Morse Functions Critical Values and Genericity

The zero set of the section $df$ is exactly the critical set of $f$, because $d{f}_p=0$ means that the differential of $f$ vanishes at $p$. Thus transversality to the zero section is vacuous away from the critical points.

Fix a critical point $p$. By [A1], in cotangent-bundle coordinates centered at $0_p$ the derivative of $df$ at $p$ induces on the fibre quotient exactly the Hessian matrix of $f$ at $p$. Therefore that quotient map is surjective if and only if ${\mathrm{Hess}}_p(f)$ is nondegenerate in the sense of [F1].

By [L1], $df$ is transverse to the zero section at $p$ if and only if that quotient map is surjective. Combining this with step 2.1 shows that $df$ is transverse to the zero section at $p$ if and only if $p$ is a nondegenerate critical point of $f$.

Since the only points at which transversality needs checking are the critical points from step 1.1, step 3.1 proves that $df$ is transverse to the zero section exactly when every critical point of $f$ is nondegenerate. By [F1], that is exactly the Morse condition.

Choose an open set $W$ with $A\subseteq W$ and $\bar{W}\subseteq V$. By [L3], fix a smooth proper exhaustion $h:M\to [0,\mathrm{\infty })$, set $K_{-1}=K_0:=\mathrm{\varnothing }$, and for $n\ge 1$ let $K_n:=h^{-1}([0,n])$. Using [A1], choose integers $r_n\ge n+1$ and positive tolerances ${\epsilon }_n$ so small that whenever smooth functions $u_n$ satisfy $\mathrm{supp}(u_n)\subseteq (\mathrm{int}{K}_{n+1}\setminus K_{n-1})\setminus \bar{W}$ and $\mathrm{\parallel }{u}_n{\mathrm{\parallel }}_{C^{r_n}(K_{n+1})}<{\epsilon }_n$ for every $n$, the locally finite sum $u:={\sum }_n{u}_n$ gives $f+u\in \mathcal{U}$. The varying orders $r_n$ dominate every derivative order imposed by the chosen strong neighbourhood on the corresponding shell.

Construct inductively $g_n$ so that $g_0=f$, the correction $u_n:=g_n-g_{n-1}$ has the support and size prescribed in step 1.1, and $d{g}_n$ is transverse to the zero section on a neighbourhood of $\bar{W}\cup K_n$. Suppose $g_{n-1}$ has been chosen. The set $C_n$ where $d{g}_{n-1}$ is not transverse on $K_n$ is compact. If $C_n=\mathrm{\varnothing }$, put $u_n:=0$ and $g_n:=g_{n-1}$; openness of the transversality locus gives the required neighbourhood of $K_n$, so all the inductive conditions hold. Hence assume that $C_n$ is nonempty. It is disjoint from both $\bar{W}$ and $K_{n-1}$: on $\bar{W}$ every earlier correction vanishes and $df$ is transverse, while transversality near $K_{n-1}$ is the inductive hypothesis. Thus $C_n$ has a finite coordinate cover whose chart closures lie in $(\mathrm{int}{K}_{n+1}\setminus K_{n-1})\setminus \bar{W}\mathrm{.}$ Using [L2], choose bump-supported coordinate functions ${\varphi }_i$ on these charts so that their differentials span every cotangent fibre on a neighbourhood of $C_n$. The resulting parameter space is positive-dimensional: a nontransverse point cannot occur when $\dim M=0$, and a nonempty coordinate cover in positive dimension supplies at least one coordinate function.

In the nonempty case of step 2.1, consider the finite-dimensional family $g_{n-1,\mathbf{a}}:=g_{n-1}+{\sum }_i{a}_i{\varphi }_i\mathrm{.}$ On a neighbourhood of $C_n$, its parameter derivatives span the cotangent fibres. On the compact remainder of $K_n$, the section $d{g}_{n-1}$ is already transverse, so after restricting to a sufficiently small parameter ball, transversality there persists for every parameter. Hence the total family is transverse to the zero section on a neighbourhood of $K_n$. By [L1], the bad slice parameters form a null set; because the parameter ball is positive-dimensional, its complement is dense. Choose ${\mathbf{a}}_n$ in that complement and close enough to $0$ that $u_n:={\sum }_i(a_n{)}_i{\varphi }_i$ satisfies the bound in step 1.1. Put $g_n:=g_{n-1}+u_n$. Transversality is open, so $d{g}_n$ is transverse on a neighbourhood of $K_n$, and it still agrees with $df$ near $\bar{W}$. Together with the empty case handled in step 2.1, this completes the induction.

