12
votes
What is decategorification?
Taking loops (or, in categorical language, endomorphisms of the monoidal unit) is commonly seen as a type of decategorification.
For example, decategorifying von Neumann algebras produces Hilbert ...
10
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 ...
9
votes
Accepted
Which spectra have a universal connective quotient?
This answer is about the $\infty$-categorical variant. This is a fancy way to say: on spaces of maps, the natural map
$$
Map(T',A) \to Map(T,A)
$$
is an equivalence for any connective $A$. Note that ...
8
votes
Accepted
Is $KU\otimes S^1_+$ isomorphic to $F(S^1_+,KU)$ as $E_\infty$ rings?
This seems to be an answer, based on discussion with Maxime Ramzi in the comments.
The space of $E_\infty$ $KU$-algebra maps from $KU\otimes S^1_+\to F(S^1_+,KU)$ is the same as the space of $E_\infty$...
7
votes
The complex $K$-theory of the Thom spectrum $MU$
You can learn about this, and more, in Part II of Adams' "Stable Homotopy and Generalised Homology" (1974), especially section 4, in the special case $E = KU$.
6
votes
Accepted
Can one bypass the geometric realization in the definition of algebraic $K$-theory?
I believe there is no good notion of homotopy groups for an arbitrary simplicial set S.
It depends on what “good” means.
Kan's original definition works for arbitrary pointed simplicial sets: $$\def\...
5
votes
Accepted
The $E$-(co)homology of $\mathrm{BGL}(R)^+$ and the algebraic $K$-theory of $R$
Let $R$ be a ring. $BGL(R)^+$ is homotopy equivalent to the $0$ component of $\Omega^\infty K(R)$, and it is stably equivalent to $BGL(R)$.
In particular, for a (co)homology theory $E$, understanding $...
5
votes
Accepted
Stabilizing conjugacy classes of integer matrices
${\rm Conj}(A)\to{\rm Conj}(A\oplus I_m)$ is not surjective, because
$$I_m\oplus A\in{\rm Conj}(A\oplus I_m) \quad\hbox{but}\quad\not\in R({\rm Conj}(A)).$$
5
votes
What are surprising examples of Model Categories?
There is Krause and Nikolaus' model structure on the category of group presentations whose homotopy category is the category of groups.
5
votes
Fibre sequence of module spectra induces a fibre sequence of $K$-theory spectra?
No, this is not true, and nothing like this is expected.
The exact sequence $\Bbb Z/2 \to \Bbb Z/4 \to \Bbb Z/2$ gives rise to a fiber sequence $H\Bbb Z/2 \to H\Bbb Z/4 \to H\Bbb Z/2$ of Eilenberg--...
4
votes
Accepted
Cohomology classes coming from algebraic K-theory
Apologies for being a bit late, but let me try to expand on my comment. An excellent source for this material is James Lewis' "user-friendly" survey article [Lew14].
First recall that for $X$...
4
votes
Is there any "deep" relation between the localization theorem of equivariant cohomology and the localization theorem of equivariant K-theory
The following is relatively standard. It is almost certainly not the "deep" reason you are asking for. I think it is useful point of view though.
Suppose I have some invariant $J$ which ...
4
votes
Accepted
$K_0$ group of an infinite factor
The following is a more streamlined version of the answer I provided in the comments:
We first observe that an infinite factor is properly infinite. (This is because a von Neumann algebra is defined ...
3
votes
The complex $K$-theory of the Thom spectrum $MU$
For a more self-contained answer, let $L$ be the tautological line bundle over $\mathbb{C}P^\infty$. This gives a class $x=[L]-[1]\in\widetilde{K}^0(\mathbb{C}P^\infty)$. It is standard that $K^0(\...
3
votes
Accepted
Equivariant sheaves on $\mathbb P^1$
Let me explain why the line bundle $\mathcal{O}(1)$ does not admit a $\mathrm{PGL}(2)$-equivariant structure. Indeed, if it does, then the vector space
$$
\mathrm{Hom}(\mathcal{O}, \mathcal{O}(1))
$$
...
3
votes
Accepted
"High-dimensional" classes in topological $K$-theory
Let $X_n = S^{2n+2}$.
Since $\operatorname{ch} : K(S^{2n+2})\otimes_{\mathbb{Z}}\mathbb{Q} \to H^{\text{even}}(S^{2n+2}; \mathbb{Q})$ is an isomorphism, there is a complex vector bundle $E \to S^{2n+2}...
3
votes
Accepted
$K_1$ of Categories for intuition
For an exact category ${\cal N}$, Dan Grayson (who occasionally shows up on MathOverflow) gave explicit generators and relations for $K_n$ here.
What follows is essentially quoted directly from that ...
3
votes
Accepted
Roadmap for Algebraic Geometry/Homotopy Theory/Algebraic $K$-Theory intersection
This sounds like a very exciting phase of your studies! As you point out, there's plenty of intersections between algebraic geometry and homotopy theory. One intersection is motivic homotopy theory, ...
2
votes
Bisimplicial spaces as a coequalizer of maps between "simpler" bisimplicial spaces
I certainly respect going back to primary sources. But, in this case, it's helpful to remember that a LOT has been written about bisimplicial sets since Quillen's 1973 paper. For example, a reference ...
1
vote
Application of higher categories in algebra
There are tons of such applications. For example, less than two years ago, they were used to prove the Redshift Conjecture, a statement about iterated algebraic $K$-theory. I previously wrote an ...
Community wiki
1
vote
Does Grayson/Quillen's "pre group completion" have a universal property?
It is the classifying category for the left action of $C$ on its product $C \times C$.
Let me elaborate on this a bit further. Let $C$ be an $E_\infty$-monoid, for example represented by a symmetric ...
1
vote
The category theory of Span-enriched categories / 2-Segal spaces
A solution to achieving the kind of enrichment you are looking for
(though I cannot say what precise comparison it would give with
$2$-Segal spaces) may be to actually remember not only the
$2$-...
1
vote
Integral group rings on which stably free modules are free
This is an old question but I thought this could be a useful answer to anyone interested in this in the future.
The only class of examples of torsion-free groups with SFC (stably-free cancellation) ...
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