# The Unapologetic Mathematician

## The Temperley-Lieb Category

Okay, after last week’s shake-ups I’m ready to get back into the swing of things. I mentioned yesterday something called the “Temperley-Lieb Category”, and it just so happens we’re right on schedule to explain it properly.

We’ve seen the category of braids and how the braided coherence theorem makes it the “free braided monoidal category on one object”. That is, it has exactly the structure needed for a braided monoidal category — no more, no less — and if we pick any object in another such category $\mathcal{C}$ we get a unique functor from $\mathcal{B}raid$ to $\mathcal{C}$.

So of course we want the same sort of thing for monoidal categories with duals. We’ll even draw the same sorts of pictures. A point on a horizontal line will be our generating object, but we also need a dual object. So specifically we’ll think of the object as being “going up through the point” and the dual as “going down through the point”. Then we can draw cups and caps to connect an upward line and a downward line and interpret it as a duality map. Notice, though, that we can’t make any curves cross each other because we have no braiding! Here’s an example of such a Temperley-Lieb diagram:

Again, we read this from bottom to top, and from left to right. On the bottom line we have a downward line followed by an upward line, which means we start at the object $X^*\otimes X$. Then we pass through a cap, which corresponds to the transformation $\epsilon_{X^*}$. Then we go through a cup ($\eta_X^*$) to get to $X\otimes X^*$, and another cup to get to $X\otimes X^*\otimes X\otimes X^*$. A cap in the middle ($\epsilon_{X^*}$) is followed by a cup ($\eta_X$), and then another pair of caps ($\epsilon_X\otimes\epsilon_X$). Then we have a cup $\eta_{X^*}$ and another $\eta_X$ to end up at $X\otimes X^*\otimes X\otimes X^*$.

We could simplify this a bit by cancelling two cup/cap pairs using the equations we imposed on the natural transformations $\eta$ and $\epsilon$. In fact, this is probably a much easier way to remember what those equations mean. The equations tell us in algebraic terms that we can cancel off a neighboring cup and cap, while the topology of diagram says that we can straighten out a zig-zag.

Incidentally, one feature that’s missing from this diagram is that it’s entirely possible to have an arc (pointing either way) start at the bottom of the diagram and leave at the top.

Now if we have any category $\mathcal{C}$ with duals and an object $C$ we can build a unique functor from the category $\mathcal{OTL}$ of oriented Temperley-Lieb diagrams to $\mathcal{C}$ sending the upwards-oriented line to the object $C$. It sends the above diagram (for example) to the morphism

$\left[(1_C\otimes\eta_C\otimes1_{C^*})\circ\eta_{C^*}\circ(\epsilon_C\otimes\epsilon_C)\circ(1_C\otimes\eta_C\otimes1_{C^*})\circ(1_C\otimes\epsilon_{C^*}\otimes1_{C^*})\circ(1_C\otimes1_C^*\otimes\eta_{C^*})\circ\eta_{C^*}\circ\epsilon_{C^*}\right]:$
$C^*\otimes C\rightarrow C\otimes C^*\otimes C\otimes C^*$.

Another useful category is the free monoidal category with duals on a single self-dual object. This is the Temperley-Lieb category $\mathcal{TL}$ which looks just the same as $\mathcal{OTL}$ with one crucial difference: since the object $X$ is its own dual, we can’t tell the difference between the two different directions a line could go. Up and down are the same thing. In the algebra this might seem a little odd, but in the diagram all it means is we get to drop the little arrows that tell us which way to go.

And now if we have any category $\mathcal{C}$ with duals and any self-dual object $C=C^*$ we have a unique functor from $\mathcal{TL}$ to $\mathcal{C}$ sending the strand to $\mathcal{C}$. This is how Temperley-Lieb diagrams are turned into (categorified) $\mathfrak{sl}_2$ representations in Khovanov homology.

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Posted by John Armstrong | Category theory, Knot theory

## 7 Comments	»

1. [...] odd. These diagrams look an awful lot like Temperley-Lieb diagrams. And in fact they are! In fact, we get a functor from to that sends to . That is, a [...]

Pingback by | July 26, 2007 | Reply

2. Nice post.

Do you have a reference for this result?

Thanks,

Bas

Comment by Bas | August 15, 2008 | Reply

3. Thanks, Bas. Unfortunately, I don’t really know a good reference offhand. I have a sneaking suspicion John Baez probably does, though.

Comment by | August 15, 2008 | Reply

4. Which result? That oriented tangles in the plane form the free braided monoidal with duals on one object?

I wrote about this result as a baby special case of the “tangle hypothesis” in a paper with James Dolan. See page 25, the paragraph beginning “Moving down the n = 1 column to k = 1…” I cite a paper by Freyd and Yetter for this result. However, if proper attribution matters to you, also check out Joyal and Street’s paper “The Geometry of Tensor Calculus I”, which contains closely related results. The never-finished Geometry of Tensor Calculus II is also worth a look.

I have not carefully read the predessors to this post, so I’m not sure how John is handling this issue, but it’s worth noting that to get all the oriented tangles in the plane, we need the object X to have an object X* which is both a left dual and a right dual of X. In other words, we need X** to be isomorphic, or perhaps equal, to X. This seems to be implicit in what John write above.

In a braided monoidal category, all this is automatic: we can use the braiding to show any left dual is a right dual. In a mere monoidal category, this is not the case. Joyal and Street handle this by introducing the notion of “pivotal” category, while I handle it by introducing duals not only for objects but also for morphisms. The latter solution seems to generalize nicely to more complicated situations.

The papers I’m giving links to explain this issue in more detail.

Comment by | August 17, 2008 | Reply

5. Okay, now I’ve gone back to John’s post on monoidal categories with duals to see how he handles the nuance I mentioned. I see he insists that in a monoidal category with duals, every object X is isomorphic to X**. One has to be a bit careful here: without a specified isomorphism between X and X**, there is no way to say which morphism corresponds to the closed loop in the above picture.

The cheapest solution is to say that X** is equal to X. This works in the tangle example above, but of course it’s not true in the category of finite-dimensional vector spaces. So, at this point one starts wanting to read the definition of “pivotal” category given on page 12 here.

Comment by | August 17, 2008 | Reply

6. Yeah, I talked about left and right duals, and I assumed before this point the isomorphism you mention.

Comment by | August 17, 2008 | Reply

7. I wrote:

Which result? That oriented tangles in the plane form the free braided monoidal with duals on one object?

I meant “oriented tangles in the plane form the free monoidal category with duals on one object.”

Comment by | August 18, 2008 | Reply

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