M obius Functions of Posets V: GCD Matrices Bruce Sagan Department - - PowerPoint PPT Presentation
M obius Functions of Posets V: GCD Matrices Bruce Sagan Department of Mathematics Michigan State University East Lansing, MI 48824-1027 sagan@math.msu.edu www.math.msu.edu/ sagan June 28, 2007 Smiths Theorem The Main Theorem Proof
Bruce Sagan Department of Mathematics Michigan State University East Lansing, MI 48824-1027 sagan@math.msu.edu www.math.msu.edu/˜sagan June 28, 2007
Smith’s Theorem The Main Theorem Proof of Smith’s Theorem
Smith’s Theorem The Main Theorem Proof of Smith’s Theorem
Details about this work can be found in GCD matrices, posets, and nonintersecting paths (with E. Altinisik, N. Tuglu), Linear and Multilinear Alg. 53 (2005) 75–84.
Details about this work can be found in GCD matrices, posets, and nonintersecting paths (with E. Altinisik, N. Tuglu), Linear and Multilinear Alg. 53 (2005) 75–84. A copy of this article and related ones can be found at http://www.math.msu.edu/˜sagan
Details about this work can be found in GCD matrices, posets, and nonintersecting paths (with E. Altinisik, N. Tuglu), Linear and Multilinear Alg. 53 (2005) 75–84. A copy of this article and related ones can be found at http://www.math.msu.edu/˜sagan Given i, j ∈ Z, let gcd(i, j) = the greatest common divisor of i and j.
Details about this work can be found in GCD matrices, posets, and nonintersecting paths (with E. Altinisik, N. Tuglu), Linear and Multilinear Alg. 53 (2005) 75–84. A copy of this article and related ones can be found at http://www.math.msu.edu/˜sagan Given i, j ∈ Z, let gcd(i, j) = the greatest common divisor of i and j. We say that i and j are relatively prime if gcd(i, j) = 1.
Details about this work can be found in GCD matrices, posets, and nonintersecting paths (with E. Altinisik, N. Tuglu), Linear and Multilinear Alg. 53 (2005) 75–84. A copy of this article and related ones can be found at http://www.math.msu.edu/˜sagan Given i, j ∈ Z, let gcd(i, j) = the greatest common divisor of i and j. We say that i and j are relatively prime if gcd(i, j) = 1. The Euler phi-function is φ(n) = #{i : 1 ≤ i ≤ n and gcd(i, n) = 1}.
Details about this work can be found in GCD matrices, posets, and nonintersecting paths (with E. Altinisik, N. Tuglu), Linear and Multilinear Alg. 53 (2005) 75–84. A copy of this article and related ones can be found at http://www.math.msu.edu/˜sagan Given i, j ∈ Z, let gcd(i, j) = the greatest common divisor of i and j. We say that i and j are relatively prime if gcd(i, j) = 1. The Euler phi-function is φ(n) = #{i : 1 ≤ i ≤ n and gcd(i, n) = 1}.
Details about this work can be found in GCD matrices, posets, and nonintersecting paths (with E. Altinisik, N. Tuglu), Linear and Multilinear Alg. 53 (2005) 75–84. A copy of this article and related ones can be found at http://www.math.msu.edu/˜sagan Given i, j ∈ Z, let gcd(i, j) = the greatest common divisor of i and j. We say that i and j are relatively prime if gcd(i, j) = 1. The Euler phi-function is φ(n) = #{i : 1 ≤ i ≤ n and gcd(i, n) = 1}.
Details about this work can be found in GCD matrices, posets, and nonintersecting paths (with E. Altinisik, N. Tuglu), Linear and Multilinear Alg. 53 (2005) 75–84. A copy of this article and related ones can be found at http://www.math.msu.edu/˜sagan Given i, j ∈ Z, let gcd(i, j) = the greatest common divisor of i and j. We say that i and j are relatively prime if gcd(i, j) = 1. The Euler phi-function is φ(n) = #{i : 1 ≤ i ≤ n and gcd(i, n) = 1}.
Theorem (H. J. S. Smith, 1876)
Let M be the n × n matrix with Mi,j = gcd(i, j).
Theorem (H. J. S. Smith, 1876)
Let M be the n × n matrix with Mi,j = gcd(i, j). Then det M = φ(1)φ(2) · · · φ(n).
Theorem (H. J. S. Smith, 1876)
Let M be the n × n matrix with Mi,j = gcd(i, j). Then det M = φ(1)φ(2) · · · φ(n).
M = 1 2 3 1 2 3 1 1 1 1 2 1 1 1 3
Theorem (H. J. S. Smith, 1876)
Let M be the n × n matrix with Mi,j = gcd(i, j). Then det M = φ(1)φ(2) · · · φ(n).
M = 1 2 3 1 2 3 1 1 1 1 2 1 1 1 3 det M = 1·2·3+1·1·1+1·1·1−1·1·1−1·1·3−1·2·1
Theorem (H. J. S. Smith, 1876)
Let M be the n × n matrix with Mi,j = gcd(i, j). Then det M = φ(1)φ(2) · · · φ(n).
M = 1 2 3 1 2 3 1 1 1 1 2 1 1 1 3 det M = 1·2·3+1·1·1+1·1·1−1·1·1−1·1·3−1·2·1 = 2.
Theorem (H. J. S. Smith, 1876)
Let M be the n × n matrix with Mi,j = gcd(i, j). Then det M = φ(1)φ(2) · · · φ(n).
M = 1 2 3 1 2 3 1 1 1 1 2 1 1 1 3 det M = 1·2·3+1·1·1+1·1·1−1·1·1−1·1·3−1·2·1 = 2. On the other hand φ(1)φ(2)φ(3)
Theorem (H. J. S. Smith, 1876)
Let M be the n × n matrix with Mi,j = gcd(i, j). Then det M = φ(1)φ(2) · · · φ(n).
