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\begin{document}
\lecture{14}{17 October 2001}{Igor Pak}{C. Goddard}

\section*{Hall Bases Continued}
Last lecture we finished with the theorem:
\begin{theorem}
\label{completeThenUniform}
Given a $\omega$ - complete word in $\overline{B} = (B_1, B_2, \ldots)$, a Hall Basis in G, then $\omega^{\overline{\alpha}}$ - uniform in $G$.
\end{theorem}

Now two lectures ago, we wanted to prove the following lemma:
\begin{lemma}
\label{stoppingStrongUniform}
$\varkappa$ for $\mathrm{U}(n,p) = \left\{\begin{pmatrix}1&\ast&\cdots&\ast\\
                           0&1&\cdots&\ast\\
                           \vdots&\vdots&\ddots&\vdots\\
                           0&0&\cdots&1\end{pmatrix}\:\ast\in\mathbb F_p\right\}$ is strong uniform.
\end{lemma}

We proved a corollary to this:
\begin{corollary}
The mixing time for a random walk on $\mathrm{U}(n,p) = \mathrm{O}(n^2 \log n)$.
\end{corollary}

Now we want to prove Theorem \ref{completeThenUniform} $\Rightarrow$ Lemma \ref{stoppingStrongUniform}.

\begin{proof}
Let $G = \mathrm{U}(n,p)$, that is the group of $n \times n$ upper triangular matrices with 1's on the diagonal.  Consider the basis: $\overline{B} = (B_1, B_2, \ldots, B_{n-1})$ where $$B_i = \left\{ \begin{pmatrix} 1 & 0 & \cdots & 0 \\
0 & 1 & 1 & \vdots \\
\vdots & & \ddots & 0  \\
0 & \cdots & 0 & 1 \\
\end{pmatrix}\: \mbox{with a 1 in the $i$th diagonal.} \right\}$$
Thus $|B_i| = n - i$.

Now we have to check $\overline{B}$ is a Hall basis for $\mathrm{U}(n,p)$.
This is ``obvious'' since we know that $\mathrm{<}\gamma_i(B_i)\mathrm{>} = H_i$ since, firstly $\mathrm{<}B_i\mathrm{>} = G_i$, where $G_i$ consists of $0$'s everywhere below the $i$th diagonal except the main diagonal, and $H_i$ is the quotient $G_{i-1} / G_{i}$, so $H_i \cong (\mathbb{Z}_p)^{n-i}$.

For the mixing time, we know $X_t = E_{i_1j_1}(\alpha_1) \cdot E_{i_2j_2}(\alpha_2) \cdot \ldots \cdot E_{i_tj_t}(\alpha_t)$ by definition since $$E_{ij}(\alpha) = \begin{pmatrix}
1 & 0 & \cdots & 0 \\
0 & 1 & \alpha & 0 \\
\vdots & \ddots & 0 & \vdots \\
0 & \cdots & 0 & 1\\
\end{pmatrix}\: \mbox{ie 1's on the diagonal and 0's elsewhere except $\alpha$ in the $ij$th position} $$

Now we want to look at $\varkappa$, which is the first time all the indices $i$, $j$ occur in this product.  So in the notation of the previous lecture, $\Lambda = \{ (i,j), 1 \leq i < j \leq n\}$.  Say that there are $N$ words that contain all $i$, $j$ and look at the complete words.  Thus,
$$\Pr(X_t = h | \varkappa = t) = \frac{1}{N} \cdot \sum_{\omega} \Pr(\omega^{\overline{\alpha}} = h)$$
where we sum over the complete words $\omega$ of length $t$ such that no shorter word is a complete word.  Therefore from Theorem \ref{completeThenUniform}, $\omega$ is uniform.  So,
$$\Pr(X_t = h | \varkappa = t) = \frac{1}{N} \cdot N \cdot \frac{1}{|G|} = \frac{1}{|G|}$$
Thus, $\varkappa$ is strong uniform.
\end{proof}

Note, we can generalise this to any nilpotent group with generators corresponding to our generators, and the mixing time $= \mathrm{O}(|\Lambda| \log|\Lambda|)$.

\section*{Brief Outline of Open Problems for Research Projects}

\subsection*{Hamilton Paths in Cayley Graphs}
There are two conflicting conjectures relating to the Hamilton paths in Cayley graphs, namely:
\begin{conjecture}{\bf (Lovasz)}
\label{lovasz}
$\forall G,\ \mathrm{<}S\mathrm{>} = G,\ S = S^{-1}$, the Cayley graph $\mathrm{\Gamma}(G, S)$ contains a Hamilton path.
\end{conjecture}

\begin{conjecture}{\bf (Babai)}
\label{babai}
$\exists \alpha > 0$ such that $\exists$ infinitely many Cayley graphs with no paths on length $> (1 - \alpha) \cdot \# vertices$.
\end{conjecture}

Aim: try and find out which one of these is true on a special groups and generating sets.

