Tannaka Duality for Geometric Stacks
... X is determined by the category QCX . In order to make a more precise statement, we must first decide what sort of object QCX is. The relevant definitions and the statement of our main result, Theorem 5.11, will be given in this section. Recall that a symmetric monoidal category is a category C equi ...
... X is determined by the category QCX . In order to make a more precise statement, we must first decide what sort of object QCX is. The relevant definitions and the statement of our main result, Theorem 5.11, will be given in this section. Recall that a symmetric monoidal category is a category C equi ...
Solution 3 - D-MATH
... This implies that the group generated by f and g has the same multiplication table as S3 , hence the homomorphism defined by sending f to (12) and g to (123) gives an isomorphism of the two groups. 2. In the additive group Rm of vectors, let W be the set of solutions of a system of homogeneous linea ...
... This implies that the group generated by f and g has the same multiplication table as S3 , hence the homomorphism defined by sending f to (12) and g to (123) gives an isomorphism of the two groups. 2. In the additive group Rm of vectors, let W be the set of solutions of a system of homogeneous linea ...
MATH 6280 - CLASS 2 Contents 1. Categories 1 2. Functors 2 3
... (6) Homology Hn : Top → Ab which sends a space S to the n’th simplicial homology group of S. (7) Cohomology H n : Topop → Ab which sends a space S to the n’th simplicial cohomology group of S. (8) The homotopy groups functors: πn : Topop → Ab (9) If F : G → H is a group homomorphism, then it gives r ...
... (6) Homology Hn : Top → Ab which sends a space S to the n’th simplicial homology group of S. (7) Cohomology H n : Topop → Ab which sends a space S to the n’th simplicial cohomology group of S. (8) The homotopy groups functors: πn : Topop → Ab (9) If F : G → H is a group homomorphism, then it gives r ...
0.3 Abelian groups - NIU Math Department
... Of course, in this case the two operations are not independent–they are connected by the distributive laws. The definition of an abelian group is also useful in discussing vector spaces and modules. In fact, we can define a vector space to be an abelian group together with a scalar multiplication sa ...
... Of course, in this case the two operations are not independent–they are connected by the distributive laws. The definition of an abelian group is also useful in discussing vector spaces and modules. In fact, we can define a vector space to be an abelian group together with a scalar multiplication sa ...
SIMPLE MODULES OVER FACTORPOWERS 1. Introduction and
... the set of idempotents of S; and by D, L, R, H, J the corresponding Green’s relations on S. Let S be a finite semigroup acting on a finite set M by (everywhere defined) transformations. Consider the power semigroup P(S), which consists of all subsets of S with the natural multiplication A · B = {a · ...
... the set of idempotents of S; and by D, L, R, H, J the corresponding Green’s relations on S. Let S be a finite semigroup acting on a finite set M by (everywhere defined) transformations. Consider the power semigroup P(S), which consists of all subsets of S with the natural multiplication A · B = {a · ...
Math 8211 Homework 1 PJW
... We say that a functor F : C → D is an isomorphism of categories if there is a functor G : D → C so that GF = 1C and F G = 1D . We say that a functor F : C → D is an equivalence of categories if there is a functor G : D → C so that GF is naturally isomorphic to the identity functor 1C and F G is natu ...
... We say that a functor F : C → D is an isomorphism of categories if there is a functor G : D → C so that GF = 1C and F G = 1D . We say that a functor F : C → D is an equivalence of categories if there is a functor G : D → C so that GF is naturally isomorphic to the identity functor 1C and F G is natu ...
HOMEWORK 1 SOLUTIONS Solution.
... Z-module homomorphism from a cyclic module to any module is determined by where a generator is sent. (See Problem 10.2.9 below.) Let φ : Z/30Z → Z/21Z be a Z-module homomorphism. Then we must have 30φ(1) = 0. The elements y ∈ Z/21Z so that 30y = 0 are y = 7k (mod 21) for k = 0, 1, 2, so there are th ...
... Z-module homomorphism from a cyclic module to any module is determined by where a generator is sent. (See Problem 10.2.9 below.) Let φ : Z/30Z → Z/21Z be a Z-module homomorphism. Then we must have 30φ(1) = 0. The elements y ∈ Z/21Z so that 30y = 0 are y = 7k (mod 21) for k = 0, 1, 2, so there are th ...
