Impossible to decompose a representation

Question

While brushing up my knowledge about representations I obtained following problem. I 'll indicate the prompts I enter in GAP but won't give the literal responses returned by GAP in order to save space. I start with two 9x9 matrices $ma$ and $mb$:

ma := [ [ 0, 0, 0, 0, 0, 0, 1, 0, 0 ], [ 0, 0, 0, 0, 1, 0, 0, 0, -1 ],
[ -1, 0, 0, 0, 0, 1, 0, 0, 0 ], [ 0, 0, 1, 0, 0, 0, 0, 0, -1 ],
[ -1, 0, 0, 0, 0, 0, 0, 1, 0 ], [ 0, 0, 0, 1, 0, 0, 0, 0, 0 ],
[ 0, 0, 0, 0, 0, 0, 0, 0, -1 ], [ 0, 1, 0, 0, 0, 0, 0, 0, 0 ],
[ -1, 0, 0, 0, 0, 0, 0, 0, 0 ] ];;
mb := [ [ 0, 0, -1, 0, 1, 0, 0, 0, 0 ], [ -1, 0, 0, 0, 0, 1, 0, 0, 0 ],
[ 0, 0, 0, -1, 0, 0, 1, 0, 0 ], [ -1, 0, 0, 0, 0, 0, 0, 1, 0 ],
[ 0, 0, 0, -1, 0, 0, 0, 0, 0 ], [ 0, 0, -1, 0, 0, 0, 0, 0, 1 ],
[ -1, 0, 0, 0, 0, 0, 0, 0, 0 ], [ 0, 0, -1, 0, 0, 0, 0, 0, 0 ],
[ 0, 1, 0, -1, 0, 0, 0, 0, 0 ] ];;

I construct the matrixgroup $g$ generated by these matrices (which is isomorphic to the alternating group on four elements):

g := Group(ma, mb);;

and the matrix algebra $a$:

a := Algebra(Rationals, [ma, mb]);;

which is, if I am not wrong, a representation of dimension 9 of $g$. If I try to decompose the algebra $a$ I obtain no non trivial subrepresentation, as there is but only one central idempotent element: the identity, shown by

Size(CentralIdempotentsOfAlgebra(a));

But

IrreducibleRepresentations(g);;

Shows that the highest dimension of an irreducible representation is $3$, so how do I continue to decompose $a$?

Answer 1

Your mistake was in believing that there is "no non trivial subrepresentation, as there is but only one central idempotent element". You end your post by "how do I continue to decompose $a$?" but you did not begin to (and rightly so!). What we can decompose is not the algebra $a$ but your representation $\pi.$

In view of its character table, $A_4$ has exactly four non-equivalent irreducible complex representations, let us call them $\alpha,\beta,\gamma$ of degree $1$ and $\rho$ of degree $3,$ hence $$\pi=a\alpha+b\beta+c\gamma+d\rho,$$ where the non-negative integers $a,b,c,d$ can be determined the following way, which exempts us from finding explicitely corresponding invariant subspaces.

• The total degree is $$a+b+c+3d=9$$

and (again in view of the character table):

• Since $ma$ (and also $mb$ but one of them is sufficient) is of order $3$ and trace $0,$ we have $a+b\mathrm j+c\mathrm j^2=0$ where $\mathrm j$ is a primitive cube root of unity. Since $a,b,c$ are real numbers, this boils down to $$a=b=c.$$
• Since $ma\,mb$ (and also $mb\,ma$ but one of them is sufficient) is of order $2$ and trace $-3,$ $$a+b+c-d=-3.$$

The solution of this system of $4$ equations on $4$ unknowns is $a=b=c=0,d=3,$ i.e. $$\pi=3\rho,$$ which means that $\Bbb C^9$ is the direct sum of three invariant $3$-dimensional subspaces $V_i$ ($i=1,2,3$), such that the restriction of $\pi$ to each $V_i$ is equivalent to $\rho.$