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In his talk "Split Octonions and the Rolling Ball," John Baez says that a Dynkin diagram describes a type of geometry, with vertices representing types of objects, and edges representing incidence relations between them. The example he gives is of $A_3$, representing projective 3-space. The first vertex represents points, the second represents lines, and the third represents planes.

How does this fit in with the usual Lie theory picture of the correspondence with the diagrams, and how does this end up actually generalizing to the other diagrams? How can the types of geometry the other diagrams correspond to be described?

RobPratt
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nph
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    I'm not super familiar with this but the terms you want to look up are "parabolic subgroup" and "flag variety," see e.g. https://en.wikipedia.org/wiki/Borel_subgroup and https://en.wikipedia.org/wiki/Generalized_flag_variety for starters. – Qiaochu Yuan Dec 31 '24 at 20:12
  • I now see that this connection between dynkin diagrams and "types of geometry" is described in Jacques Tits' "A Local Approach to Buildings" – nph Apr 17 '25 at 14:12

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As Qiaochu says, this is about the incidence geometry of generalised flag varieties.

For a reference on how Dynkin diagrams can be used to encode flag varieties see Chapter 2 here. And for a discussion of the incidence geometry see here.

I will go through the basic ideas. For a complex semisimple Lie group $G$ (resp. Lie algebra $\mathfrak{g}$) we can consider its parabolic subgroups (resp. subalgebras). These are the subgroups containing a Borel subgroup. There are several other ways to characterise them including the relevant fact that $G/P$ is not only a smooth homogeneous manifold but a complete projective variety. There are only a finite number of conjugacy classes of parabolic subgroups and we can relate them to subsets of the Dynkin diagram.

This goes briefly as follows. Each node on the Dynkin diagram corresponds to a simple root (up to some choice of Cartan subalgebra $\mathfrak{h}$). Let $\Delta$ be the set of roots, $\Phi$ a choice of simple roots with $\Delta_+$ the positive roots and $\Phi' \subset \Phi$. We define $$\mathfrak{l} := \mathfrak{h} \oplus \bigoplus_{\alpha \in \mathrm{span}(\Phi')} \mathfrak{g}^\alpha $$ and $$\mathfrak{n} := \bigoplus_{\alpha \in \Delta_+ \setminus \mathrm{span}(\Phi')} \mathfrak{g}^\alpha .$$ Then $\Phi$ corresponds to the parabolic subalgebra (and thus its conjugacy class):

$$ \mathfrak{p} := \mathfrak{l} \oplus \mathfrak{n}. $$

Note a parabolic subalgebra corresponds to a unique parabolic subgroup and vice versa. The usual notation here is to draw a Dynkin diagram with crossed nodes corresponding to our choice of simple roots (The link to Baston and Eastwood above defines the subset the other way round to me but marks the diagram the same).

Now as Baez mentions in the video parabolic subgroups/subalgebras are the stabilisers of certain subspaces or more generally of flags (chains of subspaces $V_1<V_2<V_3<\cdots <V$) in some space. To be more exact, in any representation $V$ of $G$ a parabolic subgroup $P \leq G$ stabilises a flag in $V$.

The idea of the incidence geometry is that two parabolic subalgebras are incident if their intersection is also parabolic. In projective geometry the intersection of the stabiliser of a plane and a line on that plane is something that stabilises both (i.e. a flag) and is thus itself parabolic. If the line is not on the plane, however, the stabiliser is not so neat. Effectively, two flags are incident if they combine to form another flag.

The flags for other semisimple Lie groups (or just in other representations) are a little more complicated but the ideas remain roughly the same. For example, in $\mathfrak{so}(n,\mathbb{C})$ or $\mathfrak{sp}(2n,\mathbb{C})$ the flags in their defining representation are of isotropic subspaces (with respect to their symmetric/symplectic bilinear form). Edit: Also here is a MathOverflow question where I attempted to describe some of the exceptional flag varieties (and do compute all their dimensions) but as you can see, it gets quite complicated.

Callum
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