M = 1 2 3 1 2 3 1 1 1 1 2 1 1 1 3 det M = 1·2·3+1·1·1+1·1·1−1·1·1−1·1·3−1·2·1 = 2. On the other hand φ(1)φ(2)φ(3) = 1 · 1 · 2 = 2.
Theorem (H. J. S. Smith, 1876)
Let M be the n × n matrix with Mi,j = gcd(i, j). Then det M = φ(1)φ(2) · · · φ(n).
M = 1 2 3 1 2 3 1 1 1 1 2 1 1 1 3 det M = 1·2·3+1·1·1+1·1·1−1·1·1−1·1·3−1·2·1 = 2. On the other hand φ(1)φ(2)φ(3) = 1 · 1 · 2 = 2. Many authors have extended Smith’s Theorem: Apostol,Beslin-Ligh, Bhat, Daniloff, Haukkanen, Haukkanen- Wang-Silanp¨ a¨ a, Jager, Li, Linstr¨
Theorem (H. J. S. Smith, 1876)
Let M be the n × n matrix with Mi,j = gcd(i, j). Then det M = φ(1)φ(2) · · · φ(n).
M = 1 2 3 1 2 3 1 1 1 1 2 1 1 1 3 det M = 1·2·3+1·1·1+1·1·1−1·1·1−1·1·3−1·2·1 = 2. On the other hand φ(1)φ(2)φ(3) = 1 · 1 · 2 = 2. Many authors have extended Smith’s Theorem: Apostol,Beslin-Ligh, Bhat, Daniloff, Haukkanen, Haukkanen- Wang-Silanp¨ a¨ a, Jager, Li, Linstr¨
We will prove a theorem which will have all these other results as special cases.
Theorem (H. J. S. Smith, 1876)
Let M be the n × n matrix with Mi,j = gcd(i, j). Then det M = φ(1)φ(2) · · · φ(n).
M = 1 2 3 1 2 3 1 1 1 1 2 1 1 1 3 det M = 1·2·3+1·1·1+1·1·1−1·1·1−1·1·3−1·2·1 = 2. On the other hand φ(1)φ(2)φ(3) = 1 · 1 · 2 = 2. Many authors have extended Smith’s Theorem: Apostol,Beslin-Ligh, Bhat, Daniloff, Haukkanen, Haukkanen- Wang-Silanp¨ a¨ a, Jager, Li, Linstr¨
We will prove a theorem which will have all these other results as special cases. Furthermore, this theorem is trivial to prove.
Smith’s Theorem The Main Theorem Proof of Smith’s Theorem
Recall that if P is a poset and α ∈ I(P) then there is an associated matrix Mα where Mα
x,y = α(x, y).
Recall that if P is a poset and α ∈ I(P) then there is an associated matrix Mα where Mα
x,y = α(x, y). If the rows and
columns of Mα are indexed by a linear extension L of P, then Mα is upper triangular.
Recall that if P is a poset and α ∈ I(P) then there is an associated matrix Mα where Mα
x,y = α(x, y). If the rows and
columns of Mα are indexed by a linear extension L of P, then Mα is upper triangular. The following is our main theorem.
Recall that if P is a poset and α ∈ I(P) then there is an associated matrix Mα where Mα
x,y = α(x, y). If the rows and
columns of Mα are indexed by a linear extension L of P, then Mα is upper triangular. The following is our main theorem.
Theorem (Altinisik-S-Tuglu)
Let P be a poset and L be a linear extension of P.
Recall that if P is a poset and α ∈ I(P) then there is an associated matrix Mα where Mα
x,y = α(x, y). If the rows and
columns of Mα are indexed by a linear extension L of P, then Mα is upper triangular. The following is our main theorem.
Theorem (Altinisik-S-Tuglu)
Let P be a poset and L be a linear extension of P. Suppose α, β ∈ I(P) and M has rows and columns indexed by L where Mx,y =
α(z, x)β(z, y).
Recall that if P is a poset and α ∈ I(P) then there is an associated matrix Mα where Mα
x,y = α(x, y). If the rows and
columns of Mα are indexed by a linear extension L of P, then Mα is upper triangular. The following is our main theorem.
Theorem (Altinisik-S-Tuglu)
Let P be a poset and L be a linear extension of P. Suppose α, β ∈ I(P) and M has rows and columns indexed by L where Mx,y =
α(z, x)β(z, y). Then det M =
α(z, z)β(z, z).
Recall that if P is a poset and α ∈ I(P) then there is an associated matrix Mα where Mα
x,y = α(x, y). If the rows and
columns of Mα are indexed by a linear extension L of P, then Mα is upper triangular. The following is our main theorem.
Theorem (Altinisik-S-Tuglu)
Let P be a poset and L be a linear extension of P. Suppose α, β ∈ I(P) and M has rows and columns indexed by L where Mx,y =
α(z, x)β(z, y). Then det M =
α(z, z)β(z, z).
Mx,y =
Mα
z,xMβ z,y
Recall that if P is a poset and α ∈ I(P) then there is an associated matrix Mα where Mα
x,y = α(x, y). If the rows and
columns of Mα are indexed by a linear extension L of P, then Mα is upper triangular. The following is our main theorem.
Theorem (Altinisik-S-Tuglu)
Let P be a poset and L be a linear extension of P. Suppose α, β ∈ I(P) and M has rows and columns indexed by L where Mx,y =
α(z, x)β(z, y). Then det M =
α(z, z)β(z, z).
Mx,y =
Mα
z,xMβ z,y =
(Mα)t
x,zMβ z,y
Recall that if P is a poset and α ∈ I(P) then there is an associated matrix Mα where Mα
x,y = α(x, y). If the rows and
columns of Mα are indexed by a linear extension L of P, then Mα is upper triangular. The following is our main theorem.
Theorem (Altinisik-S-Tuglu)
Let P be a poset and L be a linear extension of P. Suppose α, β ∈ I(P) and M has rows and columns indexed by L where Mx,y =
α(z, x)β(z, y). Then det M =
α(z, z)β(z, z).
