[Math] Intuition behind the Morse inequalities

at.algebraic-topologydg.differential-geometryintuitionmorse-theory

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

Best Answer

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.