Questions tagged [property-t]

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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(...
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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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16 votes
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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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1 answer
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Residual finiteness of random groups with property (T)

A well known result of A. Zuk states that for $\frac{1}{3} < d < \frac{1}{2}$, a random group $\Gamma$ with respect to Gromov's density model with density $d$ has Kazhdan's property (T) with ...
pitariver's user avatar
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7 votes
1 answer
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Properties (T) and (FA)

I have been thinking a lot recently about Property (T) and Property (FA) for discrete groups. I understand that the prior implies the latter, but not the other way around, and I have also seen one or ...
NoName's user avatar
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1 answer
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Kazhdan's property (T) for $\tilde{C}_2$-lattices

It is known that higher rank lattices have property (T) and also that lattices on 2-dimensional Euclidean buildings have property (T) provided the thickness $q+1$ of the building is large enough (...
Stefan Witzel's user avatar
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
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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
14 votes
0 answers
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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
9 votes
3 answers
754 views

Is there a one relator group with property (T)?

Is there a one-relator group with property (T)? That is, is there an $n > 2$, and some $x \in F_n$ (the free group on $n$ generators) such that the quotient of $F_n$ by the normal subgroup ...
Pablo's user avatar
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6 votes
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Do discrete groups with property $(T)$ have "modest" subgroup growth?

I saw it conjectured at http://www.mathunion.org/ICM/ICM1994.1/Main/icm1994.1.0309.0317.ocr.pdf that "discrete subgroups with property $(T)$ may have modest subgroup growth." (Page 5, directly above ...
Lorenzo'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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101 votes
4 answers
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How feasible is it to prove Kazhdan's property (T) by a computer?

Recently, I have proved that Kazhdan's property (T) is theoretically provable by computers (arXiv:1312.5431, explained below), but I'm quite lame with computers and have no idea what they actually can ...
Narutaka OZAWA's user avatar
3 votes
1 answer
214 views

Uniform bounds on Kazhdan constants in groups

Does there exist a finitely generated discrete group $G$ such that it has property (T), but for every $\varepsilon > 0$ there exists a generating set $S$ with the corresponding Kazhdan constant ...
Michal Kotowski's user avatar