Timeline for How many Lie and associative algebras over a finite field are there?
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| Sep 8, 2020 at 23:53 | comment | added | Thiago | @Qiaochu Yuan Thank you for your answer and for the references. I'll try to read it carefully later. | |
| Sep 8, 2020 at 17:09 | history | edited | Qiaochu Yuan | CC BY-SA 4.0 |
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| Sep 8, 2020 at 17:02 | history | edited | Qiaochu Yuan | CC BY-SA 4.0 |
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| Sep 8, 2020 at 10:15 | history | edited | Qiaochu Yuan | CC BY-SA 4.0 |
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| Sep 8, 2020 at 9:59 | history | edited | Qiaochu Yuan | CC BY-SA 4.0 |
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| Sep 8, 2020 at 9:57 | comment | added | YCor | By the way, the existence of a component of dimension $\ge 2n^3/27$ in the variety of Lie algebra laws in dimension $n$ was observed in: M. Vergne, Réductibilité de la variété des algèbres de Lie nilpotentes. (French) C. R. Acad. Sci. Paris Sér. A-B 263 (1966), A4–A6. | |
| Sep 8, 2020 at 9:51 | history | edited | Qiaochu Yuan | CC BY-SA 4.0 |
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| Sep 8, 2020 at 9:45 | history | edited | Qiaochu Yuan | CC BY-SA 4.0 |
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| Sep 8, 2020 at 8:33 | comment | added | Qiaochu Yuan | The $\frac{4}{27}$ confuses me also but I am very unfamiliar with the structure theory of finite-dimensional algebras. I guess such an algebra $A$ over $\mathbb{F}_q$ looks like a finite product $A/J(A)$ of matrix rings $M_n(\mathbb{F}_{q^i})$ extended by the Jacobson radical $J(A)$ which should in particular be nilpotent... I guess after two square-zero extensions $A/J^3(A) \to A/J^2(A) \to A/J(A)$ complicated stuff is happening? | |
| Sep 8, 2020 at 8:25 | comment | added | YCor | I'm more surprised by Neretin's lower bound $4/27$: it should come either from non-nilpotent Lie algebras, or nilpotent of unbounded nilpotency length (as otherwise it would provide for large $p$, as many $p$-groups of order $p^n$) | |
| Sep 8, 2020 at 8:23 | comment | added | YCor | In the Lie case, one chooses for $m<n$ the $(n-m)$-Grassmanian in $\Lambda^2(K^n)$. This gives as many 2-step $n$-dim nilpotent Lie algebras. It seems to be maximal around $m=2n/3$ (whence the mod 3 discussion), in which case this Grassmannian has dimension $(2/27)n^3+O(n^2)$. This also gives the lower bounds for $p$-groups (at least for odd $p$), yielding $p^{(2/27)n^3+O(n^2)}$ such groups (2-nilpotent of exponent $p$). The estimate in $S^2(K^n)$ instead of $\Lambda^2(K^n)$ seems roughly the same, yielding as many commutative associative nilpotent algebras (or local, if one requires a unit). | |
| Sep 8, 2020 at 3:09 | history | edited | Qiaochu Yuan | CC BY-SA 4.0 |
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| Sep 8, 2020 at 1:46 | history | edited | Qiaochu Yuan | CC BY-SA 4.0 |
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| Sep 8, 2020 at 1:31 | history | edited | Qiaochu Yuan | CC BY-SA 4.0 |
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| Sep 8, 2020 at 1:28 | history | answered | Qiaochu Yuan | CC BY-SA 4.0 |