3.4 Connecting Operator
3.4.2 Lower Bound of Entropy
In this subsection, the connecting operator Cˆx;m3;m1m2 is employed to esti-mate the lower bound of entropy and in particular, to verify the positivity of entropy.
Definition 3.22. Let X = (X1,· · · , XM)t, where Xk are N × N matrices.
Define the summation of Xk by
|X| =
Note that, (3.4.42) implies
|Xx;mˆ 1,m2,m3;α| =
2(m1−1)m2X
k=1
A(k)x;mˆ 1,m2,m3;α= Aˆx;m1,m2,m3;α. (3.4.44)
As usual, the set of all matrices with the same order can be partially ordered.
114 Pattern Generation Problems Definition 3.23. Let M = [Mij] and N = [Nij] be two M × M matrices, M≥ N if Mij ≥ Nij for all 1 ≤ i, j ≤ M.
Notably, if A2 ≥ A′2 then An ≥ A′n for all n ≥ 2. Therefore, h(A2) ≥ h(A′2). Hence, the spatial entropy as a function of A2 is monotonic with respect to the partial order ≥.
Definition 3.24. A P + 1 multiple index αp ≡ (α1α2· · · αpαP+1) (3.4.45)
is called a periodic cycle if
αP+1 = α1. (3.4.46)
It is called diagonal cycle if (3.4.46) holds and αi ∈ {2m2(i−1)+i|1≤i≤22m2} (3.4.47)
for each 1 ≤ i ≤ P + 1. For a diagonal cycle (3.4.45)
¯
αP = α1; α2; · · · ; αP (3.4.48)
and
¯
αPn= ¯αP; ¯αP; · · · ; ¯αP. (n-times) First, prove the following Lemma.
Lemma 3.25. Let m1 ≥ 2, m2 ≥ 2, P ≥ 1, αP be a diagonal cycle. Then, for any m3 ≥ 1,
ρ(Amxˆ;2×m1 2×(m3P+2)) (3.4.49)
≥ ρ(|(Sx;mˆ 3;m1m2;α1α2Sˆx;m3;m1m2;α2α3· · · Sx;mˆ 3;m1m2;αPαP+1)m3Xx;mˆ 1,m2,2;α1|).
Proof. Since αP is a periodic cycle, Theorem 3.21 implies Xˆx;m1,m2,m3P+2; ¯αPm1
(3.4.50)
= (Sx;mˆ 3;m1m2;α1α2Sx;mˆ 3;m1m2;α2α3· · · Sˆx;m3;m1m2;αPαP+1)nXx;mˆ 1,m2,2;α1. (3.4.51)
Furthermore αP is diagonal and |Xx;mˆ 1,m2,m3P+2; ¯αPm1| = Ax;mˆ 1,m2,m3P+2; ¯αPm1
lies in the diagonal part of (3.4.41) with m3+ P = m3P + 2, therefore ρ(Amx;mˆ1 1,m2,m3P+2) ≥ ρ(|Xˆx;m1,m2,m3P+2; ¯αPm1|).
(3.4.52)
Therefore, (3.4.49) follows from (3.4.50) and (3.4.52). The proof is complete.
3.4. CONNECTING OPERATOR 115 The following Lemma is valuable in studying maximum eigenvalue of (3.4.52).
Lemma 3.26. For any m1 ≥ 2, m2 ≥ 2, 1 ≤ k ≤ 2(m1−1)m2 and α1 ∈ {(i − 1)2m2 + i|1 ≤ i ≤ 2m2}, if
tr(A(k)x;mˆ 1,m2,2;α) = 0, (3.4.53)
then for all 1 ≤ ℓ ≤ 2(m1−1)m2,
(Sx;mˆ 3;m1m2;α1α2)kℓ = 0, (3.4.54)
for all α2 ∈ {(i − 1)2m2 + i|1 ≤ i ≤ 2m2}, i.e., the k-th rows of matrices Sˆx;m3;m1m2;α1α2 are zeros. Furthermore, for any diagonal cycle αP, let U =
(u1u2· · · u2m2(m1−1)) be an eigenvector of Sˆx;m3;m1m2;α1α2Sx;mˆ 3;m1m2;α2α3· · · Sˆx;m3;m1m2;αPα1, if uk 6= 0 for some 1 ≤ k ≤ 2(m1−1)m2, then tr(A(k)x;mˆ 1,m2,2;α1) > 0.
Proof. Since A(k)x;mˆ 1,m2,2;α1 can be expressed as (3.4.33). Therefore, tr(A(k)x;mˆ 1,m2,2;α1) = 0 if and only if (3.4.54) holds for all 1 ≤ ℓ ≤ 2(m1−1)m2. The second part of
the Lemma follows easily from the first part. The proof is complete.
By Lemma 3.25 and Lemma 3.26, the lower bound of entropy can be obtained as follows.
Theorem 3.27. Let α1α2· · · αPα1be a diagonal cycle. Then for any m1 ≥ 2, m2 ≥ 2,
h(Ax;2×2×2) (3.4.55)
≥ 1
m1m2P log ρ(Sx;mˆ 3;m1m2;α1α2Sx;mˆ 3;m1m2;α2α3· · · Sˆx;m3;m1m2;αPα1).
In particular, if a diagonal cycle α1α2· · · αPα1 exists and m1 ≥ 2, m2 ≥ 2
such that ρ(Sx;mˆ 3;m1m2;α1α2Sx;mˆ 3;m1m2;α2α3· · · Sx;mˆ 3;m1m2;αPα1) > 1, then h(Ax;2×2×2) >
0.
Proof. First, by the similar method in the proof of Lemma 2.10 and Lemma 2.11 and Theorem 2.12 in [5] we have
lim sup
m3→∞
1
m3(log ρ(|(Sˆx;m3;m1m2;α1α2Sx;mˆ 3;m1m2;α2α3· · · Sˆx;m3;m1m2;αPα1)nXx;mˆ 1,m2,2;α1|))
= log ρ(Sx;mˆ 3;m1m2;α1α2Sˆx;m3;m1m2;α2α3· · · Sx;mˆ 3;m1m2;αPα1).
(3.4.56) Now, show that h(Ax;2×2×2) ≥ 1
m1m2P lim sup
m3→∞
1
m3(log ρ(|(Sx;mˆ 3;m1m2;α1α2Sx;mˆ 3;m1m2;α2α3· · · Sˆx;m3;m1m2;αPα1)nXx;mˆ 1,m2,2;α1|))
116 Pattern Generation Problems Indeed, from (3.3.18) and (3.4.49),
h(Ax;2×2×2) = lim
m2m3→∞
1 (m3P + 2)m2
log ρ(Aˆx;2×m2×(m3P+2))
= lim
m2m3→∞
1
m1(m3P + 2)m2
log ρ(Amxˆ;2×m1 2×(m3P+2))
≥ 1
m1m2P lim sup
m3→∞
1
m3(log ρ(|(Sˆx;m3;m1m2;α1α2Sx;mˆ 3;m1m2;α2α3· · · Sx;mˆ 3;m1m2;αPα1)nXˆx;m1,m2,2;α1|)).
And by (3.4.56), the proof is complete.
Example 3.28. Consider
Tx;2×2×2 = ⊗(G ⊗ E)2. Then, it is easy to check that
Cˆx;m3;22;11 =
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
.
Therefore,
h(Tx;2×2×2) ≥ log 2 2 .
Moreover, in Proposition3.15it can be shown that h(Tx;2×2×2) = log g where g = 1+2√5.
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