Any Lie monomial may be written as a linear combination of "simpler" Lie monomials

Question

Some definitions:

A Lie monomial in the elemens of a set $X$ is a finite product of elements of $X$ bracketed by Lie brackets in any manner, e.g. $[[[x_3,[x_1,x_2]],x_3],[x_2,[x_1,x_1]]] $.

A simpler Lie monomial is a Lie monomial bracketed successively from left to right, that is, Lie monomials of the form: $[\cdots[x_1,x_2],x_3],x_4],\cdots,x_k], $ e.g.$[[[x_1,x_2],x_3],x_4].$

The length of a Lie monomial is the number of elements bracketed in it, e.g. $[[[x_1,x_2],x_3],x_4]$ is of length 4.

I am trying to prove the following statement:

Any Lie monomial lies in the span of simpler lie monomials

I found this statement is on Reuteneuter's book about Free Lie algebras, althought he doesn't give any demonstration. Well, I'll show what I did until now:

We proceed by induction on the lenght $|w|$ of a Lie monomial $w$. If $|w|=2,$ we have $w=[x,x']$ and there's nothing to prove. We assume for $|w|=k>1$ and consider monomial $w$ of lenght $|w|=k+1.$ Since now $|w|\geq 3, $ we may write $w=[w_1,w_2],$ where $1\leq |w|\leq k$ and $|w_2| = k+1 - |w_1|.$ We can also assume that both $w_1,w_2$ are simpler lie monomials, since their lenght is smaller than $k$ covered by the induction hyphotesis. Notice that if $|w_2| =1,$ we are done. So we may assume that $|w_2|>1$ and write $w_2 = [w_2',x]$, where $w_2'$ is also a simpler lie monomial of lenght $|w_2'| = |w_2|-1 = k-|w_1|$. Now we use the Jacobi identity:

$$w = [w_1,[w_2',x]] = [[w_1,w_2'],x]+[w_2',[w_1,x]].$$

Here comes the problem. The induction hyphotesis guarentees that $[w_1,w_2']$ is a simpler lie momomial, since its lenght is $|w_1|+|w_2'| = k$; therefore, $[[w_1,w_2'],x]$ is also a simpler lie monomial. But what about $[w_2',[w_1,x]]$? Of course $[w_1,x]$ is a simpler one. But I cannot ensure that the bracket $[w_2',[w_1,x]]$ will remain a simpler lie monomial. Also, it's lenght is just $k+1$ which is not covered by the induction hyphotesis. What now?

Any help on finding a path to prove this? Thank you.

Answer 1

1. Definitions

$\newcommand{\kk}{\mathbf{k}}$ First, let me restate your notations and definition.

Fix a Lie algebra $L$ over any commutative ring $\mathbf{k}$. The word "span" shall always mean "$\mathbf{k}$-linear span" from now on. If $U$ is a subset of $L$, then $\kk U$ shall denote the span of $U$.

If $U$ and $V$ are two subsets of $L$, then $\left[U, V\right]_0$ shall mean the set of all Lie brackets $\left[u, v\right]$ with $u \in U$ and $v \in V$. This is a subset of $L$.

If $U$ and $V$ are two $\mathbf{k}$-submodules of $L$, then $\left[U, V\right]$ shall mean the span of the set $\left[U, V\right]_0$. This is a $\mathbf{k}$-submodule of $L$ and contains $\left[U, V\right]_0$ as a subset (but is, in general, greater).

Let $X$ be a subset of $L$. Let me restate the definition of Lie monomials as follows:

Definition 1. We define a sequence $\left(B_1, B_2, B_3, \ldots\right)$ of subsets of $L$ recursively as follows: We set $B_1 = X$; then, for each $n > 1$, we set $B_n = \bigcup\limits_{a + b = n} \left[B_a, B_b\right]_0$, where the $a$ and $b$ in the big union are meant to range over positive integers.

Thus, $B_1 = X$ and $B_2 = \left[B_1, B_1\right]_0 = \left[X, X\right]_0$ and $B_3 = \left[B_1, B_2\right]_0 \cup \left[B_2, B_1\right]_0 = \left[X, \left[X, X\right]_0\right]_0 \cup \left[\left[X, X\right]_0, X\right]_0$ and $B_4 = \left[B_1, B_3\right]_0 \cup \left[B_2, B_2\right]_0 \cup \left[B_3, B_1\right]_0$ and so on.

