Timeline for Cycle index of $(S_n \times S_n) \rtimes C_2$ acting on matrix indices by row/column permutation and transposition
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Sep 11, 2018 at 15:54 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
gcd & lcm formatting
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| Sep 11, 2018 at 15:45 | vote | accept | Chris Russell | ||
| Sep 11, 2018 at 15:32 | answer | added | Ira Gessel | timeline score: 4 | |
| Sep 11, 2018 at 15:20 | history | edited | YCor | CC BY-SA 4.0 |
ltimes corrected to rtimes
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| Sep 11, 2018 at 15:07 | answer | added | Mark Wildon | timeline score: 2 | |
| Sep 11, 2018 at 14:28 | comment | added | Chris Russell | I see, the second question of mine is not a good one. Thank you for pointing this out. I am thinking I should have asked a different question. What I want this for is to calculate the number of binary matrices up to row permutations, column permutations and transposition. Perhaps I should just ask if any knows how to do that. | |
| Sep 11, 2018 at 14:16 | comment | added | Mark Wildon | The way the question is worded might suggest that the cycle index of $(\sigma,\rho,c)$ depends only on the cycle-types of $\sigma$ and $\rho$. This is not the case: for example, working with transpositions, $((12),(34),c)$ is conjugate to $(\mathrm{id}_{S_4},1,c)$, and not conjugate to $((12),(12),c)$, which is in the class with representative $(\mathrm{id}_{S_4},\mathrm{id}_{S_4},c)$. | |
| Sep 11, 2018 at 12:21 | history | edited | Chris Russell | CC BY-SA 4.0 |
edited tags; edited title
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| Sep 11, 2018 at 12:20 | comment | added | Chris Russell | Sorry again. You are right, the action I describe is not an action on $n \times n$ binary matrices. It is an action on $\{1, \dots, n\} \times \{1, \dots, n\}$. I want to use this action to enumerate binary matrices (via the action on functions from $\{1, \dots, n\} \times \{1, \dots, n\}$ into $\{0,1\}$) and that caused me to mix up ideas when writing my question. I will reformulate my question. | |
| Sep 11, 2018 at 12:07 | comment | added | Mark Wildon | I am still confused: take $\sigma = \rho = \mathrm{id}_{S_n}$. Then $k_1 = l_1 = n$, so your formula says $z (\mathrm{id}_{S_n}, \mathrm{id}_{S_n}, 1) = s_1^{n^2}$. This is the cycle type of the identity permutation on a set of size $n^2$. But there are $2^{n^2}$ binary $n \times n$ matrices. | |
| Sep 11, 2018 at 11:52 | history | edited | Chris Russell | CC BY-SA 4.0 |
added 20 characters in body; edited title
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| Sep 11, 2018 at 11:51 | comment | added | Chris Russell | Oh my apologies, I actually just want to work with the square case when considering this extension and I miswrote my question. I want to consider the action on $n \times n$ binary matrices. | |
| Sep 11, 2018 at 11:49 | comment | added | Mark Wildon | For the extension with $C_2$, do you want an action on the union of the sets of $m \times n$ binary matrices and $n \times m$ binary matrices? In the formula, is $x \ge mn$? Can't we just take $x = mn$, or $\mathrm{lcm}(m,n)$ if preferred? | |
| Sep 11, 2018 at 11:39 | history | edited | Chris Russell | CC BY-SA 4.0 |
added 318 characters in body
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| Sep 11, 2018 at 11:34 | history | asked | Chris Russell | CC BY-SA 4.0 |