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Here is a connection that I recently noticed, but I haven't quite been able to make sense of. It might follow from well-known facts; apologies, if so. This is quite far from my area.

First, in "Yang-Mills Algebra", arXiv:0206205 by Connes and Dubois-Violette, they construct the Yang-Mills algebra $\mathcal{A}$ and study some of its properties. On p. 7, they mention that $\mathcal{A}$ is the universal enveloping algebra of a certain graded Lie algebra $\bigoplus_{j \geq 1} \mathfrak{g}_j$, and give an explicit formula for the dimensions, which begins:

$$ 4, \mathbf{6}, 16, 45, 144, 440, \dots $$

This is sequence A072279 (In fact, I think the bold 6 in their paper and the OEIS entry is a typo, as their own formula yields 5). EDIT: No, as @DamienC points out, it seems as if the 6 is correct.

Second, I was recently looking at the ranks of the lower central factors of the surface group $S_2 = \langle a, b, c, d \mid [a,b][c,d]=1 \rangle$, i.e. the factor groups in the lower central series of $S_2$. Curiously, I noticed that this is given by the same formula as in the case above (though with a 5 instead of a 6). There are known ways to get there in high-powered ways, as e.g. in this answer using Koszul duality (which I do not understand). Furthermore, this question and the answer to it are of course relevant, especially given the paper linked in there by Papadima and Yuzvinsky. So I suppose everything on this side seems fairly well understood.

My question is then:

Q. Is there a deeper connection between surface groups and the (Lie algebra associated to the) Yang-Mills algebra $\mathcal{A}$?

Of course, the answer might be "no, the dimensions just satisfy the same relation", or "yes, by general facts about Koszul algebras", either of which I suppose I would be satisfied with, too. I've added the tag, just in case.

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    $\begingroup$ 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. $\endgroup$
    – YCor
    Commented Jan 11, 2023 at 11:19
  • $\begingroup$ @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? $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Jan 11, 2023 at 15:27
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    $\begingroup$ I got this experimentally for many cases (it's work with J. Giol, not published nor even written down). $\endgroup$
    – YCor
    Commented Jan 11, 2023 at 15:32
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    $\begingroup$ 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. $\endgroup$
    – DamienC
    Commented Jan 11, 2023 at 18:17
  • $\begingroup$ @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...! $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Jan 12, 2023 at 15:40

1 Answer 1

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This is a long comment, rather than an answer.

As far as I understand, on the one hand we have the Lie algebra appearing in the paper of Connes and Dubois-Violette, that is the graded Lie algebra $\mathfrak{g}$ generated by $x_0,x_1,x_2,x_3$ (in degree $1$) with cubic relations: $$ [x_i,[x_i,x_l]]+[x_j,[x_j,x_l]]+[x_k,[x_k,x_l]],\quad\mathrm{with}~\{i,j,k,l\}=\{0,1,2,3\}. $$ Dimensions of the graded pieces of $\mathfrak{g}$ are given by the sequence A072279 in the OEIS: $$ 4,6,16,45,144,440,\dots $$ $$ ~ $$ On the other hand, we have the Lie algebra $\mathfrak{L}$ generated by $a,b,c,d$ (in degree $1$) with quadratic relation: $$ [a,b]+[c,d]=0. $$ You seem to claim that dimensions of the graded pieces of $\mathfrak{L}$ are also given by the sequence A072279, but with a 5 instead of a 6: $$ 4,5,16,45,144,440,\dots $$ I couldn't find a reference for this [actually, the sequence with a 5 instead of a 6 doesn't appear in the OEIS]. $$ ~ $$ A naive way to understand such a coincidence of dimensions would be to find a surjective graded lie algebra morphism $\mathfrak{g}\to\mathfrak{L}$ that has one-dimensional kernel that sits in degree $2$.

I don't know if this is a reasonnable thing to ask or not.

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    $\begingroup$ A trivial observation one can add is that the quadratic relation for the LCS is one of the three relations of the algebra that appears in the paper of Connes and Dubois-Violette as "the quadratic anti self-duality algebra", which is a quotient of the YM algebra. So at least the two algebras discussed by the OP have the same nontrivial quotient that had already been studied. $\endgroup$
    – Vladimir Dotsenko
    Commented Jan 13, 2023 at 6:24
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    $\begingroup$ @DamienC Thank you for these comments, which are interesting. Regarding: "I couldn't find a reference for [the lower central factors of $S_2$]", this is in the paper of Papadima and Yuzvinsky mentioned above (link). You can also check this with GAP. $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Jan 13, 2023 at 12:25
  • $\begingroup$ @Carl-FredrikNybergBrodda yes indeed, I could undertand what's going on thanks to YCor's comment to your question. $\endgroup$
    – DamienC
    Commented Jan 13, 2023 at 22:18

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