Definition 2. Set $B = B_1 \cup B_2 \cup B_3 \cup \cdots$. The elements of $B$ are called the Lie monomials in $X$. For each $n \geq 1$, the elements of $B_n$ are called the Lie monomials in $X$ of length $n$.

Next, I shall define what you call "simpler monomials", but I will call them "left-bracketed Lie monomials" instead:

Definition 3. We define a sequence $\left(C_1, C_2, C_3, \ldots\right)$ of subsets of $L$ recursively as follows: We set $C_1 = X$; then, for each $n > 1$, we set $C_n = \left[C_{n-1}, X\right]_0$.

Thus, $C_1 = X$ and $C_2 = \left[C_1, X\right]_0 = \left[X, X\right]_0$ and $C_3 = \left[C_2, X\right]_0 = \left[\left[X, X\right]_0, X\right]_0$ and $C_4 = \left[C_3, X\right]_0 = \left[\left[\left[X, X\right]_0, X\right]_0, X\right]_0$ and so on.

Definition 4. Set $C = C_1 \cup C_2 \cup C_3 \cup \cdots$. The elements of $C$ are called the left-bracketed Lie monomials in $X$.

2. The claim and the proof outline

Now, your claim is the following:

Theorem 1. We have $B \subseteq \kk C$.

The proof is easiest made using the following definition:

Definition 5. We define a sequence $\left(L_1, L_2, L_3, \ldots\right)$ of $\mathbf{k}$-submodules subsets of $L$ recursively as follows: We set $L_1 = \kk X$; then, for each $n > 1$, we set $L_n = \left[L_{n-1}, \kk X\right]$.

Thus, $L_1 = \kk X$ and $L_2 = \left[L_1, \kk X\right] = \left[\kk X, \kk X\right]$ and $L_3 = \left[L_2, \kk X\right] = \left[\left[\kk X, \kk X\right], \kk X\right]$ and so on.

Now, let me split Theorem 1 into the following bite-sized pieces:

Proposition 2. We have $L_n = \kk C_n$ for each $n \geq 1$.

Proposition 3. We have $\left[L_a, L_b\right] \subseteq L_{a+b}$ for any $a \geq 1$ and $b \geq 1$.

Proposition 4. We have $B_n \subseteq L_n$ for each $n \geq 1$.

See below for the detailed proofs of these three propositions as well as the derivation of Theorem 1 from them. But first, here are hints that should suffice if you have any experience with the Lie algebra axioms:

Proposition 2 is proven by straightforward induction on $n$.

To prove Proposition 3, we proceed by induction on $a$. In the induction step, we assume that Proposition 3 is true for $a-1$, and intend to prove it for $a$. It suffices to show that $\left[\left[x, y\right], z\right] \in L_{a+b}$ for all $x \in L_{a-1}$, $y \in X$ and $z \in L_b$ (because $L_a = \left[L_{a-1}, L\right]$ is spanned by elements of the form $\left[x, y\right]$ with $x \in L_{a-1}$ and $y \in X$). But this follows by applying the Jacobi identity \begin{align} \left[\left[x, y\right], z\right] = \left[\left[x, z\right], y\right] - \left[x, \left[z, y\right]\right] \end{align} and realizing that both addends $\left[\left[x, z\right], y\right]$ and $- \left[x, \left[z, y\right]\right]$ on the right hand side belong to $L_{a+b}$ (indeed, we have $\left[x, z\right] \in \left[L_{a-1}, L_b\right] \subseteq L_{a+b-1}$ (by the induction hypothesis) and thus $\left[\left[x, z\right], y\right] \in \left[L_{a+b-1}, \kk X\right] = L_{a+b}$ (by the definition of $L_{a+b}$), and we also have $\left[z, y\right] \in \left[L_b, \kk X\right] = L_{b+1}$ (by the definition of $L_{b+1}$) and therefore $\left[x, \left[z, y\right]\right] \in \left[L_{a-1}, L_{b+1}\right] \subseteq L_{a+b}$ (by the induction hypothesis)). So Proposition 3 follows by induction.

Proposition 4 is proven by strong induction on $n$, using Proposition 3.

Combining Proposition 2 with Proposition 4, we obtain $B_n \subseteq L_n = \kk C_n$ for each $n \geq 1$. Thus, $B \subseteq \kk C$, so that Theorem 1 is proven.

