Banach-Mazur distance to complex $\ell^1$ of a space containing real $\ell^1$

Question

Consider a complex Banach space $X$ with a real subspace isometric to $\ell^1_{\mathbb R}$. What is the best constant $c$ such that $X$ contains a complex subspace $c$-isometric to $\ell^1_{\mathbb C}$?

I guess this is a very classical question, but I could not find an answer.

In this question, $\ell^1_{\mathbb R}$ (respectively $\ell^1_{\mathbb C}$) is the real (resp. complex) Banach space of summable sequences of real (resp. complex) numbers. And two Banach spaces $Y,Z$ are called $c$-isometric if there is a linear isomorphism $T:Y\to Z$ such that $\|T\| \|T^{-1}\| \leq c$.

The motivation is that I want to understand the complex Banach spaces of trivial (Rademacher) type. A famous theorem of Maurey and Pisier, usually stated for real Banach spaces, shows that $\ell^1_{\mathbb R}$ is finitely representable in any (real) space of trivial type $\mathbb R$. My question is essentially equivalent to "what happens when the field of scalars is $\mathbb C$?".

Answer 1

Schechtman and I discussed your question this morning and have these comments.

You can get $1+\epsilon$. The usual argument for improving the constant works in the complex case as well as the real case; i.e., if a complex Banach space contains a subspace complex isomorphic to $\ell_1$, then for all $\epsilon >0$ it contains a subspace $1+\epsilon$-complex isomorphic to $\ell_1$.

To see that a complex Banach space that contains a real subspace isomorphic to $\ell_1$ also contains a complex subspace that is complex isomorphic to $\ell_1$, apply Dor's extension to the complex case of Rosenthal's $\ell_1$ theorem. The full theorem says that a (real or complex) Banach space contains an isomorphic copy of $\ell_1$ iff the space contains a bounded sequence that has no weakly Cauchy subsequence.

For other values of $p$ the situation is different. By the Odell-Schlumprecht distortion theorem, for every $M$ there is a an equivalent renorming of $\ell_p$, $1<p<\infty$, s.t. no subspace of the resulting space is $M$-isomorphic to $\ell_p$.

Something we don't see is the following (not that we spent much time thinking about it). Suppose a complex Banach space contains a real subspace isomorphic to real $\ell_p$, $1<p<\infty$. Does it contain a complex subspace isomorphic to complex $\ell_p$? However, the local version looks OK; i.e., if real $\ell_p$ is $C$-finitely representable in a complex space $X$ (and hence $1+\epsilon$-finitely representable by Krivine's theorem) then complex $\ell_p$ is something like $4+\epsilon$-finitely representable in $X$. For that you use spreading model theory to get a suppression $1$ unconditional basis in a spreading model s.t. the basis is isometrically $\ell_p$ for real coefficients. It looks like with a bit of work you can change the $4+\epsilon$ to $1+\epsilon$, but we did not check it to the end. It looks like you can make the norm invariant for multiplication of expansions in every coordinate by $\pm 1$ (different $\pm$ in the various coordinates) by passing to a block basis, and multiplication of coordinates by complex numbers having modulus one should also work by passing to a block basis.

Answer 2

I think that $c$ arbitrarily close to $1$ can be realized. From Dvoretzky's theorem we now that given any $\varepsilon>0$, there is some finite $N$ such that the $N$-dimensional space with the $\ell^1$ norm, denote it by $\ell^1_{\mathbb{R}}(N)$, contains a $2$-plane $H$ which is $(1+\varepsilon)$-isometric to Euclidean plane. Write $$\ell^1_{\mathbb{R}} = \ell^1_{\mathbb{R}}(N) \oplus \ell^1_{\mathbb{R}}(N) \oplus \dots $$ where $\oplus$ is a "Banach sum" with the norm on the sum being the sum of the norms. Then the subspace obtained by taking a copy of $H$ in each copy of $\ell^1_{\mathbb{R}}(N)$ should be $(1+\varepsilon)$-isometric to $\ell^1_{\mathbb{C}}$.