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Why is the dimension of the cartan subalgebra $2n-\text{rank}(A)$ in the defintion from Kumar's book. From a few examples I can see why the defintion is the way it is, but, I would like a better understanding of why the dimension of the Cartan subalgebra is $2n-\text{rank}(A)$ and not $n$ or something else. I

To be specific why does the Lie algebra associated to the universal extention $\tilde{L}SU(2)$ not have an associated Cartan matrix while the affine Lie algebra associated to $\mathbb{T}\ltimes\tilde{L}SU(2)$ does?

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    $\begingroup$ Probably part of the problem is that "the" definition doesn't really exist, just a range of closely related definitions. The ideas come from the independent thesis work of Kac and of Moody, but later developments tweaked the notions in various ways. It's a good idea to say more explicitly what version you are looking at. $\endgroup$ Commented Sep 5, 2013 at 21:21
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    $\begingroup$ I find this dimension in every definition I have read. The Serre relations are all implemented slightly differently, but the dimension of the Cartan subalgebra is always $n+\text{corank}(A)$. $\endgroup$ Commented Sep 5, 2013 at 21:44
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    $\begingroup$ One reason why the extra dimensions are useful is that you end up with a nondegenerate inner product on the weight space. In your first example, your weight space becomes a singular plane embedded in $\mathbb{R}^{2,1}$, and it's harder to see the geometry attached to reflections. $\endgroup$
    – S. Carnahan
    Commented Sep 5, 2013 at 21:47
  • $\begingroup$ I think you have a good answer to my admittingly soft question. $\endgroup$ Commented Sep 6, 2013 at 14:04

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I'll just elaborate on my comment from last year.

A Kac-Moody algebra is defined by generators-and-relations, with starting data given by an $n \times n$ matrix $A$. It can be written as $L \rtimes D$, where $D$ is a $n - rank(A)$-dimensional commutative Lie algebra that acts on $L$ by derivations. The standard examples are affine Lie algebras, where $D$ is one dimensional. It is common to throw away $D$ when studying these objects, because it doesn't really do much for the representation theory - the only information it adds is an "energy" grading. However, $D$ can help with some geometric intuition, because the Cartan subalgebra gets a non-singular inner product. This makes the geometry of reflections from the Weyl group more clear.

The standard example is with the loop algebra of $\mathfrak{sl}_2$: if you forget about $D$, you lose the energy grading on the roots, and it is hard to draw a picture of them where they don't land on top of each other. When you add $D$, the weight space grows from a 2-dimensional space with singular inner product to the Lorentz space $\mathbb{R}^{2,1}$, the new "energy" axis is a line in the light cone (i.e., all vectors have norm zero), and the roots lie in a plane tangent to the light cone containing this axis. The Weyl group now acts by visible reflections, and the parabolic shape that shows up in characters of highest-weight representations can be seen as the intersection of a hyperboloid of constant norm with a plane parallel to the root space.

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  • $\begingroup$ Do you have a reference for this description for a general rank? Thanks! $\endgroup$ Commented May 16, 2014 at 4:36
  • $\begingroup$ Some basic properties are found in the early part of Kac's Infinite Dimensional Lie Algebras, but I am unaware of interesting results for the general theory. I seem to recall that Kyoji Saito and collaborators have some work in the case $D$ has dimension 2, relating the Lie algebras to elliptic singularities and mirror symmetry. $\endgroup$
    – S. Carnahan
    Commented May 16, 2014 at 5:21
  • $\begingroup$ Thanks. I do not know much about Kac-Moody algebras, but I feel that the generator associated to the affine level can be understood as the "extra" dimension in the Cartan subalgebra, and a representation for a corank 2 algebra can acсommodate representations of two affine algebras at different level. Do you think this intuition might be correct? $\endgroup$ Commented May 16, 2014 at 5:29

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