Timeline for Injective proof about sizes of conjugacy classes in S_n
Current License: CC BY-SA 2.5
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 3, 2009 at 21:35 | comment | added | Jonah Ostroff | One last question. Is there a quick way we can see these injections aren't surjective (for n ≥ 3)? For conjugacy classes with more or fewer than one 1-cycle, this is easy, because 1 will never be the fixed point of the resulting permutation. Is there a permutation that obviously gets missed among the classes with EXACTLY one 1-cycle? | |
| Nov 3, 2009 at 18:35 | history | edited | user631 | CC BY-SA 2.5 |
added 18 characters in body
|
| Nov 3, 2009 at 18:12 | vote | accept | Jonah Ostroff | ||
| Nov 3, 2009 at 18:12 | history | bounty ended | Jonah Ostroff | ||
| Nov 3, 2009 at 17:54 | history | edited | user631 | CC BY-SA 2.5 |
added 276 characters in body
|
| Nov 3, 2009 at 17:54 | comment | added | Jonah Ostroff | Sorry, somehow I missed a word two comments ago. In case it's not clear, what I'm showing is a way to bubble together three cycles, one of which has length > 1 and the other two having length 1. | |
| Nov 3, 2009 at 17:49 | comment | added | Jonah Ostroff | Between these two bubbling techniques, we can repeatedly bubble together the two or three largest cycles until we hit one of two walls: either we're stuck with two cycles (one of length 1 and another of length n-1, so we're done) or a big n-cycle (which we truncate to get an n-1-cycle). If people are satisfied with my second bubbling technique (which really just uses FC's ideas), I'm happy to give FC the check (and the bounty). | |
| Nov 3, 2009 at 17:47 | comment | added | Jonah Ostroff | I guess I'm uncomfortable saying that a case needs to be checked "by hand". I'd feel better if instead we could do the following: given an A-cycle for |A| > 1 and 1-cycles, bubble them together to make an A+2-cycle. A similar technique to the other bubbling should work: look at the smallest element x between A and the two 1-cycles. If it's in A, bubble things together with A first (starting at x) and the larger fixed point last. If it's a fixed point, bubble them together with the fixed points first (starting with x) and ending with the smallest element of A. Again, easy to undo. | |
| Nov 3, 2009 at 17:43 | history | edited | user631 | CC BY-SA 2.5 |
added 594 characters in body
|
| Nov 3, 2009 at 17:39 | comment | added | Sonia Balagopalan | Oh! I lost track of the fact that you are ignoring 1-cycles! | |
| Nov 3, 2009 at 17:39 | comment | added | Harrison Brown | What if we bubbled backwards, starting with the largest cycle? | |
| Nov 3, 2009 at 17:35 | comment | added | Sonia Balagopalan | @Jonah: You are right. I was being careless. | |
| Nov 3, 2009 at 17:32 | comment | added | Jonah Ostroff | Sonia, that first map can't be injective, because there are six permutations of the form (12)(3)(4) and only three of the form (12)(34). | |
| Nov 3, 2009 at 17:28 | comment | added | Harrison Brown | My earlier remark was just that you have to know |A|, |B| in advance, which you didn't mention in your post, so I was confused. I'm still a little confused, actually -- how do you bubble together a k-cycle and a 1-cycle? The same way as you do in general doesn't work, and I don't understand what you said before that. | |
| Nov 3, 2009 at 17:28 | comment | added | Sonia Balagopalan | We would have $(3)(4)(12)\mapsto (34)(12)\mapsto (1234)$ | |
| Nov 3, 2009 at 17:25 | comment | added | Jonah Ostroff | I think the implication is that your algorithm would bubble cycles together in an order solely determined by their sizes, so when inverting the injection (and when you know what conjugacy class you're inverting to), you always know |A| and |B| (and which is which) as you undo each bubbling. Where I'm confused is the issue of what happens when there are two or more 1-cycles. FC, perhaps you can illustrate for me how the injection works on the conjugacy class of S_4 containing (12)(3)(4)? | |
| Nov 3, 2009 at 17:22 | comment | added | Harrison Brown | Hmm, now that I think about it, it doesn't even work uniquely when |A| < |B|, although if you specify a conjugacy class the examples like I mentioned above do decompose uniquely. I'll have to think about it some more. | |
| Nov 3, 2009 at 17:14 | comment | added | Harrison Brown | The breakup isn't unique; (2, 4, 5, 3, 7), for instance, is the image of both (2 4 5) + (3 7) and (2 4) + (5 3 7). Maybe it's unique when |A| < |B|, but I don't think what you said is quite enough to prove this. | |
| Nov 3, 2009 at 16:58 | comment | added | Sonia Balagopalan | That's a really nice argument! | |
| Nov 3, 2009 at 16:22 | history | answered | user631 | CC BY-SA 2.5 |