3. Formal proofs

Let me prove these Propositions 2, 3 and 4 in detail, just to make sure everything is exactly as I claimed (you have most likely proven them yourself by now). This has turned out to be even duller than expected.

We will tacitly use the observation that if $U$ and $V$ are two subsets of $L$, and if $u\in U$ and $v\in V$, then $\left[ u,v\right] \in\left[ U,V\right] $. (Indeed, if $U$ and $V$ are two subsets of $L$, then the definition of $\left[ U,V\right] $ shows that $\left[ U,V\right] =\left( \text{the span of }\left[ U,V\right] _0 \right) =\kk \left( \left[ U,V\right] _0 \right) $. Thus, if $U$ and $V$ are two subsets of $L$, and if $u\in U$ and $v\in V$, then we have $\left[ u,v\right] \in\left[ U,V\right] _0 \subseteq\kk \left( \left[ U,V\right] _0 \right) =\left[ U,V\right] $.)

We shall use two simple lemmas:

Lemma 5. Let $S$ be a subset of $L$. Let $M$ be a $\kk$-submodule of $L$. If $S\subseteq M$, then $\kk S\subseteq M$.

Proof of Lemma 5. Recall that $\kk S$ is the span of $S$, and thus is the smallest $\kk$-submodule of $L$ that contains $S$ as a subset. Hence, any $\kk$-submodule of $L$ that contains $S$ as a subset must contain $\kk S$ as a subset. Applying this to the $\kk$-submodule $M$, we conclude that $M$ contains $\kk S$ as a subset if $M$ contains $S$ as a subset. In other words, if $S\subseteq M$, then $\kk S\subseteq M$. This proves Lemma 5. $\blacksquare$

Lemma 6. Let $U$ and $V$ be two subsets of $L$. Then, $\left[ \kk U,\kk V\right] =\kk \left( \left[ U,V\right] _0 \right) $.

Proof of Lemma 6. Recall that $\left[ \kk U,\kk V\right] $ is the span of the subset $\left[ \kk U,\kk V\right] _0 $ (by the definition of $\left[ \kk U,\kk V\right] $). In other words, $\left[ \kk U,\kk V\right] =\kk \left( \left[ \kk U,\kk V\right] _0 \right) $. Hence, $\left[ \kk U,\kk V\right] $ is a $\kk$-submodule of $L$ and satisfies $\left[ \kk U,\kk V\right] _0 \subseteq\left[ \kk U,\kk V\right] $.

Let $r\in\left[ \kk U,\kk V\right] _0 $ be arbitrary. We shall show that $r\in\kk \left( \left[ U,V\right] _0 \right) $.

Indeed, we have $r\in\left[ \kk U,\kk V\right] _0 $; in other words, $r=\left[ u,v\right] $ for some $u\in\kk U$ and some $v\in\kk V$ (by the definition of $\left[ \kk U,\kk V\right] _0 $). Consider these $u$ and $v$.

We have $u\in\kk U$. In other words, $u$ is a $\kk$-linear combination of the elements of $U$ (by the definition of the span $\kk U$). In other words, $u=\sum_{i\in I}\lambda_{i}u_{i}$ for some finite set $I$ and some families $\left( \lambda_{i}\right) _{i\in I} \in\kk ^{I}$ and $\left( u_{i}\right) _{i\in I}\in U^{I}$. Consider this $I$ and these families.

We have $v\in\kk U$. In other words, $v$ is a $\kk$-linear combination of the elements of $V$ (by the definition of the span $\kk V$). In other words, $v=\sum_{j\in J}\mu_{j}v_{j}$ for some finite set $J$ and some families $\left( \mu_{j}\right) _{j\in J}\in\kk ^{J}$ and $\left( v_{j}\right) _{j\in J}\in V^{J}$. Consider this $J$ and these families.

