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A generalized Cartan matrix is a square matrix $ A=(a_{ij})$ with integral entries such that

  • For diagonal entries,$a_{ii}=2$.
  • For non-diagonal entries, $a_{ij}\leq 0$.
  • $a_{ij}=0$ if and only if $a_{ji}=0$
  • $A$ can be written as $DS$, where $D$ is a diagonal matrix, and $S$ is a symmetric matrix.

From the Cartan Matrix I can recover a semisimple Lie algebra. What I'm wondering is what goes wrong if the coefficients are non integers but real? I know in Lie theory this cannot happen but I do not really grasp what goes wrong in the definition of the Lie algebra...

Dac0
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    Well, what "recovery" process would you use to get a Lie algebra out of the matrix? In Chevalley-Serre relations as per https://mathworld.wolfram.com/Chevalley-SerreRelations.html, relations 5 and 6 make little sense for non-integer $a_{ij}$. In a finite-dim semisimple LA, for $i \neq j$ the $1-a_{ij}$ measures the length of a certain root string, so $a_{ij}$ must be an integer. Compare answers to https://math.stackexchange.com/q/2509230/96384. If one relaxes that condition, the CS relations might at best give out an infinite dimensional LAs, or something that's not an LA at all. – Torsten Schoeneberg Aug 18 '22 at 20:01
  • @TorstenSchoeneberg yes I agree with you and was one of the few things I came up with. But this doesn't prevent to have a Lie algebra but only a finite dimensional one... – Dac0 Aug 22 '22 at 14:23
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    So would you refine your question to asking: What kind of object do we get, and has it been studied, if we take generators and relations 1-4 in Chevalley-Serre relations, with non-integer $a_{ij}$? I would not know an answer to this, but it seems like a valid question. – Torsten Schoeneberg Aug 22 '22 at 15:43

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The reason is related to the root system of the Lie Algebra. The entries of the Cartan matrix are defined as;

$$a_{ij}=2{(r_i,r_j)\over(r_j,r_j)}$$

Where $r_i$ are the simple roots$\space(\Phi)$ of the Lie algebra. As it turns out the simple roots have the property that if $\alpha,\beta \in \Phi$ then the projection of $\alpha$ onto $\beta$ is always an integer or a half integer multiple of $\beta$.

This will property will obviously always cause $a_{ij}$ to be an integer.

Volk
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  • Well, not really... e.g. you have non crystallographic systems such as $H_3$ and $H_4$ that do not have integers $a_{ij}$, my question is exactly what does go wrong in those cases... – Dac0 Aug 18 '22 at 09:27
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    But @Dac0, I assume you do not doubt that root systems coming from Lie algebras are crystallographic? Which is certainly proven in any good source on semisimple LAs? I am trying to understand what exactly is your question. – Torsten Schoeneberg Aug 18 '22 at 15:16
  • @TorstenSchoeneberg I think his question is essentially, "What in the definition of a Lie Algebra makes it $\textbf{necessarily}$ crystallographic?" I haven't really thought deeply about this but my intuition is that the fact that non crystallographic root systems fail to produce Weyl groups is the culprit – Volk Aug 18 '22 at 16:30
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    Humphreys explains something here. It is indeed related to the Weyl group. – Dietrich Burde Aug 18 '22 at 17:46
  • @DietrichBurde the link you posted is more an historical explanation than a mathematical one... but thank you anyway – Dac0 Aug 22 '22 at 14:25
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    Well, he explains "The Bourbaki notion "crystallographic" selects precisely the Weyl groups in this setting, as those which leave some lattice invariant in the natural representation. From the classification one finds that these are the finite Coxeter groups with products of two generators having order 1, 2, 3, 4, or 6." – Dietrich Burde Aug 22 '22 at 14:28
  • Ok thank you!!! – Dac0 Aug 22 '22 at 15:32