My talk at "Advances in Quantum and Low-Dimensional Topology", University of Iowa, March 2016.

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...Kuperberg's A2 spider

Motivation

Suppose you are interested in the representation theory of SL$(3, {\mathbb C})$.

There's a trivial 1-dimensional representation.

There's an obvious 3-dimensional representation, called the "defining" representation.

There are morphisms between representations.

There are operations on morphisms:

Kuperberg's A2 spider encodes all of this with trivalent "web" diagrams.

There are trivalent vertices with arrows either all-in or all-out. Roughly, a trivalent vertex is the cross product, and a cap is the dot product.

Linear combinations of morphisms are just formal linear combinations of diagrams. Composition and tensor product are multilinear. Taking the dual is conjugate linear.

There are relations that let you simplify bubbles, digons and squares. There's a unique consistent way to sprinkle $q$s into the defining relations. See here. So the spider for SL$(3, {\mathbb C})$ belongs to a one-parameter family of spiders for $U_q\mathfrak{sl}_3$.

I won't define $U_q\mathfrak{sl}_3$ here. Whatever it is, you can study its representation theory by drawing doodles!

New definition

Here is a new way to encode the representation theory of $U_q\mathfrak{sl}_3$ by drawing doodles.

Start with the root system for SL(3), and the weights for the defining representation:

Make a choice of positive roots: red, green and blue.

A "diagram" will be a triple of oriented Temperley-Lieb diagrams, one red, one green, and one blue, superimposed. Strands of different colors can pass through each other.

The defining representation

Instead of Kuperberg's up arrow, we use a sum of 3 pairs of arrows.

+ +

Each term corresponds to a weight. Each color corresponds to a root. Take dot products weight · root.   1 ↦ up arrow,   –1 ↦ down arrow,   0 ↦ missing color.

Morphisms

Instead of Kuperberg's vertex with arrows all-in, we have a sum of 6 terms, one for each of the 6 = 3! orders of the 3 weights.

If the 3 different pairs of edges go in to a vertex, there is a unique way to join them so the colors and orientations match.

Similarly for an all-out vertex.

Tensor = side by side. Dual = reflect vertically.

Compose by stacking. If the colors don't match you get zero.

Relations

$\circlearrowleft$ = $q$, $\circlearrowright$ = $q^{-1}$, $\stackrel{\style{display: inline-block; transform:rotate(0.5turn);} { \curvearrowleft}}{\curvearrowleft}$ = $q$ $\downarrow$ $\uparrow$ .

Same for green and blue.

Theorem

Our definition is isomorphic to Kuperberg's spider.

Proof: Just check it satisfies Kuperberg's relations.

Work in progress

Added later

I should have mentioned crossings. In my picture they have a fairly natural definition where each color does its own Kauffman bracket.

I don't know what happens if $q$ is an $n$th root of unity. Maybe allow $n$-valent vertices with arrows all-in or all-out?