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The incoming and outgoing boundaries are the regular levels zero and one, and all critical points lie in the interior. Near the boundary use a collar coordinate as the function. The Morse handle-attachment theorem produces a relative chain group freely generated by the interior critical points. Thus $\chi (W,M_0)$ is the alternating sum of their Morse indices, regardless of how their critical values are ordered.

For real $c_1<\cdots <c_n$ and $C=\mathrm{diag}(c_1,\dots ,c_n)$, the critical points are the coordinate projections $P_I$, indexed by $k$-subsets $I=\{i_1<\cdots <i_k\}$. Indeed $df([A,P])=\mathrm{tr}([P,C]A)$, which vanishes for every skew-Hermitian $A$ exactly when $[P,C]=0$. In the graph chart at $P_I$,Every coefficient is nonzero, so the Hessian matrix is nondegenerate. Each pair $j<i$ supplies two negative real directions, giving the displayed Morse index. The function is a Morse function even if distinct critical points happen to have equal critical values.

Here $N_r$ counts $k$-subsets $i_1<\cdots <i_k$ with ${\sum }_a(i_a-a)=r$, equivalently partitions of an integer $r$ contained in a $k$ by $(n-k)$ rectangle. The diagonal trace Morse function on a complex Grassmannian has only even Morse indices. Its Morse chain complex therefore has zero differentials, proving the displayed torsion-free integral homology. For $\mathrm{Gr}(2,4)$ the even-degree ranks are $(1,1,2,1,1)$ in degrees $0,2,4,6,8$.

On a complete compact Riemannian manifold, nondegenerate critical points of fixed-endpoint geodesic energy yield a CW complex model of the path space, with one cell per geodesic and dimension equal to its Morse index. Finite-dimensional broken-geodesic models prove the assertion at bounded energy, and exhaustion gives the full path space. On $S^n$ with distinct nonantipodal endpoints the cell dimensions are $j(n-1)$, one in each such dimension.

Crossing one nondegenerate critical point of Morse index $\lambda$ in a compact sublevel interval attaches a handle $D^{\lambda }\times D^{m-\lambda }$. Between critical levels a suitably normalized gradient flow identifies the sublevels. The Morse lemma supplies the local quadratic model for the attachment. Retracting each handle to its core disk gives a CW complex model with one $\lambda$-cell per critical point. Closed manifolds have finite such decompositions; a noncompact manifold requires a proper bounded-below exhaustion for the corresponding statement.

The Morse index of a fixed-endpoint geodesic equals the number of conjugate points strictly between its endpoints, counted with multiplicity. Its nullity equals the multiplicity of the terminal conjugate point. A conjugate multiplicity is the dimension of the space of Jacobi fields vanishing at the initial point and at the point in question.

Morse theory relates the critical points of a Morse function to changes in the homotopy type of its sublevel sets. Crossing a nondegenerate critical value attaches a cell or handle of dimension equal to the Morse index. The theory also applies to geodesic energy through broken-geodesic approximations.

The Morse lemma states that if $p$ is a nondegenerate critical point of a smooth real function $f$ on an $m$-dimensional smooth manifold, there are smooth coordinates $y_1,\dots ,y_m$, centered at $p$, such thatHere $\lambda$ is the Morse index, the number of negative directions of the Hessian matrix at $p$. The equality is exact on a neighborhood, not merely a second-order expansion.

To prove it, start with any coordinates putting $p=0$. Since $df(0)=0$, integral Taylor theorem givesThe matrix $A(x)=(a_{ij}(x))$ is smooth and symmetric, with $A(0)=\frac{1}{2}\mathrm{Hess}f(0)$. By Sylvester's law of inertia, a constant invertible linear coordinate change makes $A(0)=\mathrm{diag}(-I_{\lambda },I_{m-\lambda })$.

