## Chord Diagrams and Lie Algebras

Merry Christmas!  In this post, we will build on some of the previous posts about chord diagrams.  In a bit of a tangent from previous thoughts, we will explore the relationship between chord diagrams and Lie algebras.  Explicitly, last time we came up with a relation, which we will henceforth call the IHX-relation:

Remember that this was really another aspect of the 4T relation for regular chord diagrams.  We will see how this relationship is a pictorial representation of the Jacobi identity, which allows us to interpret generalized chord diagrams modulo the IHX relation as instructions on how to combine a large number of Lie brackets.

The first step is to introduce the Penrose’s tensor notation, which is a very natural tool for writing down instructions on manipulating tensor powers of vector spaces.  If we have a Lie algebra with an invariant inner product, we can turn a large class of graphs with extra data into such instructions.  This class of graphs includes chord diagrams, and we will see that in this framework, the IHX relation and the Jacobi relation are the same thing.

I should warn readers that there’s not much of a punch-line to this post.  Instead of being about building to some nifty conclusion, this is about a different perspective on chord diagrams and a series of nifty things you can do with them.  Also, this is a very long post that is a bit rambly.  I wouldn’t mind some constructive criticism on how some of this stuff could have been explained better.

0. IHX implies Anti-Symmetry

Before anything else, I should mention a useful fact that should have appeared in the previous post.  The IHX relation tells us how to change the orientation on any one vertex:

This says that switching the orientation on a vertex just multiplies that diagram by $-1$.  This relation is called ‘Antisymmetry’.

1.A. Penrose’s Tensor Notation

Penrose’s tensor notation is a way of drawing diagrams that describe how to combine simple operations on a vector space into more complicated ones.  The general idea is that these simple operations will have a certain number of inputs and outputs, and so we will draw lines connecting which outputs we want to plug into which inputs.

For example, if we have an algebra $A$, multiplication gives us a map $m:A\otimes A\rightarrow A$.  We can draw diagrams to tell us how to multiply several elements together:

This is the same information usually provided by parenthesis.

In general, we’ll have a fixed vector space $V$, and some maps we will think of as building blocks.  Each will be a map from $V^{\otimes n}\rightarrow V^{\otimes m}$, and will be denoted by some sort of distinguishing symbol with n inputs and m outputs:

Then we can combine these symbols with lines telling us what to plug into what.  The whole diagram will have some inputs and outputs, and so will define a map from $V^{\otimes i}\rightarrow V^{\otimes j}$, for some i and j.

1.B.  Lie Algebras

Ok, so what can we do with these tensor pictures if we have a Lie algebra $L$?  Well, the most basic thing we can do is the original example we used; we take the multiplication map $b:L\otimes L \rightarrow L$ and make multiplication diagrams out of it.  For instance, we can write the Jacobi identity as:

Here, a horizontal bar denotes the map $b$.

So that isn’t particularly exciting or new… but there are more interesting things availible when we have a inner product defined on $L$ that is symmetric, non-degenerate and invariant (the Killing form on a semi-simple Lie algebra, for instance).  An invariant inner product on $L$ is one such that $([a,b],c)=(a,[b, c])$.  Notice this gives us a nice 3-form $a\otimes b\otimes c \rightarrow ([a,b],c)$.  This 3-form is cyclically symmetric, because $([a,b],c)=(c,[a,b])=([c,a],b)$.

Since the inner product is non-degenerate, it defines a vector space isomorphism between $L$ and $L^*$ by $a\rightarrow (a,-)$.  This isomorphism is important for making Penrose diagrams because it will let us turn an input into an output and vice versa.  We’ve already seen an example of this.  If we want to turn the only output of $b$ into a third input, we would have to take that third input and take its inner product with the output of $b$; this is the 3-form we just described above, thought of as a map from $L\otimes L\otimes L\rightarrow k$.

