Is $\mathcal{K}(H)$ injective $\mathcal{B}(H)$-module?

Question

Does anyone know if the right Banach $\mathcal{B}(H)$-module $\mathcal{K}(H)$ is injective? This module is not dual, so standard arguments via flatness do not work.

Injectivity is understood in the following sense.

Definition #1. A bounded morphism of $A$-modules is called admissible if it admits a left inverse bounded linear operator.

Definition #2. A Banach module $J$ over Banach algebra $A$ is called injective if for any admissible bounded morphism of $A$-modules $\xi:X\to Y$ and any bounded morphism of $A$-modules $\varphi:X\to J$ there exists a bounded morphism of $A$-modules $\psi:Y\to J$ such that $\psi\xi=\varphi$. Note,

The standard tool for this type of problems is the following fact.

Fact #1. A Banach module $J$ over a unital algebra $A$ is injective iff the map $$ \rho_J:J\to\mathcal{B}(A,J):x\mapsto (a\mapsto x\cdot a) $$ admits a left inverse map $\tau$ which is a morphism of $A$-modules.

Here are my ideas that eventually failed.

Idea #1. Construct $\phi$ and $\xi$ such that $\psi$ does not exist. The problem is that $\mathcal{K}(H)$ is not complemented in $\mathcal{B}(H)$ for infinite dimensional $H$. The only "interesting" admissible morphism I'm aware of is $\xi=\rho_{\mathcal{K}(H)}$. By interesting I mean admissible morphism without obvious left inverse bounded module map.

Idea #2. Solve commutative analogue of the problem and modify solution for the non-commutative case. Strangely, the $\ell_\infty$-module $c_0$ is injective (see my proof below), but one cannot adopt this proof for $\mathcal{K}(H)$ since $\mathcal{B}(H)$ does not possess the Dunford-Pettis property.

Idea #3. Note that $\mathcal{B}(H)$-module $\mathcal{B}(H)$ is injective, and modify proof for the case of $\mathcal{K}(H)$. Indeed, $\rho_{B(H)}$ has a left inverse module map $$ \tau:\mathcal{B}(\mathcal{B}(H), \mathcal{B}(H))\to \mathcal{B}(H): T\mapsto (\xi\mapsto T(\xi\bigcirc e)(e)) $$ where $e$ is a unit vector in $H$ and $\eta\bigcirc\zeta:H\to H:\xi\mapsto\langle\xi,\zeta\rangle\eta$ is a generic rank one operator. Unfortunately, for infinite dimensional $H$ this expression for $\tau$ does not define a left inverse for $\rho_{\mathcal{K}(H)}$ since $\tau(T)=1_H\notin\mathcal{K}(H)$ for $T:\mathcal{B}(H)\to\mathcal{K}(H):a\mapsto a\circ (e\bigcirc e)$.

Idea #4. Look for more sophisticated right inverses. One can check that the linear map $$ \tau:\mathcal{B}(\mathcal{B}(H), \mathcal{K}(H))\to \mathcal{K}(H): T\mapsto \left(\xi\mapsto \sum_{i=1}^{n}\langle T(\xi\bigcirc e_i)(e_i), e_i\rangle e_i\right) $$ is a left inverse for $\rho_{K(H)}$ when $H$ is finite dimensional with orthonormal basis $(e_i)_{i=1}^n$. Unfortunately $\Vert \tau\Vert=n^{1/2}$, so this operator is not even well defined for infinite dimensional $H$.

Idea #5. Look for general description of left inverses of $\rho_{K(H)}$ when $H$ is finite dimensional and hope for new insights. Let $(e_i)_{i=1}^n$ be an orthnormal basis of $H$. Denote $$ E_{k,l,p,q}:\mathcal{B}(H)\to\mathcal{K}(H):a\mapsto \langle a(e_l),e_k\rangle(e_p\bigcirc e_q) $$ then $(E_{k,l,p,q})_{k,l,p,q=1}^n$ form a basis of $\mathcal{B}(\mathcal{B}(H),\mathcal{K}(H))$. Also denote $$ \tau_{k,l,p,q,i,j}=\langle\tau(E_{k,l,p,q})(e_j),e_i\rangle $$ For example, given this notation the map $\tau$ from idea #4 is well defined by $$ \tau_{k,l,p,q,i,j}=\delta_k^j\delta_l^i\delta_p^i\delta_q^i $$ Then one can show that $\tau$ is a left inverse module map for $\rho_{K(H)}$ iff $$ \tau_{k,l,p,q,i,j}=\delta_k^j\mu_{i,l,p,q} $$ for some array of numbers $(\mu_{i,l,p,q})_{i,l,p,q=1}^n$ satisfying $$ \sum_{l=1}^n \mu_{i,l,p,l}=\delta_p^i $$ Interesting, but not very useful.

Answer 1

This is a partial answer that solves the commutative analogue of the original question. Consider $\ell_\infty$-morphisms $$ \rho:c_0\to\mathcal{B}(\ell_\infty, c_0):x\mapsto(a\mapsto x\cdot a), $$ $$ \tau:\mathcal{B}(\ell_\infty,c_0)\to c_0:T\mapsto (T(\delta_1)(1), T(\delta_2)(2),\ldots) $$ It is really easy to check $\tau\rho=1_{c_0}$, but the only thing left to prove is that the range of $\tau$ is actually the $c_0$. Since $\ell_\infty$ is a Grothendieck space, then any $T\in\mathcal{B}(\ell_\infty, c_0)$ is weakly compact. Since $\ell_\infty$ has the Dunford-Pettis property, then any weakly compact operator is completely continuous. Therefore any $T\in\mathcal{B}(\ell_\infty, c_0)$ maps weakly null sequences to the norm null sequences. It is remains to note that $(\delta_i)_{i=1}^\infty$ is weakly null to get that $\lim_i\Vert T(\delta_i)\Vert=0$. Therefore $\operatorname{Im}(\tau)\subset c_0$.