The symmetric group has two irreducible representations of degree $3$: One is the standard representation and the other is the product of the standard representation with the (degree $1$) sign representation. I'm interested in the latter one. So $S_4$ acts on a $3$-dimensional vector space, say $V \cong \mathbb{Q}^3$ generated by $x_1, x_2, x_3$, by the product of the standard and sign irreps.

Question: Is there a simple description of this action only involving the basis vectors $x_1, x_2, x_3$?

My knowledge of representation theory is still very limited. But from what I know, $S_4$ acts on say $\mathbb{Q}^4$ just by permutation of the basis vectors $v_1, v_2, v_3, v_4$. Then $v_1 + v_2 + v_3 + v_4$ is invariant under the $S_4$-action, hence $S_4$ acts on the $3$-dimensional subspace given by the orthogonal complement of $v_1 + v_2 + v_3 + v_4$. But I really don't understand how to describe this action, let alone how to describe it in the general case above where one does not start with a $4$-dimensional space and consider a $3$-dimensional subspace.

PS: This question arose when I was looking at this mathSE question.

Thanks in advance for any help!

  • $\begingroup$ Do you know that this representation is isomorphic to some isomorphisms $\mathfrak{S}_4 \rightarrow Isom^+(C_3)$ where $C_3$ is the cube in $\mathbb{R}^3$ and $Isom^+(C_3)$ is its group of linear isometries with positive determinant? $\endgroup$ – nombre Nov 26 '16 at 0:07
  • $\begingroup$ @nombre No I haven't seen this statement before. I guess this might be helpful cause then we're already in $\mathbb{R}^3$ (or $\mathbb{Q}^3$). Would you be able to explain this a little further? $\endgroup$ – Tom Bombadil Nov 26 '16 at 9:31
  • $\begingroup$ I posted a detailed explaination as an answer. $\endgroup$ – nombre Nov 26 '16 at 10:29

Define $C_3:=\{\varepsilon_1.e_1 + \varepsilon_2.e_2 + \varepsilon_3.e_3 \ | \ \varepsilon_1,\varepsilon_2,\varepsilon_3 \in \{-1;1\}\}$ where $(e_1,e_2,e_3)$ is the canonical basis of $\mathbb{R}^3$. So $C_3$ is a cube of length $2$ centered on $(0,0,0)$.

Let $Isom^+(C_3)$ denote the set of positive (determinant) linear isometries of the euclidean space $\mathbb{R}^3$ which fix $C_3$. Let $d_1,d_2,d_3,d_4$ respectively denote $\{(1,1,1);(-1,-1,-1)\},\{(1,-1,1);(-1,1,-1)\},\{(-1,-1,1);(1,1,-1)\},\{(-1,1,-1);(1,-1,1)\}$. $d_1,d_2,d_3,d_4$ are the great diagonals of $C_3$.

Since those diagonals realize the greatest distance between points of $C_3$, they are preserved by elements of $Isom^+(C_3)$. This gives us a morphism $\varphi:Isom^+(C_3) \rightarrow \mathfrak{S}_4$ caracterised by $\forall f \in Isom^+(C_3), \forall 1 \leq i \leq 4, d_{\varphi(f)(i)} = f(d_i)$.

Claim: $\varphi$ is an isomorphism.

Indeed, for $f \in Ker(\varphi)$, $f$ fixes the great diagonals, so for each diagonal $d_i$, $f$ fixes or interchanges its elements, which means $d_i \subset E_1(f)$ or $d_i \subset E_{-1}(f)$. Now any $3$-uple of points in three different diagonals form a basis of $\mathbb{R}$, so $E_1(f)$ and $E_{-1}(f)$, whose sum is direct, cannot both contain two diagonals, and one of them has to contain $3$, but then it is of dimension $3$. $E_{-1}(f)$ cannot be of dimension $3$, otherwise $f$ would be $-id_{\mathbb{R}^3}$, which is not positive. So $\dim(E_1(f)) = 3$, and $f = id_{\mathbb{R}^3}$.

Now to prove surjectivity, one only has to find antecedents for a system of generators of $\mathfrak{S}_4$, for instance $(1 \ 2)$ and $(1 \ 2 \ 3 \ 4)$. See that $\varphi(r_{\pi}) = (1 \ 2)$ and $\varphi(r_{\frac{\pi}{2}}) = (1 \ 2 \ 3 \ 4)$ where $r_{\pi}$ is the rotation with angle $\pi$ and axis $\mathbb{R}.(e_1+e_3)$, and $r_{\frac{\pi}{2}}$ is the rotation with angle $\frac{\pi}{2}$ and axis $\mathbb{R}.e_3$.

