# How to calculate the Witten-Reshetikhin-Turaev invariants from a triangulation?

I'm interested in the Witten-Reshetikhin-Turaev invariants for 3-manifolds, and in particular, how to calculate them from a triangulation of the 3-manifold (recall that as they were first introduced, their calculation is based on a surgery diagram for the manifold). Is there any some sort of "state sum" model giving the WRT invariant for a $3$-manifold, which is based on the triangulation of the $3$-manifold?

I am well aware of the Turaev-Viro construction, and it is exactly the type of definition I'm hoping for, but of course the TV invariant is $\left|\cdot\right|^2$ of the WRT invariant, not the WRT invariant itself.

The simpler the construction, the better. Certainly one can go algorithmically from a triangulation to a surgery diagram, but I want something which is conceptually based on the triangulation.

I guess any answer I get will also include the case for a pair $(Y^3,L)$, and even perhaps allow for $M^3$ to have nonempty boundary (i.e. the full tqft). However, I will be satisfied even if I just get a construction for a closed $3$-manifold $M$.

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This is a well-known open problem. (minor point: you can go algorithmically from surgery presentations to triangulations, but not the other way so easily). –  Daniel Moskovich Jun 22 '11 at 10:33
If you let the 3-manifold bound a 4-manifold and extend the triangulation, it's not an open problem. A version of the Crane-Yetter state sum calculates the WRT invariant in that case. See my answer below. –  Kevin Walker Jun 22 '11 at 21:50

This can be done if you extend the triangulation of your 3-manifold $M$ to a triangulation of a 4-manifold $W$ whose boundary is $M$. You can find the basic idea in Chapter 9 of these draft notes on my web page, and there are very explicit details for the associated Hilbert space in this arXiv preprint (with Z. Wang). (In the latter paper we use cubes instead of simplices to make computer implementation easier.) There's also a summary in the notes from a talk.

The main idea is that the Witten-Reshetikhin-Turaev TQFT can be reinterpreted as a 3+1-dimensional TQFT, and the 3+1-dimensional version is "fully extended" (goes all the way down to points). Once you have a fully extended TQFT, standard techniques allow you to construct a state sum model. In this case, the state sum you get is a modified version of the Crane-Yetter state sum. (You would get a Turaev-Viro state sum for a fully extended 2+1 dimensional TQFT.)

EDIT:

If you don't want to wade through the above sources, here's the explicit formula.

Let $C$ be a modular ribbon category. Let $D= \sqrt{\sum_i d_i^2}$, the square root of the sum of the squares of the quantum dimensions. Choose a handle decomposition of $W^4$. Keep in mind is the case where the $i$-handles are thickenings of the $i$-cells of some cell decomposition of $W$, such as the dual cell decomposition to a triangulation. Let ${\cal H}_i$ be the set of $i$-handles. Let ${\cal L}_2$ be the set of labelings of the 2-handles by simple objects of $C$. For fixed $\alpha\in{\cal L}_2$, let ${\cal L}_1(\alpha)$ denote the set of labelings of the 1-handles by orthogonal basis elements of the associated vertex spaces. (To each 1-handle is associated a $C$-picture on the linking 2-sphere, that is, a collection $c$ of ribbon endpoints on the 2-sphere labeled by simple objects $a_1\otimes \cdots\otimes a_m$ according to $\alpha$. The "vertex space" is $\hom_C(1, a_1\otimes\cdots\otimes a_m)$ or, more canonically, the vector space associated to $(B^3; c)$, where we think of $B^3$ as the normal fiber to the core of the 1-handle.) Then $$Z(W^4) = \sum_{\alpha\in{\cal L}_2} \sum_{\beta\in{\cal L}_1(\alpha)} \prod_{h_4\in {\cal H}_4} D \prod_{h_3\in {\cal H}_3} D^{-1} \prod_{h_2\in {\cal H}_2} D^{-1} \,\text{Loop}(h_2, \alpha)\quad\quad\quad\quad$$ $$\quad\quad\quad\quad\prod_{h_1\in {\cal H}_1} D \,\text{Th}(h_1, \beta)^{-1} \prod_{h_0\in {\cal H}_0} D^{-1} \, \text{Link}(h_0, \alpha, \beta) ,$$

where

• $\text{Loop}(h_2, \alpha)$ is the loop value (quantum dimension) of the simple object $\alpha(h_2)$;
• $\text{Th}(h_1, \beta)$ is the evaluation of generalized theta graph with the two vertices labeled by $\beta(h_1)$ and $\overline{\beta(h_1)}$ (in the case of a generic cell decomposition where each 1-cell is incident to three 2-cells, this is just an ordinary theta graph); and
• $\text{Link}(h_0, \alpha, \beta)$ is the evaluation of the graph in $S^3$ corresponding to the link of the 0-handle $h_0$, labeled according to $\alpha$ and $\beta$.

