The algebra of essential relations on a finite set
... P = E/N . More precisely, P is isomorphic to a product of matrix algebras over suitable group algebras, the product being indexed by the set of all (partial) order relations on X, up to permutation. Consequently, we know the Jacobson radical J(E) and we find all the simple E-modules. The idea of pas ...
... P = E/N . More precisely, P is isomorphic to a product of matrix algebras over suitable group algebras, the product being indexed by the set of all (partial) order relations on X, up to permutation. Consequently, we know the Jacobson radical J(E) and we find all the simple E-modules. The idea of pas ...
![arXiv:math/9802122v1 [math.CO] 27 Feb 1998](http://s1.studyres.com/store/data/013469206_1-76e512f42e60a4643a93990b56cee184-300x300.png)
arXiv:math/9802122v1 [math.CO] 27 Feb 1998
... only if A = i=0 {ai } ⊕ k Āi for some complete set {a0 , a1 , . . . , ak−1 } of residues modulo k, and k sets Āi , each of which satisfies min(Ai ) = 0 and tiles the integers with translation set C/k. The decomposition is unique. We can have gcd(A) = 1 although this may not be true for the Āi . ...
... only if A = i=0 {ai } ⊕ k Āi for some complete set {a0 , a1 , . . . , ak−1 } of residues modulo k, and k sets Āi , each of which satisfies min(Ai ) = 0 and tiles the integers with translation set C/k. The decomposition is unique. We can have gcd(A) = 1 although this may not be true for the Āi . ...
Group Theory G13GTH
... The units R× For any ring R, the set of units R× is a group under multiplication. Here an element r of R is a unit (or invertible element) if there is a s ∈ R such that rs = 1. If R is a field, like when R = Fp is the field of p elements for some prime p, then R× = R \ {0}. You have seen that F× p i ...
... The units R× For any ring R, the set of units R× is a group under multiplication. Here an element r of R is a unit (or invertible element) if there is a s ∈ R such that rs = 1. If R is a field, like when R = Fp is the field of p elements for some prime p, then R× = R \ {0}. You have seen that F× p i ...
SHAPIRO`S LEMMA FOR TOPOLOGICAL K
... proper if and only if the underlying group action of G on Z is proper, i.e., if and only if the structural map G × Z → Z × Z; (g, z) 7→ (g · z, z) is proper in the sense that inverse images of compact sets are compact. A map φ : Y → Z between two X o G-spaces Y and Z is X o G-equivariant if φ is G-e ...
... proper if and only if the underlying group action of G on Z is proper, i.e., if and only if the structural map G × Z → Z × Z; (g, z) 7→ (g · z, z) is proper in the sense that inverse images of compact sets are compact. A map φ : Y → Z between two X o G-spaces Y and Z is X o G-equivariant if φ is G-e ...
On separating a fixed point from zero by invariants
... Let G be a linear algebraic group over an infinite field k of any characteristic and let X be an algebraic variety over k on which G acts. Then G acts naturally on the ring of functions k[X] by g(f ) := f ◦ g −1 for f ∈ k[X] and g ∈ G. The ring of fixed points of this action is denoted by k[X]G and ...
... Let G be a linear algebraic group over an infinite field k of any characteristic and let X be an algebraic variety over k on which G acts. Then G acts naturally on the ring of functions k[X] by g(f ) := f ◦ g −1 for f ∈ k[X] and g ∈ G. The ring of fixed points of this action is denoted by k[X]G and ...
Fermat`s and Euler`s Theorem
... largest solution is x+dm0 = x+m, which does not belong to {0, 1, . . . , m−1}. So there are exactly d solutions for x. 10. See Examples 20.14 and 20.15, page 188. Homework for Section 20 (only the starred problems will be graded): ...
... largest solution is x+dm0 = x+m, which does not belong to {0, 1, . . . , m−1}. So there are exactly d solutions for x. 10. See Examples 20.14 and 20.15, page 188. Homework for Section 20 (only the starred problems will be graded): ...
