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# Euler characteristic

In algebraic topology, the Euler characteristic is a topological invariant (in fact, homotopy invariant) defined for a broad class of topological spaces. It is usually denoted by χ.

The Euler characteristic of a 2-dimensional topological polyhedron can be calculated using the following formula: χ = F - E + V, where F,E and V are the numbers of faces, edges and vertices respectively. In particular, for any polyhedron homeomorphic to a sphere we have

χ(S2) = F - E + V = 2.

For instance, for a cube we have 6 − 12 + 8 = 2 and for a tetrahedron we have 4 − 6 + 4 = 2. The last formula is also called Euler's formula.

## Definitions and properties

For a finite CW-complex and in particular for a finite simplicial complex, the Euler characteristic can be defined as the alternating sum

$\chi=k_0-k_1+k_2-\cdots,$

where ki denotes the number of cells of dimension i.

Then, one can define the Euler characteristic of a manifold as the Euler characteristic of a simplicial complex homeomorphic to it. For example, the circle and torus have Euler characteristic 0 and solid balls have Euler characteristic 1.

The Euler characteristic of closed orientable surfaces can be calculated using their genus g

χ = 2 - 2g.

The Euler characteristic of closed non-orientable surfaces can be calculated using their (non-orientable) genus k

χ = 2 - k.

The Euler characteristic is independent of the triangulation. The formula can also be used for decompositions into arbitrary polygons.

For closed manifolds, the Euler characteristic coincides with the Euler number, i.e., the Euler class of its tangent bundle evaluated on the fundamental class of manifold.

For closed Riemannian manifolds, the Euler characteristic can also be found by integrating the curvature--see the Gauss-Bonnet theorem for two-dimensional case and generalized Gauss-Bonnet theorem for general case. A discrete analog of the Gauss-Bonnet theorem is Descartes' theorem that the "total defect" of a polyhedron, measured in full circles, is the Euler characteristic of the polyhedron; see defect (geometry).

More generally still, for any topological space, we can define the nth Betti number bn as the rank of the n-th homology group. The Euler characteristic can then be defined as the alternating sum

$\chi=b_0 - b_1 + b_2 - b_3 +\, \cdots.$

This definition makes sense if the Betti numbers are all finite and zero beyond a certain index n0.

Two topological spaces which are homotopy equivalent have isomorphic homology groups and hence the same Euler characteristic.

From this definition and Poincaré duality, it follows that Euler characteristic of any closed odd-dimensional manifold is zero.

If M and N are topological spaces, then the Euler characteristic of their product space M × N is

$\chi(M \times N) = \chi(M) \cdot \chi(N)$.

## Partially ordered set

The concept of Euler characteristic of a bounded finite poset is another generalization, important in combinatorics. A poset is "bounded" if it has smallest and largest elements, which let us call 0 and 1. The Euler characteristic of such a poset is μ(0,1), where μ is the Möbius function in that poset's incidence algebra.

## History

The first rigorous proof of Euler's formula, given by 20-year-old Cauchy, is:

Remove one face of the polyhedron. By pulling the edges of the missing face away from each other, deform all the rest into a planar network of points and curves. With no loss of generality it's possible to assume that the deformed edges remained straight line segments. Regular faces cease to be regular polygons if of course they were regular to start with. However, the number of vertices, edges and faces remained the same as those of the given polyhedron (the removed face corresponds to the exterior of the network.)

Apply repeatedly a series of additional transformations that would simplify the network without changing its Euler's number (also Euler's characteristic) FE + V.

1. If there is a polygonal face with more than three sides, we draw a diagonal. This adds one edge and one face. Continue adding edges until all the faces are triangular.
2. Remove (one at a time) all the triangles with two edges shared by the exterior of the network. This removes a vertex, two edges and one face.
3. Remove a triangle with only one edge adjacent to the exterior. This decreases the number of edges and faces by one each and does not change the number of vertices.

Carry out steps 2 and 3 repeatedly one after another until only one triangle is left. For a single triangle F = 2 (counting the exterior), E = 3, V = 3. Therefore FE + V = 2. Which proves the theorem.

03-10-2013 05:06:04