Euler Characteristic of the Projective plane and sphere?

[PsychonautQQ]

The Euler Characterist of the projective plane and sphere is given by V - E + F. V is vertices, E is edges, F is faces.

A presentation of the projective plane is {a | aa} and a presentation of the sphere is {b | bb^1}
Yet the Euler characteristic is 2 for the sphere and 2-n for the connected sum of n protective planes. So one projective plane should have euler characteristic of 1. Looking at these manifolds as equivalences on the closed disk, it seems that their Euler characteristic should be the same. They both have 2 vertices, 2 edges, and 1 face. Or perhaps since the edges are identified together there is only 1 edge, and perhaps for the sphere you have to count both side of the disc as a face...

Can somebody clarify this?

 

[lavinia]

The Euler Characterist of the projective plane and sphere is given by V - E + F. V is vertices, E is edges, F is faces.

A presentation of the projective plane is {a | aa} and a presentation of the sphere is {b | bb^1}
Yet the Euler characteristic is 2 for the sphere and 2-n for the connected sum of n protective planes. So one projective plane should have euler characteristic of 1. Looking at these manifolds as equivalences on the closed disk, it seems that their Euler characteristic should be the same. They both have 2 vertices, 2 edges, and 1 face. Or perhaps since the edges are identified together there is only 1 edge, and perhaps for the sphere you have to count both side of the disc as a face...

Can somebody clarify this?

Not sure what you mean by presentation. Can you explain it?

A sphere is a closed disk with its boundary identified to a point. This is one 2 cell and one 0 cell - no 1 cells.

A projective plane is a square(a square is homeomorphic to a closed disk) with both pairs of opposite edges identified with a half twist (double Mobius band). This leaves one 2 cell, two 0 cells and 2 edges.

I think the simplest triangulation of a sphere is a tetrahedron. This has four vertices, four faces, and six edges.

One way to triangulate the projective plane is to start with a triangulation of the sphere that is invariant under the antipodal map. Modding out by the antipodal map divides the number of vertices, edges, and faces by two.

 

[mathwonk]

an equivalent way to view lavinia's last remark is to note that since the sphere is a double cover of the projective plane, any triangulation of that plane by small triangles pulls back to a triangulation of the sphere with twice as many vertices, edges and faces. Hence the euler characteristic of the sphere must be double that of the projective plane.

 

[PsychonautQQ]

Thanks guys! I was getting confused I guess because I was trying to understand Euler's Characteristic of a sphere by looking at a picture of a closed disk with a point on top and bottom, and arrows running up either side to indicate orientation of how the edges are identified.