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Consider the complete graph $K_4$ with $4$ vertices, and choose some orientation for the edges. The symmetric group $S_4$ acts on the vertices and thereby also induces an action on the (oriented) edges. Since there are $6$ edges, I can say that this gives me a representation $\rho$ of $S_4$ on $\mathbb{Q}^6$.

I don't know much about representation theory, so my question is this:

How do I find the irreducible representations of $\rho$?

Ideas so far: I'm pretty sure that there is no non-zero element of $\mathbb{Q}^6$ which is invariant under this $S_4$-action. This means that $\rho$ has no $1$-dimensional irreducible subrepresentation, correct? Other than $1$-dimensionals, $S_4$ has one $2$-dimensional irreducible representation and two $3$-dimensionals. Since $\rho$ is $6$-dimensional, it is therefore either a direct sum of three $2$-dimensional irreducible representations or a direct sum of two $3$-dimensional irreducible representations. But how do I find which it is?

Update/Another idea: The complete graph has the shape of a tetrahedron. Name the vertices $v_1, \ldots, v_4$ and the edges $a_1, \ldots, a_6$ and fix some orientation of the edges. Let $W \cong \mathbb{Q}^6$ be the space generated by the edges (over $\mathbb{Q}$). Consider the $3$-dimensional space $U_1$ generated by three elements $x_1, x_2, x_3 \in W$, each of which is a sum of three different edges that lie on one of the sides of the tetrahedron. In my case, I've chosen $$x_1 = a_1 +a_2+a_3, \hspace{2mm} x_2 = a_3+a_4-a_6, \hspace{2mm} x_3 = a_1+a_5-a_6$$ (but I'm guessing the choice shouldn't matter). Now the $S_4$-action sends sides of the tetrahedron to sides of the tetrahedron. So this implies that the space $U_1$ is invariant under the $S_4$-action and therefore an irreducible representation.

Similarly, one can define a space $U_2$ generated by three elements $y_1, y_2,y_3 \in W$, each of which is again a sum of three different edges that are incident to one vertex, I'm going to call these the edges of a "cap" of the tetrahedron. In my case, I've chosen $$y_1 = a_1-a_2-a_5, \hspace{2mm} y_2 = a_2 -a_3-a_6, \hspace{2mm} y_3=a_4+a_5+a_6$$ (again I guess the choice shouldn't matter). The $S_4$-action again sends "caps" of the tetrahedron to other "caps", which implies that $U_2$ is invariant under the $S_4$-action and therefore an irreducible representation.

Since $x_1, x_2, x_3, y_1, y_2,y_3$ together form a basis of $W$, it follows that $W= U_1 \oplus U_2$ is a direct sum decomposition into irreducible representations.

By calculating the effect of the elements $id, (1,2), (1,2)(3,4), (1,2,3,4), (1,2,3) \in S_4$ (which represent all conjugacy classes of $S_4$) on $x_1, x_2, x_3$, one finds that the character of this representation $\chi_{U_1}$ corresponds to the character of the product of the standard and sign representation of $S_4$. Therefore the representation of $S_4$ on $U_1$ is the product of the standard and sign representation. Similarly, one finds that the representation of $S_4$ on $U_2$ is the standard representation.

Could someone comment on the above explanations? Do they make sense?

Edit: The precise edges that I've used above are the following (where $(v_i, v_j)$ denotes the edge directed from vertex $v_i$ toward vertex $v_j$): $$a_1 = (v_1, v_2), a_2 = (v_2,v_3), a_3 = (v_3, v_1), a_4 = (v_1, v_4), a_5 =(v_2, v_4), a_6=(v_3, v_4)$$ Orientation Choices

