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  Why are Witten-Reshetikhin-Turaev invariants expected to be integral?

+ 8 like - 0 dislike

A Witten-Reshetikhin-Turaev (WRT) Invariant $\tau_{M,L}^G(\xi)\in\mathbb{C}$ is an invariant of closed oriented 3-manifold $M$ containing a framed link $L$, where $G$ is a simple Lie group, and $\xi$ is a root of unity. Components of $L$ are coloured by finite dimensional $G$-modules.

Since Murakami in 1995, people have been proving integrality results for increasingly general classed of simple Lie groups $G$ (first $SO(3)$ then $SU(n)$, then any compact simple Lie group), roots of unity $\xi$ (first prime, then non-prime), and $3$--manifolds $M$ (first integral homology $3$-sphere, then rational homology $3$-sphere, then the general case). These results usually state that the WRT invariant is an algebraic integer- an element of $\mathbb{Z}[\xi]$- or the stronger result that it's the evaluation at $\xi$ of an element in the Habiro ring.

Papers on the integrality of WRT invariants usually list wonderful things that can be done once integrality properties are established (e.g. integral TQFT or categorification or representations over $\mathbb{Z}$ of the mapping class group). But why would we expect $\tau_{M,L}^G(\xi)$ to be an element of $\mathbb{Z}[\xi]$? As far as I know that's always been the case. Is there some perhaps some not-quite-rigourous construction of WRT invariants which takes place entirely over $\mathbb{Z}[\xi]$, or maybe over the Habiro ring?

Question: Why are WRT invariants expected to be algebraic integers? Is there a conceptual explanation for their integrality? Are all WRT invariants in fact expected to come from analytic functions over roots of unity (i.e. elements of the Habiro ring), and if so, why?
This post imported from StackExchange MathOverflow at 2015-01-02 16:47 (UTC), posted by SE-user Daniel Moskovich
asked Jul 19, 2014 in Theoretical Physics by Daniel Moskovich (130 points) [ no revision ]
retagged Jan 2, 2015
Does there exist a conceptual explanation for integrality even for the Jones polynomial itself, without mentioning Khovanov homology? (and does that count?)

This post imported from StackExchange MathOverflow at 2015-01-02 16:47 (UTC), posted by SE-user Sam Lewallen
For the Jones Polynomial, as you mention, the conceptual explanation might be that we have alternative (non-TQFT) constructions in which integrality is evident.

This post imported from StackExchange MathOverflow at 2015-01-02 16:47 (UTC), posted by SE-user Daniel Moskovich

1 Answer

+ 6 like - 0 dislike

My humble point of view is that the Witten-Reshetikin-Turaev invariant (at least for $G=SU(2)$ or $SO(3)$) at the root $\xi$ is (by definition) a rational function on $\xi$ which looks very much like a polynomial with integer coefficients. (Note that the function depends on $\xi$.)

When $M=S^3$, the rational function is indeed a polynomial with integer coefficients that does not depend on $\xi$: that's the (colored) Jones polynomial. So the Witten-Reshetikin-Turaev invariant of $L\subset S^3$ lies in $\mathbb Z[\xi]$ in that case. And it is natural to ask whether this holds for general $M$.

When $M\neq S^3$ the rational function is no longer a polynomial and moreover it depends of $\xi$ (but this dependence is not an issue concerning integrality). The numerator of this rational function is an arbitrary element in $\mathbb Z[\xi]$, but the denominator is not arbitrary: it is just a product of quantum integers $$[n] = \frac{\xi^n - \xi^{-n}}{\xi - \xi^{-1}} = \xi^{n-1} + \xi^{n-3} + \ldots + \xi^{-n+1}.$$ Hence the invariant is "almost" a polynomial with integer coefficients: it is an element in $\mathbb Z[\xi]$ divided by a product of quantum integers like $[2]^3[4]^2[7]$. (I am probably ignoring some renormalisation factor.)

As far as I remember, when $\xi$ is a $4r$-th root of unity and $r$ is prime, then every $[n]$ is actually invertible in $\mathbb Z[\xi]$, and hence the invariant indeed lies in $\mathbb Z[\xi]$.

This post imported from StackExchange MathOverflow at 2015-01-02 16:47 (UTC), posted by SE-user Bruno Martelli
answered Jul 20, 2014 by Bruno Martelli (60 points) [ no revision ]

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