How is $\mathcal L(E)$ graded? Question about Hilbert $B$-module I am reading G.G.KASPAROV's THE OPERATOR AT-FUNCTOR
AND EXTENSIONS OF C*-ALGEBRAS
Let $B$ be a C*-algebra and let $E$ be a $B$-right-module. Assume there is a $B$-valued inner product on $E$ which (besides the usual axioms of an inner product) satisfies $\langle x,yb\rangle =\langle x,y\rangle b$ for $x,y\in E$ and $b\in B$. Then call $E$ a pre-Hilbert $B$-module. When $E$ is complete, call it a Hilbert $B$-module.
Denote by $\mathcal L(E)$ the linear operators on $E$ which have adjoint operators. Such operators are necessarily bounded.
Assume $B=B^0\oplus B^1$ as a linear space, where $B^i$ are closed selfadjoint linear subspaces, and $B^iB^i\subset B^0$, $B^iB^{1-i}\subset B^1$. Then say $B$ is graded. Assume$$
 E=E^0\oplus E^1\\\langle E^i,E^i\rangle \subset B^0\\\langle E^i,E^{1-i}\rangle \subset B^1\\E^iB^i \subset E^0 \\E^iB^{1-i}\subset E^1
$$
Now say $E$ is graded.
proposition. $\mathcal L(E)$ is graded as a C*-algebra, and $\mathcal L(E)^iE^i\subset E^0$, $\mathcal L(E)^i E^{1-i}\subset E^1$ .

My confusion is that, it seems that $\mathcal L(E)$ is graded in a trivial way to me, in other words, $\mathcal L(E)^1=0$. Since $\mathcal L(E)$ is a C*-algebra, we can decompose any element in $\mathcal L(E)$ into linear combination of positive elements. Now Assume $A=T^*T$ is positive, then we have $\langle T^*Tx,x\rangle =\langle T^*x,T^*x\rangle \in B^0$, therefore $T^*T$ must preserve degree of $x$, which means $T^*T\in \mathcal L(E)^0$. Therefore any linear combination of positive elements is in $\mathcal L(E)^0$.
Am I wrong?
 A: The trivial grading on $\mathcal L(E)$ does not necessarily respect the grading on $E$.  For example, suppose $B$ is trivially graded,
$$E=\left\{x\in\ell^\infty(B):\sum_{n\in\mathbb N}x(n)^*x(n)\text{ converges in }B\right\},$$
and grade $E$ as follows:
\begin{align*}
E^0&=\left\{x\in E:x(n)=0\text{ when $n$ is odd}\right\},\\
E^1&=\{x\in E:x(n)=0\text{ when $n$ is even}\},
\end{align*}
Now define $s\in\mathcal L(E)$ by $sx(n)=x(n-1)$ for $n>1$ and $sx(1)=0$. Then $s$ swaps $E^0$ and $E^1$, but in the trivial grading, the only operator which does this is the zero operator.
Instead, one should consider the grading on $\mathcal L(E)$ defined as follows:  Define $u\in\mathcal L(E)$ by
$$u(x_0+x_1)=x_0-x_1,$$
for $x_i\in E^i$, $i=0,1$. It follows that $u$ is a self-adjoint unitary, so we obtain an automorphism $\alpha$ of $\mathcal L(E)$ defined by $\alpha(t)=utu$.  Now define
$$\mathcal L(E)^i=\{t\in\mathcal L(E):\alpha(t)=(-1)^it\},\qquad(i=0,1).$$
Then we have a direct sum decomposition $\mathcal L(E)=\mathcal L(E)^0\oplus\mathcal L(E)^1$, which defines a grading for $\mathcal L(E)$ satisfying $\mathcal L(E)^iE^j\subset E^{i+j}$ for $i,j\in\{0,1\}$.
