TITLE: Low-dimensional Irreducible Representations of $S_n$
QUESTION [21 upvotes]: For $n\geq 7$, I would like to show that $S_n$ has no irredicuble representations of dimension $m$ for $2\leq m\leq n-2$.
The catch is that I am not allowed to use any "machinery" (evidently, this problem should be solvable having knowledge only of chapter 1 of Fulton and Harris).
At the moment, I don't really have any idea of how to approach this, especially without the use of character theory.  Any suggestions would be greatly appreciated!

REPLY [2 votes]: Theorem. For $n \geq 7$, the only representations of $S_n$ of dimension $\leq n-1$ are direct sums of (a) the trivial rep (b) the sign rep (c) the $n-1$ dimensional rep on $n$-tuples of numbers summing to $0$ and (d) the tensor product of (b) and (c).

I'll use the induction hypothesis that the statement is true for $S_{n-1}$ and $S_{n-2},$ using $S_7$ and $S_8$ as base cases. I haven't proved the base cases but reliable references say they are true.
A similar induction was used in "Braid group representations of low degree‎" by Formanek to prove a result for the braid group.
It suffices to consider only unitary representations $\rho:S_n\to U_m,$ by the unitary trick as in the other answer. And we can assume $\rho$ is irreducible and faithful - otherwise it factors through $\det:S_n\to S_n/A_n.$
Case 1. The multiplicity of $-1$ in $\rho((n-1,n))$ is $1$ or $m-1.$
If the rank is $m-1,$ tensor with $\det$ to reduce to the case where the rank is $1.$ This case is then a representation as a reflection group: each $\rho((i,j))$ is reflection in the hyperplane perpendicular to a unit vector $v_{ij}$ which is determined up to phase.
For any subset $X\subset\{1,\dots,n\}$ with $|X|=n-1,$ the restriction to $S_X$ defines a representation $S_{n-1}\simeq S_X\to U_m.$ This representation is faithful and generated by $n-2$ reflections, so it's rep "(c)" of $S_{n-1}$ if $m=n-2,$ or the direct sum of that rep with a $1$-dim trivial rep if $m=n-1.$ This fixes the inner products between each pair of vectors $v_{ij},$ and hence determines $\rho$ up to isometry as being rep "(c)" of $S_n$ (and forces $m=n-1.$)
Case 2. The multiplicity of $-1$ in $\rho((n-1,n))$ is in $2,\dots,m-2.$
Let $V=V_{-1}\oplus V_1$ be the eigenspace decomposition with respect to $\rho((n-1,n)).$
The restriction of $\rho$ to $S_{n-2}$ (the symmetries of $\{1,\dots,n-2\}$) splits as $V_{-1}\oplus V_1,$ and each of these spaces has dimension at most $m-2\leq n-3.$ But this restriction must still be faithful. By the $S_{n-2}$ part of the induction hypothesis, this must be the (c) rep for $S_{n-2}.$
This forces $\dim V_{-1}=2.$
The $(-1)$-eigenspaces of any two disjoint transpositions in $S_{n-2}$ is a one-dimensional subspace of $V_{-1},$ which is independent of the choice of transpositions. So actually the $(-1)$-eigenspaces of any two transpositions in $S_{n-2}$ intersect in a fixed one-dimensional subspace, which is therefore a reducing subspace.