Kazhdan constant and finite index subgroups

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.

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.