So $\varphi$ is an isomorphism, and $\varphi^{-1}$ is a representation of $\mathfrak{S}_4$.

To compute the values of the associated character $\chi$, one can use the fact that the trace of a rotation in $\mathbb{R}^3$ with angle $\theta$ is $1 + 2\cos(\theta)$. This yields $\chi(1 \ 2) = -1$ and $\chi(1 \ 2 \ 3 \ 4) = 1$.

$\varphi^{-1}(1 \ 2 \ 3)$ has order $3$ in dimension $3$ so its trace $\chi(1 \ 2 \ 3)$ is $1 + j + j^2 = 0$.

$\varphi^{-1}((1 \ 2)(3 \ 4))$ has order $2$ and positive determinant so its trace $\chi((1 \ 2)(3 \ 4))$ is $1 + (-1) + (-1) = -1$.

You can check that this is the same character as the one you were looking for.

This type of proof is quite classic when dealing with representions of $\mathfrak{S}_n$. If you want a degree $p$ representation of $\mathfrak{S}_n$, try to realize it with isometries of a subset $X$ of the euclidean $\mathbb{R}^p$. To this effect, try finding $n$ things that are permutated by those isometries, and you get a morphism $Isom(X) \rightarrow \mathfrak{S}_n$. If that morphism is surjective (so if $(1 \ 2)$ and $(1 \ 2 \ ... \ n)$ are values), then $\mathfrak{S}_n$ is a quotient of $Isom(X)$ and with luck, this quotient is a subgroup of $Isom(X)$. Now $Isom(X)$ is a subgroup of $Isom(\mathbb{R}^p)$, and this is nice because elements of $Isom(X)$ are actually maps on the whole space, so they can act on many things, giving you other representations of $\mathfrak{S}_n$. For instance, $Isom^+(C_3)$ permutes $e_1,e_2,e_3$, and this gives another degree $3$ representation of $\mathfrak{S}_4$ which happens to reduce to the direct sum of the trivial representation on $\mathbb{R}.(e_1+e_2+e_3)$, and the degree $2$ subrepresentation on $Ker((e_1+e_2+e_3)^*)$, which is the degree $2$ irreductible representation of $\mathfrak{S}_4$.

Of course, you can also find representations of $\mathfrak{S}_n$ without geometric tools, and you might enjoy more equivalent algebraic tools.


If $V$ is spanned by the basis $\{v_1, v_2, v_3, v_4\}$, and $v = v_1 + v_2 + v_3 + v_4$, then (as you have observed) the action of $S_4$ on $V$ fixes $v$. Therefore, there is a natural action of $S_4$ on the three-dimensional quotient $\overline{V} = V/\langle v \rangle$. This might be easier than thinking of the action as being on the orthogonal complement.

Then $\overline{V} \otimes W$, where $W$ is a one dimensional space with sign action, gives the other three-dimensional representation of $S_4$. If you haven't come across the tensor product yet, it is the way to construct a representation whose character is the product of the characters of two given representations.

You can describe these actions directly in terms of a basis $\{x_1,x_2,x_3\}$, by choosing a basis for $\overline{V}$ (or $\overline{V} \otimes W$), and then calculating the actions of $S_4$. But it will be a bit artificial because it forces you to break the inherent symmetry of $v_1, v_2, v_3, v_4$.

  • $\begingroup$ No need to break the symmetry. Just compute the action of $S_4$ on $V \otimes W$ - the resulting space has a similar one-dimensional subrepresentation, which you can then quotient out, or take its orthogonal complement. $\endgroup$ – Dustan Levenstein Nov 24 '16 at 8:31
  • $\begingroup$ @Ted Thank you for your answer. I guess what I'm asking for is really what you call "a bit artificial". From your description it seems like it really depends on the basis right? Are there any statements I can make that are independent of the basis of $\bar{V} \otimes W$? (Im a bit unsure myself of what I'm looking for here... Again I mention this question, which is what I really want to solve.) $\endgroup$ – Tom Bombadil Nov 24 '16 at 9:05

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