We perform completion of squares with coefficients depending smoothly on the original $x$. The first pivot $a_{11}(x)$ stays nonzero and keeps its sign on a sufficiently small neighborhood. Algebraically,The remaining coefficient matrix is the Schur complement. At zero it is the remaining signed diagonal matrix, so its first pivot is also nonzero nearby. Continue recursively, shrinking the neighborhood finitely many times. Every pivot $d_i(x)$ is smooth and nonzero, with sign ${ϵ}_i=-1$ for $i\le \lambda$ and $+1$ thereafter. Absorb $|d_i(x)|$ into the square by its smooth positive square root. This produces a smooth triangular matrix $B(x)$ with nonzero diagonal such thatAll coefficients here are functions of the original $x$; no circular coordinate substitution is involved. Define $\Phi (x)=B(x)x$. Its derivative at zero is $D\Phi (0)=B(0)$, which is invertible. The inverse function theorem makes $y=\Phi (x)$ a smooth coordinate system and gives the asserted exact quadratic form. The number of negative squares agrees with the Hessian matrix signature and hence with the Morse index. This is the smooth completing-square proof of the Morse lemma.

A Morse function is a smooth real-valued function all of whose critical points have nonsingular Hessian matrices. We use the descending convention for a gradient-like vector field $V$: away from critical points, $df(V)<0$, and near a critical point of Morse index $\lambda$, in Morse lemma coordinates $f=f(p)-|u{|}^2+|v{|}^2$, the field can be chosen as $V=(2u,-2v)$.

For its flow ${\varphi }_t$, define the unstable manifold $W^u(p)=\{x:{\lim }_{t\to -\infty }{\varphi }_t(x)=p\}$ and the stable manifold $W^s(q)=\{x:{\lim }_{t\to +\infty }{\varphi }_t(x)=q\}$. Their dimensions are $\mathrm{ind}p$ and $m-\mathrm{ind}q$. The Morse-Smale transversality condition isFor a closed manifold, this lets us form the Morse-Smale complex over $\mathbb{Z}$. Orient each unstable manifold and takeThe transverse intersection for an index difference of one is one-dimensional; quotienting by flow time gives a finite set of trajectories, and $n(p,q)$ counts them with orientation signs. Compactifying the one-dimensional spaces of trajectories for an index difference of two adds broken trajectories. Their oriented boundary count is zero, giving ${\partial }^2=0$. The descending cells are the unstable manifolds, and these counts are the incidence numbers of the corresponding cellular chain complex, so the homology of this complex is $H_{\ast }(M;\mathbb{Z})$. Ambient orientability is unnecessary: orientations of the unstable cells provide the requisite coorientations and trajectory signs.

For the Klein bottle, use the quotient of ${\mathbb{R}}^2$ by $(s,t)\sim (s+2\pi ,-t)$ and $(s,t)\sim (s,t+2\pi )$. A model of the standard four-critical-point height function isIt is well defined under both identifications. Its critical points, their Morse indices, and critical values areThe Hessian matrix is diagonal with entries $-\cos s$ and $-\epsilon \cos t$, so each point is nondegenerate. The flat quotient metric gives descending equations $\dot{s}=\sin s$, $\dot{t}=\epsilon \sin t$. Their saddle separatrices are the coordinate lines; the two saddle-to-saddle intersections are empty, and all remaining stable and unstable intersections are transverse. This is a Morse-Smale gradient flow. If one insists on the exact local-model version of a gradient-like vector field above, change the two coordinate speeds near the critical points to the standard quadratic-model speeds. This preserves the separatrices, signs and transversality.

Both branches of either saddle flow to the minimum with opposite endpoint signs, so $\partial a=\partial b=0$. From the maximum to $a$, the two trajectories lie on $s=0$ and approach $t=\pi$ from opposite directions. The identification in $t$ is a translation, so their incidence signs are opposite and cancel. The two trajectories to $b$ lie on $t=0$ and approach $s=\pi$ from opposite directions. The identification in $s$ reverses the transverse $t$-coordinate, reversing one additional orientation sign. These two contributions consequently have the same sign. Choosing the orientation of $b$ gives $\partial q=2b$.

