# Definition An irreducible [[Group Representation|representation]] (irrep) cannot be expressed in terms of representations of lower [[Dimension|dimensionality]]. In other words, if by *one and the same* equivalence transformation, all the matrices in the representation of a [[group]] can be made to acquire the same block form, then the representation is said to be *reducible*, otherwise, it is *irreducible*. We can form a reducible representation $\Gamma_R$ from irreducible representations $\Gamma_i$ as: $ \Gamma_R = \begin{pmatrix} \Gamma_1 & \mathbb{0} & \mathbb{0} & \mathbb{0} & \ldots \\ \mathbb{0} & \Gamma_2 & \mathbb{0} & \mathbb{0} & \ldots \\ \mathbb{0} & \mathbb{0} & \Gamma_3 & \mathbb{0} & \ldots \\ \mathbb{0} & \mathbb{0} & \mathbb{0} & \Gamma_4 & \ldots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix} = \sum_i \Gamma_i $ where the last equality is a notational convention for listing the irreducible representations in a reducible representations. # Theorem 1 A necessary and sufficient condition for two irreducible representations to be equivalent is that the [[Character of a Representation|characters]] be the same. ### Proof *Necessary Condition:* If they are equivalent, i.e., they are related by a [[similarity transformation]], then the characters are the same by virtue of the properties of the [[trace]], through which the [[Character of a Representation|character]] is defined. *Sufficient Condition:* If the characters are the same, then the [[vector|vectors]] for each of the irreducible representations in $h$-dimensional [[Group Representation#The Vector Space of Representations|vector space of representations]] cannot be orthogonal. See the theorem on [[Orthogonality Relations for Character]]. # Theorem 2 The number of irreducible representations is equal to the number of classes. ### Proof Starting from the first [[Orthogonality Relations for Character#First Orthogonality Relation|orthogonality relation of character]], and assuming there are $k$ distinct classes, and each class $\mathcal{C}_{k'}$ contains $N_{k'}$ elements $ \begin{align} \sum_{k'=1}^k N_{k}' \left[\chi^{(\Gamma_i)}(\mathcal{C}_k')\right]^* \chi^{(\Gamma_{j})}(\mathcal{C}_{k'}) &= h \delta_{\Gamma_i, \Gamma_{j}},\\ \sum_{k'=1}^k \left[ \sqrt{\frac{N_{k'}}{h}}\chi^{(\Gamma_i)}(\mathcal{C}_k')\right]^* \left[\sqrt{\frac{N_{k'}}{h}} \chi^{(\Gamma_{j})}(\mathcal{C}_{k'})\right] &= \delta_{\Gamma_i, \Gamma_{j}}.\\ \end{align} $ Each term $ \left[\sqrt{\frac{N_{k'}}{h}} \chi^{(\Gamma_{j})}(\mathcal{C}_{k'})\right] $ is the $k'^{\text{th}}$ component of a $k$-dimensional [[vector]] corresponding to the irreducible representation $\Gamma_j$. Now, how many of these vectors are there? In a $k$-dimensional [[vector space]], there can only be $k$ [[Linear Independence|linearly independent]] vectors, each with $k$ [[components]]. If we have more than $k$ vectors, the $(k+1)^{\text{st}}$ would be linearly dependent on the rest. If we have fewer than $k$, then the number of independent vectors would not be large enough to span the entire space. Thus, we must have exactly $k$ vectors and thus $k$ irreducible representations. Consequently, we have as many irreducible representations as we do classes. The number of irreducible representations of an [[Abelian group]] is equal to the [[Group Order|order]] of the group since each [[group element|element]] in an [[Abelian group]] forms its own [[Conjugacy Class|class]]. # The Vector Space of Representations Consider the [[Wonderful Orthogonality Theorem]]: $ \sum_{g} D_{\mu\nu}^{(\Gamma_j)}(g)\left[D_{\mu'\nu'}^{(\Gamma_{j'})}(g)\right]^* = \frac{h}{\ell_j} \delta_{\Gamma_j, \Gamma_{j'}} \delta_{\mu\mu'} \delta_{\nu\nu'} $ We can consider this to be the orthogonality relation of a [[vector space]] of [[dimension]] $h$, where $h$ is the [[Group Order|order]] of the group we are working with. The representations $D_{\mu, \nu}^{(\Gamma_j)}(g)$ can be considered as elements of this [[vector space]]: $ V_{\mu, \nu}^{(\Gamma_j)} = \left[D_{\mu, \nu}^{(\Gamma_j)}(g_1), D_{\mu, \nu}^{(\Gamma_j)}(g_2), \ldots, D_{\mu, \nu}^{(\Gamma_j)}(g_h)\right] $ where each of the three indices $\mu, \nu, \Gamma_j$ labels a vector $V_{\mu, \nu}^{(\Gamma_j)}$ in this space. All distinct vectors in this space are orthogonal, and thus, two representations are orthogonal if one of their three indices is different. However, in an $h$-dimensional vector space the maximum number of [[Orthogonality|orthogonal]] vectors is $h$, so how many vectors $V_{\mu, \nu}^{(\Gamma_j)}$ can we make? For each representation $\Gamma_j$, we have $\ell_j$ choices for $\mu$ and and $\nu$, and we need to sum over all the representations, so we get: $ \sum_j \ell_j^2 \leq h $ # Theorem 3 The [[group order|order] of a [[group]] $h$ and the [[dimension|dimensionality]] $\ell_j$ of its irreducible representations $\Gamma$ are related by: $ \sum_j \ell_j = h $ ### Proof By construction, the [[Regular Representation]] has dimensionality $h$, which comes from the [[Group Multiplication Table|multiplication table]] since it contains every [[group element|element]] of the [[group]]. However, each irreducible representation is contained $\ell_j$ times in the [[Regular Representation]], as shown in the theorem [[Regular Representation#Theorem|here]]. Thus, $ \chi^{\text{reg}}(e) = h = \sum_{\Gamma_j} \alpha_j \chi^{(\Gamma_j)}(e) $ We have $\chi^{(\Gamma_j)}(e) = \ell_j$ because it's just the [[trace]] of an identity matrix of dimensions $\ell_j\times\ell_j$, and $a_j$ is the number of times $\chi^{(\Gamma_j)}$ occurs in $\Gamma^{\text{reg}}$. This is also just just $\ell_j$ as discuses and cited above. Thus, we have $ \sum_j \ell_j^2 = h. \quad \square $ # In Physics In physics, each irreducible representation describes the transformation properties of a set of [[eigenstate|eigenstates]] and corresponds to distinct energy [[Eigenvalue|eigenvalues]]. Let's assume we have an irreducible representation $\Gamma_R$ of a group $G$: $ \Gamma_R = \Gamma_1 + \Gamma_1' + \Gamma_2 $ where the indices of the irreducible representations represent their dimensionality. Let us also assume that it is an irreducible representation of some other group $G'$. This means that, some perturbation changes the symmetry group of the system from $G$ to $G'$, breaking up its eigenstates from 4 doubly degenerate bands to two non-degenerate bands, and two-degenerate bands (non-degenerate with the first pair). Group theory does not tell us anything about the ordering, values, or anything else about these energies, only their degeneracies.