Skip to main content

Questions tagged [chain-complexes]

Filter by
Sorted by
Tagged with
51 votes
8 answers
7k views

Motivating the category of chain complexes

Let $R$ be a commutative ring. For awhile I have been trying to motivate to myself more fully the definition of and various structures on the category $\text{Ch}(R)$ of chain complexes of $R$-modules (...
Qiaochu Yuan's user avatar
25 votes
2 answers
1k views

Complete the following sequence: point, triangle, octahedron, . . . in a dg-category

Let $\mathcal C$ be a pre-triangulated dg-category (or a stable $\infty$-category, if you wish). An object $X$ in $\mathcal C$ gives a "point": $$X$$ A morphism $X\xrightarrow f Y$ in $\mathcal C$ ...
John Pardon's user avatar
  • 18.7k
23 votes
6 answers
6k views

Why chain homotopy when there is no topology in the background?

Given two morphisms between chain complexes $f_\bullet,g_\bullet\colon\,C_\bullet\longrightarrow D_\bullet$, a chain homotopy between them is a sequence of maps $\psi_n\colon\,C_n\longrightarrow D_{n+...
Daniel Moskovich's user avatar
23 votes
3 answers
4k views

how to make the category of chain complexes into an $\infty$-category

I'd like to have some simple examples of quasi-categories to understand better some concepts and one of the most basic (for me) should be the category of chain complexes. Has anyone ever written down ...
Yosemite Sam's user avatar
  • 1,889
22 votes
0 answers
676 views

Are there "chain complexes" and "homology groups" taking values in pairs of topological spaces?

Throughout this question, notation of the form $(X,A)$ denotes a sufficiently nice pair of topological spaces. I think for most of what I'm saying here, it is enough to assume that the inclusion $A \...
Vidit Nanda's user avatar
  • 15.5k
20 votes
1 answer
682 views

What is the dimension of the variety of chain complexes?

Let $V_0, V_1, \dots, V_n$ denote a series of finite-dimensional vector spaces. We write $v_i : = \dim V_i$ for $i=0, 1, \dots, n$. I am thinking of these as real vector spaces, but I think the answer ...
Matthew Kahle's user avatar
17 votes
1 answer
612 views

Is there a notion of a chain complex with corners?

Roughly speaking, algebraic topology works by reducing questions about topological objects such as manifolds and cell to questions about chain complexes. On the topological side, although in the PL ...
Daniel Moskovich's user avatar
16 votes
1 answer
818 views

Lift chain complex from $\mathbb{F}_2$ to $\mathbb{Z}$

We start with a finite dimensional chain complex over $\mathbb{F}_2$, equipped with a basis. That is, we have finitely many finite dimensional $\mathbb{F}_2$-vector spaces $C_0,\dots,C_k$ with bases $...
Sucharit Sarkar's user avatar
13 votes
3 answers
2k views

What are normalized singular chains good for?

One of the common definitions of homology using the singular chains, i.e. maps from the simplex into your space. The free abelian group on these can be made into a chain complex and one can take the ...
skupers's user avatar
  • 8,167
13 votes
0 answers
680 views

Singular chains generated by manifolds with corners --- does it really work?

Usually we define singular homology via the complex of singular chains: $$C_\ast(X)=\bigoplus_{n\geq 0}\mathbb Z\langle\sigma:\Delta^n\to X\rangle$$ where the right hand side denotes the free abelian ...
John Pardon's user avatar
  • 18.7k
12 votes
3 answers
2k views

Is the tensor product of chain complexes a Day convolution?

Recently, Jade Master asked whether the tensor product of chain complexes could be viewed as a special case of Day convolution. Noting that chain complexes may be viewed as $\mathsf{Ab}$-functors from ...
Emily's user avatar
  • 11.8k
12 votes
1 answer
2k views

Explicit description of the "simplicial tensor product" of chain complexes

Recall that there is an equivalence of categories (Dold-Kan) $$N:\mathrm{s}\mathbf{Ab}\simeq \operatorname{Ch}_{\geq 0}(\mathbf{Ab}):\Gamma$$ between simplicial abelian groups and (connective) chain ...
Harry Gindi's user avatar
  • 19.6k
12 votes
1 answer
448 views

Are totally degenerate chains null-homologous?