Now, \begin{align*} r & =\left[ u,v\right] =\left[ \sum_{i\in I}\lambda_{i}u_{i},\sum_{j\in J}\mu_{j}v_{j}\right] \qquad\left( \text{since }u=\sum_{i\in I}\lambda _{i}u_{i}\text{ and }v=\sum_{j\in J}\mu_{j}v_{j}\right) \\ & =\sum_{i\in I}\lambda_{i}\sum_{j\in J}\mu_{j}\left[ u_{i},v_{j}\right] \qquad\left( \text{since the Lie bracket is }\kk \text{-bilinear} \right) \\ & =\sum_{\left( i,j\right) \in I\times J}\lambda_{i}\mu_{j} \underbrace{\left[ u_{i},v_{j}\right] }_{\substack{\in\left[ U,V\right] _0 \\\text{(since }u_{i}\in U\text{ and }v_{j}\in V\text{)}}}\in\sum_{\left( i,j\right) \in I\times J}\lambda_{i}\mu_{j}\left[ U,V\right] _0 . \end{align*} Thus, $r$ is a $\kk$-linear combination of elements of $\left[ U,V\right] _0 $. In other words, $r$ belongs to the span of $\left[ U,V\right] _0 $. In other words, $r\in\kk \left( \left[ U,V\right] _0 \right) $ (because $\kk \left( \left[ U,V\right] _0 \right) $ is the span of $\left[ U,V\right] _0 $).

Now, forget that we fixed $r$. We thus have proven that $r\in\kk \left( \left[ U,V\right] _0 \right) $ for each $r\in\left[ \kk U,\kk V\right] _0 $. In other words, $\left[ \kk U,\kk V\right] _0 \subseteq\kk \left( \left[ U,V\right] _0 \right) $. Thus, Lemma 5 (applied to $M=\kk \left( \left[ U,V\right] _0 \right) $ and $S=\left[ \kk U,\kk V\right] _0 $) yields that $\kk \left( \left[ \kk U,\kk V\right] _0 \right) \subseteq\kk \left( \left[ U,V\right] _0 \right) $. Now, recall that $\left[ \kk U,\kk V\right] =\kk \left( \left[ \kk U,\kk V\right] _0 \right) \subseteq\kk \left( \left[ U,V\right] _0 \right) $.

On the other hand, let $q\in\left[ U,V\right] _0 $. We shall show that $q\in\left[ \kk U,\kk V\right] $.

In fact, we have $q\in\left[ U,V\right] _0 $; in other words, $q=\left[ \widetilde{u},\widetilde{v}\right] $ for some $\widetilde{u}\in U$ and some $\widetilde{v}\in V$ (by the definition of $\left[ U,V\right] _0 $). Consider these $\widetilde{u}$ and $\widetilde{v}$. From $\widetilde{u}\in U\subseteq\kk U$ and $\widetilde{v}\in V\subseteq\kk V$, we obtain $\left[ \widetilde{u},\widetilde{v}\right] \in\left[ \kk U,\kk V\right] $. Thus, $q=\left[ \widetilde{u},\widetilde{v}\right] \in\left[ \kk U,\kk V\right] $.

Forget that we fixed $q$. We thus have proven that $q\in\left[ \kk U,\kk V\right] $ for each $q\in\left[ U,V\right] _0 $. In other words, $\left[ U,V\right] _0 \subseteq\left[ \kk U,\kk V\right] $. Thus, Lemma 5 (applied to $M=\left[ \kk U,\kk V\right] $ and $S=\left[ U,V\right] _0 $) yields that $\kk \left( \left[ U,V\right] _0 \right) \subseteq\left[ \kk U,\kk V\right] $. Combining this with $\left[ \kk U,\kk V\right] \subseteq\kk \left( \left[ U,V\right] _0 \right) $, we obtain $\left[ \kk U,\kk V\right] =\kk \left( \left[ U,V\right] _0 \right) $. This proves Lemma 6. $\blacksquare$

Proof of Proposition 2. We shall prove Proposition 2 by induction on $n$:

Induction base: The definition of $C_{1}$ yields $C_{1}=X$. The definition of $L_{1}$ yields $L_{1}=\kk \underbrace{X}_{=C_{1}}=\kk C_{1}$. In other words, Proposition 2 holds for $n=1$. This completes the induction base.

Induction step: Fix a positive integer $m>1$. Assume that Proposition 2 holds for $n=m-1$. We must prove that Proposition 2 holds for $n=m$.

We have assumed that Proposition 2 holds for $n=m-1$. In other words, $L_{m-1}=\kk C_{m-1}$. The recursive definition of the $C_{n}$ yields $C_{m}=\left[ C_{m-1},X\right] _0 $. Hence, $\kk C_{m}=\kk \left( \left[ C_{m-1},X\right] _0 \right) $. But Lemma 6 (applied to $U=C_{m-1}$ and $V=X$) yields $\left[ \kk C_{m-1},\kk X\right] =\kk \left( \left[ C_{m-1},X\right] _0 \right) $. Comparing these two equalities, we obtain $\kk C_{m}=\left[ \kk C_{m-1} ,\kk X\right] $.

