Proposition 1

Let $\varrho_1$ and $\varrho_2$ be unitary representations of a topological group G in Hilbert spaces $E_1$ and $E_2$. The mapping from Hom$_G(\varrho_1, \varrho_2)$ into Sesq$_G(\varrho_2, \varrho_1)$ which to $u$ associates the sesquilinear form $q_u$ defined by $q_u(x, y) =\langle x|u(y)\rangle$ is an isomorphism of vector spaces.

If $u$ is a G-morphism, then one has

for all $g\in G$ and all $(x, y)\in E_2\times E_1$, hence the indicated mapping is a linear mapping from Hom$_G(\varrho_1, \varrho_2)$ into Sesq$_G(\varrho_2, \varrho_1)$. By EVT, V, p. 16, Corollary 2, it is injective.

Conversely, let $q$ be a G-invariant sesquilinear form and $u$ the unique linear mapping from $E_1$ into $E_2$ such that $q(x, y) =\langle x|u(y)\rangle$ for all $(x, y)\in E_2\times E_1($loc. cit.). For every $g$ in G and every $(x, y)$ in $E_2\times E_1$, one has

hence $u=\varrho_2(g)\circ u\circ \varrho_1(g^{-1})$. Consequently, $u\in$ Hom$_G(\varrho_2, \varrho_1)$. The proposition follows.

Let $\varrho$ be a unitary representation of a topological group G in a Hilbert space E. If $K =\mathbf{C}$, let $\overline{E}$ denote the conjugate space of E (EVT, V, p. 6). If $K =\mathbf{R}$, put $\overline{E}= E$. The conjugate representation $\overline{\varrho}$ is the representation of G in $\overline{E}$ defined by $\overline{\varrho}(g) =\varrho (g)$ for all $g\in G$. It is a unitary representation of G; by definition, a subspace of $\overline{E}$ is a subrepresentation of $\overline{E}$ if and only if it is a subrepresentation of E.