Let $X$ be a CW complex. Suppose $\gamma\in C_n(X)$ is a cycle which is a sum of maps $\sigma:\Delta^n\to X$ which factor as $\Delta^n\to\mathbb R^{n-1}\to X$. Does it follow that $\gamma$ is null-...
John Pardon's user avatar
  • 18.7k
12 votes
1 answer
367 views

Can we use unparameterized chains to calculate singular homology?

Most models of singular chains on a topological space $X$ use maps from some particular collection of "nice" objects, such as the standard simplices $\Delta^n$, the standard cubes $[0,1]^n$, etc. ...
John Pardon's user avatar
  • 18.7k
11 votes
1 answer
502 views

Is there a practical criterion to determine whether the limit of a diagram of real chain complexes is also a homotopy limit?

Consider a diagram D: I→ChR of real connective chain complexes. In the example I have in mind all chain complexes are concentrated in some fixed degree n. There is a canonical map lim D → holim D ...
Dmitri Pavlov's user avatar
11 votes
0 answers
204 views

Is it possible to take "limits up to homotopy"?

Before I begin, an apology: it's been a while since I've done much analysis, so I might be misusing (or just missing) terminology. I have a chain complex $(V_\bullet,\partial)$ of topological vector ...
Theo Johnson-Freyd's user avatar
9 votes
3 answers
2k views

Module category equivalent to graded module category?

Main Question Let $R$ be a graded ring, graded by the nonnegative integers. Denote by $\mathrm{gr}R-\mathrm{Mod}$ the category of $\mathbb{Z}$-graded left $R$-modules with morphisms that preserve ...
MTS's user avatar
  • 8,559
9 votes
3 answers
2k views

Reference for homotopy (co)limits of (co)chain complexes via totalization of double complexes

It seems to be a well-known fact that homotopy (co)limits of (co)simplicial diagrams of nonnegatively graded (co)chain complexes in (Grothendieck) abelian categories can be computed by using the Dold-...
Dmitri Pavlov's user avatar
9 votes
1 answer
887 views

Left adjoint of totalization?

There is a functor from bicomplexes to chain complexes sending a bicomplex to its associated total chain complex. Does this functor have a left adjoint, and if so, what is it?
Ryan Greenwood's user avatar
8 votes
1 answer
399 views

Is the category of chain complexes a reflexive or coreflexive subcategory of the category of functors?

Let $A$ be an abelian category (you can assume additional conditions for its goodness). Let $\mathrm{Seq}(A) = \mathrm{Func}(\mathbb{Z}, A)$, where $\mathbb{Z}$ is the standard order category on ...
Arshak Aivazian's user avatar
8 votes
1 answer
623 views

Is there a good computer package for working with complexes over non-commutative rings?

I'm interested in doing computations with certain non-commutative rings, most of which involve taking derived tensor products. Does anyone know of a computer algebra package which will find ...
Ben Webster's user avatar
  • 44.7k
8 votes
1 answer
288 views

Injective model structure on sheaves of bounded complexes of $A$-modules

The following might be very well known for people who works with model categories, but I do not find the answer. Let $A$-be a ring. Denote $\mathbf{Ch}_+(A)$ the category of positive degree cochain ...
Tintin's user avatar
  • 2,871
8 votes
1 answer
369 views

Fiber product of spaces and cohomology

Let $Y \to X \leftarrow Z$ be a cospan of topological spaces and let $W = Y \times_X^h Z$ be their homotopy fiber product. I am interested in sufficient conditions on the map $Y \to X$ that ensure ...
Matthias Ludewig's user avatar
7 votes
2 answers
3k views

On the difference between a projective chain complex and a level-wise projective chain complex

Let R be an associative ring with a unit, and consider the standard projective model structure of non-negatively graded (left) R-module, $Ch_R$. A map $f:M\to N$ in $Ch_R$ is a weak equivalence if it ...
Shlomi A's user avatar
  • 583
7 votes
2 answers
460 views

Trees in chain complexes

$\DeclareMathOperator{\Ch}{\mathit{Ch}}$Let $\Ch_\mathbb{Q}$ denote the model category of chain complexes over rational numbers. Let $T_\ast$ be a tree in $\Ch_{\mathbb{Q}}$ with $n$ vertices. How to ...
Surojit Ghosh's user avatar
7 votes
2 answers
1k views

Resolutions of unbounded complexes and homotopy (co)limits.