Topological Methods in Combinatorics
... (c) K|k+1 is a retract of Σ(V (K))|k+1 . Lemma 2.11 Let K be a simplicial complex and U ⊆ V (K). (a) If K is k-connected and K ∩ Σ(U ) is (k − 1)-connected, then K ∪ Σ(U ) is k-connected. (b) If K ∪ Σ(U ) and K ∩ Σ(U ) are k-connected, then K is k-connected. Lemma 2.12 Let K1 and K2 be two k-connect ...
... (c) K|k+1 is a retract of Σ(V (K))|k+1 . Lemma 2.11 Let K be a simplicial complex and U ⊆ V (K). (a) If K is k-connected and K ∩ Σ(U ) is (k − 1)-connected, then K ∪ Σ(U ) is k-connected. (b) If K ∪ Σ(U ) and K ∩ Σ(U ) are k-connected, then K is k-connected. Lemma 2.12 Let K1 and K2 be two k-connect ...
Chapter 7
... An integer n is said to be a divisor of an integer i if i is an integer multiple of n; i.e., i = qn for some integer q. Thus all integers are trivially divisors of 0. The integers that have integer inverses, namely ±1, are called the units of Z. If u is a unit and n is a divisor of i, then un is a d ...
... An integer n is said to be a divisor of an integer i if i is an integer multiple of n; i.e., i = qn for some integer q. Thus all integers are trivially divisors of 0. The integers that have integer inverses, namely ±1, are called the units of Z. If u is a unit and n is a divisor of i, then un is a d ...
Base change for unit elements of Hecke algebras
... b( fE) have matching orbital integrals for all fE ~ HE. The main result of this paper is that fE, b( fE) have matching orbital integrals if fE is the unit element of HE, namely, the characteristic function of KE (recall that we normalized the measure on G(E) so that KE has measure 1). In this case b ...
... b( fE) have matching orbital integrals for all fE ~ HE. The main result of this paper is that fE, b( fE) have matching orbital integrals if fE is the unit element of HE, namely, the characteristic function of KE (recall that we normalized the measure on G(E) so that KE has measure 1). In this case b ...
DEHN FUNCTION AND ASYMPTOTIC CONES
... 3.B. Subgroups of G with contracting elements. An easy way to prove quadratic filling is the use of elements whose action by conjugation on the unipotent part is contracting. Although G itself does not contain such elements, we will show that it contains large enough such subgroups. More precisely, ...
... 3.B. Subgroups of G with contracting elements. An easy way to prove quadratic filling is the use of elements whose action by conjugation on the unipotent part is contracting. Although G itself does not contain such elements, we will show that it contains large enough such subgroups. More precisely, ...
Factoring by Grouping
... In other words, when we write our factorization we need to verify that each polynomial in the factorization is non-factorable (or prime). Sometimes it may be necessary to factor more than once to produce this “complete” factorization. Factor completely by Grouping. ...
... In other words, when we write our factorization we need to verify that each polynomial in the factorization is non-factorable (or prime). Sometimes it may be necessary to factor more than once to produce this “complete” factorization. Factor completely by Grouping. ...
Birkhoff's representation theorem
This is about lattice theory. For other similarly named results, see Birkhoff's theorem (disambiguation).In mathematics, Birkhoff's representation theorem for distributive lattices states that the elements of any finite distributive lattice can be represented as finite sets, in such a way that the lattice operations correspond to unions and intersections of sets. The theorem can be interpreted as providing a one-to-one correspondence between distributive lattices and partial orders, between quasi-ordinal knowledge spaces and preorders, or between finite topological spaces and preorders. It is named after Garrett Birkhoff, who published a proof of it in 1937.The name “Birkhoff's representation theorem” has also been applied to two other results of Birkhoff, one from 1935 on the representation of Boolean algebras as families of sets closed under union, intersection, and complement (so-called fields of sets, closely related to the rings of sets used by Birkhoff to represent distributive lattices), and Birkhoff's HSP theorem representing algebras as products of irreducible algebras. Birkhoff's representation theorem has also been called the fundamental theorem for finite distributive lattices.