2 Random Cayley graphs of general groups
Let \(A\) be a real symmetric matrix with eigenvalues \(|\mu _0| \ge |\mu _1| \ge \dots \ge |\mu _{n-1}|\). Then for every natural number \(m\),
By Lemma 2, \(\operatorname {Tr}(A^{2m}) = \sum _{i=0}^{n-1}\mu _i^{2m}\). All terms are nonnegative (even powers), so \(\operatorname {Tr}(A^{2m}) - \mu _0^{2m} = \sum _{i \ge 1}\mu _i^{2m} \ge \mu _1^{2m}\). Taking \(2m\)-th roots gives \(|\mu _1| \le (\operatorname {Tr}(A^{2m}) - \mu _0^{2m})^{1/2m}\).
For the random Cayley graph \(X(G,S)\) with \(|S| = c\log _2 n\) elements chosen randomly in \(G\), let \(P_{2m}\) denote the probability that a fixed walk of length \(2m\) in the graph (starting at a given vertex, with prescribed sequence of generator-steps) is closed, i.e. returns to its start.
Over the probability space of all normalized adjacency matrices of Cayley graphs \(X(G,S)\) with \(|S| = c\log _2 n\) random elements, for every positive integer \(m\), writing \(n = |G|\),
Let \(A = A^{*}\) be the normalized adjacency matrix; its largest eigenvalue is \(\mu _0 = 1\) (the all-ones eigenvector). By Lemma 12, \(|\mu _{1}^{*}| \le (\operatorname {Tr}(A^{2m}) - 1)^{1/2m}\). Since \(y \mapsto y^{1/2m}\) is concave, Jensen’s inequality (Lemma 3) gives \(\mathbb {E}(|\mu _{1}^{*}|) \le (\mathbb {E}(\operatorname {Tr}(A^{2m})) - 1)^{1/2m}\). The diagonal entry \((A^{2m})_{gg}\) is the probability that a length-\(2m\) walk from \(g\) is closed; by vertex transitivity (Lemma 8) this is the same for every \(g\), so \(\operatorname {Tr}(A^{2m}) = n\, P_{2m}\) and \(\mathbb {E}(\operatorname {Tr}(A^{2m})) = n\, \mathbb {E}(P_{2m})\). Substituting yields (1).
Let \(U\) be a random word of length \(2t\) in the free monoid \(M_{2d}\) generated by \(d\) letters and their inverses. Then
where an identity sequence is one that reduces to the empty word in the free group \(F_d\).
In the dynamic process producing the random walk, let \(A\) be the event that the reduced word \(W\) in the free group \(F_{c\log _2 n}\) has length \(l(W) {\lt} l(w) := 2m(1 - \ln \ln 2m / \ln 2m)\). Then
The number of ways to choose which positions of the \(2m\)-letter word form the sub-block that reduces to the identity is at most \(2^{2m}\). This block has length at least \(2m\ln \ln 2m / \ln 2m\), and with \(d = c\log _2 n\) distinct letters the probability that such a block is an identity sequence is, by Lemma 15, at most \((2/d)^{(\text{block length})/2} = (2/c\log _2 n)^{m\ln \ln 2m / \ln 2m}\). Multiplying the count of placements by this probability gives (3).
Let \(B\) be the event that the reduced word \(W\) has length at least \(2m(1 - \ln \ln 2m / \ln 2m)\) and no letter appears exactly once in it. Then, writing \(l(W)\) for the reduced length,
where \(l(W) \ge 2m(1 - \ln \ln 2m / \ln 2m)\).
If no letter appears exactly once then the number of distinct letters in \(W\) is at most \(l(W)/2\). Expose the letters of \(W\) in two stages: first the first occurrence of each distinct letter (at most \(2^{l(W)}\) ways to place this sub-block of size at most \(l(W)/2\)), then the remaining occurrences. Each remaining letter must coincide with one already exposed, an event of probability at most \(l(W)/2c\log _2 n\), and there are at least \(l(W)/2\) such remaining letters. Hence \(\operatorname {Pr}(B) \le 2^{l(W)}(l(W)/2c\log _2 n)^{l(W)/2}\). Since \(l(W) \le 2m\), the second inequality follows.
Let \(C\) be the event that neither \(A\) nor \(B\) holds, yet after assigning to each letter a random element of \(G\) the reduced word evaluates to the identity. Then
Since neither \(A\) nor \(B\) holds, the reduced word has a letter \(\tau \) that appears exactly once. Expose the assignments of all letters except \(\tau \); write \(x(\tau )\) for the (still random) assignment of \(\tau \). The event that the word evaluates to the identity becomes a single equation \(g\, x(\tau )\, h = 1\) with \(g,h \in G\) known. As \(x(\tau )\) is uniform over the at least \(n - 2m\) group elements not yet used, this equation has probability at most \(1/(n-2m) = 1/n + O(m/n^2)\).
