Chern Character, Loop Spaces and Derived Algebraic Geometry
... Perhaps it is also not surprising that our original motivation of understanding which are the geometric objects classified by elliptic cohomology, will lead us below to loop spaces (actually a derived version of them, better suited for algebraic geometry: see below). In fact, the chromatic picture o ...
... Perhaps it is also not surprising that our original motivation of understanding which are the geometric objects classified by elliptic cohomology, will lead us below to loop spaces (actually a derived version of them, better suited for algebraic geometry: see below). In fact, the chromatic picture o ...
de Rham cohomology
... The topological invariance of the de Rham groups suggests that there should be some purely topological way of computing them. In fact, there exists such a way and the connection between the de Rham groups and topology was first proved by Georges de Rham himself in the 1931. In these notes we give a ...
... The topological invariance of the de Rham groups suggests that there should be some purely topological way of computing them. In fact, there exists such a way and the connection between the de Rham groups and topology was first proved by Georges de Rham himself in the 1931. In these notes we give a ...
Finitely generated abelian groups, abelian categories
... Presheaves Suppose X is a topological space. Let UX be the set of all open subsets of X. We can view UX as a category, in which the morphisms are inclusion maps. A presheaf on X is a contravariant functor F : UX → A = abelian groups such that F (∅) = {0}. One typically thinks of F (U ) as a set of ...
... Presheaves Suppose X is a topological space. Let UX be the set of all open subsets of X. We can view UX as a category, in which the morphisms are inclusion maps. A presheaf on X is a contravariant functor F : UX → A = abelian groups such that F (∅) = {0}. One typically thinks of F (U ) as a set of ...
Splitting of short exact sequences for modules
... In words, the map g reduces the first component modulo a and shifts each of the other components over one position. Both f and g are Z-linear (i.e., they are additive), f is obviously injective, g is obviously surjective, and im f = aZ ⊕ 0 = ker g, so (1.1) with these modules M , N , P and linear ma ...
... In words, the map g reduces the first component modulo a and shifts each of the other components over one position. Both f and g are Z-linear (i.e., they are additive), f is obviously injective, g is obviously surjective, and im f = aZ ⊕ 0 = ker g, so (1.1) with these modules M , N , P and linear ma ...
on end0m0rpb3sms of abelian topological groups
... (ii) X generates F(X) algebraically, and (hi) for every continuous map tpoiX into any topological group G such that
... (ii) X generates F(X) algebraically, and (hi) for every continuous map tpoiX into any topological group G such that
LECTURE 2 1. Finitely Generated Abelian Groups We discuss the
... Theorem 1.5. If A is a finitely generated torsion-free abelian group that has a minimal set of generators with q elements, then A is isomorphic to the free abelian group of rank q. Proof. By induction on the minimal number of generators of A. If A is cyclic (that is, generated by one non-zero elemen ...
... Theorem 1.5. If A is a finitely generated torsion-free abelian group that has a minimal set of generators with q elements, then A is isomorphic to the free abelian group of rank q. Proof. By induction on the minimal number of generators of A. If A is cyclic (that is, generated by one non-zero elemen ...
Cohomology of cyro-electron microscopy
... [37, 36], and has relations to profound problems in computational complexity [1] and operator theory [2]. This article examines the problem from an algebraic topological angle — we will show that the problem of cryo-EM is a problem of cohomology, or, more specifically, the Čech cohomology of a simp ...
... [37, 36], and has relations to profound problems in computational complexity [1] and operator theory [2]. This article examines the problem from an algebraic topological angle — we will show that the problem of cryo-EM is a problem of cohomology, or, more specifically, the Čech cohomology of a simp ...
HOMOTOPICAL ENHANCEMENTS OF CYCLE CLASS MAPS 1
... (With some work we may choose this complex functorially.) Regrade this homologically so that Z is in degree 2d and the differential has degree −1. Let H2d (X, ZD (d)) denote the homotopy sheaf associated to this complex. Then (1) The sheafification of π0 (H2d (X, ZD (d))) is the functor of points of ...
