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(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.

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. …
1 vote

What can be said about $A$ and $B$ given the exact sequence $0 \to R^p \to A \to R^r \to R^q...

Assume $R$ is a PID. Clearly, there is a short exact sequence $0\to M_2\to A\to M_1\to 0$, where $M_1\subset R^r$ is the image of the map $A\to R^r$, so a free module of rank $d\le r$, and $M_2\cong …
Vladimir Dotsenko's user avatar
1 vote

Hilbert Syzygy Theorem - Induction step

A proof using Gröbner bases is in Using algebraic geometry by David A. Cox, John B. Little, Donal O'Shea, Theorem 2.1. However, I was always sure that there should be (at least in the graded case) an …
Vladimir Dotsenko's user avatar
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 …
Vladimir Dotsenko's user avatar
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).
Vladimir Dotsenko's user avatar
1 vote

Whitehead's second Lemma and invariants of exterior square

The way the question is formulated, it is trivial. If $V\ne\mathfrak{g}$ ($\mathfrak{g}$ here being the adjoint module, in which case you already know everything), the corresponding sum is clearly d …
Vladimir Dotsenko's user avatar
2 votes

Analogy of Gerstenhaber algebra

One possible answer is contained in the paper of Victor Ginzburg and Travis Schedler, "Free products, cyclic homology, and Gauss-Manin connection", https://arxiv.org/abs/0803.3655. You will be in part …
Vladimir Dotsenko's user avatar
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 .
Vladimir Dotsenko's user avatar
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 …
Vladimir Dotsenko's user avatar
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 …
Vladimir Dotsenko's user avatar
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 …
Vladimir Dotsenko's user avatar
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.
Vladimir Dotsenko's user avatar
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 …
Vladimir Dotsenko's user avatar
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 …
Vladimir Dotsenko's user avatar
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 …
Vladimir Dotsenko's user avatar

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