![arXiv:math/0403252v1 [math.HO] 16 Mar 2004](http://s1.studyres.com/store/data/017716978_1-976b7d80197f8d34f66860694a7698bb-300x300.png)
arXiv:math/0403252v1 [math.HO] 16 Mar 2004
... field. If it has an additional argument t, it is called a time-dependent vector field. Let v be the value of function (2.1) at the point A in a river. Then vector v is a bound vector. It represents the velocity of the water jet at the point A. Hence, it is bound to point A. Certainly, one can transl ...
... field. If it has an additional argument t, it is called a time-dependent vector field. Let v be the value of function (2.1) at the point A in a river. Then vector v is a bound vector. It represents the velocity of the water jet at the point A. Hence, it is bound to point A. Certainly, one can transl ...
K(n)-COMPACT SPHERES H˚ akon Schad Bergsaker Contents
... up −1 . The coefficient rings π∗ (K(n)) etc. are often written as K(n)∗ (or K(n)∗ = π−∗ (K(n)) in cohomology). The above is the classical construction of these theories. Now we would like to consider products. This used to be rather complicated, but with the new foundations for stable homotopy theor ...
... up −1 . The coefficient rings π∗ (K(n)) etc. are often written as K(n)∗ (or K(n)∗ = π−∗ (K(n)) in cohomology). The above is the classical construction of these theories. Now we would like to consider products. This used to be rather complicated, but with the new foundations for stable homotopy theor ...
EQUIVALENCE OF QUOTIENT HILBERT MODULES 1
... Pm K(·,¯ w) to K̃(·, w) extends linearly to an isometric module map. Therefore, i,j=1 ∂i ∂j log K(z, z)dzi ∧ dz̄j is a complete invariant for the module M The second approach is to normalise the reproducing kernel K, that is, define the kernel K0 (z, w) = ψ(z)K(z, w)ψ(w), where ψ(z) = K(z, w0 )−1 K( ...
... Pm K(·,¯ w) to K̃(·, w) extends linearly to an isometric module map. Therefore, i,j=1 ∂i ∂j log K(z, z)dzi ∧ dz̄j is a complete invariant for the module M The second approach is to normalise the reproducing kernel K, that is, define the kernel K0 (z, w) = ψ(z)K(z, w)ψ(w), where ψ(z) = K(z, w0 )−1 K( ...
Fourier analysis on finite abelian groups 1.
... to C× . However, an equally important use is for the trace of a group homomorphism ρ : G → GLn (k) from G to invertible n-by-n matrices with entries in a field k. In the latter sense, ...
... to C× . However, an equally important use is for the trace of a group homomorphism ρ : G → GLn (k) from G to invertible n-by-n matrices with entries in a field k. In the latter sense, ...
The Zero-Sum Tensor
... Department of Game Design Uppsala University Visby, Sweden mikael.fridenfalk@speldesign.uu.se Abstract—The zero-sum matrix, or in general, tensor, reveals some consistent properties at multiplication. In this paper, three mathematical rules are derived for multiplication involving such entities. The ...
... Department of Game Design Uppsala University Visby, Sweden mikael.fridenfalk@speldesign.uu.se Abstract—The zero-sum matrix, or in general, tensor, reveals some consistent properties at multiplication. In this paper, three mathematical rules are derived for multiplication involving such entities. The ...
HOMOLOGY ISOMORPHISMS BETWEEN ALGEBRAIC GROUPS MADE DISCRETE
... Hp (Rδ , Z) is the p’th exterior power of R (see the proof of Lemma 9 below). 3. Proof of Theorem 1 Let V be an abelian group which carries the structure of a rational vector space, that is, the Z scalar multiplication extends to one of Q. Let α be an endomorphism of the group V . Then it is readily ...
... Hp (Rδ , Z) is the p’th exterior power of R (see the proof of Lemma 9 below). 3. Proof of Theorem 1 Let V be an abelian group which carries the structure of a rational vector space, that is, the Z scalar multiplication extends to one of Q. Let α be an endomorphism of the group V . Then it is readily ...
solutions - Cornell Math
... with an homogeneous of degree n and an = 0 for almost all n. Example 1. Let A be a polynomial ring k[x1 , . . . , xm ]. Then A is a graded ring with An equal to the set of homogeneous polynomials of degree n in the usual sense; in other words, An is the k-span of the set of monomials of degree n. St ...
... with an homogeneous of degree n and an = 0 for almost all n. Example 1. Let A be a polynomial ring k[x1 , . . . , xm ]. Then A is a graded ring with An equal to the set of homogeneous polynomials of degree n in the usual sense; in other words, An is the k-span of the set of monomials of degree n. St ...
The Functor Category in Relation to the Model Theory of Modules
... pp-formulas = finitely generated subfunctors of representables: If φ is a pp-formula, Fφ is a finitely generated subfunctor of a representable functor in fp(mod(R), Ab). If F is a finitely generated subfunctor of a representable functor, then there exists pp-formula φ such that F ∼ = Fφ . ...
