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I am trying to find an example of an algebra over a field of characteristic p (prime) which satisfies anti-symmetry and Jacobi identity but is not a lie algebra. i.e., [x,x] is not zero.

Can one provide a pattern or a general method to modify existing Lie algebras in order to save anti-symmetry and Jacobi but not $[x,x]=0$ ?

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  • $\begingroup$ Maybe in characteristic 2 try $[x,x]=y, [x,y]=0, [y,y]=0$? $\endgroup$
    – Ben McKay
    Jun 10, 2015 at 10:28

1 Answer 1

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You must have $p=2$ otherwise antisymmetry implies the alternate identity $[x,x]=0$.

The minimal example taylored on the pattern below I can provide is the (multiplicative) subalgebra (over the field $\mathbb{Z}/2\mathbb{Z}$) generated by $$ E_{12}+E_{23}= \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{pmatrix}\ ;\ E_{13}= \begin{pmatrix} 0 & 0 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix} $$
then, you can check you have your algebra.

Now, if, for some reason, you need to preserve some identities or relations of an existing Lie algebra (always in characteristic $2$ for the reason said before), proceed as follows

  1. take you Lie algebra $(L,[-,-]_L)$ over $k$ (of characteristic $2$)
  2. add two dimensions by forming $L^{(1)}=L\oplus k.e_1\oplus k.e_2$
  3. extend the bracket $[x,y]$ to $L^{(1)}$ by
    1. $[x,y]_L$ if $x,y\in L$
    2. $[x,e_i]=[e_i,x]=0$ if $x\in L$
    3. $[e_1,e_1]=e_2;\ [e_2,e_i]=[e_i,e_2]=0$

then $L^{(1)}$ admits $L$ as a sector (direct summand and quotient), fulfills anti-symmetry and Jacobi, but not the alternate identity ($[x,x]=0$).

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