Questions tagged [property-t]

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What is the current status of the question of whether or not the mapping class group has Kazhdan's Property (T)?

$\DeclareMathOperator\Mod{Mod}$Let $\Mod(S)$ be the mapping class group of a closed oriented surface $S$ of genus at least $3$. My question is easy to state: is it currently known whether or not $\...
Thomas's user avatar
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14 votes
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On uniform Kazhdan's property (T)

For a finitely generated group $\Gamma$ and its finite generating subset $S$, the Kazhdan constant $\kappa(\Gamma,S)$ is defined to be $$\kappa(\Gamma,S)=\inf_{\pi,v} \max_{g\in S} \| v - \pi_g v \|,$...
Narutaka OZAWA's user avatar
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Does anybody know if the Fourier algebra of SL(3,Z) has an approximate identity?

(Note to those who like to tidy LaTeX, or ${\rm \LaTeX}$: I kindly request that you don't put any LaTeX in the title of this question, nor change the bolds below to blackboard bold.)$\newcommand{\FA}{{...
Yemon Choi's user avatar
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3 votes
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Property $(T)$ for $\mathrm{SL}_2(\mathbb{Z}) \ltimes \mathbb{Z}^2$

(This is in part a request for references and in part a somewhat pedagogical question.) I gave a course on expanders seven years ago, and I am giving a course on expanders again now. We will soon do ...
H A Helfgott's user avatar
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3 votes
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Kazhdan Property T of semisimple Lie groups

I am reading the paper [Margulis, G. A.; Nevo, A.; Stein, E. M., Analogs of Wiener's ergodic theorems for semisimple Lie groups. II. Duke Math. J. 103 (2000), no. 2, 233–259] (MSN). I want to ...
A beginner mathmatician's user avatar
3 votes
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Relative property (T) and normal closure

I am in a situation where a discrete, finitely generated group $H$ satisfies property (T), and was wondering if I was able to conclude anything about the pair $(G,H^G)$, where $G$ is a finitely ...
Anonymous's user avatar
2 votes
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Strong converse of Kazhdan's property (T)

In his 1972 paper Sur la cohomologie des groupes topologiques II, Guichardet proved$^\ast$ that (non-abelian) free groups satisfy the following strong converse of property (T): The $1$-cohomology $H^1(...
MaoWao's user avatar
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