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Question

Assume a vector space $V$ and a binary operation $*$ that satisfies for all $u, v, w \in V$:

  1. $u * v \in V,$
  2. $u * v = - v * u\,,$
  3. $u * (v * w) = (u * v) * w.$

Can this operation satisfy the Jacobi identity

$u * (v * w) + w * (u * v) + v * (w * u) = 0 \quad \forall u, v, w \in V$?

I believe the answer is no, but I cannot prove it.

Attempted solution

Assuming the Jacobi identity holds, I tried to arrive at a contradiction.

\begin{align} 0 &= \phantom{-} u * (v * w) + w * (u * v) + v * (w * u) \\ &= -(v * w) * u + w * (u * v) + v * (w * u) \\ &= -v * (w * u) + w * (u * v) + v * (w * u) \\ &= w * (u * v). \end{align}

I.e., we have $w * (u * v) = 0$ for all $u, v, w \in V$. This doesn't look like a contradiction to me. It is as far as I could get.

Maybe a different approach makes more sense?

user3749105
  • 127

2 Answers2

1

You have lots of possibilities. In what follows I will assume that you want to product to be bilinear.

What you proved is that a product of $3$ elements is always zero. Let us write $U\subset V$ the subspace of elements $u\in V$ such that $u*v=0$ for all $v\in V$. Then for any $u,v\in V$, $u*v\in U$.

Conversely, to define a product which satisfies your conditions, choose a subspace $U\subset V$, and a basis $(e_1,\dots,e_n)$ such that $(e_1,\dots,e_r)$ is a basis of $U$. You can then define the product by $$ e_i*e_j = \sum_{k=1}^r a_{i,j,k}e_k$$ for all $1\leqslant i,j\leqslant n$ with the only conditions that $a_{i,j,k}=-a_{j,i,k}$, and $a_{i,j,k}=0$ if $i\leqslant r$.

This means you have to choose (freely) some scalars $a_{i,j,k}$ for $r<i<j\leqslant n$ and $1\leqslant k\leqslant r$ (if the charateristic of the base field is $2$, you also have to include $i=j$).

Captain Lama
  • 27,658
1

A Lie algebra $L$ is also associative if it is $2$-step nilpotent, i.e., if $[x,[y,z]]=0$ for all $x,y,z\in L$. The converse is also true. See the posts at MSE:

Associative Lie algebra

Does there exist a non-trivial, associative Lie algebra?

So your computation arriving at $u\cdot (v\cdot w)=0$ for the algebra product is completely correct. It is not a contradiction, on the contrary, it precisely describes the class of such algebras, being skew-commutative, satisfying the Jacobi identity, being associative, and $2$-step nilpotent as a Lie algebra.

Dietrich Burde
  • 140,055
  • But is $w * (u * v) = 0$ for all $u, v, w$ a contradiction? In other words, does that imply that an anti-symmetric and associative operation cannot satisfy the Jacobi identity? – user3749105 Apr 19 '23 at 18:40
  • @user3749105 No, all $2$-step nilpotent Lie algebras satisfy $w\cdot (u\cdot v)=0$ (see the answer), so there is no contradiction. Except, if the Lie algebra is not $2$-step nilpotent. Then we obtain a contradiction. Note that people usually write $x\cdot y$ as $[x,y]$ for Lie algebras. – Dietrich Burde Apr 19 '23 at 18:43