Does localisation of a faithful retraction induce an isomorphism between adjointables of Hilbert modules?

Question

The title is quite a mouthful, so let me develop some context. All of this is from the book on Hilbert Modules by C. Lance.

If $A$ is a $C^*$-algebra, $M(A)$ its multiplier algebra and $B$ a sub-algebra of $M(A)$ then a positive linear map $\tau: A\to B$ is called a retraction if:

1. For all $a\in A, b\in B$: $\tau(ab)= \tau(a)b$
2. $\tau(A)$ is dense in $B$ relative to the strict topology.
3. There is an approximate identity $e_\alpha$ in $A$ so that $\tau(e_\alpha)$ converges to a projection in $B$.

A retraction is additionally called faithful if $\tau(a)>0$ for all positive $a>0$ in $A$.

If $E$ is a Hilbert $A$-module and $\tau: A\to B$ is a faithful retraction one can also give $E$ the structure of a Hilbert $B$ module via: $$x\cdot b := \lim_\alpha x\cdot e_\alpha \cdot b,\qquad \langle x,y\rangle_\tau := \tau(\langle x,y\rangle) \quad\text{for all $x,y\in E$, $b\in B$}.$$ Any adjointable map (wrt $\langle \cdot,\cdot\rangle$) $t:E\to E$ is also adjointable wrt $\langle\cdot,\cdot\rangle_\tau$, giving a $*$-morphism $$\pi_\tau:\mathcal L_A(E, \langle \cdot,\cdot\rangle)\to \mathcal L_B(E, \langle\cdot,\cdot\rangle_\tau),$$ $\pi_\tau$ is called the localisation of $\tau$. In the above case of a faithful retraction this map is clearly injective. On page 58 of his book Lance remarks without comment that it is actually an isomorphism.

Is the localisation $\pi_\tau$ also surjective for faithful $\tau$?

I think a simple (finite dimensional) counter-example is possible, but the above book is quite well regarded and it would be strange to have such an error, so I do not trust my counter-example.

Answer 1

I think I agree (like you, I find it hard to think that Lance was wrong...) that there is a problem here. For convenience of other readers, let me state that your counter-example is a special case of an example which Lance considers on page 58; see also page 39. Here I'll follow page 39. (A related question with I believe an incorrect answer).

Let $E$ be a Hilbert $A$-module, and consider $E^n$ as a Hilbert $M_n(A)$ module for the action and inner-product $$ x\cdot a = \big(\sum_i x_i \cdot a_{ij}\big)_{j=1}^n \quad \langle x,y\rangle = \big( \langle x_i, y_j \rangle \big)_{i,j=1}^n \qquad (x=(x_i)\in E^n, a=(a_{ij})\in M_n(A)). $$

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Let $T$ be a linear map $E^n\rightarrow E^n$ say $T(x) = \big( T_i(x) \big)$ for some linear maps $T_i:E^n\rightarrow E$, say $T_i(x) = \sum_{j=1}^n T_{i,j}(x_j)$ for linear maps $T_{i,j}:E\rightarrow E$. If $T$ is adjointable, it is in particular $M_n(A)$-linear (and bounded), so \begin{align*} T(x\cdot a) = T(x)\cdot a \quad & \implies\quad \big( T_i(x\cdot a) \big)_i = \big( T_j(x) \big)_j \cdot a \\ & \implies\quad \forall\,i\quad \sum_j T_{i,j}((x\cdot a)_j) = \sum_j T_j(x) \cdot a_{ji} \\ & \implies\quad \forall\,i\quad \sum_{j,k} T_{i,j}(x_k\cdot a_{kj}) = \sum_{j,k} T_{j,k}(x_k) \cdot a_{ji} \end{align*} If $x_i = \delta_{i,i_0} x$ and $a_{ij} = \delta_{i,i_1}, \delta_{j,j_0} a$ then we obtain $\delta_{i_0,i_1} T_{i,j_0}(x\cdot a) = \delta_{i,j_0} T_{i_1,i_0}(x) \cdot a$ for all $i$. So $T$ is "diagonal", and with $i_0=i_1, i=j_0$ we get $T_{j_0,j_0}(x\cdot a) = T_{i_0,i_0}(x) \cdot a$. So there is an $A$-linear, bounded map $T'$ say, with $T_{i,j} = \delta_{i,j} T'$.

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Let us check this with a different calculation, using adjoints. Let $T$ be as before, and let $S$ be similar, so $$ \langle T(x), y \rangle = \big( \langle T_i(x), y_j \rangle \big)_{i,j}, \qquad \langle x, S(y) \rangle = \big( \langle x_i, S_j(y) \rangle \big)_{i,j}. $$ We would like these to be equal, for all $x$ and $y$, so $$ \forall\, i,j \quad \sum_k \langle T_{i,k}(x_k), y_j \rangle = \sum_k \langle x_i, S_{j,k}(y_k) \rangle. $$ Again, set $x_i = \delta_{i,i_0}x, y_j = \delta_{j,j_0} y$ to obtain $\langle T_{i,i_0}(x), y \rangle \delta_{j,j_0} = \langle x, S_{j,j_0}(y) \rangle \delta_{i,i_0}$ for all $i,j$. So again $T,S$ are diagonal, and $\langle T_{i_0,i_0}(x), y \rangle = \langle x, S_{j_0,j_0}(y) \rangle$ for all $i_0, j_0$. So there is $T\in\mathcal L(E)$ with $T_{i,j} = \delta_{i,j} T$ and $S_{i,j} = \delta_{i,j} T^*$.