The shell supports are locally finite, so [L2] makes $u:={\sum }_{n\ge 1}{u}_n$ smooth. Put $g:=f+u$. Step 1.1 gives $g\in \mathcal{U}$. Every correction is supported outside $\bar{W}$, so $g=f$ on the neighbourhood $W$ of $A$ and $g-f$ is supported in $M\setminus A$. Fix $p\in M$ and choose $n$ with $p\in K_n$. Step 3.1 makes $d{g}_{n+1}$ transverse near $p$, while every $u_m$ with $m\ge n+2$ has support disjoint from $K_n$ and therefore vanishes near $p$. Thus $g$ agrees near $p$ with $g_{n+1}$, so $dg$ is transverse there. Since $p$ was arbitrary, $dg$ is transverse everywhere.

By [F1], the final function $g$ is Morse. This proves the relative strong-topology density statement.

By [L1], choose a smooth proper exhaustion $h:M\to [0,\mathrm{\infty })$ and write $K_n:=h^{-1}([0,n])$. For each $n\ge 1$, let ${\mathcal{O}}_n$ be the set of smooth functions $g$ such that $dg$ is transverse to the zero section on some open neighbourhood of $K_n$. If $g$ is Morse, then [F1] gives $g\in {\mathcal{O}}_n$ for every $n$. Conversely, if $g\in {\bigcap }_n{\mathcal{O}}_n$, then every point of $M$ lies in some $K_n$, so $dg$ is transverse near that point; [F1] then makes $g$ Morse. Thus the Morse functions are exactly ${\bigcap }_{n\ge 1}{\mathcal{O}}_n$.

Each ${\mathcal{O}}_n$ is open by [A1].

Each ${\mathcal{O}}_n$ is dense. Indeed, given any smooth function $f$ and any strong neighbourhood $\mathcal{U}$ of $f$, [L2] produces a Morse function $g\in \mathcal{U}$, and step 1.1 shows that every Morse function belongs to every ${\mathcal{O}}_n$.

Step 1.1 identifies the Morse functions with a countable intersection of the open dense sets ${\mathcal{O}}_n$ from steps 2.1 and 2.2. Therefore the Morse functions form a residual subset of $C^{\mathrm{\infty }}(M,\mathbb{R})$ in the strong topology.

By [L1], the critical set of $f$ is finite; write it as $\{p_1,\dots ,p_r\}$. Choose pairwise disjoint coordinate charts ${\phi }_i:U_i^{\mathrm{\prime }}\to B_i\subseteq {\mathbb{R}}^{n_i}$ with ${\phi }_i(p_i)=0$, and then shrink to relatively compact subcharts $U_i⋐U_i^{\mathrm{\prime }}$ that are still pairwise disjoint.

If $r=0$, take $\eta :=1$; there are no neighbourhood conditions to check, so the conclusion is vacuous. If instead $\dim M=0$, then $M$ is discrete and each $U_i:=\{p_i\}$ is a coordinate neighbourhood. Again take $\eta :=1$. The displayed matrix inequality is vacuous on the zero vector space, every point is critical and nondegenerate, and every function on $U_i$ has exactly the one critical point $p_i$. Thus the conclusion holds in either boundary case. Henceforth assume $r\ge 1$ and $\dim M\ge 1$.

Write $F_i:=\mathrm{\nabla }(f\circ {\phi }_i^{-1})$ on $B_i$. Since $p_i$ is nondegenerate by [F1], the derivative $D{F}_i(0)$ is the Hessian matrix at $p_i$ and is invertible. By [L2], after shrinking $U_i$ again we may assume $F_i$ is a diffeomorphism from ${\phi }_i(U_i)$ onto an open neighbourhood of $0$.

Continuity of the Hessian matrices on the compact closures $\overline{U_i}$ lets us shrink the $U_i$ so that every Hessian matrix $H_i(x)$ of $f$ on $\overline{U_i}$ stays within half the least singular value of $D{F}_i(0)$. Hence there is ${\eta }_i>0$ such that $\mathrm{\parallel }{H}_i(x)v\mathrm{\parallel }\ge 2{\eta }_i\mathrm{\parallel }v\mathrm{\parallel }$ for all $x\in \overline{U_i}$ and all coordinate vectors $v$. Let $\eta :={\min }_i{\eta }_i>0$.

