Morse Critical Points Hessians and Indices — Examples

If $x\in S^n$ is not a pole, let $v:=e_{n+1}-x_{n+1}x$. Then $x\cdot v=0$, so $v\in T_x{S}^n$, and $d{h}_x(v)=v_{n+1}=1-x_{n+1}^2\mathrm{\ne }0$. Hence only the poles can be critical.

Near the north pole, write the upper hemisphere as $u\mapsto (u,\sqrt{1-\mathrm{\parallel }u{\mathrm{\parallel }}^2})$. Then $h(u)=\sqrt{1-\mathrm{\parallel }u{\mathrm{\parallel }}^2}=1-\frac{1}{2}\mathrm{\parallel }u{\mathrm{\parallel }}^2+O(\mathrm{\parallel }u{\mathrm{\parallel }}^4)$, so the Hessian at $u=0$ is $-I_n$ and the north pole has index $n$.

Near the south pole, use $u\mapsto (u,-\sqrt{1-\mathrm{\parallel }u{\mathrm{\parallel }}^2})$. Then $h(u)=-\sqrt{1-\mathrm{\parallel }u{\mathrm{\parallel }}^2}=-1+\frac{1}{2}\mathrm{\parallel }u{\mathrm{\parallel }}^2+O(\mathrm{\parallel }u{\mathrm{\parallel }}^4)$, so the Hessian at $u=0$ is $I_n$ and the south pole has index $0$.

Steps 1.1-2.2 give exactly two nondegenerate critical points with distinct critical values $-1$ and $1$. Therefore $h$ is Morse and excellent by [F1].

The partial derivatives satisfy $\mathrm{\partial }{q}_{\lambda }\mathrm{/}\mathrm{\partial }{x}_i=-2x_i$ for $1\le i\le \lambda$ and $\mathrm{\partial }{q}_{\lambda }\mathrm{/}\mathrm{\partial }{x}_i=2x_i$ for $\lambda <i\le n$, so all first derivatives vanish exactly at $x=0$.

The Hessian matrix at the origin is $\mathrm{diag}(\underset{\lambda }{\underbrace{-2,\dots ,-2}},\underset{n-\lambda }{\underbrace{2,\dots ,2}})$, so the critical point is nondegenerate and has exactly $\lambda$ negative directions. By [F1], its index is $\lambda$.

The displayed formula is already the Morse normal form from [L1], including the endpoint cases $\lambda =0$ and $\lambda =n$.

The partial derivatives are $\mathrm{\partial }f\mathrm{/}\mathrm{\partial }x=-2\pi \sin (2\pi x)$ and $\mathrm{\partial }f\mathrm{/}\mathrm{\partial }y=-2\pi \sin (2\pi y)$, so a point is critical exactly when $x,y\in \{0,\frac{1}{2}\}$ mod $\mathbb{Z}$. Hence there are exactly four critical points.

The Hessian matrix is $\mathrm{diag}(-4{\pi }^2\cos (2\pi x),-4{\pi }^2\cos (2\pi y))$. At $(0,0)$ it is negative definite, at $(\frac{1}{2},\frac{1}{2})$ it is positive definite, and at $(0,\frac{1}{2})$ and $(\frac{1}{2},0)$ it has one positive and one negative eigenvalue. Therefore the four critical points have indices $2,0,1,1$ respectively, and all are nondegenerate.

Hence $f$ is Morse by [F1]. Its critical values are $2,-2,0,0$, so the two saddles share the value $0$. Therefore $f$ is not excellent.

One has $f^{\mathrm{\prime }}(x)=4x^3$, so $f^{\mathrm{\prime }}(x)=0$ only at $x=0$. Thus $0$ is an isolated critical point.

Also $f^{\mathrm{\prime }\mathrm{\prime }}(x)=12x^2$, so the Hessian at the critical point is ${\mathrm{Hess}}_0(f)=0$. Therefore the critical point is degenerate.

Hence an isolated critical point need not be nondegenerate.

The differential is $d{f}_{(x,y)}=(2x,0)$ in the standard coordinates, so $d{f}_{(x,y)}=0$ exactly when $x=0$. Thus the whole line $\{(0,y):y\in \mathbb{R}\}$ is critical.

The Hessian matrix is constant, namely $\mathrm{Hess}(f)=\left(\begin{array}{cc}2& 0\\ 0& 0\end{array}\right)$. Its kernel contains the $y$-axis, so every critical point on the line from step 1.1 is degenerate.

Since the critical set contains an entire line, it is not discrete and its points are not isolated. Therefore degenerate critical sets can be nonisolated.

If $M=\mathrm{\varnothing }$, then there are no points to test. So $f$ has no critical points, and the Morse condition is vacuous.

If $M$ is a nonempty $0$-manifold, [F1] says that every point of $M$ is a nondegenerate critical point of index $0$. Hence every smooth function on $M$ is Morse.

In the same $0$-dimensional case, excellence is exactly the condition that distinct points have distinct values, again by [F1].

Therefore the empty and zero-dimensional boundary cases agree with the stated Morse conventions.