Mx,y =
Mα
z,xMβ z,y =
(Mα)t
x,zMβ z,y
So M = (Mα)tMβ,
Recall that if P is a poset and α ∈ I(P) then there is an associated matrix Mα where Mα
x,y = α(x, y). If the rows and
columns of Mα are indexed by a linear extension L of P, then Mα is upper triangular. The following is our main theorem.
Theorem (Altinisik-S-Tuglu)
Let P be a poset and L be a linear extension of P. Suppose α, β ∈ I(P) and M has rows and columns indexed by L where Mx,y =
α(z, x)β(z, y). Then det M =
α(z, z)β(z, z).
Mx,y =
Mα
z,xMβ z,y =
(Mα)t
x,zMβ z,y
So M = (Mα)tMβ, implying det M = det Mα det Mβ.
Recall that if P is a poset and α ∈ I(P) then there is an associated matrix Mα where Mα
x,y = α(x, y). If the rows and
columns of Mα are indexed by a linear extension L of P, then Mα is upper triangular. The following is our main theorem.
Theorem (Altinisik-S-Tuglu)
Let P be a poset and L be a linear extension of P. Suppose α, β ∈ I(P) and M has rows and columns indexed by L where Mx,y =
α(z, x)β(z, y). Then det M =
α(z, z)β(z, z).
Mx,y =
Mα
z,xMβ z,y =
(Mα)t
x,zMβ z,y
So M = (Mα)tMβ, implying det M = det Mα det Mβ. By triangularity of Mα, Mβ we have det M =
z α(z, z)β(z, z).
In the Main Theorem, the entries of M are sums, Mx,y =
α(z, x)β(z, y). while the factors of det M are individual terms.
In the Main Theorem, the entries of M are sums, Mx,y =
α(z, x)β(z, y). while the factors of det M are individual terms. In Smith’s Theorem, the entries of M are individual terms, while the factors of det M are sums:
In the Main Theorem, the entries of M are sums, Mx,y =
α(z, x)β(z, y). while the factors of det M are individual terms. In Smith’s Theorem, the entries of M are individual terms, while the factors of det M are sums: if we let S = {i : 1 ≤ i ≤ n and gcd(i, n) = 1} then we have φ(n) = #S
In the Main Theorem, the entries of M are sums, Mx,y =
α(z, x)β(z, y). while the factors of det M are individual terms. In Smith’s Theorem, the entries of M are individual terms, while the factors of det M are sums: if we let S = {i : 1 ≤ i ≤ n and gcd(i, n) = 1} then we have φ(n) = #S =
1.
In the Main Theorem, the entries of M are sums, Mx,y =
α(z, x)β(z, y). while the factors of det M are individual terms. In Smith’s Theorem, the entries of M are individual terms, while the factors of det M are sums: if we let S = {i : 1 ≤ i ≤ n and gcd(i, n) = 1} then we have φ(n) = #S =
1. To switch the role of sum and individual term we need M¨
inversion.
In the Main Theorem, the entries of M are sums, Mx,y =
α(z, x)β(z, y). while the factors of det M are individual terms. In Smith’s Theorem, the entries of M are individual terms, while the factors of det M are sums: if we let S = {i : 1 ≤ i ≤ n and gcd(i, n) = 1} then we have φ(n) = #S =
1. To switch the role of sum and individual term we need M¨
restrictions:
In the Main Theorem, the entries of M are sums, Mx,y =
α(z, x)β(z, y). while the factors of det M are individual terms. In Smith’s Theorem, the entries of M are individual terms, while the factors of det M are sums: if we let S = {i : 1 ≤ i ≤ n and gcd(i, n) = 1} then we have φ(n) = #S =
1. To switch the role of sum and individual term we need M¨
restrictions: α(z, x), β(z, y) = 0 implies z ≤ x and z ≤ y.
In the Main Theorem, the entries of M are sums, Mx,y =
α(z, x)β(z, y). while the factors of det M are individual terms. In Smith’s Theorem, the entries of M are individual terms, while the factors of det M are sums: if we let S = {i : 1 ≤ i ≤ n and gcd(i, n) = 1} then we have φ(n) = #S =
1. To switch the role of sum and individual term we need M¨
restrictions: α(z, x), β(z, y) = 0 implies z ≤ x and z ≤ y. To use M¨
In the Main Theorem, the entries of M are sums, Mx,y =
α(z, x)β(z, y). while the factors of det M are individual terms. In Smith’s Theorem, the entries of M are individual terms, while the factors of det M are sums: if we let S = {i : 1 ≤ i ≤ n and gcd(i, n) = 1} then we have φ(n) = #S =
1. To switch the role of sum and individual term we need M¨
restrictions: α(z, x), β(z, y) = 0 implies z ≤ x and z ≤ y. To use M¨
collapse the two restrictions to one, we specialize to the case of a meet semi-lattice.
A meet (respectively, join) semi-lattice is a poset P where every x, y ∈ P have a meet (respectively, join).
A meet (respectively, join) semi-lattice is a poset P where every x, y ∈ P have a meet (respectively, join).
Proposition
(a) A finite meet semi-lattice having a ˆ 1 is a lattice.
A meet (respectively, join) semi-lattice is a poset P where every x, y ∈ P have a meet (respectively, join).
Proposition
(a) A finite meet semi-lattice having a ˆ 1 is a lattice. (b) A finite join semi-lattice having a ˆ 0 is a lattice.
A meet (respectively, join) semi-lattice is a poset P where every x, y ∈ P have a meet (respectively, join).
Proposition
(a) A finite meet semi-lattice having a ˆ 1 is a lattice. (b) A finite join semi-lattice having a ˆ 0 is a lattice. Proof of (a).
A meet (respectively, join) semi-lattice is a poset P where every x, y ∈ P have a meet (respectively, join).
