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Stream: learning: questions

Topic: Of Q-constructions and other demons

Federica Pasqualone (May 21 2025 at 14:46):

Hi everyone, I have the following question: Is the Q-construction over an exact category $\mathcal{E}$ yielding a triangulated category? Or not always? I just wanna make sure I got it right. Thanks!

Kevin Carlson (May 21 2025 at 17:26):

No, I think you have a level shift there. Algebraic K-theory of an exact (or Waldhausen or stable $\infty-$) category is a single spectrum, not a whole triangulated category.

Federica Pasqualone (May 21 2025 at 17:51):

Thanks @Kevin Carlson ! Actually I was wondering about $\Sigma$, i.e. the shift. Because one takes equivalence classes of spans of admissible monic epic ...

Federica Pasqualone (May 21 2025 at 17:54):

Well the K-theory comes later, it is the loop space of the geometric realization of the nerve of the Quillen construction, i.e. $K\left(\mathcal{E}\right) := \Omega|NQ\mathcal{E}|$.

Kevin Carlson (May 21 2025 at 17:58):

Ah, I see what you were thinking; I tend to think of the Q-construction as going all the way to the algebraic K-theory. I don't remember basically any details about the category $Q\mathcal E$ but I'd be a bit shocked if it were triangulated. Do you know any of the appropriate properties?

Federica Pasqualone (May 21 2025 at 18:02):

I am meditating on the shift :woman_in_lotus_position: :rolling_on_the_floor_laughing:

Federica Pasqualone (May 21 2025 at 18:03):

It should be that one takes an exact category and defines the morphisms as equivalence classes of spans of admissible mono-epis (the objects are the same as the original exact category). Here you are $Q\mathcal{E}$.

Federica Pasqualone (May 21 2025 at 18:09):

For the triangulated category, if this is what you were referring to, we need to define in fact a triangulation ($\Sigma$ and a class of distinguished triangles) satisfying certain axioms.

Federica Pasqualone (May 21 2025 at 18:13):

Add to the list of Federica's thoughts that the derived category of an Abelian one is indeed triangulated...

Antonio Lorenzin (May 21 2025 at 18:29):

Federica Pasqualone said:

Add to the list of Federica's thoughts that the derived category of an Abelian one is indeed triangulated...

You actually don't need Abelian here, you can define derived categories also for exact categories, and these are triangulated as well! This is covered in "Exact categories" by Bühler.

Federica Pasqualone (May 21 2025 at 18:31):

Then we are back to the question about $Q\mathcal{E}$ ... (it's a loop space :rolling_on_the_floor_laughing: ) Thanks @Antonio Lorenzin !

Federica Pasqualone (May 22 2025 at 20:33):

And since you mentioned them @Kevin Carlson , $S_{\bullet} \mathcal{C}$ of a Waldhausen is Waldhausen, but we have to switch to Reedy cofibrations. If we take levelwise cofibrations, what goes wrong? Well ...

Federica Pasqualone (May 22 2025 at 20:39):

$\mathcal{S}_{\bullet}\mathcal{C} \subseteq \textbf{Fun}\left(\textbf{Ar}\left[\bullet\right], \mathcal{C}\right)$ is a full subcategory inclusion with objects such that the tail of the lattice progressively reduces to a unique zero. Now the latching object, ...