@MikeMiller ku-homology or cohomology?

Segal has a lovely interpretation of the ku-homology in "K-homology theory and algebraic K-theory" that's analogous to the symmetric product interpretation of homology

instead of the symmetric product assigning a multiplicity to a list of points, and then adding them when points collide, you assign complex vector subspaces of some infinite-dimensional V to each point, assuming they're orthogonal, and take sums when the points collide

writing that out, i feel like i may have tried to make this sales job to the chatroom before

yep. chat.stackexchange.com/transcript/message/19165632#19165632

@archipelago so unfortunately i'm pretty unfamiliar with this stuff; do you know offhand whether at least the map $S^1 \to Emb(S^1,\mathbb{R}^2)$ is a homotopy equivalence (i can kinda-sorta convince myself that it should be), and perhaps more generally if the space $Emb(S^1,\mathbb{R}^k)$ is understandable?

@Tyler I was thinking cohomology but I am quite happy with homology; I really like this description

cohomology I don't know, mainly because I don't understand a geometric interpretation of bundles whose structure group is a highly connected cover of U

@AaronMazel-Gee Diff(R^2) acts on Emb(S^1, R^2) transitively (Schoenflies thm), and thus has the same stabilizer above every point, Diff(R^2,S^1). We get a fiber sequence Diff(R^2, S^1) -> Diff(R^2) -> Emb(S^1, R^2). Diff(R^2, S^1) is equivalent to Diff(R^2, R^2 \ D^2) (cone off towards infinity), which is equivalent to Diff(D^2, S^1), which is contractible (Smale). Diff(R^2) has the homotopy type of O(2) (also a consequence of Smale). Thus Emb(S^1, R^2) has the homotopy type of O(2).

The appropriate generalization in higher dimensions is Emb(S^{n-1}, R^n), but you're going to run into issues from both whether or not there are Schoenflies theorems (there are topological obstructions coming from exotic spheres one dimension up), and then the fact that Diff(D^n, S^{n-1}) is nasty in high dimensions (to my understanding the understanding of this comes from Waldhausen K-theory of spaces)

I don't know much about Emb(S^1, R^n); I think this is a totally different story above n=2. for n=3 of course pi_0 is isomorphism classes of oriented knots, and the homotopy type more generally is completely understood for a fixed connected component I think due to work of Hatcher and Budney. for larger n probably the Goodwillie calculus people understand the story better... pi_0 is easy but presumably higher pi_i are hard.

@TylerLawson what's the story modulo that point, if it's not hard to say?

oh! so one consequence of Bott periodicity is that the space representing KU^{2k} is the (2k-1)-connected cover of BU. so KU^{2k}(X) could be interpreted as bundles with structure group the (2k-2)-connected cover of U

e.g. KU^2(X) is about stable SU(n)-bundles for X finite

and further up they become analogues of stable spin-bundles and string-bundles

but I don't know a geometric interpretation of the 5-connected or 7-connected covers of the unitary group

that's pretty compelling

oops, all of those KU's should have been ku's, I should be more careful

I got you :)

Sanity check: If I have an abelian category A and consider D(A) , the derived category of A, is D^+(A) and D^-(A) D^b(A) full subcategories of D(A)? D^+ is the subcategory consisting of upper bounded complexes, D^-(A) lower bounded complexes and D^b(A) the bounded complexes.

I do not see how this not can be true, but I might be surprised.

@Dedalus I guess it depends on whether you want the mapping complexes of $D^+$ etc. to be upper/lower bounded as well?

@EspenNielsen With mapping complexes, I suppose you are referring to the internal hom?

The reason I am asking is that sometimes I have seen derived homs between two (upper bounded, lower bounded, bounded) objects been computed in D^+/D^-, D^b. How does this derived Hom, calculated in D^+/D^-/D^b, relate to the derived Hom in the full derived category?

@Dedalus yes, that's true. in any of the derived categories, a morphism is some kind of equivalence class of zigzags between regular maps going forward and quasi-isomorphisms going backward. if you have two complexes which are bounded (of some type) and such a zigzag of maps of unbounded complexes, you can use the truncation functors tau^{\leq n} and tau^{\geq n} (which preserve quasi-isos) to turn it into an equivalent zigzag of maps of bounded complexes, and similarly to show faithfulness

if you ask about the chain homotopy category instead of the derived category, I think you get a much more complicated answer that depends on things like: enough projectives / injectives? AB4 / AB4*? do objects have finite projective / injective dimension?

(sorry, what I meant was whether the derived category could obtained somehow from the chain homotopy category.)

@TylerLawson Thanks! Those comments answered my questions.

If $C$ is a symmetric monoidal 2-groupoid, the stable homotopy hypothesis shows that its classifying space deloops to a stable 2-type $|C|$. If $C = \mathsf{Alg}_{\mathbb C}^\times$ is the 2-groupoid of units of the Morita 2-category of $\mathbb C$-algebras, Wedderburn theory implies $|C|\simeq \Sigma^2 H\mathbb C^\times$. Has this been written down anywhere?

I'm also interested in the next level up: is there some analogue of Wedderburn theory for suitably finite $\mathbb C$-linear monoidal categories? (I'd also be happy to learn what the stable 3-type $|\mathsf{MonCat}_{\mathbb C}^\times|$ is, but would be surprised if this has been computed.)

@ArunDebray That the picard category of the Morita category is $\Sigma^2 H\mathbb C^\times$ is well-known. For example in the appendix of ArXiv:1509.06811 we show that the fully-duaizable part is the same as Kapronov-Voevodsky 2-vector spaces and in the last sections of arXiv:1712.08029 I explicitly mention that he Morita category of this 2-Vect is $\Sigma^2 H\mathbb C^\times$.

Also if you restrict to rigid tensor categories, then the Picard category of the 3-category of tensor categories is known to be $\Sigma^3 H\mathbb C^\times$. This follows by looking at the global dimension. It is multiplicative under tensor product, invariant under Morita equivalence, and the only tensor category with global dimension 1 is Vect. This implies that Vect is the only invertible tensor category. I don't think this changes if you allow non-rigid tensor categories, but I'm not certain

@ChrisSchommer-Pries Awesome, thank you!

@ChrisSchommer-Pries Is it known what the answer is if we replace vector spaces with super vector spaces?