But the recursive definition of the $L_{n}$ yields $L_{m}=\left[ \underbrace{L_{m-1}}_{=\kk C_{m-1}},\kk X\right] =\left[ \kk C_{m-1},\kk X\right] $. Comparing these two equalities, we obtain $L_{m}=\kk C_{m}$. In other words, Proposition 2 holds for $n=m$. This completes the induction step. Thus, Proposition 2 is proven. $\blacksquare$

Proof of Proposition 3. We shall prove Proposition 3 by induction on $a$:

Induction base: Let $b\geq1$ be an integer. We shall prove that $\left[ L_{1},L_{b}\right] \subseteq L_{1+b}$.

Indeed, the recursive definition of the $L_{n}$ yields $L_{b+1}=\left[ L_{b},\kk X\right] $ and $L_{1}=\kk X$.

Let $r\in\left[ L_{1},L_{b}\right] _0 $ be arbitrary. We shall show that $r\in L_{b+1}$.

Indeed, we have $r\in\left[ L_{1},L_{b}\right] _0 $; in other words, $r=\left[ u,v\right] $ for some $u\in L_{1}$ and some $v\in L_{b}$ (by the definition of $\left[ L_{1},L_{b}\right] _0 $). Consider these $u$ and $v$.

We have $\left[ \underbrace{v}_{\in L_{b}},\underbrace{u}_{\in L_{1}}\right] \in\left[ L_{b},\underbrace{L_{1}}_{=\kk X}\right] =\left[ L_{b},\kk X\right] =L_{b+1}$ (since $L_{b+1}=\left[ L_{b} ,\kk X\right] $). But $r=\left[ u,v\right] =-\left[ v,u\right] $ (since the Lie bracket is antisymmetric). Hence, $r=-\underbrace{\left[ v,u\right] }_{\in L_{b+1}}\in-L_{b+1}\subseteq L_{b+1}$ (since $L_{b+1}$ is a $\kk$-submodule of $L$).

Now, forget that we fixed $r$. We thus have proven that $r\in L_{b+1}$ for each $r\in\left[ L_{1},L_{b}\right] _0 $. In other words, $\left[ L_{1},L_{b}\right] _0 \subseteq L_{b+1}$. Hence, Lemma 5 (applied to $M=L_{b+1}$ and $S=\left[ L_{1},L_{b}\right] _0 $) yields that $\kk \left( \left[ L_{1},L_{b}\right] _0 \right) \subseteq L_{b+1} $. But $\left[ L_{1},L_{b}\right] $ is the span of $\left[ L_{1} ,L_{b}\right] _0 $ (by the definition of $\left[ L_{1},L_{b}\right] $); in other words, $\left[ L_{1},L_{b}\right] =\kk \left( \left[ L_{1},L_{b}\right] _0 \right) $. Hence, $\left[ L_{1},L_{b}\right] =\kk \left( \left[ L_{1},L_{b}\right] _0 \right) \subseteq L_{b+1}=L_{1+b}$.

Now, forget that we fixed $b$. We thus have proven that $\left[ L_{1} ,L_{b}\right] \subseteq L_{1+b}$ for all $b\geq1$. In other words, Proposition 3 holds for $a=1$. This completes the induction base.

Induction step: Fix an integer $c>1$. Assume that Proposition 3 holds for $a=c-1$. We must prove that Proposition 3 holds for $a=c$.

Let $b\geq1$ be an integer. We shall show that $\left[ L_{c},L_{b}\right] \subseteq L_{c+b}$.

Proposition 2 (applied to $n=b$) yields $L_{b}=\kk C_{b}$. Proposition 2 (applied to $n=c$) yields $L_{c}=\kk C_{c}$. Now, \begin{align*} \left[ \underbrace{L_{c}}_{=\kk C_{c}},\underbrace{L_{b}} _{=\kk C_{b}}\right] =\left[ \kk C_{c},\kk C_{b}\right] =\kk \left( \left[ C_{c},C_{b}\right] _0 \right) \end{align*} (by Lemma 6, applied to $U=C_{c}$ and $V=C_{b}$).

Now, let $r\in\left[ C_{c},C_{b}\right] _0 $ be arbitrary. We shall show that $r\in C_{c+b}$.

