4
$\begingroup$

Let $S$ be a locally compact Hausdorff space. By $C_0(S)$ we denote the space of continuous functions vanishing at infinity. Let $\mu$ be a finite Borel regular measure om $S$, then consider $L_\infty(S,\mu)$ as a left Banach $C_0(S)$-module.

Definition 1:

A left Banach $A$-module $F$ is called strictly flat if for any isometric morphism $\xi:X\to Y$ of right $A$-modules the linear operator $\xi\widehat{\otimes}_{A} 1_F:X\widehat{\otimes}_{A}F \to Y\widehat{\otimes}_{A} F$ is bounded below.

Definition 2:

A right Banach $A$-module $J$ is called strictly injective if for any isometric morphism $\xi:X\to Y$ of right $A$-modules and any continuous morphism of right $A$-modules $\phi:X\to J$ there exists a continuous morphism of right $A$-modules $\psi:Y\to J$ such that $\phi=\psi\xi$.

One can show that $F$ is strictly flat iff $F^*$ is strictly injective.

Question:

What is a criterion for strict flatness of the left $C_0(S)$-module $L_\infty(S,\mu)$? In other words when the right $C_0(S)$-module $L_\infty(S,\mu)^*$ is strictly injective. Clearly, I would like to see a criterion in terms of $S$ and $\mu$.

Motivation:

One can show that $C_0(S)^*$ is a strictly injective $C_0(S)$-module. Hence so is its retract $L_1(S,\mu)$. One can hope that the same holds true for $L_1$'s bigger counterpart $L_\infty(S,\mu)^*\simeq_1 L_1(S,\mu)^{**}$ which is an $L_1$-space too. But I believe that almost always this is not the case, due to the presence of non-countably additive measures in $L_\infty(S,\mu)^*$.

Observations:

  1. Necessary condition: $\operatorname{supp}(\mu)$ is compact.
  2. Sufficient condition: $L_\infty(S,\mu)$ and $C_0(S)$ are isomorphic as $C^*$-algebras.

I believe that a sufficient condition is close to a criterion. If so, the question must be hard, but I would be glad to hear your thoughts anyway.

$\endgroup$

0

You must log in to answer this question.