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4 votes
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Is there an ∞-categorical interpretation of the Quillen S⁻¹S construction?

A published reference for the claim (right after the question in boldface) is the proof given by Thomason on pages 1657-1658 of "First quadrant spectral sequences in algebraic K-theory via ...
John Rognes's user avatar
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2 votes

Is there an ∞-categorical interpretation of the Quillen S⁻¹S construction?

Consider an $E_n$-monoid X. We can deloop $X$ to an $\infty$-category $\mathbf{B}X$. There's a natural functor $X^\circlearrowleft : \mathbf{B}X \rightarrow \text{Spc}$ given by the left action of $X$ ...
Georg Lehner's user avatar
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4 votes
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Homotopy coherence datum for composition of Becker-Gottlieb transfers

$\newcommand{\S}{\mathcal S} \newcommand{\Top}{\mathrm{Top}} \newcommand{\Z}{\mathbb Z} \newcommand{\Span}{\mathrm{Span}} \newcommand{\Fin}{\mathrm{Fin}} \newcommand{\Fun}{\mathrm{Fun}}$ This is a bit ...
Maxime Ramzi's user avatar
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0 votes

f.g. modules vs. f.g. projective modules

Let $R = \mathbb Z_4$ (by which I mean integers modulo $4$). Then the abelian group $K_0(R)$ is freely generated by $[\mathbb Z_4]$ and $G(R)$ is freely generated by $[\mathbb Z_2]$, and we have $[\...
Blazej's user avatar
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4 votes

Algebraic K-theory of a ring

The comments point out that this question is very similar to another question, but an attempt to close as a duplicate failed. So, rather than leave this question on the unanswered queue, I will try to ...
David White's user avatar
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5 votes
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Grothendieck group and an almost localization

Not in general, no. Let $S$ be any triangulated category, and consiser the inclusion of the left (say) summand $S \to S \times S$. You can choose the functor $F$ to be the direct sum functor $\oplus : ...
Maxime Ramzi's user avatar
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5 votes
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The action of the Grothendieck group on higher K-theory groups

No this is not true in general. Let $R$ be a flat $k$-algebra with the two maps $K_1(R) \to K_1(R\otimes_k R)$ being different. For example, $R = k[t^{\pm 1}]$ for some nonzero ring $k$ and then the ...
Maxime Ramzi's user avatar
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3 votes
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Example of triangulated category with vanishing $K_0$

The answer is no in general: let $R= \mathbb Z$ and $E= \mathbb Z/p$. Then $[E] = 0$ because of the co/fiber sequence $\mathbb Z\to \mathbb Z\to \mathbb Z/p$. However, by the theorem of the heart, $...
Maxime Ramzi's user avatar
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9 votes

Besides $F_q$, for which rings $R$ is $K_i(R)$ completely known?

In the ten years since this question was asked, there has been a lot of progress in algebraic $K$-theory. For example, Achim Krause, Ben Antieau, and Thomas Nikolaus came up with an algorithm to ...
David White's user avatar
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