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Algebraic and topological K-theory, relations with topology, commutative algebra, and operator algebras
9
votes
Accepted
Grothendieck group of triangulated categories
Let $A$ be a triangulated category, and let $B=A\times A$, with $A$ regarded as a full triangulated subcategory of $B$ via the embedding $u(X)=(X,0)$, and let $f:B\to A$ be the functor $f(X,Y)=X\oplus …
11
votes
Swan K-theory of Z/4
There seems to be a classification of representations for the example you mention (and more generally for representations of $C_p$ over $\mathbb{Z}/p^s\mathbb{Z}$) in
V. S. Drobotenko, E. S. Drobo …
1
vote
Accepted
On countable homotopy colimits in (the derived categories of) AB3 abelian categories
$\textrm{AB4}$ (and a fortiori $\textrm{AB3}$) is not enough, as it's not true for the opposite category of a module category, which is $\textrm{AB4}$.
Let $R$ be any ring, and consider the inverse s …
13
votes
Bass' stable range of $\mathbf Z[X]$
There's a comment at the top of page 993 of
L N Vaseršteĭn, A A Suslin, "Serre's problem on projective modules over polynomial rings, and algebraic K-theory", Math. USSR Izv., 1976, 10 (5), 937–1001, …
8
votes
Accepted
Induced map in K-theory by a "trivial" bimodule
No. Let $R=\mathbb{Z}\times\mathbb{Z}$, let $P$ and $Q$ be the projective modules $\mathbb{Z}\times0$ and $0\times\mathbb{Z}$, and let
$$P_\bullet=\dots\longrightarrow0\longrightarrow P\otimes_\mathb …
2
votes
Accepted
When the restriction of derived equivalence to a summand is a derived equivalence as well
It's certainly not true in general, as it could be that all the Grothendieck groups are zero, in which case the condition $[T](\mathcal{K}(\mathcal{A}')) = \mathcal{K}(\mathcal{B}')$ gives no informat …
7
votes
Accepted
Split cofibrations up to quasi-isomorphism
No. You can construct counterexamples by taking projective resolutions of modules in a nonsplit short exact sequence. For example, from the short exact sequence $0\to\mathbb{Z}\stackrel{2}{\to}\mathbb …