Proposition
(a) A finite meet semi-lattice having a ˆ 1 is a lattice. (b) A finite join semi-lattice having a ˆ 0 is a lattice. Proof of (a). Every x, y ∈ P have a meet, so every nonempty Q ⊆ P has a meet. (Induct on |Q| which must be finite.)
A meet (respectively, join) semi-lattice is a poset P where every x, y ∈ P have a meet (respectively, join).
Proposition
(a) A finite meet semi-lattice having a ˆ 1 is a lattice. (b) A finite join semi-lattice having a ˆ 0 is a lattice. Proof of (a). Every x, y ∈ P have a meet, so every nonempty Q ⊆ P has a meet. (Induct on |Q| which must be finite.) We need to prove that any x, y ∈ P have a join. Let Q = {z : z ≥ x and z ≥ y}.
A meet (respectively, join) semi-lattice is a poset P where every x, y ∈ P have a meet (respectively, join).
Proposition
(a) A finite meet semi-lattice having a ˆ 1 is a lattice. (b) A finite join semi-lattice having a ˆ 0 is a lattice. Proof of (a). Every x, y ∈ P have a meet, so every nonempty Q ⊆ P has a meet. (Induct on |Q| which must be finite.) We need to prove that any x, y ∈ P have a join. Let Q = {z : z ≥ x and z ≥ y}. Then Q = ∅ because ˆ 1 ∈ Q.
A meet (respectively, join) semi-lattice is a poset P where every x, y ∈ P have a meet (respectively, join).
Proposition
(a) A finite meet semi-lattice having a ˆ 1 is a lattice. (b) A finite join semi-lattice having a ˆ 0 is a lattice. Proof of (a). Every x, y ∈ P have a meet, so every nonempty Q ⊆ P has a meet. (Induct on |Q| which must be finite.) We need to prove that any x, y ∈ P have a join. Let Q = {z : z ≥ x and z ≥ y}. Then Q = ∅ because ˆ 1 ∈ Q. So ∧Q exists and is the join:
A meet (respectively, join) semi-lattice is a poset P where every x, y ∈ P have a meet (respectively, join).
Proposition
(a) A finite meet semi-lattice having a ˆ 1 is a lattice. (b) A finite join semi-lattice having a ˆ 0 is a lattice. Proof of (a). Every x, y ∈ P have a meet, so every nonempty Q ⊆ P has a meet. (Induct on |Q| which must be finite.) We need to prove that any x, y ∈ P have a join. Let Q = {z : z ≥ x and z ≥ y}. Then Q = ∅ because ˆ 1 ∈ Q. So ∧Q exists and is the join:
A meet (respectively, join) semi-lattice is a poset P where every x, y ∈ P have a meet (respectively, join).
Proposition
(a) A finite meet semi-lattice having a ˆ 1 is a lattice. (b) A finite join semi-lattice having a ˆ 0 is a lattice. Proof of (a). Every x, y ∈ P have a meet, so every nonempty Q ⊆ P has a meet. (Induct on |Q| which must be finite.) We need to prove that any x, y ∈ P have a join. Let Q = {z : z ≥ x and z ≥ y}. Then Q = ∅ because ˆ 1 ∈ Q. So ∧Q exists and is the join:
A meet (respectively, join) semi-lattice is a poset P where every x, y ∈ P have a meet (respectively, join).
Proposition
(a) A finite meet semi-lattice having a ˆ 1 is a lattice. (b) A finite join semi-lattice having a ˆ 0 is a lattice. Proof of (a). Every x, y ∈ P have a meet, so every nonempty Q ⊆ P has a meet. (Induct on |Q| which must be finite.) We need to prove that any x, y ∈ P have a join. Let Q = {z : z ≥ x and z ≥ y}. Then Q = ∅ because ˆ 1 ∈ Q. So ∧Q exists and is the join:
A meet (respectively, join) semi-lattice is a poset P where every x, y ∈ P have a meet (respectively, join).
Proposition
(a) A finite meet semi-lattice having a ˆ 1 is a lattice. (b) A finite join semi-lattice having a ˆ 0 is a lattice. Proof of (a). Every x, y ∈ P have a meet, so every nonempty Q ⊆ P has a meet. (Induct on |Q| which must be finite.) We need to prove that any x, y ∈ P have a join. Let Q = {z : z ≥ x and z ≥ y}. Then Q = ∅ because ˆ 1 ∈ Q. So ∧Q exists and is the join:
A meet (respectively, join) semi-lattice is a poset P where every x, y ∈ P have a meet (respectively, join).
Proposition
(a) A finite meet semi-lattice having a ˆ 1 is a lattice. (b) A finite join semi-lattice having a ˆ 0 is a lattice. Proof of (a). Every x, y ∈ P have a meet, so every nonempty Q ⊆ P has a meet. (Induct on |Q| which must be finite.) We need to prove that any x, y ∈ P have a join. Let Q = {z : z ≥ x and z ≥ y}. Then Q = ∅ because ˆ 1 ∈ Q. So ∧Q exists and is the join:
A meet (respectively, join) semi-lattice is a poset P where every x, y ∈ P have a meet (respectively, join).
Proposition
(a) A finite meet semi-lattice having a ˆ 1 is a lattice. (b) A finite join semi-lattice having a ˆ 0 is a lattice. Proof of (a). Every x, y ∈ P have a meet, so every nonempty Q ⊆ P has a meet. (Induct on |Q| which must be finite.) We need to prove that any x, y ∈ P have a join. Let Q = {z : z ≥ x and z ≥ y}. Then Q = ∅ because ˆ 1 ∈ Q. So ∧Q exists and is the join:
A meet (respectively, join) semi-lattice is a poset P where every x, y ∈ P have a meet (respectively, join).
