Example of Leibniz algebra that is not a Lie algebra.

Question

I need an example of Leibniz algebra that is not a Lie algebra.

I found the following example but it seems that is not true:

$L $ is a $k $-vector space and { $e_1,e_2 $} a basis and the billinear form is defined as follows

$[e_1,e_2]=[e_2,e_2]=e_1$ and $[e_1,e_1]=[e_2,e_1]=0$.

But $[[e_1,e_1],e_2] \neq [[e_1,e_2],e_1] +[[e_1,[e_1,e_2]] $ because

RS=$[0,e_2] $ and LS=0.

Dietrich Burde · Accepted Answer · 2018-01-22T19:52:02.330

1

The right hand side equals the left hand side in this case, because $[0,e_2]=0$, since it is bilinear. So it should work.

edited Jan 22 '18 at 19:52

answered Jan 22 '18 at 19:45

Dietrich Burde

Can you argument why [$0,e_2]$=0? – Problemsolving Jan 22 '18 at 19:53
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I already did. It is bilinear. What does this mean? In particular we have $f(0,y)=0\cdot f(0,y)=0$, from $f(\lambda x,y)=\lambda f(x,y)$. – Dietrich Burde Jan 22 '18 at 19:53

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