Is the distance between all points on the event horizon zero? These are the Kruskal-Szekeres coordinates of Schwarzschild spacetime:

It is isomorphic to the split-complex plane. But on the split-complex plane the distance between all points on the null diagonals, as measured by the norm of the difference between two spacetime points, is zero.
Does this imply that the distance between all points on the event horizon is also zero?
 A: If this were a 1+1 dimensional spacetime then you'd be correct, but there are two other spatial dimensions not shown on the chart. Each point on the chart is really a sphere whose radius is a function of $X^2-T^2$.
The event horizon is a 3-cylinder (direct product of a 2-sphere and a line), but with a degenerate metric along the length of the cylinder. If you put $(θ,\phi,z)$ coordinates on it in the obvious way, then the metric is $ds^2 = r_s^2 (dθ^2 + \sin^2 θ\,d\phi^2) \; [+\;0dz^2]$. The distance between points at the same position on the horizon at different "times" is zero, but between different points it's nonzero and spacelike.
Note that these coordinates are not the Schwarzschild coordinates restricted to $r=r_s$. Schwarzschild coordinates are singular at $r=r_s$ and don't cover the horizon. But these coordinates are Eddington-Finkelstein infalling coordinates restricted to $r=r_s$, and with $t$ renamed to $z$.
A: Let me try to add some comments to the already given explanation, which focuses on there being more dimensions in actual Schwarzschild spacetime.
So we first have to polish the question a bit. The statement claiming that the Schwarzschild space-time (independent from the coordinates used to describe it) is isomorphic to the complex-split plane is false in general, even if we just take into account 1+1 dimensions.
So I would in turn ask you what sort of isomorphism you have in mind.
The following discussion is what I thing you are going at.
Let us work with the assumption there is only 1 spatial dimension, so the question is whether a 1+1 Minkowski space is the "same" as the split-complex plane. The reason is, the complex-split plane has a "metric" that is not space-time dependent, so it cannot possibly be isomorphic with Schwarschild at least in the differential geometric sense.
Now, the answer depends on the type of isomorphism you have in mind. If you consider the above only as metric spaces the answer is yes. We have an immediate isomorphism, which sends each axis to itself. And the light cones are sent into diagonals no problem.
If the question is about algebraic properties, then the answer is no. Minkowski spacetime does not have a multiplication operation, it makes no sense to multiply spacetime points with themselves.
As soon as you introduce the full geometry of Schwarzschild, you lose any sort of isomorphism. The only remaining possibility is local isomorphism which becomes true given the properties of a manifold, that is there is always a neighborhood to a point where things are "flat". This in the case of Schwarzschild means, if you get very close to a point it does reduce to Minkowski, however this then does not allow you to compare arbitrary points in the horizon.
I hope I have clarified some of the confusion, let me know in the comments if something is not clear or if this does not exactly pertain the question.
