𝑘−ℓ-,--1+𝑘||||−1JS^𝑘𝑛ℓ,(6.28) with spectrum eigenvalue₂𝐹₀occurringwithmu𝛼lti−plℓic,i−ty𝛼-+-𝛼𝑛-𝑛−−ℓ+𝛼−1𝑛||||1−1,,(6.29) by Corollary 4.3 and Proposition 4.2.

Example 6.14. For 𝑛=5, ℓ=3, we have (𝐼+𝐽)^∨3=⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣11111118886111111111111188816111111111111888116111111111118881161111111111188811161111111111888111161111111118881116111111111188811111161111111888161111111111118881161111⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦,(6.30) with characteristic polynomial (𝜆−106)(𝜆−11)⁴(𝜆−2)⁵.(6.31) As for the case of derangements, the Johnson basis can be read off directly from the matrix (𝐼+𝐽)^∨ℓ.

Appendix

Generalized Derangement Numbers and Integer Sequences
The first two columns of the 𝐷_𝑛,ℓ triangle, 𝐷_𝑛,0 and 𝐷_𝑛,1, give sequences A000166 and A000255 in the On-Line Encyclopedia of Integer Sequences [7]. The comments for A000255 do not contain our combinatorial interpretation.
The first two columns of the 𝐴_𝑛,ℓ triangle, 𝐴_𝑛,0 and 𝐴_𝑛,1, give sequences A000522 and A001339. The comments contain our combinatorial interpretation. The next two columns, 𝐴_𝑛,2 and 𝐴_𝑛,3, give sequences A001340 and A00134; here, our combinatorial interpretation is not mentioned in the comments.

Generalized Derangement Triangles
ℓ=0 is the leftmost column. The rows correspond to n from 0 to 9. Values of 𝐷_𝑛,ℓ
  ⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣1113244836810129454534936112436181501130613139892791162385164002468044220186612284117730006228898630242223054925200000061044265462232671790220000020040080023457036290000002306020023585070200000042808003420230000000052600362000000000880⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦.(A.1)Values of 𝐴_𝑛,ℓ
  ⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣19110833169962712565564607010108911752166149013673819014101397818142003310021189875210082867819147316000760405744285606672195522408300084744040445565187672004700004002000045597179000210006602080044545080000300045606004400300003000022000362000000000880⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦.(A.2)

Exponential polynomials ℎ_𝑛,𝑚(𝑠,𝑡)
Note that, as is common for matrix indexing, we have dropped the commas in the numerical subscripts𝑛=0ℎ₀₀=1,ℎ₀₁=𝑡,ℎ₀₂=2𝑡²,ℎ₀₃=6𝑡³,ℎ₀₄=24𝑡⁴,(A.3)𝑛=1ℎ₁₀=𝑠+𝑡,ℎ₁₁=𝑠𝑡+2𝑡²,ℎ₁₂=2𝑠𝑡²+6𝑡³,ℎ₁₃=6𝑠𝑡³+24𝑡⁴,ℎ₁₄=24𝑠𝑡⁴+120𝑡⁵,(A.4)𝑛=2ℎ₂₀=𝑠ℎ²+₂₃2𝑠𝑡+2𝑡²,ℎ₂₁=𝑠²𝑡+4𝑠𝑡²+6𝑡³,ℎ₂₂=2𝑠²𝑡²+12𝑠𝑡³+24𝑡⁴,=6𝑠²𝑡³+48𝑠𝑡⁴+120𝑡⁵,ℎ₂₄=24𝑠²𝑡⁴+240𝑠𝑡⁵+720𝑡⁶,(A.5)𝑛=3ℎ₃₁ℎ₃₀=𝑠³+3𝑠²𝑡+6𝑠𝑡²+6𝑡³,ℎ₃₃=𝑠³𝑡+6𝑠²𝑡²+18𝑠𝑡³+24𝑡⁴,ℎ₃₂=2𝑠³𝑡²+18𝑠²𝑡³+72𝑠𝑡⁴+120𝑡⁵,=6𝑠³𝑡³+72𝑠²𝑡⁴+360𝑠𝑡⁵+720𝑡⁶,ℎ₃₄=24𝑠³𝑡⁴+360𝑠²𝑡⁵+2160𝑠𝑡⁶+5040𝑡⁷,(A.6)𝑛=4ℎ₄₀=𝑠⁴+4𝑠³𝑡+1ℎ2𝑠²₄₂𝑡²+24𝑠𝑡³+24𝑡⁴,ℎ₄₁=𝑠⁴𝑡+8𝑠³𝑡²+36𝑠²𝑡³+96𝑠𝑡⁴+120𝑡⁵,ℎ₄₃=2𝑠⁴𝑡²+24𝑠³𝑡³+144𝑠²𝑡⁴+480𝑠𝑡⁵+720𝑡⁶,ℎ₄₄=6𝑠⁴𝑡³+96𝑠³𝑡⁴+720𝑠²𝑡⁵+2880𝑠𝑡⁶+5040𝑡⁷,=24𝑠⁴𝑡⁴+480𝑠³𝑡⁵+4320𝑠²𝑡⁶+20160𝑠𝑡⁷+40320𝑡⁸.(A.7)

Acknowledgment

The authors would like to thank Stacey Staples for discussions about zeons and trace formulas.