Math 614, Fall 2015 Problem Set #1: Solutions 1. (a) Since every
... B, i.e., iff both A and B are in its complement, which is equivalent to the condition that the prime be disjoint from AB. Hence, there exists a prime contained both P and Q unless 0 ∈ AB. But if a ∈ A and b ∈ B are such that ab = 0, the D(a)∩D(b) = D(ab) = D(0) = ∅, and P ∈ D(a) while Q ∈ D(b), so ...
... B, i.e., iff both A and B are in its complement, which is equivalent to the condition that the prime be disjoint from AB. Hence, there exists a prime contained both P and Q unless 0 ∈ AB. But if a ∈ A and b ∈ B are such that ab = 0, the D(a)∩D(b) = D(ab) = D(0) = ∅, and P ∈ D(a) while Q ∈ D(b), so ...
Note on Nakayama`s Lemma For Compact Λ
... X/IX is finite then X is actually a torsion Λ module. This is immediate from the structure theorem for Λ modules that we have in this case. This result does not, however, extend to other pro-p groups in general. We first note that the concept of a torsion module is only useful when Λ has no zero div ...
... X/IX is finite then X is actually a torsion Λ module. This is immediate from the structure theorem for Λ modules that we have in this case. This result does not, however, extend to other pro-p groups in general. We first note that the concept of a torsion module is only useful when Λ has no zero div ...
Multiplying Expressions - Key Resources for GCSE Mathematics
... Now consider the two questions below. You may wish to refer to your copy of the algebra strand of the Revised learning objectives for mathematics for Key Stages 3 and 4. • Where could activities based on a multiplication grid, as in Resources 1a, 1b and 1c, fit into mathematics lessons in Years 7 to ...
... Now consider the two questions below. You may wish to refer to your copy of the algebra strand of the Revised learning objectives for mathematics for Key Stages 3 and 4. • Where could activities based on a multiplication grid, as in Resources 1a, 1b and 1c, fit into mathematics lessons in Years 7 to ...
TENSOR PRODUCTS II 1. Introduction Continuing our study of
... for a linear map, ϕ and ψ belong to the R-modules HomR (M, M 0 ) and HomR (N, N 0 ), so one could ask if the actual elementary tensor ϕ ⊗ ψ in HomR (M, M 0 ) ⊗R HomR (N, N 0 ) is related to the linear map ϕ ⊗ ψ : M ⊗R N → M 0 ⊗R N 0 . Theorem 2.5. There is a linear map HomR (M, M 0 ) ⊗R HomR (N, N 0 ...
... for a linear map, ϕ and ψ belong to the R-modules HomR (M, M 0 ) and HomR (N, N 0 ), so one could ask if the actual elementary tensor ϕ ⊗ ψ in HomR (M, M 0 ) ⊗R HomR (N, N 0 ) is related to the linear map ϕ ⊗ ψ : M ⊗R N → M 0 ⊗R N 0 . Theorem 2.5. There is a linear map HomR (M, M 0 ) ⊗R HomR (N, N 0 ...
Math 31 – Homework 5 Solutions
... Thus D3 acts on the set X = {1, 2, 3} of vertices by permuting them. Determine the orbit and stabilizer of each vertex under this action. Solution. Let’s start with the vertex 1. First note that if r1 denotes counterclockwise rotation by 120 degrees, r1 · 1 = 2, so 2 ∈ orb(1). Also, r2 · 1 = 3, so ...
... Thus D3 acts on the set X = {1, 2, 3} of vertices by permuting them. Determine the orbit and stabilizer of each vertex under this action. Solution. Let’s start with the vertex 1. First note that if r1 denotes counterclockwise rotation by 120 degrees, r1 · 1 = 2, so 2 ∈ orb(1). Also, r2 · 1 = 3, so ...
MULTILINEAR ALGEBRA: THE EXTERIOR PRODUCT This writeup
... Vk That is, the equality shows that any monomial in A M is a linear combination of {ei1 ∧· · ·∧eik }. By skew symmetry, we may assume that i1 < · · · < ik . Since any Vk element of A M is a linear combination of monomials in turn, we are done. ...
