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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!

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This is not true. The tensor product of the sign representation and the standard has dimension $n-1$, and is not isomorphic to the standard. – David Speyer Oct 2 '11 at 23:55
Thanks David. I'll edit the post to take this into account. – Jonathan Gleason Oct 2 '11 at 23:58
The case $n=7$ seems easy. Look at a 7-cycle. If it has a primitive seventh root as an eigenvalue, then it must have all six of them as eigenvalues, because it is conjugate to its non-trivial powers. That alone forces the dimension to be at least 6. OTOH if 1 is the sole eigenvalue, then all the 7-cycles would act trivially. Hence so would the subgroup that they generate, which is $A_7$... Alas, I cannot generalize this argument to higher $n$. – Jyrki Lahtonen Oct 3 '11 at 4:46
What about proving by induction that the only small irreducible reps of $S_n$ are the listed ones: two 1-dimensional, and two $(n-1)$-dimensional ones. So if $n\ge8$ is the smallest value with a counterexample, then the rep must remain irreducible when restricted to any of the $n$ conjugate 'point stabilizer' subgroups isomorphic to $S_{n-1}$. The restriction cannot be a sum of 1-dimensional reps, because then the copies of $A_{n-1}$ (and hence $A_n$) would act trivially. The remaining problem would be to show that neither of the $n-2$ reps of $S_{n-1}$ can be a restriction of a rep of $S_n$. – Jyrki Lahtonen Oct 3 '11 at 20:36
The result is true as stated for complex representations, but it is not true in general in other characteristics. For example, $S_n$ has a faithful irreducible representation of degree at most $n-2$ in characteristic $p$ whenever $p$ is a prime divisor of $n$ and $n >4.$ Take the obvious permutation module. Then $n$-long column vectors with zero sum form a submodule, but this submodule contains the span of the all-1 vector, which is invariant. I realise that characteristic 0 was the intent of the question, but it illustrates that the question is subtle. – Geoff Robinson Dec 20 '11 at 16:43

Following Jyricki's excellent suggestion, I will show that, for $n \neq 4$, the $n-2$ dimensional representations of $S_{n-1}$ do not extend to representations of $S_n$. Let $V$ be the vector space of $(n-1)$-tuples of real numbers adding up to $0$, with $S_{n-1}$ acting by permutation. I will show that this does not extend to an action of $S_n$.

Suppose for the sake of contradiction that there is an action of $S_n$ on $V$ extending the action of $S_{n-1}$. Since this is an action of a finite group in characteristic zero, it preserves a symmetric bilinear form. There is only one (up to scaling) $S_{n-1}$-invariant symmetric bilinear form on $V$ -- namely, the one induced from the standard inner product on $\mathbb{R}^{n-1}$. So $S_n$ must preserve this inner product.

For $1 \leq i, j < n$, the transposition $(i j)$ acts on $V$ by orthogonal reflection in the vector $e_i - e_j$. Since $(i n)$ and $(i j)$ are conjugate in $S_n$, we know that $(i n)$ acts by orthogonal reflection in some vector; call it $v_i$.

For $1 \leq j,k < n$, with $i \neq j$, $k$, the reflections $(i n)$ and $(j k)$ commute. This means that $e_j -e_k$ and $v_i$ must be orthogonal. So $v_i$ must be of the form $$v_i = a \cdot \left( \sum_{j \neq i} e_j - (n-2) e_i \right)$$ for some scalar $a$.

Now recall that ${\Large (} (ij) (in) {\Large )}^3 = \mathrm{Id}$. So the angle between $v_i$ and $e_i - e_j$ must be $\pi/3$ or $2 \pi /3$. We compute $$|e_i - e_j| = \sqrt{2}$$ $$|v_i| = a \sqrt{(n-2)+(n-2)^2 } = a \sqrt{(n-1)(n-2)}$$ $$\langle e_i - e_j, v_i \rangle = a \left( -(n-2) - 1 \right) = - a (n-1).$$ So the angle between $v_i$ and $e_i -e _j$ is $$\cos^{-1} \left( \frac{-(n-1)}{\sqrt{2(n-1)(n-2)}} \right).$$

When $n=4$, this is $\cos^{-1} (-\sqrt{3}/2) = 2 \pi /3$. And, indeed, when $n=4$, the action does extend! But for $n>4$, $\frac{(n-1)}{\sqrt{2(n-1)(n-2)}} \neq \pm \sqrt{3}/2$, so the action does not extend.

Now, a quick sketch of how to finish the problem from here. We prove by induction the statement

For $n \geq 5$, the only representations of $S_n$ of dimension $<n$ 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).

The base case is left to you; starting by considering the eigenvalues of the $5$-cycle is a good idea.

Let $W$ be a rep of $S_n$ of dimension $<n$. Consider the restriction of $W$ to $S_{n-1}$. If $\dim W < n-2$, then induction shows that it is a direct sum of (a)'s and (b)'s, and I leave it as an exercise to show that $W$ is the same direct sum as a representation of $S_n$.

If $\dim W = n-2$ and $W$ is not made up of (a)'s and (b)'s, then $W$ is either type (c) or type (d). The above result gives a contradiction in the first case, and tensoring with the sign rep and then applying the above result gives a contradiction in the second case.

Finally, we must deal with the possibilities when $\dim V = n-1$. Details are left to you, but I think modifying the above argument will do this case as well.

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