One can check the coefficient directly from the cell attachment of the top cell: its boundary word is $ab{a}^{-1}b$, with exponent sums zero on the base loop $a$ and two on the fiber loop $b$. Thus the integral Morse complex of the Klein bottle isandCounting trajectories without signs, or working only modulo two, would miss the order-two class.

Work over $\mathbb{Q}$. A closed connected orientable surface of genus $g$ has Betti numbers $(1,2g,1)$; the circle has Betti numbers $(1,1)$. The Künneth theorem therefore givesWe use the following precise consequence of the Morse handle-attachment theorem: a closed manifold with a Morse function has the homotopy type of a finite CW complex with one index-$j$ cell for each index-$j$ critical point. Its degree-$j$ rational cellular chain group has dimension $c_j$, so its homology has dimension at most $c_j$. Consequently $c_j\ge b_j$, the weak Morse inequalities. Summing yieldsThis counts every critical point, regardless of whether some critical values coincide.

To attain the bound, construct a surface Morse function $h:{\Sigma }_g\to \mathbb{R}$ from the standard orientable handle decomposition: start with a disk, attach $2g$ orientable bands in the usual paired pattern forming the genus-$g$ surface with one boundary component, and cap that boundary with a disk. Give the first disk a minimum, each band one saddle with local form $-x^2+y^2$, and the final cap a maximum. Regular collar coordinates join successive levels without additional critical points. This constructs $h$ with counts $(1,2g,1)$.

On the product takeIts critical points are exactly the pairs of a critical point of $h$ with $\theta =0,\pi$. The Hessian matrix is block diagonal, with surface block $\mathrm{Hess}h$ and circle entry $-\delta \cos \theta$. It is nonsingular and its Morse index is the sum of the two factor indices. Thus this product of Morse functions has countsand exactly $4g+4$ critical points. If distinct critical values are desired, choose $\delta$ outside the finite set producing collisions after first choosing distinct critical values for $h$. The minimal Morse critical count on a circle times an orientable surface is therefore attained for every $g\ge 0$.

Here projection means Hermitian orthogonal projection: $P=P^{\ast }=P^2$ with $\mathrm{tr}P=k$, as indicated by the unitary conjugation in the question. These matrices represent the complex Grassmannian as orthogonal projections, a compact manifold of complex dimension $k(n-k)$. The constants $c_i$ are real. Let $A^{\ast }=-A$ and use the given tangent directions $[A,P]$. Cyclicity of the matrix trace givesThe commutator $B=[P,C]$ is skew-Hermitian. If this derivative vanishes for every skew-Hermitian $A$, choose $A=B$ to obtainso $B=0$. Conversely $[P,C]=0$ makes every tangent derivative zero. Because the diagonal entries of $C$ are distinct, a commuting matrix $P$ is diagonal. Its idempotent entries are zero or one. Thus the critical projections are exactlywith $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ critical points and values $f(P_I)={\sum }_{i\in I}{c}_i$.

To compute the Hessian matrix, write $E=\mathrm{span}\{e_i:i\in I\}$ and $E^{\perp }=\mathrm{span}\{e_j:j\notin I\}$. A neighboring plane is the graph of a complex linear map $Z:E\to E^{\perp }$. With $W=\left(\genfrac{}{}{0ex}{}{I}{Z}\right)$, its orthogonal projection is $P(Z)=W(I+Z^{\ast }Z{)}^{-1}{W}^{\ast }$. Writing $C_E,C_{E^{\perp }}$ for the diagonal blocks, we getExpanding $(I+Z^{\ast }Z{)}^{-1}=I-Z^{\ast }Z+O(\Vert Z{\Vert }^4)$ yieldsEach complex coordinate has two real components, each with Hessian coefficient $2(c_j-c_i)$. Every coefficient is nonzero, proving that $f$ is a Morse function. Its negative directions correspond to $j<i$. For $I=\{i_1<\cdots <i_k\}$ there are $i_a-a$ complementary indices below $i_a$, givingThis is the diagonal trace Morse function on a complex Grassmannian. Its critical values need not be distinct; ties between sums do not make its critical points degenerate.

