Show that the Lie algebra generated by x, y with relations $ad(x)^2(y) = ad(y)^5(x) = 0$ is infinite dimensional and construct a basis This is a problem from Introduction to Representation Theory by Etingof et al. I have no idea how to approach it. My friend wrote a program that uses Gaussian elimination to find a basis, which starts with x, y, [x, y], [y, [x, y]], ... but freezes once it gets to elements with 9 or more occurrences of x, y.
I expect that there is some way to characterize the nonzero elements of this Lie algebra (i.e., elements that are "immune" to being shown to be zero through repeated application of the Jacobi identity) which could be turned into a way to enumerate them. I don't see what it is, though.
It's also totally unclear to me how to show that some set we know spans the Lie algebra (proven using some criterion as described in the previous paragraph) is actually linearly independent, unless we find a faithful representation... but I have no idea how to find a faithful representation for an infinite-dimensional Lie algebra.
Hints would be greatly appreciated.
Edit: After looking at the partial output, we can form the "straightforward" conjecture that:


*

*If $n \geq 4$ is even, then there is one basis element of grade $n$, namely $\mathrm{ad}([x, y])^{(n-4)/2}[y, [y, [y, x]]]$. If $n \geq 5$ is odd, then there are two basis elements of grade $n$, namely $[x, z]$ and $[y, z]$ where $z$ is the single basis element of grade $n-1$.


I also have a different conjecture, which is mutually exclusive with the above: that if we flatten out all nonzero elements into the form ${\rm ad}(e_1) {\rm ad}(e_2) \ldots {\rm ad}(e_{n-1}) e_n$ using the Jacobi identity, then in the sequences $e_1, \ldots, e_{n-1}, e_n$ and $e_1, \ldots, e_n, e_{n-1}$, any span containing two x's must contain at least two y's and any span containing five y's must contain at least two x's, otherwise we can use the Jacobi identity to rearrange the element somehow to show that it vanishes. If this is true, then the long-range behaviour must be:


*

*For $n \geq 8$ there is exactly one basis element of grade $n$, namely $[... [y, [x, [y, [y, [x, y]]]]] ...]$ (the pattern $y x y$ repeats).


I still don't see an approach to proving that none of the elements in this sequence actually vanishes, though.
Edit 2: Here is the hint given by Etingof in an email:

No, there are no published solutions, since these problems are used as
  a homework for classes. This is indeed a nontrivial problem.
This is the positive part of the affine Kac-Moody algebra of type
  A_2^2, which is the twisted loop algebra for sl(2). So what you need
  to do is to relate this Lie algebra to loops into sl(2). I see from
  your post that you are on the right track.

