- #1
StaticZero
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- TL;DR
- How to split a non-simply reducible representation?
Suppose we have a group ##G = D_3## and its reducible representation ##T = E \oplus E## where ##E## is the two-dimensional irrep. In its own basis, this representation can be written like this (I provide the operator only for ##C_3^+## group element which is counter-clockwise rotation by ##2\pi/3## around ##z##)
$$
T(C_3^+) = \begin{pmatrix}
-\frac12 & -\frac{\sqrt3}2 & 0 & 0 \\
\frac{\sqrt3}{2}&-\frac12 & 0 & 0 \\
0 & 0 & -\frac12 & -\frac{\sqrt3}{2} \\
0 & 0 & \frac{\sqrt3}{2}&-\frac12
\end{pmatrix}
$$
Suppose now that we have done a basis change and forgot what was that old basis which delivered a block-diagonal form for the representation ##T##, so in our new basis all matrices of ##T## are of generic form.
How could we find a basis in which the given representation returns to block-diagonal form?
In the case of simply reducible representation (=that consists of only copies of irreps) it can be easily done by observing the character table and constructing the projector
$$
P^{(\alpha)} = \frac{s_\alpha}{|G|} \sum_g \chi^{(\alpha)}_{g'} T_g
$$
where we sum with respect to group elements and ##s_\alpha## is the dimension of the irrep ##\alpha## and ##g' \equiv g^{-1}##.
Unfortunately, this operator does not help in case of multiple copies of irreps as in the example being investigated where ##T = E \oplus E## because it can 'filter out' only those irreps that have distinct character.
Two other projectors which I can recall for
$$
P^{(\alpha)}_i = \frac{s_\alpha}{|G|} \sum_g T^{(\alpha)}_{g', ii} T_g
$$
$$
P^{(\alpha)}_{ik} = \frac{s_\alpha}{|G|} \sum_g T^{(\alpha)}_{g', ik} T_g
$$
are neither helpful, too, by the similar reasoning.
Could you please explain whether these multiple copies of the same irreps can be 'filtered out'?
$$
T(C_3^+) = \begin{pmatrix}
-\frac12 & -\frac{\sqrt3}2 & 0 & 0 \\
\frac{\sqrt3}{2}&-\frac12 & 0 & 0 \\
0 & 0 & -\frac12 & -\frac{\sqrt3}{2} \\
0 & 0 & \frac{\sqrt3}{2}&-\frac12
\end{pmatrix}
$$
Suppose now that we have done a basis change and forgot what was that old basis which delivered a block-diagonal form for the representation ##T##, so in our new basis all matrices of ##T## are of generic form.
How could we find a basis in which the given representation returns to block-diagonal form?
In the case of simply reducible representation (=that consists of only copies of irreps) it can be easily done by observing the character table and constructing the projector
$$
P^{(\alpha)} = \frac{s_\alpha}{|G|} \sum_g \chi^{(\alpha)}_{g'} T_g
$$
where we sum with respect to group elements and ##s_\alpha## is the dimension of the irrep ##\alpha## and ##g' \equiv g^{-1}##.
Unfortunately, this operator does not help in case of multiple copies of irreps as in the example being investigated where ##T = E \oplus E## because it can 'filter out' only those irreps that have distinct character.
Two other projectors which I can recall for
$$
P^{(\alpha)}_i = \frac{s_\alpha}{|G|} \sum_g T^{(\alpha)}_{g', ii} T_g
$$
$$
P^{(\alpha)}_{ik} = \frac{s_\alpha}{|G|} \sum_g T^{(\alpha)}_{g', ik} T_g
$$
are neither helpful, too, by the similar reasoning.
Could you please explain whether these multiple copies of the same irreps can be 'filtered out'?