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I am wondering if there is some general relation between Kazhdan constants of a group and it finite index subgroups?

Let $G$ be a finitely generated group with a generating set $\Sigma$ that satisfies Kazhdan property (T) with constant $\kappa(G,\Sigma)$.

1) if $\Gamma$ is a finite index subgroup of $G$, is there a generating set $\Theta$ of $\Gamma$ for which $\kappa(\Gamma,\Theta)$ can be estimated in terms of $\kappa(G,\Sigma)$?

2) if $G$ is a finite index subgroup in $H$, is there a generating set $\Theta$ of $H$ such that $\kappa(H,\Theta)$ can be estimates in terms of $\kappa(G,\Sigma)$?

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  • $\begingroup$ It can depend how you define $\kappa$. One option is "every rep with a $\kappa$-invariant unit vector has an invariant unit vector". Another is "every rep with $\kappa$-invariant unit vector $\xi$ has an invariant unit vector $\kappa$-close to $\xi$" (or something of this kind). The first is shorter but the second is more natural and more flexible if you ask these kind of questions (esp. for 2). $\endgroup$ – YCor Feb 27 '18 at 8:39
  • $\begingroup$ The first one, that is the standard one. Regardless of the choice, can one say something for either of these? $\endgroup$ – duh Feb 27 '18 at 9:28
  • $\begingroup$ Then using the standard one will give a less natural statement for (2). Good luck! $\endgroup$ – YCor Feb 27 '18 at 10:26
  • $\begingroup$ Wait, so what is the natural statement for (2) and the second definition? $\endgroup$ – duh Feb 27 '18 at 12:09
  • $\begingroup$ It's suggested in my first comment (but I'm not sure of an optimal formulation). This little issue with dealing whether the invariant should be found close to the almost invariant vectors is a technical issue that occurs quite systematically when dealing with Property T (e.g., proving the equivalence with the the condition in terms of convergence of positive definite functions). $\endgroup$ – YCor Feb 27 '18 at 13:11
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If $n:=[G:H]$, then $\mathbb C[G] \subset M_n \mathbb C[H]$, where $g \in G$ maps to a permutation matrix decorated with elements from $H$ and the embedding depends essentially only on a choice of a transveral of the quotient map $G \to G/H$. One can arrange things, so that generators $S \subset G$ map to permutations with decorations of length at most $2[G:H]+1$. Those $[G:H]|S|$ decorations generate $H$, this is essentially the content of Schreier's lemma. Let's call this generating set $\Sigma$. We can also arrange that $\mathbb C[H] \subset \mathbb C[G] \subset M_n \mathbb C[H]$ is of the form $h \mapsto h \oplus h^\perp$, where the exact form of $h^{\perp} \in M_{n-1}(\mathbb C[H])$ depends on the situation.

The following proposition is well-known, the book of Bekka-de la Harpe-Valette is an excellent reference for all this.

Proposition 1.1.9 (Bekka-de la Harpe-Valette) If $(S,\varepsilon)$ is a Kazhdan pair for $G$ and $\delta>0$, then every $(S,\varepsilon\delta)$-invariant vector $\xi$ admits a $G$-invariant vector at distance less than $\delta\|\xi\|$.

We claim that if $(S,\varepsilon)$ is a Kazhdan pair for $G$, then $(\Sigma,[G:H]^{-1/2}\varepsilon)$ is a Kazhdan pair for $H$.

(1) Suppose that $G$ is a Kazhdan group and $(S,\varepsilon)$ be a Kazhdan pair. Let $\mathcal H$ be a unitary representation of $H$ and $\xi \in \mathcal H$ and assume that $\|h\xi-\xi\|<[G:H]^{-1/2}\varepsilon \|\xi\|$ for all $h \in \Sigma$. Then, we obtain a $G$-representation $\mathcal H^{\oplus n}$ from above and a vector $\eta = (\xi,...,\xi)$ that is $[G:H]^{-1/2}\varepsilon$-fixed by $S$. Thus, there must be a $G$-fixed vector $\eta_0 =(\eta_0^1,...,\eta_0^n)$ at distance less than $[G:H]^{-1/2} \|\eta\|$. It follows that $\eta_0^1$ is non-zero and $H$-fixed.

