Intuition behind the Morse inequalities?

Question

Forgive me if this is sort of a vague question, but can someone supply me with their intuition behind the Morse inequalities?

Answer 1

If you understand intuition behind the fact that the Euler characteristic is the alternating sum of the betti numbers, then I think you can grasp the Morse inequalities. A Morse function gives rise to a CW structure on the manifold, by considering the unstable manifolds of index $k$ critical points as giving $k$-cells attached to the $k-1$-skeleton. The Morse inequalities apply more generally to compact CW complexes.

The Euler characteristic of the $\gamma$-skeleton $M^{(\gamma)}$ of a compact CW complex $M$ is given by the alternating sum of the betti numbers of the $\gamma$-skeleton, and by the alternating sum of the number of $j$-cells $C^j$ of the $\gamma$-skeleton, $j\leq \gamma$, which is the same as the corresponding sum for $M$. The betti numbers of $M$ will agree with those of $M^{(\gamma)}$ except possibly in dimension $\gamma$, where one has $b_\gamma(M^{(\gamma)})\geq b_\gamma(M)$, because some $\gamma+1$-cells might kill cellular homology generated by $\gamma$-cells. This translates directly into the Morse inequality of index $\gamma$. Following the notation of the Wikipedia page on Morse theory, one has

$$C^\gamma - C^{\gamma-1} + \cdots +(-1)^\gamma C^0 =(-1)^{\gamma} \chi(M^{(\gamma)}) = b_\gamma(M^{(\gamma)}) - b_{\gamma-1}(M^{(\gamma)}) +\cdots +(-1)^\gamma b_0(M^{(\gamma)}) \geq b_\gamma(M) -b_{\gamma-1}(M) + \cdots +(-1)^\gamma b_0(M).$$

Thus, one may recall the inequality $b_\gamma(M^{(\gamma)})\geq b_\gamma(M)$, which has a fairly intuitive explanation in terms of cellular homology, and derive the Morse inequalities as a consequence. There is also a relative version which may be easier to understand.