... (With some work we may choose this complex functorially.) Regrade this homologically so that Z is in degree 2d and the differential has degree −1. Let H2d (X, ZD (d)) denote the homotopy sheaf associated to this complex. Then (1) The sheafification of π0 (H2d (X, ZD (d))) is the functor of points of ...
THE EULER CLASS OF A SUBSET COMPLEX 1. Introduction Let G
... Theorem 2.5. Let G be a finite group and X be a finite G-set. Then the Ext class ζX is equal to the Euler class e(IX ) of the augmentation module IX under the canonical isomorphism ExtnZG (Z, Z̃) ∼ = H n (G, Z̃). Proof. Fix an ordering of elements in X so that we have a fixed orientation throughout. ...
... Theorem 2.5. Let G be a finite group and X be a finite G-set. Then the Ext class ζX is equal to the Euler class e(IX ) of the augmentation module IX under the canonical isomorphism ExtnZG (Z, Z̃) ∼ = H n (G, Z̃). Proof. Fix an ordering of elements in X so that we have a fixed orientation throughout. ...
Selected Exercises 1. Let M and N be R
... or x = 0. Show that a torsion-free divisible R-module is injective. Conclude that K is an injective R-module, for any field K containing R. 20. Let R be a Noetherian commutative ring and Q an injective R-module. Fix an ideal I ⊆ R and set ΓI (Q) := {x ∈ Q | I n x = 0, for some n ≥ 0}. Show that ΓI ( ...
... or x = 0. Show that a torsion-free divisible R-module is injective. Conclude that K is an injective R-module, for any field K containing R. 20. Let R be a Noetherian commutative ring and Q an injective R-module. Fix an ideal I ⊆ R and set ΓI (Q) := {x ∈ Q | I n x = 0, for some n ≥ 0}. Show that ΓI ( ...
Some definitions that may be useful
... • A 1-morphism (A, F ) → (B, G) is a K-enriched functor E : A → B and a natural isomorphism α:F ∼ = G ◦ E, i.e. it is a triangle. • A 2-morphism is a cone on a bigon. Now again I’ll work over K = K-mod for K a ring. Pick algebras A, B and A = A-mod and B = B-mod, and take the forgetful maps as the f ...
... • A 1-morphism (A, F ) → (B, G) is a K-enriched functor E : A → B and a natural isomorphism α:F ∼ = G ◦ E, i.e. it is a triangle. • A 2-morphism is a cone on a bigon. Now again I’ll work over K = K-mod for K a ring. Pick algebras A, B and A = A-mod and B = B-mod, and take the forgetful maps as the f ...
Orbifolds and their cohomology.
... can be extended to orbifolds. Most importantly for us, there is a notion of an orbifold vector bundle (and in particular, of a tangent and cotangent bundle to an orbifold) and of de Rham cohomology. The general principle when defining the orbifold analogues of such concepts is that one should specif ...
... can be extended to orbifolds. Most importantly for us, there is a notion of an orbifold vector bundle (and in particular, of a tangent and cotangent bundle to an orbifold) and of de Rham cohomology. The general principle when defining the orbifold analogues of such concepts is that one should specif ...
Topological Field Theories
... to turn Catk into a topological category as well: given D, D ∈ Catk , discard all noninvertible morphisms in the category HomCatk (D, D0 ) = Fun(D, D0 ), obtaining a groupoid, and consider its classifying space (i.e., the geometric realization of its nerve). We are now left with two problems: • Our ...
... to turn Catk into a topological category as well: given D, D ∈ Catk , discard all noninvertible morphisms in the category HomCatk (D, D0 ) = Fun(D, D0 ), obtaining a groupoid, and consider its classifying space (i.e., the geometric realization of its nerve). We are now left with two problems: • Our ...
Equivariant cohomology and equivariant intersection theory
... compute any homogeneous component of HG∗ (X) by finite approximations of EG . This will be the starting point for the definition of equivariant Chow groups, in Section 3. Examples 1) G = S 1 (the multiplicative group of complex numbers of modulus one). Let G act on Cn by scalar multiplication; then ...
... compute any homogeneous component of HG∗ (X) by finite approximations of EG . This will be the starting point for the definition of equivariant Chow groups, in Section 3. Examples 1) G = S 1 (the multiplicative group of complex numbers of modulus one). Let G act on Cn by scalar multiplication; then ...
