Why do we introduce a realisation of a generalized Cartan matrix?

Question

When introducing the generalized Cartan matrix, one also introduces the the corresponding realisation. Why is this necessary?

I've read that if we didn't introduce the realisation, then we might not be sure to have linearly independence between the simpe roots in the finite dimensional case. However I'm not sure why? Here are my thoughts:

I'm thinking the trouble may occur when $\det(A)>0$ is removed. Then we may have $\det(A)=0$ so we have a non-zero vector $x\in \mathbb{C}^n$ st. $Ax=0$ or equivalently $\sum_{i=1}^nx_iA_{ij}=0$ for all $1\leq j\leq n$. This may somehow give a linear dependence between the roots. I know that $A_{ij}=2\frac{\langle \alpha_i,\alpha_j\rangle}{\langle\alpha_i,\alpha_i\rangle}$ for simple roots $\alpha_k$ ($1\leq k\leq n$) but I'm having a hard time convincing myself that $\sum_{i=1}^nx_iA_{ij}=0$ gives $\sum_{i=1}^nx_i\alpha_i=0$?

Answer 1

I've just this week had a similar confusion and found your post. The conclusion I have come to is pretty much what the person in the comment describes, but to expand:

Just as finite simple Lie algebras can be characterised by a Cartan matrix, Kac-Moody algebras can also be completely characterised by a generalised Cartan matrix. However the procedure for recovering the algebra from the matrix is not the same. Or rather, the gCM recipe specialises to the CM recipe but is less intuitive.

This is most striking in the dimensionality of $H$. In the finite case, $H = \mathbb{C}^r = \mathbb{C}^n =$ span$\Pi^V$ (and the Killing form is nondegenerate on $H$).

In the KM case, if we naively took $H = \mathbb{C}^n$, we would find that we cannot find a set $\Pi^V$ such that $\langle \alpha_i, \alpha_j^V\rangle = A_{ji}$ and $\Pi^V$ is linearly independent in $H$, as the previous answer states. So that can't be the way to extract the algebra. The recipe is instead: take $H = \mathbb{C}^{2n -r}$ and define extra basis vectors (which can be canonically obtained), so that $\Pi^V \cup \{$extra vectors$\}$ is linearly independent in this enlarged space.

From now on let's talk about Aff$^{(1)}$ only.

Note that we are not expanding the roots. We still have $\Delta \subset $ span$_{\mathbb{Z}}\{\alpha_i\}$, $i = 0,...,k$. This includes the imaginary root $\delta = \sum_i a^i \alpha_i$, equivalently the zeroth root $\alpha_0 \notin \overset{o}{\Delta}$, where $\overset{o}{\Delta}$ is the root set of the finite subalgebra. It's just that the coroots are unable to distinguish between $e_{\alpha} \in \mathfrak{g}_{\alpha}$ and $e_{\alpha + n \delta} \in \mathfrak{g}_{\alpha + n \delta}$. If we want to lift this degeneracy we must add in an extra generator $d$ to $H$ which can distinguish. That doesn't mean we add another associated root. <As far as I understand, the corresponding element $\Lambda_0 \in H^*$ is not a root.>

In the end we are not really talking about the Cartan matrix, we are talking about the algebra, which is characterised by the CM in a similar but different way to how finite algebras are. Explicitly, this algebra is

$$ \begin{aligned} \mathfrak{g} &= \left(\bigoplus_{\beta \in \Delta} \mathfrak{g}_{\beta}\right) \oplus H\\ \mathfrak{g}_{\beta} &= \left\{ x_{\beta} \in \mathfrak{g} \; | \; \left[h, x_{\beta}\right] = \langle \beta, h\rangle x_{\beta} = \beta(h) x_{\beta}\; \forall \; h \in H \right\}\\ H &= \text{span}\Pi^V \cup \{d\} \end{aligned}$$ with $$ \begin{aligned} \Pi &= \{\alpha_i, \; i = 0,...,k\}\\ \Pi^V &= \{\alpha_i^V, \; i = 0,...,k\}\\ \Delta &\subset \text{span}\{\Pi\}\\ \Delta &= \{ \alpha + n \delta, \; \; \alpha \in \overset{o}{\Delta}, n \in \mathbb{Z}\} \cup \{n \delta, n \in \mathbb{Z}\backslash 0\}\\ \end{aligned} $$

$\langle \alpha_i, \alpha_j^V\rangle$ is specified by the Cartan matrix and $d$ is defined to be that element which, under the action of the maps $\alpha_i \in H^*$, is mapped to $\langle \alpha_j, d \rangle = \delta_{j, 0}$, regardless of which member of Aff$^{(1)}$ we are looking at.

I made the mistake of confusing the matrix with the algebra, and it then bothered me that we had introduced extra generators that don’t correspond to roots. In fact the algebra has the generator by its definition – it just happens that a large class of algebras have the same extra generator, and differ only by the dual contractions between roots and coroots.

I hope that this way of thinking about it helps you or someone (and that it's correct!) Please let me know if I'm getting things wrong or missing stuff out.

Answer 2

take for example $A = \begin{pmatrix} 2 & -2 \\ -2 & 2\end{pmatrix}$. We cannot find any realization with $\{\alpha_1^\vee, \alpha_2^\vee\}\subset \mathfrak{h}$ and $\{\alpha_1, \alpha_2\}\subset \mathfrak{h}^*$ if $ \mathfrak{h}$ is a vector space of $\dim \mathfrak{h}=2$. In fact, you would get $\langle\alpha_1^\vee,\alpha_1\rangle = -\langle\alpha_1^\vee,\alpha_2\rangle$ and $\langle\alpha_2^\vee,\alpha_1\rangle = -\langle\alpha_2^\vee,\alpha_2\rangle$. So, if $\alpha_1^\vee, \alpha_2^\vee$ are a basis, this forces $\alpha_2$ to be $-\alpha_1$. But we would like that both simple roots and simple coroots are linearly independent!

That's why you need to look in dimension $3$ instead.