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  • $\begingroup$ Wait a minute, a permutation of the vertices may reverse an edge. How do you deal with that? $\endgroup$ – Andreas Caranti Nov 22 '16 at 9:33
  • $\begingroup$ @AndreasCaranti I'm not sure I see your point. What do you think is the problem? $\endgroup$ – Tom Bombadil Nov 22 '16 at 9:35
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    $\begingroup$ First of all, if you consider $K_4$ with oriented edges, then you do not get an action of $S_4$ on the edges since swapping two vertices reverses the orientation of the edge connecting them. Secondly, in general if a finite group $G$ acts on a finite set $M$ and you take the free $R$-module $RM$ with basis $M$, then $G$ always has a fixed point, namely the sum of all elements in $M$. $\endgroup$ – Matthias Klupsch Nov 22 '16 at 9:36
  • $\begingroup$ OK, let's put it this way: why are you orienting the edges? $\endgroup$ – Andreas Caranti Nov 22 '16 at 9:36
  • $\begingroup$ @AndreasCaranti I'm orienting the edges because the problem comes from this question I have about cellular cohomology. $\endgroup$ – Tom Bombadil Nov 22 '16 at 9:41
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Let $V$ be the four-dimensional vector space spanned by the vertices. The six dimensional space spanned by the oriented edges, subject to the convention that flipping the orientation corresponds to multiplying by $-1$ can be viewed as the exterior power $\Lambda^2V$. The general theory of representations of the symmetric group with special emphasis of computing exterior and symmetric powers of representations is described very well in the warmly recommended book 'Representation Theory: A First Course' by Fulton and Harris.

Edited in: yes this makes a lot of sense. I started writing comments about this, but the system threatened to move everything to chat, so I write here. $U_2$ being standard is correct, see last two comments. $U_1$ being standard times sign is also correct. I wasn't sure first that it would actually be a representation at all. That is: I wasn't sure that any group element that maps a certain face to itself would always map the corresponding vector to either itself or its negative and not to one of the other six possibilities. However, after staring a bit at the nice picture you edited in I see that this is indeed the case: your $U_1$ is indeed a representation and then you can use characters as you did to see which one.

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    $\begingroup$ In this example, $\Lambda^2V$ decompsoes as the sum of the two $3$-dimensional modules for $S_4$. $\endgroup$ – Derek Holt Nov 22 '16 at 10:30
  • $\begingroup$ @DerekHolt Okay but which two? I'm guessing it should be twice the same (probably twice the standard representation). And is there a way to find a basis of $\Lambda^2V$ in which I the direct sum decomposition is easily visible? $\endgroup$ – Tom Bombadil Nov 22 '16 at 13:04
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    $\begingroup$ Sorry, I thought "the sum of the two $3$-dimensional modules for $S_4$" was unambiguous, since there are precsiely two of them up to isomophism. You can use character theory to deduce this. I am not sure how to make it explicit- I need to think about that. $\endgroup$ – Derek Holt Nov 22 '16 at 13:59
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    $\begingroup$ continued... So inside the dual of the standard we have 4 vectors (let's say $\alpha_1, \alpha_2, \alpha_3, \alpha_4$) spanning the rep and they are subject to the relation $\alpha_1 + \alpha_2 + \alpha_3 + \alpha_4 = 0$. Now if I'm not mistaken (but I didn't check carefully, you should check it) you can embed this rep into yours by identifying $\alpha_1$ with the sum of the four edges pointing out of vertex 1, idenitfying $\alpha_2$ with the sum of the four edges pointing out of vertex 2, etc. Certainly these four sum to zero as you have every edge twice, in opposite direction. Ctd below $\endgroup$ – Vincent Nov 22 '16 at 15:19
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    $\begingroup$ Ok, yes, staring at the picture some more I see that you are right. The rep on the caps is indeed the standard, representation. You can see it by characters as you did, or by looking at the map from what I call $V$ above to the rep in question, sending each vertex to the corresponding cap. Geometrically it is clearly a homomorphism and since the kernel of this map is clearly the sum of all vertices, the rep $U_2$ is isomorphic to V modulo its subrep spanned by the sum of vertices which is the definition of the standard rep. $\endgroup$ – Vincent Nov 24 '16 at 8:57

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