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
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.