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There is a picture in the article of Harlow which (in particular) represents light cone in $AdS_3$. Note that point X has nonzero radial coordinate.

enter image description here

If point X has zero radial coordinate it is quite easy to show that light cone is given double cone. In particular it is formed only by null geodesics with zero angular momentum(straight lines). Yet this won't be the case if X has nonzero radial coordinate.
So my question is whether it is still going to look like double cone?
I have found some relevant information about the matter here. I hope it will allow me to plot the graph explicitly. However before spending possibly significant amount of time on this affair I would like my intuition to be tested.

So here are my arguments. First of all let us remind ourselves that only those null geodesic in $AdS_3$ which have zero momentum can go through point with $r=0$. Secondly every null geodesic that starts on some point on the boundary will terminate at the antipodal point.

enter image description here

This being said I state that in the light cone presented in the article of Harlow there will be only two null geodesics with zero angular momentum. Rest of the cone(I suspect it will still look like double cone) will be formed by null geodesics with different nonzero values of J all intersecting given point X.

enter image description here

Is it really how this thing works? Or am I still missing something?

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Yes it seems right. AdS is maximally symmetric so the light cone associated with this point off centre is just a shifted version of the light cone for the origin that you talked about. This double cone intersects the world line r=0 at only two points.

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