Counting degrees of freedom in Lie algebra structure constants (aka why are there any nontrivial Lie algebras of dim >5?)

Question

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!

Answer 1

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.

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 — 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 $\leqslant 7$

Gorbatsevich - Some properties of the space of n-dimensional Lie algebras (where in particular your observation on linear independence is proved)

Answer 2

It is simply not true that if you impose $k$ polynomial constraints on $n$ variables then the result has dimension $n-k$, even if the constraints "look independent," and this is itself an example. When this happens the resulting affine variety is said to be a complete intersection, and as far as I know this is a pretty rare condition.

Here's a similar but simpler example where we can see what's going on more explicitly. Consider the set of $n \times m$ matrices of rank at most $1$. It's not hard to show that this is an affine variety cut out by the equations

$$x_{ij} x_{k \ell} - x_{i \ell} x_{k j} = 0$$

given by the vanishing of all $2 \times 2$ minors. This imposes ${n \choose 2} {m \choose 2}$ polynomial constraints on $nm$ variables, and of course the former is larger than the latter as soon as $n, m \ge 4$, yet nonzero matrices of rank at most $1$ obviously exist, and in fact this variety clearly has dimension $n + m - 1$ (a nonzero matrix of rank $1$ is an outer product of two nonzero vectors, but we have the freedom to scale either vector). Removing zero and quotienting by scalar multiplication gets us a projective variety isomorphic to $\mathbb{P}^{n-1} \times \mathbb{P}^{m-1}$ embedded into $\mathbb{P}^{mn-1}$; this is the Segre embedding.

This is similar to the Lie algebra example in that we are imposing a system of homogeneous quadratic equations, but differs in that it is much easier in this case to compute the actual dimension of the variety. I have no idea how to compute the dimension in the Lie algebra case.

In this case it's not hard to see explicitly that many of the constraints are redundant most of the time: for example, if $x_{11} \neq 0$ we actually only need to impose the constraint that minors of the form $x_{11} x_{k \ell} - x_{1 \ell} x_{k 1}$ vanish, and there are $(n - 1)(m - 1)$ of these, which gives a naive dimension count of

$$nm - (n - 1)(m - 1) = n + m - 1$$

which is actually correct. On the other hand, if, say, the entire first row vanishes then these constraints are trivially satisfied, so in that case we need the others. So, loosely speaking, this variety is covered by many "patches" on which a much smaller set of equations suffices to cut it out, but no single such set works on every patch, and we need the entire much larger set to cut out the whole thing.

As another perspective on how "independent" the constraints really are, we can rephrase the above argument more algebraically: starting from the constraints $x_{11} x_{k \ell} - x_{1 \ell} x_{k 1} = 0$ we can write

$$x_{11} x_{k \ell} = x_{1 \ell} x_{k 1}$$ $$x_{11} x_{i j} = x_{1 j} x_{i 1}$$ $$x_{11} x_{kj} = x_{1 j} x_{k 1}$$ $$x_{11} x_{i \ell} = x_{1 \ell} x_{i 1}$$

from which we deduce that

$$x_{11}^2 x_{k \ell} x_{ij} = x_{1 \ell} x_{k1} x_{1 j} x_{i 1} = x_{11}^2 x_{kj} x_{i \ell}$$

so we see very explicitly that if $x_{11} \neq 0$ then we could deduce all ${n \choose 2} {m \choose 2}$ constraints from just the $(n - 1)(m - 1)$ constraints involving $x_{11}$ above. But since $x_{11}$ could be zero we can get almost but not all the way to deducing the rest of the constraints from these ones.