All Questions
Tagged with algebraic-k-theory triangulated-categories
7 questions
3
votes
1
answer
244
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Grothendieck group and an almost localization
Let $T$ be a small triangulated category and let $S\subset T$ be a full triangulated subcategory. We denote this embedding by $I: S\rightarrow T$.
Let $F: T\rightarrow S$ be a triangulated functor ...
1
vote
1
answer
127
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Example of triangulated category with vanishing $K_0$
Let $R$ be a ring, let $\operatorname{Perf}(R)$ the category of perfect modules over $R$. Suppose we have $E$ an perfect $R$-module (concentrated in degree $0$) such that its class $[E]\in K_0(R)$ is ...
1
vote
2
answers
306
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How to compute the higher $K$-theory of a triangulated category having a semi-orthogonal decomposition?
I am starting to learn the $K$-theory of triangulated categories and is stuck with the following.
Let $\mathcal{T}$ be a triangulated category having a semi-orthogonal decomposition $\langle \mathcal{...
4
votes
1
answer
410
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Can higher G-theory of Noetherian schemes be computed by derived categories?
Recently I learned from the Stacks project that for every abelian category ${\mathcal A}$, there is a natural isomorphism $K_0({\mathcal A})\cong K_0(D^{b}(\mathcal A))$.
When we set $\mathcal A$ to ...
2
votes
1
answer
331
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Grothendieck group of triangulated categories
Let $A$ be a full triangulated subcategory of $B$, $u:A\rightarrow B$ the corresponding embedding. Let $f:B\rightarrow A$ be a triangulated functor
satisfying:
$f\circ u = id$
Let $b \in B $, if $f(b)...
2
votes
0
answers
178
views
construction of $K_0$-group and Karoubian completion
Let $A$ be a ring. The $K_0$ group of $A$ can be defined in most
old fashioned way as the Grothendieck group of the set of isomorphism classes
of its finitely generated projective $R$ modules, ...
23
votes
0
answers
647
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Is this a model for $K$-theory of a triangulated category?
The recent question Complete the following sequence: point, triangle, octahedron, . . . in a dg-category reminded me of something I wanted to clarify long time ago; most likely this is now well known ...