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I was given a basis and a bracket denoting a bunch of relations on this basis of a Lie algebra. I am trying to prove that i have in fact been given a Lie algebra and thus that the bracket upholds the Jacobi identity.

The example i have is 7 dimensional thus i have a bunch of cases to check although i can eleminate a couple since i'm working with a nilpotent space. I was wondering if there is a 'smart' method to do this or if there are some sufficiant conditions, instead of going over every possible 3-tuple of basis elements.

Morieris
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  • We only need to check certain $3$-tuples, since the Jacobi identity is automatically satisfied if two basis vectors in a triple coincide. A $7$-dimensional example can be checked quickly - what are your brackets? – Dietrich Burde Jul 05 '19 at 18:38
  • The brackets are given by \begin{align} [E_1, E_2] &= E_3,[E_1, E_3]= E_4,[E_1, E_4]= E_5,[E_1, E_5]= E_6,[E_1, E_6]= E_7,\ [E_2, E_3] &= E_5,[E_2, E_4]= E_6,[E_3, E_4]= E_7, \end{align} – Morieris Jul 05 '19 at 18:57
  • Yes, this is a well-known filiform nilpotent Lie algebra. The Jacobi identity holds. It is $\mathfrak{g}_I(1)$, see my paper here, page $5$. – Dietrich Burde Jul 05 '19 at 19:00
  • I believe to prove that the Jacobi identity holds we can exclude tuples of which the sum of the subscripts are $>7$, contain $E_7$ and such to reduce the amount of tuples, so yes, it probably isn't that much effort to check them all manually.

    I was just wondering if there is a more efficient way for perhaps higher dimensional cases.

     – Morieris Jul 05 '19 at 19:03
  • No, in general there is no shortcut. The polynomial equations in the structure constants arising from the Jacobi identity are difficult in general. – Dietrich Burde Jul 05 '19 at 19:29
  • Alright, thank you for your help! – Morieris Jul 05 '19 at 20:13

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