Morse Critical Points Hessians and Indices

Write $g:=f\circ x^{-1}$, $h:=f\circ y^{-1}$, $a:=x(p)$, $b:=y(p)$, and $\varphi :=x\circ y^{-1}$. Then $h=g\circ \varphi$, and the matrix $J$ in the statement is $D{\varphi }_b$.

Because $p$ is critical, the first derivative of $g$ at $a$ is zero, so differentiating $h=g\circ \varphi$ twice at $b$ gives $D^2{h}_b(u,v)=D^2{g}_a(Ju,Jv)+D{g}_a(D^2{\varphi }_b(u,v))=D^2{g}_a(Ju,Jv)$ for all $u,v\in {\mathbb{R}}^n$.

Step 2.1 is exactly the matrix identity $H_y(f,p)=J^{\mathsf{T}}{H}_x(f,p)J$, so the two chart Hessians are congruent.

Fix a coordinate chart and let $(g_{ij})$ be the metric matrix there. Since $g$ is positive definite, $(g_{ij})$ is invertible at each point; write $(g^{ij})$ for the inverse matrix, define ${\mathrm{\Gamma }}_{ijk}:=\frac{1}{2}({\mathrm{\partial }}_i{g}_{jk}+{\mathrm{\partial }}_j{g}_{ik}-{\mathrm{\partial }}_k{g}_{ij})$ and ${\mathrm{\Gamma }}_{ij}^k:={\sum }_{\mathrm{\ell }}{g}^{k\mathrm{\ell }}{\mathrm{\Gamma }}_{ij\mathrm{\ell }}$, and use [F1] to obtain a local cotangent-bundle connection.

The formula for ${\mathrm{\Gamma }}_{ijk}$ is symmetric in $i$ and $j$, so ${\mathrm{\Gamma }}_{ij}^k={\mathrm{\Gamma }}_{ji}^k$. Hence the local connection is symmetric.

Adding ${\mathrm{\Gamma }}_{ijk}$ and ${\mathrm{\Gamma }}_{ikj}$ gives ${\mathrm{\Gamma }}_{ijk}+{\mathrm{\Gamma }}_{ikj}={\mathrm{\partial }}_i{g}_{jk}$, so the local connection is metric-compatible with $g$ in the sense of [F1].

Let $x=(x^i)$ and $y=(y^a)$ be overlapping charts, and let ${\mathrm{\nabla }}^x,{\mathrm{\nabla }}^y$ be the local connections from step 1.1 on those chart domains. On the overlap, write ${\mathrm{\partial }}_{y^b}={\sum }_i\frac{\mathrm{\partial }{x}^i}{\mathrm{\partial }{y}^b}{\mathrm{\partial }}_{x^i},d{y}^a={\sum }_k\frac{\mathrm{\partial }{y}^a}{\mathrm{\partial }{x}^k}d{x}^k\mathrm{.}$ Using the connection formula from [F1] and the Leibniz rule gives ${\mathrm{\nabla }}_{{\mathrm{\partial }}_{y^b}}^{x}d{y}^a=-{\sum }_c{\tilde{\mathrm{\Gamma }}}_{bc}^{a}d{y}^c,$ where ${\tilde{\mathrm{\Gamma }}}_{bc}^a={\sum }_{i,j,k}\frac{\mathrm{\partial }{y}^a}{\mathrm{\partial }{x}^k}{\mathrm{\Gamma }}_{ij}^k\frac{\mathrm{\partial }{x}^i}{\mathrm{\partial }{y}^b}\frac{\mathrm{\partial }{x}^j}{\mathrm{\partial }{y}^c}+{\sum }_k\frac{\mathrm{\partial }{y}^a}{\mathrm{\partial }{x}^k}\frac{{\mathrm{\partial }}^2{x}^k}{\mathrm{\partial }{y}^b\mathrm{\partial }{y}^c}\mathrm{.}$ Thus ${\mathrm{\nabla }}^x$ is also a cotangent-bundle connection in the $y$-chart, with coefficients ${\tilde{\mathrm{\Gamma }}}_{bc}^a$.

