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In particular, if the Lie algebra $\mathfrak{g}$ is nilpotent, is there a tight (tighter than $\frac{n(n+1)}{2}$) upper bound on how many relations are allowed among the basis vectors?

Another question, if a Lie algebra $\mathfrak{g}$ is nilpotent, is it guaranteed that there exists some change of basis such that all of the lie brackets have the form

$[e_i,e_j]=ce_k$

where $e_i,e_j,e_k$ are all basis vectors and $c \in F$? i.e. there are no sums:

$[e_i,e_j]=e_m+e_n$

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    $\begingroup$ I don't understand the first question. What does "number of allowable Lie brackets" mean? $\endgroup$ – Qiaochu Yuan Jul 16 '17 at 21:05
  • $\begingroup$ For example, in a 4 dimension Lie algebra $\mathfrak{g}$ , we have these as potential brackets: $[e1, e2]$, $[e1, e3]$, $[e1, e4]$, $[e2, e3]$, $[e2, e4]$, $[e3, e4]$ where $\{e1,e2,e3,e4\}$ is a basis for $\mathfrak{g}$. However, there are no 4-dimensional Lie algebras with 6 nontrivial relations on basis elements. So there must be some 'upper bound' on how many there can be (due to Jacobi). Furthermore, there are even less nontrivial relations when $\mathfrak{g}$ is nilpotent. So I believe there is an even lower upper bound. $\endgroup$ – user376249 Jul 18 '17 at 15:39
  • $\begingroup$ The sentence with "however" is not true. We can write down all six brackets nontrivially, so that the Jacobi identity is satisfied. $\endgroup$ – Dietrich Burde Jul 18 '17 at 18:16
  • $\begingroup$ I still don't understand the question. An upper bound on how many what? $\endgroup$ – Qiaochu Yuan Jul 18 '17 at 18:48
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No, we cannot assume that such a basis exists for a nilpotent Lie algebra. For the title question, "number of allowable Lie brackets" could mean different things. One interpretation is the dimension of $A_n$, the variety of $n$-dimensional Lie algebra laws. Neretin has shown that the bound on this dimension in general is roughly $\frac{2}{27}n^3. $ We have $n^3$ possibilities for defining Lie bracket vectors on an $n$-dimensional vector space, and thus we obtain the trivial bound $n^3/2$ by skew-symmetry. The variety of $n$-dimensional Lie algebra structures on a vector space over $K$ consists of tuples of structure constants in $K^{n^3}$ satisfying skewsymmetry and the Jacobi identity.

The second interpretation is, how many brackets $[e_i,e_j]$ for $i<j$ in the basis vectors we may assume to be zero in an appropriate basis. For example, in dimension 3, we need to have all three brackets in the basis vectors nonzero, as the Lie algebra $\mathfrak{sl}_2$ shows. For the Heisenberg Lie algebra we can find a basis such that only one bracket in basis elements is nonzero. For nilpotent Lie algebras in general, I don't know what the bound is. Certainly there is not always a basis as above.

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  • $\begingroup$ How is it possible that there is a cubic? Is that bound including relations between vectors that are not basis vectors? If so, my question was how many nontrivial relations can there be that involve basis vectors. $\endgroup$ – user376249 Jul 16 '17 at 18:47
  • $\begingroup$ Take two basis vectors $e_1,e_2$. Then we have 8 parameters for the brackets, and after skewsymmetry and Jacobi still 2 parameters, i.e., $[e_1,e_2]=ae_1+be_2$. So the maximum number of allowable Lie brackets is 2 here, starting from $2^3$. $\endgroup$ – Dietrich Burde Jul 16 '17 at 20:10
  • $\begingroup$ Why do we have 8 parameters? $\endgroup$ – user376249 Jul 18 '17 at 16:07
  • $\begingroup$ As elements of the vector space we have all four brackets $[e_i,e_j]$ with $1\le i,j\le 2$ in two basis vectors, so 8 parameters. Because of skewsymmetry some parameters are given. $\endgroup$ – Dietrich Burde Jul 18 '17 at 18:12
  • $\begingroup$ Dietrich, would you reformulate the question you want to answer? As Qiaochu, I don't know what it purports to mean. $\endgroup$ – YCor Jul 18 '17 at 21:19

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