Two more questions on Kontsevich's "Noncommutative Identities" (Derivations on $\mathbb{C}\langle X,Y \rangle$) [Solved] The following two questions regard once more the following article: arXiv:1109.2469.
In the second chapter we are dealing with the Lie Algebra $\mathfrak{g}$ of derivations $\delta$ of $\mathcal{A}:=\mathbb{C}\langle X,Y \rangle$ satisfying the properties

$(1)$ $\delta(X)=[D,X]$ for some $D\in \mathcal{A}$
$(2)$ $\delta(Y)=0$
$(3)$ For all $t\in\mathbb{C},\  \exists D_t\in\mathcal{A}[t]$ s.t. $\delta(X+tY) \bigg(=\delta(X)\bigg)=[D_t,X+tY]$.

He then claims that:
$(i)$ a linear basis for $\mathfrak{g}$ is given by:
$$\delta_{n,m}=[c_{n,m},X], \ \delta_{n,m}(Y)=0 \quad n\ge0, m\ge1$$
where for any $n,m\ge0$ we define:
$$c_{n,m}:=\sum_{{n+m \choose n} \text{ shuffles } w}w$$
i.e. the sum of all words in $X,Y$ containing $n$ letters $X$ and $m$ letters $Y$.
Elements $D_t\in\mathcal{A}$ corresponding to the derivation $\delta_{n,m}$ are given by
$$D_t=\sum_{0\le k\le n}c_{n-k,m+k}t^k$$
EDIT [09.06] Solved, see below
$(ii)$ $\mathfrak{g}$ is commutative
EDIT [13.06] Solved, see answer.

My "progress":
(i) So far I've only been able to prove that the above tuples $(\delta_{n,m},D_{n,m})$ are indeed elements of $\mathfrak{g}$, and in the argumentation I've shown that
$$[c_{l-1,r+1},X]= [Y,c_{l,r}].$$
Then my try goes on like:
Let now $\sigma\in \mathfrak{g}$ be any (non trivial) element. We decompose the associated $D$ in its homogeneous components $D_{n,m}$ where $n$ indicates the number of $X$'s and ditto $m$ for $Y$. We do this because no cancellation can take place between terms in which these do not match, when evaluating the lie Bracket $[D,X]$, so we can work with each of these components separately.
So let $D_{n,m}$ be a non zero homogeneous component of $D$. We claim that:
$$D_{n,m}=\lambda \cdot c_{n,m} \quad \lambda \in \mathbb{C}.$$
Assume not, then, up to removing the homogeneous components, we have that there must be at least one missing homogeneous polynomial. We will show that this will lead to a contradiction to the existance of a $D_t$ of the desired form.
Let $D_t:=\sum_{k\ge0}^N w_kt^k$, where $w_k$ is just some sum of words in $X$ and $Y$ as always, be a generic candidate. Do to the definition of the lie algebra, we ask ourselves which property must $D_t$ have to be able to satisfy the given equation:
\begin{align*} [D_{n,m},X] & = [D_t,X+tY] = \left[\sum_{k=0}^N w_kt^k,X+tY\right]\\
& = [w_0,X]+\sum_{k=0}^{N-1}([w_{k+1},X]+[w_k,Y])t^{k+1}+[w_N,Y]t^{N+1}\end{align*}
Since the field we are working on is infinite, and since the term on the left hand side is of degree 0 and has trivial centralizer, we see that two straightforward conditions which must be satisfied are:
$$w_0-D_{n,m}=\sum_{k=0}^M \alpha_k X^k \ \text{and} \ w_N=\sum_{k=0}^M \beta_k Y^k \quad \alpha_k,\beta_l \in \mathbb{C} $$
$$[\dots]$$
EDIT [09.06] I've solved this part. One just needed to start from the $w_N=\sum_{k=0}^M \beta_k Y^k$ term, reduce to the case $\beta_k=\delta_{ik}$ and then using the previous result we get the claim.
$(ii)$ My try:
Since we have found a basis and because of the Leibniz property our task reduces to show the following:
$$\forall k,m\ge 0; n,l\ge 1 : \quad [\delta_{n,m},\delta_{k,l}]\equiv 0 \iff [\delta_{n,m},\delta_{k,l}](X)=0 \text{ and } [\delta_{n,m},\delta_{k,l}](Y)=0$$
Clearly the second equality is a direct consequence of the definition of the $\delta$'s.
For the first equality one needs to work a little bit more:
$$[\delta_{n,m},\delta_{k,l}](X) = [\text{ some manipulations... }]= [X,\delta_{k,l}(c_{n,m})-\delta_{n,m}(c_{k,l})-[c_{n,m},c_{k,l}]]$$
I've now checked computationally that the the expression in the right hand side of the Bracket is zero for a few values of $(k,l),(n,m)$; but how to prove this in general?
EDIT [13.06] Solved, see answer.
Any help is greatly appreciated! Many thanks in advance
 A: So, for those interested, here is the solution to the whole:
Part $(i)$:
Lemma:
The equation
$$\sum_{k=0}^{N-1}([w_{k+1},X]+[w_k,Y])t^{k+1}+[w_N,Y]t^{N+1}=0$$
has a unique solution $(w_0,\dots,w_{N-1},Y^{N+m})\in(\mathbb{C}\langle X,Y \rangle\setminus \mathbb{C}\langle Y\rangle)^{N}\times \mathbb{C}\langle Y\rangle$ iff $m\ge 1$.\
In particular the solution to the above is given by $(w_k)_{k=0}^{N}=(c_{n-k,m+k})_{k=0}^{N}$
Proof:
As we know from our above calculations the given $N$-tuple is indeed a solution to the system for $m\ge 0$. Due to the trivial centralizer each $w_k$ depends uniquely on $w_{k+1}$ up to a polynomial in $Y$. As we fixed a starting $w_N$, and crossed the monomial in $Y$ out, the solution is unique. By our previous argumentation we know that in order for the complete cancellation to take place, we need $m$ to satisfy the above condition. Thus the claim follows.

