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32 votes
3 answers
4k views

Replacing triangulated categories with something better

Gelfand and Manin in their 1988 book on homological algebra write that the non-functoriality of cones means that "something is going wrong in the axioms of a triangulated category. Unfortunately at ...
Hugh Thomas's user avatar
  • 6,282
28 votes
2 answers
3k views

Why not define triangulated categories using a mapping cone functor?

Recall that the usual definition of a triangulated category is an additive category equipped with a self equivalence called $[1]$ in which certain diagrams, of the form $X \to Y \to Z \to X[1]$ are ...
David E Speyer's user avatar
15 votes
5 answers
3k views

Tate Cohomology via stable categories

Situation Let $G$ be a finite group and provide $G\text{-mod} := {\mathbb Z}G\text{-mod}$ with the Frobenius structure of ${\mathbb Z}$-split short exact sequences. Denote by $\underline{G\text{-mod}}...
Hanno's user avatar
  • 2,756
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
12 votes
0 answers
688 views

Kontsevich's derived noncommutative geometry and Rosenberg's noncommutative 'spaces'

It appears to me (though I may be wrong) that the common opinion is that the main difference between derived noncommutative geometry and Rosenberg's noncommutative 'spaces' is that Rosenberg's version ...
Doelt_k's user avatar
  • 439
9 votes
1 answer
1k views

How to write down the determinant of a quasi-isomorphism?

This question about the determinant of a perfect complex reminded me of an old question that I had. The construction of the determinant (as in MR1914072 or MR0437541) is a difficult piece of ...
jlk's user avatar
  • 3,284