We have $r\in\left[ C_{c},C_{b}\right] _0 $. In other words, $r=\left[ u,z\right] $ for some $u\in C_{c}$ and some $z\in C_{b}$ (by the definition of $\left[ C_{c},C_{b}\right] _0 $). Consider these $u$ and $z$.

We have $u\in C_{c}=\left[ C_{c-1},X\right] _0 $ (by the recursive definition of the $C_{n}$, since $c>1$). In other words, $u=\left[ x,y\right] $ for some $x\in C_{c-1}$ and some $y\in X$ (by the definition of $\left[ C_{c-1},X\right] _0 $). Consider these $x$ and $y$.

The recursive definition of the $L_{n}$ yields $L_{c+b}=\left[ L_{c+b-1} ,\kk X\right] $ and $L_{b+1}=\left[ L_{b},\kk X\right] $.

We can apply Proposition 3 to $c-1$ instead of $a$ (since we have assumed that Proposition 3 holds for $a=c-1$). We thus obtain $\left[ L_{c-1} ,L_{b}\right] \subseteq L_{\left( c-1\right) +b}=L_{c+b-1}$.

We can apply Proposition 3 to $c-1$ and $b+1$ instead of $a$ and $b$ (since we have assumed that Proposition 3 holds for $a=c-1$). We thus obtain $\left[ L_{c-1},L_{b+1}\right] \subseteq L_{\left( c-1\right) +\left( b+1\right) }=L_{c+b}$.

Proposition 2 (applied to $n=c-1$) yields $L_{c-1}=\kk C_{c-1}$. From $x\in C_{c-1}\subseteq\kk C_{c-1}=L_{c-1}$ and $z\in C_{b} \subseteq\kk C_{b}=L_{b}$, we obtain $\left[ \underbrace{x}_{\in L_{c-1}},\underbrace{z}_{\in L_{b}}\right] \in\left[ L_{c-1},L_{b}\right] \subseteq L_{c+b-1}$. Combining this with $y\in X\subseteq\kk X$, we obtain $\left[ \underbrace{\left[ x,z\right] }_{\in L_{c+b-1} },\underbrace{y}_{\in\kk X}\right] \in\left[ L_{c+b-1},\kk X\right] =L_{c+b}$.

From $z\in L_{b}$ and $y\in X\subseteq\kk X$, we obtain $\left[ \underbrace{z}_{\in L_{b}},\underbrace{y}_{\in\kk X}\right] \in\left[ L_{b},\kk X\right] =L_{b+1}$. Hence, $\left[ \underbrace{x}_{\in L_{c-1}},\underbrace{\left[ z,y\right] }_{\in L_{b+1}}\right] \in\left[ L_{c-1},L_{b+1}\right] \subseteq L_{c+b}$.

But recall that \begin{align*} r & =\left[ \underbrace{u}_{=\left[ x,y\right] },z\right] =\left[ \left[ x,y\right] ,z\right] =\underbrace{\left[ \left[ x,z\right] ,y\right] }_{\in L_{c+b}}-\underbrace{\left[ x,\left[ z,y\right] \right] }_{\in L_{c+b}}\\ & \qquad\qquad\left( \begin{array} [c]{c} \text{since the Jacobi identity yields}\\ \left[ \left[ x,z\right] ,y\right] =\left[ \left[ x,y\right] ,z\right] +\left[ x,\left[ z,y\right] \right] \end{array} \right) \\ & \in L_{c+b}-L_{c+b}\subseteq L_{c+b}\qquad\left( \text{since } L_{c+b}\text{ is a }\kk \text{-submodule of }L\right) . \end{align*}

Now, forget that we fixed $r$. We thus have shown that $r\in L_{c+b}$ for each $r\in\left[ C_{c},C_{b}\right] _0 $. In other words, $\left[ C_{c} ,C_{b}\right] _0 \subseteq L_{c+b}$.

Hence, Lemma 5 (applied to $M=L_{c+b}$ and $S=\left[ C_{c},C_{b}\right] _0 $) yields that $\kk \left( \left[ C_{c},C_{b}\right] _0 \right) \subseteq L_{c+b}$. Now, recall that $\left[ L_{c},L_{b}\right] =\kk \left( \left[ C_{c},C_{b}\right] _0 \right) \subseteq L_{c+b}$.