Conversely, let ${\tilde{\mathrm{\Gamma }}}_{ij}^k$ be the coefficients of any symmetric metric-compatible cotangent connection in this chart, and write ${\tilde{\mathrm{\Gamma }}}_{ijk}:={\sum }_{\mathrm{\ell }}{g}_{k\mathrm{\ell }}{\tilde{\mathrm{\Gamma }}}_{ij}^{\mathrm{\ell }}$. Symmetry and metric compatibility give ${\mathrm{\partial }}_i{g}_{jk}={\tilde{\mathrm{\Gamma }}}_{ijk}+{\tilde{\mathrm{\Gamma }}}_{ikj}$, ${\mathrm{\partial }}_j{g}_{ik}={\tilde{\mathrm{\Gamma }}}_{jik}+{\tilde{\mathrm{\Gamma }}}_{jki}$, and ${\mathrm{\partial }}_k{g}_{ij}={\tilde{\mathrm{\Gamma }}}_{kij}+{\tilde{\mathrm{\Gamma }}}_{kji}$; substituting ${\tilde{\mathrm{\Gamma }}}_{ijk}={\tilde{\mathrm{\Gamma }}}_{jik}$ and ${\tilde{\mathrm{\Gamma }}}_{ikj}={\tilde{\mathrm{\Gamma }}}_{kij}$ yields $2{\tilde{\mathrm{\Gamma }}}_{ijk}={\mathrm{\partial }}_i{g}_{jk}+{\mathrm{\partial }}_j{g}_{ik}-{\mathrm{\partial }}_k{g}_{ij}$, hence ${\tilde{\mathrm{\Gamma }}}_{ijk}={\mathrm{\Gamma }}_{ijk}$ and ${\tilde{\mathrm{\Gamma }}}_{ij}^k={\mathrm{\Gamma }}_{ij}^k$.

The coefficients from step 2.3 are symmetric in $b,c$: the first term is symmetric because ${\mathrm{\Gamma }}_{ij}^k={\mathrm{\Gamma }}_{ji}^k$ by step 2.1, and the second is symmetric by equality of mixed partials. If ${\widehat{g}}_{cd}={\sum }_{r,s}{g}_{rs}\frac{\mathrm{\partial }{x}^r}{\mathrm{\partial }{y}^c}\frac{\mathrm{\partial }{x}^s}{\mathrm{\partial }{y}^d}$ are the metric coefficients in the $y$-chart, then differentiating this identity and using step 2.2 in the $x$-chart yields $\frac{\mathrm{\partial }{\widehat{g}}_{cd}}{\mathrm{\partial }{y}^b}={\tilde{\mathrm{\Gamma }}}_{bcd}+{\tilde{\mathrm{\Gamma }}}_{bdc},$ where ${\tilde{\mathrm{\Gamma }}}_{bcd}:={\sum }_a{\widehat{g}}_{da}{\tilde{\mathrm{\Gamma }}}_{bc}^a$. So ${\mathrm{\nabla }}^x$ is symmetric and metric-compatible in the $y$-chart as well.

Step 3.1 now applies in the $y$-chart: the $y$-coefficients of ${\mathrm{\nabla }}^x$ are exactly the Christoffel symbols computed from the metric in that chart. But that is how ${\mathrm{\nabla }}^y$ was defined in step 1.1, so ${\mathrm{\nabla }}^x={\mathrm{\nabla }}^y$ on the overlap. Therefore the local operators glue to a global cotangent-bundle connection, and the same chartwise uniqueness proves global uniqueness.

Thus $g$ has a unique symmetric metric-compatible cotangent-bundle connection, whose coordinate coefficients are the displayed Christoffel symbols.

Choose a smooth chart $x$ around $p$, put $a:=x(p)$ and $h:=f\circ x^{-1}$. By [F2], the coordinate matrix of $({\mathrm{\nabla }}^{2}f{)}_p$ is $({\mathrm{\partial }}_{ij}h(a)-{\sum }_k{\mathrm{\Gamma }}_{ij}^k(a){\mathrm{\partial }}_{k}h(a){)}_{ij}$.