Proposition
(a) A finite meet semi-lattice having a ˆ 1 is a lattice. (b) A finite join semi-lattice having a ˆ 0 is a lattice. Proof of (a). Every x, y ∈ P have a meet, so every nonempty Q ⊆ P has a meet. (Induct on |Q| which must be finite.) We need to prove that any x, y ∈ P have a join. Let Q = {z : z ≥ x and z ≥ y}. Then Q = ∅ because ˆ 1 ∈ Q. So ∧Q exists and is the join:
1.
A meet (respectively, join) semi-lattice is a poset P where every x, y ∈ P have a meet (respectively, join).
Proposition
(a) A finite meet semi-lattice having a ˆ 1 is a lattice. (b) A finite join semi-lattice having a ˆ 0 is a lattice. Proof of (a). Every x, y ∈ P have a meet, so every nonempty Q ⊆ P has a meet. (Induct on |Q| which must be finite.) We need to prove that any x, y ∈ P have a join. Let Q = {z : z ≥ x and z ≥ y}. Then Q = ∅ because ˆ 1 ∈ Q. So ∧Q exists and is the join:
π = B1/ . . . /Bk and σ = C1/ . . . /Cl have a meet,
A meet (respectively, join) semi-lattice is a poset P where every x, y ∈ P have a meet (respectively, join).
Proposition
(a) A finite meet semi-lattice having a ˆ 1 is a lattice. (b) A finite join semi-lattice having a ˆ 0 is a lattice. Proof of (a). Every x, y ∈ P have a meet, so every nonempty Q ⊆ P has a meet. (Induct on |Q| which must be finite.) We need to prove that any x, y ∈ P have a join. Let Q = {z : z ≥ x and z ≥ y}. Then Q = ∅ because ˆ 1 ∈ Q. So ∧Q exists and is the join:
π = B1/ . . . /Bk and σ = C1/ . . . /Cl have a meet, namely the partition whose blocks are the nonempty Bi ∩ Cj for 1 ≤ i ≤ k and 1 ≤ j ≤ l.
A meet (respectively, join) semi-lattice is a poset P where every x, y ∈ P have a meet (respectively, join).
Proposition
(a) A finite meet semi-lattice having a ˆ 1 is a lattice. (b) A finite join semi-lattice having a ˆ 0 is a lattice. Proof of (a). Every x, y ∈ P have a meet, so every nonempty Q ⊆ P has a meet. (Induct on |Q| which must be finite.) We need to prove that any x, y ∈ P have a join. Let Q = {z : z ≥ x and z ≥ y}. Then Q = ∅ because ˆ 1 ∈ Q. So ∧Q exists and is the join:
π = B1/ . . . /Bk and σ = C1/ . . . /Cl have a meet, namely the partition whose blocks are the nonempty Bi ∩ Cj for 1 ≤ i ≤ k and 1 ≤ j ≤ l. By the proposition, Πn is a lattice.
Let P be a meet semi-lattice.
Let P be a meet semi-lattice. So Mx,y =
α(z, x)β(z, y)
Let P be a meet semi-lattice. So Mx,y =
α(z, x)β(z, y) =
α(z, x)β(z, y).
Let P be a meet semi-lattice. So Mx,y =
α(z, x)β(z, y) =
α(z, x)β(z, y). Let g : P → R be arbitrary. Substituting α(z, x) = g(z) and β(z, y) = ζ(z, y) into the Main Theorem, we obtain
Let P be a meet semi-lattice. So Mx,y =
α(z, x)β(z, y) =
α(z, x)β(z, y). Let g : P → R be arbitrary. Substituting α(z, x) = g(z) and β(z, y) = ζ(z, y) into the Main Theorem, we obtain Mx,y =
g(z)
Let P be a meet semi-lattice. So Mx,y =
α(z, x)β(z, y) =
α(z, x)β(z, y). Let g : P → R be arbitrary. Substituting α(z, x) = g(z) and β(z, y) = ζ(z, y) into the Main Theorem, we obtain Mx,y =
g(z) and det M =
α(z, z)β(z, z)
Let P be a meet semi-lattice. So Mx,y =
α(z, x)β(z, y) =
α(z, x)β(z, y). Let g : P → R be arbitrary. Substituting α(z, x) = g(z) and β(z, y) = ζ(z, y) into the Main Theorem, we obtain Mx,y =
g(z) and det M =
α(z, z)β(z, z) =
g(z).
Let P be a meet semi-lattice. So Mx,y =
α(z, x)β(z, y) =
α(z, x)β(z, y). Let g : P → R be arbitrary. Substituting α(z, x) = g(z) and β(z, y) = ζ(z, y) into the Main Theorem, we obtain Mx,y =
g(z) and det M =
α(z, z)β(z, z) =
g(z). Define f : P → R by f(z) =
w≤z g(w)
Let P be a meet semi-lattice. So Mx,y =
α(z, x)β(z, y) =
α(z, x)β(z, y). Let g : P → R be arbitrary. Substituting α(z, x) = g(z) and β(z, y) = ζ(z, y) into the Main Theorem, we obtain Mx,y =
g(z) and det M =
α(z, z)β(z, z) =
g(z). Define f : P → R by f(z) =
w≤z g(w) so
Mx,y = f(x ∧ y)
Let P be a meet semi-lattice. So Mx,y =
α(z, x)β(z, y) =
α(z, x)β(z, y). Let g : P → R be arbitrary. Substituting α(z, x) = g(z) and β(z, y) = ζ(z, y) into the Main Theorem, we obtain Mx,y =
g(z) and det M =
α(z, z)β(z, z) =
g(z). Define f : P → R by f(z) =
w≤z g(w) so
Mx,y = f(x ∧ y) and det M =
w≤z
µ(w, z)f(w) .
Let P be a meet semi-lattice. So Mx,y =
α(z, x)β(z, y) =
α(z, x)β(z, y). Let g : P → R be arbitrary. Substituting α(z, x) = g(z) and β(z, y) = ζ(z, y) into the Main Theorem, we obtain Mx,y =
g(z) and det M =
α(z, z)β(z, z) =
g(z). Define f : P → R by f(z) =
w≤z g(w) so
Mx,y = f(x ∧ y) and det M =
w≤z
µ(w, z)f(w) .
