Recall that if P is a poset and α ∈ I ( P ) then there is an associated matrix M α where M α x , y = α ( x , y ) . If the rows and columns of M α are indexed by a linear extension L of P , then M α 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 � M x , y = α ( z , x ) β ( z , y ) . z ∈ P Then � det M = α ( z , z ) β ( z , z ) . z ∈ P Proof. Let t denote transposition. Then � M α z , x M β M x , y = z , y z
Recall that if P is a poset and α ∈ I ( P ) then there is an associated matrix M α where M α x , y = α ( x , y ) . If the rows and columns of M α are indexed by a linear extension L of P , then M α 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 � M x , y = α ( z , x ) β ( z , y ) . z ∈ P Then � det M = α ( z , z ) β ( z , z ) . z ∈ P Proof. Let t denote transposition. Then � M α z , x M β � ( M α ) t x , z M β M x , y = z , y = z , y z z
Recall that if P is a poset and α ∈ I ( P ) then there is an associated matrix M α where M α x , y = α ( x , y ) . If the rows and columns of M α are indexed by a linear extension L of P , then M α 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 � M x , y = α ( z , x ) β ( z , y ) . z ∈ P Then � det M = α ( z , z ) β ( z , z ) . z ∈ P Proof. Let t denote transposition. Then � M α z , x M β � ( M α ) t x , z M β M x , y = z , y = z , y z z So M = ( M α ) t M β ,
Recall that if P is a poset and α ∈ I ( P ) then there is an associated matrix M α where M α x , y = α ( x , y ) . If the rows and columns of M α are indexed by a linear extension L of P , then M α 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 � M x , y = α ( z , x ) β ( z , y ) . z ∈ P Then � det M = α ( z , z ) β ( z , z ) . z ∈ P Proof. Let t denote transposition. Then � M α z , x M β � ( M α ) t x , z M β M x , y = z , y = z , y z z So M = ( M α ) t M β , implying det M = det M α det M β .
Recall that if P is a poset and α ∈ I ( P ) then there is an associated matrix M α where M α x , y = α ( x , y ) . If the rows and columns of M α are indexed by a linear extension L of P , then M α 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 � M x , y = α ( z , x ) β ( z , y ) . z ∈ P Then � det M = α ( z , z ) β ( z , z ) . z ∈ P Proof. Let t denote transposition. Then � M α z , x M β � ( M α ) t x , z M β M x , y = z , y = z , y z z So M = ( M α ) t M β , implying det M = det M α det M β . By triangularity of M α , M β we have det M = � z α ( z , z ) β ( z , z ) .
In the Main Theorem, the entries of M are sums, � M x , y = α ( z , x ) β ( z , y ) . z ∈ P while the factors of det M are individual terms.
In the Main Theorem, the entries of M are sums, � M x , y = α ( z , x ) β ( z , y ) . z ∈ P 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, � M x , y = α ( z , x ) β ( z , y ) . z ∈ P 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, � M x , y = α ( z , x ) β ( z , y ) . z ∈ P 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 . i ∈ S
In the Main Theorem, the entries of M are sums, � M x , y = α ( z , x ) β ( z , y ) . z ∈ P 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 . i ∈ S To switch the role of sum and individual term we need M¨ obius inversion.
In the Main Theorem, the entries of M are sums, � M x , y = α ( z , x ) β ( z , y ) . z ∈ P 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 . i ∈ S To switch the role of sum and individual term we need M¨ obius inversion. The sums in the Main Theorem have two implicit restrictions:
In the Main Theorem, the entries of M are sums, � M x , y = α ( z , x ) β ( z , y ) . z ∈ P 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 . i ∈ S To switch the role of sum and individual term we need M¨ obius inversion. The sums in the Main Theorem have two implicit restrictions: α ( z , x ) , β ( z , y ) � = 0 implies z ≤ x and z ≤ y .
In the Main Theorem, the entries of M are sums, � M x , y = α ( z , x ) β ( z , y ) . z ∈ P 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 . i ∈ S To switch the role of sum and individual term we need M¨ obius inversion. The sums in the Main Theorem have two implicit restrictions: α ( z , x ) , β ( z , y ) � = 0 implies z ≤ x and z ≤ y . To use M¨ obius inversion we need a single restriction z ≤ w .
