Morse Inequalities and the Handle Chain Complex — Examples

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

Near the north pole write the upper hemisphere as $u\mapsto (u,\sqrt{1-\mathrm{\parallel }u{\mathrm{\parallel }}^2})$, so that $h(u)=\sqrt{1-\mathrm{\parallel }u{\mathrm{\parallel }}^2}=1-\frac{1}{2}\mathrm{\parallel }u{\mathrm{\parallel }}^2+O(\mathrm{\parallel }u{\mathrm{\parallel }}^4)$; the Hessian at $u=0$ is $-I_n$, hence the north pole is a nondegenerate critical point of index $n$. Near the south pole write $u\mapsto (u,-\sqrt{1-\mathrm{\parallel }u{\mathrm{\parallel }}^2})$, so that $h(u)=-\sqrt{1-\mathrm{\parallel }u{\mathrm{\parallel }}^2}=-1+\frac{1}{2}\mathrm{\parallel }u{\mathrm{\parallel }}^2+O(\mathrm{\parallel }u{\mathrm{\parallel }}^4)$; the Hessian at $u=0$ is $I_n$ and the south pole has index $0$. Therefore $\mathrm{Crit}(h)=\{\pm e_{n+1}\}$ and, by [F2], $M_h(t)=1+t^n,\mathrm{\#}\mathrm{Crit}(h)=2.$

By [L1] read in unreduced form, $b_0(S^n;F)=b_n(S^n;F)=1$ and $b_k(S^n;F)=0$ for $k\mathrm{\notin }\{0,n\}$; hence by [F3] $P_{S^n,F}(t)=1+t^n=M_h(t)\mathrm{.}$

The correction polynomial of the Morse polynomial identity is unique [F4]; since $Q=0$ satisfies $M_h=P_{S^n,F}+(1+t)Q$, the actual correction polynomial is $Q=0$ and $m_k(h)=b_k(S^n;F)$ for every $k$. By [F5] the height function is $F$-perfect, for every field $F$, and every weak inequality $m_k\ge b_k$ is an equality.

On the torus the gradient of $f$ is $-2\pi (\sin 2\pi x,\sin 2\pi y)$, which vanishes exactly when both coordinates lie in $\{0,\frac{1}{2}\}$; at such a point the Hessian is $\mathrm{diag}(-4{\pi }^2\cos 2\pi x,-4{\pi }^2\cos 2\pi y)$, whose diagonal entries are both negative at $(0,0)$, of opposite signs at the two mixed points, and both positive at $(\frac{1}{2},\frac{1}{2})$. Hence the critical points are these four points, all nondegenerate, with indices $2,1,1,0$, and by [F2], $M_f(t)=1+2t+t^2\mathrm{.}$

By [F2] the index-ordered presentation of $f$ has $m_0=1$ handle of index $0$, $m_1=2$ of index $1$ and $m_2=1$ of index $2$; its handle chain complex over $F$ has ${\dim }_F{C}_0=1$, ${\dim }_F{C}_1=2$, ${\dim }_F{C}_2=1$.

First boundary: the $1$-handles are attached to the single $0$-handle along two feet on its boundary circle $S^1$, and the attaching $0$-sphere $S^0$ is oriented as the boundary of the attaching interval, so the two feet contribute with opposite signs and their intersection numbers with the belt circle cancel; by the endpoint coefficient clause of [F3] this gives zero coefficients, so ${\mathrm{\partial }}_1=0$. Equivalently, each $1$-handle is a band gluing the two marked points with opposite orientations, and the two feet lie in the same component of the connected boundary.

Second boundary: just below the maximum $(0,0)$, the sublevel is the torus with an open disk removed. Each $1$-handle is an untwisted band in this oriented surface. Its outgoing sides $D^1\times S^0$ are both in the boundary, and its belt sphere consists of their two midpoints. The remaining $2$-handle caps this boundary circle, which traverses the two sides of each band in opposite core directions: this follows from the boundary orientation of the rectangle $D^1\times D^1$. With the belt-point orientations compatible with the oriented core, the two local intersection signs are opposite. Thus every coefficient of ${\mathrm{\partial }}_2$ is zero by [F3], integrally and over every field; modulo two the two points likewise cancel.

