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!



Counting degrees of freedom in Lie algebra structure constants (aka why are there any nontrivial Lie algebras of dim >5?)
Linear independence does not really say much.


This algebraic variety is discussed in some detail in an old paper of Kirillov and Neretin: [The variety $A\_n$ of $n$-dimensional Lie algebra structures](https://www.mat.univie.ac.at/%7Eneretin/kiri87a.pdf).


The case of $n=4$ which already shows many interesting phenomena (but already is nontrivial from the computer algebra viewpoint) is analysed in detail in the recent preprint [Manivel, Sturmfels, and Sverrisdóttir - Four-Dimensional Lie Algebras Revisited](https://arxiv.org/abs/2208.14631) — you might find it enlightening.


Update: I checked the MathSciNet review of the paper of Kirillov and Neretin and found two other relevant references:


[Carles, Diakité - les variétés d'Algèbres de Lie de dimension $\leqslant 7$](https://www.sciencedirect.com/science/article/pii/002186938490125X)


[Gorbatsevich - Some properties of the space of n-dimensional Lie algebras](https://iopscience.iop.org/article/10.1070/SM2009v200n02ABEH003991) (where in particular your observation on linear independence is proved)