All Morse indices are even. Choose a Morse-Smale gradient flow and form its integral Morse chain complex. The odd-degree chain groups vanish, so every differential vanishes. Equivalently the Morse handle-attachment theorem supplies only even-dimensional cells. The integral homology of complex Grassmannians is thereforewhere $N_r$ counts the subsets with ${\sum }_a(i_a-a)=r$, and $0\le r\le k(n-k)$. In particular there is no torsion. The nondecreasing sequence $i_a-a$ lies between zero and $n-k$, so reversing it identifies these subsets with partitions of an integer $r$ fitting in a $k$ by $(n-k)$ rectangle. The Poincaré polynomial is explicitlyFor the permitted example $\mathrm{Gr}(2,4)$, the subsets $12,13,14,23,24,34$ have indices $0,2,4,4,6,8$. ThusThe general calculation above applies throughout $2\le k\le n-2$; it does not reduce to any of the excluded rank-one cases.

Use tangent vectors, rather than the affine tangent plane, to differentiate. Every skew-symmetric matrix $A$ gives a curve $X\exp (tA)$ in the special orthogonal group. Its first derivative under $f$ is $\mathrm{tr}(CXA)$. The trace pairing annihilates every skew-symmetric matrix precisely when $CX$ is symmetric: choosing $A$ with only entries $a_{ij}=-a_{ji}$ tests each off-diagonal antisymmetric entry. Thus criticality is equivalent to $Y=CX$ being symmetric.

Since $X$ is orthogonal, $Y{Y}^T=C^2$, so at a critical point $Y^2=C^2$. Consequently $Y$ commutes with $C^2$. The diagonal entries $c_i^2$ are distinct, and the equation $Y{C}^2=C^{2}Y$ gives $(c_j^2-c_i^2)y_{ij}=0$ for $i\ne j$. Hence $Y$ and $X$ are diagonal. Orthogonality and the determinant constraint give exactlyConversely each such $D$ makes $CD$ symmetric, so this list is complete.

To calculate the Hessian matrix, use local coordinates $X=\exp (A)D$ with $A^T=-A$; the matrix exponential is a local coordinate map near zero. Its quadratic term givesbecause $(A^2{)}_{ii}=-{\sum }_{j\ne i}{a}_{ij}^2$. There are no mixed quadratic terms. For $i<j$, $c_j>c_i>0$, so $c_i{\epsilon }_i+c_j{\epsilon }_j$ has the sign of ${\epsilon }_j$ and is never zero. Every listed critical point is therefore nondegenerate, with Morse indexThis is the trace height on the special orthogonal group calculation; the rotations in the hint select the individual coordinates $a_{ij}$.

For $n=3$, the signs $(+--),(-+-),(--+),(+++)$ have indices $0,1,2,3$ respectively. A Morse function on a compact closed manifold has Euler characteristic equal to the alternating number of critical points, by its Morse handle-attachment theorem and resulting finite CW complex. Hence

The smooth h-cobordism theorem states: if $(W;M_0,M_1)$ is a compact smooth h-cobordism, $\dim W\ge 6$, and its connected boundary manifolds are simply connected, then there is a diffeomorphism $W\cong M_0\times [0,1]$ restricting to the prescribed identification on $M_0$. In particular $M_0$ and $M_1$ are diffeomorphic. Being an h-cobordism means that both inclusions $M_i\hookrightarrow W$ are homotopy equivalences; their fundamental groups then also identify with that of $W$.