Now, forget that we fixed $b$. We thus have proven that $\left[ L_{c} ,L_{b}\right] \subseteq L_{c+b}$ for all $b\geq1$. In other words, Proposition 3 holds for $a=c$. This completes the induction step. Hence, Proposition 3 is proven by induction. $\blacksquare$

Proof of Proposition 4. We shall prove Proposition 4 by strong induction on $n$. Thus, we fix an integer $m\geq1$. We assume that Proposition 4 holds for all $n<m$. We want to prove that Proposition 4 holds for $n=m$ as well.

Let $r\in B_{m}$. We shall prove that $r\in L_{m}$.

If $m=1$, then this holds for easy reasons (in fact, if $m=1$, then $B_{m}=B_{1}=X\subseteq\kk X=L_{1}=L_{m}$ (since $1=m$), and therefore $r\in B_{m}\subseteq L_{m}$). Thus, for the rest of this proof, we WLOG assume that we don't have $m=1$. Hence, $m>1$.

Thus, the recursive definition of $B_{m}$ yields $B_{m}=\bigcup\limits_{a+b=m} \left[ B_{a},B_{b}\right] _0 $. Hence, $r\in B_{m}=\bigcup\limits_{a+b=m} \left[ B_{a},B_{b}\right] _0 $. In other words, $r\in\left[ B_{a} ,B_{b}\right] _0 $ for some pair $\left( a,b\right) $ of positive integers satisfying $a+b=m$. Consider this pair $\left( a,b\right) $. We have $a>0$ (since $a$ is a positive integer) and thus $a+b>b$, so that $b<a+b=m$. Similarly, $a<m$.

We have assumed that Proposition 4 holds for all $n<m$. Hence, we can apply Proposition 4 to $n=a$ (since $a<m$). We thus obtain $B_{a}\subseteq L_{a}$. The same argument (applied to $b$ instead of $a$) yields $B_{b}\subseteq L_{b}$.

But $r\in\left[ B_{a},B_{b}\right] _0 $. In other words, $r=\left[ u,v\right] $ for some $u\in B_{a}$ and some $v\in B_{b}$ (by the definition of $\left[ B_{a},B_{b}\right] _0 $). Consider these $u$ and $v$. We have $u\in B_{a}\subseteq L_{a}$ and $v\in B_{b}\subseteq L_{b}$. Now, $r=\left[ \underbrace{u}_{\in L_{a}},\underbrace{v}_{\in L_{b}}\right] \in\left[ L_{a},L_{b}\right] \subseteq L_{a+b}$ (by Proposition 3). In view of $a+b=m$, this rewrites as $r\in L_{m}$.

Now, forget that we fixed $r$. We thus have proven that $r\in L_{m}$ for each $r\in B_{m}$. In other words, $B_{m}\subseteq L_{m}$. In other words, Proposition 4 holds for $n=m$. This completes the induction proof. Thus, Proposition 4 is proven by strong induction. $\blacksquare$

Proof of Theorem 1. Let $r\in B$. Thus, $r\in B=B_{1}\cup B_{2}\cup B_{3}\cup\cdots$. In other words, $r\in B_{n}$ for some integer $n\geq1$. Fix this $n$.

Now, $r\in B_{n}\subseteq L_{n}$ (by Proposition 4), so that $r\in L_{n}=\kk C_{n}$ (by Proposition 2). But $C_{n}\subseteq C_{1}\cup C_{2}\cup C_{3}\cup\cdots=C$ (since $C=C_{1}\cup C_{2}\cup C_{3}\cup\cdots$), so that $r\in\kk \underbrace{C_{n}}_{\subseteq C}\subseteq\kk C$.

Now, forget that we fixed $r$. We thus have proven that $r\in\kk C$ for each $r\in B$. In other words, $B\subseteq\kk C$. This proves Theorem 1. $\blacksquare$

Answer 2

Write $[x_1, \dots, x_n] = [\cdots[x_1, x_2], \dots x_n]$. By the Jacobi formula, $$[x, [y, z]] = [x, y, z] - [x, z, y].$$ Induction on the length of $w$ then shows the statement holds for any bracket $[x, w]$ with $x$ a simpler monomial, as $[x_1, \dots, x_n] = [[x_1, \dots, x_k], x_{k+1}, \dots, x_n]$. But $$[[x, y], z] = -[z, [x, y]] + [y, [x, z]],$$ so another induction on the length of $x$ proves the result for an arbitrary bracket $[x, w]$.