Basics of the theory of cyclic rook polynomials and cyclic permanents of rectangular matrices

Abstract

Combinatorial formulas expressing cyclic rook polynomials and cyclic permanents of rectangular matrices in terms of expansions along rows are presented Comment: 3 pages, no figures

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arXiv:0904.4109v2 [math.CO] 16 Jul 2009
Basics of the theory of cyclic rook polynomials
and cyclic permanents of rectangular matrices
A. M. Kamenetskii
Let A be a k × m matrix, n ≤ k, and let j(i),j(i)
Then, by definition, A[Nn|j(i)⊕?pi
l
column is A[Nn|j(i)⊕?pi
Theorem 1 Let 0 ≤ r ≤ n, n ≥ 1, k ≥ 1, t ≥ 0. Then
1,...,j(i)
pi ∈ Nm, 1 ≤ i ≤ q.
l=1A[Nn|j(i)
l=1⊕j(i)
l=1⊕j(q)
l].
l] = A[Nn|j(i)]+?pi
l
l] and
A[Nn|j(1)⊕?p1
l=1⊕j(1)
,...,j(q)⊕?pq
] is a n × q matrix, whose i th
l=1⊕j(i)
?
t
?
i=0
aiP−r+i
n
?
⊗ Jk
=
??
t
?
i=0
atT(i)
n+t
?
⊗ Jk
?
[Nnk|(k(r + 1) ⊕
?
1≤i≤t
i≡(r+1)modn
⊕k(n + i))?k?,
(k(r + 2) ⊕
?
1≤i≤t
i≡(r+2)modn
⊕k(n + i))?k?,...,(kn ⊕
?
1≤i≤t
i≡nmodn
⊕k(n + i))?k?,
(k ⊕
?
i≡1modn
?
i≡rmodn
1≤i≤t
⊕k(n + i))?k?,(2k ⊕
?
i≡2modn
1≤i≤t
⊕k(n + i))?k?,...,
(kr ⊕
1≤i≤t
⊕k(n + i))?k?].
An injective mapping ϕ of a set A induces the following bijective mapping ϕ∗
of the set ϕ(A) \ (ϕ(A) ∩ A) onto A \ (A ∩ ϕ(A)): if j ∈ ϕ(A) \ (ϕ(A) ∩ A),
then ϕ∗(j) = jm, where jmis uniquely determined by the conditions j0= j and
j1,j2,...,jm−1∈ ϕ(A), jl= ϕ−1(jl−1), 1 ≤ l ≤ m, jm∈ A \ (ϕ(A) ∩ A). Let
¯ α = (x1,...,xn) be a sequence, A ⊆ Nn, ϕ ∈ Inj(A,Nn). Denote by ¯ ϕ(¯ α) the
sequence obtained from ¯ α in the following way. The components xj, j ∈ ϕ(A),
are deleted. And if j ∈ ϕ(A) \ (ϕ(A) ∩ A), then the deleted component xj is
replaced by xϕ∗(j).
Theorem 2 (Expansion of the cyclic rook polynomial of a rectangular
matrix along the last k rows) Let A = (ai,j) 1≤i≤m
a commutative ring with unity, m ≤ n, 1 ≤ k ≤ m − 1. Then
1≤j≤nbe an m×n matrix over
R(x;z;A)=
?
S⊆Nm\Nm−k,S?=∅
+R(x;z;A[Nm−k|Nn]).
?
ϕ∈Inj(S,Nn)
z|ϕ|(
?
i∈S
ai,ϕ(i))R(x;z;A[Nm−k|¯ ϕ(Nn)]x|S|
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If k > 1, then such an expansion along arbitrary k rows, for an arbitrary
matrix, does not exist. On the other hand, if k = 1, then such an expansion
along arbitrary row exists.
Theorem 3 Let A = (ai,j) 1≤i≤m
1≤j≤nand m ≤ n, 1 ≤ i ≤ m. Then
R(x;z;A)=
ai,ixzR(x;z;A[Nm\ (i)|Nn\ (i)]
?
+R(x;z;A[Nm\ (i)|Nm\ (i),i,m + 1,...,n].
+
j∈Nn\(i)
ai,jxR(x;z;A[Nm\ (i)|1,2,...,i − 1,i + 1,...,j − 1,i,j + 1,...,n]
Theorem 4 (Expansion of the cyclic rook polynomial along arbitrary k
rows) Let A = (ai,j) 1≤i≤m
1≤j≤n
m ≤ n, 1 ≤ k ≤ m − 1,¯β ∈ Qk,m. Then
per(z;A) =
?
ϕ∈Inj({¯β},Nn)
z|ϕ|(
?
i∈{¯β}
ai,ϕ(i))per(z;A[Nm\¯β|¯ ϕ(Nn)].
Theorem 5 Let A = (ai,j) and B = (bi,j) be m×n matrices over a commutative
ring with unity, m ≤ n, R(x;z;A) =?m
m
?
l?
m
?
m
?
¯
β∈Qs,n
l=0rl(z;A)xl, r0(z;A) = 1, R(x;A) =
?m
R(x;z;A + B)
l=0rl(A)xl, r0(A) = 1. Then
=
s=0
?
¯ α∈Qs,m
?
ϕ∈Inj({¯ α},Nn)
z|ϕ|

