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I am starting to study some lecture notes about metric geometry and I would appreciate it if someone could some questions regarding the notion of the tangent cone.

Consider 3 half lines joined by their point of origin. You get a network of 3 roads and a junction point. This is an Aleksandrov space of non-positive curvature. It is even geodesically complete. Now, I am having trouble defining the tangent cone at this junction point:

  • Is it isometric or BiLipschitz to $R^k$ for some $k>0$ ?
  • What is the Hausdorff dimension of the space at this point ? is it 1 ?
  • is the junction point the boundary of the Aleksandrov space ?

These questions might be trivial so I apologize in advance but I just didn't find enough examples that talk about this.

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    $\begingroup$ It looks like an exercise. If $X$ is your space and $o$ the joining point (I assume the metric is the geodesic one), it's obvious that $X$ is isometric to its tangent cone at $0$. More generally, if $X$ is a proper metric space and and has a 1-parameter subgroup of non-isometric dilations fixing a point $o$, then every tangent cone of $(X,o)$ is isometric to $(X,o)$ as pointed metric space. $\endgroup$
    – YCor
    Jun 17 '20 at 16:05
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Loosely speaking, if you are standing at the origin then there are only three directions you can travel. So the space of directions $S_o$ is only 3 points. Taking the product of the space of directions with $[0,\infty)$ and identifying $S_o \times \{0\}$ to a point gives you the tangent cone, which in this case is isometric to your starting space.

If you are looking for resources on tangent cones, I would see Nikolaev's paper 'The tangent cone of an Aleksandrov space of curvature $\leq k$' or Halbeisen's 'On tangent cones of Alexandrov spaces with curvature bounded below'

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