Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
Algebraic and topological K-theory, relations with topology, commutative algebra, and operator algebras
1
vote
Linearized Waldhausen $K$-Theory
Let me make some guesses, although I worry I'll only say some formal things you already know (or that are wrong).
One description of the Waldhausen K-theory of a pointed connected space $X$ is that …
5
votes
Accepted
Complexification of real k-theory gives index $2$ subgroup of complex k-theory
This is the same as the induced map $\pi_4(KO) \to \pi_4(KU)$. Via Bott periodicity (see also this answer), this is the same as the induced map $\pi_0(KSp) \to \pi_0(KU)$ coming from the map sending a …
18
votes
Accepted
Brauer Groups and K-Theory
I'm a bit late to the party, but here's what I suspect the answer should look like. Forgive me for working somewhat to very heuristically throughout.
To star things off, here's a silly question: why …
23
votes
0
answers
584
views
What is the symmetric monoidal functor from Clifford algebras to invertible K-module spectra?
There ought to be a symmetric monoidal functor from the symmetric monoidal $2$-groupoid whose objects are Morita-invertible real superalgebras (precisely the Clifford algebras), morphisms are invertib …
18
votes
0
answers
875
views
What is operator tmf?
One of the many wonderful things about K-theory, relative to other generalized cohomology theories, is that it can be defined for not-necessarily-commutative C*-algebras. The resulting construction, i …
23
votes
Magic behind idempotent-complete categories a.k.a. why (sometimes) be Karoubian is sexier th...
I don't know if this is going to answer your question but here's some relevant background. Splitting idempotents has a very special property from a categorical point of view: it is an absolute colimit …
6
votes
Simplest explicit counterexample for $Vect(BG) \ne Rep(G)$ as monoids
Complex line bundles over $BG$ are classified by $H^1(BG, \mathbb{C}^{\times}) \cong H^2(BG, \mathbb{Z})$. On the other hand, $1$-dimensional complex representations are classified by $\text{Hom}(G, \ …
13
votes
Accepted
What sort of ring-theoretic properties does the representation ring of a compact Lie group p...
Assume that $G$ is connected. Let $T$ be a maximal torus of $G$. Restriction induces a map $R(G) \to R(T)$. Note that $R(T)$ is a Laurent polynomial ring in $r$ variables where $r$ is the rank. Becaus …
33
votes
2
answers
2k
views
What are the "correct" conventions for defining Clifford algebras?
I have three related questions about conventions for defining Clifford algebras.
1) Let $(V, q)$ be a quadratic vector space. Should the Clifford algebra $\text{Cliff}(V, q)$ have defining relati …
6
votes
Accepted
Morita equivalence and isomorphisms in cohomology theories
The conceptual point is that all of these invariants are Morita invariant because they can be defined directly in terms of the category of modules. Explicitly:
Starting from the category of modules …
13
votes
What (if anything) unifies stable homotopy theory and Grothendieck's six functors formalism?
I feel like we gave up too quickly on parameterized spectra as a reasonable (if necessarily incomplete) answer to this question; we don't get the two formalisms in the question as special cases of an …