Part $(ii)$:
We first show that the restriction the standard Lie bracket of derivations makes this into a Lie-subalgebra.
Since the Lie Algebra is spanned by the $\delta_{i,j}$ we can restrict our attention to those.
Let thus $\delta_{n,m},\delta_{k,l}\in\mathfrak{g}$ be any two basis elements. Clearly the second condition is a direct consequence of the definition of $\mathfrak{g}$:
$$[\delta_{n,m},\delta_{k,l}](Y) = \delta_{n,m}\circ\delta_{k,l}(Y)-\delta_{k,l}\circ \delta_{n,m}(Y)=\delta_{n,m}(0)-\delta_{k,l}(0)=0.$$
For the first condition one needs to work a little bit more. Using that for any linear map $\delta:\mathfrak{g}\rightarrow \mathfrak{g}$ obeying the Leibniz' law it holds that $\delta([x,y])=[\delta(x),y]+[x,\delta(y)]$ for all $x,y\in\mathfrak{g}$, we have that:
\begin{align*}
[\delta_{n,m},\delta_{k,l}](X) & = \delta_{n,m}\circ\delta_{k,l}(X)-\delta_{k,l}\circ \delta_{n,m}(X) \\
& = \delta_{n,m}([c_{k,l},X])-\delta_{k,l}([c_{n,m},X])\\
& = [\delta_{n,m}(c_{k,l}),X]+[X,\delta_{k,l}(c_{n,m})]+[[c_{k,l},X],c_{n,m}]+[c_{k,l},[c_{n,m},X]]\\
& = [\delta_{n,m}(c_{k,l}),X]+[X,\delta_{k,l}(c_{n,m})]+[X,[c_{n,m},c_{k,l}]]\\
& = [\delta_{n,m}(c_{k,l})-\delta_{k,l}(c_{n,m})-[c_{n,m},c_{k,l}],X]
\end{align*}
Where the third last equality is a consequence of the Jacobi identity.
And lastly for the third condition we have, via a similar calculation:
\begin{align*}
[\delta_{n,m},\delta_{k,l}](X+tY) &= \delta_{n,m}\circ\delta_{k,l}(X+tY)-\delta_{k,l}\circ \delta_{n,m}(X+tY)\\
& = \delta_{n,m}([D_t^{(k,l)},X+tY])-\delta_{k,l}([D_t^{(n,m)},X+tY]) \\
& = [\delta_{n,m}(D_t^{(k,l)})-\delta_{k,l}(D_t^{(n,m)}),X+tY]\\ & \ +[D_t^{(n,m)},[D_t^{(k,l)},X+tY]]-[D_t^{(k,l)},[D_t^{(n,m)},X+tY]] \\
& = [\delta_{n,m}(D_t^{(k,l)})-\delta_{k,l}(D_t^{(n,m)})-[D_t^{(n,m)},D_t^{(k,l)}],X+tY].
\end{align*}
To show that the Lie algebra is abelian we need to show that the Lie-bracket is equivalent to the zero derivation, and because of the Leibniz property our task reduces to show that the action of the bracket on $X$ and $Y$ is the zero map. For $Y$ it is clear by definition. For $X$ we note that the term $\delta_{n,m}(c_{k,l})-\delta_{k,l}(c_{n,m})-[c_{n,m},c_{k,l}]$ is a homogeneous polynomial of degree $n+k$ in $X$ and $m+l$ in $Y$. But we know that for each total degree $(i,j)$ there exists exactly one element of the basis, $\delta_{i,j}$, corresponding to it. Thus it must hold
$$\delta_{n,m}(c_{k,l})-\delta_{k,l}(c_{n,m})-[c_{n,m},c_{k,l}]=\lambda\cdot c_{n+k,m+l} \ , \lambda \in \mathbb{C}$$
But we clearly see that the elements $X^{n+k}Y^{m+l}$ and $Y^{m+l}X^{n+k}$ come with a factor $0$ in the left hand side, for any quadruple $(k,l,m,n)$. For example in the former case, the terms contributing to $X^{n+k}Y^{m+l}$ are given by (assuming $n,k\ge 1$, otherwise a similar argumentation leads the same result):
\begin{align*}[X^{n+k}Y^{m+l}](\delta_{n,m}(c_{k,l})-\delta_{k,l}(c_{n,m})-[c_{n,m},c_{k,l}]) & = [X^{n+k}Y^{m+l}](X^{k-1}[c_{n,m},X]Y^l-X^{n-1}[c_{k,l},X]Y^m)\\ & = X^{k-1}(-X\cdot X^nY^m)Y^l-X^{n-1}(-XX^kY^l)Y^m=0
\end{align*}
hence for the two sides to be equal we need $\lambda=0$ and hence the left hand side is zero as well, lastly giving the claim the algebra is abelian.
