I am reading Infinite dimensional lie algebras by Kac. He starts with a $n \times n$ GCM (Generalized Cartan Matrix) $A$ of rank $l$, then he defines the realization associated with the matrix $A$ which is of dimension $d=2n-l$. I know that in the simple Lie algebra case this dimension is $n$ as $A$ is invertible


I don't understand why we are taking the space of dimension $2n-l$?


Please explain, When the GCM is not of finite type, what we are getting extra by this definition of Lie algebra?


Thanks for your time.



Realisation of Kac-Moody Lie algebras
An equivalent definition of a *realisation* of a GCM $A=(a\_{ij})\_{1\leq i,j\leq n}$ of rank $\ell$ is as follows: it is a triple $(\mathfrak h, \Pi, \Pi^{\vee})$ where $\mathfrak h$ is a complex vector space, $\Pi=\{\alpha\_i \ | \ 1\leq i\leq n\}\subseteq\mathfrak h^\*$ and $\Pi^{\vee}=\{\alpha\_i^{\vee} \ | \ 1\leq i\leq n\}\subseteq\mathfrak h$ are indexed subsets of $\mathfrak h^\*$ and $\mathfrak h$, respectively, such that


(R1) $\Pi$ and $\Pi^{\vee}$ are linearly independent subsets;


(R2) $\langle \alpha\_j,\alpha\_i^{\vee}\rangle= a\_{ij}$ for all $i,j\in\{1,\dots,n\}$;


(R3) $\mathfrak h$ has minimal dimension for these properties.


In other words, the conditions (R1) and (R2) force the dimension of $\mathfrak h$ to be at least $2n-\ell$ (this can be infered from Kac's proof of the existence of a realisation for $A$, see [1, Proposition 1.1] (see also [2, Section 3.5])).


The condition (R3) is thus just there to avoid extra unnecessary dimensions, but does not play any role in the theory: one can in fact define a Kac-Moody algebra $\mathfrak g\_{\mathcal D}$ from a weaker notion of "realisation of $A$" (called a "Kac-Moody root datum" $\mathcal D$), which only keeps the above condition (R2) (see [3, Chapitre 7], or [2, Section 7.3]). If $\mathcal D$ satisfies (R1) and (R2) but $\mathrm{dim} (\mathfrak h)>2n-\ell$, then $\mathfrak g\_{\mathcal D}$ is just a trivial central extension of $\mathfrak g(A)$ that adds the missing dimensions in $\mathfrak h$.


On the other hand, there are very good reasons to keep the condition (R1); here are two of them:


1) One can define a gradation of $\mathfrak g(A)$ (or $\mathfrak g\_{\mathcal D}$) by the free abelian group $Q:=\bigoplus\_{i=1}^n\mathbb Z\alpha\_i$: denoting by $e\_i,f\_i$ the Chevalley generators, one sets $\mathrm{deg}(e\_i)=\alpha\_i$, $\mathrm{deg}(f\_i)=-\alpha\_i$, and $\mathrm{deg}(\mathfrak h)=0$, and one then extends as a Lie algebra gradation. However, if $\Pi$ is not linearly independent, then this abstract gradation need not correspond to an eigenspace decomposition for the adjoint action of $\mathfrak h$, as in Kac's book (see [3, Chapitre 7], or [2, Section 3.5]).


2) For every $\lambda\in\mathfrak h^\*$, the Kac\_Moody algebra $\mathfrak g(A)$ (or $\mathfrak g\_{\mathcal D}$) acts on an irreducible highest-weight module $L(\lambda)$ with highest weight $\lambda$ (see [1, Chapter 9]). Moreover, if $\lambda$ is dominant integral (i.e. $\lambda(\alpha\_i^{\vee})\in\mathbb N$ for all $i$), then $L(\lambda)$ is integrable, that is, it can be integrated to a representation of the Kac-Moody group associated to $\mathfrak g(A)$. However, if $\Pi^{\vee}$ is not linearly independent, then such a dominant integral weight might not exist.


[1] V. Kac, Infinite-dimensional Lie algebras, 3rd edn., Cambridge University Press, Cambridge, 1990.


[2] T. Marquis, [An introduction to Kac-Moody groups over fields](http://dx.doi.org/10.4171/187), EMS Textbooks in Mathematics, European Mathematical Society (EMS), Zürich, 2018


[3] B. Rémy, [Groupes de Kac–Moody déployés et presque déployés](http://smf4.emath.fr/Publications/Asterisque/2002/277/html/smf_ast_277.html), Astérisque
(2002), no. 277, viii+348.