The homotopy-equivalence condition is necessary for a trivial cobordism: each end of a cylinder is a deformation retract, by $(x,t)\mapsto (x,(1-s)t)$ for the incoming end and $(x,t)\mapsto (x,(1-s)t+s)$ for the outgoing end. Its relative homology with respect to either end is consequently zero. Compactness is also necessary for a product with a closed compact end. The dimensional bound and simple connectivity are not necessary for an individual cobordism to be a product. The cylinder $S^2\times [0,1]$ is a simply connected trivial cobordism of dimension three; and, in the theorem's dimension range, $(S^1\times S^4)\times [0,1]$ is a six-dimensional trivial cobordism with fundamental group $\mathbb{Z}$. These prove the two non-necessity assertions explicitly, while the inclusion argument proves necessity of the h-condition.

For the remaining claims, use a cobordism Morse function: its two regular boundary levels are zero and one, and all critical points are interior. An arbitrary function with unrelated boundary values would not have the relative handle-count property. The relative Morse chain complex for $(W,M_0)$ has one generator per critical point, in its Morse index. ThusA trivial cobordism has zero relative homology, hence zero relative Euler characteristic. Modulo two the displayed sum is the number of critical points. An odd number makes it odd, and therefore nonzero. This proves an odd-critical-point cobordism is not trivial.

With precisely two critical points of indices $\lambda \le \mu$, a trivial cobordism requires the two-generator relative complex to be acyclic over $\mathbb{Z}$. Equal indices leave a rank-two homology group; a gap of more than one leaves both generators with zero possible differential. Therefore $\mu =\lambda +1$ is necessary. The remaining two-term complex isChoose a transverse gradient-like vector field. A nonzero differential requires the index-$\mu$ point to be above the index-$\lambda$ point; otherwise no descending connecting trajectories exist. In a common regular level between them, the belt sphere $S_U(p)$ has dimension $d-\lambda -1$, and the attaching sphere $S_L(q)$ has dimension $\mu -1=\lambda$, where $d=\dim W$. Their dimensions add to $d-1$, the level's dimension, so their transverse intersection consists of points. Each intersection is one descending trajectory from $q$ to $p$, modulo time translation. Orient the attaching sphere and use an orientation of a vector space on the handle core to coorient the belt sphere; the resulting signed count is the cellular boundary coefficientThe two homology groups are the kernel and cokernel of multiplication by $a$. They vanish precisely when $a$ is a unit of $\mathbb{Z}$. Consequentlyare necessary. This proves the two-critical-point obstruction to a trivial cobordism; it makes an algebraic incidence statement and does not assume that an algebraic count of one automatically supplies geometric handle cancellation.

For an explicit nontrivial cobordism from $S^3$ to itself, attach two zero-framed two-handles to the outgoing end of $S^3\times [0,1]$ along the standard Hopf link. Its two components are unknots with linking number one. This realizes two successive $(2,2)$ surgeries, as suggested by the hint. The outgoing boundary is again $S^3$, and this can be checked geometrically: the Hopf-link exterior is $T^2\times [0,1]$, and each component's zero-framing longitude is the other's meridian. Filling along these two slopes glues two solid tori with meridians intersecting once, the standard genus-one Heegaard decomposition of $S^3$. Thus zero surgery on the Hopf link does return the three-sphere, not just an unspecified homology sphere.

The trace has only the two added index-two handles. Its relative cellular complex has $C_2(W,S^3)={\mathbb{Z}}^2$ and all other groups zero, soIt cannot be a product cylinder. Equivalently this trace is $S^2\times S^2$ with two disjoint open four-balls removed: the usual handle presentation of $S^2\times S^2$ has one zero-handle, the two zero-framed Hopf-linked two-handles, and one four-handle. Removing the bottom and top balls exposes the two three-sphere ends. The relative-homology calculation alone already proves the requested cobordism is nontrivial.

Along the downward gradient flow, the gradient-flow dissipation identity, the chain rule and the defining property of the Riemannian gradient giveA smooth vector field on a compact manifold is complete, so the trajectory exists for every $t\ge 0$. In particular, its function value never increases.

