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!

  • $\begingroup$ A trivial point: you don't really mean no solutions, only no non-$0$ solutions (i.e., no non-Abelian Lie algebras). $\endgroup$
    – LSpice
    Jan 17, 2023 at 17:45
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    $\begingroup$ What do you mean by "no solutions?" This is a system of homogeneous polynomial equations, hence, it has at least the zero solution, i.e. a commutative Lie algebra. $\endgroup$ Jan 17, 2023 at 17:45
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    $\begingroup$ I edited it to refer to nontrivial solutions instead of just solutions. Apologies --- I figured since I was discussing Lie Algebras it would be clear I meant nontrivial ones. $\endgroup$
    – Kensmosis
    Jan 17, 2023 at 20:59

2 Answers 2


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)

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    $\begingroup$ The Kirillov-Neretin paper notes that already in dimension 3, there exists a component that is not a complete intersection. This fact partially addresses your question of why we can't simply compute the dimension by subtracting the number of equations from the number of variables. $\endgroup$ Jan 18, 2023 at 13:26
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    $\begingroup$ Thanks for the additional references! A proof of linearity is much more satisfying than some quick and dirty computer program, and I'm looking forward to checking it out. $\endgroup$
    – Kensmosis
    Jan 19, 2023 at 3:11
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    $\begingroup$ The Kirillov paper does give me some sense of the issue involved (and Qiaochu Yuan's example below also helped me grasp it more clearly). It's clear I need to learn a bit about algebraic varieties before being able to think sensibly about the Jacobi equations. My thanks to you (and everyone else who chimed in) for pointing me in the right direction! $\endgroup$
    – Kensmosis
    Jan 19, 2023 at 3:21
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    $\begingroup$ @Kensmosis If you're new to the concept of a variety that is not a complete intersection, the textbook example is the twisted cubic. It's an illuminating exercise to try (unsuccessfully, of course) to express it using only two equations, and to explain to yourself why having three equations doesn't reduce it to a zero-dimensional object. $\endgroup$ Jan 19, 2023 at 13:45
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    $\begingroup$ @TimothyChow Thanks for the suggestion. I am indeed new to varieties. I'd heard of them, but never realized they were this interesting or useful. Coincidentally, I just encountered the twisted cubic at the beginning of Harris' intro book on the subject. It's definitely clear I was being naive in my approach to the Jacobi eqs, and now I'm trying to get a basic flavor for how one does go about tackling such problems in general. $\endgroup$
    – Kensmosis
    Jan 20, 2023 at 15:04

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.

  • $\begingroup$ Thanks, that's a very illuminating example (and way easier for me to visualize than the Lie Algebra case). Sounds like I need to study algebraic varieties a bit to get a better handle on what's happening in the Lie Algebra case. $\endgroup$
    – Kensmosis
    Jan 19, 2023 at 3:13
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    $\begingroup$ @Kensmosis For pedagogical purposes, it is also useful to consider the example $x_ix_j=0$ with $(i,j)\ne (1,1)$. There are $n$ unknowns and $\binom{n+1}{2}-1$ linearly independent equations, and still there is a nontrivial solution $(1,0,\ldots,0)$. $\endgroup$ Jan 19, 2023 at 10:37
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    $\begingroup$ @VladimirDotsenko Thanks for the distilled example! I've begun reading a simple intro to the subject (Harris' Invitation to Algebraic Geometry). If you have suggestions for a more directly-relevant (but still introductory) treatment of the specific topics I'd need to tackle these sorts of problems, I'd be very interested. From what I gather, it's called intersection theory --- but that seems to be a fairly broad area of algebraic geometry. $\endgroup$
    – Kensmosis
    Jan 20, 2023 at 14:55
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    $\begingroup$ @Kensmosis no, intersection theory is much deeper than things used in discussions here. A good headline to begin with is dimension theory. In particular, "complete intersection" is an intersection where adding each of the equations reduces the dimension by one, and to understand it well, it is of utmost importance to have full clarity on different ways of how one can think of dimension in algebraic geometry. $\endgroup$ Jan 20, 2023 at 16:49
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    $\begingroup$ @printf: in that example the conditions you're imposing are all linear so you only need to check that they're linearly independent. The point of this discussion is that for nonlinear constraints the relevant notion of "independence" is considerably more subtle and hard to check. $\endgroup$ Jan 22, 2023 at 8:27

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