Lie algebra generated by elements of the symmetric group $S_N$ For an arbitrary group $G$, the group algebra $\mathbb{C}(G)$ is defined as the set of all formal linear combinations of the elements of $G$: $\mathbb{C}(G)=\{c_1g_1+c_2g_2+\ldots+c_n g_n|c_i\in \mathbb{C},g_i\in G\}$. Multiplication in $\mathbb{C}(G)$ is induced by the group multiplication of $G$ and extends to $\mathbb{C}(G)$ by linearity. In a similar fashion, we can define a "group Lie algebra" $\mathfrak{L}(G)=\{c_1g_1+c_2g_2+\ldots+c_n g_n|c_i\in \mathbb{C},g_i\in G\}$, which is the same as $\mathbb{C}(G)$ as a linear vector space, but in $\mathfrak{L}(G)$ only the Lie bracket $[,]:\mathfrak{L}(G)\times \mathfrak{L}(G)\to \mathfrak{L}(G)$ is defined, to be the commutator in $\mathbb{C}(G)$, i.e. $[l_1,l_2]=l_1\cdot l_2-l_2\cdot l_1$, where "$\cdot$" is the multiplication in $\mathbb{C}(G)$. My problem is to determine the structure of the Lie algebra $\mathfrak{L}(G)$ for the symmetric group $G=S_N$.
Let us take $S_3$ as an example. Since $|S_3|=6$, the group algebra $\mathbb{C}(S_3)$ of $S_3$ is 6-dimensional, spanned by $\{1,P_{12},P_{23},P_{13},P_{12}P_{23},P_{23}P_{12}\}$. It is easy to see that $\{1,I,I^2\}$ are central elements (i.e. commute with everything in $\mathbb{C}(S_3)$), where $I=P_{12}+P_{23}+P_{13}$. The 3-dimensional subspace orthogonal to these central elements is spanned by $\{P_{12}-P_{23},P_{12}-P_{13},[P_{12},P_{23}]\}$, which forms an $\mathfrak{su}(2)$ Lie algebra, their relation to the spin generators are
$$P_{12}-P_{23}=\sqrt{6}(s_x-s_y),~~P_{12}-P_{13}=\sqrt{6}(s_x-s_z),~~[P_{12},P_{23}]=2i(s_x+s_y+s_z),$$
where $[s_i,s_j]=i\epsilon_{ijk}s_k$. Therefore we conclude that $\mathfrak{L}(S_3)=\mathfrak{u}(1)^3\oplus \mathfrak{su}(2)$.
But when trying to determine the structure of $\mathfrak{L}(S_N)$ for arbitrary $N$,  this kind of brute force calculation seems hopeless. I don't  even know the structure of $\mathfrak{L}(S_4)$. For large $N$, I want to know this: if $\mathfrak{L}(S_N)$ is a direct sum of simple Lie algebras, let $d_N$ be the dimension of its largest irreducible(simple) component (e.g. $d_3=3$). Does $d_N$ grow algebraically fast or exponentially fast with $N$?
Any hints, suggestions, or relevant references will be welcomed. Thanks.
 A: $\newcommand{\fru}{\mathfrak{u}}
\newcommand{\frsu}{\mathfrak{su}}
\newcommand{\frgl}{\mathfrak{gl}}
\newcommand{\frsl}{\mathfrak{sl}}
\newcommand{\frL}{\mathfrak{L}}
\newcommand{\End}{\operatorname{End}}
\newcommand{\KG}{K\left[G\right]}
\newcommand{\CC}{\mathbb{C}}
\newcommand{\LG}{\mathfrak{L}\left(G\right)}$
Yes. For each $N \geq 0$, we have $\frL\left(S_N\right) \cong \bigoplus\limits_{\lambda \vdash N} \left( \frsu\left(\dim V_\lambda\right) \oplus \fru\left(1\right) \right)$, where the direct sum is over all partitions $\lambda$ of $N$, and where $V_\lambda$ denotes the Specht module corresponding to the partition $\lambda$.
But this isn't really specific to the symmetric groups. More generally, we can answer this question for any finite group $G$ if we know its representation theory.
