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.


1 Answer 1


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, EMS Textbooks in Mathematics, European Mathematical Society (EMS), Zürich, 2018

[3] B. Rémy, Groupes de Kac–Moody déployés et presque déployés, Astérisque (2002), no. 277, viii+348.

  • $\begingroup$ nice answer. thanks. Well explained and the nice references. $\endgroup$
    – GA316
    Sep 3, 2018 at 11:42

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