A real tree is a metric space $(X,d)$ such that for any points $x,y\in X$ there is a unique path from $x$ to $y$, and which is a geodesic. Equivalently, it is a $0$-hyperbolic space.
A simplicial tree (by this I mean a graph that is a tree) can be seen as a real tree, but not all real trees come from simplicial tree.
Now, given a metric space $(X,d)$, an ultrafilter $\omega$ on $\mathbb{N}$ non-principal, a sequence $u=(u_n)\in \mathbb{R}$ and a sequence of points $p=(p_n)\in X$, there is a way to define the ultralimite $X_\infty(d,u,p)$: we consider the set of sequence $(x_n)$ of points of $X$, such that the sequence $\frac{d(x_n,p_n)}{u_n}$ is bounded, we define a pseudo-distance $d(x,y)=\underset{\omega}{\lim} \frac{d(x_n,y_n)}{u_n}$ and we take a quotient, with $x\sim y$ iff $d(x,y)=0$ to get a metric space called the ultralimit of $X$.
A few things I know:
- If $X$ is a real tree, so is the limit, because if $X$ is $0$-hyperbolic the ultralimit is too.
- In fact, in $u_n\to \infty$, then if $X$ is only $\delta$-hyperbolic, the ultralimit is $0$-hyperbolic thus a real tree.
- From what I have seen, the case $u_n\to \infty$ seems to be often considered an is called the asymptotic cone of $X$
Now my question is, if $X$ is a simplicial tree, we can see it as a real tree and consider an ultralimit $X_\infty$, and so this metric space is a real tree. Are there cases when we can say this real tree comes from a simplicial tree ?
I have a feeling that if $u_n$ does not go to $\infty$, it might be the case - and maybe this is why it is not considered, because we do not get "more complicated space" ?) but although seeing a simplicial tree as a real tree is not that hard, I find it hard to do the converse and given a real tree, understand wether it comes from a simplicial tree or not.
