of scalars

Consider a $\Lambda$–tree $T$, and an embedding $\Lambda\hookrightarrow \Tilde\Lambda$ (for example, one may think of ${{\mathbb{Z}}}\subset{{\mathbb{R}}}$). Then $T$ may be viewed as a $\Tilde\Lambda$–metric space, but it is not a $\Tilde\Lambda$–tree if $\Lambda$ is not convex in $\Tilde\Lambda$: as a matter of fact, $T$ is not geodesic as a $\Tilde\Lambda$–metric space (there are holes in the geodesics). However, there is a natural way to *fill the holes*:

There exists a $\Tilde\Lambda$–tree $\Tilde T$ and an isometric embedding $T\hookrightarrow \Tilde T$ which is canonical in the following sense: if $T'$ is another $\Tilde\Lambda$–tree with an isometric embedding $T\hookrightarrow T'$, then there is a unique $\Tilde\Lambda$–isometric embedding $\Tilde T{\rightarrow}T'$ commuting with the embeddings of $T$ in $\Tilde T$ and $T'$.

For example, take $T$ to be the ${{\mathbb{Z}}}$–tree corresponding corresponding to the set of vertices of a simplicial tree $S$. Then the embedding ${{\mathbb{Z}}}\subset {{\mathbb{R}}}$ gives an ${{\mathbb{R}}}$–tree $\Tilde T$ which is isometric to the geometric realization of $S$.

The proposition also holds if one only assumes that $T$ is $0$–hyperbolic. In this case, taking $\Tilde\Lambda=\Lambda$, one gets a natural $\Lambda$–tree containing $T$.

Subtrees
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A *subtree* $Y$ of a $\Lambda$–tree $T$ is a convex subset of $T$, [i.e. ]{}such 