Similarly, we can turn inputs into outputs.  To do this, pick an orthonormal basis $\{x_i\}$.  Then, we can send any Lie algebra element $a$ to $\sum x_i\otimes [x_i,a]$, this is a map from $L\rightarrow L\otimes L$.  We can get a map with no inputs and three outputs in the same way, this sends a scalar $\lambda$ to $\sum\sum \lambda x_i\otimes [x_i,x_j]\otimes x_j$.

Now we have four different basic maps with which we can build Penrose diagrams.  However, these are all facets of the same map, and we can use this to our advantage.  Let’s take a planar graph with some degree 1 vertices labelled Input or Output, and all the rest trivalent vertices.  We turn this into a Penrose diagram by choosing a direction on every edge; there is no restriction on this choice except that the direction must go out of every Input vertex and go into each Output vertex.  Then, each trivalent vertex will have some number of incoming edges, which determines which of the above four simple maps to use.  So, this might look like:

Here, all four maps are denoted by horizontal bars, and they are distinguished by the number of inputs.

There is some ambiguity in which inputs/outputs to use in which slot, but the planar embedding of the graph gives each trivalent vertex a cyclic orientation of its edges.  This orientation is enough to resolve this ambiguity, because as we noticed before, the 3-form $([a,b],c)$ is cyclically symmetric.

Thus, by making some choices, we can turn the planar graph into a Penrose diagram, and subsequently get a map from $L^{\otimes}\rightarrow L^{\otimes}$.  Important fact: this map is independant of the choices we made!  We won’t prove this, but its a series of simple computations to see that reversing a single arrow gives the same map.

We can still do this for diagrams that aren’t quite planar.  So what did we need?  We needed a graph with

1) a bunch of ordered inputs

2) a bunch of ordered outputs

3) all other vertices trivalent, with a choice of cyclic orientation of the edges

Any such graph will determine an appropriate map.  For the sake of simplicity, lets call such graphs Jacobi diagrams.  This might differ from the usual use of this term by allowing inputs and outputs, rather than being closed.  In that case, a better name for them might be ‘open Jacobi diagrams’.

1.C. Using the Lie Algebra Axioms.

We haven’t yet used the fact that $L$ is a Lie algebra, we have only used that it is an algebra with an invariant trace.  What are the consequences of anti-symmetry and the Jacobi relation?  (Hopefully you know, since I’ve said it like, four times already)

Anti-symmetry is easy.  Any trivalent vertex can be thought of as a Lie bracket with the ordering of the inputs determined by the orientation of that vertex.  Since swapping the two inputs picks up a minus sign, we see that reversing the orientation of any vertex in a Jacobi diagram will change the corresponding map by a negative sign.  Pictorially:

We have already written down what the Jacobi relation looks like in Penrose notation:

We can turn this into a graph by just contracting the horizontal bars to vertices.  Moreso, now that we know how to change the orientation of a vertex, we can turn the diagram into a more familiar one:

As promised, it is the IHX relation.  What does this mean?  It means that if we have an Jacobi diagram with a pair of vertices like the left hand side of the above equation, that the map corresponding to it is the difference of the map corresponding to the two Jacobi diagrams on the right hand side.

2.A. Chord Diagrams.

This gives us an immediate source of Jacobi diagrams.  We can take any chord diagram or generalized chord diagram modulo the 4T or IHX relation and think of it as a Jacobi diagram with no inputs and outputs.  It will then give us a well defined linear map from $k\rightarrow k$, and so its given by multiplication by a constant.  This gives us a method for assigning to each chord diagram an element in $k$, which is well-defined modulo the 4T relation.

Note, though, that the 1T relation is nowhere to be found.  In fact, this relation will be violated unless the Killing form on $L$ is identically zero.  Unfortunately, this reduces the relevance of these techniques for knot theory.

2.B. Arc diagrams

No reason to stop here, let’s come up with some other Jacobi diagrams.  If we take any generalized chord diagram, and a point on the outer circle, we can break the circle apart there, and treat one endpoint as an input and one endpoint as an output:

This gives us a Jacobi diagram, which we will call an arc diagram.  Any such arc diagram gives us a linear map from $L$ to itself.

