Euler Characteristics and Mayer-Vietoris Sequence

Euler Characteristic Revisited

Rank of a Group

The rank of a finitely generated abelian group is the number of \newcommand{\Z}{\mathbb{Z}}\Z summands in its decomposition.

e.g. $\text{rank}(\mathbb{Z}\oplus \mathbb{Z}\oplus \mathbb{Z}/69\mathbb{Z})=2$.

Homological Euler Characteristic

The Euler characteristic of a finite CW complex $X$ can be computed from its homology groups: $\chi (X)={\sum }_{n\ge 0}(-1{)}^n\text{rank}(H_n(X)).$

$\chi (X)$ is a topological invariant and does not depend on the specific CW structure chosen for $X$.

The Euler characteristic

Proof Let $C_n=C_n^{\text{CW}}(X)$, $Z_n=\ker (d_n)$, and $B_n=\text{im}(d_{n+1})$. These groups are related by two short exact sequences for each $n$:$\begin{array}{r}0\to Z_n\to C_n\to B_{n-1}\to 0\\ 0\to B_n\to Z_n\to H_n\to 0\end{array}$From a lemma on short exact sequences of finitely generated abelian groups, we know that ranks are additive. Thus:

1. $\text{rank}(C_n)=\text{rank}(Z_n)+\text{rank}(B_{n-1})$
2. $\text{rank}(Z_n)=\text{rank}(B_n)+\text{rank}(H_n)$

Now, we compute $\chi (X)$: $\begin{array}{rl}\chi (X)& =\sum _n(-1{)}^n\text{rank}(C_n)\\ & =\sum _n(-1{)}^n(\text{rank}(Z_n)+\text{rank}(B_{n-1}))\\ & =\sum _n(-1{)}^n\text{rank}(Z_n)+\sum _n(-1{)}^n\text{rank}(B_{n-1})\\ & =\sum _n(-1{)}^n(\text{rank}(B_n)+\text{rank}(H_n))+\sum _n(-1{)}^n\text{rank}(B_{n-1})\\ & =\sum _n(-1{)}^n\text{rank}(H_n)+\sum _n(-1{)}^n\text{rank}(B_n)+\sum _n(-1{)}^n\text{rank}(B_{n-1}).\end{array}$The last two sums cancel each other out via a change of index. Therefore, we are left with the desired result: $$\chi(X) = \sum_n (-1)^n \text{rank}(H_n(X)) $$$\square$

e.g.

• Sphere: $\chi (S^n)=\{\begin{array}{ll}2& n\text{ even}\\ 0& n\text{ odd}\end{array}$.
• Real Projective Space: $\chi ({\mathbb{RP}}^n)=\{\begin{array}{ll}1& n\text{ even}\\ 0& n\text{ odd}\end{array}$.
• Klein Bottle: $\chi (K)=0$.
• Genus $g$ Surface: $\chi (M_g)=2-2g$.
• Complex Projective Space: $\chi ({\mathbb{CP}}^n)=n+1$.

Mayer-Vietoris Sequence

Mayer-Vietoris Sequence

Suppose a space $X$ is the union of the interiors of two subspaces $A$ and $B$, i.e., $X=\text{int}(A)\cup \text{int}(B)$. Then there is a long exact sequence in homology: $\cdots \to H_n(A\cap B)\to H_n(A)\oplus H_n(B)\to H_n(X)\stackrel{{\partial }_{\ast }}{\to }{H}_{n-1}(A\cap B)\to \cdots$

Proof This sequence arises from the short exact sequence of chain complexes: $0\to C_n(A\cap B)\stackrel{x\mapsto (x,-x)}{\to }{C}_n(A)\oplus C_n(B)\stackrel{(x,y)\mapsto x+y}{\to }{C}_n(A+B)\to 0$ Here $C_n(A+B)$ represents chains in $A\cup B$, and by the Excision Theorem, its homology is $H_n(X)$. $\square$

e.g. The suspension $SX$ of a space $X$ is given by $SX=C_{+}X{\cup }_X{C}_{-}X$, where $C_{+}X$ and $C_{-}X$ are two cones on $X$ joined at their base. Let $A=C_{+}X$ and $B=C_{-}X$. Then $A\cap B=X$ and $A\cup B=SX$. Both $A$ and $B$ are contractible, so their reduced homology groups are zero, i.e., ${\tilde{H}}_k(C_{\pm }X)=0$ for all $k$. The reduced Mayer-Vietoris sequence contains the segment: $\cdots \to {\tilde{H}}_n(C_{+}X)\oplus {\tilde{H}}_n(C_{-}X)\to {\tilde{H}}_n(SX)\stackrel{{\partial }_{\ast }}{\to }{\tilde{H}}_{n-1}(X)\to {\tilde{H}}_{n-1}(C_{+}X)\oplus {\tilde{H}}_{n-1}(C_{-}X)\to \dots$Since the cone homology groups are zero, this simplifies to:$\cdots \to 0\to {\tilde{H}}_n(SX)\stackrel{{\partial }_{\ast }}{\to }{\tilde{H}}_{n-1}(X)\to 0\to \dots$This shows that the connecting homomorphism ${\partial }_{\ast }$ is an isomorphism: ${\tilde{H}}_n(SX)\cong {\tilde{H}}_{n-1}(X).$