Lagmental Vicfred

Morse Inequalities Compare Critical Points with Homology by Vicfred

Last updated: Sun 04 December 2022

A Morse function needs at least as many index-k critical points as the kth Betti number. I want the notation, the mechanism, and the failure mode visible at the same time.

Start locally

A symplectic manifold \((M^{2n},\omega)\) has a closed nondegenerate two-form. A Hamiltonian \(H:M\to\mathbf R\) determines a vector field through contraction with \(\omega\).

$$ m_k=\#\{p:df_p=0,\ \operatorname{index}_p(f)=k\} $$

A reliable calculation names domain and codomain. The notation \(\mathsf{data}\mapsto\mathsf{claim}\) is harmless only after both \(\operatorname{dom}\) and \(\operatorname{cod}\) have been fixed.

$$ m_k\ge b_k(M),\qquad\sum_k(-1)^km_k=\chi(M) $$

Compute before generalising

A worked instance is useful here because it exposes every index that the compressed statement hides.

$$ \sum_{k=0}^{n}m_kt^k-\sum_{k=0}^{n}b_kt^k=(1+t)Q(t),\qquad Q(t)\in\mathbf Z_{\ge0}[t] $$

The global view

The two-row display is also a debugging tool: if the conclusion changes when only notation changes, some hidden choice has entered the argument.

$$ \begin{aligned} \mathsf{D}\;&:\quad m_k=\#\{p:df_p=0,\ \operatorname{index}_p(f)=k\},\\[5pt] \mathsf{C}\;&:\quad m_k\ge b_k(M),\qquad\sum_k(-1)^km_k=\chi(M). \end{aligned} $$

Edge conditions

Symplectic geometry has no preferred notion of distance. Nondegeneracy of a two-form is not positive definiteness of a metric.

$$ \boxed{\begin{gathered} \text{compact conclusion}\\[-2pt] m_k\ge b_k(M),\qquad\sum_k(-1)^km_k=\chi(M) \end{gathered}} $$

The notation is dense, but it is doing honest work: every delimiter records scope and every index records dependence. Removing one should require a mathematical reason.

This article was posted on Sun 23 October 2022. Facts and circumstances may have changed since publication.
Please contact me before jumping to conclusions if something seems wrong or unclear.