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\begin{example*}
  This is an example to show that the ``\gls{irrelalt}''
  property is hard to satisfy.

  With three voters, a natural way to set up the voting system is to say that
  candidates award 2 points to their favourite, 1 point to their second
  favourite, and no points to the least favourite.
  Suppose that $33$ people vote $A > B > C$, $33$ vote $C > A > B$ and $34$ vote
  $B > C > A$.
  Then one can check that $B$ ends up higher than $A$ if we run the voting
  system described above.
  But if we change the $C > A > B$ voters to $A > C > B$, then we would get that
  $A$ scores higher than $B$, which shows that this voting system does not have
  the ``\gls{irrelalt}'' property.
\end{example*}

Given some pairwise rankings, we say that there is a \emph{Condorcet winner} if
they form a transitive relation (i.e., they induce a total ordering).

\begin{remark*}
  If there is always a Condorcet winner, then necessarily $u = v = w$.
  To see this, let $x \in \{-1, 1\}^n$ be arbitrary, let $y = -x$, and let $z =
  (u(x), u(x), \ldots, u(x))$.
  Since $y = -x$, this is a valid vote (i.e. $(x, y, z) \in X$).
  The outcome is $(u(x), v(-x), u(x))$ (where $w(z) = u(x)$ by weak
  monotonicity).
  For this to be a valid ranking, we must have $v(-x) = -u(x)$.
  That is true for all $x$.
  By symmetry, we also have $v(-x) = -w(x)$ for all $x$, so $u(x) = w(x)$, so by
  symmetry $u = v = w$.
\end{remark*}

\begin{proof}[Proof of \nameref{thm:3.1}]
  Suppose we have three candidates $A$, $B$, $C$ and $n$ voters.
  Encode a vote as a sequence $(x, y, z) \in \{-1, 1\}^{3n}$ where for each $i$,
  $x_i = 1$ if and only if voter $i$ prefers $A$ to $B$, $y_i = 1$ if and only
  if voter $i$ prefers $B$ to $C$, and $z_i = 1$ if and only if voter $i$
  prefers $C$ to $A$.

  For this to be a vote, $(x_i, y_i, z_i) \in \{-1, 1\}^3 \setminus \{(1, 1, 1),
  (-1, -1, -1)\}$.
  Let $X$ be the set of all such $(x, y, z)$.
  A voting rule is then a function $f : X \to \{-1, 1\}^n \setminus \{(1, 1, 1),
  (-1, -1, -1)\}$.
  If axiom 3 holds (``\gls{irrelalt}''), then we can
  write $f(x, y, z) = (u(x), v(y), w(z))$ for three Boolean functions $u, v, w :
  \{-1, 1\}^n \to \{-1, 1\}$.

  Let us assume that $u = v = w$, as shown in the above Remark.
  Then we have a Condorcet winner if and only if for every $(x, y, z)$, it is
  not the case that $u(x) = u(y) = u(z)$.
  But $u(x) = u(y) = u(z)$ is false if and only if $u(x) u(y) + u(y)u(z) +
  u(z)u(x) = -1$.
  So if there is always a Condorcet winner, then
  \[
    \Ebb_{(x, y, z) \in X} (u(x)u(y) + u(y)u(z) + u(z)u(x))
    = -1
  ,\]
  so $\Ebb_{(x, y, z) \in X} u(x)u(y) = -\third$.

  If we choose $(x, y, z)$ uniformly from $X$, then $x$ is uniform in $\{-1,
  1\}^n$.
  Having picked $x$, the possibilities for $(y_i, z_i)$ are $(-x_i, x_i)$,
  $(x_i, -x_i)$ and $(-x_i, -x_i)$ with all three equally likely.
  So $y_i = x_i$ with probability $\third = \frac{1 + \left( -\third
  \right)}{2}$.
  Therefore,
  \[
    \Ebb_{(x, y, z) \in X} u(x) u(y)
    = \stab_{-\third} u
  .\]
  We can conclude that $\stab_{-\third} u = -\third$.
  By the Fourier formula for $\stab$, it follows that
  \[
    \sum_A \left( -\third \right)^{|A|} \ft{u}(A)^2
    = -\third
  .\]
  By \gls{parseval}, $\sum_A \ft{u}(A)^2 = 1$.
  It follows that $\ft{u}(A)^2 = 0$ when $|A| \neq 1$, or in other words that
  $u$ is linear (with zero constant term).
  By \cref{lemma:3.2}, $u$ is a dictator, violating axiom 2.
\end{proof}

We will show that if we relax the above theorem to consider voting rules where
with probability $\eps$ we fail to give a Condorcet winner, then there is an
``almost dictator'' (someone who determines the vote most of the time).

Note that if $p = \Pbb(\text{Condorcet winner})$, then
\[
  \Ebb_{(x, y, z) \in X} (u(x)u(y) + u(y)u(z) + u(z)u(x))
  = -p + 3(1 - p)
  = 3 - 4p
.\]
\glssymboldefn{W}%
So $3\stab_{-\third} u = 3 - 4p$, so $p = \frac{3}{4}(1 - \stab_{-\third} u)$.
Therefore, if $p \ge 1 - \eps$, then $1 - \eps \le \frac{3}{4} (1 -
\stab_{-\third} u)$ so $\stab_{-\third} u \le -\third + \frac{4}{3} \eps$.
But writing $W_1(f)$ for $\sum_{|A| = 1} \ft{f}(A)^2$,
\[
  \stab_{-\third} u
  \ge -\third W_1(u) - \frac{1}{27} (1 - W_1(u))
,\]
so
\[
  -\third W_1(u) - \frac{1}{27} (1 - W_1(u))
  \le -\third + \frac{4}{3} \eps
.\]
If $W_1(u) = 1 - \delta$ then
\[
  -\third + \frac{\delta}{3} - \frac{\delta}{27}
  \le -\third + \frac{4}{3}\eps
,\]
$\delta \le \frac{9}{2} \eps$, i.e. $W_1(u) \ge 1 - \frac{9}{2} \eps$.

To show that there is an almost dictator we will need to introduce a lot more
theory.

\newpage
\section{Hypercontractivity}

% TODO: ??
We begin with an important lemma of Bonami.

\begin{fclemma}[Bonami's Lemma]
  % Lemma 4.1
  \label{lemma:4.1}
  Let $f : \{-1, 1\}^n \to \Rbb$ be a function of degree $\le k$.
  Then $\|f\|_4 \le 3^{k / 2} \|f\|_2$.
\end{fclemma}

This statement is natural: to have $\|f\|_4$ much larger than $\|f\|_2$, $f$
must be `spiky', because if it is roughly constant, then these norms are roughly
the same; but if $f$ is `spiky', then we expect it to not be low degree
polynomial.

\begin{example*}
  Consider $f(0) = 1$, $f(x) = 0$ otherwise.
  $\|f\|_2 = 2^{-n / 2}$, $\|f\|_4 = 2^{-n / 4}$.
  So $\|f\|_4$ is much larger than $\|f\|_2$ in this case, and indeed $f$ is not
  a low degree polynomial.
\end{example*}

\begin{proof}
  Induction on $n$.
  $n = 1$ is easy.
  Let $f : \{-1, 1\}^n \to \Rbb$.
  Then $f = g + x_n h$ where $g(x) = E_n f = \half (f(x_{n \subst 1}) + f(x_{n
  \subst -1}))$ and $h(x) = \D_n x = \half (f(x_{n \subst 1}) - f(x_{n \subst
  -1}))$.
  Note also that $g$ and $x_n h$ are orthogonal.