Recursive linear orders given by Kleene-Brouwer order

Question

The Kleene-Brouwer order $<_{KB}$ transforms a computable tree $T$ on $\omega^{< \omega}$ into a computable linear order (mapping infinite paths through $T$ into infinite descending sequences). What recursive linear orders are isomorphic to the Kleene-Brouwer order on some tree?

Define as in [1] an adequate linear order to be one with least element $0$, every element has an immediate successor and every element is either finitely above $0$ or finitely above a limit point (and may as well demand the field be equal to $\omega$). Is every adequate linear order represented by $<_{KB}$ on some computable tree? What about adequate orders with no hyperarithmetic descending sequences?

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The motivation here (and I guess a second question if the answer to the main question is negative) is that in [1] Friedman proves there are some adequate computable linear orders that don't support any jump-hierarchy. In other words, you can't define sets that look like $0^{\alpha}$ on those orders (i.e. with $\alpha$ ranging over elements in the ordering not true notations). Does that also mean that there is a computable tree with some branches but no hyperarithmetic ones such that $<_{KB}$ on that tree but doesn't support a jump-hierarchy? What if $<_{KB}$ on that tree produces an adequate order?

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1: Friedman, UNIFORMLY DEFINED DESCENDING SEQUENCES OF DEGREES

Answer 1

The computable linear orders isomorphic to the Kleene-Brouwer order of a computable tree are precisely the orders with a greatest element.

In one direction, the root of the tree is always greatest in the KB-ordering.

In the other direction, given a computable linear order $L$ with a greatest element, non-uniformly fix the greatest element and WLOG assume it is the first element of the domain. We will build a tree $T$ and a map $f: L \to T$ such that $f$ is an isomorphism between $L$ and the KB-ordering on $T$.

At stage $s=0$, add the root to $T$ and define $f$ to send the greatest element of $L$ to the root.

At stage $s+1$, let $x$ be the next element of the domain of $L$. Let $y$ be the least element of $L$ which is greater than $x$ and already an element of the domain (such element must exist, because we have already dealt with the greatest element). Add a new child to $f(y)$ greater than any existing child, and define $f(x)$ to be this new child.

One verifies by induction that $f$ is order-preserving between $L$ and the KB-ordering of $T$.