TITLE: Impartiality axiom in Terry Tao's Arrow's Theorem proof
QUESTION [2 upvotes]: The short expository paper is here.
On page 2, 

The notion of a quorum is well-defined; it is not possible for such a group to be able to force a vote some of the time and not at other times because this would contradict the independence axiom. Also, it doesn't matter who $A$ and $B$ are becase of the impartiality axiom.

I don't quite understand the bolded setence. Also, what difference would it make to the proof if we don't assume the impartiality axiom?
Thanks.

REPLY [1 votes]: The bolded sentence means that if a group $G$ of voters is a quorum for one pair of candidates, then $G$ is a quorum for every pair of candidates. Specifically, suppose that $A$ and $B$ are candidates, and $G$ is a quorum for the pair $\langle A,B\rangle$. This means that if everyone in $G$ votes $A$ ahead of $B$, and everyone not in $G$ votes $B$ ahead of $A$, then the voting system places $A$ ahead of $B$. Now let $C$ and $D$ be any two candidates; impartiality means that the voting system treats the pair $\langle C,D\rangle$ exactly as it treats the pair $\langle A,B\rangle$, so if everyone in $G$ votes $C$ ahead of $D$, and everyone not in $G$ votes $D$ ahead of $C$, then the voting system places $C$ ahead of $D$. In other words, $G$ is also a quorum for the pair $\langle C,D\rangle$.
If this were not the case, the notion of quorum would not be well-defined: we might even have a group $G$ of voters who were a quorum for $\langle A,B\rangle$ but not for $\langle B,A\rangle$! That is it might happen that if everyone in $G$ votes $A$ ahead of $B$, and everyone not in $G$ votes $B$ ahead of $A$, then the voting system places $A$ ahead of $B$, but the voting system also places $A$ ahead of $B$ if everyone in $G$ votes $B$ ahead of $A$, and everyone not in $G$ votes $A$ ahead of $B$.
The rest of the argument depends on the fact that the notion of quorum is well-defined, so that a quorum for one pair of candidates is a quorum for every pair.
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