I want to understand once and for all what the resolution of an unbounded complex is. I've been trying to read 'Homotopy limits in triangulated categories' by Marcel Bokstedt and Amnon Neeman and can'...
Richard Jennings's user avatar
7 votes
0 answers
647 views

Understanding the higher stack of perfect complexes

One of the most famous examples of higher Artin stacks is the stack of perfect complexes. I recall here the basic stuff: We fix a function $b: \mathbb{Z} \rightarrow \mathbb{N}$ which is zero ...
Martin Hurtado's user avatar
7 votes
0 answers
668 views

Is there an obstruction which classifies "quasi-isomorphism but not chain equivalence"?

Fix a ring $R$ and let $C_\bullet$, $D_\bullet$ be (possibly unbounded) chain complexes of $R$-modules. Assume that $f_\bullet:C_\bullet \to D_\bullet$ is a quasi-isomorphism: that is to say, $f$ is a ...
Vidit Nanda's user avatar
  • 15.5k
6 votes
2 answers
963 views

Projective objects in the derived category of chain complexes

I have been trying to understand projective objects in the derived category of chain complexes of modules over a ring. If we stick to the category of chain complexes, the only projective objects are ...
Sofía Marlasca Aparicio's user avatar
6 votes
0 answers
138 views

Does homotopy equivalence between deformations of cochain complexes imply gauge equivalence?

Let $(C^{\bullet},d)$ be a cochain complex over a field $k$ with char$k$=0. We fix an Artin local $k$-algebra $(A,m)$. A deformation of $d$ is a new operator $d+\epsilon: C^{\bullet}\otimes_k m\to C^{\...
Zhaoting Wei's user avatar
  • 9,009
6 votes
0 answers
555 views

When is the dual of a limit the same as the colimit of the duals?

We all know that the dual of the colimit of a diagram in the category of chain complexes (and similar categories) is the limit of the duals diagram. This follows immediately from the general fact that ...
Daniel Robert-Nicoud's user avatar
6 votes
0 answers
180 views

(Reference Request) Tensor product of chain complexes in terms of strict $\infty$-categories

(note: this question is essentially a reference request for the tensor product described at the end. the rest is context) It is well known that the category of chain complexes (in positive degree, ...
Simon Henry's user avatar
  • 42.4k
5 votes
2 answers
683 views

Model categories and chain complexes

I'm fairly new to thinking about homological algebra and chain complexes in their own right, i.e outside of isolated examples such as for constructing simplicial homology, or for computing $Ext$ ...
asd's user avatar
  • 168
5 votes
1 answer
540 views

Sign in May’s General algebraic approach to Steenrod operations

In the first section of J. P. May’s General algebraic approach to Steenrod operations, May defines for $\pi\subseteq\Sigma_r$ an integer $q\in\mathbb{Z}$ and a commutative ring $\Lambda$, the $\Lambda\...
FKranhold's user avatar
  • 1,623
5 votes
1 answer
411 views

Homology of singular chain complex modulo subdivision

Let $S_p(X)$ be the $p$-th singular chain group and $\mathcal S(X)$ be the singular chain complex of a topological space $X$. There is a barycentric subdivision operator (which is also a chain map) $\...
Zhang Yuhan's user avatar
5 votes
1 answer
359 views

Model structures on the category of unbounded chain complexes

In his book "Model Categories" Mark Hovey constructs both projective and injective model structures on unbounded chain complexes of $R$-modules. For what kinds of abelian categories does ...
Victor's user avatar
  • 1,875
5 votes
1 answer
2k views

Cohomology of derived tensor product of complexes and Künneth spectral sequence

Let $R$ be any commutative ring, let $V^\bullet$ and $W^\bullet$ be (co)chain complexes of $R$-modules, indexed cohomologically. We can also assume that they have both cohomology in nonpositive ...
Francesco Genovese's user avatar
5 votes
1 answer
363 views