Since ${\mathrm{\partial }}_1={\mathrm{\partial }}_2=0$, the handle chain complex equals its homology: $H_0\cong F$, $H_1\cong F^2$, $H_2\cong F$; by [F2] the same holds for $H_{\ast }(T^2;F)$, so $P_{T^2,F}(t)=1+2t+t^2=M_f(t)\mathrm{.}$ Comparing in the Morse polynomial identity [F4], the correction polynomial is $Q=0$ and $m_k(f)=b_k(T^2;F)$ for every $k$: the function $f$ is $F$-perfect, for every field $F$.

Euler check: ${\sum }_k(-1{)}^k{m}_k(f)=1-2+1=0$, and by the Euler identity of [F4] this equals $\chi (T^2)$; the same alternating sum of the Betti numbers $1-2+1$ is zero.

The formula $f([x])={\sum }_i(i+1)x_i^2$ is well defined on the quotient because replacing $x$ by $-x$ leaves every squared coordinate unchanged; the quotient ${\mathbb{R}\mathbb{P}}^n$ is the familiar closed smooth $n$-manifold with the standard charts $u_j=x_j\mathrm{/}{x}_i$ around the axis $[e_i]$.

Write ${\lambda }_j=j+1$. A tangent vector $v$ to the unit sphere at $x$ satisfies $x\cdot v=0$, and the differential of the lifted function is $2{\sum }_j{\lambda }_j{x}_j{v}_j$. It vanishes on all such $v$ exactly when $({\lambda }_j{x}_j{)}_j$ is a scalar multiple of $x$. Since the ${\lambda }_j$ are distinct, at most one coordinate of a critical point is nonzero. Thus the critical classes are exactly the axes $[e_i]$. In the chart $u_j=x_j\mathrm{/}{x}_i$ for $j\mathrm{\ne }i$, $f(u)=\frac{{\lambda }_i+{\sum }_{j\mathrm{\ne }i}{\lambda }_j{u}_j^2}{1+{\sum }_{j\mathrm{\ne }i}{u}_j^2}\mathrm{.}$ Expanding at zero gives $f(u)={\lambda }_i+{\sum }_{j\mathrm{\ne }i}({\lambda }_j-{\lambda }_i)u_j^2+O(\mathrm{\mid }u{\mathrm{\mid }}^4)$, so the Hessian is $\mathrm{diag}(2(j-i){)}_{j\mathrm{\ne }i}$, with exactly $i$ negative entries. Every critical point is nondegenerate of index $i$, and $M_f(t)=1+t+\cdots +t^n$.

Over ${\mathbb{F}}_2$ the cellular complex of [F3] reads $C_j={\mathbb{F}}_2$ with $d_j=2=0$ for all $j$, so $H_j({\mathbb{R}\mathbb{P}}^n;{\mathbb{F}}_2)={\mathbb{F}}_2$ for $0\le j\le n$ and zero otherwise, by [F4]; hence $b_k=1=m_k$ for all $k$ and $f$ is ${\mathbb{F}}_2$-perfect with correction polynomial $Q=0$.

Over a field $F$ of characteristic different from two the same complex reads $C_j=F$ with $d_j=2$ invertible for positive even $j$ and $d_j=0$ for odd $j$; taking kernels modulo images gives $H_0=F$, $H_j=0$ for $0<j<n$, and $H_n=F$ if $n$ is odd and $H_n=0$ if $n$ is even. Thus for $n\ge 2$ we have $b_1({\mathbb{R}\mathbb{P}}^n;F)=0<1=m_1(f)$, and $f$ is not $F$-perfect by [F5].

Finally, for every field the Euler identity holds: ${\sum }_{k=0}^n(-1{)}^k{m}_k(f)={\sum }_{k=0}^n(-1{)}^k$ equals $1$ for even $n$ and $0$ for odd $n$, while $\chi ({\mathbb{R}\mathbb{P}}^n)=1$ for even $n$ and $0$ for odd $n$ by [L1] applied to the cell counts of [F3] (one cell in each dimension $0,\dots ,n$). The alternating sums of the Betti numbers agree with the same value in both characteristics, consistent with the Euler identity.

For $h(x)=x_3$ on $S^2$, a point away from the poles has tangent vector $v=e_3-x_{3}x$ with $dh(v)=1-x_3^2>0$, so it is not critical. In pole charts $h(u)=\pm \sqrt{1-\mathrm{\mid }u{\mathrm{\mid }}^2}$ has Hessian $\mp I_2$ at $u=0$; thus the south and north poles have indices $0,2$, and $M_h=1+t^2$. Sphere homology gives $P_{S^2,F}=1+t^2$, so the correction polynomial is $Q=0$.

