Timeline for Equalizer in Free groups
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 20, 2015 at 11:42 | comment | added | მამუკა ჯიბლაძე | Decided to sketch Wu's construction since it is technical and in fact the idea is very simple. The kernel is generated by all those commutators which nontrivially involve all variables. To find free generating subset of such commutators, pass to the associated Lie ring which is the free Lie ring on $n$ generators. There are several known monomial bracket bases of free Lie rings (Lyndon, Shirshov, etc.) and one then just has to take the subset of one of these bases consisting of those brackets which involve each variable at least once. | |
| Jan 19, 2015 at 13:53 | comment | added | მამუკა ჯიბლაძე | More about the image - it is the intersection of the diagonal $F_{n-1}\subset F_{n-1}^n$ with the image of $(d_1,...,d_n):F_n\to F_{n-1}^n$ and I believe one can compute it using known methods. | |
| Jan 19, 2015 at 13:44 | comment | added | მამუკა ჯიბლაძე | Wu has free generators for the kernel of $E_n\to F_{n-1}$; it is true that one still has to determine the image, but I think this also can be done using Wu's calculations. | |
| Jan 19, 2015 at 11:57 | comment | added | Zuriel | Thanks @მამუკაჯიბლაძე for the reference! But I do not see how Wu's result answer my question. | |
| Jan 19, 2015 at 9:44 | comment | added | მამუკა ჯიბლაძე | (Well in fact Wu has a free generating set for the intersection of the kernels of the $d_i$ but I think this can give what you need). | |
| Jan 19, 2015 at 9:31 | comment | added | მამუკა ჯიბლაძე | This has been worked out by Jie Wu in "Combinatorial description of homotopy groups of certain spaces", see Theorem 3.4 there. Probably there are other versions too. | |
| Jan 19, 2015 at 9:14 | comment | added | Zuriel | Thanks @EricWofsey! If it is unknown for general $n$, would it be an interesting open problem? | |
| Jan 19, 2015 at 9:13 | history | edited | Zuriel | CC BY-SA 3.0 |
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| Jan 19, 2015 at 9:09 | comment | added | Eric Wofsey | It's not hard to show that $E_2$ is freely generated by the elements $x_1^kx_2^k$ for $k\in\mathbb{Z}\setminus\{0\}$, but things seem much more complicated for $n>2$. | |
| Jan 19, 2015 at 8:53 | comment | added | Zuriel | Yes @EricWofsey, thanks!! Then is there any hope finding a minimal generating set of $E_n$? | |
| Jan 19, 2015 at 8:47 | comment | added | Eric Wofsey | For $n=2$ your elements do not generate all of $E_2$; for instance, they do not generate $x_1^2x_2^2$. | |
| Jan 19, 2015 at 8:38 | comment | added | Zuriel | @EricWofsey, you are exactly right!! If I restrict my $\sigma$ to the identity and order-reversing permutations, could the list be a generating set? | |
| Jan 19, 2015 at 8:34 | comment | added | Eric Wofsey | I don't think your elements are actually in $E_n$ unless $\sigma$ is either the identity or the order-reversing permutation. | |
| Jan 19, 2015 at 8:19 | history | asked | Zuriel | CC BY-SA 3.0 |