Decomposing arbitrary representations

In summary: It is then easy to check that this invariant has the desired property: it is a basis for which the given representation of ##D_3## returns to block-diagonal form.
  • #1
StaticZero
2
0
TL;DR Summary
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'?
 
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  • #2
Is this question ill-stated, am I missing something trivial or are there other problems with it?
 
  • #3
If you're just interested in the case ##G=D_3##, here's a possibility: let ##r## and ##s## be the usual generators of ##D_3##, viewed as linear operators on your ##4##-dimensional representation. Then ##r-s## has a nontrivial kernel (since in the ##2##-dimensional standard representation of ##D_3,## there is vertex on an equilateral triangle where these group elements act identically). Now if ##0\neq x\in\text{ker}(r-s),## note that ##\text{Span}(x,rx)## is a ##2##-dimensional invariant subspace. Apply the usual procedure to find an invariant complement.
 
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FAQ: Decomposing arbitrary representations

1. What is "decomposing arbitrary representations"?

"Decomposing arbitrary representations" refers to a mathematical process of breaking down a complex representation into smaller, more manageable parts. This is often done in the field of group theory, where representations of groups can be decomposed into simpler representations.

2. Why is it important to decompose arbitrary representations?

Decomposing arbitrary representations allows for a better understanding of the underlying structure and properties of a representation. It also makes it easier to analyze and manipulate the representation for various applications, such as in physics or computer science.

3. How is decomposing arbitrary representations done?

The process of decomposing arbitrary representations can vary depending on the specific representation and the desired outcome. However, it often involves using techniques from linear algebra, such as diagonalization or finding eigenvalues and eigenvectors.

4. What are the applications of decomposing arbitrary representations?

Decomposing arbitrary representations has various applications in mathematics, physics, and computer science. It is commonly used in the study of symmetry and group theory, as well as in quantum mechanics and signal processing.

5. Are there any limitations to decomposing arbitrary representations?

Yes, there can be limitations to decomposing arbitrary representations, particularly when the representation is highly complex or has a large number of dimensions. In these cases, it may be difficult to find a complete decomposition or the resulting decomposition may not be easily interpretable or useful.

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