The Euler characteristic of a manifold

Question

The Euler characteristic of a manifold is the alternating sum of the number of critical points of a Morse function on it.

$$ \chi(X) := \sum_{k=0}^{n} (-1)^k b_k = \sum_{k=0}^{n} (-1)^k c_k. $$

My question: Could you please tell me the way to see how can we get the equality? I though that it should be $$ \sum_{k=0}^{n} (-1)^k b_k \leq \sum_{k=0}^{n} (-1)^k c_k,$$ as in the case $k=n$ in the Morse inequalities:

Let $b_k$ denote $k^{th}$ Betti number of $\mathbb{M}$, i.e. the dimension of the $k^{th}$ homology group $H_k\mathbb{M}$ and $c_k$ denote the number of index $k$ critical points of $f$. If all critical points of $f$ are non-degenerate, then for every $k$, $$ c_k - c_{k-1} + c_{k-2} - \dots + (-1)^k c_0 \geq b_k - b_{k-1} + b_{k-2} + \dots + (-1)^k b_0. $$

Answer 1

The answer is that the inequality is an equality for the dimension of the manifold.

Let me introduce you to another way of writing the Morse inequalities. Denote by $P_t=\sum_{k=0}^n b_kt^k$ and $M_t=\sum_{k=0}^nc_kt^k$ the Poincaré polynomial and the Morse polynomial. Then there exists a polyomial $Q_t$, whose coefficients are not negative, such that $$ M_t=P_t+(1+t)Q_t $$ This result is pure homological algebra and follows from the fact that the Morse complex is finitely generated and computes the homology of the manifold. Plugging in $t=-1$ will give you the formula for the Euler characteristic.

To get the other inequality you have to expand the denomiator of $\frac{M_t-P_t}{1+t}=(M_t-P_t)(1-t+t^2-\ldots\pm t^n)$ and collect terms.

Answer 2

While the answer of Thomas Rot is perfectly fine and gives more information than required, I would like to give a low technology approach.

Let $M^n$ be a closed manifold and let us consider the Morse complex associated to a Morse function on $M$:

$$0\overset{\partial_{n+1}}{\longrightarrow}C_n\overset{\partial_n}{\longrightarrow}C_{n-1}\overset{\partial_{n-1}}{\longrightarrow}\cdots\overset{\partial_2}{\longrightarrow}C_1\overset{\partial_1}{\longrightarrow}C_0\overset{\partial_0}{\longrightarrow}0,$$ then for all $k\in\{0,\ldots,n\}$, using the rank-nullity theorem, one has: $$c_k=\dim(\ker(\partial_k))+\dim(\operatorname{im}(\partial_k))\tag{1}.$$ Furthermore, recall that the homology of this complex computes the singular homology of $M$ with coefficients in $\mathbb{Z}/2\mathbb{Z}$, so that: $$b_k=\dim(\ker(\partial_k))-\dim(\operatorname{im}(\partial_{k+1}))\tag{2}.$$ According to the equalities $(1)$ and $(2)$, one gets: $$b_k+\dim(\operatorname{im}(\partial_{k+1}))=c_k-\dim(\operatorname{im}(\partial_k)).$$ Hence, noticing that $\partial_0=0=\partial_{n+1}$, one gets the result summing the above equalities.