Direct sums and tensor products of finite-dimensional unitary representations

On $V_1\oplus V_2$ with the inner product $⟨(v_1,v_2),(w_1,w_2)⟩=⟨v_1,w_1⟩+⟨v_2,w_2⟩$ define $({\pi }_1\oplus {\pi }_2)(k)(v_1,v_2)=({\pi }_1(k)v_1,{\pi }_2(k)v_2)$; the group law holds componentwise, and $\mathrm{\parallel }({\pi }_1\oplus {\pi }_2)(k)(v_1,v_2){\mathrm{\parallel }}^2=\mathrm{\parallel }{\pi }_1(k)v_1{\mathrm{\parallel }}^2+\mathrm{\parallel }{\pi }_2(k)v_2{\mathrm{\parallel }}^2=\mathrm{\parallel }(v_1,v_2){\mathrm{\parallel }}^2$ shows that each operator is an isometry, bijective with inverse $({\pi }_1\oplus {\pi }_2)(k^{-1})$, hence unitary by [F7], while strong continuity follows from $\mathrm{\parallel }({\pi }_1\oplus {\pi }_2)(k)(v_1,v_2)-({\pi }_1\oplus {\pi }_2)(k_0)(v_1,v_2){\mathrm{\parallel }}^2=\mathrm{\parallel }{\pi }_1(k)v_1-{\pi }_1(k_0)v_1{\mathrm{\parallel }}^2+\mathrm{\parallel }{\pi }_2(k)v_2-{\pi }_2(k_0)v_2{\mathrm{\parallel }}^2\to 0$. The disjoint union of bases of $V_1$ and $V_2$ is a basis of $V_1\oplus V_2$, so the dimension is $d_1+d_2$ by [F5], and expanding the inner product gives $c_{(v_1,v_2),(w_1,w_2)}^{{\pi }_1\oplus {\pi }_2}(k)=⟨{\pi }_1(k)v_1,w_1⟩+⟨{\pi }_2(k)v_2,w_2⟩=c_{v_1,w_1}^{{\pi }_1}(k)+c_{v_2,w_2}^{{\pi }_2}(k)$ for all vectors and all $k$, which is (1).

Fix orthonormal bases $(e_i{)}_{i\le d_1}$ of $V_1$ and $(f_j{)}_{j\le d_2}$ of $V_2$ [F2]; the elementary tensors $e_i\otimes f_j$ span $V_1\otimes V_2$ because $v\otimes w={\sum }_{i,j}⟨v,e_i⟩⟨w,f_j⟩e_i\otimes f_j$ by the expansion of $v$ and $w$, and they are linearly independent because a relation ${\sum }_{i,j}{\lambda }_{ij}{e}_i\otimes f_j=0$ returns ${\lambda }_{i_0{j}_0}=0$ when one applies the linear functional induced by the bilinear form $(v,w)\mapsto ⟨v,e_{i_0}⟩⟨w,f_{j_0}⟩$ through [F4]; hence they form a basis and $\dim (V_1\otimes V_2)=d_1{d}_2$ by [F5]. Declaring this basis orthonormal makes $V_1\otimes V_2$ a finite-dimensional complex Hilbert space, and the same expansions give $⟨v\otimes w,v^{\mathrm{\prime }}\otimes w^{\mathrm{\prime }}⟩={\sum }_{i,j}⟨v,e_i⟩⟨w,f_j⟩\overline{⟨v^{\mathrm{\prime }},e_i⟩⟨w^{\mathrm{\prime }},f_j⟩}=⟨v,v^{\mathrm{\prime }}⟩⟨w,w^{\mathrm{\prime }}⟩$ for all elementary tensors, so such an inner product exists and is unique with this property because the elementary tensors span.

