Generalized Ehrhart Polynomials Nan Li (MIT) with Sheng Chen (HIT) - PowerPoint PPT Presentation

Generalized Ehrhart Polynomials Nan Li (MIT) with Sheng Chen (HIT) and Steven Sam (MIT) Aug 5, 2010 Fpsac

Outline • Ehrhart Theorem • Generalized Ehrhart polynomials by examples • Main theorem in three equivalent versions • Proof: “writing in base n” trick • Further questions

Ehrhart Theorem Let P ⊂ R d be a polytope with rational vertices. (0 , n ) • � � (0 , 1) ����������� • � � � � � � � � nP � P � � • • � � ( 1 (0 , 0) 2 , 0) � • • � ( 1 (0 , 0) 2 n , 0) � 4 n 2 + n + 1 1 n even #( nP ∩ Z 2 ) = 4 n 2 + n + 3 1 n odd 4 Definition We call f ( n ) a quasi-polynomial, if f ( n ) = f i ( n ) ( n ≡ i mod T ), for some T ∈ N and polynomials f i ( n )’s. Theorem (Ehrhart) i ( P , n ) = #( nP ∩ Z d ) is a quasi-polynomial. In particular, if P has integral vertices, i ( P , n ) is a polynomial, called the Ehrhart polynomial .

Example: non homogenous “dilation”  2 x + y ≤ n + 1   (0 , n + 1) ����������� • x ≥ 0  y ≥ 0    2 x + y = n + 1 2 x + y ≤ n   • •   ( n 2 + 1 = + (0 , 0) 2 , 0) x ≥ 0 x ≥ 0   y ≥ 0 y ≥ 0   • (0 , n + 1) � � � � � � � � � � • The number of integer points on this diagonal is � ( n 2 + 1 2 , 0) f ( n ) = # { ( x , y ) ∈ Z 2 ≥ 0 | 2 x + y = n + 1 }

Example: non homogenous “dilation”  2 x + y ≤ n + 1   (0 , n + 1) ����������� • x ≥ 0  y ≥ 0    2 x + y = n + 1 2 x + y ≤ n   • •   ( n 2 + 1 = + (0 , 0) 2 , 0) x ≥ 0 x ≥ 0   y ≥ 0 y ≥ 0   • (0 , n + 1) � � � � � � � � � � • The number of integer points on this diagonal is � ( n 2 + 1 2 , 0) f ( n ) = # { ( x , y ) ∈ Z 2 ≥ 0 | 2 x + y = n + 1 }

Example: non homogenous “dilation”  2 x + y ≤ n + 1   (0 , n + 1) ����������� • x ≥ 0  y ≥ 0    2 x + y = n + 1 2 x + y ≤ n   • •   ( n 2 + 1 = + (0 , 0) 2 , 0) x ≥ 0 x ≥ 0   y ≥ 0 y ≥ 0   • (0 , n + 1) � � � � � � � � � � • The number of integer points on this diagonal is � ( n 2 + 1 2 , 0) f ( n ) = # { ( x , y ) ∈ Z 2 ≥ 0 | 2 x + y = n + 1 }

Theorem (Popoviciu’s Formula) The number of nonnegative integer solutions ( x , y ) to ax + by = m, where a , b are coprime integers, is given by the formula � ma ′ � � mb ′ � m ab − − + 1 , b a where { r } = r − ⌊ r ⌋ and a ′ and b ′ are any integers satisfying aa ′ + bb ′ = 1 . Example For 2 x + y = n + 1, we have a = 2 , b = 1 , a ′ = 1 , b ′ = − 1 Then � n +3 � n + 1 � � − ( n + 1) � n + 1 n odd 2 − − + 1 = 2 1 2 n 2 + 1 n even

