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Hi,

This question came up to me when reading the paper of Cartier on Vassiliev invariants, but it can probably be turned into a more general question.

Let $T$ be the category whose objects are finite sequences of $\{+,-\}$ (including the empty one) and whose morphisms are framed, oriented tangles. In particular, $End(\emptyset)$ is the set of framed oriented links. It's quite well known that $T$ is the free ribbon category on one object, hence if $C$ is a ribbon category and $V \in C$ then there is a functor $F:T\rightarrow C$ mapping $+$ to $V$ and $-$ to $V^*$. In particular, $F$ associate to any link viewed as an element of $End(\emptyset)$ an automorphism of the unit object of $C$ (with respect to the monoidal structure of $C$).

Cartier applies this construction in the case $C$ is the ribbon category having the same objects as $T$, whose morphisms are $k[[\hbar]]$-linear combinations of chords diagrams, and whose ribbon structure comes essentially from a Drinfeld associator. The point is that he doesn't apply the above construction to $End(\emptyset)$ but rather to $End(+)$, and claims that its restriction to knots is a universal finite type invariant.

It's quite clear that the closure operation identifies $End(+)$ with the set of links with a distinguished component. In particular it also contains the set of knots, which explains that his claim makes sense. But still, I'm wondering

Why did he make this particular choice ?

There is, of course, an obivous answer. The product on $End(+)$ can be identified with the connected sum of links along the distinguished component. Hence, it contains the set of knots as a sub-monoid and not only as a subset. As a nice consequence, the invariant he constructs is compatible with the connected sum of knots, which is of course a desirable feature.

It seems quite obvious to me that this construction applied to $End(\emptyset)$ also gives a universal finite type invariant, just with a different normalization. I would also say that it's a general fact that for an arbitrary ribbon category these invariants are more or less the same.

There is still a problem if you want to realize this invariant. For example if $\mathfrak g$ is a simple Lie algebra, a choice of $t \in S^2(\mathfrak g)^{\mathfrak g}$ leads to an infinitesimal symmetric monoidal category structure on $\mathfrak g$-mod. Putting again a Drinfeld associator you get a ribbon category structure on $\mathfrak g[[\hbar]]$-mod, hence a functor from $T$ for each $V \in \mathfrak g$-mod which factor through Cartier's category of chord diagramm.

Now the usual RT construction leads to a numerical invariant, while Cartier's one leads to an endomorphism of $V$. It's not a big deal, one can take its quantum trace in order to get something numerical, but it's somehow artificial since this endomorphism is already an invariant, although I'm pretty convinced that no information is lost in doing that, for the reason explained above. But taking the usual trace instead of the "quantum" one, for example, also leads to a (less natural, but still) invariant in that case...

Well, it's probably not a real subtle problem but rather a technical detail, but I'm wondering if there is a natural way to handle/understand this, or a "philosophical" picture explaining the relations between these invariants.

Edit: This paper of Nathan Geer deals with the situations mentionned by Theo and explain that the cut open invariant may be non trivial even if the quantum trace vanish and that's an interesting problem to see how it can be turned into an invariant which doesn't depends on the choice of the distinguished component.

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Two quick comments. First, if V is simple (which it often is in applications), then End(V) is canonically identified with the scalars via Schur's lemma. Second, sometimes the quantum dimension of V is 0, in which case you get substantially more information from the cut open invariant. –  Noah Snyder Oct 27 '11 at 16:35
    
Thanks, I agree that if V is simple it works fine, and in that case you get a numerical invariant is itself compatible with the connected sum. Note however that in the context of finite type invariant you want to prove something like "every weight system can be integrated to a knot invariant", and it seems to me that you also need non simple object. –  Adrien Oct 27 '11 at 17:33
    
For your second comment, the quantum dimension being 0 or not, it's always true that an endormophism could potentially carry more information than a scalar, but it's not obvious to me that it's indeed the case in this context. Do you have an example ? –  Adrien Oct 27 '11 at 17:35
    
@Adrien: Note that if a simple module has zero (quantum) dimension, then all traces are $0$, and so the closed invariant has no information at all. Standard examples of simple zero-dimensional modules arrise when working in characteristic $p$, when working with quantum groups at roots of unity, and when working with (maybe quantum) super Lie algebras. I don't know the details, but I do know that some of the theory is developed by Nathan Geer. Development is important, because you do want an invariant that doesn't care about which component is distinguished. –  Theo Johnson-Freyd Oct 27 '11 at 21:14
    
@Theo: I agree of course that all trace are 0, it was just not obvious to me that in such a situation the cut open invariant is non constant. I'm aware of the situations you mention, but don't know so much about them.. –  Adrien Oct 27 '11 at 21:43
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