This site is being phased out.

Euler characteristic

From Mathematics Is A Science
(Redirected from Euler characteristic in topology)
Redirect page
Jump to navigationJump to search

Euler observed that for a convex polyhedron, $$\# {\rm vertices} - \# {\rm edges} + \# {\rm faces} = 2.$$

Euler formula examples.jpg

A remarkable result, especially when you realize that it has nothing to do with convexity.

Let's try adding pyramids:

EulerChar.png

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):

Cubic torus.png

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.

Euler characteristic examples.jpg

Exercise. Compute the Euler characteristic of

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?

Euler characteristic of sphere.jpg

Solution: Euler characteristic of sphere solution.jpg

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.