Skip to main content
Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Saves in:saves
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options not deleted user 124549

A branch of algebraic topology concerning the study of cocycles and coboundaries. It is in some sense a dual theory to homology theory. This tag can be further specialized by using it in conjunction with the tags group-cohomology, etale-cohomology, sheaf-cohomology, galois-cohomology, lie-algebra-cohomology, motivic-cohomology, equivariant-cohomology, ...

6 votes
1 answer
324 views

Steenrod squares in terms of chain maps

$\DeclareMathOperator\Sq{Sq}$The Steenrod squares $\Sq^i: H^n({-};\mathbb{F}_2) \to H^{n+i}({-};\mathbb{F}_2)$ are fundamental cohomological operations. By the Yoneda lemma, they induce a map between …
Student's user avatar
  • 5,230
4 votes

An intuitive explanation for group cohomology via cochains?

And again can you connect the modern treatment of nonabelian cohomology with this classical picture? …
Student's user avatar
  • 5,230
31 votes
2 answers
3k views

A natural construction of real numbers?

Summary Someone claims $\mathbb{R}$ can be constructed as the following intriguing quotient, which is related to Gromov's bounded cohomology. I want to find out if it is true. … Related Category-theoretic description of the real numbers (Mathematics Stack Exchange) Gromov's bounded cohomology, see Ivanov - Notes on the bounded cohomology theory and the 9th page of A'Campo's paper …
Student's user avatar
  • 5,230
1 vote
1 answer
222 views

Steenrod operations from the delooping viewpoint

Let $F$ be a finite field, $Sq^i$ be the $i$-th Steenrod operation $$ H^*(-;F) \to H^{*+i}(-;F).$$ By Yoneda lemma, such operation is a map $\phi_i: B^{*}F \to B^{*+i} F$, where $B$ denotes the delo …
Student's user avatar
  • 5,230
1 vote
0 answers
144 views

Simplicial realization of the circle action on the free loop space

Given a simply connected topological space $X$, it is well known that its free loop space $LX$ has cohomology being the Hochschild homology of the singular cochains [1]: $$HH_\bullet(S^\star X) \simeq …
Student's user avatar
  • 5,230
0 votes
0 answers
299 views

nPOV: Cohomology and derived functors

In the nPOV, cohomology is realized as the connected component of the derived hom space [1]. … References [1] nLab: cohomology [2] nLab: derived functor …
Student's user avatar
  • 5,230
1 vote

Unifying "cohomology groups classify extensions" theorems

There's definitely a deep relation among obstruction theory, extension problem, cohomology theory, deformation theory, .. etc Some relating urls could be helpful: Deligne's letter to Miller (mathoverflow.net … (mathieu.anel.free.fr/mat/doc/…) John Baez's lecture note "n-category and cohomology" (http://www.math.uchicago.edu/~may/IMA/BaezShulman.pdf) In particular, roughly speaking: (1) Deligne pointed out …
Student's user avatar
  • 5,230
4 votes
0 answers
186 views

Cohomology and higher structures

Given a Lie algebra $\frak{g}$, a linear representation $V$, and a 3-cohomology class $\alpha \in H^3(\frak{g}$$, M)$ we can construct a Lie 2-algebra. … Namely, do $(n+1)$-th cohomology classifies Lie n-algebras, if any? If so, how does do we connect this back to the classical extension theory? …
Student's user avatar
  • 5,230
4 votes
2 answers
528 views

How much do characteristic classes fail to characterize bundles?

As $B$ varies, I believe the question is equivalent to Question': To what extent does the cohomology ring $H^\star(BG;\mathbb{Z})$ fail to characterize $BG$ up to homotopic equivalence. … It is well-known that taking cohomology does forget much information. And for good enough spaces still, one needs to consider the cochain complex as an $E_\infty$ algebra. …
Student's user avatar
  • 5,230
3 votes
0 answers
164 views

Obstruction to delooping

In this case, the monoidal structures and the braided structures (up to suitable equivalence) can be classified by some cohomology classes. …
Student's user avatar
  • 5,230
2 votes
0 answers
266 views

Road map: beyond Artin-Wedderburn theorem

For a noncommutative semisimple ring $R$, its structure and its category of representations can be largely understood using Artin-Wedderburn theorem. Such structure theory is useful, for example, in t …
Student's user avatar
  • 5,230
6 votes
0 answers
350 views

Cohomology without comonad?

From this, one derives a cohomology theory of this algebraic theory. … This subsumes group cohomology, Lie algebra cohomology, Hochschild cohomology, and Harrison's cohomology for commutative algebras [2, chapter 6+7]. …
Student's user avatar
  • 5,230