This is a question about the true number of constraints imposed by the Jacobi identity on the structure constants of a Lie algebra.

For an $n$-dimensional Lie algebra, there are $\frac{n^2(n-1)}{2}$ structure constants $f_{ab}^c$, where I've accounted for antisymmetry but not the Jacobi identity. Accounting for obvious symmetries, the Jacobi identity $f_{ad}^e f_{bc}^d +f_{bd}^e f_{ca}^d + f_{cd}^e f_{ab}^d= 0$ seems to impose $\frac{n^2(n-2)(n-1)}{6}$ constraints.

For $n>5$, this exceeds the total degrees of freedom of the structure constants. The problem appears to be overspecified and there should (in all likelihood) be no nontrivial solutions. I.e. by such a naive counting of constraints there should be no finite-dimensional Lie algebras with $n>5$. Obviously, this is not the case.

It's certainly possible for an apparently overconstrained system of quadratic equations to have nontrivial solutions. However, it's natural to ask whether that's more than mere coincidence and there is some other mechanism at play — either a less obvious relationship between the equations or a reason they were formed in a manner which admits solutions.

I wrote a simple computer program to analyze how many of the Jacobi equations are trivial or redundant in a couple of simple cases ($n=3$ to $n=12$), and none are. In fact, they all are linearly independent (i.e. if we regard distinct quadratic terms as basis elements in some vector space, then the rank of the resulting $\big[\frac{n^2(n-2)(n-1)}{6}\big] \times \big[\frac{1}{2}\frac{n^2(n-1)}{2}\big(\frac{n^2(n-1)}{2}+1\big)\big]$ sparse matrix is the same as the number of Jacobi equations). The equations may be related in more complicated ways than mere linear independence, of course.

So am I missing something obvious? There must be some additional reduction in the number of independent equations that I'm missing. Somehow I must be massively overcounting the Jacobi identity equations.

I tried posting this on math.stackexchange a few years ago and got no responses, and now I'm revisiting the subject and realized it still puzzles me — so I figured I'd try posting it here.

Presumably, the set of solutions (modulo an overall scale factor) would provide an alternate approach to classifying Lie algebras of a given dimension over the relevant field.

Most likely, a similar issue arises for the structure constants of an associative algebra, though I haven't gone through the motions there.

Thanks in advance for your help!

nosolutions, only nonon-$0$solutions (i.e., no non-Abelian Lie algebras). $\endgroup$