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We know that certain structures in a Coxeter-Dynkin diagram will guarantee that it never encodes a (semi)simple Lie algebra, for example a quadruple node or the diagram $E_n,n>8$, which is "too long". My question is: What will result?

Does such a "lie-ing" (sorry for the pun) diagram define an algebra at all? If yes, uniquely so? If yes, which property of a Lie algebra gets violated? (Yes, I already looked up "affine Lie algebra", but that gives only a partial answer - some Coxeter-Dynkin diagrams now additionally are legit.)

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    $\begingroup$ You get a Kac-Moody algebra: en.wikipedia.org/wiki/Kac%E2%80%93Moody_algebra $\endgroup$
    – Sam Hopkins
    Commented Nov 7, 2022 at 14:38
  • $\begingroup$ To expand on Sam's comment affine Lie algebras are just the nicest family of Kac-Moody Lie algebras. You can extend on to the hyperbolic Lie algebras and then on to even more general ones. I think any generalised Cartan matrix and thus any "pseudo-Dynkin" diagram gives rise to a Kac-Moody Lie algebra (although these will be infinite dimensional in general). See also this question $\endgroup$
    – Callum
    Commented Nov 7, 2022 at 14:58
  • $\begingroup$ Ah, that question in principle has an useful answer, nr. 2, there (and as usual wasn't spawned by the "intelligent" dupe finder...). Whereas technically no dupe, I think my question can be closed, since I can look up myself the definitions. (As I suspected, the Cartan matrix also generalizes, so I don't have to ask that either.) $\endgroup$
    – Hauke Reddmann
    Commented Nov 7, 2022 at 18:29

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