Example: nonlinear “dilation” (0 , n 2 + 2 n ) • • • • � �   � � � � � = 1 � �  �  + � �   � 2 �   � � � � � � • • • • • � � (0 , 0) (2 n + 4 , 0) The number of integer points on the diagonal is ≥ 0 | ( n 2 + 2 n ) x + (2 n + 4) y = (2 n + 4)( n 2 + 2 n ) } f ( n ) = # { ( x , y ) ∈ Z 2 We can not use Popoviciu’s Formula directly, so we need the following generalizations: • Replace GCD ( a , b ) = 1, a , b ∈ Z by GCD ( a ( n ) , b ( n )) = 1 for all n ∈ Z , where a ( n ) , b ( n ) ∈ Z [ n ] � n − 1 � � a ( n ) � • Replace by function , a ( n ) , b ( n ) ∈ Z [ n ]. 2 b ( n )

Example: nonlinear “dilation” (0 , n 2 + 2 n ) • • • • � �   � � � � � = 1 � �  �  + � �   � 2 �   � � � � � � • • • • • � � (0 , 0) (2 n + 4 , 0) The number of integer points on the diagonal is ≥ 0 | ( n 2 + 2 n ) x + (2 n + 4) y = (2 n + 4)( n 2 + 2 n ) } f ( n ) = # { ( x , y ) ∈ Z 2 We can not use Popoviciu’s Formula directly, so we need the following generalizations: • Replace GCD ( a , b ) = 1, a , b ∈ Z by GCD ( a ( n ) , b ( n )) = 1 for all n ∈ Z , where a ( n ) , b ( n ) ∈ Z [ n ] � n − 1 � � a ( n ) � • Replace by function , a ( n ) , b ( n ) ∈ Z [ n ]. 2 b ( n )

Example: nonlinear “dilation” (0 , n 2 + 2 n ) • • • • � �   � � � � � = 1 � �  �  + � �   � 2 �   � � � � � � • • • • • � � (0 , 0) (2 n + 4 , 0) The number of integer points on the diagonal is ≥ 0 | ( n 2 + 2 n ) x + (2 n + 4) y = (2 n + 4)( n 2 + 2 n ) } f ( n ) = # { ( x , y ) ∈ Z 2 We can not use Popoviciu’s Formula directly, so we need the following generalizations: • Replace GCD ( a , b ) = 1, a , b ∈ Z by GCD ( a ( n ) , b ( n )) = 1 for all n ∈ Z , where a ( n ) , b ( n ) ∈ Z [ n ] � n − 1 � � a ( n ) � • Replace by function , a ( n ) , b ( n ) ∈ Z [ n ]. 2 b ( n )

Example � � n 2 +2 n and GCD ( n 2 + 2 n , 2 n + 4). Compute 2 n +4 • n = 2 m , n 2 + 2 n = 4 m 2 + 4 m = m (4 m + 4). � � = 0 and GCD ( n 2 + 2 n , 2 n + 4) = 2 n + 4. n 2 +2 n 2 n +4 • n = 2 m + 1, n 2 + 2 n = 4 m 2 + 8 m + 3 = m (4 m + 6) + (2 m + 3). � � � � n 2 +2 n m + 2 m +3 = 2 m +3 4 m +6 = 1 = 2 . 2 n +4 4 m +6 For GCD , we apply Euclidean algorithm and get 4 m + 6 = 2(2 m + 3), So GCD ( n 2 + 2 n , 2 n + 4) = 2 m + 3 = n + 2 Therefore, � n 2 + 2 n � � � 0 n even 2 n + 4 n even and GCD ( n 2 +2 n , 2 n +4) = = 1 2 n + 4 n odd n + 2 n odd 2

Example � � n 2 +2 n and GCD ( n 2 + 2 n , 2 n + 4). Compute 2 n +4 • n = 2 m , n 2 + 2 n = 4 m 2 + 4 m = m (4 m + 4). � � = 0 and GCD ( n 2 + 2 n , 2 n + 4) = 2 n + 4. n 2 +2 n 2 n +4 • n = 2 m + 1, n 2 + 2 n = 4 m 2 + 8 m + 3 = m (4 m + 6) + (2 m + 3). � � � � n 2 +2 n m + 2 m +3 = 2 m +3 4 m +6 = 1 = 2 . 2 n +4 4 m +6 For GCD , we apply Euclidean algorithm and get 4 m + 6 = 2(2 m + 3), So GCD ( n 2 + 2 n , 2 n + 4) = 2 m + 3 = n + 2 Therefore, � n 2 + 2 n � � � 0 n even 2 n + 4 n even and GCD ( n 2 +2 n , 2 n +4) = = 1 2 n + 4 n odd n + 2 n odd 2