... pp-formulas = finitely generated subfunctors of representables: If φ is a pp-formula, Fφ is a finitely generated subfunctor of a representable functor in fp(mod(R), Ab). If F is a finitely generated subfunctor of a representable functor, then there exists pp-formula φ such that F ∼ = Fφ . ...
Tensors, Vectors, and Linear Forms Michael Griffith May 9, 2014
... The trouble with this new definition is that it leaves us without a sense of how to actually going about adding vectors, or really comparing them in any meaningful way. In stripping out the geometry, we have also removed much of the methodology with which we are familiar. In order to reconstruct th ...
... The trouble with this new definition is that it leaves us without a sense of how to actually going about adding vectors, or really comparing them in any meaningful way. In stripping out the geometry, we have also removed much of the methodology with which we are familiar. In order to reconstruct th ...
Explicit tensors - Computational Complexity
... between V1 ⊗ · · · ⊗ Vn and k d1 ⊗ · · · ⊗ k dn . There is also a way of defining tensor products without choosing bases at all by the universal property of turning multilinear mappings into linear ones. 2.2. Basic properties Let π ∈ Sn be a permutation of {1, . . . , n}. If vi ∈ Vi , then vπ(1) ⊗· ...
... between V1 ⊗ · · · ⊗ Vn and k d1 ⊗ · · · ⊗ k dn . There is also a way of defining tensor products without choosing bases at all by the universal property of turning multilinear mappings into linear ones. 2.2. Basic properties Let π ∈ Sn be a permutation of {1, . . . , n}. If vi ∈ Vi , then vπ(1) ⊗· ...
Three Directions In Waring Ranks and Apolarity
... What is r (F ) if F is a product of linear forms, possibly with repetition (defining equation of a hyperplane multiarrangement)? In joint work with Alex Woo (University of Idaho) we address this when the hyperplane arrangement arises as the set of mirrors of a finite reflection group. Example Q The ...
... What is r (F ) if F is a product of linear forms, possibly with repetition (defining equation of a hyperplane multiarrangement)? In joint work with Alex Woo (University of Idaho) we address this when the hyperplane arrangement arises as the set of mirrors of a finite reflection group. Example Q The ...
10. Modules over PIDs - Math User Home Pages
... with endomorphism T being multiplication by x (on the quotient). Choice of k-basis that illuminates the action of T is more ambiguous now. Still, there are not very many plausible natural choices. Let d = deg f . Then take k-basis consisting of (the images in the quotient of) 1, x, x2 , . . . , xd−1 ...
... with endomorphism T being multiplication by x (on the quotient). Choice of k-basis that illuminates the action of T is more ambiguous now. Still, there are not very many plausible natural choices. Let d = deg f . Then take k-basis consisting of (the images in the quotient of) 1, x, x2 , . . . , xd−1 ...
Splitting of short exact sequences for modules
... In Section 2 we will give two ways to characterize when a short exact sequence of Rmodules splits. Section 3 will discuss a few consequences. Before doing that, we want to stress that being split is not just saying there is an isomorphism M → N ⊕ P of Rmodules, but how the isomorphism works with the ...
... In Section 2 we will give two ways to characterize when a short exact sequence of Rmodules splits. Section 3 will discuss a few consequences. Before doing that, we want to stress that being split is not just saying there is an isomorphism M → N ⊕ P of Rmodules, but how the isomorphism works with the ...
No Slide Title
... Hilbert spaces • A Hilbert space is a vector space in which the scalars are complex numbers, with an inner product (dot product) operation : H×H C – Definition of inner product: Black dot is an inner xy ...
... Hilbert spaces • A Hilbert space is a vector space in which the scalars are complex numbers, with an inner product (dot product) operation : H×H C – Definition of inner product: Black dot is an inner xy ...
Sheaf Cohomology
... with no non-trivial global sections this then equals M = 0; Which is sufficient because the sheaf induced by the module M = 0 is the trivial sheaf. The sheaf in question is simply (the sheaf associated to the presheaf defined by): ...
... with no non-trivial global sections this then equals M = 0; Which is sufficient because the sheaf induced by the module M = 0 is the trivial sheaf. The sheaf in question is simply (the sheaf associated to the presheaf defined by): ...
Frobenius algebras and monoidal categories
... in V (called associativity and unital constraints), such that the pentagon, involving the five ways of ...
... in V (called associativity and unital constraints), such that the pentagon, involving the five ways of ...
representable functors and operations on rings
... are rings, then the set Hom^(J5,i2) of ring homomorphisms does not, in general, have a ring structure (unlike, for instance, the case where G, H are abelian groups, in which case Homg(G,H) is naturally an abelian group). However, we shall show in § 1 that, if B also has a 'co-ring' structure (in whi ...
... are rings, then the set Hom^(J5,i2) of ring homomorphisms does not, in general, have a ring structure (unlike, for instance, the case where G, H are abelian groups, in which case Homg(G,H) is naturally an abelian group). However, we shall show in § 1 that, if B also has a 'co-ring' structure (in whi ...
