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# Homological Algebra - Science topic

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Questions related to Homological Algebra
Question
I'm looking a good technical and university notes for commutative algebra and homological Algebra that have good examples for any subjects. Please hint and guide me if you have In my point of view.
Kaplansky "Commutative Rings" has lots of exercises. The book is geared towards Number Theory.
Question
I need to know whether the ring k[x]/(x^2) with k field is von Neumann regular ring or not.
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.
Question
The problem is:
Let Z denote the set of all integers.
Consider Z/nZ = Zn as trivial G-module.
Show that there is a isomorphism between the First Homology Group H1(G, Zn) and the factor G/G'Gn;
Where G' is the commutator (derived) subgroup G'=[G,G] and Gn is the subgroup of G generated by all its n-th Powers.
• With this isomorphism proved, we want to conclude that if G is a finite p-Group, then H1(G,Fp) is a Fp-module (vector space) with dimension equal the number of least generators of G.
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 ... --> AG --> BG --> CG --> 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 IG 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/IGM . Consider the natural exact sequence 0 --> IG --> 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, IG) --> H0(G,A) -->... But A is a free A-module, hence H1(G, A) = 0, and the image of H0(G, IG) --> H0(G,A) is null practically by definition, so that H1(G, Z) = IG/ IG2 , 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 Fp instead of Z/p. After 1), we know that G/G'Gp = H1(G, Fp). Both sides are finite abelian groups killed by p, so they can be viewed (when written additively) as Fp 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'Gp ), and this last quantity is = dim G/G'Gp by elementary linear algebra.
The reverse inequality holds true, but its proof is somewhat independent from the previous homological considerations. Write G*=G/G'Gp. 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'Gp is known as the Burnside basis theorem.
Question
Let Z denote the set of all integers, and let G be a finite cyclic Group.
For every ZG-module A, and n=1,2,3...
Show that:
Hn(G,A) is isomorphic to Hn+1(G,A).
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
Question
I have got a problem where other conditions satisfy except the openness. Can I invoke the theorem where X is closed subset of Rn. Detail question will follow the below link.
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.
Question
In the light of the presentations exist in the literature for the general and special linear group GLn(q) , SLn(q) [based on the construction given by Chevalley and Steinberg as well as on the concept of BN-pair syetem] such as    http://dl.dropbox.com/u/5188175/slnpresentations.pdf  , is there a convenient way for  listing  the elements of the general (or special) linear group GLn(q)  over finite field  in general or at least for GL2(q) or SL2(q) ?
how is the Groebner - Shirshov basis for the classification using diamond lemma
Question
Let F denote a finite field and let A denote a n times n matrix over F. How can one compute efficiently all the invariant subspaces of A with dimension less than or equal to n_{0} (where n_{0}<<n). Also, is there any method to build (and cover all) these subspaces from smaller subspaces ?
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 Lv is the smallest A-invariant subspace that contains v.
Question
How Tor_1(M,F) = 0 for all finitely presented modules F implies M is flat?
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.
Question
As the title suggests, how do i see that for any n, the covering map S^{2n} → RP^{2n} induces 0 in integral homology and cohomology, except in dimension 0?
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
Question
Considering G a group and H its subgroup. <br />F a field, and F(G) a group ring. Since F(G) is a ring, we can consider also that F(H) is a subring, and then F(G) as a left or right F(H)-module.<br /><br /><strong>*</strong>I thought about writting an isomorphism between F(G)/F(H) and G/H. If one is finitely generated, I could say rank[F(G)/F(H)] = rank[G/H] = [G : H].<br />I am not sure about this proposal.
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
Question
Knowing that R is a commutative ring and R[X] is a free R-module.
It seems obvious because any R-module can be written as quotient of a free R-module and any of its submodules.
As R[X²] is a submodule of R[X] over R, then it is an ideal of R.
It turn to simply show a bijective homomorphism, but I encounter problems describing the elements in this relations.
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
Question
For what type of separable  C* algebras, there is no any integer n and any nontrivial morphism  from A to the Cuntz algebra O_{n}? In the other word  for  what type of  C* algebras, Hom(A, O_{n}) is trivial for all n?
The motivation:  I was thinking to consider a NC analogy for singular homology as follows:
A singular object is  a continuous map from Delta^{n}\to X. So We have a natural morphism from C(X)\to  C(\Delta^{n}).
With some (or  a lot of) abuse of notations  and equations, the equation \sum t_{i}=1,  which defines the standard simplex  \Delta^{n},  could be considered as a commutative picture of universal property of Cuntz algebra  O_{n}. This is  a motivation to construct   a  complex  as  followos:   We  put  C_{n}(A)= Free abelian group generated by Hom(A,, O_{n}). Now we can define  a  complex
.......  C_{n}(A) \to  C_{n+1}(A) \to....
using n+1 embeddings O_{n} to O_{n+1}
Does this idea leads to triviality?For what type of  C* algebras, this  construction is useful?