... Vk That is, the equality shows that any monomial in A M is a linear combination of {ei1 ∧· · ·∧eik }. By skew symmetry, we may assume that i1 < · · · < ik . Since any Vk element of A M is a linear combination of monomials in turn, we are done. ...
dmodules ja
... The category of modules over the Weyl algebra is a well-studied algebraic object and we hope to study -modules on X by using these methods. In particular, effective algorithms have been developed for -modules on affine space; for example, see the work of Oaku [12], Walther [16], Saito et al. [15], ...
... The category of modules over the Weyl algebra is a well-studied algebraic object and we hope to study -modules on X by using these methods. In particular, effective algorithms have been developed for -modules on affine space; for example, see the work of Oaku [12], Walther [16], Saito et al. [15], ...
Math 594. Solutions 3 Book problems §5.1: 14. Let G = A1 × A2
... (since g0 has order n and χ0 gives an isomorphism of hg0 i with µn (C)), so that χ(G) = µn (C). Now we claim that H ∩ K = 1. Indeed, if h ∈ H ∩ K, then χ(h) = 1 since h ∈ K, but χ induces an isomorphism χ0 : H → µn (C), so χ(h) = χ0 (h) = 1 forces h = 1 since, in particular, χ0 is injective. We conc ...
... (since g0 has order n and χ0 gives an isomorphism of hg0 i with µn (C)), so that χ(G) = µn (C). Now we claim that H ∩ K = 1. Indeed, if h ∈ H ∩ K, then χ(h) = 1 since h ∈ K, but χ induces an isomorphism χ0 : H → µn (C), so χ(h) = χ0 (h) = 1 forces h = 1 since, in particular, χ0 is injective. We conc ...
Math 412: Problem Set 2 (due 21/9/2016) Practice P1 Let {V i∈I be
... 4. Write Mn (F) for the space of n×n matrices with entries in F. Let sln (F) = {A ∈ Mn (F) | Tr A = 0} and let pgln (F) = Mn (F)/F · In (matrices modulu scalar matrices). Suppose that n is invertible in F (equivalently, that the characteristic of F does not divide n). Show that the quotient map Mn ( ...
... 4. Write Mn (F) for the space of n×n matrices with entries in F. Let sln (F) = {A ∈ Mn (F) | Tr A = 0} and let pgln (F) = Mn (F)/F · In (matrices modulu scalar matrices). Suppose that n is invertible in F (equivalently, that the characteristic of F does not divide n). Show that the quotient map Mn ( ...
EXAMPLE SHEET 1 1. If k is a commutative ring, prove that b k
... is a C-comodule, write ρpxq “ ni“1 xi b ci with the ci linearly independent, and proceed in a similar way to the proof of the fundamental theorem of coalgebras). 16. Suppose that k is a field, and consider the functor Setop f Ñ Vect given on objects X by X ÞÑ k (Setf is the category of finite sets). ...
... is a C-comodule, write ρpxq “ ni“1 xi b ci with the ci linearly independent, and proceed in a similar way to the proof of the fundamental theorem of coalgebras). 16. Suppose that k is a field, and consider the functor Setop f Ñ Vect given on objects X by X ÞÑ k (Setf is the category of finite sets). ...
Symmetric Spectra Talk
... complex vector bundle EU (n)+ ×U (n) Cn over BU (n) = EU (n)/U (n). EU (n) is the canonical U (n)bundle over BU (n). To construct an associated vector bundle, we can take the standard representation, that is, take EU (n) × Cn and mod out by the diagonal action of the unitary group. (U (n) acts on Cn ...
... complex vector bundle EU (n)+ ×U (n) Cn over BU (n) = EU (n)/U (n). EU (n) is the canonical U (n)bundle over BU (n). To construct an associated vector bundle, we can take the standard representation, that is, take EU (n) × Cn and mod out by the diagonal action of the unitary group. (U (n) acts on Cn ...
