Who colored in my Dynkin diagrams? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T11:25:25Z http://mathoverflow.net/feeds/question/49222 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/49222/who-colored-in-my-dynkin-diagrams Who colored in my Dynkin diagrams? Vivek Shende 2010-12-13T07:58:55Z 2010-12-13T18:10:59Z <p>Many of you will recognize these as the <a href="http://en.wikipedia.org/wiki/ADE_classification" rel="nofollow">ADE diagrams</a>, festively colored for the holidays!</p> <p><img src="http://math.princeton.edu/~vshende/ColoredDynkins.png" alt="alt text"></p> <blockquote> <blockquote> <p>Does anyone know a mathematical interpretation for these diagrams, when colored like this?</p> </blockquote> </blockquote> <p><strong>Edit:</strong> 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?</p> http://mathoverflow.net/questions/49222/who-colored-in-my-dynkin-diagrams/49225#49225 Answer by Bugs Bunny for Who colored in my Dynkin diagrams? Bugs Bunny 2010-12-13T08:20:53Z 2010-12-13T08:20:53Z <p>It is the bypartite graph structure on the corresponding Coxeter diagram.</p> http://mathoverflow.net/questions/49222/who-colored-in-my-dynkin-diagrams/49232#49232 Answer by Bruce Westbury for Who colored in my Dynkin diagrams? Bruce Westbury 2010-12-13T09:05:43Z 2010-12-13T09:05:43Z <p>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.</p> http://mathoverflow.net/questions/49222/who-colored-in-my-dynkin-diagrams/49234#49234 Answer by Philippe Nadeau for Who colored in my Dynkin diagrams? Philippe Nadeau 2010-12-13T09:17:58Z 2010-12-13T09:17:58Z <p>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.</p> <p>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 <em>bipartite Coxeter element</em>. Note that it is essentially unique since $RL=(LR)^{-1}$.</p> <p>These special Coxeter elements have figured prominently in recent years in the theory of noncrossing partitions in general Coxeter groups, cf. the <a href="http://arxiv.org/abs/math/0611106" rel="nofollow">memoir</a> by Drew Armstrong for a nice survey, with a combinatorial approach; see also the articles of Brady &amp; Watt for a more geometric approach.</p> http://mathoverflow.net/questions/49222/who-colored-in-my-dynkin-diagrams/49250#49250 Answer by follower for Who colored in my Dynkin diagrams? follower 2010-12-13T12:13:47Z 2010-12-13T12:13:47Z <p>Those are trees, and trees are bipartite as demonstrated by your coloring.</p> http://mathoverflow.net/questions/49222/who-colored-in-my-dynkin-diagrams/49264#49264 Answer by Marcel Bischoff for Who colored in my Dynkin diagrams? Marcel Bischoff 2010-12-13T14:36:49Z 2010-12-13T14:36:49Z <p>They are bipartite graphs. Some application is in the book </p> <p>Goodman, l'Harpe, Jones - "Coxeter Graphs and Towers of Algebras"</p> <p>where they describe how these graphs correspond to Matrices over $\mathbb{Z}$ with norm smaller than 1. </p> <p>Cf. Theorem 1.1.2 which says there is a 1-1 correspondence between </p> <p>1) Indecomposable Matrices with entries $\lbrace 0, 1\rbrace$ up to pseudo equivalence</p> <p>2) Irreducible Coxeter Graphs with bicoloration of type A,D,E</p> <p>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 $&lt;4$ of the index. See also </p> <p>Jones Sunders - "Introduction to Subfactors" (eg. chapter 3.3)</p> http://mathoverflow.net/questions/49222/who-colored-in-my-dynkin-diagrams/49268#49268 Answer by Jim Bryan for Who colored in my Dynkin diagrams? Jim Bryan 2010-12-13T15:36:42Z 2010-12-13T15:36:42Z <p>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: </p> <p><a href="http://arxiv.org/pdf/0803.3766v2" rel="nofollow">http://arxiv.org/pdf/0803.3766v2</a></p> http://mathoverflow.net/questions/49222/who-colored-in-my-dynkin-diagrams/49288#49288 Answer by Allen Knutson for Who colored in my Dynkin diagrams? Allen Knutson 2010-12-13T18:10:59Z 2010-12-13T18:10:59Z <p>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.</p> <p>Less naively, there is indeed a way of distinguishing them in all other cases: the <em>affine</em> 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.)</p> <p>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 $A_{even}$. </p>