What is the structure of a Banach space $X$ when $Y$ and $X/Y$ are hereditarily indecomposable?

Question

Assume that $X$ is a separable Banach space and $Y$ a closed subspace such that $Y$ and $X/Y$ are hereditarily indecomposable (HI). The general question is what is the possible structure of $X$. Obviously $X$ could be a HI space. Also $X$ could be the direct sum of $Y$ and $Z$ with both infinite dimensional HI subspaces.

Question I : Are the two alternatives the only possible answers?

Note that the space $X$ is always HI saturated (i.e., every infinite-dimensional closed subspace contains a HI subspace).

A weaker version of the above question is the following

Question * : Assume that $X$ is indecomposable is then HI.

Question II : If the second alternative occurs and W a closed subspace of $X$ such that $W$ and $X/W$ are HI, is $W$ essentially isomorphic to $Y$ or $Z$ (i.e. there are further subspaces of finite codimension which are isomorphic ) and $X/W$ essentially isomorphic to the other one? Namely the pair $W$, $X/W$ admits two alternatives when both are HI.

Answer 1

The answer to Question II is negative. In particular the following general holds.

Theorem: Every separable HI space $Y$ is the quotient of the direct sum of two HI spaces $X$, $Z$ such that $Y$ is not essentially isomorphic to any of $X$, $Z$ and the kernel of the quotient map $Q$ is HI.

This result is a consequence of the known theorem that every separable Banach space $Y$ is the quotient of a separable HI space $X$ such that the quotient map is strictly singular.

Given this result we start with $Y$ a separable HI space and set $X$ the HI space that is mapped onto $Y$ through a strictly singular map $Q$. Next take a $Z$ subspace of $Y$ with infinite dim and codim. Consider the map $Q'$ that sends $(x,z)$ to $Qx+z$ which is onto the space $Y$ from the direct sum of $X$ and $Z$.

The space $W = \mathrm{Ker}Q'$ consists of the pairs $(x,z)$ satisfying the equation $z = -Qx$. The space $W$ is HI. Indeed in every $V$ subspace of $W$ there exists a normalized element $(x,z)$ with norm of $z$ very small. Therefore $(x,z)$ is very close to the space $X$ which is HI. Hence $W$ is also HI.

Answer 2

There exists a separable reflexive Banach space $\hat X^*$ which is indecomposable but not HI, and admits a subspace $Y$ such that both $Y$ and $\hat X^*/Y$ are HI. This example provides a negative answer to Question I and Question $^*$.

The space $\hat X^*$ is obtained in Proposition 23 of a paper of Valentin Ferenczi [CJM-99], using the spaces provided by Proposition 25 in the same paper.

Note that $\hat X^*$ is not HI by construction, and it is indecomposable because its predual $\hat X$ is HI and reflexive (see Proposition 23).

A Banach space $X$ is called QHI if every infinite dimensional quotient of each subspace of $X$ is indecomposable. Since a quotient of a subspace is a subspace of a quotient, subspaces and quotients of a QHI space are QHI. Moreover, if $X$ is reflexive and QHI then the dual $X^*$ is QHI.

Proposition 25 provides two reflexive QHI Banach spaces $X_1$ and $X_2$ with a ``common" subspace $Z$ (some additional conditions are also satisfied), and the predual of $\hat X^*$ is defined in Proposition 23 as $\hat X =(X_1\oplus X_2)/D$ with $D=\{(z,-z): z\in Z\}$.

The expression $Jx_1 =(x_1,0)+D$ defines an embedding of $X_1$ into $\hat X$ such that $\hat X/J(X_1)$ is a quotient of $X_2$: $$ \hat X/J(X_1) = (X_1\oplus X_2)/(X_1+D)\simeq \frac{X_2}{(X_1+D)\cap X_2}. $$ Thus both $J(X_1)$ and $\hat X/J(X_1)$, as well as their dual spaces, are QHI.

Taking as $Y$ the annihilator $J(X_1)^\perp \equiv (\hat X/J(X_1))^*$, both $J(X_1)^\perp$ and $\hat X^*/J(X_1)^\perp$ are QHI, hence they are HI.