Theorem (Wilf, 1968)
Let f : P → R where P is a meet semi-lattice and let M be the matrix with Mx,y = f(x ∧ y). Then det M =
g(z) where g(z) =
w≤z µ(w, z)f(w).
Mx,y = f(x ∧ y) = ⇒ det M =
g(z) where g(z) =
w≤z µ(w, z)f(w).
Mx,y = f(x ∧ y) = ⇒ det M =
g(z) where g(z) =
w≤z µ(w, z)f(w).
q
x
❆ ❆ ❆ q
y
✁ ✁ ✁ qz
.
Mx,y = f(x ∧ y) = ⇒ det M =
g(z) where g(z) =
w≤z µ(w, z)f(w).
q
x
❆ ❆ ❆ q
y
✁ ✁ ✁ qz
. M = x y z x y z f(x) f(x) f(x) f(x) f(y) f(x) f(x) f(x) f(z) .
Mx,y = f(x ∧ y) = ⇒ det M =
g(z) where g(z) =
w≤z µ(w, z)f(w).
q
x
❆ ❆ ❆ q
y
✁ ✁ ✁ qz
. M = x y z x y z f(x) f(x) f(x) f(x) f(y) f(x) f(x) f(x) f(z) . det M = f(x)f(y)f(z) + 2f(x)3 − f(x)3 − f(x)2f(y) − f(x)2f(z).
Mx,y = f(x ∧ y) = ⇒ det M =
g(z) where g(z) =
w≤z µ(w, z)f(w).
q
x
❆ ❆ ❆ q
y
✁ ✁ ✁ qz
. M = x y z x y z f(x) f(x) f(x) f(x) f(y) f(x) f(x) f(x) f(z) . det M = f(x)f(y)f(z) + 2f(x)3 − f(x)3 − f(x)2f(y) − f(x)2f(z). On the other hand
Mx,y = f(x ∧ y) = ⇒ det M =
g(z) where g(z) =
w≤z µ(w, z)f(w).
q
x
❆ ❆ ❆ q
y
✁ ✁ ✁ qz
. M = x y z x y z f(x) f(x) f(x) f(x) f(y) f(x) f(x) f(x) f(z) . det M = f(x)f(y)f(z) + 2f(x)3 − f(x)3 − f(x)2f(y) − f(x)2f(z). On the other hand g(x) = µ(x, x)f(x) = f(x),
Mx,y = f(x ∧ y) = ⇒ det M =
g(z) where g(z) =
w≤z µ(w, z)f(w).
q
x
❆ ❆ ❆ q
y
✁ ✁ ✁ qz
. M = x y z x y z f(x) f(x) f(x) f(x) f(y) f(x) f(x) f(x) f(z) . det M = f(x)f(y)f(z) + 2f(x)3 − f(x)3 − f(x)2f(y) − f(x)2f(z). On the other hand g(x) = µ(x, x)f(x) = f(x), g(y) = µ(y, y)f(y) + µ(x, y)f(x) = f(y) − f(x),
Mx,y = f(x ∧ y) = ⇒ det M =
g(z) where g(z) =
w≤z µ(w, z)f(w).
q
x
❆ ❆ ❆ q
y
✁ ✁ ✁ qz
. M = x y z x y z f(x) f(x) f(x) f(x) f(y) f(x) f(x) f(x) f(z) . det M = f(x)f(y)f(z) + 2f(x)3 − f(x)3 − f(x)2f(y) − f(x)2f(z). On the other hand g(x) = µ(x, x)f(x) = f(x), g(y) = µ(y, y)f(y) + µ(x, y)f(x) = f(y) − f(x), g(z) = µ(z, z)f(z) + µ(x, z)f(x) = f(z) − f(x).
Mx,y = f(x ∧ y) = ⇒ det M =
g(z) where g(z) =
w≤z µ(w, z)f(w).
q
x
❆ ❆ ❆ q
y
✁ ✁ ✁ qz
. M = x y z x y z f(x) f(x) f(x) f(x) f(y) f(x) f(x) f(x) f(z) . det M = f(x)f(y)f(z) + 2f(x)3 − f(x)3 − f(x)2f(y) − f(x)2f(z). On the other hand g(x) = µ(x, x)f(x) = f(x), g(y) = µ(y, y)f(y) + µ(x, y)f(x) = f(y) − f(x), g(z) = µ(z, z)f(z) + µ(x, z)f(x) = f(z) − f(x). g(x)g(y)g(z) = f(x) [f(y) − f(x)] [f(z) − f(x)]
Mx,y = f(x ∧ y) = ⇒ det M =
g(z) where g(z) =
w≤z µ(w, z)f(w).
q
x
❆ ❆ ❆ q
y
✁ ✁ ✁ qz
. M = x y z x y z f(x) f(x) f(x) f(x) f(y) f(x) f(x) f(x) f(z) . det M = f(x)f(y)f(z) + 2f(x)3 − f(x)3 − f(x)2f(y) − f(x)2f(z). On the other hand g(x) = µ(x, x)f(x) = f(x), g(y) = µ(y, y)f(y) + µ(x, y)f(x) = f(y) − f(x), g(z) = µ(z, z)f(z) + µ(x, z)f(x) = f(z) − f(x). g(x)g(y)g(z) = f(x) [f(y) − f(x)] [f(z) − f(x)] = det M.
Smith’s Theorem The Main Theorem Proof of Smith’s Theorem
Theorem
For all n ≥ 1: n =
φ(d).
Theorem
For all n ≥ 1: n =
φ(d).
fractions have been reduced to lowest terms.