In the Main Theorem, the entries of M are sums, � M x , y = α ( z , x ) β ( z , y ) . z ∈ P 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 . i ∈ S To switch the role of sum and individual term we need M¨ obius inversion. The sums in the Main Theorem have two implicit restrictions: α ( z , x ) , β ( z , y ) � = 0 implies z ≤ x and z ≤ y . To use M¨ obius inversion we need a single restriction z ≤ w . To 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: 1. We have z ≥ x for all z ∈ Q so ∧ Q ≥ x .
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. We have z ≥ x for all z ∈ Q so ∧ Q ≥ x . Similarly ∧ Q ≥ 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 . So ∧ Q exists and is the join: 1. We have z ≥ x for all z ∈ Q so ∧ Q ≥ x . Similarly ∧ Q ≥ y . 2. If 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 . So ∧ Q exists and is the join: 1. We have z ≥ x for all z ∈ Q so ∧ Q ≥ x . Similarly ∧ Q ≥ y . 2. If z ≥ x and z ≥ y then z ∈ 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: 1. We have z ≥ x for all z ∈ Q so ∧ Q ≥ x . Similarly ∧ Q ≥ y . 2. If z ≥ x and z ≥ y then z ∈ Q . So z ≥ ∧ 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: 1. We have z ≥ x for all z ∈ Q so ∧ Q ≥ x . Similarly ∧ Q ≥ y . 2. If z ≥ x and z ≥ y then z ∈ Q . So z ≥ ∧ Q . Example. Π n 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). 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. We have z ≥ x for all z ∈ Q so ∧ Q ≥ x . Similarly ∧ Q ≥ y . 2. If z ≥ x and z ≥ y then z ∈ Q . So z ≥ ∧ Q . Example. Π n is a lattice: Π n is finite and has a ˆ 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: 1. We have z ≥ x for all z ∈ Q so ∧ Q ≥ x . Similarly ∧ Q ≥ y . 2. If z ≥ x and z ≥ y then z ∈ Q . So z ≥ ∧ Q . Example. Π n is a lattice: Π n is finite and has a ˆ 1. Also, any π = B 1 / . . . / B k and σ = C 1 / . . . / C l 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: 1. We have z ≥ x for all z ∈ Q so ∧ Q ≥ x . Similarly ∧ Q ≥ y . 2. If z ≥ x and z ≥ y then z ∈ Q . So z ≥ ∧ Q . Example. Π n is a lattice: Π n is finite and has a ˆ 1. Also, any π = B 1 / . . . / B k and σ = C 1 / . . . / C l have a meet, namely the partition whose blocks are the nonempty B i ∩ C j 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: 1. We have z ≥ x for all z ∈ Q so ∧ Q ≥ x . Similarly ∧ Q ≥ y . 2. If z ≥ x and z ≥ y then z ∈ Q . So z ≥ ∧ Q . Example. Π n is a lattice: Π n is finite and has a ˆ 1. Also, any π = B 1 / . . . / B k and σ = C 1 / . . . / C l have a meet, namely the partition whose blocks are the nonempty B i ∩ C j for 1 ≤ i ≤ k and 1 ≤ j ≤ l . By the proposition, Π n is a lattice.
Let P be a meet semi-lattice.