Since $p$ is critical, $d{f}_p=0$, and [L1] therefore forces every coefficient ${\mathrm{\partial }}_{k}h(a)$ to vanish.

Step 1.1 reduces to the matrix $({\mathrm{\partial }}_{ij}h(a){)}_{ij}$, which is exactly the matrix of ${\mathrm{Hess}}_p(f)$ by [F1]. Therefore $({\mathrm{\nabla }}^{2}f{)}_p={\mathrm{Hess}}_p(f)$.

By [F1], any two chart matrices of ${\mathrm{Hess}}_p(f)$ are congruent real symmetric matrices.

Therefore [L1] and [L2] give the same triple $(\text{positive},\text{negative},\text{zero})$ for every chart matrix.

For one diagonal normal form, the zero count is the kernel dimension, the negative count is the maximal dimension of a negative-definite subspace, and the positive count is the maximal dimension of a positive-definite subspace, by [F2] and [L2]. Hence nullity, index, and coindex are the same in every chart.

Thus the Morse nullity, index, and coindex are intrinsic.

If $\dim M=0$, then $\{p\}$ is open in $M$, so it already contains no other point and hence no other critical point.

Assume $\dim M=n>0$. Choose a chart $x:U\to {\mathbb{R}}^n$ with $x(p)=0$, write $g:=f\circ x^{-1}$, and define $G(u):=({\mathrm{\partial }}_{1}g(u),\dots ,{\mathrm{\partial }}_{n}g(u))$. By [L1], for $q\in U$ one has $d{f}_q=0$ exactly when $G(x(q))=0$. [L1, given, assume-case[ positive-dimension], construct]

The derivative $DG(0)$ is the Hessian matrix of $g$ at $0$, and [F1] makes it invertible because $p$ is nondegenerate.

Applying [L2] to $G$ at $0$ gives a neighbourhood $W$ of $0$ in which $G^{-1}(0)=\{0\}$. Therefore the corresponding neighbourhood $x^{-1}[W]\subseteq U$ contains no critical point except $p$.

The zero-dimensional case is step 1.1, and the positive-dimensional case is step 3.1. Hence every nondegenerate critical point is isolated.

For each critical point $p$, [F1] and [L1] give an open neighbourhood $U_p$ containing no critical point other than $p$.

If $q$ is not critical, choose a chart $x:U\to {\mathbb{R}}^n$ around $q$ and write $g:=f\circ x^{-1}$. By [L3], some partial derivative $\mathrm{\partial }g\mathrm{/}\mathrm{\partial }{x}^i$ is nonzero at $x(q)$; continuity keeps it nonzero on a smaller open neighbourhood $V_q\subseteq U$, so $V_q$ contains no critical point.

The family of all $U_p$ together with all $V_q$ covers $M$. By [L2], it has a finite subcover. Only finitely many sets of the form $U_p$ occur in that subcover, and each such set contains exactly one critical point by step 1.1. Therefore $f$ has finitely many critical points.

Hence every Morse function on a compact manifold has finitely many critical points.

If $m=0$, define $A(u):={\int }_0^1(1-t)F^{\mathrm{\prime }\mathrm{\prime }}(tu)dt$. Then $F(u)=u^{2}A(u)$, $A(0)=\frac{1}{2}{F}^{\mathrm{\prime }\mathrm{\prime }}(0)\mathrm{\ne }0$, and after shrinking the domain one has $\epsilon A(u)>0$ for $\epsilon =\mathrm{sgn}(F^{\mathrm{\prime }\mathrm{\prime }}(0))$. Putting $u^{\mathrm{\prime }}:=u\sqrt{\epsilon A(u)}$ gives $F(u)=\epsilon (u^{\mathrm{\prime }}{)}^2$.

