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How to prove that a generalized Cartan matrix whose diagram contains no cycles is symmetrizable?

Any hint would be sufficient.

Thanks in Advance.

Hanno
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GA316
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  • What is a generalized Cartan matrix? What is its diagram? What is a cycle in such diagram? Happy New Year! – Robert Lewis Dec 26 '14 at 03:59
  • @RobertLewis Should I include all the definitions in my post? – GA316 Dec 26 '14 at 04:06
  • Well, it would certainly help me! ;) You might, since this seems to me to be a somewhat specialized area. Or maybe give a link to a web-based reference. Best of luck with it! – Robert Lewis Dec 26 '14 at 04:10
  • @RobertLewis Sorry, I cant find any referenec. I am reading Kac's book on infinite dimensional lie algebras. This is the best possible reference I manage to get http://en.wikipedia.org/wiki/Cartan_matrix. Thanks for your interest. – GA316 Dec 26 '14 at 04:16
  • Well, that's a good start. I know it's a lot of work to pull all this stuff together . . . – Robert Lewis Dec 26 '14 at 04:19

1 Answers1

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Hint

A size $n$ generalized Cartan Matrix $A = (A_{ij})_{1\leq i,j\leq n}$ is symmetrizable precisely if for all sequences $i_1,i_2,...,i_k$ of indices in $\{1,2,...,n\}$ you have $A_{i_1 i_2} A_{i_2 i_3} ... A_{i_k i_1} = A_{i_2 i_1} A_{i_3 i_2} ... A_{i_1 i_k}$.

Since $A_{ij}=0$ if there is no edge between $i$ and $j$ in the Dynkin diagram $\Gamma$ of $A$, we may assume that $i_1\to i_2\to ...\to i_k\to i_1$ constitutes a path in $\Gamma$. Then, since $\Gamma$ has no cycles by definition, any edge must be traversed in one direction precisely as often as it is traversed in the other direction.

Hanno
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