Let P be a meet semi-lattice. So � M x , y = α ( z , x ) β ( z , y ) z ≤ x , z ≤ y
Let P be a meet semi-lattice. So � � M x , y = α ( z , x ) β ( z , y ) = α ( z , x ) β ( z , y ) . z ≤ x , z ≤ y z ≤ x ∧ y
Let P be a meet semi-lattice. So � � M x , y = α ( z , x ) β ( z , y ) = α ( z , x ) β ( z , y ) . z ≤ x , z ≤ y z ≤ x ∧ 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 � � M x , y = α ( z , x ) β ( z , y ) = α ( z , x ) β ( z , y ) . z ≤ x , z ≤ y z ≤ x ∧ 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 � M x , y = g ( z ) z ≤ x ∧ y
Let P be a meet semi-lattice. So � � M x , y = α ( z , x ) β ( z , y ) = α ( z , x ) β ( z , y ) . z ≤ x , z ≤ y z ≤ x ∧ 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 � � M x , y = g ( z ) and det M = α ( z , z ) β ( z , z ) z ≤ x ∧ y z ∈ P
Let P be a meet semi-lattice. So � � M x , y = α ( z , x ) β ( z , y ) = α ( z , x ) β ( z , y ) . z ≤ x , z ≤ y z ≤ x ∧ 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 � � � M x , y = g ( z ) and det M = α ( z , z ) β ( z , z ) = g ( z ) . z ≤ x ∧ y z ∈ P z ∈ P
Let P be a meet semi-lattice. So � � M x , y = α ( z , x ) β ( z , y ) = α ( z , x ) β ( z , y ) . z ≤ x , z ≤ y z ≤ x ∧ 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 � � � M x , y = g ( z ) and det M = α ( z , z ) β ( z , z ) = g ( z ) . z ≤ x ∧ y z ∈ P z ∈ P Define f : P → R by f ( z ) = � w ≤ z g ( w )
Let P be a meet semi-lattice. So � � M x , y = α ( z , x ) β ( z , y ) = α ( z , x ) β ( z , y ) . z ≤ x , z ≤ y z ≤ x ∧ 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 � � � M x , y = g ( z ) and det M = α ( z , z ) β ( z , z ) = g ( z ) . z ≤ x ∧ y z ∈ P z ∈ P Define f : P → R by f ( z ) = � w ≤ z g ( w ) so M x , y = f ( x ∧ y )
Let P be a meet semi-lattice. So � � M x , y = α ( z , x ) β ( z , y ) = α ( z , x ) β ( z , y ) . z ≤ x , z ≤ y z ≤ x ∧ 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 � � � M x , y = g ( z ) and det M = α ( z , z ) β ( z , z ) = g ( z ) . z ≤ x ∧ y z ∈ P z ∈ P Define f : P → R by f ( z ) = � w ≤ z g ( w ) so � � . M x , y = f ( x ∧ y ) and det M = µ ( w , z ) f ( w ) z w ≤ z
Let P be a meet semi-lattice. So � � M x , y = α ( z , x ) β ( z , y ) = α ( z , x ) β ( z , y ) . z ≤ x , z ≤ y z ≤ x ∧ 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 � � � M x , y = g ( z ) and det M = α ( z , z ) β ( z , z ) = g ( z ) . z ≤ x ∧ y z ∈ P z ∈ P Define f : P → R by f ( z ) = � w ≤ z g ( w ) so � � . M x , y = f ( x ∧ y ) and det M = µ ( w , z ) f ( w ) z w ≤ z Theorem (Wilf, 1968) Let f : P → R where P is a meet semi-lattice and let M be the matrix with M x , y = f ( x ∧ y ) . Then � det M = g ( z ) z ∈ P where g ( z ) = � w ≤ z µ ( w , z ) f ( w ) .
� M x , y = f ( x ∧ y ) = ⇒ det M = g ( z ) z ∈ P where g ( z ) = � w ≤ z µ ( w , z ) f ( w ) .
� M x , y = f ( x ∧ y ) = ⇒ det M = g ( z ) z ∈ P where g ( z ) = � w ≤ z µ ( w , z ) f ( w ) . y q z q ❆ ✁ Example. Let P = . ❆ ✁ ❆ ✁ x q
� M x , y = f ( x ∧ y ) = ⇒ det M = g ( z ) z ∈ P where g ( z ) = � w ≤ z µ ( w , z ) f ( w ) . y q z ❆ q ✁ Example. Let P = . ❆ ✁ ❆ ✁ x x y z q x f ( x ) f ( x ) f ( x ) M = . y f ( x ) f ( y ) f ( x ) z f ( x ) f ( x ) f ( z )
� M x , y = f ( x ∧ y ) = ⇒ det M = g ( z ) z ∈ P where g ( z ) = � w ≤ z µ ( w , z ) f ( w ) . y q z ❆ q ✁ Example. Let P = . ❆ ✁ ❆ ✁ x x y z q x f ( x ) f ( x ) f ( x ) M = . y f ( x ) f ( y ) f ( x ) z f ( x ) f ( x ) f ( z ) det M = f ( x ) f ( y ) f ( z ) + 2 f ( x ) 3 − f ( x ) 3 − f ( x ) 2 f ( y ) − f ( x ) 2 f ( z ) .