Homology Group - Computer Science, Stony Brook University
... Suppose M and N are simplicial complexes, f : M → N is a continuous map, ∀σ ∈ M, σ is a simplex, f (σ ) is a simplex. For each simplex, we can add its gravity center, and subdivide the simplex to multiple ones. The resulting complex is called the gravity center subdivision. Theorem Suppose M and N a ...
... Suppose M and N are simplicial complexes, f : M → N is a continuous map, ∀σ ∈ M, σ is a simplex, f (σ ) is a simplex. For each simplex, we can add its gravity center, and subdivide the simplex to multiple ones. The resulting complex is called the gravity center subdivision. Theorem Suppose M and N a ...
the orbit spaces of totally disconnected groups of transformations on
... pn(x*, M/G) =0 or 1. Thus the space M/G, in this case, would be a space that closely resembles an w-gm. If M were an orientable w-gm over Z and the group G did not reverse the orientation of M then M would be an orientable w-gm over every field L and G would act trivially on LV¿(M; L). In the case a ...
... pn(x*, M/G) =0 or 1. Thus the space M/G, in this case, would be a space that closely resembles an w-gm. If M were an orientable w-gm over Z and the group G did not reverse the orientation of M then M would be an orientable w-gm over every field L and G would act trivially on LV¿(M; L). In the case a ...
A11
... (c) Explain why q(x) splits completely in M(β). (d) Explain why M(β)/M is a Galois extension. Let G = Gal(M(β)/M). Since β is a generator of M(β)/M, to understand how σ ∈ G acts on M(β) it is enough to understand what σ does to β. (e) Explain why σ(β) = β · ζ n for some n ∈ {0, . . . , m − 1}. Let u ...
... (c) Explain why q(x) splits completely in M(β). (d) Explain why M(β)/M is a Galois extension. Let G = Gal(M(β)/M). Since β is a generator of M(β)/M, to understand how σ ∈ G acts on M(β) it is enough to understand what σ does to β. (e) Explain why σ(β) = β · ζ n for some n ∈ {0, . . . , m − 1}. Let u ...
pdf file
... 1.9 Define a basis Nk of neighborhoods of 0 in the completion M̂ by: P ∈ Nk if there exists an N such that pn ∈ ak M for all n > N . The collection of sets P + Nk where P ∈ M̂ is a basis for a topology on M̂ . The module operations and the map φ are continuous. 1.10 Let k be a field. Then k[[h]] is ...
... 1.9 Define a basis Nk of neighborhoods of 0 in the completion M̂ by: P ∈ Nk if there exists an N such that pn ∈ ak M for all n > N . The collection of sets P + Nk where P ∈ M̂ is a basis for a topology on M̂ . The module operations and the map φ are continuous. 1.10 Let k be a field. Then k[[h]] is ...
Boundary manifolds of projective hypersurfaces Daniel C. Cohen Alexander I. Suciu
... is to study the Milnor fibration f : Cℓ+1 \ {f (x) = 0} → C∗ . Of course, the different approaches are interrelated. For example, if the degree of f is n, then the Milnor fiber F = f −1 (1) is a cyclic n-fold cover of X. Consequently, knowledge of the cohomology groups of X with coefficients in cert ...
... is to study the Milnor fibration f : Cℓ+1 \ {f (x) = 0} → C∗ . Of course, the different approaches are interrelated. For example, if the degree of f is n, then the Milnor fiber F = f −1 (1) is a cyclic n-fold cover of X. Consequently, knowledge of the cohomology groups of X with coefficients in cert ...
Algebra 411 Homework 5: hints and solutions
... group, so we only get six different subgroups. I have also stated above what each subgroup is isomorphic to, although this was not part of the question. There is also the group G itself, which is a subgroup. Finally, we must consider groups generated by two or more elements. The only possible new gr ...
... group, so we only get six different subgroups. I have also stated above what each subgroup is isomorphic to, although this was not part of the question. There is also the group G itself, which is a subgroup. Finally, we must consider groups generated by two or more elements. The only possible new gr ...