Theorem
For all n ≥ 1: n =
φ(d).
fractions have been reduced to lowest terms. Example. If n = 6 then S = 1 6, 2 6, 3 6, 4 6, 5 6, 6 6
1 6, 1 3, 1 2, 2 3, 5 6, 1 1
Theorem
For all n ≥ 1: n =
φ(d).
fractions have been reduced to lowest terms. For d|n, let Sd ⊆ S be the fractions with denominator d. Example. If n = 6 then S = 1 6, 2 6, 3 6, 4 6, 5 6, 6 6
1 6, 1 3, 1 2, 2 3, 5 6, 1 1
Theorem
For all n ≥ 1: n =
φ(d).
fractions have been reduced to lowest terms. For d|n, let Sd ⊆ S be the fractions with denominator d. Example. If n = 6 then S = 1 6, 2 6, 3 6, 4 6, 5 6, 6 6
1 6, 1 3, 1 2, 2 3, 5 6, 1 1
S1 = 1 1
1 2
1 3, 2 3
1 6, 5 6
Theorem
For all n ≥ 1: n =
φ(d).
fractions have been reduced to lowest terms. For d|n, let Sd ⊆ S be the fractions with denominator d. Then |Sd| = φ(d) since c/d ∈ Sd iff 1 ≤ c ≤ d and gcd(c, d) = 1. Example. If n = 6 then S = 1 6, 2 6, 3 6, 4 6, 5 6, 6 6
1 6, 1 3, 1 2, 2 3, 5 6, 1 1
S1 = 1 1
1 2
1 3, 2 3
1 6, 5 6
Theorem
For all n ≥ 1: n =
φ(d).
fractions have been reduced to lowest terms. For d|n, let Sd ⊆ S be the fractions with denominator d. Then |Sd| = φ(d) since c/d ∈ Sd iff 1 ≤ c ≤ d and gcd(c, d) = 1. Also S = ⊎d|nSd Example. If n = 6 then S = 1 6, 2 6, 3 6, 4 6, 5 6, 6 6
1 6, 1 3, 1 2, 2 3, 5 6, 1 1
S1 = 1 1
1 2
1 3, 2 3
1 6, 5 6
Theorem
For all n ≥ 1: n =
φ(d).
fractions have been reduced to lowest terms. For d|n, let Sd ⊆ S be the fractions with denominator d. Then |Sd| = φ(d) since c/d ∈ Sd iff 1 ≤ c ≤ d and gcd(c, d) = 1. Also S = ⊎d|nSd so n = |S| Example. If n = 6 then S = 1 6, 2 6, 3 6, 4 6, 5 6, 6 6
1 6, 1 3, 1 2, 2 3, 5 6, 1 1
S1 = 1 1
1 2
1 3, 2 3
1 6, 5 6
Theorem
For all n ≥ 1: n =
φ(d).
fractions have been reduced to lowest terms. For d|n, let Sd ⊆ S be the fractions with denominator d. Then |Sd| = φ(d) since c/d ∈ Sd iff 1 ≤ c ≤ d and gcd(c, d) = 1. Also S = ⊎d|nSd so n = |S| =
|Sd| Example. If n = 6 then S = 1 6, 2 6, 3 6, 4 6, 5 6, 6 6
1 6, 1 3, 1 2, 2 3, 5 6, 1 1
S1 = 1 1
1 2
1 3, 2 3
1 6, 5 6
Theorem
For all n ≥ 1: n =
φ(d).
fractions have been reduced to lowest terms. For d|n, let Sd ⊆ S be the fractions with denominator d. Then |Sd| = φ(d) since c/d ∈ Sd iff 1 ≤ c ≤ d and gcd(c, d) = 1. Also S = ⊎d|nSd so n = |S| =
|Sd| =
φ(d). Example. If n = 6 then S = 1 6, 2 6, 3 6, 4 6, 5 6, 6 6
1 6, 1 3, 1 2, 2 3, 5 6, 1 1
S1 = 1 1
1 2
1 3, 2 3
1 6, 5 6
Inverting n =
d|n φ(d) gives
Inverting n =
d|n φ(d) gives
Corollary
φ(n) =
µ(d, n)d.
Inverting n =
d|n φ(d) gives
Corollary
φ(n) =
µ(d, n)d. Mx,y = f(x∧y) = ⇒ det M =
g(z), g(z) =
µ(w, z)f(w).
Inverting n =
d|n φ(d) gives
Corollary
φ(n) =
µ(d, n)d. Mx,y = f(x∧y) = ⇒ det M =
g(z), g(z) =
µ(w, z)f(w).
Theorem (H. J. S. Smith, 1876)
If M is n × n with Mi,j = gcd(i, j) then det M = φ(1)φ(2) · · · φ(n).
Inverting n =
d|n φ(d) gives
Corollary
φ(n) =
µ(d, n)d. Mx,y = f(x∧y) = ⇒ det M =
g(z), g(z) =
µ(w, z)f(w).
Theorem (H. J. S. Smith, 1876)
If M is n × n with Mi,j = gcd(i, j) then det M = φ(1)φ(2) · · · φ(n).
Inverting n =
d|n φ(d) gives
Corollary
φ(n) =
µ(d, n)d. Mx,y = f(x∧y) = ⇒ det M =
g(z), g(z) =
µ(w, z)f(w).
Theorem (H. J. S. Smith, 1876)
If M is n × n with Mi,j = gcd(i, j) then det M = φ(1)φ(2) · · · φ(n).
q
1
❅ ❅ q
2
q3
q
4
.
Inverting n =
d|n φ(d) gives
Corollary
φ(n) =
µ(d, n)d. Mx,y = f(x∧y) = ⇒ det M =
g(z), g(z) =
µ(w, z)f(w).
Theorem (H. J. S. Smith, 1876)
If M is n × n with Mi,j = gcd(i, j) then det M = φ(1)φ(2) · · · φ(n).
Define f : En → R by f(d) = d.
q
1
❅ ❅ q
2
q3
q
4
.
