I am reading THE OPERATOR K-FUNCTOR AND EXTENSIONS OF $C^*$-ALGEBRAS To cite this article: G G Kasparov 1981 Math. USSR Izv. 16 513
Definition 1.1. $G$ isa fixed compact group satisfying the second axiom of countability which acts as an automorphism group on all algebras under consideration.
Definition 1.11. The Hilbert space over $B$ is defined as the Hilbert direct sum $\mathcal H_B=\bigoplus _{i=1}^\infty (V_i\otimes _C B)$ where $V_i$ is a countable collection of finite-dimensional Euclidean spaces in which all irreducible unitary representations of $G$ are realized. Each of the $V_i$ has to be repeated infinitely many times in the collection under consideration. The inner product on $V_i\otimes B$ is defined by $\langle x\otimes a,y\otimes b\rangle =\langle x,y\rangle a^*b$.
Definition 2.3. A canonically graded Hilbert space is defined as the direct sum $\mathcal H_B\oplus \mathcal H_B$, where the gradings of the two summands are opposite and the grading of the first one is defined in such a way that for all $j$, all $x\in V_j$, and all $b\in B$, $\text{deg}(x\otimes b)=\text{deg}(b)$.
Definition 4.2. We identify $\mathcal H_B\oplus \mathcal H_B$ with $\mathcal H_B$ by mean of an $G$-invariant isometry of degree $0$.
My question is, why does such isometry exist? Consider the restriction on direct summands like $((V\otimes B)\oplus (V\otimes B)\oplus ...)\oplus ((V\otimes B)\oplus (V\otimes B)\oplus ...)\to (V\otimes B)\oplus (V\otimes B)\oplus ...$. If we let $\phi((\alpha_1\oplus \alpha_2\oplus ...)\oplus (\beta_1\oplus \beta_2\oplus ...))=(\alpha_1\oplus \beta_1\oplus\alpha_2\oplus \beta_2\oplus ...)$, then $\phi$ is an isomorphism of Hilbert $B$-modules. However, such map may not preserve degree. For instance, assume $B$ is trivially graded so that $B^0=B$ and $B^1={0}$, then $\mathcal H_B$ is also graded trivially, while $\mathcal H_B\oplus \mathcal H_B$ is not, since the grading on two summands are opposite.
Did I get it wrong?