4
$\begingroup$

Given a finite group $G$ and its irreducible representation $\pi$ I want to find explicit elements of the group algebra $\mathbb{C}[G]$ lying in components of the left regular representation isomorphic to $\pi$ (there are $(\dim \pi)^2$ linearly independent such elements.) It should be some standard `canonical' way to do this, but I shamefully do not see it.

Particular case of $S_4$ and 2-dimensional representation is of interest, but general procedure is appreciated as well.

$\endgroup$
2
  • $\begingroup$ By the way, the $2$-dimensional irreducible representation of $S_{4}$ is "really" a representation of $S_{3}$, because it has the normal Klein $4$-subgroup of $S_{4}$ in its kernel. $\endgroup$ Mar 24, 2015 at 12:18
  • $\begingroup$ Do you mean "isomorphic to $\pi$"? $\endgroup$
    – Sasha
    Mar 24, 2015 at 20:52

3 Answers 3

8
$\begingroup$

The space that you seek is the two-sided ideal in $\mathbb C[G]$ generated by the character of $\pi$ (see details below).

This follows from the explicit Wedderburn decomposition of $\mathbb C[G]$. If we write $$ \mathbb C[G] = \bigoplus_{\pi \in \hat G} M_{d_\pi}(\mathbb C), $$ then the identity element of the $\pi$th summand is $$ \epsilon_\pi = \sum_{g\in G} \frac{d_\pi}{|G|}\chi_\pi(g^{-1})g, $$ where $\chi_\pi$ is the character of $\pi$. See for example Theorem 1.7.9 in my book, the relevant part of which is also available online here (see Theorem 7.9).

$\endgroup$
5
  • $\begingroup$ Here character $\chi$ is understood as a group algebra element $\sum_g \chi(g) g$, yes? $\endgroup$ Mar 24, 2015 at 7:55
  • 3
    $\begingroup$ It is actually usual to associate the character $\chi$ to the element $\sum_{g \in G} \chi(g^{-1})g \in \mathbb{C}G.$ $\endgroup$ Mar 24, 2015 at 11:38
  • 3
    $\begingroup$ And this is consistent with the fact that the two sided ideal usually associated to the character $\chi$ is $e_{\chi} \mathbb{C}G$, where $e_{\chi}$ is the central idempotent $\frac{\chi(1)}{|G|} \sum_{g \in G} \chi(g^{-1})g.$ $\endgroup$ Mar 24, 2015 at 11:49
  • $\begingroup$ @GeoffRobinson You are right; otherwise it seems you end up with the contragredient representation. In the case of the symmetric group, this does not matter, because every element is conjugate to its inverse. I have edited my answer. $\endgroup$ Mar 24, 2015 at 11:58
  • $\begingroup$ I like to distinguish $1_g$ (element of group algebra) from $g$ (element of group), but perhaps the latter notation is more common. Thanks for pointing out the corrections. $\endgroup$ Mar 24, 2015 at 12:10
4
$\begingroup$

To supplement the previous answer, it is often not so straightforward to explicitly find the primitive central idempotents of $\mathbb{C}G$ for large groups $G$, though there are computer packages (GAP, MAGMA, etc.), which do this efficiently. In the case of the symmetric group, there are explicit combinatorial constructions ( in terms of Young Tableaux, etc) which realise the irreducible representations (even over $\mathbb{Q}$), but this explicit construction of the actual representations (rather than just their characters) is the exception, rather than the rule.

$\endgroup$
2
$\begingroup$

If you fix a basis for the irreducible representation $\pi$, then the following elements of the group algebra span the $(\dim\pi)^2$ dimensional space you are interested in. \begin{equation} f_{i,j}=\sum_g \pi_{i,j}(g^{-1}) g \,. \end{equation} This is a generalization of the fact that characters (or central idempotents) span the group algebra of an abelian group. So if you find the matrix elements of the two dimensional irreducible representation (the one with two boxes in each of its two rows) of $S_4$, you can plug it into this and obtain the group algebra elements. These matrix elements must have been calculated, but if you want to do it yourself, you could do the following.

Consider the $6$ dimensional combinatorial representation obtained by permuting 4 objects two of which are the same and the other two are the same. A convenient basis is $\{i,j\}$ i.e., choosing two positions out of four. The matrix elements in this basis are not hard to find.

This is actually a representation of $S_3$ as Geoff Robinson pointed out, but we will treat it as a representation of $S_4$ and continue (since as a representation of $S_4$, it is multiplicity free). This $6$ dimensional representation has three irreducible representations of $S_4$: the one dimensional trivial representation, the three dimensional standard representation and the two dimensional one we are interested in. The trivial and the standard span a four dimensional space isomorphic to the defining representation of $S_4$. This defining representation is spanned by the $4$ vectors $\sum_{j\neq 1}\{1,j\}$, $\sum_{j\neq 2}\{2,j\}$, $\sum_{j\neq 3}\{3,j\}$ and $\sum_{j\neq 4}\{4,j\}$. Finding the two dimensional space orthogonal to this is not hard. Then you can find the matrix elements easily and plug it into the above equation to find the group algebra elements.

$\endgroup$
3
  • 1
    $\begingroup$ You can construct the two dimensional representation of $S_4$ "on the nose" as follows: firstly note that $S_4$ is the group of rigid motions that map the vertex set of a regular (Platonic) octahedron onto itself. Thus $S_4$ also acts on the space of all functions on the vertex-set of this octahedron whose values add up to $0$ on each face. This vector space is the two-dimensional representation of $S_4$. Once you know this, it is easy to write down the matrices in this representation. $\endgroup$ Mar 24, 2015 at 16:03
  • $\begingroup$ I see. Thanks. On the 6 dimensional vertex space, the action of $S_4$ seems to be the same as the one I mentioned. I didn't realize that it can be visualized on an octahedron. It looks like we need only 4 faces to determine a basis for the functions that add up to zero on each face. The other four will be linearly dependent on these. Unless I'm mistaken, it seems to me that the two dimensional representation of $S_4$ is the space orthogonal to this. $\endgroup$
    – Hari Krovi
    Mar 24, 2015 at 17:49
  • $\begingroup$ A function on the vertex-set which sums to $0$ on each face is completely determined by its value at two adjacent vertices. The sum-zero condition will allow you to fix all the other values of this function uniquely and consistently. $\endgroup$ Mar 25, 2015 at 3:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.