If $g$ is sufficiently $C^2$-close to $f$ on $\overline{U_i}$, then the gradient map $\mathrm{\nabla }(g\circ {\phi }_i^{-1})$ is $C^1$-close to $F_i$. Because $F_i$ is a local diffeomorphism carrying $0$ to $0$ and the Hessian gap from step 3.1 keeps the derivative uniformly invertible on $\overline{U_i}$, the inverse-function argument persists under sufficiently small $C^1$ perturbation: the perturbed gradient has a unique zero in ${\phi }_i(U_i)$, and at that zero its derivative is still invertible. Thus $g$ has exactly one nondegenerate critical point in $U_i$.

Steps 1.1, 3.1, and 4.1 give the claimed finite critical set, disjoint critical neighbourhoods, uniform Hessian gap, and local persistence statement.

By [L1], the compact set $K:=M\setminus {\bigcup }_i{U}_i$ contains no critical point of $f$. Hence [F1] gives $d{f}_x\mathrm{\ne }0$ for every $x\in K$.

By [A1], the continuous function $x\mapsto \mathrm{\parallel }d{f}_x\mathrm{\parallel }$ attains a positive minimum $m$ on $K$. Put $\delta :=m\mathrm{/}2$.

If $g$ satisfies $\mathrm{\parallel }dg-df{\mathrm{\parallel }}_{C^0}<\delta$ on $K$, then for every $x\in K$ one has $\mathrm{\parallel }d{g}_x\mathrm{\parallel }\ge \mathrm{\parallel }d{f}_x\mathrm{\parallel }-\mathrm{\parallel }d{g}_x-d{f}_x\mathrm{\parallel }>m-\delta =\delta >0$. Therefore $d{g}_x\mathrm{\ne }0$ on $K$, so [F1] shows that $g$ has no critical point in $K$.

Since $K=M\setminus {\bigcup }_i{U}_i$, this says that sufficiently small $C^1$ perturbations create no new critical points outside the chosen critical neighbourhoods.

Density is immediate from [L1]: given any smooth function and any $C^{\mathrm{\infty }}$ neighbourhood of it, there is a Morse function in that neighbourhood. Since the $C^{\mathrm{\infty }}$ topology is finer than the $C^2$ topology on the same compact manifold, this already implies density in the $C^2$ topology as well.

Let $f$ be Morse. Apply [L2] to choose pairwise disjoint critical neighbourhoods $U_1,\dots ,U_r$ for the critical points of $f$, with the stated persistence and Hessian-gap properties. Apply [L3] to the complement $M\setminus {\bigcup }_i{U}_i$. Then every function $g$ sufficiently close to $f$ in the $C^2$ topology has exactly one critical point in each $U_i$, all those critical points are nondegenerate by [L2], and there are no further critical points outside the $U_i$ by [L3].

Therefore every critical point of such a nearby function $g$ is nondegenerate. Hence $g$ is Morse. This proves that the Morse functions are open in the $C^2$ topology.

Combining steps 1.1 and 2.1 gives that the Morse functions form an open dense subset in the $C^2$ topology. Because the identity map from the $C^{\mathrm{\infty }}$ topology to the $C^2$ topology is continuous on a compact source, the same set is also open and dense in the $C^{\mathrm{\infty }}$ topology.

By [L1], the critical points of $f$ are $p_1,\dots ,p_r$. Choose pairwise disjoint open neighbourhoods $V_i⋐U_i$ of the $p_i$ such that each $U_i$ contains no critical point other than $p_i$, and by [L2] choose smooth functions ${\rho }_i$ with ${\rho }_i=1$ on $V_i$ and $\mathrm{supp}({\rho }_i)\subseteq U_i$.

Let $K:=M\setminus {\bigcup }_{i=1}^r{V}_i$. This compact set contains no critical point of $f$, so [A1] gives a constant $m>0$ with $\mathrm{\parallel }d{f}_x\mathrm{\parallel }\ge m$ for every $x\in K$. Using [A2], choose real numbers ${\lambda }_i$ such that the shifted numbers $f(p_i)+{\lambda }_i$ are pairwise distinct, the finite sum $u:={\sum }_{i=1}^r{\lambda }_i{\rho }_i$ satisfies $f+u\in \mathcal{U}$, and ${\sum }_i\mathrm{\mid }{\lambda }_i\mathrm{\mid }\mathrm{\parallel }d{\rho }_i{\mathrm{\parallel }}_{C^0(K)}<m\mathrm{/}2.$ Define $g:=f+u$.

