How to put roughly equidistan points on the surface of a sphere, each with 4 neighbors? Say I want to play a game like Go or Go Bang or Chess on a the surface of a sphere (see here for a question about examples).
The grid on the sphere should have the following characterisitcs:


*

*all nodes more or less equidistant

*each node connected to four neighbors

*connections don't intersect
I've tried around with constructing such a grid by adding nodes on the edges of an octahedron, or by adding nodes to a bucky-ball like grid, both don't work.
 A: A quick calculation using Euler's $V-E+F=2$, and the fact that since each edge joins two vertices and we have four edges at each vertex, $E=2V$ gives $F-V=2$ ie that there are two more faces than vertices.
Now suppose there are $f_r$ faces with $r$ vertices. We know that each vertex adjoins 4 faces, so that $$\sum_{r=2}^\infty rf_r=4V=4F-8$$
Now since $F=\sum f_r$ we can rewrite this as $$\sum(r-4)f_r=-8$$
If we disallow faces with two sides we have $$-f_3+f_5+2f_6+3f_7+\dots=-8$$
So we can have as many quadrilateral faces as we like (they count zero to the sum), but we need at least eight triangles, and for every face with more than $4$ sides we need more triangles.
So you need to work with this constraint. You could, for example put pyramids on two opposite sides of a cube. Then you could divide the other sides in half along the "equator". You can keep on cutting slices off the cube as long as you like.
This doesn't give you a full answer, but it does suggest some ways to go. 
Note: When you have three edges meeting at each vertex, you can just cut off corners. But that is not an option here - you need to keep the corners on and put extra edges/vertices where the cut would go. So another construction would be to draw edges round each vertex of an octahedron to get roughly equal spacing of vertices. Then edges around the vertices of the resulting figure etc.
A: I found one way to do it myself: I took a buckyball-like structure, the grid consists of hexagons that are surrounded by (alternating) hexagons and pentagons, so that they share sides. Now, if I define each shared edge of hexagons as a node, each new node is connected to four neighbors.The whole thing looks like triangles, each node is a corner of two triangles and they are arranged so that five triangles form a circle. This grid satisfies all my conditions lined out aboce, as fas as I can see.
I'll leave the question open, because there may be other ways and even if not someone else may provide a clearer asnwer, or a more general one.
Edit to add: this grid does not satisfy all requirements - while all edges are the same length, there are huge pentagon shaped holes between, so one could argue that the nodes are not equidistant.
A: Option 1: Give up on a sphere and use a torus.  I think the answer is trivial there: imagine a 2-D chess or go board with the left stitched to the right, and the top stitched to the bottom.  (The same way old 2-D video games work: fly off the "top" of the screen, and reappear at the bottom at the same horizontal offset, traveling along the same vector.)
Option 2: Give up on a sphere and go 3-D.  What you ask for is nearly met by a 3-D tetrahedral lattice/matrix.  (The way carbon atoms form bonds in a diamond.)  This meets "equidistant, non-intersecting, and exactly four neighbors"... just not "on a sphere".
Now, these conditions are met in the "body" of the lattice.  You'd just need to decide what to do at the boundary of the volume.  Topologically, you could envision a cube as boundary, and "connect" the bonds from the front face to the back, the right face to the left, and the top to the bottom.  (Sounds like a 3-D version of a torus... I'm sure there is a name for that.)
Option 3: If you loosen your notion of "equidistant", doesn't an octahedron give you a good, basic place to start?  Or, using a globe as an analogy:


*

*Draw the equator, and four meridian lines at 0°, 90°, 180°, and 270°.

*Now you have 6 intersection points, all with four neighbors, and they are equidistant.

*Now add any number of additional "parallels": each "latitude line" intersects four meridians, and each newly added intersection has four neighbors.


If you are really angling for the "edges" to be similar in length, then maybe you can start with this mesh and use some iterative computational techniques to spread out the intersections more evenly.
update: I think my "Option 3" is equivalent to @MarkBennet's suggestion "You could, for example put pyramids on two opposite sides of a cube."
A: The general problem of "how do I pack N points on a sphere, where I maximize the minimum distance between any two points" is known as "finding spherical codes".
More info on Spherical Codes from:
Wikipedia
Mathworld
Neil Sloane -- general discussion, plus solved values for values of N up to 130.
Now this approach gives you the POINTS, but getting the "neighbors" is a bit subtle.  I assume you'd want to create a set of polyhedral faces as the convex hull of those sets of points, then consider "neighbors" as points that are connected by an edge.  I think qhull (or its wrappers like pyhull) should be able to do this for you.
Alas, this doesn't exactly answer the problem statement, as such collections of points usually have 3 "closest" neighbors (not 4 as the OP requested).  And for certain values of N, there are other anomalies that creep in; the OP would have to decide if this is fit for his purpose.
But this sort of "game board" may be of interest to the OP or others looking at this question.
