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Let $\mathfrak{g}$ be a simple finite-dimensional Lie algebra. Let $\mathfrak{h}$ be a Cartan subalgebra, $C$ the Cartan matrix, and $R$ a system of simple roots $\left(\alpha_{1},\cdots,\alpha_{n}\right)$.

Using the following fact

The Killing form $Tr\left(ad_{X}\circ ad_{Y}\right)$ induces a bilinear symmetric non-degenerate form on $\mathfrak{h}^{*}$

how can it be shown that there exists a symmetric matrix $A$ such that $$ A=diag\left(r_{1},\cdots,r_{n}\right)C $$ where $r_{j}$ are prime and strictly positive $\forall j$?

Note: This question is related to exam prep for an advanced graduate course "Lie Theory and Representations 2". So far, I figured that if $\mathfrak{h}$ is considered in diagonalized form, $ad$ can be represented by a $diag$ matrix, and I thought about applying Serre relations $$ \left(ad_{e_{i}}\right)^{1-C_{i,j}}\left(e_{j}\right)=0\forall i\neq j $$ and $$ \left(ad_{f_{j}}\right)^{1-C_{i,j}}\left(f_{j}\right)=0\forall i\neq j $$ to get the $C$ part, but so far, I could not get it to stick. I think I may be on the right path?

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migrated from mathoverflow.net Nov 17 '14 at 15:51

This question came from our site for professional mathematicians.

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    $\begingroup$ Lemma 4.6 of Kac "Infinite dimensional Lie algebras" gives the simmetrizability of GCM of finite type with no reference to the structure of Lie algebra. $\endgroup$ – Andrei Smolensky Nov 15 '14 at 8:02
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    $\begingroup$ Does it mean this is a homework which we are supposed to do? $\endgroup$ – Venkataramana Nov 15 '14 at 8:56
  • $\begingroup$ @AndreiSmolensky Thanks; indeed, and I found something similar to this lemma. I think I should find a way to solve this problem as suggested by the statement, especially that it is the first in a series of interrelated questions. $\endgroup$ – Jake Nov 15 '14 at 10:36
  • $\begingroup$ @Venkataramana The course has no homework. In my case, I only have the course notes and some exam prep material such as past exams. It is quite a Spartan course setup. $\endgroup$ – Jake Nov 15 '14 at 10:41
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    $\begingroup$ @Jake: Probably it's not the Killing form as such that matters here, but just the fact that the Cartan matrix is nonsingular. This is what makes Kac-Moody algebras more complicated, when you start with a "generalized" Cartan matrix and sometimes have to treat the "symmetrizable" case separately. $\endgroup$ – Jim Humphreys Nov 15 '14 at 15:04
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Am I missing something here? If I take $r_i=(\alpha _i, \alpha _i)$, and $D$ the diagonal matrix with entries $r_i$, then clearly $A=DC$ is symmetric, where $C$ is the Cartan matrix.

Now, if $\mathfrak g$ is a simple Lie algebra, then the numbers $r_i$ are either all $1$, or exactly one of them is $2$, a prime, or exactly one of them is $3$ (a prime). The condition that the $r_i$ be prime, seems strange in this context.

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  • $\begingroup$ I think you are quite right. I also thought about trying something like that (but did not get to it yet). The condition of $r_{i}$ being prime is indeed automatically satisfied, so perhaps it is a hint. $\endgroup$ – Jake Nov 16 '14 at 14:46
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    $\begingroup$ I don't know: $1$ is usually not considered a prime! $\endgroup$ – Venkataramana Nov 16 '14 at 14:48

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