Representation Theory of Lie Algebras, Lecture 11

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1 Generalized Cartan Matrices

Recall that a simple finite–dimensional Lie algebra can be constructed from the Cartan matrix of a Dynkin diagram. Here, we define a generalization.

Let $A$ be a $n \times n$ matrix. A generalized Cartan matrix is a matrix $A$ satisfying:

The diagonal entries of $A$ are all equal to 2; in other words, $a_{ii}=2 \text { for } i=1,2,\ldots ,n$
The off–diagonal entries of $A$ are non–positive.
If $a_{ij}=0$ then $a_{ji}=0$.

Note that the usual Cartan matrices all have full rank $n.$ This is no longer the case for generalized Cartan matrices, so the corresponding root system has a different definition. Namely, we still have a set $\Pi $ of $n$ linearly independent simple co–roots $ \check {\alpha }_1,\ldots ,\check {\alpha }_n$ in a vector space $\mathfrak {h}$, but now the dimension of $\mathfrak {h}$ is $2n - \mathrm {rank}(A)$.

The realization of the Lie algebra is also different. Define $\tilde {\mathfrak {g}}(A)$ to be the Lie algebra generated by $e_i,f_i,1\leq i \leq n$ and $\mathfrak {h}$ with relations \begin{align*} [e_i,f_j] &= \delta _{ij}\check {\alpha }_i \\ [h,h']&=0 \text { for all } h,h' \in \mathfrak {h}\\ [h,e_i] &= \alpha _i(h)e_i \\ [h,f_i] &= -\alpha _i(h)f_i \ \end{align*}

We will need a root space decomposition (like before). Let $\tilde {\mathfrak {n}}_+$ denote the subalgebra generated by $e_1,\ldots ,e_n$. Similarly for $ \tilde {\mathfrak {n}}_-$. Let $Q$ be the root lattice \[ Q = \sum _{i=1}^n \mathbf {Z}\alpha _i \] and \[ Q_+ = \sum _{i=1}^n \mathbf {Z}_+\alpha _i \] Let $\tilde {\omega }$ be the unique involution on $\tilde {\mathfrak {g}}$ that maps \[ e_i \mapsto -f_i, \quad f_i \mapsto -e_i, \quad h\mapsto -h. \]

Theorem 1.2 in Kac’s book.

(a): We have the direct sum of vector spaces \[ \tilde {\mathfrak {g}}(A) = \tilde {\mathfrak {n}}_+ \oplus \mathfrak {h} \oplus \tilde {\mathfrak {n}}_- \]
(b): The subalgebra $\tilde {\mathfrak {n}}_+$ is freely generated by $e_1,\ldots ,e_n$. Similarly for $ \tilde {\mathfrak {n}}_- $
(c): There is the root space decomposition: \[ \tilde {\mathfrak {g}} = \left ( \bigoplus _{\alpha \in Q_+, \alpha \neq 0} \tilde {\mathfrak {g}}_{\alpha }\right ) \oplus \mathfrak {h} \oplus \left ( \bigoplus _{\alpha \in Q_+, \alpha \neq 0} \tilde {\mathfrak {g}}_{-\alpha }\right ) \] where \[ \tilde {\mathfrak {g}}_{\alpha } = \{x \in \tilde {\mathfrak {g}} : [h,x] = \alpha (h)x \text { for all } h \in \mathfrak {h}\}. \] Furthermore, each of these is finite–dimensional and is contained in $\tilde {n}_{\pm }$.

Proof (sketch). Let $V$ be a vector space of dimension $n$ with a basis $v_1,\ldots ,v_n$ and let $\lambda $ be a linear function on $\mathfrak {h}$. We define a representation of $\tilde {\mathfrak {g}}$ on the tensor algebra $T(V)$ by \begin{align*} f_i(a) &= v_i \otimes a \text { for } a \in T(V), \\ h(1) &= \lambda (h)1\\ h(v_j \otimes a) &= -\alpha _j(h)v_j \otimes a + v_j \otimes h(a) \text { for } a \in T^{s-1}(V).\\ e_i(1)&= 0 \\ e_i (v_j \otimes a) &= \delta _{ij}\check {\alpha }_i(a) + v_j \otimes e_i(a) \text { for } e_i \in T^{s-1}(V) \end{align*}

Intuitively, the actions of $f_i,h$ are defined in an “obvious” way (since $h$ acts diagonally) and then $e_i$ is defined to make the relations hold. I will skip the calculations that the relations hold.