With \(l(w) = 2m(1 - \ln \ln 2m / \ln 2m)\),
Following [BS], realize the random pair (set \(S\), walk) by a dynamic process: (a) choose a random word of length \(2m\) in the free monoid \(M_{2c\log _2 n}\); (b) reduce it in the free group \(F_{c\log _2 n}\); (c) assign to each letter a random element of \(G\). This is equivalent to first choosing the random Cayley graph with \(|S| = c\log _2 n\) and then a random walk of length \(2m\). The event “the walk of length \(2m\) is closed” is contained in \(A \cup B \cup C\), so
Substituting the bounds (3), (4), (5) from Lemmas 16, 17, 18 gives the claim. (When the right-hand side of (4) is at most \(1\) it is monotone decreasing in \(l(W)\), so its maximum is at the minimal \(l(W) = l(w)\); otherwise the estimate is trivial.)
For every \(1 {\gt} \delta {\gt} 0\) there is a \(c(\delta ) {\gt} 0\) such that the following holds. Let \(G\) be a group of order \(n\) and let \(S\) be a set of \(c(\delta )\log _2 n\) random elements of \(G\). Then
Moreover for \(0 {\lt} \delta {\lt} 1 - \tfrac 1e\) one may take \(c(\delta ) \le (1 + o(1))\, 2e^4\ln 2 / [(1-\delta )e - 1]^2\).
Combine Lemma 19 with the trace bound (1) of Lemma 14:
Set \(2m \sim \ln n / b\) for an absolute constant \(b\). For large \(n\) the upper bound becomes
For every fixed \(1 {\gt} \delta {\gt} 0\) one can choose \(b\) and \(c\) so this is smaller than \(1 - \delta \); in particular if \(\delta {\lt} 1 - \tfrac 1e\) then \(b = 1\) and \(c \ge (1+o(1))\, 2e^4\ln 2 / [(1-\delta )e - 1]^2\) suffice, giving \(\mathbb {E}|\mu _{1}^{*}| \le 1 - \delta \).
For every \(1 {\gt} \delta {\gt} 0\) there is a \(c(\delta ) {\gt} 0\) such that the following holds. Let \(K\) be a complete graph on \(n\) vertices whose edges are coloured by \(n - 1\) colours so that each colour class is a perfect matching. Let \(H\) be the subgraph of \(K\) consisting of all edges whose colours belong to a randomly chosen set of \(c(\delta )\log _2 n\) colours. Then \(\mathbb {E}(|\mu _{1}[H]|) {\lt} (1-\delta )\, c(\delta )\log _2 n\).
Let \(G\) be a group and let \(S\) be a random set of \(d\) elements in \(G\). Put \(\mu _{1}= \mu _{1}[X(G,S)]\). Then for every \(c {\gt} 0\),
Choose the elements \(x_1, \dots , x_d\) of \(S\) one by one in order of index, and let \(d = \lambda _0 \ge \lambda _1 \ge \dots \ge \lambda _{n-1}\) be the eigenvalues of \(X(G,S)\). Fix \(j\) with \(1 \le j \le n-1\) and let \(F_i^{j} = \mathbb {E}(\lambda _j(X(G,S)) \mid x_1, \dots , x_i)\) be the conditional expectation given the first \(i\) generators. Then \((F_0^{j}, \dots , F_d^{j})\) is a martingale, and by the variational (min–max) characterization of the \(j\)-th eigenvalue, changing a single generator moves \(\lambda _j\) by at most \(2\), so \(|F_{i+1}^{j} - F_i^{j}| \le 2\). Azuma’s inequality (Lemma 4) gives \(\operatorname {Pr}(|\lambda _j - \mathbb {E}\lambda _j| \ge 2cd^{1/2}) = \operatorname {Pr}(|F_d^{j} - F_0^{j}| \ge 2cd^{1/2}) \le 2e^{-c^2/2}\). The second largest eigenvalue in absolute value is either \(\lambda _1\) or \(\lambda _{n-1}\); applying the bound to both and taking a union bound yields the factor \(4\).
For every \(1 {\gt} \varepsilon {\gt} 0\) there exists a \(c(\varepsilon ) {\gt} 0\) such that the following holds. Let \(G\) be a group of order \(n\) and let \(S\) be a random set of \(c(\varepsilon )\log _2 n\) elements of \(G\). Then the Cayley graph \(X(G,S)\) is an \(\varepsilon \)-expander almost surely, i.e. the probability that it is such an expander tends to \(1\) as \(n \to \infty \).
By Theorem 20, with the appropriate \(c\), the expected normalized second eigenvalue is at most \(1 - \delta \) for a \(\delta = \delta ( \varepsilon )\). By the concentration Lemma 22, \(\mu _{1}^{*}\) deviates from its mean by more than \(o(1)\) with probability tending to \(0\); hence almost surely \(\mu _{1}^{*}[X(G,S)] \le 1 - \delta + o(1) \le 1 - \delta /2\). The easy direction of the eigenvalue/expansion connection (Lemma 11) then makes \(X(G,S)\) an \(\varepsilon \)-expander almost surely.