Timeline for Categorical interpretation of disjoint cycle notation for tracing permutations
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Jun 2, 2014 at 21:24 | vote | accept | David Spivak | ||
| Jun 2, 2014 at 3:10 | vote | accept | David Spivak | ||
| Jun 2, 2014 at 19:24 | |||||
| May 31, 2014 at 0:10 | answer | added | Todd Trimble | timeline score: 7 | |
| May 30, 2014 at 20:20 | history | edited | David Spivak | CC BY-SA 3.0 |
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| May 30, 2014 at 19:31 | comment | added | darij grinberg | As you see, I didn't know it either. :) Is there some categorical literature on the trace in the $\mathcal{B}ij$ setting? | |
| May 30, 2014 at 19:29 | comment | added | David Spivak | @darijgrinberg, thanks for the Okounkov/Vershik reference (arxiv.org/pdf/math/0503040v3.pdf). Looks like they didn't know it was called a trace. | |
| May 30, 2014 at 19:24 | history | edited | David Spivak | CC BY-SA 3.0 |
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| May 30, 2014 at 19:19 | comment | added | darij grinberg |
@DavidSpivak: oh, my apologies -- I forgot that you had taken the skeleton. Then it makes sense. These trace maps, for $U$ being $[1]$, occur in Okounkov/Vershik arXiv:math/0503040v3 p. 25, and I recall implementing them in sage (for general $U$) under the name retract_okounkov_vershik. Checking that the image is a permutation using my definition is not too hard: the inverse of the image of $\sigma$ is the image of $\sigma^{-1}$ (although the trace map is not a group morphism).
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| May 30, 2014 at 19:10 | history | edited | David Spivak | CC BY-SA 3.0 |
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| May 30, 2014 at 19:09 | comment | added | David Spivak | @darijgrinberg. Yes, $\mathcal{Bij}$ is the skeleton of the category of finite sets, so every bijection is an automorphism. As for your suggestion about $\sigma^i$, that is correct, but let me pose a question. Clearly, when you simply "remove the elements of $U$ from the cycle notation" it's obvious that you get a new permutation. If I use your formula, "send every $x\in X$ to $\sigma^i(x)$ where $i$....", how do I prove that the result always gives a new permutation? | |
| May 30, 2014 at 19:02 | comment | added | David Spivak | @BenjaminSteinberg, that's true. A permutation can be understood as a functor $\sigma:{\mathbb Z}\to{\bf Set}$, where ${\mathbb Z}$ is the free group on one generator, considered as a category with one object. If $F:{\mathbb Z}\to\{*\}$ is the unique functor to the terminal category, then the left Kan extension $F_!\sigma$ is the set of cycles, and the unit map $\sigma\to F^*F_!\sigma$ is in some sense the cycle decomposition of $\sigma$. However, this does not really help me because I don't know how to use it to compose elements of the group $S_n$, nor to see the trace map. | |
| May 30, 2014 at 17:58 | comment | added | Benjamin Steinberg | The category of finite sets with a Z-action has a coproduct decomposition of each object into indecomposables. This is what gives cycle decomp. | |
| May 30, 2014 at 17:42 | comment | added | darij grinberg | 2) I think you need $X = Y$ for your definition of the cycle map. How do you decompose a bijection between two different sets into cycles? One way to define your trace map without referring to cycle notation is by saying that $Tr^{U}_{X,X}\left(\sigma\right)$ is the map which sends every $x \in X$ to $\sigma^i\left(x\right)$ where $i$ is the smallest positive integer such that $\sigma^i\left(x\right) \in X$. Not sure how categorical this is. | |
| May 30, 2014 at 17:41 | comment | added | darij grinberg | 1) This is an interesting, if not really well-posed, question, and I'd much prefer an answer which gives a formal underpinning to the intuitive observation that disjoint cycle decomposition is a kind of $\mathbb{F}_1$-analogue of the Jordan canonical form. | |
| May 30, 2014 at 17:10 | history | asked | David Spivak | CC BY-SA 3.0 |