One can also argue similarly for (2), but it is less clear what the optimal bound would be.

(2) Now suppose that $H$ is a Kazhdan group and let $\mathcal H$ be a unitary $G$-representation and $\xi \in \mathcal H$ with $\|g\xi - \xi\|<\varepsilon \|\xi\|$ for all $g \in S$. It follows that $\|h\xi - \xi\|< (2[G:H]+1)\varepsilon$ and hence, if $\varepsilon>0$ is sufficiently small, there must exist non-zero $H$-invariant vectors nearby. Moreover, the finite group $G/\cap_{g \in G}H^g$ (which is of size at most $[G:H]!$) acts on the space of $H$-invariant vectors. Thus, if $\varepsilon>0$ is small enough, there must exist a $G$-invariant vector. I would guess that the spectral gap of a finite group $L$ is at least of the order $|L|^{-1}$, so that this also gives a quantitative bound.

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  • $\begingroup$ Beware that the $G$-invariant vector outputted in the proposition is not a unit vector (and thus can be zero when $\delta>1$, in which case this reformulation of Prop 1.1.9 is vacuous). And "less than" is $\le$ in the book but improved to $<$ in Remark 1.10 under the assumption that $S$ is finite. $\endgroup$ – YCor Mar 12 '18 at 15:24
  • $\begingroup$ $\delta=[G:H]^{-1/2}$ and I was only thinking about finite $S$. $\endgroup$ – Andreas Thom Mar 12 '18 at 15:33
  • $\begingroup$ In your proposition $\delta$ is not yet specified, so I took a while reading both you and BHV to see that everything matches. The goal of my comment was to spare potential other readers from losing 2 minutes feeling confused. $\endgroup$ – YCor Mar 12 '18 at 17:20
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This paper by Uzy Hadad shows that there is no connection between Kazhdan constant of $SL_n(\mathbb{Z})$ and Kazhdan constants of its finite index subgroups.

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    $\begingroup$ Your conclusion that "there is no connection" is a personal, if not simplistic, interpretation of Hadad's results. It's quite predictable that the estimates asked by the OP (especially in 1) are likely to involve the index $[G:\Gamma]$, which is not involved in Hadad's results. $\endgroup$ – YCor Mar 1 '18 at 10:22
  • $\begingroup$ Unless I misunderstand the result, it says that for an infinite sequence of finite index subgroups (hence for arbitrary large index) the constant is $>\epsilon$ and for an infinite sequence of subgroups, the constant is $< \epsilon$. So there does not seem to be a (monotone) dependence on the index. $\endgroup$ – Mark Sapir Mar 1 '18 at 10:30
  • $\begingroup$ Yes of course, there's something to say from this relevant reference, which is not that there is "no connection". For (1) it's my expectation that we have, for all $G,\Sigma,\Lambda$, the existence of a finite generating subset $\Sigma'$ of $\Lambda$ (depending on $\Sigma$) with a reasonable lower bound on $\kappa(\Lambda,\Sigma')$ in terms of $\kappa(\Gamma,\Sigma)$ an $[\Gamma:\Lambda]$. To obtain this, one should play with both the constructive proof to produce finite generating subsets on finite index subgroups, and induction of unitary representations from $\Lambda$ to $\Sigma$. $\endgroup$ – YCor Mar 1 '18 at 10:39
  • $\begingroup$ And by the way there is no "Kazhdan constant of $SL_n(\mathbb{Z})$" or Kazhdan constant for a given finite index subgroup. The Kazhdan constant is a function of a pair of a group with a subset (typically but not necessarily, finite an generating). $\endgroup$ – YCor Mar 1 '18 at 10:41
  • $\begingroup$ I tend to agree with @YCor, Hadad's paper does not exclude the possibility of an estimate between the constants. $\endgroup$ – duh Mar 1 '18 at 11:28

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