If the trajectory is still in $C_{[a,b]}$ at time $t$, its monotone function value has stayed in $[a,b]$ throughout $[0,t]$. ConsequentlyThus a uniform exit bound isThe trajectory is outside the closed band for every $t>T$. Once it leaves below level $a$, monotonicity prevents its return. The argument also covers the zero-width band $a=b$.

For the total occupation-time bound for a gradient flow, let $m={\min }_{M}f$ and $F={\max }_{M}f$, which exist by compactness. For every $R>0$, the same gradient flow identity givesTaking $R\to \infty$, by monotone convergence theorem, gives the total occupation-time boundThis allows arbitrarily many visits to $C_{\epsilon }$; it bounds their total duration. Neither of these two estimates needs the Morse function hypothesis.

For exponential convergence of a Morse gradient flow, first observe that a limiting point $z=x_{\infty }$ is a critical point. Indeed, writing ${\varphi }_s$ for the flow and using continuous dependence on initial data,Differentiation at $s=0$ gives $\nabla f(z)=0$.

We use the Morse lemma: near a nondegenerate critical point of Morse index $\lambda$, there are coordinates $(u,v)\in {\mathbb{R}}^{\lambda }\times {\mathbb{R}}^{n-\lambda }$ in whichChoose a Riemannian metric equal to the Euclidean metric on a smaller ball in these coordinates. Such a global Riemannian metric exists by a partition of unity: patch this metric to any background metric outside a slightly larger ball. One can do this simultaneously at every critical point, since a Morse function on a compact manifold has only finitely many critical points. The metric is chosen before forming its gradient flow; changing the metric can change trajectories.

For this metric, any trajectory converging to $z$ eventually remains in the smaller coordinate ball. From some time $t_0$ onward its equations are exactlyConvergence forces $u(t_0)=0$, and thenThe straight radial segment stays in the coordinate ball, so its length bounds the Riemannian distance from above:Increase $C$ beyond both $e^{2t_0}|v(t_0)|$ and ${\max }_{0\le t\le t_0}{e}^{2t}d({\gamma }_x(t),z)$. Then the requested strict estimate holds for all $t\ge 0$:This proof uses the Morse lemma, existence and uniqueness for smooth ordinary differential equations, continuous dependence on initial data, and the existence of a partition of unity. The expanding coordinates explain why convergence to a saddle occurs only along its stable manifold.

For a counterexample, use an arc coordinate $x$ around a point of the circle, its flat local metric, and a globally smooth function that equals $x^4$ on this arc. Such a function can be made by a smooth cutoff, agreeing with a positive constant outside a larger arc. For $x_0>0$ sufficiently small the downward gradient flow remains in the arc and isIt converges to the degenerate critical point $0$, and its Riemannian distance from $0$ is $x(t)$ for sufficiently large $t$. For every $k>0$, $e^{kt}x(t)\to \infty$, so no exponential bound is possible. This counterexample is on a compact manifold as well.

The critical points of $F$ are exactly pairs of critical points of its factors. Its Hessian matrix is their block direct sum, so their Morse indices add and nondegeneracy is preserved. The critical-count polynomial is the product of those of the factors. This construction also gives products of arbitrary smooth functions, with the critical-point count still multiplying even when some Hessians are degenerate.

For positive distinct diagonal entries $c_1<\cdots <c_n$, the critical points of this function are diagonal sign matrices $D=\mathrm{diag}({\epsilon }_1,\dots ,{\epsilon }_n)$ with $\prod {\epsilon }_i=1$. Criticality forces $CX$ symmetric; its square is $C^2$, so it commutes with the distinct-spectrum diagonal matrix $C^2$ and is diagonal. In the skew coordinates $a_{ij}$ its Hessian is $-{\sum }_{i<j}(c_i{\epsilon }_i+c_j{\epsilon }_j)a_{ij}^2$. The sign of each parenthesis is that of ${\epsilon }_j$, giving Morse index ${\sum }_{j:{\epsilon }_j=1}(j-1)$.