Transgression in terms of k-invariant for chain complexes

I am looking for a reference for the following. Say we have a $G$-space $X$ whose homology groups (in field coefficients $k$) are non-zero only in dimension zero and for a fixed $n>0$. Let $M$ ...
Nick's user avatar
  • 235
5 votes
0 answers
129 views

The interaction between differentials on a graded ring and chain-homotopy equivalences

I am wondering about the following question: Given a differential graded algebra $A$, how many other differentials can we put on the underlying graded ring of $A$, which are also chain-homotopy ...
Mo Behzad Kang's user avatar
5 votes
0 answers
723 views

Sign problem for the shift functor on DG modules

Let $\mathcal{A}$ be a Differential graded $k$ category, where $k$ is a commutative ring. I want to extend the shift functor $[1]$ on chain complexes to dg modules over $\mathcal{A}$. Now a right dg ...
Sondre's user avatar
  • 345
4 votes
1 answer
267 views

A particular morphism being zero in the singularity category

Let $R$ be a commutative Noetherian ring and $D^b(R)$ be the bounded derived category of finitely generated $R$-modules. Let $D_{sg}(R)$ be the singularity category, which is the Verdier localization $...
strat's user avatar
  • 361
4 votes
1 answer
558 views

derived tensor product and finite projective dimension

Let $(R, \mathfrak m)$ be a Noetherian local ring. Let $M,N$ be non-zero finitely generated $R$-modules. Is it known that $M\otimes_R^{\mathbf L} N$ has finite projective dimension if and only if $M$ ...
strat's user avatar
  • 361
4 votes
1 answer
193 views

Split cofibrations up to quasi-isomorphism

$R$ a ring $(1\neq 0)$, $\mathbf{Perf}(R)$ is the category of perfect complexes (of right $R$-modules). Suppose that $A_{\bullet}\rightarrow B_{\bullet}\rightarrow B_{\bullet}/A_{\bullet}$ a short ...
Let's user avatar
  • 511
4 votes
1 answer
475 views

de Rham's Theorem using atlases and colimit preservation

$\DeclareMathOperator\DR{DR}\newcommand\SmoothManifold{\mathrm{SmoothManifold}}\newcommand\ChainComplexes{\mathrm{ChainComplexes}}\DeclareMathOperator\Sing{Sing}\newcommand\DifferentialGradedRAlgebras{...
user avatar
4 votes
1 answer
257 views

Homotopy totalization and chains - reference

Simple case. Take $X_{\bullet}$ a cosimplicial space. Is it true that the chain complex of its homotopy totalization is quasi-isomorphic to the homotopy totalization of its chain complex? Because of ...
Andrea Marino's user avatar
4 votes
1 answer
191 views

Homotopy coherent space maps induces homotopy coherent chain complex morphisms

It is an elementary fact that a map $f:X \to Y$ between spaces induces a chain complex morphism $f_* : C_*(X) \to C_*(Y) $. This allows one to transfer 1-category theoretic arguments from spaces to ...
Andrea Marino's user avatar
4 votes
1 answer
814 views

A-infinity modules

Using: https://arxiv.org/pdf/math/9910179.pdf as a reference... My question involves spelling out explicitly the comment in 4.2 - "Equivalently, the datum of an $A_\infty$-structure on a graded ...
Matthew Levy's user avatar
4 votes
1 answer
594 views

Smith Normal Form for block matrices over the integers

Are there any known results on the Smith Normal Form for block matrices over the integers? In particular, I am interested in matrices of size $kr \times ks$ made of square blocks of size $k$ such that ...
user53075's user avatar
4 votes
1 answer
228 views

Gluing a manifold along its boundary, via chain complexes

Given closed oriented $n$-manifolds $M, M', M''$ and bordisms $W, W'$ with $\partial W = M \sqcup - M'$ and $\partial W' = M' \sqcup - M''$, we can collar-glue them to obtain a bordism from $M$ to $M''...
Markus Zetto's user avatar
4 votes
0 answers
207 views

Is the category of cochain complexes with terms in an additive category a 2-category?

$\def\hom{\operatorname{Hom}} \def\bbZ{\mathbb{Z}}$This question is a follow-up to this other one. There the OP asks whether "the category of chain complexes" (can be interpreted in several ...
Elías Guisado Villalgordo's user avatar