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I am searching in the literature (so far unsuccessfully) the correct notion of "abstract Dynkin diagram" as the one that characterises intrinsically graphs that arise as the Dynkin diagram of a semisimple Lie algebra. In other words: what properties/extra data must a graph have to guarantee that it is a Dynking diagram?

My attempt is that such a graph must have weights $(w_i)$ attached to each vertex, at most 3 edges joining any two given vertices, and an extra condition making sure that the Cartan matrix it (potentially) comes from is symmetrisable. Namely, if $e_{ij}$ denotes the number of edges between vertices $i$ and $j$, then put $d_i := +\sqrt{|w_i|}$ and $c_{ij}:= - \sqrt{|w_j e_{ij}/w_j|}$, then the $c_{ij}$ are integers and $DCD^{-1}$ is a symmetric positive definite matrix, where $D=\mathrm{diag}(d_1, \ldots, d_n)$ and $C=(c_{ij})$. I presume that this is correct, but I am not entirely sure. And if so, this is certainly not the most elegant way to phrase it.

I assume that, in this context, an isomorphism of abstract Dynkin diagrams should be a graph isomorphism that respects the weights up to a common factor on every connected component. I don't think a preserving condition on the symmetric matrix $DCD^{-1}$ is needed.

Of course, the correct notion of abstract Dynkin diagram and isomorphism should be the one with the property that isomorphism classes of abstract Dynkin diagrams are in bijection with isomorphism classes of complex semisimple Lie algebras (or abstract Cartan matrices, or abstract root systems).

Does anyone know what the correct notions are, or a reference where this is done? Samelson briefly mentions "abstract Dynkin diagram" but does elaborate on this idea (page 68), and Knapp only defines "abstract Dynkin diagrams" that come from an abstract Cartan matrix.

Minkowski
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    It makes sense to focus on the connected components of "such a graph", i.e., it suffices to answer the title question for simple Lie algebras. – Dietrich Burde Aug 20 '24 at 18:50

1 Answers1

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The more common term you'll find for these is "admissible". From the perspective of the graph we must have following properties. Firstly the graph is a partially directed multigraph.

  1. Removing a node from the diagram and edges attached to it must still give an admissible diagram (so any induced subgraph must be admissible).
  2. Collapsing a pair of points connected by a single line into one point must give an admissible diagram.
  3. There can be no cycles.
  4. A node may have at most three edges connected to it.
  5. Only a multiple edge can have a direction.

I believe these are enough to classify the admissible Dynkin diagrams. You can see this in Humphreys' Introduction to Lie algebras Theorem 11.4 although he doesn't lay it out quite like that.

Edit: on reflection (ha!) you need something more to restrict graphs with a central node and three single chain branches to only $D_n$ and $E_n$ types. Humphreys does this by looking at the root system so I don't know the most efficient condition on the diagram to achieve this. It's a bit clunky but this would do it:

  1. Around a node with three different attached nodes, if the lengths of each branch from the central node (including the central node each time) are $p,q,r$ respectively, then $\frac{1}{p}+\frac{1}{q}+\frac{1}{r} > 1$
Callum
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