?

i∈{¯ α}
i∈{¯ α}
ai,ϕ(i)

xsR(x;z;B[Nm\ ¯ α|¯ ϕ(Nn)]);

rl(z;A + B)=
s=0
?
¯ α∈Qs,m
?
ϕ∈Inj({¯ α},Nn)
z|ϕ|
?

i∈{¯ α}
ai,ϕ(i)
rl−s(z;B[Nm\ ¯ α|¯ ϕ(Nn)]);

per(z;A + B)=
s=0
?
¯ α∈Qs,m
?
ϕ∈Inj({¯ α},Nn)
z|ϕ|
?
ai,ϕ(i)
per(z;B[Nm\ ¯ α|¯ ϕ(Nn)]);
R(x;A + B)=
s=0
?
¯ α∈Qs,m
per(A[¯ α|¯β])xsR(x;B(¯ α|¯β));
rl(A + B)=
l?
s=0
?
β∈Qs,n
¯ α∈Qs,m
¯
per(A[¯ α|¯β])rl−s(B(¯ α|¯β));
per(A + B)=
m
?
s=0
?
β∈Qs,n
¯ α∈Qs,m
¯
per(A[¯ α|¯β])per(B(¯ α|¯β)).
Theorem 6 Let A be an m × n matrix over a commutative ring with unity,
m ≤ n, R(x;z;A) =?m
l?
l=0rl(z;A)xl, r0(z;A) = 1, (z)(k)=?k−1
?m − s
i=0(z+i), k ≥ 1,
(z)(0)= 1. Then
rl(z;yJm,n−A) =
s=0
(−1)s
l − s
?
rs(z;A)(n−l+z)(l−s)yl−s,
0 ≤ l ≤ m.
2

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In particular,
per(z;yJm,n− A) =
m
?
s=0
(−1)srs(z;A)(n − m + z)(m−s)ym−s.
Theorem 7
per(z;(a0In+ a1Pn) ⊗ Jk) =
k
?
s=0
(z)(s)
s!
[s!
?k
s
?
ak−s
0
as
1(s + z)(k−s)]n
for all n ≥ 1.
Theorem 8 Let r ≥ t + 1. Then for all n ≥ 1 we have
R(x;z;(
t
?
i=0
aiP−r+i
n
)⊗Jk) =
?
(l1,...,lt)
0≤li≤k,1≤i≤t
?
M
f(z;
?v?l1?
1
,v?l2?
2
¯ γ \ (1?k−l1?,2?k−l2?,...,t?k−lt?)
,...,v?lt?
t
,v?k?
t+1,v?k?
t+2,...,v?k?
r
?
·
(K?0,r?
[k],t(a0x,a1x,...,atx;z))n[L(1?k−l1?,v?l1?
1
,2?k−l2?,v?l2?
2
,...,t?k−lt?,v?lt?
t
,v?k?
t+1,v?k?
t+2,...,v?k?
r )|L(¯ γ)],
where the summation is carried over the domain M defined by the conditions
¯ γ
∈∪rk
d=0G?0,r?
{1?k−l1?,v?l1?
[k],d,t,
{¯ γ} \ ({¯ γ} ∩ {w?rk?})=
1
,2?k−l2?,v?l2?
2
,...,t?k−lt?,v?lt?
t
,v?k?
t+1,v?k?
t+2,...,v?k?
r }.
Lemma 1 Let ¯ α = (x1,...,xm) and ¯ γ = (y1,...,yn) be two sequences, m ≤ n,
{¯ α} ⊆ {¯ γ}, {¯ α} = {b?l1?
1
,...,b?ld?
d
}, Ai= {j ∈ Nm|xj= bi}, Bi= {j ∈ Nn|yj=
bi}, 1 ≤ i ≤ d. Then, for all i,1 ≤ i ≤ d, we have
f(z;
?¯ α
¯ γ
?
) = (
?|Bi|
|Ai|
?
(|Ai|!))−1
?
ϕ∈Inj(Ai,Bi)
z|ϕ|f(z;
?¯ α \ (b?li?
i
)
¯ ϕ(¯ γ)
?
).
The author expresses his deep gratitude to S. K.Lando for invaluable support
and attention to the work.
References
[1] A. M. Kamenetskii, Russian Mathematical Surveys, 60:3 (2005), 177–178
[2] A. M. Kamenetskii, Russian Mathematical Surveys, 62:6 (2007), 175–176
[3] A. M. Kamenetskii, Russian Mathematical Surveys, 63:4 (2008), 187–188
e-mail:kamenetsky a@list.ru
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