Inverting n =
d|n φ(d) gives
Corollary
φ(n) =
µ(d, n)d. Mx,y = f(x∧y) = ⇒ det M =
g(z), g(z) =
µ(w, z)f(w).
Theorem (H. J. S. Smith, 1876)
If M is n × n with Mi,j = gcd(i, j) then det M = φ(1)φ(2) · · · φ(n).
Define f : En → R by f(d) = d. Then in Wilf’s Theorem Mi,j = f(i ∧ j)
q
1
❅ ❅ q
2
q3
q
4
.
Inverting n =
d|n φ(d) gives
Corollary
φ(n) =
µ(d, n)d. Mx,y = f(x∧y) = ⇒ det M =
g(z), g(z) =
µ(w, z)f(w).
Theorem (H. J. S. Smith, 1876)
If M is n × n with Mi,j = gcd(i, j) then det M = φ(1)φ(2) · · · φ(n).
Define f : En → R by f(d) = d. Then in Wilf’s Theorem Mi,j = f(i ∧ j) = f(gcd(i, j))
q
1
❅ ❅ q
2
q3
q
4
.
Inverting n =
d|n φ(d) gives
Corollary
φ(n) =
µ(d, n)d. Mx,y = f(x∧y) = ⇒ det M =
g(z), g(z) =
µ(w, z)f(w).
Theorem (H. J. S. Smith, 1876)
If M is n × n with Mi,j = gcd(i, j) then det M = φ(1)φ(2) · · · φ(n).
Define f : En → R by f(d) = d. Then in Wilf’s Theorem Mi,j = f(i ∧ j) = f(gcd(i, j)) = gcd(i, j).
q
1
❅ ❅ q
2
q3
q
4
.
Inverting n =
d|n φ(d) gives
Corollary
φ(n) =
µ(d, n)d. Mx,y = f(x∧y) = ⇒ det M =
g(z), g(z) =
µ(w, z)f(w).
Theorem (H. J. S. Smith, 1876)
If M is n × n with Mi,j = gcd(i, j) then det M = φ(1)φ(2) · · · φ(n).
Define f : En → R by f(d) = d. Then in Wilf’s Theorem Mi,j = f(i ∧ j) = f(gcd(i, j)) = gcd(i, j). g(n) =
µ(d, n)f(d)
q
1
❅ ❅ q
2
q3
q
4
.
Inverting n =
d|n φ(d) gives
Corollary
φ(n) =
µ(d, n)d. Mx,y = f(x∧y) = ⇒ det M =
g(z), g(z) =
µ(w, z)f(w).
Theorem (H. J. S. Smith, 1876)
If M is n × n with Mi,j = gcd(i, j) then det M = φ(1)φ(2) · · · φ(n).
Define f : En → R by f(d) = d. Then in Wilf’s Theorem Mi,j = f(i ∧ j) = f(gcd(i, j)) = gcd(i, j). g(n) =
µ(d, n)f(d) =
µ(d, n)d
q
1
❅ ❅ q
2
q3
q
4
.
Inverting n =
d|n φ(d) gives
Corollary
φ(n) =
µ(d, n)d. Mx,y = f(x∧y) = ⇒ det M =
g(z), g(z) =
µ(w, z)f(w).
Theorem (H. J. S. Smith, 1876)
If M is n × n with Mi,j = gcd(i, j) then det M = φ(1)φ(2) · · · φ(n).
Define f : En → R by f(d) = d. Then in Wilf’s Theorem Mi,j = f(i ∧ j) = f(gcd(i, j)) = gcd(i, j). g(n) =
µ(d, n)f(d) =
µ(d, n)d = φ(n).
q
1
❅ ❅ q
2
q3
q
4
.
Inverting n =
d|n φ(d) gives
Corollary
φ(n) =
µ(d, n)d. Mx,y = f(x∧y) = ⇒ det M =
g(z), g(z) =
µ(w, z)f(w).
Theorem (H. J. S. Smith, 1876)
If M is n × n with Mi,j = gcd(i, j) then det M = φ(1)φ(2) · · · φ(n).
Define f : En → R by f(d) = d. Then in Wilf’s Theorem Mi,j = f(i ∧ j) = f(gcd(i, j)) = gcd(i, j). g(n) =
µ(d, n)f(d) =
µ(d, n)d = φ(n). ∴ det [gcd(i, j)] = det
q
1
❅ ❅ q
2
q3
q
4
.
Inverting n =
d|n φ(d) gives
Corollary
φ(n) =
µ(d, n)d. Mx,y = f(x∧y) = ⇒ det M =
g(z), g(z) =
µ(w, z)f(w).
Theorem (H. J. S. Smith, 1876)
If M is n × n with Mi,j = gcd(i, j) then det M = φ(1)φ(2) · · · φ(n).
Define f : En → R by f(d) = d. Then in Wilf’s Theorem Mi,j = f(i ∧ j) = f(gcd(i, j)) = gcd(i, j). g(n) =
µ(d, n)f(d) =
µ(d, n)d = φ(n). ∴ det [gcd(i, j)] = det
g(d)
q
1
❅ ❅ q
2
q3
q
4
.
Inverting n =
d|n φ(d) gives
Corollary
φ(n) =
µ(d, n)d. Mx,y = f(x∧y) = ⇒ det M =
g(z), g(z) =
µ(w, z)f(w).
Theorem (H. J. S. Smith, 1876)
If M is n × n with Mi,j = gcd(i, j) then det M = φ(1)φ(2) · · · φ(n).
Define f : En → R by f(d) = d. Then in Wilf’s Theorem Mi,j = f(i ∧ j) = f(gcd(i, j)) = gcd(i, j). g(n) =
µ(d, n)f(d) =
µ(d, n)d = φ(n). ∴ det [gcd(i, j)] = det
g(d) = φ(1) · · · φ(n).
q
1
❅ ❅ q
2
q3
q
4
.