On each neighbourhood $V_i$ one has $g=f+{\lambda }_i$, because ${\rho }_i=1$ there and every ${\rho }_j$ with $j\mathrm{\ne }i$ vanishes there. Therefore $p_i$ is still a critical point of $g$, and the Hessian of $g$ at $p_i$ equals the Hessian of $f$ there.

For $x\in K$, step 2.1 gives $\mathrm{\parallel }d{g}_x-d{f}_x\mathrm{\parallel }\le {\sum }_i\mathrm{\mid }{\lambda }_i\mathrm{\mid }\mathrm{\parallel }d{\rho }_i{\mathrm{\parallel }}_{C^0(K)}<m\mathrm{/}2,$ so $\mathrm{\parallel }d{g}_x\mathrm{\parallel }\ge \mathrm{\parallel }d{f}_x\mathrm{\parallel }-\mathrm{\parallel }d{g}_x-d{f}_x\mathrm{\parallel }>m\mathrm{/}2>0$. Hence $g$ has no critical point on $K$. Since every critical point of $f$ lies in some $V_i$, step 3.1 shows that the critical set of $g$ is exactly $\{p_1,\dots ,p_r\}$.

The critical values of $g$ are $g(p_i)=f(p_i)+{\lambda }_i$, which are pairwise distinct by step 2.1. The same step also gives $g\in \mathcal{U}$. Therefore $g$ has the same critical points and Hessians as $f$, while all of its critical values are distinct.

Thus one can separate all repeated critical values by disjoint local perturbations without changing any critical Hessian.

To prove density, let $\mathcal{U}$ be a $C^{\mathrm{\infty }}$ neighbourhood of an arbitrary smooth function $f$. By [L1], choose a Morse function $h\in \mathcal{U}$. Because $\mathcal{U}$ is itself a $C^{\mathrm{\infty }}$ neighbourhood of the compact Morse function $h$, [L2] yields a function $g\in \mathcal{U}$ with the same critical points and Hessians as $h$, but with pairwise distinct critical values. By [F1], this $g$ is excellent Morse.

Let $f$ be excellent Morse. By [L3], choose pairwise disjoint critical neighbourhoods $U_1,\dots ,U_r$ around the critical points $p_1,\dots ,p_r$ of $f$ such that every sufficiently small $C^2$ perturbation has exactly one nondegenerate critical point in each $U_i$ and none outside ${\bigcup }_i{U}_i$. Because the critical values $f(p_i)$ are pairwise distinct, choose pairwise disjoint open intervals $I_i$ with $f(p_i)\in I_i$ and $f(U_i)\subseteq I_i$.

If $g$ is sufficiently close to $f$ in the $C^2$ topology, then step 1.2 gives exactly one critical point $q_i$ of $g$ in each $U_i$ and no others. The $C^0$ closeness keeps $g(q_i)$ inside the same interval $I_i$, so the critical values of $g$ are pairwise distinct because the intervals are disjoint. Since each $q_i$ is nondegenerate, [F1] shows that $g$ is excellent Morse.

Step 1.1 gives density in the $C^{\mathrm{\infty }}$ topology, hence also in the coarser $C^2$ topology, and step 2.1 gives openness in the $C^2$ topology. Every $C^2$-open subset is also $C^{\mathrm{\infty }}$-open on a compact manifold, so the same set of excellent Morse functions is open in the $C^{\mathrm{\infty }}$ topology as well. Thus the excellent Morse functions are open dense in both topologies.

If $N=1$, then every embedded compact submanifold of $\mathbb{R}$ is zero-dimensional, hence finite. The restriction of a nonzero linear functional to a finite manifold has all points critical and nondegenerate in the zero-dimensional Morse convention, so it is Morse.

Assume now that $N\ge 2$. Define a smooth family of cotangent sections by $\mathcal{D}:M\times S^{N-1}\to T^{\ast }M,\mathcal{D}(x,u)=d(h_u{)}_x\mathrm{.}$ By [L2], its target zero section is an embedded submanifold. At any zero $(x,u)$, [A1] says that the parameter derivative in tangent directions to $S^{N-1}$ spans the whole fibre $T_x^{\ast }M$, so $\mathcal{D}$ is transverse to the zero section.