Derived funcors, Lie algebra cohomology and some first applications
... If M is trivial, all inner derivations are zero, hence: Corollary 5.1. If M is a trivial g-modul H 1 (g, M ) ∼ = Der(g, M ) ∼ = Homk (gab , M ). Considering k as a trivial g-module yields, for semisimple and finite dimensional g and char(k) = 0: Corollary 5.2. If g is finite-dimensional and semisim ...
... If M is trivial, all inner derivations are zero, hence: Corollary 5.1. If M is a trivial g-modul H 1 (g, M ) ∼ = Der(g, M ) ∼ = Homk (gab , M ). Considering k as a trivial g-module yields, for semisimple and finite dimensional g and char(k) = 0: Corollary 5.2. If g is finite-dimensional and semisim ...
Algebraic Geometry 3-Homework 11 1. a. Let O be a noetherian
... 1. a. Let O be a noetherian local unique factorization domain (UFD). Show that O is a normal ring. Show that O is a discrete valuation ring (DVR) if O has Krull dimension one. b. Let X be an integral k-scheme. X is called locally factorial if each local ring OX,x is a UFD (for example, a smooth k-sc ...
... 1. a. Let O be a noetherian local unique factorization domain (UFD). Show that O is a normal ring. Show that O is a discrete valuation ring (DVR) if O has Krull dimension one. b. Let X be an integral k-scheme. X is called locally factorial if each local ring OX,x is a UFD (for example, a smooth k-sc ...
Homework 5
... 2. In this exercise we assume that A, B, C are complexes of abelian groups, although everything works for complexes of modules over any ring R. We denote by Kom(Z) the category of complexes of abelian groups and their morphisms. (a) Given a morphism f : A −→ B of complexes, explain how to define com ...
... 2. In this exercise we assume that A, B, C are complexes of abelian groups, although everything works for complexes of modules over any ring R. We denote by Kom(Z) the category of complexes of abelian groups and their morphisms. (a) Given a morphism f : A −→ B of complexes, explain how to define com ...
Part III. Homomorphisms and Factor Groups
... Note. By Theorem 13.12 part (4), we know that for K < G0 , where K = {e0}, we have φ−1 [K] < G. This subgroup φ−1 [K] includes all elements of G mapped under φ to e0 . You encountered a similar idea in linear algebra when considering an m × n matrix A as a linear transformation from Rn to Rm. Rn and ...
... Note. By Theorem 13.12 part (4), we know that for K < G0 , where K = {e0}, we have φ−1 [K] < G. This subgroup φ−1 [K] includes all elements of G mapped under φ to e0 . You encountered a similar idea in linear algebra when considering an m × n matrix A as a linear transformation from Rn to Rm. Rn and ...
Noncommutative Lp-spaces of W*-categories and their applications
... 2.2 Intuitively, *-categories, Banach *-categories, C*-categories, and W*-categories should be thought of as many-objects versions (horizontal categorifications) of *-algebras, Banach *-algebras, C*-algebras, and von Neumann algebras. In particular, in the case when the set of objects has one elemen ...
... 2.2 Intuitively, *-categories, Banach *-categories, C*-categories, and W*-categories should be thought of as many-objects versions (horizontal categorifications) of *-algebras, Banach *-algebras, C*-algebras, and von Neumann algebras. In particular, in the case when the set of objects has one elemen ...
1. Grassmann Bundles. Note if you have a smooth embedded curve
... Lemma 1.4. The map π above makes γkn = γ n (Rn+k ) into a smooth vector bundle of rank k. Proof. We just need to construct trivializations of π over each chart ΨV ◦ ΦV : UV → Rnk from the previous proof. Here ΨV involved choosing a basis for V and V ⊥ . Hence we have a natural isomorphism φ : V → Rn ...
... Lemma 1.4. The map π above makes γkn = γ n (Rn+k ) into a smooth vector bundle of rank k. Proof. We just need to construct trivializations of π over each chart ΨV ◦ ΦV : UV → Rnk from the previous proof. Here ΨV involved choosing a basis for V and V ⊥ . Hence we have a natural isomorphism φ : V → Rn ...
Solutions to Practice Quiz 6
... 5. Let G = Z4 ⊕ Z4 , and consider the subgroups H = {(0, 0), (2, 0), (0, 2), (2, 2)} and K = h(1, 2)i. Identify the following groups (as direct products of cyclic groups of prime order): (a) H and G/H. Clearly, H ∼ = Z2 ⊕ Z2 . Moreover, G/H ∼ = Z2 ⊕ Z2 ; indeed, the assignment (0, 1) 7→ (0, 1) and ( ...
... 5. Let G = Z4 ⊕ Z4 , and consider the subgroups H = {(0, 0), (2, 0), (0, 2), (2, 2)} and K = h(1, 2)i. Identify the following groups (as direct products of cyclic groups of prime order): (a) H and G/H. Clearly, H ∼ = Z2 ⊕ Z2 . Moreover, G/H ∼ = Z2 ⊕ Z2 ; indeed, the assignment (0, 1) 7→ (0, 1) and ( ...
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.