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Results tagged with homological-algebra
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user 1306
(Co)chain complexes, abelian Categories, (pre)sheaves, (co)homology in various (possibly highly generalized) settings, spectra, derived functors, resolutions, spectral sequences, homotopy categories. Chain complexes in an abelian category form the heart of homological algebra.
10
votes
Accepted
Poincaré duality for (co)homology of Lie algebras?
First, let me expand on the reply of Dietrich Burde: I got hold of the paper of Hazewinkel, and can now be more precise about what is and what is not there (last time I saw it was some years ago).
…
- 16.9k
9
votes
Is $Tor_A(k,k)$ a bicommutative Hopf algebra?
This is not true. Consider the algebra $A=T(V)/V^{\otimes 2}$, it is a commutative algebra whose augmentation ideal has zero multiplication. We have $\mathrm{Tor}_A(k,k)\cong T(V[1])$ with the shuffle …
- 16.9k
9
votes
Why is the Hochschild homology of k[t] just k[t] in degrees 0 and 1?
Is there something about k[t] that ensures it's concentrated in degrees 0 and 1.
Yes: it's a free associative algebra - that's why you only have 0th and 1st homology.
- 16.9k
7
votes
What should I call a "differential" which cubes, rather than squares, to zero?
The situation similar to what you are describing happens when people talk about the so-called N-Koszul algebras, originally defined by R. Berger in his paper "Koszulity for nonquadratic algebras" (J. …
6
votes
Accepted
A potential resolution of $R/r$
This complex (in simultaneously a more general setting, where you have several elements $t_1,\ldots,t_p$, and a more special setting, because only the case of $R$ being a free algebra was studied then …
- 16.9k
6
votes
Accepted
Commutator of finite global dimension algebras
Yes. See the result of Section 2.5 of a wonderful paper of Bernhard Keller :
https://webusers.imj-prg.fr/~bernhard.keller/publ/ilc.pdf
(and the references therein).
- 16.9k
5
votes
Whitehead lemmas in Lie algebra cohomology for non-algebraically closed fields
Cohomology does not change under field extensions, so just extend everything to the algebraic closure to prove your result in char 0 in general. Of course, field extensions do not preserve irreducibil …
- 16.9k
5
votes
Accepted
Infinity-homotopies
I don't know if you found an answer since you posted the question, but I will write this just in case: there is a "cute" (easy) definition in case of nonsymmetric operads which generalises the A-infin …
- 16.9k
5
votes
Projective resolutions for commutative monoids
Homological algebra for monoids have been done by a lot of people in theoretical computer science. What is done for associative algebras over a field in a brilliant paper of David Anick (http://www.js …
- 16.9k
4
votes
Accepted
Second cohomology group of the contact Lie algebra $K_3$
Yes, it is true. In fact, it is true that $H^i(K_{2n+1},F)=0$ for $0<i\le 2n$. This can be deduced from the theorem of Feigin sketched in
Feigin, B.L. Cohomology of contact Lie algebras.
(Russian) C. …
- 16.9k
4
votes
Hochschild cohomology of certain local algebras
The right term to look for is "truncated quiver algebras".
There are two relevant references which I believe lead to a complete answer to your question in the non-commutative case.
First, Section …
- 16.9k
3
votes
Accepted
Homotopy transfer in the opposite direction
Let us denote by $p\colon Y\to X$ and $i\colon X\to Y$ the maps of your SDR. Since $pi=\mathop{\mathrm{id}}\nolimits_X$, the map $i$ is injective, and is an isomorphism with its image. The map $\pi=i\ …
- 16.9k
3
votes
Quadratic algebras and Koszul algebras
A useful reference for answering your questions at least partially is Theorem 1.7 (especially part (5) of it) in the notes http://inmabb.criba.edu.ar/revuma/pdf/v48n2/v48n2a05.pdf .
- 16.9k
3
votes
Two definitions of minimal models
The two definitions are the same. The thesis of Lefèvre-Hasegawa does not require the differential to be zero, it requires the component $m_1$ of the differential to be equal to zero: minimality trans …
- 16.9k
3
votes
Accepted
Differential of the Twisted complex for algebraic operads
Since $\alpha$ is of degree $-1$, these terms come in pairs appearing with opposite signs that cancel each other. In other words, the (co)associativity for (co)operads has a sequential axiom and a par …
- 16.9k