Proof of (a). From the relations defining $\tilde {\mathfrak {g}}$, we can show by induction on $s$ that any bracket of $s$ elements of $e_i,f_i$ is in $ \tilde {n}_- + \mathfrak {h} + \tilde {n}_+$. Let $u = n_- + h + n_+ = 0$; we need to show that $n_-=h=n_+=0$. In the representation on $T(V)$, we thus have \[ u(1) = n_-(1) + \lambda (h)1 = 0 \] using that $e_i(1)=0$. By considering the homogeneous grading, we thus have that $\lambda (h)=0$ for all $\lambda \in \mathfrak {h}^*$ and so $h=0$. It just remains to prove that $n_-=0$. We can take $V$ to be the subspace spanned by $f_1,\ldots ,f_n$, so that the tensor algebra $T(V)$ is isomorphic to the universal enveloping algebra $ U(\tilde {n}_-)$, with the map $\tilde {\mathfrak {n}}_- \mapsto \tilde {\mathfrak {n}}_-(1)$ being the canonical embedding in its universal enveloping algebra. Thus $n_-=0$ as needed.

Proof of (b). This follows from the Poincare–Birkhoff–Witt theorem and the automorphism $ \tilde {\omega } $.

Proof of (c). Using the last two defining relations we have \[ \tilde {n}_{\pm } = \bigoplus _{\alpha \in Q_+, \alpha \neq 0} \tilde {\mathfrak {g}}_{\pm \alpha }. \] Furthermore there is the trivial estimate \[ \dim \tilde {\mathfrak {g}}_{\alpha } \leq n^{\vert \mathrm {ht} \alpha }. \] Here, \[ \mathrm {ht}\left (\sum _i k_i\alpha _i \right ) = \sum _i k_i \] Together with (a) this proves (c).

The Lie algebra $\tilde {\mathfrak {g}}$ is not the Lie algebra you’re looking for. We will need to modulo out by an ideal, since here $\mathfrak {h}$ is in general larger than we are expecting it to be. First, here is a general statement about gradings.

Let $\mathfrak {h}$ be an abelian Lie algebra, and let $V$ be a representation with decomposition \[ V = \bigoplus _{\lambda \in \mathfrak {h}^*} V[\lambda ] \] where \[ V[\lambda ] = \{v \in V: hv=\lambda (h)v \text { for all} h \in \mathfrak {h}\}. \] Then for any subrepresentation $U$ there is the decomposition \[ U = \bigoplus _{\lambda \in \mathfrak {h}^*} (V_[\lambda ]\cap U) \]

Let $u \in U$; then $ u = \sum _{k=1}^m u_k$ for some $v_k \in V[\lambda _k].$ We need to show each $v_k\in U$. Pick $h \in \mathfrak {h}$ such that $\lambda _1(h),\ldots ,\lambda _m(h)$ are distinct. Then \[ h^j(v) = \sum _{k=1}^m \lambda _k(h)^j v_k \in U \] for $j=0,1,\ldots ,m-1$. This is a linear system of $m$ equations in $v_1,\ldots ,v_m$ with nondegenerate matrix, so each $v_k$ is in U.

Among the ideals of $\tilde {\mathfrak {g}}$ which intersect $\mathfrak {h}$ trivially, there is a unique maximal ideal $\mathfrak {i}$, and it furthermore satisfies \[ \mathfrak {i} = (\mathfrak {i} \cap \tilde {n}_- ) \oplus (\mathfrak {i} \cap \tilde {n}_+ ) \] where this is a direct sum of vector spaces.

