# Theorem The reduction of any [[Irreducible Representation|reducible repressentation]] into its [[Irreducible Representation|irreducible]] constituents is unique. Thus, if $\chi(\mathcal{C}_k)$ is the [[Character of a Representation|character]] for some [[conjugacy class]] $\mathcal{C}_k$ in a reducible representation, this theorem says that we can write the characters for the reducible representation $\chi(\mathcal{C}_k)$ as a linear combination of the characters for the *irreducible* representations of the [[group]] $\chi^{(\Gamma_i)}(\mathcal{C}_k)$ $ \chi(\mathcal{C}_k) = \sum_{\Gamma_i} \alpha_i \chi^{(\Gamma_i)}(\mathcal{C}_k) $ where the $\alpha_i$ are non-negative integers representing the number of times the irreducible representation $\Gamma_i$ is contained in the reducible representation. Furthermore, the theorem states that the $\alpha_i$ are unique. ## Proof We prove here that the $\alpha_i$ are unique. ==Is this enough to prove the whole theorem?== Consider the following sum over classes: $ \sum_k N_k [\chi^{(\Gamma_j)}(\mathcal{C}_k)]^* \chi(\mathcal{C}_k) \equiv S_j $ Since $\chi(\mathcal{C}_k)$ is reducible, we can write it as a linear combination as stated in the theorem's statement above. Plugging in, $ \begin{align} S_j &= \sum_k N_k [\chi^{(\Gamma_j)}(\mathcal{C}_k)]^* \sum_{\Gamma_i} \alpha_i \chi^{(\Gamma_i)}(\mathcal{C}_k), \\ &= \sum_{\Gamma_i} \alpha_i \left[ \sum_k N_k [\chi^{(\Gamma_j)}(\mathcal{C}_k)]^* \chi^{(\Gamma_i)}(\mathcal{C}_k) \right] \end{align} $ Applying the [[Orthogonality Relations for Character#First Orthogonality Relation|first orthogonality relation for character]] $ S_j = \sum_{\Gamma_i} \alpha_i h \delta_{\Gamma_i \Gamma_j} = \alpha_j h $ So we end up with $ \alpha_j = S_j/h =\frac{1}{h} \sum_k N_k [\chi^{(\Gamma_j)}(\mathcal{C}_k)]^* \chi(\mathcal{C}_k). \quad \square $ Thus, the coefficients $\alpha_i$ in the decomposition are uniquely determined. Thus, we can directly obtain the number of times the various irreducible representations occur in a reducible representation directly from the [[Character Table]].