Lecture notes on Witt vectors
... is a natural transformation of functors from rings to rings. Proof. Let A be the polynomial ring Z[an , bn | n ∈ S]. Then the unique ring homomorphism φp : A → A that maps an to apn and bn to bpn satisfies that φp (f ) = f p modulo pA. Let a and b be the sequences (an | n ∈ S) and (bn | n ∈ S). Sinc ...
... is a natural transformation of functors from rings to rings. Proof. Let A be the polynomial ring Z[an , bn | n ∈ S]. Then the unique ring homomorphism φp : A → A that maps an to apn and bn to bpn satisfies that φp (f ) = f p modulo pA. Let a and b be the sequences (an | n ∈ S) and (bn | n ∈ S). Sinc ...
Multilinear spectral theory
... rank-1 and rank-(R1, . . . , RN ) approximation of higher-order tensors,” SIAM J. Matrix Anal. Appl., 21 (4), 2000, pp. 1324–1342. Same equations first appeared in the context of rank-1 tensor approximations. Our study differs in that we are interested in all critical values as opposed to only the m ...
... rank-1 and rank-(R1, . . . , RN ) approximation of higher-order tensors,” SIAM J. Matrix Anal. Appl., 21 (4), 2000, pp. 1324–1342. Same equations first appeared in the context of rank-1 tensor approximations. Our study differs in that we are interested in all critical values as opposed to only the m ...
contact email: donsen2 at hotmail.com Contemporary abstract
... Let a ∈ Aut(Z). Then, we can find two Automorphism; identity Automorphism and Automorphism with a(n) = −n. Note that Z is a cyclic group since every nonzero integer can be written as a finite sum 1 + 1 + · · · or (−1) + (−1) + · · · (−1) and any Automorphism( which including Isomorphisms) has a mapp ...
... Let a ∈ Aut(Z). Then, we can find two Automorphism; identity Automorphism and Automorphism with a(n) = −n. Note that Z is a cyclic group since every nonzero integer can be written as a finite sum 1 + 1 + · · · or (−1) + (−1) + · · · (−1) and any Automorphism( which including Isomorphisms) has a mapp ...
31 Semisimple Modules and the radical
... examine modules over arbitrary algebras A having this property, i.e., that they are direct sums of simple modules. We assume as before that A is a finite dimensional algebra over the field F and that all modules are finitely generated (and thus finite dimensional) left A-modules. Furthermore, the le ...
... examine modules over arbitrary algebras A having this property, i.e., that they are direct sums of simple modules. We assume as before that A is a finite dimensional algebra over the field F and that all modules are finitely generated (and thus finite dimensional) left A-modules. Furthermore, the le ...
(A SOMEWHAT GENTLE INTRODUCTION TO) DIFFERENTIAL
... Theorem 1.1. Let (R, m) → (S, n) be a flat local ring homomorphism, that is, a ring homomorphism making S into a flat R-module such that mS ⊆ n. Then S is Gorenstein if and only if R and S/mS are Gorenstein. Moreover, there is an equality of Bass series I S (t) = I R (t)I S/mS (t). (See Definition A ...
... Theorem 1.1. Let (R, m) → (S, n) be a flat local ring homomorphism, that is, a ring homomorphism making S into a flat R-module such that mS ⊆ n. Then S is Gorenstein if and only if R and S/mS are Gorenstein. Moreover, there is an equality of Bass series I S (t) = I R (t)I S/mS (t). (See Definition A ...
Tensor product of modules
In mathematics, the tensor product of modules is a construction that allows arguments about bilinear maps (e.g. multiplication) to be carried out in terms of linear maps (module homomorphisms). The module construction is analogous to the construction of the tensor product of vector spaces, but can be carried out for a pair of modules over a commutative ring resulting in a third module, and also for a pair of a right-module and a left-module over any ring, with result an abelian group. Tensor products are important in areas of abstract algebra, homological algebra, algebraic topology and algebraic geometry. The universal property of the tensor product of vector spaces extends to more general situations in abstract algebra. It allows the study of bilinear or multilinear operations via linear operations. The tensor product of an algebra and a module can be used for extension of scalars. For a commutative ring, the tensor product of modules can be iterated to form the tensor algebra of a module, allowing one to define multiplication in the module in a universal way.