{\it Examples:}
\smallskip
\noindent
$1)$ \, Try Hall's 19 (up to automorphisms) Cayley graphs of
$A_5$ with $2$ generators (aim for negative answer.)

\noindent
$2)$ \, Try $S_n$ and conjugacy classes (aim for positive answer.)

\noindent
$3)$ \, Try general nilpotent groups (positive.)

\noindent
$4)$ \, Try three involutions in general groups (positive.) \,
NB: every finite simple group can be generated by three
involutions.

\noindent
$5)$ \, Try wreath and semidirect product of finite groups
(positive; easy for direct products.)

\subsection*{Diameter Problem}

Suppose we have $A_n$, $S_n$ and $\mathrm{<}S\mathrm{>} = A_n$, where $S$ is a set of generators.

\begin{conjecture}
$\mathrm{diameter}(A_n, S) < cn^2$, $c$ - constant.  Also works for $S_n$.
\end{conjecture}

Look at the following weaker versions of this:
\begin{enumerate}
\item For the worst case when $|S| = 2$, we have the following:
\begin{theorem}{\bf (Babai, Hetyei)}
$\mathrm{diam} < e^{\sqrt{n} \log n(1 + \mathrm{o}(1))}$.  This gives a bound of the maximum order of permuations in $S_n$.
\end{theorem}

Aim: Find something similar for $\mathrm{SL}(n, p)$.

\begin{conjecture}
$G$ - simple $\Rightarrow \mathrm{diam} = \mathrm{O}((\log |G|)^c)$.
\end{conjecture}
So $\mathrm{diam} \leq (n^2 \log p)^2$ which would be hard to prove, but $e^{n}$ may be manageable.

\item Average Case.

\begin{theorem}{\bf (Dixon)}
$\mathrm{<}\sigma_1, \sigma_2\mathrm{>} = A_n$ with $\Pr \rightarrow 1$ as $n \rightarrow \infty$.
\end{theorem}
\begin{theorem}{\bf (Babai-Seress)}
$\mathrm{diam}(\mathrm{\Gamma}(A_n, \{\sigma_1, \sigma_2\}))
= n^{\mathrm{O}(\log n)}$ \, w.h.p.
\end{theorem}


Aim: get something close for $\mathrm{PSL}(n,p)$.

\item Problem.

\begin{conjecture}{\bf (Kantor)}
$\mathrm{diam}(\mathrm{\Gamma}(A_n, \{\sigma_1, \sigma_2\})) =
\mathrm{O}(n \log n)$ w.h.p.
\end{conjecture}

Some people believe this is not true.

Question:  True or False?

Weaker version: Prove that $\mathrm{\Gamma}(A_n, \{\sigma_1,
\sigma_2\})$ are NOT exanders w.h.p.

\end{enumerate}

\subsection*{Random Graphs vs Random Cayley Graphs}
\begin{enumerate}
\item \begin{theorem} {\bf (Ramsey Theory)} \
In random undirected graph $\mathrm{\Gamma}$ with $n$ vertices,
there exists a $m=c \cdot \log n$ complete subgraph in $\mathrm{\Gamma}$
and a $m=c \cdot \log n$ complete subgraph
in $\mathrm{\overline{\Gamma}}$~w.h.p.
\end{theorem}
Now suppose $\mathrm{\Gamma}$ is a random Cayley graph over a fixed
group $G$. People believe the same is true.

Aim: prove it (N. Alon proved the result with $m=c\sqrt{\log n}$.)

\item \begin{theorem}
$\mathrm{\Gamma}$ - random graph on $n$ vertices
$\Rightarrow \mathrm{Aut}(\Gamma) = 1$
with high probability.
\end{theorem}

NB: Erd\H os and R\'enyi proved that one has to remove
$\theta(n^2)$ edges before a nontrivial automorphism appears.

Question: if $\mathrm{\Gamma}$ - random Cayley graph, is
$\mathrm{Aut}(\Gamma) = G$ with high probability?

L. Goldberg and M. Jerrum conjecture this for $G = \mathbb{Z}_n$.
\end{enumerate}


\subsection*{Percolation on finite Cayley graphs }

Fix a Cayley graph $\mathrm{\Gamma}$ and probability $p$.
Delete edges with $\Pr = (1 - p)$ independently and look
at the connected components.
Say $\mathrm{\Gamma} \supset$ large cluster if $\exists$
connected component $> \frac{1}{2} |\Gamma|$.

\begin{conjecture}{\bf (Benjamini)}
If $\mathrm{diam}(\Gamma) < c \cdot \frac{|G|}{\log^2 |G|}$, then
Cayley graph
$\Gamma$ contains large cluster with $\Pr > \frac{1}{2}$ for
$p < 1 - \varepsilon$
where $\varepsilon$ is independent of the size of the graph.
\end{conjecture}

Itai Benjamini confirms the conjecture for abelian groups.



Question: Is this true for G nilpotent? What about $S_n$?
\end{document}