There is a well-studied action of $\operatorname{Gal}(\bar{\mathbb Q}/\mathbb Q)$ on (some version of) the $E_2$ operad; see for example this MO question.

Modular tensor categories are examples of $E_2$-algebra objects in $\mathrm{Cat}$, the 2-category of linear categories.

How does the $\operatorname{Gal}(\bar{\mathbb Q}/\mathbb Q)$-action on $E_2$ act on the set of MTCs? (I assume it does preserve modularity.)

To be pedantic, I mean the following. Suppose $(\mathcal C,\otimes,\dots)$ is an MTC. Then we have an action of $E_2$ on the underlying category $\mathcal C$, which I could write as $(\otimes,\dots) : E_2 \to \mathrm{End}(\mathcal C)$. Now choose $\gamma \in \operatorname{Gal}(\bar{\mathbb Q}/\mathbb Q)$, and use it to define an automorphism $\gamma : E_2 \to E_2$. Define a new MTC $\mathcal C^\gamma$ by declaring that the underlying category $\mathcal C$ is unchanged, but the braided tensor structure is modified by $\gamma$ to $(\otimes,\dots) \circ \gamma : E_2 \to E_2 \to \mathrm{End}(\mathcal C)$. I assume that in general $\mathcal{C} \not\cong \mathcal{C}^\gamma$ as braided monoidal categories. So what is the braided monoidal structure on $\mathcal{C}^\gamma$?

In case it makes it easier, I care most about the case where $\gamma$ is in the cyclotomic Galois group $\operatorname{Gal}(\mathbb Q^{cyc}/\mathbb Q)$. Actually, most (all?) MTCs over $\mathbb C$ in fact can be defined over $\mathbb Q^{cyc}$. So it wouldn't surprise me if $\mathcal{C}^\gamma$ only depends on the image of $\gamma$ along the map $\operatorname{Gal}(\bar{\mathbb Q}/\mathbb Q) \to \operatorname{Gal}(\mathbb Q^{cyc}/\mathbb Q)$.

  • 5
    $\begingroup$ As far as I know, there is such an action for the profinite and pro-$\ell$ completion of $E_2$. It's unclear to me how to get an interesting action on MTC's from that. $\endgroup$
    – Adrien
    Jun 26, 2018 at 18:30
  • 1
    $\begingroup$ There are some weird "profinite" facts about MTCs, for example that the mapping class group action of the torus factors through a finite group and that the twists are roots of unity. On the other hand, the braid group images are typically infinite. $\endgroup$ Jun 26, 2018 at 18:53
  • 1
    $\begingroup$ @NoahSnyder Are you suggesting that, even though the Galois group acts merely on a completion of $E_2$, that's OK because that completion has a good chance of acting on MTCs? $\endgroup$ Jun 26, 2018 at 19:24
  • 2
    $\begingroup$ As Adrien says there is no Galois action on E_2 itself, you need to complete it somehow. What acts in a transparent manner on k-linear MTCs is the Grothendieck-Teichmuller group GT(k). If $k=\mathbb Q_\ell$ then the absolute Galois group sits inside GT(k) so you get a Galois action in this case. I'm not sure what the profinite Grothendieck-Teichmuller group acts on (braided monoidal categories enriched in profinite sets?). $\endgroup$ Jun 26, 2018 at 19:46
  • 1
    $\begingroup$ If the action of the Galois group on some MTC factors through the cyclotomic character, then the action must be very simple: the action does not affect the associativity isomorphism of the braided structure at all, only the symmetry. Specifically, the "full twist" is multiplied with a scalar given by the cyclotomic character. $\endgroup$ Jun 26, 2018 at 19:48

1 Answer 1


The Galois action on the profinite or pro-$\ell$ completion of $E_2$ comes, as Dan says, from an action of the corresponding version of the Grothendieck-Teichmüller group. However, to the best of my knowledge none of those acts on MTC's. Rather, they act on braided tensor categories which are themselves pro-finite or pro-unipotent in an appropriate sense. However, even in that case although the action might be higly non-trivial this still produces a braided tensor category equivalent to the one you started with (at least if your category is actually ribbon, but I thin this is true in general).

For the pro-algebraic version $GT(k)$, $k$ a field a characteristic 0, those can roughly speaking be identified with those braided tensor categories over $k[[\hbar]]$ which are symmetric monoidal modulo $\hbar$. This includes, e.g., finite dimensional modules over the formal version $U_{\hbar}(\mathfrak g)$ of quantum groups, but typically not their rational or specialized versions.

The Galois action you get for $k=\mathbb{Q}_\ell$ is discussed in some extent at the end of Kassel-Rosso-Turaev's book "Quantum groups and knot invariants", and in Furusho's paper Galois action on knots II: Proalgebraic string links and knots.

  • $\begingroup$ Thanks. I'll take a look at those references. I'm slightly confused by your first paragraph. I don't know what a pro-finite or pro-unipotent braided tensor category is --- MTCs are very finite, but I take it that they are finite in some other way? And it seems strange that the action would take all braided tensor categories to equivalent ones --- this seems to conflict with the claim that the action is nontrivial? $\endgroup$ Jun 29, 2018 at 1:43
  • $\begingroup$ Sorry if this was unclear, I just meant something fairly tautological, essentially by definition it means the representation of the pure braid group you get lands in a (pro-)finite or (pro-)unipotent group. MTC's are very finite-dimensional but that's something different. $\endgroup$
    – Adrien
    Jun 29, 2018 at 8:41
  • $\begingroup$ As for the non-triviality of the action, basically in the pro-unipotent case this comes from the GT action on Drinfeld associator, which is free and transitive (which is what I mean by highly non-trivial). Yet any two associators are twist-equivalent if you are allowed string links and not just braids (this is an old result of Drinfeld and Le-Murakami, explained in Furusho's paper). So I'd say the action on the set of equivalences classes of pro-unipotent ribbon categories is trivial, but the action on an individual one is not. $\endgroup$
    – Adrien
    Jun 29, 2018 at 8:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.