# Low Dimensional Topology

## July 19, 2010

### Combinatorics of Jacobi diagrams

Filed under: Combinatorics,Knot theory,Quantum topology — dmoskovich @ 2:25 pm

A Jacobi diagram is surely the most tantalizing object in all of quantum topology. Its definition is combinatorial and easy to visualize, to understand, and to manipulate; and the smallest insight into the structure of the graded space of Jacobi diagrams would lead to huge rewards in quantum topology and in the study of Lie algebras. And yet, despite the concerted efforts of some of the finest minds in mathematics (for instance and in no particular order: Greg Kuperberg, Dror Bar-Natan, Alex Stoimenow, Oliver Dasbach, Pierre Vogel, Jerry Levine, David Yetter, Tomotada Ohtsuki, Justin Sawon, David Broadhurst, Maxim Kontsevich, and many others), they remain forever teasing, tempting, but forever maddeningly beyond reach.

A primitive Jacobi diagram is a graph (actually a multigraph) whose vertices have valence 1 or 3 and whose trivalent vertices are oriented, i.e. a cyclic ordering of 3 edges around each trivalent vertex is fixed (clockwise or counterclockwise), and whose legs are labeled by elements of some finite set $S$. The degree of a Jacobi diagram is defined to be half the total number of vertices of the diagram. A Jacobi diagram is a formal sum of primitive Jacobi diagrams over a ring $R$. Here’s an example of a Jacobi diagram over the matrix ring $M_2(\mathbb{Z}/2\mathbb{Z})$, whose legs are labeled by elements of the set $S =\{1,2,3,4,5\}$

Embarassingly, almost nothing is known about Jacobi diagrams when $R$ is not the field of rational numbers $\mathbb{Q}$. The space of Jacobi diagrams is the vector space over $\mathbb{Q}$ spanned by Jacobi diagrams subject to the AS and IHX relations, which are local moves between Jacobi diagrams which differ inside a dotted circle as indicated below. The space of Jacobi diagrams is graded by degree. A Jacobi diagram is said to be $n$-loop if its Euler number is $1-n$, i.e. if its first Betti number is $n$. (A Jacobi diagram of the type we have just defined is sometimes called an open Jacobi diagram, and the space of these Jacobi diagrams is sometimes denoted $\mathcal{B}$ in the literature.)

Jacobi diagrams are important, I suppose, because finite-type invariants can be constructed by evaluating a weight system (a map $W: \mathcal{B}\longrightarrow \mathbb{Q}$ satisfying some relations) on a Jacobi diagram. So information about Jacobi diagrams is information about finite type invariants. For more stuff along these lines on the blogosphere, see this Everything Seminar post.

### Tree-like Jacobi diagrams

Information about Jacobi diagrams is also information about Lie algebras. To see this most clearly, think of Serre’s procedure to write non-associative words as rooted trees. So write

Then, wonder of wonders, the AS and IHX relations are seen to be anti-symmetry and the Jacobi relation of the Lie bracket. So you see that the space of Jacobi diagrams is some sort of a generalization of a free Lie algebra- specifically, a free Lie algebra $\mathcal{L}(n)$ over $n$ letters is a space of Jacobi diagrams over the set $S=\{0,1,\ldots,n\}$ whose underlying graphs are trees, with one special leg chose to be the root whose label is $0$, and all of whose other legs are labeled by elements of $\{1,\ldots,n\}$.

What about the whole space of tree-like Jacobi diagrams with $m$ legs $B^{\text{tree}}_m(S)$? There’s actually a tremendously useful identification, which I think is actually due to Shigeyuki Morita. It tells us that, over $\mathbb{Q}$, we have the following short exact sequence:

$B^{\text{tree}}_m(S)\overset{\alpha}{\longrightarrow} \mathcal{L}_{m-1}(S)\oplus S \overset{\beta}{\longrightarrow} \mathcal{L}_m(S)\longrightarrow 0.$

