Corollary 6

Let $G_1$ and $G_2$ be topological groups. Let $\varrho_1$ be an irreducible unitary representation of $G_1$ in a Hilbert space $E_1$ and $\varrho_2$ an irreducible unitary representation of $G_2$ in a Hilbert space $E_2$. The external tensor product $\varrho_1\boxtimes \varrho_2$ is an irreducible unitary representation of $G_1\times G_2$ in $E_1\widehat{\otimes}_2E_2$.

The external tensor product $\varrho =\varrho_1\boxtimes \varrho_2$ is a unitary representation of $G = G_1\times G_2$ in the space $E = E_1\widehat{\otimes}_2E_2$ by Lemma 8 of V, p. 384.

Let $q$ be a $(G_1\times G_2$)-invariant sesquilinear form on E. For every pair $(x_1, y_1)\in E^2_1$, the mapping $(x_2, y_2)\mapsto q(x_1\otimes x_2, y_1\otimes y_2)$ belong to Sesq$_{G_2}(\varrho_2, \varrho_2)$. Let us denote by $b(x_1, y_1)$ the unique complex number such that

for every $(x_2, y_2)\in E^2_2$ (Cor. 3). Let $\varepsilon \in E_2$ be of norm 1, so that $b(x_1, y_1) =q(x_1\otimes \varepsilon , y_1\otimes \varepsilon )$ for every $(x_1, y_1)\in E^2_1$. This formula implies that the mapping $b$ is sesquilinear on $E_1$. Moreover

for every $(x_1, y_1)\in E^2_1$ (TVS, V, p. 26, formula (5)), hence $b$ is continuous.

Let $g\in G$ and $(x_1, y_1)\in E^2_1$. Since $\varrho_2$ is unitary and $q$ is invariant, we get

so that $b\in$ Sesq$_{G_1}(\varrho_1, \varrho_1)$. There therefore exists a unique $\lambda \in \mathbf{C}$ such that $b(x_1, y_1) =\lambda \langle x_1|y_1\rangle$ for every $(x_1, y_1)\in E^2_1$ (Cor. 3), that is to say

for every $(x_1, y_1, x_2, y_2)\in E^2_1\times E^2_2$. It follows that the space Sesq$_{G_1\times G_2}(\varrho , \varrho )$ is of dimension 1, which implies that the representation $\varrho =\varrho_1\boxtimes \varrho_2$ is irreducible (loc. cit.).

We recall that $\mathbf{U}$ denotes the group of complex numbers of modulus 1.