If $X$ is triangulated, the Euler characteristic of the triangulation is the alternating sum $f_0 − f_1 + f_2-…$ where the number $f_n$ counts the n-simplices in the triangulation.
I know that any triangulation of a topological space must satisfy the topological invariant that is the Euler Characteristic.
However is the converse true. For example: five vertices, nine line segments and six triangles would give an Euler Characteristic of 2. Does this imply that a sphere, which has characteristic 2, could have a triangulation of this form?
It would imply having triangulations with almost arbitrarily large vertices, lines and triangles for a sphere, so I think the answer could be no but I am struggling to prove/disprove it.
Best Answer
It is not true, The Euler characteristic of an odd dimensional sphere and an odd dimensional torus is $0$, the triangulation with one zero cell and one $2n+1$-cell is the $2n+1$-dimensional sphere, not the torus.