All Questions
Tagged with derived-categories triangulated-categories
80
questions
4
votes
1
answer
257
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 $...
- 351
3
votes
1
answer
114
views
Thick subcategory containment in bounded derived category vs. singularity category
Let $R$ be a commutative Noetherian ring, and $D^b(\operatorname{mod } R)$ the bounded derived category of the abelian category of finitely generated $R$-modules. Let me abbreviate this as $D^b(R)$. ...
- 481
3
votes
1
answer
176
views
Image, upto direct summands, of derived push-forward of resolution of singularities
Let $\mathcal C$ be a full subcategory (closed under isomorphism also) of an additive category $\mathcal A$. Then, $\text{add}(\mathcal C)$ is the full subcategory of $\mathcal A$ consisting of all ...
- 481
4
votes
1
answer
446
views
Exact sequences in Positselski's coderived category induce distinguished triangles
I am learning about Positselski's co- and contraderived categories. We know that short exact sequences do not generally induce distinguished triangles in the homotopy category but they do in the usual ...
- 43
6
votes
1
answer
229
views
Comparing stabilization of stable category modulo injectives and a Verdier localization
Let $\mathcal A$ be an abelian category with enough injectives. Let $\mathcal I$ be the collection of injective objects. Let $\mathcal A/\mathcal I$ be the quotient category whose objects are same as ...
- 374
1
vote
0
answers
148
views
When is a functor of chain complexes triangulated?
Let $\textsf{A}, \textsf{B}$ be abelian categories.
Let $F: \operatorname{Ch}(\textsf{A}) \to \operatorname{Ch}(\textsf{B})$ be an additive functor of chain complexes. If $F$ preserves chain ...
- 1,350
3
votes
0
answers
80
views
Examples of tensor-triangulated categories not satisfying the local-to-global principle
From now on, we will consider only rigid-compactly generated tensor-triangulated categories. Let $(\mathcal{T}, \otimes, 1)$ be one of these categories, it is known that the thick tensor ideals of ...
- 767
3
votes
0
answers
105
views
Multiplication map by a ring element on an object vs. all its suspensions in singularity category
Let $R$ be a commutative Noetherian ring, consider the bounded derived category of finitely generated $R$-modules $D^b(R)$ and consider the singularity category $D_{sg}(R):=D^b(R)/D^{perf}(R)$. Let $r\...
- 280
7
votes
1
answer
297
views
structure in triangulated category similar to t-structure
It’s well known that the heart of a t-structure is an abelian category. My question is that can we find some structure on a triangulated category which can “produce” an exact category in analogy with ...
- 73
3
votes
1
answer
245
views
"Essential injectivity" of Balmer spectra
Let $(\mathcal T, \otimes)$ be a tensor tringulated (tt-)category. Balmer defined a functor from the category of tt-categories to the category of locally ringed spaces, called the Balmer spectra or tt-...
- 256
4
votes
2
answers
228
views
Moral reason for negative sign in rotation axiom for triangulated categories
I would like to know if there is a "moral" reason why in the definition of triangulated categories the "rotation axiom" TR2 requires that we have to add a negative sign to an arrow ...
- 555
2
votes
1
answer
204
views
Literature request: $K^b(\text{proj} A)$ Krull-Schmidt for $\text{gl dim}A = \infty$ and general results about its Grothendieck group
I'm interested in the Grothedieck group of the triangulated category $K^b(\text{proj}A)$ when $A$ is a finite dimensional algebra over a field of infinite global dimension.
For this purpose, It would ...
- 23
4
votes
1
answer
221
views
Decompose an unbounded (cochain) complex in the homotopy category
Let $\mathcal{A}$ be an abelian category, it is known that any complex $A^{\bullet}$ admits a distinguished triangle
$$B^{\bullet}\rightarrow A^{\bullet}\rightarrow C^{\bullet}\rightarrow B^{\bullet}[...
anon
3
votes
0
answers
381
views
Mapping cone is a functor
It is a well-known general fact that in a triangulated category, the cone $Z$ of a morphism $X \longrightarrow Y$ (that means there exists a distinguished triangle $X \longrightarrow Y \longrightarrow ...
- 823
0
votes
0
answers
167
views
Cone of morphism induced by Serre duality
For a smooth projective variety $X$, Serre duality gives an exact autoequivalence on the derived category :
$$
S_X : D^\flat(X) \to D^\flat(X), \hspace{3em} S_X(-) = - \otimes \omega_X[\dim X]
$$
...
- 275