Counting edges in embeddable CW-complexes in R^3 Using Euler's formula ($V-E+F = 2$ where $V$, $E$ and $F$ are the number of vertices, edges and faces), we can easily count the number of edges in maximal graphs that are embeddable in plane: 3n-6. I have a similar question but for CW-complexes that are realizable in $R^3$ (let us call them spacial complexes). 
For spacial complexes, Euler's formula dictates $V-E+F-C = 0$ where $C$ is the number of 3-D cells. I can also see that, for spacial complexes to be maximal (by adding faces until if we add any other face in it then can not be embedded in $R^3$), they should be tetrahedralized. 
For each cell (tetrahedron) we have 4 faces and for each face we have 2 cells, therefore $4C = 2F$. That is, the number of cells is half the number of faces but still I do not know how to find the maximum number of edges as a function of the number of vertices.
 A: By coincidence I just happened to read the required information to answer this question in a recent preprint, arXiv:math.1308.5798, "Neighborly inscribed poytopes and Delaunay trisngulations", Gonska and Padrol.   For references to the following facts (which go back to the late 70's) see this preprint.   If a simplicial  d+1 polytope has its vertices on the unit sphere, then an appropriate stereographic projection realizes the combinatorial type of the boundary of the polytope as a Delaunay triangulation in d-space.  Since cyclic-4 polytopes can be realized with vertices on the unit sphere, it is possible to embed complexes with edges between any pair of vertices.    The same argument shows that in dimension d it is possible to embed complexes with the same number of faces in each dimension as cyclic (d+1)-polytopes.  So for i at most half the dimension you can get all possible i-faces.  The maximum number of higher dimensional faces is unclear.  To begin with,  I do not see any reason to assume the complex is a ball.  As I recall there are two-neighborly seven vertex embeddings of the solid torus in R^3. 
