# First Orthogonality Relation This follows from the [[Wonderful Orthogonality Theorem]]. This orthogonality relation states that the [[Primitive Character|primitive characters]] of an [[Irreducible Representation]] obey the relation: $ \sum_g \chi^{(\Gamma_j)}(g) \chi^{(\Gamma_{j'})}(g^{-1}) = h \delta_{\Gamma_j \Gamma_{j'}} $ or $ \sum_g \chi^{(\Gamma_j)}(g) \chi^{(\Gamma_{j'})}(g)^* = h \delta_{\Gamma_j \Gamma_{j'}} $ where $\Gamma_j$ is the [[Irreducible Representation]] $j$ with [[Dimension|dimensionality]] $\ell_j$. These can also be written as: $ \begin{align} \sum_{k} N_k \chi^{(\Gamma_j)}(\mathcal{C}_k) \left[\chi^{(\Gamma_{j'})}(\mathcal{C}_k)\right]^* &= h \delta_{\Gamma_j, \Gamma_{j'}},\\ \sum_{k} N_k \left[\chi^{(\Gamma_j)}(\mathcal{C}_k)\right]^* \chi^{(\Gamma_{j'})}(\mathcal{C}_k) &= h \delta_{\Gamma_j, \Gamma_{j'}}, \end{align} $ where $N_k$ is the number of elements in a [[conjugacy class]] $\mathcal{C}_k$. Thus, unless the representations are identical or equivalent, the characters are orthogonal in $h$-dimensional space, where $h$ is the [[group order|order]] of the [[group]]. This orthogonality relation is between the *rows* of the [[Character Table]]. ## Proof Starting from the [[Wonderful Orthogonality Theorem]]: $ \sum_{g} D_{\mu\nu}^{(\Gamma_j)}(g)D_{\nu'\mu'}^{(\Gamma_{j'})}(g^{-1}) = \frac{h}{\ell_j} \delta_{\Gamma_j, \Gamma_{j'}} \delta_{\mu\mu'} \delta_{\nu\nu'} $ Take the diagonal elements on both sides, i.e. $\mu = \nu$ and $\mu' = \nu'$ $ \sum_{g} D_{\mu\mu}^{(\Gamma_j)}(g)D_{\mu'\mu'}^{(\Gamma_{j'})}(g^{-1}) = \frac{h}{\ell_j} \delta_{\Gamma_j, \Gamma_{j'}} \delta_{\mu\mu'} \delta_{\mu\mu'} $ Now sum over $\mu$ and $\mu'$ to get the [[trace]]: $ \begin{align} \sum_{g} \sum_\mu D_{\mu\mu}^{(\Gamma_j)}(g) \sum_{\mu'}D_{\mu'\mu'}^{(\Gamma_{j'})}(g^{-1}) &= \frac{h}{\ell_j} \delta_{\Gamma_j, \Gamma_{j'}} \sum_{\mu\mu'}\delta_{\mu\mu'} \delta_{\mu\mu'}, \\ \sum_{g} \chi^{(\Gamma_j)}(g) \chi^{(\Gamma_{j'})}(g^{-1}) &= \frac{h}{\ell_j} \delta_{\Gamma_j, \Gamma_{j'}} \sum_{\mu} \delta_{\mu\mu}, \\ \sum_{g} \chi^{(\Gamma_j)}(g) \chi^{(\Gamma_{j'})}(g^{-1}) &= h \delta_{\Gamma_j, \Gamma_{j'}} \quad \square \end{align} $ Since any arbitrary representation is [[Theorem on the Unitarity of Representations|equivalent to a some uintary representation]], and the [[trace]] (and thus the character) is invariant under [[Similarity Transformation|similarity transformations]], then we can rewrite the theorem as $ \sum_{g} \chi^{(\Gamma_j)}(g) \left[\chi^{(\Gamma_{j'})}(g)\right]^* = h \delta_{\Gamma_j, \Gamma_{j'}}. $ Essentially, we have used the fact that for unitary representations, we can write $\chi^{\Gamma}(g^{-1}) = [\chi^{\Gamma}(g)]^*$, which we can prove easily. For unitary representations, we have: $ D^\Gamma(g)^\dagger = D^\Gamma(g)^{-1} = D^\Gamma(g^{-1}) $ where the last step follows from the properties of [[Group Representation|matrix representations]]. Then, taking the [[trace]]: $ \text{trace}[D^\Gamma(g)^\dagger] = \text{trace}[D^\Gamma(g)]^* = [\chi^\Gamma(g)]^* = \text{trace}[D^\Gamma(g^{-1}] = \chi^\Gamma(g^{-1}) $ Moreover, since [[Character of a Representation|character]] is the same for each [[group element|element]] of the same [[conjugacy class]], we can rewrite the theorem as a sum over classes $ \sum_{k} N_k \chi^{(\Gamma_j)}(\mathcal{C}_k) \left[\chi^{(\Gamma_{j'})}(\mathcal{C}_k)\right]^* = h \delta_{\Gamma_j, \Gamma_{j'}}, $ Since the right hand side is real, we also get: $ \sum_{k} N_k \left[\chi^{(\Gamma_j)}(\mathcal{C}_k)\right]^* \chi^{(\Gamma_{j'})}(\mathcal{C}_k) = h \delta_{\Gamma_j, \Gamma_{j'}}, $ where $N_k$ is the number of elements in a [[conjugacy class]] $\mathcal{C}_k$. # Second Orthogonality Relation The following orthogonality relation holds for irreducible representations: $ \sum_{\Gamma_j} N_k \chi^{(\Gamma_j)}(\mathcal{C}_k) \left[\chi^{(\Gamma_{j})}(\mathcal{C}_{k'})\right]^* = h \delta_{k k' },\\ $ This orthogonality relation is between the *columns* of the [[Character Table]]. ## Proof We write down the following two matrices: $ \begin{align} Q_{ij} &= \chi^{(\Gamma_i)}(\mathcal{C}_j), \\ Q'_{ij} &= \frac{N_i}{h}[\chi^{(\Gamma_j)}(\mathcal{C}_i)]^* \end{align} $ Note that both of these matrices are square since the number of classes is equal to the number of irreducible representations (see proof [[Irreducible Representation#Theorem 2|here]]). The $ij$ matrix element of the product is then (summing over classes): $ \begin{align} (Q Q')_{ij} = \sum_{k} \frac{N_k}{h}\chi^{(\Gamma_i)}(\mathcal{C}_k)\left[\chi^{(\Gamma_j)}(\mathcal{C}_k)\right]^* = \delta_{\Gamma_i\Gamma_j} \end{align} $ where the last equality follows from the [[#First Orthogonality Relation|first orthogonality relation]]. Thus, we have: $ Q Q' = I \iff Q' = Q^{-1} \iff Q = (Q')^{-1} \iff (Q Q') = I $ Let's write $(Q' Q)$ again and take the $k k'$ element (summing over irreducible representations): $ \begin{align} (Q' Q)_{k k'} &= \frac{N_k}{h} \sum_{\Gamma_i} [\chi^{(\Gamma_i)}(\mathcal{C}_k)]^* \chi^{(\Gamma_i)}(\mathcal{C}_{k'}) = \delta_{k k'}, \\ &= \frac{N_k}{h} \sum_{\Gamma_i} \chi^{(\Gamma_i)}(\mathcal{C}_k) [\chi^{(\Gamma_i)}(\mathcal{C}_{k'})]^* = \delta_{kk'} \end{align} $ And we end up with $ \sum_{\Gamma_i} N_k \chi^{(\Gamma_i)}(\mathcal{C}_k) [\chi^{(\Gamma_i)}(\mathcal{C}_{k'})]^* = h\delta_{kk'}. \quad \square $