Minimality properties of James' space

I am interested in the following question about James' quasi-reflexive Banach space $$\mathcal{J}$$:

Does there exists a non-Hilbertian subspace $$X$$ of $$\mathcal{J}$$ such that $$X$$ isomorphically embeds into every non-Hilbertian subspace of itself?

Here, by "subspace" I mean "closed, infinite-dimensional vector subspace", and by "Hilbertian" I mean "isomorphic to $$\ell_2$$".

I vaguely recall having found, one year ago, a paper proving that the answer to this question was no, or at least giving a similar/partial result suggesting that the answer should be no. Problem is, I don't manage to find this paper again, I don't even remember who were the authors and what was the exact result they proved. Do some of you recall having seen something like that?

• It's a pitty that the ask-johnson-tag does not exist anymore. – Jochen Wengenroth Feb 3 '20 at 17:16

A weaker `block version' is true for the conditional spreading basis (the summing basis) of $$\mathcal{J}$$: Every seminormalized block basis of the spreading basis has a subsequence either equivalent to an unconditional basis ($$\ell_2$$) or a convex block sequence equivalent to the basis itself. The result holds in general in spaces with a convex block homogeneous conditional spreading basis. See the section 5 of the following paper
• @Bunyamin Sari: In this paper: jstor.org/stable/2041285?seq=1#metadata_info_tab_contents, the authors prove that James' space has uncountably many pairwise nonequivalent unconditional basic sequences. This contradicts your claim that the only one is $\ell_2$. Actually, $\mathcal{J}$ itself cannot have the property I'm asking about: the same paper shows it has non-Hilbertian reflexive subspaces, hence it cannot isomorphically embed into these subspaces. – N. de Rancourt Feb 5 '20 at 12:49
• Let $X:=\left(\sum_n \oplus E_n \right)_2$ with $d_n(X):=\sup d(E,\ell_2^n)\to\infty$. Is there $Y:=\left(\sum_n \oplus F_n \right)_2\subset X$ so that $d_n(Y)\to \infty$ in much slower pace (so that $X$ doesn't embed into $Y$). This is surely true, and would answer the OP's question in negative. – Bunyamin Sari Feb 8 '20 at 23:52