# almost projective Banach space, complex scalars

It is well-known that if a real Banach space $E$ is "almost metrically projective" then $E$ is isometrically isomorphic to some $\ell^1(\Gamma)$. We say $E$ is "almost metrically projective" if whenever $T$ is a bounded linear map from $E$ into a quotient Banach space $X/Y$, then for every $\epsilon > 0$ there is an almost norm preserving lifting of $T$, i.e., an operator $\tilde{T}:E\to X$ such that $q\tilde{T}=T$ and $\| \tilde{T}\|<=(1+\epsilon)\| T \|$, where $q:X \to X/Y$ is the canonical quotient map.

This result is due to Grothendieck (Canadian J. M., 1955), and proofs can be found in the books of Semadeni (Banach Spaces of Continuous Functions, 1971) and Lacey (Isometric Theory of Classical Banach Spaces, 1974). However, these proofs deal only with real scalars. Grothendieck states (Remark 6, p. 559) "There is little doubt ... that the theorems in this article are valid for complex scalars as well as real scalars. It is not clear, however, that this extension is immediate."

Has a complete proof for complex scalars ever appeared in the literature?

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I don't recall seeing anything for the complex case for this question or your one on injective spaces. This one looks easier to me. You get that the inclusion from $E$ into its second dual isometrically factors through some $L_1$ space and that $E$ is almost isometric to $\ell_1(\Gamma)$. Is that enough to get $E$ contractively complemented in some some $L_1$ space? That would do it, yes? Aren't the contractively complemented subspaces of $L_1$ spaces treated in Lacey's book? I am travelling and cannot easily check. – Bill Johnson Jun 3 '13 at 13:12
Well, for example from $E^{*}$ being one injective you get that $E^{**}$ is contractively complemented in an $L_1$ space, which gives what I said. – Bill Johnson Jun 4 '13 at 18:23
(Sorry editing conflict.) Clearly the lifting property of E implies the embedding E into its second dual factors through some $L_1$ space, as would be true for any T:E→F, since any F is a quotient of some $L_1$. Don't see why E would then be almost isometric to $\ell_1(\Gamma)$. The proofs in Semadeni and Lacey start with the observation that if E is almost metrically projective then its dual $E^*$ is a $P_{1+\epsilon}$ space for all ϵ>0. This is OK for complex scalars. – Fred Dashiell Jun 4 '13 at 18:42
Also $E^* \in P_{1+\epsilon}$ for all $\epsilon > 0$ implies $E^* \in P_1$ ok for complex scalars OK in complex case for dual spaces. So I see what you were saying. – Fred Dashiell Jun 4 '13 at 18:49
It seems that what we need is the complex version of the fact that if $E^*$ is 1-injective then $E$ is an abstract $L$ space (in fact $E$ is the space of normal measures on some hyperstonean space). – Fred Dashiell Jun 4 '13 at 18:59