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Suppose we are given a vector space $V$ equipped with a bilinear form $[,]:V\times V\to V$ such that

$$[x,y]=h, \quad \text{and} \quad [x,h]=0=[y,h]$$ for any $x,y$ and $h$ in $V$.

How can we show that this bilinear form defines a Lie bracket on $V$?

A Lie bracket has to be antisymmetric meaning that $[x,y]=-[y,x]$ and satisfies in the Jacobi identity $ [x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0$.

Many thanks!

user510716
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  • @Gae. S Yes, sorry. We can say there exists $h$. My main problem is with the antisymmetric part. – user510716 Oct 08 '19 at 16:28
  • Your conditions don't make sense "for any $x,y,h$" (if you mean e.g. "for all $x,y,h$ we have ..." then that forces $V=0$). Maybe instead you mean to use $h$ just as an abbreviation for $[x,y]$? – Torsten Schoeneberg Oct 08 '19 at 17:41

1 Answers1

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The Jacobi identity here is trivial, because each term in it is zero. We have $[x,[y,z]]=0$ for all $x,y,z$, since the Heisenberg Lie algebra is $2$-step nilpotent - see your previous question and the answers to it:

Associative Lie algebra

So we have $$ [x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0+0+0.$$

By convention, only the Lie bracket $[x,y]=h$ is given, with the implicit understanding that $[y,x]=-h$. Here $(x,y,h)$ is a basis of the Heisenberg Lie algebra.

Dietrich Burde
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  • Thank you very much for your answer! But can you please explain a bit more about the antisymmetric part? – user510716 Oct 08 '19 at 16:33
  • There is nothing more to explain. Since the Lie bracket is antisymmetric, it is enough to say what $[x,y]$ is. Then we know that $[y,x]=-[x,y]$. It is enough to specify this for a basis of the Lie algebra. For the Heisenberg Lie algebra, a basis is $(x,y,h)$ with Lie bracket(s) $[x,y]=h$. We even omit zero brackets. – Dietrich Burde Oct 08 '19 at 18:03