# Generating the partners in a multi-dimensional irreducible representation.

I am trying to block diagonalize a Hermitian matrix using the irreducible representations of its symmetry group.

Using the group's character table, it is straightforward to generate a set of projection matrices that will project an arbitrary vector onto the irreducible sub-space, as follows

$$\mathbf{P}^{(\Gamma_n)} = \frac{l_n}{h} \sum_R \chi^{(\Gamma_n)}(R) P_R$$

where $\Gamma_n$ is the representation, $l_n$ is its dimensionality, $h$ is the size of the group, $\chi^{(\Gamma_n)}(R)$ is the character for the $R^\text{th}$ operation in the representation, and $P_R$ is the $R^\text{th}$ matrix operator. For example, looking at the 2-dimensional $E$ representation of the symmetry of the square in $\mathbb{R}^3$ gives

$$\mathbf{P}^E = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{array} \right)$$

For this simple case, it is apparent that given $(\begin{array}{ccc}1&0&0\end{array})$ it's partner in the representation is $(\begin{array}{ccc}0&1&0\end{array})$. But, for cases where you have multiple representations of the same type, it is not clear how to generate the basis for each one as $\mathbf{P}^{(\Gamma_n)}$ projects onto all of the representations of $\Gamma_n$ present. For instance, consider the case of the symmetry of the triangle where d-orbitals (quadratic functions) form two $E$ representations.

Both Tinkham and Dresselhaus, et al., give the formula

$$\mathrm{P}^{(\Gamma_n)}_{kl} = \frac{l_n}{h} \sum_R D^{(\Gamma_n)}(R)^\ast_{kl}P_R$$

where

$$D^{(\Gamma_n)}(R)^\ast_{kl} = \mathbf{e}^{(\Gamma_n)}_k P_R \mathbf{e}^{(\Gamma_n)\ast}_l$$

given that $\mathbf{e}^{(\Gamma_n)}_k$ is the $k^\text{th}$ vector in the irreducible representation $\Gamma_n$ which is related to $\mathbf{P}^{(\Gamma_n)}$ via

$$\mathbf{P}^{(\Gamma_n)} = \sum_k \mathrm{P}^{(\Gamma_n)}_{kk}.$$

When $\mathrm{P}^{(\Gamma_n)}_{kl}$ is applied to $k^\text{th}$ vector gives the $l^\text{th}$ vector in the representation, in other words, given a vector in the representation you can generate all its partners.

My question is how do I generate $D^{(\Gamma_n)}(R)^\ast_{kl}$ when I only know $\mathbf{P}^{(\Gamma_n)}$? This seems like a chicken and egg problem.

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Since this has received no traction here, I've asked over at mathoverflow. –  rcollyer Jun 1 '13 at 1:50