# 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.

• The center of the group algebra has dimension equal to the number of conjugacy classes of the group. This might be helpful to compute one component but I do not know a general method to compute the rest of the components. Nov 20, 2018 at 3:14

## 1 Answer

$$\newcommand{\fru}{\mathfrak{u}} \newcommand{\frsu}{\mathfrak{su}} \newcommand{\frgl}{\mathfrak{gl}} \newcommand{\frsl}{\mathfrak{sl}} \newcommand{\frL}{\mathfrak{L}} \newcommand{\End}{\operatorname{End}} \newcommand{\KG}{K\left[G\right]} \newcommand{\CC}{\mathbb{C}} \newcommand{\LG}{\mathfrak{L}\left(G\right)}$$ 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 $$$$\frgl\left(m\right) \cong K^- \oplus \frsl\left(m\right) \label{darij1.eq.gl-sl} \tag{1}$$$$ (an isomorphism of $$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.