Timeline for Yang-Mills algebra and lower central series of surface groups
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 13, 2023 at 0:21 | answer | added | DamienC | timeline score: 2 | |
| Jan 12, 2023 at 15:41 | history | edited | Carl-Fredrik Nyberg Brodda | CC BY-SA 4.0 |
added 67 characters in body
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| Jan 12, 2023 at 15:40 | comment | added | Carl-Fredrik Nyberg Brodda | @DamienC I see, that's interesting -- thanks for checking this! I suppose the paper only mentions the formula holds for $j > 2$, but they then give no indication for how $j=1$ and $j=2$ are obtained...! | |
| Jan 11, 2023 at 18:17 | comment | added | DamienC | The bold 6 is not a typo in their paper. They have a Lie algebra with 4 generators, and relations are cubic. Generators being of degree one, the degree 2 part of the Lie algebra doesn't "see" any relation. Its dimension is thus the one of $\wedge^2(k^4)$. That is indeed 6. | |
| Jan 11, 2023 at 15:32 | comment | added | YCor | I got this experimentally for many cases (it's work with J. Giol, not published nor even written down). | |
| Jan 11, 2023 at 15:27 | comment | added | Carl-Fredrik Nyberg Brodda | @YCor Re: “Sequences of dimensions of lower central factors often satisfy such kind of identities”, is there a reason for this, other than that (say) those are the only ones we can reasonably compute? | |
| Jan 11, 2023 at 11:19 | comment | added | YCor | It should definitely be $a(2)=5$ instead of $6$. As indicated in OEIS, this sequence (with $a(2)=6$) satisfies $\prod (1-x^n)^{a(n)} = (1-x^2)(1-4x+x^2)$. With $a(2)$ redefined as $5$, this gives the smoother formula $\prod (1-x^n)^{a(n)} = 1-4x+x^2$ (note that $a(n)$ is defined by this formula). Sequences of dimensions of lower central factors often satisfy such kind of identities. | |
| Jan 11, 2023 at 10:52 | history | asked | Carl-Fredrik Nyberg Brodda | CC BY-SA 4.0 |