� M x , y = f ( x ∧ y ) = ⇒ det M = g ( z ) z ∈ P where g ( z ) = � w ≤ z µ ( w , z ) f ( w ) . y q z ❆ q ✁ Example. Let P = . ❆ ✁ ❆ ✁ x x y z q x f ( x ) f ( x ) f ( x ) M = . y f ( x ) f ( y ) f ( x ) z f ( x ) f ( x ) f ( z ) det M = f ( x ) f ( y ) f ( z ) + 2 f ( x ) 3 − f ( x ) 3 − f ( x ) 2 f ( y ) − f ( x ) 2 f ( z ) . On the other hand
� M x , y = f ( x ∧ y ) = ⇒ det M = g ( z ) z ∈ P where g ( z ) = � w ≤ z µ ( w , z ) f ( w ) . y q z ❆ q ✁ Example. Let P = . ❆ ✁ ❆ ✁ x x y z q x f ( x ) f ( x ) f ( x ) M = . y f ( x ) f ( y ) f ( x ) z f ( x ) f ( x ) f ( z ) det M = f ( x ) f ( y ) f ( z ) + 2 f ( x ) 3 − f ( x ) 3 − f ( x ) 2 f ( y ) − f ( x ) 2 f ( z ) . On the other hand g ( x ) = µ ( x , x ) f ( x ) = f ( x ) ,
� M x , y = f ( x ∧ y ) = ⇒ det M = g ( z ) z ∈ P where g ( z ) = � w ≤ z µ ( w , z ) f ( w ) . y q z q ❆ ✁ Example. Let P = . ❆ ✁ ❆ ✁ x x y z q x f ( x ) f ( x ) f ( x ) M = . y f ( x ) f ( y ) f ( x ) z f ( x ) f ( x ) f ( z ) det M = f ( x ) f ( y ) f ( z ) + 2 f ( x ) 3 − f ( x ) 3 − f ( x ) 2 f ( y ) − f ( x ) 2 f ( 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 ) ,
� M x , y = f ( x ∧ y ) = ⇒ det M = g ( z ) z ∈ P where g ( z ) = � w ≤ z µ ( w , z ) f ( w ) . y q z q ❆ ✁ Example. Let P = . ❆ ✁ ❆ ✁ x x y z q x f ( x ) f ( x ) f ( x ) M = . y f ( x ) f ( y ) f ( x ) z f ( x ) f ( x ) f ( z ) det M = f ( x ) f ( y ) f ( z ) + 2 f ( x ) 3 − f ( x ) 3 − f ( x ) 2 f ( y ) − f ( x ) 2 f ( 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 ) .
� M x , y = f ( x ∧ y ) = ⇒ det M = g ( z ) z ∈ P where g ( z ) = � w ≤ z µ ( w , z ) f ( w ) . y q z ❆ q ✁ Example. Let P = . ❆ ✁ ❆ ✁ x x y z q x f ( x ) f ( x ) f ( x ) M = . y f ( x ) f ( y ) f ( x ) z f ( x ) f ( x ) f ( z ) det M = f ( x ) f ( y ) f ( z ) + 2 f ( x ) 3 − f ( x ) 3 − f ( x ) 2 f ( y ) − f ( x ) 2 f ( 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 )]
� M x , y = f ( x ∧ y ) = ⇒ det M = g ( z ) z ∈ P where g ( z ) = � w ≤ z µ ( w , z ) f ( w ) . y q z ❆ q ✁ Example. Let P = . ❆ ✁ ❆ ✁ x x y z q x f ( x ) f ( x ) f ( x ) M = . y f ( x ) f ( y ) f ( x ) z f ( x ) f ( x ) f ( z ) det M = f ( x ) f ( y ) f ( z ) + 2 f ( x ) 3 − f ( x ) 3 − f ( x ) 2 f ( y ) − f ( x ) 2 f ( 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 .
Outline Smith’s Theorem The Main Theorem Proof of Smith’s Theorem
Theorem For all n ≥ 1 : � n = φ ( d ) . d | n
Theorem For all n ≥ 1 : � n = φ ( d ) . d | n Proof. Consider the set S = { 1 / n , 2 / n , . . . , n / n } where the fractions have been reduced to lowest terms.
