Rigidity of some quantum representations of mapping class groups

I am attending the geometry seminar. The speaker is Pierre Godfard.

The abstract is here.

The topics to be covered are

Mapping class groups and rigidity conjectures
Quantum represnetations and main results
Twisted moduli space of curves (no time)

We take a surface $S_{g}^{n}$ of genus $g$ and $n$ components. The mapping class group is \[\begin{equation} \mathrm{Mod}(S_{g}^{n}) = \pi_{0}( \mathrm{Homeo}^{+}(S_{g}^{n}, \underbrace{\delta S_{g}^{n}}_{\text{Fix the boundary pointwise}})). \end{equation}\]

When $g \geq 1$ and $n=0$, we have \[\begin{equation} \mathrm{Mod}(S_{g}^{n}) = \mathrm{Out}^{+}(\pi_{1}(S_{g})) \simeq \pi_{1}(\underbrace{\mathcal{M}_{g}}_{\text{ Moduli of curces }}) \end{equation}\]

$\mathrm{Mod}(S_{0}^{n})$ is a quotient of $\underbrace{PB_{n}}_{\text{ pure braid group }} \times \mathbb{Z}^{n}$

We have analogies with $\mathop{\mathrm{SL}}_{n}(\mathbb{Z})$ and $\mathrm{Aut}(\underbrace{F_{n}}_{\text{ Free group on $m$ generators }})$.

We’ll be concerned with representations of $\mathrm{Mod}(S_{g}^{n})$. Unlike for $\mathop{\mathrm{SL}}_{n}(\mathbb{Z})$, the representation theory is more wild because those quantum representations exist

For a group $G$, Kahzdan’s property (T) is the following.

For all unitary representations $\rho$ of $G$ in a Hilbert space $H^{1}(G,W) = 0$.

Here are some known results.

$\mathop{\mathrm{SL}}_{n}(\mathbb{Z}), n \geq 3$ proved by Kahzdan in 1967
$\mathrm{Aut}(F_{n})$, $n \geq 5$ proved by Kaluba et al 2019, 2021. The proof is computer assisted. The method is due to Ozawa.
$\mathrm{Mod}(S_{g}^{n})$ does not have property (T) for $g \leq 2$. For $\geq 3$, it is open and some people expect it.

What would property (T) say about finite dimensional representations?

The most studied consequence of property (T) is the following.

(Ivanov)

For $g \geq 3$, for all $\rho$ finite image, one has $H^{1}( \mathrm{Mod}(S_{g}^{n}), \rho) = 0$.

This is known for $\ker \rho$ containing some known subgroups of $\mathrm{Mod}(S_{g}^{n})$ by Ershov-He’ 17.

For a representations $\rho: G \rightarrow \mathop{\mathrm{GL}}_{n}(\mathbb{C})$, $H^{1}(G, \mathrm{ad}\rho)$ classifies deformations of $\rho$. If it is trivial, then \[\begin{equation} \text{ for all } \rho_{\varepsilon} \in G \times (-\varepsilon,\varepsilon) \rightarrow \mathop{\mathrm{GL}}_{m}(\mathbb{C}) \end{equation}\] with $\rho_{0}=\rho$, $\rho_{\varepsilon}$ is conjugate to $\rho$ for all $\varepsilon$.

Property (T) will imply more generally the finite dimensional property (T) conjecture. This says that \[\begin{equation} \text{ for all } \rho: \mathrm{Mod}(S_{g}^{n}) \rightarrow \mathbb{U}_{d}, \end{equation}\] for all $g \geq 3$, $H^{1}(\mathrm{Mod}(S_{g}^{n}),\rho)=0$.

In particular, \[\begin{equation} \text{ for all } \rho: \mathrm{Mod}(S_{g}^{n}) \rightarrow P\mathbb{U}_{d}, g \geq 3 \end{equation}\] we would expect $H^{1}(\mathrm{Mod}(S_{g}^{n}), \mathrm{ad}\rho)=0$.

Then some strange correspondence happens.

Fusion rules

For $S_{g'}^{n} \subseteq S_{g}^{n}$ , one has \[\begin{equation} \mathrm{Mod}(S_{g'}^{n'}) \rightarrow^{P} \mathrm{Mod}(S_{g}^{n}) , \end{equation}\] then $\rho^{*} P g(\lambda_{1},\dots,\lambda_{n})$ is a $\oplus$ of representations of the form $\rho_{g'}( u_{1}',\dots,u'_{n'})$ in a prescribed manner.

The data of the $\rho_{g}(\lambda_{1},\dots,\lambda_{n})_{g, \lambda_{1},\dots, \lambda_{n}}$ together with the fusion rules is called a modular function. This data is equivalent to that of a modular category.

Let $G$

be a finite group and let $\mathcal{E} = \mathcal{Z}(\underbrace{\mathop{\mathrm{Rep}}G}_{\text{ finite dimensional rep of $G$}})$. Then $\rho_{g}$ is $\mathbb{C}[\underbrace{\mathrm{Hom}(\pi_{1}(S_{g}),G)/G}_{\text{ upto conjugation }}]$

Very different class of examples
$SU(2)$ modular categories.

For every $l \geq 3$, the object $\mathcal{E}_{\mathop{\mathrm{SU}}(2),l}$ are related to quantum representations. When $l$ is a prime number, all the $\rho_{g}(\lambda_{1},\dots,\lambda_{m})$ are irreducible. This is due to Roberts’ 01. For $g \geq 2$, $\rho_{g}$ has an infinite image.

Here is now the main result:

(G 2025)

For $\mathcal{E} = \mathcal{E}_{\mathop{\mathrm{SU}}(2),l}$ and $l$ prime, for $g \geq 7$, $\rho_{g}$ is rigid and $H^{1}(\mathrm{Mod}(S_{g}, \mathrm{ad}\rho_{g}) = 0$.

The rest of the talk was the sketch of this theorem. I will write some keywords that I could pick up.

Use deformation of representations. $\{ \text{ deformations of $\varphi$ }\}$ upto isometry are trivial.
Ocneanu rigidity. What does it mean?: It means that quantum representation do not have deformations that are quantum representations. In terms of fusion rules, for any colection $\rho_{g}(\lambda_{1},\dots,\lambda_{n})_{g, \lambda_{1},\dots, \lambda_{n}}$ satisfying the fusion rules, one has \[\begin{equation} \tilde{\rho}_{g}(\lambda_{1},\dots,\lambda_) \simeq \rho_{g}(\lambda_{1},\dots,\lambda_{n}). \end{equation}\]

Then we follow the following picture.

Comments:

Why only $\mathop{\mathrm{SU}}(2)$? Because irreducibility of the $\rho_{g}(\lambda_{1},\dots,\lambda_{n})$ is used?
We used partial homological stability of degree 1. What about other degrees?

This page was updated on November 28, 2025.
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