So let $G$ be a finite group, and let $K$ be a field of characteristic $0$ over which $G$ splits. (It is sufficient, but not necessary, to take $K = \CC$.)
For each $K$-algebra $A$, we let $A^-$ be the Lie algebra on the $K$-vector space $A$ whose Lie bracket is the commutator of $A$. Thus, what you call $\frL\left(G\right)$ is just $\left(\CC\left[G\right]\right)^-$.
Since the field $K$ has characteristic $0$, we have
\begin{equation}
\frgl\left(m\right) \cong K^- \oplus \frsl\left(m\right)
\label{darij1.eq.gl-sl}
\tag{1}
\end{equation}
(an isomorphism of $K$-Lie algebras) for each positive integer $m$.
Let $V_i$ (with $i$ running over some finite indexing set $I$) be all irreducible representations of $G$ over $K$ (up to isomorphism, without repetitions). Theorem 4.1.1 (ii) in Pavel Etingof, Oleg Golberg, Sebastian Hensel, Tiankai Liu, Alex Schwendner, Dmitry Vaintrob, and Elena Yudovina, Introduction to representation theory, Student Mathematical Library #59, AMS 2011 shows that $\KG \cong \bigoplus\limits_{i \in I} \End \left(V_i\right)$ as $K$-algebras. Thus,
\begin{align}
\left(\KG\right)^-
&\cong \left(\bigoplus\limits_{i \in I} \End \left(V_i\right)\right)^-
\cong \bigoplus\limits_{i \in I} \underbrace{\left(\End \left(V_i\right)\right)^-}_{= \frgl\left(V_i\right) \cong \frgl\left(\dim V_i\right) } \\
&\cong \bigoplus\limits_{i \in I} \underbrace{\frgl\left(\dim V_i\right) }_{\substack{\cong K^- \oplus \frsl\left(\dim V_i \right) \\ \text{(by \eqref{darij1.eq.gl-sl})}}}
\cong \bigoplus\limits_{i \in I} \left( K^- \oplus \frsl\left(\dim V_i \right) \right) .
\label{darij1.eq.gen-form}
\tag{2}
\end{align}
When $G$ is the symmetric group $S_N$, we can take $I$ to be the set of all partitions $\lambda$ of $N$ (this is a well-known fact from the representation theory of symmetric group; see, e.g., §5.12 in op. cit.), and the corresponding irreducible representations $V_\lambda$ are the so-called Specht modules. For each partition $\lambda$ of $N$, the dimension $\dim V_\lambda$ has a nice expression called the hook-length formula (§5.17 in op. cit.), and there is a basis of $V_\lambda$ indexed by the standard tableaux of shape $\lambda$ (see, e.g., Mark Wildon, Representation theory of the symmetric group, 2014).
In light of this, \eqref{darij1.eq.gen-form} (applied to $G = S_N$) becomes
\begin{align}
\left(K\left[S_N\right] \right)^-
\cong \bigoplus\limits_{\lambda \vdash N} \left( K^- \oplus \frsl\left(\dim V_\lambda \right) \right) .
\end{align}
When $K = \CC$, this further rewrites as
\begin{align}
\left(\CC \left[S_N\right]\right)^-
\cong \bigoplus\limits_{\lambda \vdash N} \left( \underbrace{\CC^-}_{\cong \fru\left(1\right)} \oplus \underbrace{\frsl\left(\dim V_\lambda \right)}_{\cong \frsu\left(\dim V_\lambda\right)} \right)
\cong \bigoplus\limits_{\lambda \vdash N} \left( \fru\left(1\right) \oplus \frsu\left(\dim V_\lambda\right) \right) .
\end{align}
Thus,
\begin{align}
\frL\left(S_N\right)
= \left(\CC \left[S_N\right]\right)^-
\cong \bigoplus\limits_{\lambda \vdash N} \left( \fru\left(1\right) \oplus \frsu\left(\dim V_\lambda\right) \right) 
\cong \bigoplus\limits_{\lambda \vdash N} \left( \frsu\left(\dim V_\lambda\right) \oplus \fru\left(1\right) \right) .
\end{align}
This is exactly my claim.