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Thus we seem to have proved that $\mathcal{L}_{M_n(A)}(E^n) \cong \mathcal{L}_A(E)$, acting diagonally on $E^n$. Note that the only thing which seems incorrect on pages 57--58 is that $\pi_\tau$ (in the notation of the OP) need not be surjective (exactly as the OP questioned).

Update: I am trying to see where Lance goes wrong. If $\tau$ is faithful then the null space $N_\tau$ is $\{0\}$ and so $E$ (and not just a quotient) is a $B$-module. The $B$ module action is just the restriction of the $M(A)$ action (in the example, already $M(A)=A$). I guess the mistake is then believing that $\mathcal{L}_{M(A)}(E)\rightarrow \mathcal{L}_B(E)$ is onto, because it's the same $E$. This isn't true because, informally, $B$ being "smaller than" $M(A)$ means that it's "easier" to be adjointable for $B$, and thus the codomain of $\pi_\tau$ might well be larger than the domain.

A further thought: Lance's own example on page 58, involving a state, is also a counter-example! If $\rho$ is a state on $A$ then we obtain a retraction $\rho:A\rightarrow B=\mathbb C1\subseteq M(A)$, and $E_\rho$ is a Hilbert space. If $A$ is separable, we can pick a faithful $\rho$, and so the claim would be that $\pi_\rho:\mathcal{L}_A(E)\rightarrow \mathcal{L}(E_\rho)$ is onto, where the right-hand side is all bounded linear maps. This seems hugely unlikely!

Answer 2

I'll try to outline the possible "counter-example" I had in mind, if somebody could remark on where the error is I would be glad: Here we will take $A=M_{2}(\Bbb C)$, $\tau(a) = \frac12\mathrm{Tr}(a)\Bbb1$, $B=\Bbb C\cdot \Bbb1_2$, $E=\Bbb C^2$ with inner product $$\langle (x_1,x_2) , (y_1,y_2)\rangle = \begin{pmatrix} \overline{x_1}y_1& \overline{x_1}y_2\\ \overline{x_2}y_1&\overline{x_2}y_2\end{pmatrix}=\begin{pmatrix}\overline{x_1}\\ \overline{x_2}\end{pmatrix}\cdot \begin{pmatrix}y_1&y_2\end{pmatrix}$$ and right-action by $A$: $$\begin{pmatrix}x_1& x_2\end{pmatrix}\cdot\begin{pmatrix} a_{11}& a_{12}\\ a_{21}& a_{22}\end{pmatrix} = \begin{pmatrix} a_{11} x_1+a_{21}x_2 & a_{12}x_1 + a_{22}x_2\end{pmatrix}$$ (ie by matrix multiplication).

Then the inner product on the localisation is $\langle x , y\rangle_\tau = \frac12 (\overline{x_1}y_1 + \overline{x_2}y_2)$ while multiplication with elements of $B$ is just regular scalar multiplication with $\Bbb C$. As such $\mathcal L_{\Bbb C}(\Bbb C^2, \langle\cdot,\cdot\rangle_\tau) \cong M_2(\Bbb C)$ follows immediately.

Any $A$-linear map must also be a linear map, so there is an injective embedding $\mathcal L_A(\Bbb C^2, \langle\cdot,\cdot\rangle)\to M_2(\Bbb C)$. Next we show that this embedding is not surjective, whence $$\dim_\Bbb C(\mathcal L_A(\Bbb C^2, \langle\cdot,\cdot\rangle))<4= \dim_\Bbb C(M_2(\Bbb C))= \dim_\Bbb C(\mathcal L_\Bbb C(\Bbb C^2, \langle\cdot,\cdot\rangle_\tau))$$ follows and $\pi_\tau$ cannot be surjective.

To see that not every linear map is adjointable look at the linear map: $$a:\Bbb C^2\to\Bbb C^2, (x_1, x_2)\mapsto (x_2,0)$$ Suppose it is adjointable, then there must be a linear map $a^*$ for which $$\langle a^* x, y\rangle = (a^* \cdot x)\cdot y^T\overset!=\langle x, a y\rangle = x\cdot (a\cdot y)^T = x\cdot (y^T\cdot a^T)$$ must hold for all $x,y\in\Bbb C^2$. Now take a look at $x=(0,1)=y$ in the above equation to get: $$a^*\cdot x\cdot y^T=\begin{pmatrix}0& (a^*)_{12}\\ 0&(a^*)_{22}\end{pmatrix}\overset!=x\cdot y^T \cdot a^T = \begin{pmatrix}0&0\\ 1&0\end{pmatrix}$$ which cannot be true.