Apply cancellation in [F1] to the affected connected component of the intermediate stage, keeping other components fixed; for a $(0,1)$-pair its second foot lies on that existing component by the one-intersection condition. It supplies a diffeomorphism of the new presentation's underlying manifold with the old one relative to the incoming face; transport each later attaching embedding across its boundary restriction; hence the modified presentation still presents $M$, and its handle counts are those of the old presentation increased by one in index $k$ and one in index $k+1$.

Let $f,f^{\mathrm{\prime }}$ be the adapted Morse functions inducing the two presentations by [F2]. Their Morse numbers count the handles by index, so $m_j^{\mathrm{\prime }}=m_j+{\delta }_{jk}+{\delta }_{j,k+1}$ for every $j$, and therefore, by [F3], $M_{f^{\mathrm{\prime }}}(t)=M_f(t)+t^k+t^{k+1}=M_f(t)+(1+t)t^k$.

The diffeomorphism of step 1.2 induces homology isomorphisms by [F6]. Thus the Betti numbers, and hence the Poincare polynomials defined in [F3], agree: $P_{M^{\mathrm{\prime }},F}(t)=P_{M,F}(t)$ for every field $F$.

Apply [F4] to both functions, whose Poincare polynomials coincide by step 2.2: $M_f=P_{M,F}+(1+t)Q$ and $M_{f^{\mathrm{\prime }}}=P_{M,F}+(1+t)Q^{\mathrm{\prime }}$. By step 2.1, the nonnegative polynomial $Q+t^k$ also satisfies the second identity. Uniqueness in [F4] therefore gives $Q^{\mathrm{\prime }}=Q+t^k$.

Model case: for $M=S^2$ the two-critical-point presentation of the height function has Morse polynomial $1+t^2$ by [F5]; inserting a cancelling $(0,1)$-pair gives $M_{f^{\mathrm{\prime }}}(t)=1+t^2+t^0+t^1=2+t+t^2$ and $Q^{\mathrm{\prime }}=0+1=1$ by steps 2.1 and 3.1. The Euler sum is preserved: $2-1+1=2=1-0+1$, consistent with the Euler characteristic identity.

For $h(x)=x_3$ on $S^2$, a point away from the poles has tangent vector $v=e_3-x_{3}x$ with $dh(v)=1-x_3^2>0$, so it is not critical. In pole charts $h(u)=\pm \sqrt{1-\mathrm{\mid }u{\mathrm{\mid }}^2}$ has Hessian $\mp I_2$ at $u=0$; thus the south and north poles have indices $0,2$, and $M_h=1+t^2$. Sphere homology gives $P_{S^2,F}=1+t^2$, so the correction polynomial is $Q=0$.

Apply [F1] at the disk stage, whose outgoing circle is nonempty, and transport the original final $2$-handle attaching map across the supplied boundary diffeomorphism. The inserted $(0,1)$-pair is cancelling and the modified presentation still presents $S^2$; its handles are the original $0$-handle and $2$-handle together with the new $0$-handle and $1$-handle, so the presentation has handles of indices $0,0,1,2$.

Let $f^{\mathrm{\prime }}$ be a Morse function inducing the modified presentation. By [F2] its Morse numbers equal the handle counts by index, that is $m_0(f^{\mathrm{\prime }})=2$, $m_1(f^{\mathrm{\prime }})=1$, $m_2(f^{\mathrm{\prime }})=1$, and all other Morse numbers vanish; hence $M_{f^{\mathrm{\prime }}}(t)=2+t+t^2$.

Over every field $F$, [F3] gives $b_0(S^2;F)=1$, $b_1(S^2;F)=0$ and $b_2(S^2;F)=1$, so $P_{S^2,F}(t)=1+t^2$. Comparing with step 2.1, $m_0(f^{\mathrm{\prime }})=2>1=b_0(S^2;F)$, and by [F4] the function $f^{\mathrm{\prime }}$ is not $F$-perfect, for any field $F$.

The correction polynomial is computed by the identity of [F5]: $2+t+t^2=(1+t^2)+(1+t)\cdot 1$, so the unique correction polynomial is $Q=1\mathrm{\ne }0$.

Finally the Euler equality holds: ${\sum }_k(-1{)}^k{m}_k(f^{\mathrm{\prime }})=2-1+1=2$, and by the Euler identity of [F5] this equals $\chi (S^2)$; the same alternating sum computed from the Betti numbers is $1-0+1=2$. Thus the Euler characteristic identity is satisfied while perfectness fails and $Q\mathrm{\ne }0$, refuting the displayed false claim.