Tensoring Polytopes

Fix a Kac-Moody Algebra .

For every dominant (resp. anti-dominant) weight There is a collection of MV polytopes that forms a basis for the irreducible representation, , of highest (resp. lowest) weight . Furthermore this basis respects the weight decomposition of the . ( I’d really like to get a firm grasp on how acts on this basis, but for now I only have a vague idea. )

For the tensor product of two representations, then, we can take as a basis, ordered pairs of MV-cycles — one part of the basis for the first and one part of the basis for the second. Since polytopes have a highest and lowest vertex, thinking of these like “head” and “tail” we draw these pairs in a manner analogous to summing vectors. This process basically gives the Minkowski sum.

But lately I’m beginning to think that a better method is to turn the second polytope “upside down” and draw them head to head. ( I hoped to have some pictures justifying this earlier this week, but at best I may have them up by Monday.) In words, if we associate to each polytope a path in the crystal (understanding that some paths are equivelant) then putting two polytopes head to tail is like concatenating these paths, head to head concatenating one path with the reverse of the other. The new path also corresponds to a polytope (two if you consider its reverse). I have no abstract justification for doing this other than it can be done — but the results for the handful of calculations I’ve done so far are very interesting, by which I mean indicative of symmetries.

The description by Anderson of can also be thought of as adding two polytopes (head to tail or head to head) and in terms of concatenating paths through the crystal may have some parallels (it certainly does for what might be called “balanced” paths) but I’m not to excited about those.

Let me summarize these two methods, let be the polytopes forming a basis for then a basis for can be given either by

or by
.
The second one is a theorem due to Anderson — I’m recalling it off the top of my head so I may revise it later.

Current calculation.

Currently reading through Kazhdan-Lusztig and trying to make the following calculation.
Let be representations with the same central weight. K-L define a vector space where is whose action on is not the usual action but a sort of anti-diagonal:

This pairing has a second definition which generalizes: . And satisfies the formula:

where is the tensor product that preserves central character.

Picture of Anderson Calculation

These calculations are done ala Anderson (see previous post). Up until the 2’s appear I have verified the calculations ala Fullton-Harris. To find polytopes inside specified region I don’t have to calculate the full crystal, only the paths of the crystal that lie in the polytope (I meant to mark the paths, just connect the blue dots — red dots arise from reflections.)

Besides being stunningly faster, with this method we easily see that any tensor product of irreps will have infinitely many direct summands.

Tensor products of Lsl2 Reps

Adapted from Anderson (arXiv:math/0110225v1):

If and are irreducible representations with dominant weights and , and is any dominant weight, then the multiplicity of in the direct sum decomposition of into irreps is the number of MV-polytopes of weight   inside the intersection

Notice that the weights of polytopes always have zero central direction so   must have central part equal to the sum of and .  will be an up-pointing parabola and will be dow- pointing.

Pictures of this process will be up soon.

A little more on Lsl_3

Red points are the root lattice. Alpha’s are the rows of the cartan matrix. The Double bold outline around the indicates that it is displaced in one extraplanar dimension. The dashed border around the points label Lambda indicates displacement in the other extraplanar dimension. The lambda’s are the fundamental weights.

The reflection associated to each α will be reflection, except for which will be a shear-reflection.

Chamber weights, that is the image of the Lambda’s under the reflections and shear reflection will have have dashed displacement of one and double-bolded displacement forming a paraboloid shape.

Recall that in the case of we also add an imaginary fundamental weight. Pseudo-Weyl polytopes are given by associating a number to each chamber weight or imaginary weight. The number associated to the imaginary weight defines the level of the polytope, the other numbers tell how far hyperplanes are displaced to cut faces in the three dimensional polytope:

These will be unboinded, paraboloid like shapes.

We will want to cap these, as I did for , with parabolas going the opposite direction.

standard tropical Plucker relations for Lsl_3

Since the Cartan matrix for is -1 at every off diagonal we can apply the usual tropical Plucker relation. This post I begin to explore what it means for the normal (affine)Weyl group, but I suspect that eventually it will need to be interpreted for my extended Weyl group (with conjugation).

First we associate to each element of the Weyl group a simple triangle in the regular hexagonal lattice of the plane. Left multiplication by a simple reflection corresponds to reflecting over one of three universal lines, and right multiplication corresponds to reflecting over one side of the triangle.

Left Multiplication done by reflecting over colored lines, notice that the leading factor on the outside of any line is all the same (or there is a representative with that leading factor).

Right multiplication is done by flipping over colored edges.

Then the tropical plucker relation that applies states that:

For each triple such that ,; we have

Each triple corresponds to a simple hexagon. This triangle furthest from is w; the sides of that triangle interior to the hexagon correspond to i,j; and the corners of the hexagon are the subscripts of M appearing in the tropical Plucker equation.

Special case reflecting onto a vertical edge

Here is another example of a polytope construction, using the same colour scheme as last time. (There should be an arrow on the bottom row, sorry that’s missing.)

But this time we see something funny happen on the last move (into the purple box).  If you simply follow steps described in the previous post then the right side of that figure should have a red dot, a dark blue dot (reflected from the lower left corner) and two light blue dots (from the previous figure).  What happens instead is that the light blue dot from the previous diagram slides along the vertical edge.

Why does this happen?  Well returning to universal enveloping algebras, unless .  So really we shouldn’t strictly mark any points on the vertical edges but instead label those edges with partions. Then points that reflect onto a side merely specify something about the number of pieces in the partition.

I’m still unclear precisely how this works.  I’m looking into what happens further in the crystal, but calculations there take longer to verify.

More on the “longest element”: Pictures

Here we have a diagram illustrating my augmented set of chamber weights.  The green dots are the new guys.  The red dots are the two real fundamental weights.  Some edges between two chamber weights are labeled with the Weyl group element that they share (that is and are connected by w).  At least thats how the bottom is labeled.  The labels on top are just a first guess.  is meant to generalize the “longest element”.

Here we have a polytope.  Edges have become dots and dots have become edges.  The red edges now correspond to the fundamental weights and the green edges correspond to the green dots.  The teal dots with the question marks are mystery new Weyl elements that transition between the black/red and green portions.

Note: where I put (mislabeled as w) in this diagram does not correspond to my first labeling — but it is where I want it on the polytope (opposite e).

Note also: Verticle edges seem absent from these polytopes.

Calculation

I had to do this one three times before it agreed with the formula in “Loop Groups”.  I hope its right this time.  I’m looking for a good program help me do these algebraic calculations faster and more accurately.  (Any suggestions?).

	4	2	-4	0
	2	6	0	8
	-4	0	8	0
	0	8	0	16

Rank: 3

Calculations

0

Rank:0

1

Rank:1

	2	2
	2	2

Rank:1

	0	0	0
	0	2	-2
	0	-2	2

Rank:1

	0	0	0	0	0	0
	0	4	-2	-4	0	-4
	0	-2	3	2	4	2
	0	-4	2	4	0	4
	0	0	4	0	8	0
	0	-4	2	4	0	4

Rank:2

Unlabeled vertical columns match the rows e.g. the second row is so the second column is . The number in the corresponding box is as studied before. The rank of each matrix gives the dimension that weight space in the irreducible quotient.