Assume $m>0$. Put $G(u,y):=\mathrm{\partial }F\mathrm{/}\mathrm{\partial }u(u,y)$. Since $\mathrm{\partial }G\mathrm{/}\mathrm{\partial }u(0,0)={\mathrm{\partial }}^{2}F\mathrm{/}\mathrm{\partial }{u}^2(0,0)\mathrm{\ne }0$, [L1] gives a smooth function $\varphi$ near $0\in {\mathbb{R}}^m$ with $\varphi (0)=0$ and $G(\varphi (y),y)=0$. [L1, given, assume-case[ positive-parameter], construct]

Set $\tilde{F}(s,y):=F(s+\varphi (y),y)-F(\varphi (y),y)$. Then $\tilde{F}(0,y)=0$ and $\mathrm{\partial }\tilde{F}\mathrm{/}\mathrm{\partial }s(0,y)=0$ for $y$ near $0$.

Define $A(s,y):={\int }_0^1(1-t){\mathrm{\partial }}^2\tilde{F}\mathrm{/}\mathrm{\partial }{s}^2(ts,y)dt$. The integral formula gives $\tilde{F}(s,y)=s^{2}A(s,y)$, and $A(0,0)=\frac{1}{2}{\mathrm{\partial }}^{2}F\mathrm{/}\mathrm{\partial }{u}^2(0,0)\mathrm{\ne }0$. After shrinking, the sign $\epsilon :=\mathrm{sgn}({\mathrm{\partial }}^{2}F\mathrm{/}\mathrm{\partial }{u}^2(0,0))$ satisfies $\epsilon A(s,y)>0$ everywhere.

Put $\beta (s,y):=\sqrt{\epsilon A(s,y)}$, define $u^{\mathrm{\prime }}:=\beta (s,y)s$, and let $H(y):=F(\varphi (y),y)$. Then $F(s+\varphi (y),y)=H(y)+\epsilon (u^{\mathrm{\prime }}{)}^2$, and $\mathrm{\partial }{u}^{\mathrm{\prime }}\mathrm{/}\mathrm{\partial }s(0,0)=\beta (0,0)\mathrm{\ne }0$, so [L2] makes $(s,y)\mapsto (u^{\mathrm{\prime }},y)$ a local diffeomorphism at $(0,0)$.

Composing the translation $(u,y)\mapsto (u-\varphi (y),y)$ from step 1.2 with the coordinate change from step 4.1 yields the required local coordinates $(u^{\mathrm{\prime }},y)$, and the case $m=0$ is already covered by step 1.1.

Because the coordinate change fixes $(0,0)$, the coordinate representative $\tilde{F}$ also satisfies $d{\tilde{F}}_{(0,0)}=0$. Setting $s=0$ in the displayed decomposition gives $\tilde{F}(0,y)=F(0,0)+H(y)$. Differentiating at $y=0$ therefore shows $d{H}_0=0$, so $0$ is a critical point of $H$.

In the coordinates $(s,y)$, the function $\tilde{F}$ has no mixed $sy$ term and no term linear in $s$, so its Hessian matrix at $(0,0)$ has block form $\left(\begin{array}{cc}2\epsilon & 0\\ 0& {\mathrm{Hess}}_0(H)\end{array}\right)$. By [F1], this is the Hessian of $F$ at $(0,0)$ in the $(s,y)$ chart, and the lower-right block is exactly its restriction to the $y$-coordinate subspace.

If ${\mathrm{Hess}}_0(H)$ had a nonzero kernel vector $v$, then $(0,v)$ would lie in the kernel of the block matrix from step 2.1. Hence a nondegenerate Hessian for $F$ forces ${\mathrm{Hess}}_0(H)$ to be nondegenerate.

Thus splitting one signed square preserves the residual critical Hessian.

If $n=0$, the manifold is locally a point, so $f$ is locally constant at $p$. The Hessian acts on the zero vector space, hence $\lambda =0$ by [F1], and the displayed formula is exactly $f=f(p)$ with both sums empty.

