Krdytkyu

Let $G=A_4$ be the alternating group on 4 letters, and let $R = mathbb{C}[G]$. Then

$$mathbb{C}[G] = Uoplus U' oplus U'' oplus V^{oplus 3},$$

where $U,U',U''$ are the three 1-dimensional irreducible representations and $V$ the one arising from the standard representation of $S_4$.

We want to:

• find idempotents $epsilon$ for each of $U,U',U''$ that realize these representations as left ideals of $mathbb{C}[G]$.


• find a subrepresentation $Wsubset R$ which is isomorphic to $V$, and express $W = mathbb{C}[G]epsilonsubset R,$ where $epsilon$ is an idempotent.

I know how to use Young diagrams to find idempotents for each irreducible representation of $S_4$. But how can we use the Young diagrams to find these things for $A_4$?

For the first part, note that $A_4$ has a unique nontrivial proper normal subgroup $$K_4:=big{(),(1;2)(3;4),(1;3)(2;4),(1;4)(2;3)big},.$$ There is a group isomorphism $varphi:(A_4/K_4)to C_3$, where $C_k$ is the cyclic group of order $k$. A representation $rho:C_3to text{GL}(X)$ of $C_3$ can be made into a representation $tilde{rho}: A_4to text{GL}(X)$ of $A_4$ by setting
$$tilde{rho}:=rhocircvarphicirckappa,,$$
where $kappa:A_4to (A_4/K_4)$ is the canonical projection. Note that $C_3$ has three nonisomorphic irreducible representations, all of which are $1$-dimensional. The irreducible representations of $C_3=langle grangle$ are $chi_j:C_3totext{GL}(C)$ sending $gmapsto expleft(frac{2pitext{i}j}{3}right)$ for $j=0,1,2$. We may identify $C_3$ with the subgroup $biglangle (1;2;3)bigrangle$ of $A_4$, and set $g:=(1;2;3)$.

We now know what to look for. The three nonisomorphic $1$-dimensional representations of $A_4$ must be given by $tilde{chi}_j$ for $j=0,1,2$. First, the left ideal $U$ which is given by the representation $tilde{chi}_0$ of $R$ is clearly generated by $sumlimits_{sigmain A_4},sigma$. We must rescale this element to get an idempotent $epsilon_Uin R$, which is
$$epsilon_U:=frac{1}{|A_4|},sumlimits_{sigmain A_4},sigma=frac{1}{12},sumlimits_{sigmain A_4},sigma,.$$

Let $U'$ and $U''$ be the $1$-dimensional left ideal of $R$ corresponding to the representations $tilde{chi}_1$ and $tilde{chi}_2$, respectively, with the corresponding idempotent elements $epsilon_{U'}$ and $epsilon_{U''}$. Now, note that $K_4$ must act trivially on $U'$ and $U''$. Ergo, we have
$$epsilon_{U'}=a_1,s+b_1,(1;2;3),s+c_1,(1;3;2),s$$
and
$$epsilon_{U''}=a_2,s+b_2,(1;2;3),s+c_2,(1;3;2),s$$
for some $a_1,a_2,b_1,b_2,c_1,c_2inmathbb{C}$. Here,
$$s:=sum_{sigmain K_4},sigma,.$$
We clearly have $a_j=expleft(frac{2pitext{i}j}{3}right),b_j$ and $b_j=expleft(frac{2pitext{i}j}{3}right),c_j$ since $g=(1;2;3)$ acts on $U'$ and $U''$ via multiplications by $expleft(frac{2pitext{i}}{3}right)$ and $expleft(frac{4pitext{i}}{3}right)$, respectively. That is,
$$epsilon_{U'}=c_1,Biggl(expleft(frac{4pitext{i}}{3}right),s+expleft(frac{2pitext{i}}{3}right),(1;2;3),s+(1;3;2),sBiggr)$$
and
$$epsilon_{U''}=c_2,Biggl(expleft(frac{2pitext{i}}{3}right),s+expleft(frac{4pitext{i}}{3}right),(1;2;3),s+(1;3;2),sBiggr),.$$
Because $epsilon_{U'}$ and $epsilon_{U''}$ are idempotent, $$c_j=frac{1}{|K_4|,Biggl(3,expleft(frac{3pitext{i}j}{3}right)Biggr)}=frac{expleft(frac{2pitext{i}j}{3}right)}{12}text{ for }jin{1,2},.$$
Ergo,
$$epsilon_{U'}=frac{1}{12},Biggl(1+expleft(frac{4pitext{i}}{3}right),(1;2;3)+expleft(frac{2pitext{i}}{3}right),(1;3;2)Biggr),sum_{sigmain K_4},sigma$$
and
$$epsilon_{U''}=frac{1}{12},Biggl(1+expleft(frac{2pitext{i}}{3}right),(1;2;3)+expleft(frac{4pitext{i}}{3}right),(1;3;2)Biggr),sum_{sigmain K_4},sigma,.$$
(Observe that $U$, $U'$, and $U''$ are in fact two-sided ideals of $R$ isomorphic to $mathbb{C}$.)

For the second part, we can take $W$ to be generated by $$p:=1+(1;2)(3;4)-(1;3)(2;4)-(1;4)(2;3)=big(1+(1;2)(3;4)big),big(1-(1;3)(2;4)big),.$$
Let $q:=(1;2;3),p$ and $r:=(1;3;2),p$. Note that $K_4$ acts on $p$ by scalar multiple (the scalars involved are $pm1$). Since $A_4=C_3K_4$ and $K_4$ is normal in $A_4$, $W$ is indeed spanned by $p$, $q$, and $r$, whence it is a $3$-dimensional left ideal of $R$. As $p^2=4p$, we may take the idempotent $epsilon_W$ to be $$epsilon_W:=frac{1}{4}p=frac{1}{4},big(1+(1;2)(3;4)big),big(1-(1;3)(2;4)big)$$ so that $W=Repsilon_W$.

We can also define $W'$ and $W''$ to be the left ideals generated by $p'$ and $p''$, respectively, where
$$p':=1+(1;3)(2;4)-(1;2)(3;4)-(1;4)(2;3)=big(1+(1;3)(2;4)big),big(1-(1;2)(3;4)big)$$
and
$$p'':=1+(1;4)(2;3)-(1;2)(3;4)-(1;3)(2;4)=big(1+(1;4)(2;3)big),big(1-(1;2)(3;4)big),.$$
Then, $W'$ is spanned by $p'$, $q':=(1;2;3),p'$, and $r':=(1;3;2),p'$, whereas $W''$ is spanned by $p''$, $q'':=(1;2;3),p''$, and $r'':=(1;3;2),p''$. The left ideals $W'$ and $W''$ are generated by the idempotents
$$epsilon_{W'}:=frac{1}{4},p'=frac{1}{4},big(1+(1;3)(2;4)big),big(1-(1;2)(3;4)big)$$
and $$epsilon_{W''}:=frac{1}{4},p''=frac{1}{4},big(1+(1;4)(2;3)big),big(1-(1;2)(3;4)big),,$$
respectively. We can also see that $W$, $W'$, and $W''$ are isomorphic irreducible $3$-dimensional representations of $A_4$. While each of $W$, $W'$, and $W''$ is a left ideal of $R$, the sum $Woplus W'oplus W''$ is a simple two-sided ideal of $R$ isomorphic to $text{Mat}_{3times 3}(mathbb{C})$.