Who colored in my Dynkin diagrams? - MathOverflow

Who colored in my Dynkin diagrams?

2010-12-13T07:58:55Z

Many of you will recognize these as the ADE diagrams, festively colored for the holidays!

Does anyone know a mathematical interpretation for these diagrams, when colored like this?

Edit: the answers below seem to treat the set of red nodes and the set of green nodes as essentially equivalent. However in the combinatorics relevant to me, the red and green ones play slightly different roles; does this also happen in the Coxeter group story?

Answer by Bugs Bunny for Who colored in my Dynkin diagrams?

2010-12-13T08:20:53Z

It is the bypartite graph structure on the corresponding Coxeter diagram.

Answer by Bruce Westbury for Who colored in my Dynkin diagrams?

2010-12-13T09:05:43Z

The nodes correspond to generators of the Weyl group. The red nodes are a commuting set of involutions and so the product is an involution. Similarly for the green nodes. These two involutions generate a dihedral subgroup. The order of the product is the Coxeter number.

Answer by Philippe Nadeau for Who colored in my Dynkin diagrams?

2010-12-13T09:17:58Z

In the associated Coxeter/Weyl groups there is such an interpretation. Recall that a Coxeter element is the product of the generators ($=$ nodes) in a certain order.

Now red generators commute between themselves, let $L$ be their product; similarly let $R$ be the product of the green generators. Then the product $LR$ is a special Coxeter element called unsurprisingly a bipartite Coxeter element. Note that it is essentially unique since $RL=(LR)^{-1}$.

These special Coxeter elements have figured prominently in recent years in the theory of noncrossing partitions in general Coxeter groups, cf. the memoir by Drew Armstrong for a nice survey, with a combinatorial approach; see also the articles of Brady & Watt for a more geometric approach.

Answer by follower for Who colored in my Dynkin diagrams?

2010-12-13T12:13:47Z

Those are trees, and trees are bipartite as demonstrated by your coloring.

Answer by Marcel Bischoff for Who colored in my Dynkin diagrams?

2010-12-13T14:36:49Z

They are bipartite graphs. Some application is in the book

Goodman, l'Harpe, Jones - "Coxeter Graphs and Towers of Algebras"

where they describe how these graphs correspond to Matrices over $\mathbb{Z}$ with norm smaller than 1.

Cf. Theorem 1.1.2 which says there is a 1-1 correspondence between

1) Indecomposable Matrices with entries $\lbrace 0, 1\rbrace$ up to pseudo equivalence

2) Irreducible Coxeter Graphs with bicoloration of type A,D,E

The graphs correspond to Bratelli diagrams of inclusions of finite von Neumann algebras. From these one can construct subfactors with index $4 \cos^2(\pi/n)$ $n=2,3,\ldots$ the index is exactly the square of the associated matrix. These are all possible values $<4$ of the index. See also

Jones Sunders - "Introduction to Subfactors" (eg. chapter 3.3)

Answer by Jim Bryan for Who colored in my Dynkin diagrams?

2010-12-13T15:36:42Z

Via the McKay correspondence, the nodes correspond to representations of $G$, a finite subgroup of $SU(2)$. If we color the nodes into two groups depending on whether or not the representation pulls back from a representation of $\hat{G}\subset SO(3)$, the image of $G$ under the double cover map $SU(2)\to SO(3)$, we get the coloring pattern that you see. You do need to swap red for green on a couple of your diagrams if you want (for example) green to always be those that don't pull back from $SO(3)$ ("binary" nodes). See Figure 1 here:

http://arxiv.org/pdf/0803.3766v2

Answer by Allen Knutson for Who colored in my Dynkin diagrams?

2010-12-13T18:10:59Z

Naively, there can be no reasonable way of distinguishing the red nodes from green in the case $A_{even}$, as the Dynkin diagram automorphism switches them.

Less naively, there is indeed a way of distinguishing them in all other cases: the affine Dynkin diagram is also bipartite, and the affine vertex can be taken to be a fixed color. (Unfortunately in your $D_n$ example it would be red, whereas in your $E_6$ example it would be green, so if you're set on those choices I can't help you.)

As others mentioned, you can multiply reds then greens and get a Coxeter element. If you raise it to half its order, you get the long element $w_0$. Of course you can only do this if the Coxeter number is even, which is again all cases except $A_{even}$.