Example � � n 2 +2 n and GCD ( n 2 + 2 n , 2 n + 4). Compute 2 n +4 • n = 2 m , n 2 + 2 n = 4 m 2 + 4 m = m (4 m + 4). � � = 0 and GCD ( n 2 + 2 n , 2 n + 4) = 2 n + 4. n 2 +2 n 2 n +4 • n = 2 m + 1, n 2 + 2 n = 4 m 2 + 8 m + 3 = m (4 m + 6) + (2 m + 3). � � � � n 2 +2 n m + 2 m +3 = 2 m +3 4 m +6 = 1 = 2 . 2 n +4 4 m +6 For GCD , we apply Euclidean algorithm and get 4 m + 6 = 2(2 m + 3), So GCD ( n 2 + 2 n , 2 n + 4) = 2 m + 3 = n + 2 Therefore, � n 2 + 2 n � � � 0 n even 2 n + 4 n even and GCD ( n 2 +2 n , 2 n +4) = = 1 2 n + 4 n odd n + 2 n odd 2

Example � � n 2 +2 n and GCD ( n 2 + 2 n , 2 n + 4). Compute 2 n +4 • n = 2 m , n 2 + 2 n = 4 m 2 + 4 m = m (4 m + 4). � � = 0 and GCD ( n 2 + 2 n , 2 n + 4) = 2 n + 4. n 2 +2 n 2 n +4 • n = 2 m + 1, n 2 + 2 n = 4 m 2 + 8 m + 3 = m (4 m + 6) + (2 m + 3). � � � � n 2 +2 n m + 2 m +3 = 2 m +3 4 m +6 = 1 = 2 . 2 n +4 4 m +6 For GCD , we apply Euclidean algorithm and get 4 m + 6 = 2(2 m + 3), So GCD ( n 2 + 2 n , 2 n + 4) = 2 m + 3 = n + 2 Therefore, � n 2 + 2 n � � � 0 n even 2 n + 4 n even and GCD ( n 2 +2 n , 2 n +4) = = 1 2 n + 4 n odd n + 2 n odd 2

Generalized division and GCD Definition Let f ( n ), g ( n ) be polynomial functions Z → Z . Define functions q , the quotient , r , the remainder and ggcd , the generalized GCD of f and g as follows: for each n ∈ Z , � f ( n ) � f ( n ) � � q ( n ) = , r ( n ) = g ( n ) , and ggcd ( n ) = GCD ( f ( n ) , g ( n )) . g ( n ) g ( n ) Theorem (Chen, L., Sam) Functions q, r, ggcd are quasi-polynomials for n sufficiently large.

Back to example: nonlinear dilation (0 , n 2 + 2 n ) • � � � � # { P ( n ) ∩ Z 2 } � P ( n ) = � 2 (( n 2 + 2 n + 1)(2 n + 5) + f ( n )), where � = 1 � � � • • � (0 , 0) (2 n + 4 , 0) ≥ 0 | ( n 2 + 2 n ) x + (2 n + 4) y = (2 n + 4)( n 2 + 2 n ) } . f ( n ) = # { ( x , y ) ∈ Z 2 Now by generalized division and GCD, we can apply Popoviciu’s Formula and get � 2 n + 5 n even f ( n ) = n + 3 n odd

Back to example: nonlinear dilation (0 , n 2 + 2 n ) • � � � � # { P ( n ) ∩ Z 2 } � P ( n ) = � 2 (( n 2 + 2 n + 1)(2 n + 5) + f ( n )), where � = 1 � � � • • � (0 , 0) (2 n + 4 , 0) ≥ 0 | ( n 2 + 2 n ) x + (2 n + 4) y = (2 n + 4)( n 2 + 2 n ) } . f ( n ) = # { ( x , y ) ∈ Z 2 Now by generalized division and GCD, we can apply Popoviciu’s Formula and get � 2 n + 5 n even f ( n ) = n + 3 n odd