Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: learning: questions

Topic: Kan extensions of M^⊗n along 𝐁sgn: 𝐁Σₙ → 𝐁ℤ/2

Emily (Sep 17 2021 at 23:57):

The nth symmetric power $\mathrm{Sym}^n(M)$ of a module $M$ is the colimit of the functor $\mathbf{B}\Sigma_n\to\mathsf{Mod}_R$ picking $M^{\otimes n}$ and acting on it via permutations. This is the same thing as the left Kan extension of $\mathbf{B}\Sigma_n\to\mathsf{Mod}_R$ along the terminal functor $\mathbf{B}\Sigma_n\to\mathbf{B}*$. So a natural question arises: what about picking monoids or groups other than $*$?

One particular instance of this we could consider is taking left or right Kan extensions along the sign map $\mathbf{B}\mathrm{sgn}\colon\mathbf{B}\Sigma_n\to\mathbf{B}\mathbb{Z}/2$, obtaining a pair $(M',\sigma)$ with

• $M'$ a module;
• $\sigma\colon M' \to M'$ an involution (i.e. order 2 automorphism) of $M'$.
  that is (co/)universal among all such pairs.

Via the usual formula for computing Kan extensions, we can show that $M'$ is the co/limit of the diagram

image.png

in $\mathsf{Mod}_R$. Is there a nice description of it?

Emily (Sep 17 2021 at 23:58):

For $n=0$ it is $R\oplus R$, while for $n=1$ it is $M\oplus M$. At first I thought that for $n=2$ we would get $(M\otimes M)\oplus(M\otimes M)$ quotiented out by the ideal generated by $(a\otimes b,c\otimes d) = (d\otimes c, b\otimes a)$ (I made a mistake here, and Sarah (Griffith) corrected it on Twitter; thanks!).

@algebraicgeome1 (a⊗b,c⊗d) = (d⊗c, b⊗a) I think

- Picard Group Sarah 🏳️‍⚧| Defund the Police (@SC_Griffith)

Emily (Sep 17 2021 at 23:58):

However, Sarah (zrf) pointed that in fact it should be a quotient of $M\otimes M$, since the map from the second factor is determined by the first from the universal property, so I'm currently unsure about what it should be... Again, is there a nice description of it?

@SC_Griffith @algebraicgeome1 Finally occurred to me: You should certainly get a quotient of M^{⊗n} just by examination of the universal property—a map out of the colimit is the same as a certain pair of maps out of M^{⊗n}s, but knowing one of the 2 uniquely determines the other because isos in the diagram

- D-filtered colimit of D-presentable sarahzrfs (@sarah_zrf)