Timeline for Waldhausen $K$-theory before group completion
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 26, 2018 at 20:34 | vote | accept | Tim Campion | ||
| Jun 26, 2018 at 20:34 | comment | added | Tim Campion | Fibrant replacement of Segal spaces should model taking geometric realization weighted by the canonical functor $\Delta \to Cat$ (i.e. a certain $(\infty,2)$-colimit). In the non-$\infty$-case, this is shown to preserve products in Example 4.5 here. | |
| Jun 26, 2018 at 20:27 | comment | added | Tim Campion | One just has to keep in mind that $S_\bullet$ "implicitly deloops" as Tyler put it above. And taking endomorphism objects, as I suggested above still maps into spaces, begging the question. So maybe one should argue that $K^1 C$ is modeled by $\widetilde{S_\bullet C}^{\mathbb{BN}}$ where $\widetilde{(-)}$ denotes fibrant replacement in complete Segal spaces, and $\mathbb{BN}$ is the walking endomorphism, and now we're using the exponential of $\infty$-categories. Can fibrant replacement of complete Segal spaces be done in a product-preserving way, as geometric realization can? | |
| Jun 25, 2018 at 7:51 | comment | added | Yonatan Harpaz | To deduce that the reflection of $K^1_{\oplus}$ (or of $K^2_{\oplus}$ for that matter) gives a group-like $\infty$-groupoid you need to also make sure that this reflection preserves semi-additive functors. For this you need to argue that the reflection can in both cases be done with the S-construction (and use the fact that geometric realization commutes with products). This was in fact one of your questions. Well, for $K^2_{\oplus}$ this is true since you can think of the codomain as being spaces, and I'm pretty sure that it's true for $K^1_{\oplus}$ as well. | |
| Jun 25, 2018 at 7:40 | comment | added | Yonatan Harpaz | Yes, I agree. Indeed, first a similar argument shows that any semi-additive additive functor from stable $\infty$-categories to symmetric monoidal $\infty$-categories takes values in group like symmetric monoidal $\infty$-categories. Then one can show that $F(\Sigma): F(C) \to F(C)$ provides a covariant inverse functor which means that $F(C)$ is in fact an $\infty$-groupoid. | |
| Jun 25, 2018 at 3:01 | comment | added | Tim Campion | Sorry, where I said above that $K^1_\oplus(\mathcal C)$ is grouplike, I meant to say that the additive reflection of $K^1_\oplus$ applied to $\mathcal C$ is grouplike, and hence by the linked argument trivial. | |
| Jun 24, 2018 at 19:50 | comment | added | Tim Campion | Yonatan's observation is actually reminiscent of John Berman's discussion of what happens when you have a covariantly functorial negation operation on a symmetric monoidal category. I think the key here is that a stable $\infty$-category has a covariant endofunctor (namely $\Sigma$) which gets identitifed with the negation operation by the $K$-theory functor. | |
| Jun 24, 2018 at 19:13 | comment | added | Tim Campion | The same argument also shows that $K^1_\oplus(\mathcal C)$ is grouplike. I once gave an argument that if $Id \Rightarrow F$ is a strong monoidal transformation of monoidal endofunctors of symmetric monoidal categories, and if $C$ is a monoid in semiadditive categories, and if $F(C)$ is grouplike, then $F(C)$ is trivial -- in particular, this should apply to $F = K^1_\oplus$, partially answering Question 1. Do you know of weaker conditions implying that $K^1_\oplus(C)$ is trivial? | |
| Jun 24, 2018 at 19:13 | comment | added | Tim Campion | Thanks - this is beautiful! So you're observing that $K^2_\oplus = K^3_\oplus$ when we restrict the domain to stable $\infty$-categories, which basically answers Question 2. | |
| Jun 24, 2018 at 18:17 | history | answered | Yonatan Harpaz | CC BY-SA 4.0 |