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Algebraic and topological K-theory, relations with topology, commutative algebra, and operator algebras
13
votes
1
answer
431
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Does Grayson/Quillen's "pre group completion" have a universal property?
Algebraic $K$-theory of a symmetric monoidal category $C$ is defined in two steps: first take geometric realization, then group complete: $K(C) = \Omega B |C|$.
In HAK II, Grayson (following Quillen …
1
vote
0
answers
111
views
Lengths and additive invariants which preserve positivity
The length of a module is well-known to be an additive invariant of finite-length modules. That is, if $R$ is a ring and $Art(R)$ its category of finite-length modules, then $length : Ob (Art(R)) \to …
14
votes
1
answer
1k
views
What is decategorification?
A decategorification is, roughly, some procedure $\Phi$ which inputs some sort of $n$-categorical data and outputs some sort of $(n-1)$-categorical data. Whatever this means, categorification is under …
13
votes
2
answers
696
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The category theory of Span-enriched categories / 2-Segal spaces
The category $\mathsf{Span}$ of spans of sets is symmetric monoidal closed under $\times$ (the cartesian product from $\mathsf{Set}$, which is not the categorical product in $\mathsf{Span}$), comple …
5
votes
0
answers
128
views
Solving polynomial equations in $K(h)$-local or $T(h)$-local spectra?
This is the same question as an earlier question of mine, except in a different category.
Let $Spt_{T(h)}^{fin}$ be the category of finite $T(h)$-local spectra. Let $K_0^\oplus(Spt_{T(h)}^{fin})$ be t …
11
votes
2
answers
849
views
Solving polynomial equations in spectra?
Let $M$ be the mod-$p$ Moore spectrum where $p \geq 3$ is a (power of) a prime. Then $M$ satisfies the "polynomial equation" $M \wedge M \cong M \oplus \Sigma M$. Is this a general phenomenon, or is i …
12
votes
0
answers
500
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The homotopy theory presented by a Waldhausen category
Waldhausen introduced his categories for the purposes of defining algebraic $K$-theory of suitable categories. From a modern perspective, it looks like he was really doing two things at once:
Waldha …
12
votes
2
answers
1k
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$K$-theory backwards
Let $R$ be a ring. The $K$-theory of $(Mod(R)^{f.g.proj},\oplus)$ is obtained by first throwing out non-isomorphisms and then group completing. What happens if these steps are reversed?
That is, cons …
11
votes
0
answers
264
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Direct proof of the equivalence of symmetric monoidal $K$-theory and exact sequence $K$-theory?
When all exact sequences split in $C$, we have $\Omega B C \simeq K(C):=\Omega Q(C)$. Heuristically, this is because the space of upper-triangular matrices is contractible. Can this be made precise? I …