Hereditarily primary Banach spaces A Banach space $X$ is said to be prime if every infinite dimensional complemented subspace is isomorphic to the space $X$. The space $X$ is primary if it has an infinite dimensional subspace $Y$ such that every complemented subspace is either isomorphic to $X$ or to the subspace $Y$. The space $X$ is quasiprime if it is primary and the only decomposition of $X$, as a direct sum, into two infinite dimensional subspaces is $X +Y$. Prime spaces are quasiprime which are primary.
Also there are examples that separate the above classes.
A Banach space $X$ is hereditarily prime (primary, quasiprime) if every infinite dimensional subspace is prime (primary, quasiprime). Hilbert spaces are hereditarily prime. Also there exists a non-Hilbertian space $X$ which is hereditarily quasiprime.
Question I Is every hereditarily prime space isomorphic to a Hilbert space?
Question II Do there exist subspaces of $\ell_p$, which are not primary?
 A: I think that the following provides a partial positive answer to  Question I.
Fact If X is  separable, Hereditarily Prime and Decomposable then it is isomorphic to a Hilbert space.
The proof goes as follows. We write X  as  V + W with both of infinite dimension.Also both are isomorphic to the space X. Let Z be a subspace of V and  Y = Z + W. Then W is a complemented subspace of Y hence Y is isomorphic to W. The subspace Z is complemented in Y hence isomorphic to Y which in turn is isomorphic to the subspace V. Hence V is isomorphic to any of its subspaces and from Gowers' theorem it is isomorphic to a Hilbert space.
The remaining case does not seem easy.
Notice that any hereditarily prime space must be $l_2 $ saturated. Indeed it does not contain HI subspace hence it is unconditionally saturated and the above yields that all these subspaces are isomorphic to $ l_2 $.
A space satisfying the following properties provides a counterexample.

*

*The space X is indecomposable and $ l_2 $ saturated.
(The existence of such a space is still open but I think could exist. )


*Every subspace is isomorphic to all further subspaces with finite codimension.


*Every subspace is either indecomposable or isomorphic to $l_2$.
A: Actually the above three conditions yielding a counterexample are satisfied by any counterexample and this is easy to be checked. However such a space has extremely peculiar structure. In particular it does not contain any subspace of the form Y + Z with Y Hilbertian and Z indecomposable. This follows from the previous positive partial answer. That means that for every indecomposable subspace Z and every Hilbertian subspace Y dist( $S_Z$ , $S_Y$ )= 0. Further the same holds for all Y subspaces of X. The last follows from the fact that X is $l_2$ saturated. It is hard to think how such a space is defined.
However if the following has a positive answer then the original question I
has also a positive one.
Question III Let X be a separable  non Hilbertian and $l_2$ saturated Banach space.Does the space X contain Y +Z with Y Hilbertian and Z non Hilbertian?
