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Euler characteristic
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Euler observed that for a convex polyhedron, $$\# {\rm vertices} - \# {\rm edges} + \# {\rm faces} = 2.$$
A remarkable result, especially when you realize that it has nothing to do with convexity.
Let's try adding pyramids:
It is easy to verify that the cube with indentation also satisfies the formula. In a similar fashion one can add indentations to any other convex polyhedron and it will still satisfy the formula.
The example suggests that the meaning of the formula is topological: it holds for any cell complex representation of the sphere.
But it doesn't work with this anymore (a cubic torus):
Definition. The Euler characteristic $\chi (K)$ of $n$-dimensional cell complex $K$ is defined as the alternating sum of the number of cells in $K$ for each dimension: $$\chi (K) = \# 0{\rm -cells} - \# 1{\rm -cells} + \# 2{\rm -cells} - \ldots \pm \# n{\rm -cells}.$$
Example. The Euler characteristic is computed below for the circle, the cylinder, the Mobius band, and the torus.
Exercise. Compute the Euler characteristic of
- all $0$-dimensional complexes,
- the Klein bottle.
More formally, suppose $C_k(K)$ is the set of $k$-cells in a finite cell complex $K$, then $$\chi (K) = \sum_{k} (-1)^k |C_k(K)|.$$
Theorem. The Euler characteristic is a topological invariant, i.e., if the realizations of complexes $K$ and $L$ are homeomorphic, then $\chi (K) = \chi (L)$.
Proof. Use the fact that the homology groups, and the Betti numbers, of a cell complex are topological invariants and apply the Euler-Poincare formula. $\blacksquare$
The converse of this theorem is not true, i.e., the Euler characteristic is not a complete topological invariant. Indeed there are (above) non-homeomorphic complexes with the same Euler characteristic.
Exercise. How come?
This well-known "inclusion-exclusion formula" for sets, $A,B \subset X$, $$|A \cup B| = |A| + |B| - |A \cap B|,$$
has an analogue for the Euler characteristic.
Theorem. Suppose $K, L,$ and $K \cap L$ are subcomplexes of finite complex $A \cup B$, then $$\chi (K \cup L) = \chi (K) + \chi (L) - \chi (K \cap L).$$
Proof. Suppose $C_k(K)$ is the set of $k$-cells in a finite cell complex $K$. Then, by the above formula, we have: $$\begin{array}{l} |C_0(K) \cup C_0(L)| = |C_0(K)| + |C_0(L)| - |C_0(K) \cap C_0(L)|, \\ |C_1(K) \cup C_1(L)| = |C_1(K)| + |C_1(L)| - |C_1(K) \cap C_1(L)|, \\ |C_2(K) \cup C_2(L)| = |C_2(K)| + |C_2(L)| - |C_2(K) \cap C_2(L)|, \\ \vdots \\ |C_n(K) \cup C_n(L)| = |C_n(K)| + |C_n(L)| - |C_n(K) \cap C_n(L)|. \end{array}$$
Since
$$\begin{array}{l} C_n(K) \cup C_n(L) = C_n(K \cup L) {\rm and} \\ C_n(K) \cap C_n(L) = C_n(K \cap L), \end{array}$$
the alternating sum of the above equations gives us the formula. QED
Theorem (Product formula). For two finite cell complexes $K$ and $L$: $$\chi (K \times L) = \chi (K) \chi(L).$$
The Euler characteristic can be used to compute the genus of surface.
A far-reaching generalization is the Lefschetz number of a continuous function.
For more see also
In the same fashion we can define the Euler characteristic of pairs, $\chi(X,A)$ for $A \subset X$.
Another article on the subject: Euler number.