Because ${\pi }_1(k)$ and ${\pi }_2(k)$ are unitary, the elementary-tensor formula of step 1.2 and the action [F4] give $⟨({\pi }_1\otimes {\pi }_2)(k)(v\otimes w),({\pi }_1\otimes {\pi }_2)(k)(v^{\mathrm{\prime }}\otimes w^{\mathrm{\prime }})⟩=⟨{\pi }_1(k)v,{\pi }_1(k)v^{\mathrm{\prime }}⟩⟨{\pi }_2(k)w,{\pi }_2(k)w^{\mathrm{\prime }}⟩=⟨v,v^{\mathrm{\prime }}⟩⟨w,w^{\mathrm{\prime }}⟩=⟨v\otimes w,v^{\mathrm{\prime }}\otimes w^{\mathrm{\prime }}⟩$ for all elementary tensors; both sides are sesquilinear and the elementary tensors span, so the identity holds on all of $V_1\otimes V_2$, each $({\pi }_1\otimes {\pi }_2)(k)$ is a bijective linear isometry, hence unitary by [F7], and $c_{v\otimes w,v^{\mathrm{\prime }}\otimes w^{\mathrm{\prime }}}^{{\pi }_1\otimes {\pi }_2}(k)=c_{v,v^{\mathrm{\prime }}}^{{\pi }_1}(k)c_{w,w^{\mathrm{\prime }}}^{{\pi }_2}(k)$ on elementary tensors. For finite expansions $u={\sum }_a{\alpha }_a{v}_a\otimes w_a$ and $u\mathrm{\prime }={\sum }_b{\beta }_b{v}_b\mathrm{\prime }\otimes w_b\mathrm{\prime }$, sesquilinearity instead gives $c_{u,u\mathrm{\prime }}^{{\pi }_1\otimes {\pi }_2}(k)={\sum }_{a,b}{\alpha }_a\overline{{\beta }_b}{c}_{v_a,v_b\mathrm{\prime }}^{{\pi }_1}(k)c_{w_a,w_b\mathrm{\prime }}^{{\pi }_2}(k)$, a finite sum of products. For strong continuity write $u={\sum }_a{c}_a{v}_a\otimes w_a$ as a finite sum; then [F3] gives $\mathrm{\parallel }({\pi }_1\otimes {\pi }_2)(k)u-({\pi }_1\otimes {\pi }_2)(k_0)u\mathrm{\parallel }\le {\sum }_a\mathrm{\mid }{c}_a\mathrm{\mid }(\mathrm{\parallel }{\pi }_1(k)v_a\mathrm{\parallel }\mathrm{\parallel }{\pi }_2(k)w_a-{\pi }_2(k_0)w_a\mathrm{\parallel }+\mathrm{\parallel }{\pi }_1(k)v_a-{\pi }_1(k_0)v_a\mathrm{\parallel }\mathrm{\parallel }{\pi }_2(k_0)w_a\mathrm{\parallel })\to 0$ as $k\to k_0$, using $\mathrm{\parallel }x\otimes y\mathrm{\parallel }=\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$ from step 1.2 and the strong continuity of ${\pi }_1$ and ${\pi }_2$; hence ${\pi }_1\otimes {\pi }_2$ is strongly continuous, which completes (2).

The trivial representation $1_K(k)z=z$ on $\mathbb{C}$ is a continuous unitary representation whose matrix coefficient at the unit vector $1$ is the constant function $1$. Now let $\pi$ be finite-dimensional continuous unitary on $V$ with orthonormal basis $e_1,\dots ,e_d$, and let $\sigma (k)$ be the linear operator whose matrix in this basis is the entrywise conjugate of that of $\pi (k)$, that is $⟨\sigma (k)e_i,e_j⟩=\overline{⟨\pi (k)e_i,e_j⟩}$ for all $i,j$; then $\sigma (k)=J\pi (k)J$, where $J({\sum }_i{a}_i{e}_i)={\sum }_i\overline{a_i}{e}_i$ and $J^2=I$, so $\sigma (kh)=J\pi (k)\pi (h)J=(J\pi (k)J)(J\pi (h)J)=\sigma (k)\sigma (h)$, so $\sigma$ is a homomorphism, and $\sigma (k)$ is unitary because its matrix is the conjugate of the unitary matrix of $\pi (k)$; each matrix entry $k\mapsto ⟨\sigma (k)e_i,e_j⟩$ is continuous as the conjugate of a continuous function, so $\sigma$ is strongly continuous, since $\mathrm{\parallel }(\sigma (k)-\sigma (k_0))x\mathrm{\parallel }\le {\sum }_i\mathrm{\mid }{x}_i\mathrm{\mid }\mathrm{\parallel }(\sigma (k)-\sigma (k_0))e_i\mathrm{\parallel }\to 0$ for $x={\sum }_i{x}_i{e}_i$ by [F3]; expanding coefficients in the basis gives $\overline{c_{v,w}^{\pi }(k)}={\sum }_{i,j}\overline{⟨v,e_i⟩\stackrel{‾}{⟨w,e_j⟩}⟨\pi (k)e_i,e_j⟩}={\sum }_{i,j}\overline{⟨v,e_i⟩}⟨w,e_j⟩\overline{⟨\pi (k)e_i,e_j⟩}=⟨\sigma (k)Jv,Jw⟩=c_{Jv,Jw}^{\sigma }(k)$ for all $v,w\in V$ and all $k$; this proves (3).