If $W$ has a boundary $M$ then we can place a labeled link $L$ in the boundary, and the labels of the link play a role similar to the labels of the 2-handles above. In this case the state sum computes $WRT(M, L)$.

For a generic cell decomposition of $W$ (i.e. dual to a triangulation), the $\text{Link}(h_0, \alpha, \beta)$ factor is an evaluation of a labeled ribbon graph which looks like the 1-skeleton of a 4-simplex. If we resolve the five 4-valent vertices oin this graph into pairs of 3-valent vertices, then this becomes the "15-j symbol" used by Crane and Yetter. (Keep in mind that Crane and Yetter use a different normalization which obscures the relation to the WRT invariant.)

On the other hand, if we choose a handle decomposition of $W$ with a single 0-handle and several 2-handles (no 1- 3- or 4-handles), then it is easy to see that the above state sum reduces to the Reshetikhin-Turaev surgery formula. The factor $\text{Link}(h_0, \alpha, \beta)$ is the generalized Jones polynomial (evaluated at a root of unity) in this case.

A similar variant yields the Turaev "shadow" state sum.

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Is finding a triangulated 4-manifold with your 3-manifold as boundary substantially easier than just presenting it as surgery on a link, which is presumably what the author is what the author is trying to avoid? (Going from the former to the latter is just picking a Morse function...) –  Ben Webster Jun 22 '11 at 23:17
@Ben: No, not substantially easier so far as I know. But I didn't think that ease of calculation was the motivation for the question. There could be other reasons to prefer thinking in terms of triangulations. Locality, for example. –  Kevin Walker Jun 22 '11 at 23:26
@Turion: See the draft notes I link to in the answer above, especially the last paragraph of Chapter 6 (for the general case) and Chapters 8 and 9 (more specific examples). –  Kevin Walker Jul 25 '13 at 16:44
For $W^4$ closed, $Z(W^4) = C^{\sigma(W)}$, where $C$ is the exponentiated central charge (a root of unity) and $\sigma(W)$ is the signature of $W$. I'm assuming that the input is a modular category (rather than merely a premodular category) and that the Euler characteristic normalization has been chosen correctly. Crane and Yetter used a different normalization, which lead to an additional factor of $a^{\chi(W)}$. –  Kevin Walker Sep 19 '13 at 14:58
Yes, that's correct. $C$ depends only on $c$ modulo 8. –  Kevin Walker Sep 19 '13 at 18:55

What you're asking for is essentially equivalent to making Witten-Reshetikhin-Turaev theory into an extended topological quantum field theory (= "local field theory").

Modulo a whole bunch of stuff, this is what we have achieved in my recent paper with A. Bartels and C. Douglas. Of course, there is still A LOT to do before those results can be turned into an effective procedure for computing 3-manifold invariants. Also (and quite importantly), we don't know yet that the local field theories constructed in that paper are really the same as Witten-Reshetikhin-Turaev...

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Does an extended TQFT really give a way of computing invariants from triangulations? Don't you need an open-closed extended TQFT? Just think about 2 dimensions. –  Noah Snyder Jun 22 '11 at 20:51
Isn't "open-closed" the same thing as "extended"?... (please tell me what you have in mind, because I'm not quite following your comment) The real problem is that you need to have framings around (or p_1-structures). If we were just talking about oriented TQFT's then triangulations should be just fine as input data. –  André Henriques Jun 22 '11 at 21:40
Like André, I thought that an open-closed 2d TQFT was the same thing as a fully extended 2d TQFT. And I'm not sure what "open-closed" would mean in higher dimensions. I'm curious to hear what Noah had in mind. –  Kevin Walker Jun 22 '11 at 23:31
"Open-closed" means you allow a different kind of boundary which you're not allowed to glue along. This is definitely a different concept than extending. It's more closely related to allowing boundaries between two different TFTs, but where one of the TFTs is trivial. I'm still a little confused though about the relationship of each of these notions (extended and open-closed) to triangulations, so I'm going to think some more about that before saying anything. –  Noah Snyder Jun 23 '11 at 18:19
Thanks Noah. Some part of my brain knew that already, but apparently that part was not in communication with the parts which were doing the typing yesterday. So an extended 2d TQFT, plus some representations of the 1-category associated to a point (to label non-gluing boundaries), gives rise to an open-closed 2d TQFT. I suppose all open-closed TQFTs might arise this way, depending on the details of one's definition of an open-closed TQFT. –  Kevin Walker Jun 23 '11 at 20:14