Applying [L1] to the family $\mathcal{D}$ shows that the set $E\subseteq S^{N-1}$ of directions for which $d(h_u)$ fails to be transverse to the zero section is null. For every $u\mathrm{\notin }E$, [F1] makes $h_u$ a Morse function on $M$.

Step 1.1 handles $N=1$, and step 2.1 handles $N\ge 2$. Therefore restricted linear heights are Morse for generic directions.

Define a smooth family of cotangent sections by $\mathcal{D}:M\times {\mathbb{R}}^N\to T^{\ast }M,\mathcal{D}(x,p)=d(d_p{)}_x\mathrm{.}$ By [L2], the zero section is an embedded submanifold of the target bundle. At any zero $(x,p)$, the parameter-derivative description in [A1] spans the full fibre $T_x^{\ast }M$, so $\mathcal{D}$ is transverse to the zero section.

Apply [L1] to the family $\mathcal{D}$. The bad centers $p$ for which $d(d_p)$ is not transverse to the zero section form a null subset $E\subseteq {\mathbb{R}}^N$. For every $p\mathrm{\notin }E$, [F1] turns this transversality conclusion into the statement that $d_p$ is Morse.

Therefore squared-distance functions are Morse for generic centers in the ambient Euclidean space.

By [L1], choose a smooth embedding $M\hookrightarrow {\mathbb{R}}^N$. Then [L2] gives a Morse height function $h:M\to \mathbb{R}$ on that embedding.

Apply [L3] to the Morse function $h$. Since excellent Morse functions are dense on the compact manifold $M$, some excellent Morse function lies arbitrarily close to $h$ and in particular exists on $M$.

Therefore every compact smooth manifold admits an excellent Morse function.

The family of supports is locally finite, so [L1] makes the sum $u={\sum }_n{u}_n$ a smooth function.

For every $x\in M$, the pointwise estimate gives $\mathrm{\mid }u(x)\mathrm{\mid }\le {\sum }_{n\ge 1}\mathrm{\mid }{u}_n(x)\mathrm{\mid }\le {\sum }_{n\ge 1}{2}^{-n-2}=\frac{1}{4}$ by [A1]. Hence $(h+u)(x)\ge h(x)-\frac{1}{4}$ for all $x$.

If $(h+u)(x)\le c$, then step 2.1 gives $h(x)\le c+\frac{1}{4}$. Therefore $\{x:(h+u)(x)\le c\}\subseteq \{x:h(x)\le c+\frac{1}{4}\}\mathrm{.}$ The right-hand side is compact because $h$ is proper, and the left-hand side is closed because $h+u$ is continuous. By [A2], the left-hand side is compact.

Thus $h+u$ is proper, and the shellwise perturbation preserves properness.

Choose a smooth proper exhaustion $h:M\to [0,\mathrm{\infty })$ by [L1], and let $\mathcal{U}:=\{g\in C^{\mathrm{\infty }}(M,\mathbb{R}):\mathrm{\mid }g(x)-h(x)\mathrm{\mid }<\frac{1}{4}\text{ for every }x\in M\}\mathrm{.}$ By [A1], this is a strong neighbourhood of $h$.

Apply [L2] to the function $h$, the closed set $A:=\mathrm{\varnothing }$, and the strong neighbourhood $\mathcal{U}$. It yields a smooth function $g\in \mathcal{U}$ whose differential is transverse to the zero section.

Since $g\in \mathcal{U}$, one has $g\ge h-\frac{1}{4}$. Thus, for every real $c$, $\{x:g(x)\le c\}\subseteq \{x:h(x)\le c+\frac{1}{4}\}\mathrm{.}$ The left-hand side is closed and the right-hand side is compact because $h$ is proper, so every sublevel set of $g$ is compact. Also $g\ge -\frac{1}{4}$, hence the inverse image under $g$ of every compact subset of $\mathbb{R}$ is a closed subset of one of these compact sublevel sets. Therefore $g$ is proper. Since step 2.1 gives $dg$ transverse to the zero section everywhere, [F1] makes $g$ Morse.

Hence $g$ is a proper Morse function on $M$.