Homological Algebra - Science topic

A ring R is called von Neumann regular if for every x in R there is x' such that xx'x=x. Matrix ring is of course an example.

The difficulty depends on your view of the (co)homology groups. Generally speaking, explicit formulas involving (co)cycles are not of great use. The best approach is to go back to the fundamental property of the (co)homology functor, as is done in Serre's "Local Fields", chap.VII, which is in my opinion the best introduction to the subject.

Let us look e.g. at homology. Given a group G and its group algebra A = Z[G], an exact sequence of Z[G]-modules o --> A --> B --> C --> 0 gives rise to an exact sequence of co-invariants ... --> A_{G -->} B_G --> C_G --> 0, where H₀ (G,M) _:= (M)_G denotes the maximal quotient of M on which G acts as identity. But this sequence is not exact on the left, where it extends to an a priori infinite exact sequnce of homology groups ... --> H₁(G,A) --> H₁(G,B) --> H₁(G,C) --> H₀(G,A) --> ... The philosophy is that the homology functor is an automatic machinery which transforms a short exact sequence into an infinite exact sequence on the left. Think of the Taylor development of a function f at a point M, which you can write down without thinking, but the terms of the expansion give you valuable information on f itself in the neighbourhood of M.

1) Let us apply this to your specific problem. Denote by I_G the augmentation ideal of A , i.e. the ideal generated by all the elements 1-g when g runs through G. Obviously H₀(G, M) = M/I_GM . Consider the natural exact sequence 0 --> I_G --> A --> Z --> 0, where G acts by conjugation on G and as identity on Z, and the rightmost map consists in taking the sum of the coefficients of an element of the group algebra A. Taking homology gives ... --> H₁(G, A) --> H₁(G, Z) --> H₀(G, I_G) --> H₀(G,A) -->... But A is a free A-module, hence H₁(G, A) = 0, and the image of H₀(G, I_G) --> H₀(G,A) is null practically by definition, so that H₁(G, Z) = I_G/ I_G² , and one can check "by hand" that the map g --> 1-g induces an isomorphism H₁(G, Z) = G/G' (remember that G acts by conjugation on itself).

Consider now the exact sequence 0 --> Z --> Z --> Z/n --> 0, where the leftmost map is multiplication by n , hence is injective.Taking homology gives H₁(G, Z) --> H₁(G, Z) --> H₁(G, Z/n) --> Z --> Z . Applying our preliminary result H₁(G, Z) = G/G' , we get immediately that G/G'Gⁿ = H₁(G, Z/n) as desired.

2) Suppose now that G is a finite p-group, where p is a prime, and write F_p instead of Z/p. After 1), we know that G/G'G^p = H₁(G, F_p). Both sides are finite abelian groups killed by p, so they can be viewed (when written additively) as F_p finite - dimensional vector spaces. Let d(G) denote the minimal number of generators of G. Passing to the quotient, we get obviously d(G) >= d(G/G'G^p ), and this last quantity is = dim G/G'G^p by elementary linear algebra.

The reverse inequality holds true, but its proof is somewhat independent from the previous homological considerations. Write G*=G/G'G^p. A theorem of M. Hall in group theory implies that a homomorphism f : G₁ --> G₂ is surjective iff the induced map f* : G₁* --> G₂* is surjective. The equality d(G) = dim G/G'G^p is known as the Burnside basis theorem.

I think that you can use the article

FROM HOMOLOGICAL ALGEBRA TO GROUP COHOMOLOGY

Semester Project By

Maximilien Holmberg-Péroux

Responsible Professor

prof. Jacques Thévenaz

Supervisor

Rosalie Chevalley

Academic year : 2013-2014

Spring Semester

Sorry for delay in my answer. I meant that it is sufficient to define for every point x₀ of X an open neighborhood U₀ of x₀ in Rⁿ in such a way that the theorem is applicable for the product of Omega and U₀. This neighborhood can be small (not necessarily containing all the set X), but you need some U₀ (depending on x₀) for each x₀.

how is the Groebner - Shirshov basis for the classification using diamond lemma

Recall the following fact.  Let A be a linear operator on a vector space V (over any field), and let v be a nonzero vector in V. Then A-cyclic subspace L_v is the smallest A-invariant subspace that contains v.

  For n dimension the group Tor_1(M,F)=0 for all modules F ( F is flat, amenable,...) certain classes of Bananch and c* modules.

Maybe you would like to see geometrically how the n-th homology of  S^n can be killed when projected to RP^n? Take a triangulation of the n-sphere which is invariant under the antipodal map  -I , e.g. the generalized octahedron (which exists in every dimension). The projection is a triangulation of the projective space RP^n, and each of its simplices is hit twice by the projection map S^n \to RP^n. However, when n is even, the antipodal map -I on R^{n+1} reverses orientation (determinant -1) but preserves the exterior normal vector field of S^n, thus it reverses the orientation on S^n. Hence each simplex contributes with both signs and you get extinction, while in odd dimensions you obtain a factor 2 (the mapping degree of the projection map).

Best regards

Jost

Hello, Kelvin,

consider the (right) transversal T for H in G, i.e. G=disjoint union_{t \in T} H.t  of left cosets, |T|=|G:H|. It is known that FG=direct sum_{t \in T} FH.t, one of the summands being simply FH (for, say, t=1). Each summand is a free FH-module isomorphic to FH. If you take the quotient FG/FH you get direct sum_{t \in T, t \ne 1} FH.t of free FH-modules of the cardinality (if finite) |T|-1=|G:H|-1. For instance, consider the case G=H, the quotient FG/FG is trivially 0. Best wishes, Yuri

Hi, if I correctly understood the problem,

you have a direct sum R[x] =R[x^2]+x*R[x^2] of R-modules, second summand consisting of sums of odd powers of x. Both summands are obviously isomorphic, and both are isomorphic to R[x] as R-modules (substitute t for x^2). The quotient R[x]/R[x^2] is isomorphic as R-module to

x*R[x^2], hence to R[x^2]  and consequently to R[x]. Best regards, Yuri 

A separable C*-algebra embeds in the Cuntz C*-algebra \mathcal{O}_2 if and only if it is an exact C*-algebra (a result by E. Kirchberg)

Just a remark. (I write it for singular cohomology, but it should be similar for simplicial homology).

On the cochain level we have:  0-> C^*(X,A)->C^*(X)->C^*(A)->0, (where C^*(X,A)=Hom(Delta_*(X)/Delta_*(A),\Z) etc.). A relative cocycle (from C^*(X,A)) is represented in C^*(X) as an absolute cocycle that vanishes on A. 

So what is 'absolute cocycles modulo relative boundaries'?

A special case gives an idea of what the result should be: If A is a component of  X, every absolute cocycle can be uniquely written as the sum of (the image of) of a relative cocycle and cocycle on A; taken the quotient with respect to the relative co-boundaries, we get an element of H^*(X,A) + Z^*(A),  i.e. the sum of a relative cohomology group and a cocycle group.

In general (if A is not a component of X) I would expect something like H^*(X,A) + some subgroup of Z^*(A) (didn't think about it, though).

Thus the quotient you are looking for is not a true 'cohomology'; in the setting of simplicial cohomology it would depend on the cell decomposition.