Theorem For all n ≥ 1 : � n = φ ( d ) . d | n Proof. Consider the set S = { 1 / n , 2 / n , . . . , n / n } where the fractions have been reduced to lowest terms. Example. If n = 6 then � 1 � � 1 � 6 , 2 6 , 3 6 , 4 6 , 5 6 , 6 6 , 1 3 , 1 2 , 2 3 , 5 6 , 1 S = = . 6 1
Theorem For all n ≥ 1 : � n = φ ( d ) . d | n Proof. Consider the set S = { 1 / n , 2 / n , . . . , n / n } where the fractions have been reduced to lowest terms. For d | n , let S d ⊆ S be the fractions with denominator d . Example. If n = 6 then � 1 � � 1 � 6 , 2 6 , 3 6 , 4 6 , 5 6 , 6 6 , 1 3 , 1 2 , 2 3 , 5 6 , 1 S = = . 6 1
Theorem For all n ≥ 1 : � n = φ ( d ) . d | n Proof. Consider the set S = { 1 / n , 2 / n , . . . , n / n } where the fractions have been reduced to lowest terms. For d | n , let S d ⊆ S be the fractions with denominator d . Example. If n = 6 then � 1 � � 1 � 6 , 2 6 , 3 6 , 4 6 , 5 6 , 6 6 , 1 3 , 1 2 , 2 3 , 5 6 , 1 S = = . 6 1 � 1 � � 1 � � 1 3 , 2 � � 1 6 , 5 � S 1 = , S 2 = , S 3 = , S 6 = . 1 2 3 6
Theorem For all n ≥ 1 : � n = φ ( d ) . d | n Proof. Consider the set S = { 1 / n , 2 / n , . . . , n / n } where the fractions have been reduced to lowest terms. For d | n , let S d ⊆ S be the fractions with denominator d . Then | S d | = φ ( d ) since c / d ∈ S d iff 1 ≤ c ≤ d and gcd ( c , d ) = 1. Example. If n = 6 then � 1 � � 1 � 6 , 2 6 , 3 6 , 4 6 , 5 6 , 6 6 , 1 3 , 1 2 , 2 3 , 5 6 , 1 S = = . 6 1 � 1 � � 1 � � 1 3 , 2 � � 1 6 , 5 � S 1 = , S 2 = , S 3 = , S 6 = . 1 2 3 6
Theorem For all n ≥ 1 : � n = φ ( d ) . d | n Proof. Consider the set S = { 1 / n , 2 / n , . . . , n / n } where the fractions have been reduced to lowest terms. For d | n , let S d ⊆ S be the fractions with denominator d . Then | S d | = φ ( d ) since c / d ∈ S d iff 1 ≤ c ≤ d and gcd ( c , d ) = 1. Also S = ⊎ d | n S d Example. If n = 6 then � 1 � � 1 � 6 , 2 6 , 3 6 , 4 6 , 5 6 , 6 6 , 1 3 , 1 2 , 2 3 , 5 6 , 1 S = = . 6 1 � 1 � � 1 � � 1 3 , 2 � � 1 6 , 5 � S 1 = , S 2 = , S 3 = , S 6 = . 1 2 3 6
Theorem For all n ≥ 1 : � n = φ ( d ) . d | n Proof. Consider the set S = { 1 / n , 2 / n , . . . , n / n } where the fractions have been reduced to lowest terms. For d | n , let S d ⊆ S be the fractions with denominator d . Then | S d | = φ ( d ) since c / d ∈ S d iff 1 ≤ c ≤ d and gcd ( c , d ) = 1. Also S = ⊎ d | n S d so n = | S | Example. If n = 6 then � 1 � � 1 � 6 , 2 6 , 3 6 , 4 6 , 5 6 , 6 6 , 1 3 , 1 2 , 2 3 , 5 6 , 1 S = = . 6 1 � 1 � � 1 � � 1 3 , 2 � � 1 6 , 5 � S 1 = , S 2 = , S 3 = , S 6 = . 1 2 3 6
Theorem For all n ≥ 1 : � n = φ ( d ) . d | n Proof. Consider the set S = { 1 / n , 2 / n , . . . , n / n } where the fractions have been reduced to lowest terms. For d | n , let S d ⊆ S be the fractions with denominator d . Then | S d | = φ ( d ) since c / d ∈ S d iff 1 ≤ c ≤ d and gcd ( c , d ) = 1. Also S = ⊎ d | n S d so � n = | S | = | S d | d | n Example. If n = 6 then � 1 � � 1 � 6 , 2 6 , 3 6 , 4 6 , 5 6 , 6 6 , 1 3 , 1 2 , 2 3 , 5 6 , 1 S = = . 6 1 � 1 � � 1 � � 1 3 , 2 � � 1 6 , 5 � S 1 = , S 2 = , S 3 = , S 6 = . 1 2 3 6
