11
$\begingroup$

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.

$\endgroup$

1 Answer 1

10
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ nice answer. thanks. Well explained and the nice references. $\endgroup$
    – GA316
    Sep 3, 2018 at 11:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.