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For a bilinear, antisymmetric, alternating operator to be a Lie bracket, it must satisfy the Jacobi identity. I assume this is because a bilinear, antisymmetric, alternating operator does not always satisfy the Jacobi identity.

If I consider this operator without Jacobi identity axiom to be defined on a finite dimensional vector space, I have $$[A_i,A_j]=C_{ij} ^kA_k.$$

I want to find $C_{ij}^k$ such that the Jacobi identity $$\text{Alt}\left([A_i,[A_j,A_k]]\right)=0$$ is not satisfied.

Although I am quite sure that there must exist such structure coefficient, I cannot find one. Can someone shed some light on this? Or, is it the case that (although I do not think so) antisymmetry is enough to yield Jacobi identity?

Nugi
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1 Answers1

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For the bracket defined on generators by $\{x,y\} = x$ and $\{x,z\}=y$ and $\{y,z\}=0$, the Jacobiator $$\{x,\{y,z\}\}+\{y,\{z,x\}\}+\{z,\{x,y\}\}=\{x,0\}+\{y,-y\}+\{z,x\}=-y$$ is nonzero.

More generally you can expand the Jacobi identity as a system of quadratic equations, e.g. in dimension $3$ for $$\begin{align*}\{x,y\}&=a_1x+a_2y+a_3z,\\\{x,z\}&=b_1x+b_2y+b_3z,\\\{y,z\}&=c_1x+c_2y+c_3z\end{align*}$$ you get the system of equations $$\begin{align*}-a_{2} c_{1} - b_{3} c_{1} + a_{1} c_{2} + b_{1} c_{3}&=0,\\ a_{2} b_{1} - a_{1} b_{2} - b_{3} c_{2} + b_{2} c_{3} &=0,\\ a_{3} b_{1} - a_{1} b_{3} + a_{3} c_{2} - a_{2} c_{3} &=0\end{align*}$$ and it easy to find a non-solution; the one above corresponds to $a_1=b_2=1$ and the rest equal to zero.

Ricardo Buring
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  • Does this mean that using this method, for Lie algebra with $n$ generators, we can always find such non-solutions for the Jacobi identity? – Nugi May 24 '20 at 08:44
  • @Nugi Yes, for $n \geqslant 3$. For instance you can just extend the same example by setting all the other brackets to zero. For $n \leqslant 2$ there are no nontrivial equations (only $0=0$) so there are no non-solutions. – Ricardo Buring May 24 '20 at 08:53