Theorem For all n ≥ 1 : � n = φ ( d ) . d | n Proof. Consider the set S = { 1 / n , 2 / n , . . . , n / n } where the fractions have been reduced to lowest terms. For d | n , let S d ⊆ S be the fractions with denominator d . Then | S d | = φ ( d ) since c / d ∈ S d iff 1 ≤ c ≤ d and gcd ( c , d ) = 1. Also S = ⊎ d | n S d so � � n = | S | = | S d | = φ ( d ) . d | n d | n Example. If n = 6 then � 1 � � 1 � 6 , 2 6 , 3 6 , 4 6 , 5 6 , 6 6 , 1 3 , 1 2 , 2 3 , 5 6 , 1 S = = . 6 1 � 1 � � 1 � � 1 3 , 2 � � 1 6 , 5 � S 1 = , S 2 = , S 3 = , S 6 = . 1 2 3 6
Inverting n = � d | n φ ( d ) gives
Inverting n = � d | n φ ( d ) gives Corollary � φ ( n ) = µ ( d , n ) d . d | n
Inverting n = � d | n φ ( d ) gives Corollary � φ ( n ) = µ ( d , n ) d . d | n � � M x , y = f ( x ∧ y ) = ⇒ det M = g ( z ) , g ( z ) = µ ( w , z ) f ( w ) . w ≤ z z ∈ P
Inverting n = � d | n φ ( d ) gives Corollary � φ ( n ) = µ ( d , n ) d . d | n � � M x , y = f ( x ∧ y ) = ⇒ det M = g ( z ) , g ( z ) = µ ( w , z ) f ( w ) . w ≤ z z ∈ P Theorem (H. J. S. Smith, 1876) If M is n × n with M i , 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 . d | n � � M x , y = f ( x ∧ y ) = ⇒ det M = g ( z ) , g ( z ) = µ ( w , z ) f ( w ) . w ≤ z z ∈ P Theorem (H. J. S. Smith, 1876) If M is n × n with M i , j = gcd ( i , j ) then det M = φ ( 1 ) φ ( 2 ) · · · φ ( n ) . Proof. Let E n = { 1 , 2 . . . , n } partially ordered by i ≤ E n j iff i | j .
Inverting n = � d | n φ ( d ) gives Corollary � φ ( n ) = µ ( d , n ) d . d | n � � M x , y = f ( x ∧ y ) = ⇒ det M = g ( z ) , g ( z ) = µ ( w , z ) f ( w ) . w ≤ z z ∈ P Theorem (H. J. S. Smith, 1876) If M is n × n with M i , j = gcd ( i , j ) then det M = φ ( 1 ) φ ( 2 ) · · · φ ( n ) . Proof. Let E n = { 1 , 2 . . . , n } partially ordered by i ≤ E n j iff i | j . 4 q 6 q � � Example. E 6 = . 2 q 3 q 5 q ❅ � ❅ � 1 q
Inverting n = � d | n φ ( d ) gives Corollary � φ ( n ) = µ ( d , n ) d . d | n � � M x , y = f ( x ∧ y ) = ⇒ det M = g ( z ) , g ( z ) = µ ( w , z ) f ( w ) . w ≤ z z ∈ P Theorem (H. J. S. Smith, 1876) If M is n × n with M i , j = gcd ( i , j ) then det M = φ ( 1 ) φ ( 2 ) · · · φ ( n ) . Proof. Let E n = { 1 , 2 . . . , n } partially ordered by i ≤ E n j iff i | j . Define f : E n → R by f ( d ) = d . 4 q 6 q � � Example. E 6 = . 2 q 3 q 5 q ❅ � ❅ � 1 q
Inverting n = � d | n φ ( d ) gives Corollary � φ ( n ) = µ ( d , n ) d . d | n � � M x , y = f ( x ∧ y ) = ⇒ det M = g ( z ) , g ( z ) = µ ( w , z ) f ( w ) . w ≤ z z ∈ P Theorem (H. J. S. Smith, 1876) If M is n × n with M i , j = gcd ( i , j ) then det M = φ ( 1 ) φ ( 2 ) · · · φ ( n ) . Proof. Let E n = { 1 , 2 . . . , n } partially ordered by i ≤ E n j iff i | j . Define f : E n → R by f ( d ) = d . Then in Wilf’s Theorem M i , j = f ( i ∧ j ) 4 q 6 q � � Example. E 6 = . 2 q 3 q 5 q ❅ � ❅ � 1 q
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