It might seem like the choice of breakpoint above was significant.  However, there is a theorem of Kontsevich’s which says that for any two breakpoints, the corresponding arc diagrams are equivalent under the IHX relation.  I won’t go over the proof (though its short), but I will mention the key lemma of the proof:

Key Lemma: Pictorially,

It says that bracketing with an input before doing anything else is the same as bracketing with that input after doing everything else.  This lemma has a nice conceptual proof.  We can think of the IHX relation as telling us that, if we try to move an edge past a vertex, we get the sum of the two possible ways we could move that edge past the vertex:

Therefore, to prove the lemma, we just push the initial input edge down the diagram, and it will split apart as it hits some vertices, and combine as it hits other vertices.  However, at the end, it will always have completely recombined into one lone diagram, which is the one that was needed.

This means that the linear map from $L$ to itself coming from the arc diagram is determined entirely by the chord diagram, and so its a stronger invariant of a chord diagram than the element of $k$ from before.  In fact, this number is the trace of the endomorphism of $L$.

2.C. The Chord Algebra.

Kontsevich’s theorem has another consequence, as well.  Since we can break apart chord diagrams in an essentially unique way, it means we can define a multiplication on them as follows.  Take two chord diagrams, break each of them apart anywhere, and stick them together, making sure to match the orientation on the outer circles.  This multiplication is commutative, and makes the vector space of chord diagrams modulo 4T but not 1T into an algebra, which we will denote $A$.

Notice that the above method for turning chord diagrams into endomorphism of $L$ is now an algebra homomorphism, because concatenating arc diagrams will become composition of maps.

2.D. Modules and Enveloping Algebras.

If we pick a module $M$ of $L$, we can do a similar trick to assign an endomorphism of $M$ to each chord diagram.  I won’t go into the details, but the idea is to think of the part of the arc diagram that used to be the outer circle as the ‘time line’ of the module, and all the other edges as copies of the Lie algebra.  This again gives an algebra map from the chord algebra $A$ to $End(M)$.

Instead of doing this whole procedure over again for each module, it’s probably wiser to do it once in a universal way.  That is, we want to find a map from $A$ to the universal enveloping algebra $\mathcal{U}(L)$ of $L$ such that its composition with the representation map $\mathcal{U}(L)\rightarrow End(M)$ is the map we computed in the previous paragraph.

This is possible, by taking a chord diagram, and just throwing away the outer circle.  Then, take all the free ends and treat them as outputs.  They are cyclically ordered, so this just depends on a choice of which one is first (and this turns out to be irrelevant).  This is a Jacobi diagram with no inputs and many outputs, so it determines a map from $k$ to  $L^{\otimes n}$ for some n.  By looking at the image of 1, this defines an element in $L^{\otimes n}$.  This element naturally sits inside the universal enveloping algebra.  Thus, we can assign an element to each chord diagram.

It is a corollary of the Key Lemma from before that this homomorphism sends $A$ to the center of $\mathcal{U}(L)$.  Also a notable fact is that the theta chord diagram (the one shaped like $\Theta$) goes to the Casimir element.

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### 3 Responses to “Chord Diagrams and Lie Algebras”

1. Predrag Cvitanovic Says:

I am very glad to see diagrammatic methods getting some traction, and hate to sound like a nagging wife, but you are sure this is exclusively Penrose notation? I am a great fan of Roger Penrose myself, but if you have seen the diagrammatic Jacobi identity in the form you write here in a publication earlier than 1976, I would love to see it.

I have written down what I know about this in
Section 4.9 A brief history of birdtracks
of birdtracks.eu [mirror birdtracks.dk]

happy chordin’
Predrag

2. stostawlecerne Says:

nice work, dude

3. math for kids Says:

math for kids…

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