LEVEL EULERIAN POSETS 1. Introduction It is the instinct of every ...

22 RICHARD EHRENBORG, GÁBOR HETYEI, AND MARGARET READDY
Proof. Let Γ be a solution to the two equations (8.2) and (8.3). Write Γ as the sum
m≥0 Γm where
the entries of the matrix Γm are homogeneous of degree m. By induction on m we will prove that Γm
is equal to Ψm, the mth homogeneous component of the matrix Ψ. The base case m = 0 is as follows.
Γ0 = Γ| a=b=0 = K(t)| t=0 = M = Ψ0.
Now assume the statement is true for all values less than m. Observe that the mth component of
equation (8.3) is
∆(Γm) =
m−1
i=0
Γi · t · Γm−1−i =
m−1
i=0
Ψi · t · Ψm−1−i = ∆(Ψm).
Hence the difference Γm − Ψm is a constant matrix N times the ab-polynomial (a − b) m . However,
the matrix N is zero since Γm| b=0 = Ψm| b=0 , proving that Γm is equal to Ψm, completing the
induction.
Example 8.2. Consider the level Eulerian poset in Figure 1. We claim that its cd-series matrix Ψ is
given by
⎛
1
1−c−d
⎜ 1
⎜ 1−c−d
Ψ = ⎜ 1
⎝
c · 1−c−d
1
1−c−d · c + 1
1
1−c−d · c
1 c · 1−c−d
1
1−c−d
1
1−c−d
1
1−c−d − 1
1
1−c−d
· c + 1 c · 1
1−c−d
⎞
⎟ .
1 c · 1−c−d + 1⎟
⎠
1
1−c−d
1
1−c−d · c
1
1−c−d
1
1−c−d
It is straightforward to check the first condition in Theorem 8.1:
⎛
1
1
1
1−t 1−t 1−t
⎜ 1 1
1
⎜ 1−t 1−t − 1 1−t
Ψ| a=t,b=0 = Ψ| c=t,d=0 = ⎜ 1
1
1
⎝ 1−t − 1 1−t − t 1−t − 1
⎞
1
1−t − 1
⎟
1 ⎟
1−t ⎟ = K(t).
1 ⎟
1−t ⎠
1
1−t
1
1−t − 1 1
1−t
1
1−t
To verify the second condition, define the four vectors
⎛ ⎞ ⎛ ⎞
1
0
⎜
x = ⎜1⎟
⎜
⎟
⎝c⎠
, y = ⎜0
⎟
⎝1⎠
1
0
, z = 1 c 1 1
and the matrix
⎛
0
⎜
A = ⎜0
⎝0
1
0
1
0
0
0
⎞
−1
0 ⎟
1 ⎠
0 0 0 0
.
We have the following relations between this matrix and these vectors:
and w = 0 1 0 0 ,
z · t · x = 2 · t + ct + tc = ∆(c + d), A · x = 2 · y, z · A = 2 · w and A · t · A = 0.