From the proposition, we have that the sum of ideals which intersect $\mathfrak {h}$ trivially, will itself intersect $\mathfrak {h}$ trivially, and so the sum of all ideals with this property is the unique maximal ideal with this property. So we have the result, as a direct sum of vector spaces. Kac’s book claims this is also a direct sum of ideals but I didn’t follow this step.

Finally, we define \[ \mathfrak {g}(A) = \tilde {\mathfrak {g}(A)}/\mathfrak {i}, \] where $\mathfrak {i}$ is the ideal in the corollary. This modding out is necessary because $ \dim \mathfrak {h} $ is no longer $n$ in general. We will call $ \mathfrak {h} \subset \mathfrak {g}(A)$ the Cartan subalgebra and $e_i,f_i$ the Chevalley generators. The Lie algebra itself is called a Kac–Moody Lie algebra.

The next computationally useful lemma uses this modding out.

Let $a \in \mathfrak {n}_+ \subseteq \mathfrak {g}(A)$ such that $[a,f_i]=0$ for all $i$. Then $a=0.$ An identical statement holds for $ e_i$.

Let $\mathfrak {g}_{-1}$ be the vector subspace spanned by $f_1,\ldots ,f_n$. Let $a$ satisfy the assumptions of the lemma. Then \[ \sum _{i,j\geq 0} (\mathrm {ad} \mathfrak {g}_{-1})^i (\mathrm {ad} \mathfrak {h})^j a \] is an ideal in $\mathfrak {g}(A)$ which intersects $\mathfrak {h}$ trivially. Therefore it is the zero ideal, so $a=0$.

So far, I have mentioned that the rank of $A$ need not be $n$ but have not used anything about the actual value. The next proposition does use the rank.

The center of $\mathfrak {g}(A)$ is equal to \[ \mathfrak {c} := \{ h \in \mathfrak {h} : \alpha _i(h)=0 \text { for all } i=1,2,\ldots ,n\}. \] In particular, $ dim \mathfrak {c} = \dim (\mathfrak {h})-n=n-\mathrm {rank}(A).$

One direction is simple; if $h \in \mathfrak {h}$ satisfies $\alpha _i(h)=0$ for all $i$ then it commutes with all $e_i,f_i$. Since $\mathfrak {h}$ is abelian, this means $h$ is central.

In the other direction, we assume that $c$ is central and write it as $ c=\sum _i c_i $, where the sum is over the grading on $\mathfrak {g}$. (Since the root spaces $\mathfrak {g}_{\alpha }$ can be nonzero for $\alpha $ over the entire root lattice). By the lemma, $c_i=0$ if $i \neq 0$. Thus $c\in \mathfrak {h}$. So $[c,e_i] = \alpha _i(c)e_i$, implying that $c$ is in that set.

If $\mathfrak {g}$ is simple, then $\det (A)\neq 0$.

A simple Lie algebra must have a trivial center.

2 Classification of Kac–Moody Algebras

There is a classification of Kac–Moody algebras, which I will not prove this late in the Ross program.

Write $u=(u_1,\ldots ,u_n)>0$ if each $u_i$ is positive (similarly for $\geq 0$).

Every generalized Cartan matrix $A$ satisfies exactly one of:

(a): (Finite) $\det (A)\neq 0$; there exists $u>0$ such that $Au >0$; if $Av\geq 0$ then $v>0$ or $v$ is the zero vector.
(b): (Affine) $A$ has corank 1. There exists $u>0$ such that $Au>0$; $Av \geq 0$ imples $Av$ is the zero vector.
(c): (Indeterminate) there exists $u>0$ such that $Au<0$; $Av\geq 0$ and $v \geq 0$ imply that $v$ is the zero vector.

Indeterminate Kac–Moody algebras are not well understood. Finite Kac–Moody algebras I already classified. Affine Kac–Moody algebras have one of the following Dynkin diagrams (I might use GenAI to Tikz this for me haha)

[Insert Image Later].

2025-07-17 15:41:48