To go from $B^{\text{tree}}_m(S)$ to $\mathcal{L}_{m-1}(S)\oplus S$, for each primitive tree-like Jacobi diagram choose a leg $l$ to be the root (so now you have an element of $\mathcal{L}_{m-1}(S)$ by the construction of Serre above), and tensor with its label $s$. The map $\alpha$ is the sum over all choices of legs to be the root. You have to sum over all choices, because there is no way of choosing a canonical leg for a general tree-like Jacobi diagram. The map $\mathcal{L}_{m-1}(S)\oplus S \overset{\beta}{\longrightarrow} \mathcal{L}_m(S)$ is the left-bracketing map. The map $\alpha$ is an isomorpism onto its image (its injective)- to see this, construct its inverse by “re-labeling” $l$ by $s$ and forget it was the root. So, starting from a primitive Jacobi diagram, you get a sum of $m$ primitive elements in $\mathcal{L}_{m-1}(S)\oplus S$, which maps back to $m$ copies of the Jacobi diagram which you started with. We are working over $\mathbb{Q}$, so we can divide by $m$, so mapping a Jacobi diagram to $m$ copies of itself is an isomorphism.

Things are more interesting over $\mathbb{Z}$, where you can’t just divide by $m$. This is the subject of the final paper of Jerry Levine, published 4 months after he died. You still have the exact sequence

$B^{\text{tree}}_m(S)\overset{\alpha}{\longrightarrow} \mathcal{L}^\prime_{m-1}(S)\otimes S \overset{\beta}{\longrightarrow} \mathcal{L}^\prime_m(S)\longrightarrow 0,$

but it’s no longer clear that $\alpha$ is an injective (what’s is $\mathcal{L}^\prime_{m}(S)$? That’s the key subtlety- AS is anti-symmetry $[a,b]=-[b,a]$ of the Lie bracket, but $[a,a]=0$, which holds true in a Lie algebra, is a strictly stronger relation in a ring where 2 fails to be invertible. Thus, over $\mathbb{Z}$, we want to map to a quasi-Lie algebra in which, indeed, the Lie bracket is anti-symmetric, but it may be that $[a,a]\neq 0$). It looks like it should be dead easy… I worked on related stuff pretty hard though when I was starting out as a grad student (before Levine’s paper), and I can testify that it’s not, at least not for me. This is a tantalizing conjecture, and it’s maddening not to know whether or not it’s true.

### Planarity

Here’s another maddening conjecture, due to Sergei Duzhin. Is every Jacobi diagram equivalent under the IHX relation to a Jacobi diagram which is a sum of primitive Jacobi diagrams whose underlying graphs are planar? This looks like it should be moronically easy (by Kuratowski’s Theorem all you would have to prove is that any copy of $K_{3,3}$ as a subgraph can be killed), but it’s wide open. I actually suspect it’s false, but I don’t know.

Why is this an conjecture important? Because it leads to estimates for the growth rate of $\mathcal{B}$ as the degree increases.

### Vanishing of Jacobi diagrams with an odd number of legs

I suppose that perhaps one of the more interesting open problems about the space of Jacobi diagrams is that, for $S=\{pt\}$ (all legs share the same label), every Jacobi diagram of odd degree vanishes (over $\mathbb{Q}$, or over $\mathbb{Z}[\frac{1}{2}]$ if you want more fun). This is an important open problem, because it would imply that no finite-type invariant can distinguish between a knot and its inverse, where the inverse of a knot is that knot with the opposite orientation. Consequences of that are discussed in this paper by Greg Kuperberg.

This is easy for 0-loop Jacobi diagrams (tree-like Jacobi diagrams), because they all vanish for $S=\{pt\}$. The AS relation applied to any trivalent vertex gives back the same tree-like Jacobi diagram you started with, with a minus sign, so the Jacobi diagram equals minus itself, and because 2 is invertible this means that the Jacobi diagram vanishes.
Good. On to 1-loop Jacobi diagrams. The AS relation applies to all vertices gives back the same Jacobi diagram, with sign $-1^m$ where $m$ is the degree. The same trick works for 2-loop Jacobi diagrams. The fun starts for 3-loop Jacobi diagrams, where applying an AS relation to all vertices might just give you something else (you could tell the difference between a Jacobi diagram and its reflection).

Not to toot my own horn, but Tomotada Ohtsuki and I managed to prove the 3-loop case in this paper. I suppose that with quite a bit of effort we could do the 4-loop case, but the 5-loop case should already by rather interesting… prove it and I’ll throw you a party!