 A: A representation of the algebra in question (called $\mathfrak{g}_4$ in the text) is given by the vector space $\mathbb{C}[t]^3$ (3-component vectors where each component is a polynomial in $t$) with
\begin{equation}
\rho(x) = X = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ t & 0 & 0 \end{pmatrix},
\quad
\rho(y) = Y = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}
\end{equation}
It is readily verified that $[X, [X, Y]] = [Y, [Y, [Y, [Y, [X, Y]]]]] = 0$, so the defining relations are satisfied. Furthermore
\begin{equation}
[Y, [Y, [X, [Y, [Y, [X, Y]]]]]] = \begin{pmatrix} 0 & -3t^2 & 0 \\ 0 & 0 & -3t^2 \\ 0 & 0 & 0 \end{pmatrix} = -3t^2 Y
\end{equation}
from which the infinite dimensionality follows.
Define
\begin{align}
B_1 &= \{b_0, c_0\} = \{x, y\} \\
B_2 &= \{d_0\} = \{[x, y]\} \\
\end{align}
and for $k \in \mathbb{N}$, define
\begin{align}
B_{6k + 3} &= \{e_k\} = \{[y, B_{6k+2}]\} \\
B_{6k + 4} &= \{f_k\} = \{[y, [y, B_{6k+2}]]\} \\
B_{6k + 5} &= \{g_k, h_k\} = \{[x, [y, [y, B_{6k+2}]]], [y, [y, [y, B_{6k+2}]]]\} \\
B_{6k + 6} &= \{a_{k+1}\} = \{[y, [x, [y, [y, B_{6k+2}]]]]\} \\
B_{6k + 7} &= \{b_{k+1}, c_{k+1}\} = \{-\frac{1}{2}[x, [y, [x, [y, [y, B_{6k+2}]]]]], [y, [y, [x, [y, [y, B_{6k+2}]]]]]\} \\
B_{6k + 8} &= \{c_{k+1}\} = \{[x, [y, [y, [x, [y, [y, B_{6k+2}]]]]]]\} \\
\end{align}
In the representation previously given, $\rho(B_i)$ for $i = 1, 2, \ldots$ are pairwise disjoint and their union is linearly independent. Therefore $B= \bigcup_{i=1}^\infty B_i$ is a disjoint union and is linearly independent.  We claim that, in fact, $B$ is a basis of $\mathfrak{g}_4$ (i.e,, our representation is faithful).
According to Etingof, the necessary calculations are "tedious". The result, which may be proven by induction on the grade of the result, is:
\begin{align}
[a_k, b_\ell] &= 2b_{k+\ell} \\
[a_k, c_\ell] &= -c_{k+\ell} \\
[a_k, d_\ell] &= d_{k+\ell} \\
[a_k, e_\ell] &= 0 \\
[a_k, f_\ell] &= -f_{k+\ell} \\
[a_k, g_\ell] &= f_{k+\ell} \\
[a_k, h_\ell] &= -2g_{k+\ell} \\
[b_k, c_\ell] &= d_{k+\ell} \\
[b_k, d_\ell] &= 0 \\
[b_k, e_\ell] &= 0 \\
[b_k, f_\ell] &= g_{k+\ell} \\
[b_k, g_\ell] &= 0 \\
[b_k, h_\ell] &= 2a_{k+\ell+1} \\
[c_k, d_\ell] &= e_{k+\ell} \\
[c_k, e_\ell] &= f_{k+\ell} \\
[c_k, f_\ell] &= h_{k+\ell} \\
[c_k, g_\ell] &= a_{k+\ell+1} \\
[c_k, h_\ell] &= 0 \\
[d_k, e_\ell] &= g_{k+\ell} \\
[d_k, f_\ell] &= a_{k+\ell+1} \\
[d_k, g_\ell] &= -2b_{k+\ell+1} \\
[d_k, h_\ell] &= -2c_{k+\ell+1} \\
[e_k, f_\ell] &= 3c_{k+\ell+1} \\
[e_k, g_\ell] &= 3d_{k+\ell+1} \\
[e_k, h_\ell] &= 0 \\
[f_k, g_\ell] &= 3e_{k+\ell+1} \\
[f_k, h_\ell] &= 0 \\
[g_k, h_\ell] &= -2f_{k+\ell+1} \\
\end{align}
These identities are proven by repeated applications of the Jacobi identity (in both the base and inductive cases).
For example
\begin{align}
[a_k, d_\ell] &= [[y, g_k], d_\ell] \\
&= [y, [g_k, d_\ell]] - [g_k, [y, d_\ell]] \\
&= 2[y, b_{k+\ell+1}] - [g_k, e_\ell] \\
&= -2d_{k+\ell+1} - [[x, f_k], e_\ell] \\
&= -2d_{k+\ell+1} - [x, [f_k, e_\ell]] + [f_k, [x, e_\ell]] \\
&= -2d_{k+\ell+1} + [x, [e_\ell, f_k]] \\
&= -2d_{k+\ell+1} + 3[x, c_{k+\ell+1}] \\
&= d_{k+\ell+1}
\end{align}
I believe some of the cases are difficult; I haven't actually tried most of them.
Since an arbitrary element of $\mathfrak{g}_4$ can be built up by taking commutators of $b_0$ and $c_0$, these identities show that an arbitrary element can be written as a linear combination of the $B = \bigcup_{k=0}^\infty \{a_k, b_k, c_k, d_k, e_k, f_k, g_k, h_k\}$.
If someone can provide a shorter elementary proof, one that is not too long to write out explicitly, I would appreciate it.
Prof. Etingof points out that the result of this problem is a specific case of the Gabber-Kac theorem. If one has all of the theory of Kac-Moody algebras at one's disposal, one need simply construct the root system for this Lie algebra, and immediately get the basis from the root system. But such theory is not covered in the text before the point where this problem appears.
A: Too long for a comment - here is the complete exercise by Etingof. He says that it is hard.
Problem 2.48. (hard!) Consider the Lie algebra $L_n$ generated by two elements $x,y$ with the defining relations ${\rm ad}(x)^2(y)= {\rm ad}(y)^{n+1}(x)=0$ for $n\ge 1$. 
(a) Show that the Lie algebras $L_1,L_2,L_3$ are finite-dimensional and find their dimensions. 
(b) (harder!) Show that the Lie algebra $L_4$ has infinite dimension. Construct explicitly a basis of this algebra.
Solutions are available, I think  (see MIT OpenCourseWare).
