20.92 (2022)
OpenThis question is about existence of an analogue of the Lazard correspondence for pre-Lie algebras and braces. A pre-Lie algebra $A$ is a vector space with a bilinear operation $(x, y) \to xy$ satisfying $(xy)z - x(yz) = (yx)z - y(xz)$ for every $x, y, z \in A$. A pre-Lie algebra $A$ is said to be left nilpotent if, for some $n \in \mathbb{N}$, $A \cdot (A \cdot (A \dots A)) = 0$ where $A$ appears $n$ times in the product. Recall that a set $A$ with binary operations $+$ and $\circ$ is a left brace if $(A, +)$ is an abelian group, $(A, \circ)$ is a group, and $a \circ (b+c) + a = a \circ b + a \circ c$ for every $a, b, c \in A$. Let $p$ be a prime, and $\mathbb{F}_p$ the field of $p$ elements. A left brace $A$ is called an $\mathbb{F}_p$-brace if its additive group is an $\mathbb{F}_p$-vector space such that $a \ast (\alpha b) = \alpha(a \ast b)$ for all $a, b \in A$, $\alpha \in \mathbb{F}_p$, where $a \ast b = a \circ b - a - b$. The idea of a connection between braces and pre-Lie algebras comes from a paper by W. Rump (2014).
a) Let $A$ be an $\mathbb{F}_p$-brace of cardinality $p^k$ for some $k$. Is it true that when $p$ is sufficiently large relative to $k$, the set $A$ with the same additive operation $+$ and with the operation $\cdot$ defined as $a \cdot b = -\sum_{i=0}^{p-2} \frac{1}{2^i} ((2^i a) \ast b)$ is a pre-Lie algebra?
(b) Let $k$ be a natural number, and let $p$ be a prime number such that $p > 2^k$. Is there a bijective correspondence between $\mathbb{F}_p$-braces of cardinality $p^k$ and left nilpotent pre-Lie algebras over $\mathbb{F}_p$ of cardinality $p^k$?
An affirmative answer to any of the above questions would have consequences for the theory of set-theoretic solutions of the Yang–Baxter equation and for the theory of Hopf–Galois extensions.
Progress
(a) Comment of 2025: If $2 \equiv \xi^{p^{n-1}}$ when $\xi$ is a primitive root modulo $p$, and if $A$ is strongly nilpotent of nilpotency index less than $p$, then the result follows by (A. Smoktunowicz, Adv. Math., 409, part B (2022), 108683). It is not known if the result follows when $A$ is not strongly nilpotent.
(b) Yes, there is (S. Trappeniers, Preprint, 2024, https://arxiv.org/abs/2406.02475).
Proof claims
No proof claims yet.
Log in to claim a proof.
Comments
No comments yet. Be the first to comment.
Log in to post a comment.