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 $\operatorname{Gal}(\bar{\mathbb Q}/\mathbb Q) \to \operatorname{Gal}(\mathbb Q^{cyc}/\mathbb Q)$.

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ 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
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    $\begingroup$ 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

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