Assume the theorem proved in dimensions $<n$, where $n>0$. Choose local coordinates $u=(u^1,\dots ,u^n)$ centered at $p$ and write $g:=f\circ u^{-1}-f(p)$. By [L1], after a linear change of the $u$-coordinates the Hessian matrix of $g$ at $0$ is diagonal with entries in $\{1,-1,0\}$. Since $p$ is nondegenerate and has index $\lambda$, [F1] and [L2] force exactly $\lambda$ negative diagonal entries, exactly $n-\lambda$ positive diagonal entries, and no zero entry. Reorder the coordinates so the first diagonal entry is negative when $\lambda >0$ and positive when $\lambda =0$; in particular ${\mathrm{\partial }}^{2}g\mathrm{/}\mathrm{\partial }(u^1{)}^2(0)\mathrm{\ne }0$. [F1, L1, L2, given, assume-case[ positive-dimension], construct]

Apply [L3] to the first coordinate $u^1$, taking the remaining variables as parameters. After shrinking the chart there are new coordinates $(v^1,y)$ with $g(v^1,y)=\epsilon (v^1{)}^2+H(y)$, where $\epsilon \in \{\pm 1\}$, $y\in {\mathbb{R}}^{n-1}$, and $0$ is a critical point of $H$.

By [L2], the Hessian of $g$ in the $(v^1,y)$ chart still has index $\lambda$. By [L4], the Hessian of $H$ at $0$ is the restriction to the $y$-coordinates, and the split $v^1$-direction contributes one negative square exactly when $\epsilon =-1$. Therefore ${\mathrm{Hess}}_0(H)$ is nondegenerate, with index $\lambda -1$ when $\epsilon =-1$ and index $\lambda$ when $\epsilon =1$.

Apply the induction hypothesis to $H$ on ${\mathbb{R}}^{n-1}$. It yields local coordinates $(v^2,\dots ,v^n)$ putting $H$ into its Morse normal form, and adjoining $v^1$ contributes one additional negative square exactly when $\epsilon =-1$. Therefore the full expression for $g$ has exactly $\lambda$ negative squares and $n-\lambda$ positive squares.

Combining steps 2.1 and 4.1 proves the displayed normal form for dimension $n$, and step 1.1 covers the base case.

By [L1], choose local coordinates centered at $p$ in which $f-f(p)=-{\sum }_{i=1}^{\lambda }(x^i{)}^2+{\sum }_{i=\lambda +1}^n(x^i{)}^2$.

If $\lambda =0$, the first sum is empty, so $f-f(p)={\sum }_{i=1}^n(x^i{)}^2$, which is strictly positive for every nearby point other than $p$; hence $p$ is a strict local minimum. If $n=0$, this same formula is $f-f(p)=0$, so the local minimum and maximum clauses coincide.

If $\lambda =n$, the second sum is empty, so $f-f(p)=-{\sum }_{i=1}^n(x^i{)}^2$, which is strictly negative away from $p$. Hence $p$ is a strict local maximum.

If $0<\lambda <n$, then along the $x^1$-axis one has $f-f(p)=-(x^1{)}^2<0$ for nearby nonzero points, while along the $x^n$-axis one has $f-f(p)=(x^n{)}^2>0$ for nearby nonzero points. Therefore every neighbourhood of $p$ contains points where $f<f(p)$ and points where $f>f(p)$, so $p$ is a saddle.

The three cases above exhaust the possible values of $\lambda$.

Because differentiation is linear, ${\mathrm{Hess}}_p(-f)=-{\mathrm{Hess}}_p(f)$.

Multiplication by $-1$ does not change the kernel of a bilinear form, so the nullity is unchanged.

A subspace is negative definite for $-{\mathrm{Hess}}_p(f)$ exactly when it is positive definite for ${\mathrm{Hess}}_p(f)$, and similarly with "positive" and "negative" exchanged. Therefore the index and coindex swap by [F1].

Hence negating $f$ preserves nullity and exchanges index with coindex.

By [L1], the equation $f=f(p)$ becomes ${\sum }_{i=1}^{\lambda }(x^i{)}^2={\sum }_{i=\lambda +1}^n(x^i{)}^2$.

If $\lambda =0$, the left-hand side is the empty sum $0$, so the equation is $0={\sum }_{i=1}^n(x^i{)}^2$, which forces $x=0$. If $\lambda =n$, the right-hand side is the empty sum and the same conclusion follows.

Therefore the local critical level is the stated quadratic cone, with the index-$0$ and index-$n$ cases collapsing to the single critical point.