Direct Sum of Group Representations
Introduction
Let two vector spaces $V_{1}$ and $V_{2}$ be given. Denote their direct sum by $V = V_{1} \oplus V_{2}$. Then $v \in V$ can be uniquely expressed as $v = v_{1} + v_{2} (v_{1} \in V_{1}, v_{2} \in V_{2})$.
Now consider a representation $(\rho_{1}, V_{1})$ of the group $G$.
$$ \rho_{1} : G \to \operatorname{GL}(V_{1}) $$
$\rho_{1}$ sends $g \in G$ to the linear transformation (= matrix) $\rho_{1}(g) : V_{1} \to V_{1}$. Similarly, for the representation $(\rho_{2}, V_{2})$ we have $\rho_{2}(g) : V_{2} \to V_{2}$. Based on this, we can define the direct sum of the two representations as follows.
Definition
The direct sum of two representations $(\rho_{1}, V_{1})$ and $(\rho_{2}, V_{2})$ of the group $G$ is denoted by $(\rho, V) = (\rho_{1} \oplus \rho_{2}, V_{1} \oplus V_{2})$ and is defined as follows.
$$ \begin{align*} \rho : G &\to \operatorname{GL}(V_{1} \oplus V_{2}) \\ g &\mapsto \rho(g) : V \to V \end{align*} $$
$$ \rho(g)(v) = \rho_{1}(g)(v_{1}) + \rho_{2}(g)(v_{2}), \quad \forall v = v_{1} + v_{2} \in V $$
Generalization
The direct sum $(\rho, V) = (\bigoplus_{i} \rho_{i}, \bigoplus V_{i})$ of the representations $(\rho_{i}, V_{i})$ is defined as follows.
$$ \rho(g) v = \sum_{i} \rho_{i}(g)(v_{i}), \quad \forall v = \sum_{i} v_{i} \in V $$
Explanation
Conversely, for a representation $(\rho, V)$ of the group $G$, suppose $V = \bigoplus_{i} V_{i}$ and that each $V_{i}$ is a subspace which is $\rho$-invariant. Then $\rho$ can be expressed as the direct sum of the representations $\rho_{i}(g) = \rho(g)|_{V_{i}}$ on the $V_{i}$.
$$ \rho = \bigoplus_{i} \rho_{i} = \bigoplus_{i} \rho(g)|_{V_{i}} $$
If we represent $\rho(g)$ as a matrix, it has the form below.
$$ \rho : g \mapsto \begin{bmatrix} [\rho_{1}(g)] & O \\ O & [\rho_{2}(g)] \end{bmatrix} $$
$$ [\rho(g)] = \begin{bmatrix} [\rho_{1}(g)] & O \\ O & [\rho_{2}(g)] \end{bmatrix} $$
In fact, this is the same as the direct sum of matrices.
Properties
(a) The direct sum of two representations of $G$ is also a representation of $G$.
Proof
(a)
It suffices to show $\rho(gh) = \rho(g)\rho(h)$. Viewing it in matrix form makes this clear. Since the product of diagonal matrices is a diagonal matrix,
$$ \begin{align*} \rho(g) \rho(h) &= \begin{bmatrix} \rho_{1}(g) & O \\ O & \rho_{2}(g) \end{bmatrix} \begin{bmatrix} \rho_{1}(h) & O \\ O & \rho_{2}(h) \end{bmatrix} \\ &= \begin{bmatrix} \rho_{1}(g)\rho_{1}(h) & O \\ O & \rho_{2}(g)\rho_{2}(h) \end{bmatrix} \\ &= \begin{bmatrix} \rho_{1}(gh) & O \\ O & \rho_{2}(gh) \end{bmatrix} \\ &= \rho(gh) \end{align*} $$
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