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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.

An index-$\lambda$ handle in dimension $d$ is this product of disks, attached along $S^{\lambda -1}\times D^{d-\lambda }$. Its core disk is $D^{\lambda }\times \{0\}$, and its belt sphere is $\{0\}\times S^{d-\lambda -1}$. The complementary boundary piece is $D^{\lambda }\times S^{d-\lambda -1}$. Crossing a critical point in the Morse handle-attachment theorem produces precisely this local change.

The free chain group in degree $i$ has one oriented generator for each index-$i$ critical point. A cellular chain complex obtained from the Morse handle-attachment theorem computes the integral homology. Under the Morse-Smale gradient flow condition its differential can instead be obtained by signed counts of connecting trajectories. Over a field, if $r_i$ is the rank of ${\partial }_i$, then $c_i=b_i+r_i+r_{i+1}$. This identity implies both weak and strong Morse inequalities.

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 usual compact closed-manifold hypothesis is needed for the first assertion, or a proper bounded-below Morse exhaustion in an appropriate noncompact version. As literally printed without any such condition, $f(x)=x$ on $\mathbb{R}$ is a Morse function with no critical points, while $b_0=1$. We prove the intended closed-manifold statement, which applies to the surface product later in the question.

By the Morse handle-attachment theorem, crossing an index-$i$ critical point attaches an $i$-handle, and no topological change occurs between critical levels. Retraction to the core disks of handles gives a CW complex model with $c_i$ cells in dimension $i$. Its cellular chain complex has $C_i\cong {\mathbb{Z}}^{c_i}$. Tensor with $\mathbb{Q}$, which computes the ranks $b_i$ of integral homology. Since homology is a quotient of a vector subspace of $C_i\otimes \mathbb{Q}$, $b_i\le c_i$. This argument has not used the Morse inequalities.

The strong Morse inequalities and top-dimensional equality areTheir weak form is $c_i\ge b_i$. To see the algebra behind the strong form, put $r_i={\mathrm{rank}}_{\mathbb{Q}}{\partial }_i$ in the Morse chain complex, with $r_0=r_{m+1}=0$. The kernel-image dimensions give $c_i=b_i+r_i+r_{i+1}$, so the alternating difference at $j$ is exactly $r_{j+1}\ge 0$. Conversely, given the strong inequalities, denote those alternating differences by $S_j$, with $S_{-1}=0$. Then $c_j-b_j=S_j+S_{j-1}\ge 0$, proving that they imply every weak inequality.

They are not equivalent as numerical restrictions. For the Betti numbers $b=(1,0,0,1)$ of $S^3$, the formal counts $c=(2,0,0,2)$ satisfy all weak inequalities and the Euler equality, but at $j=1$ give $c_1-c_0=-2<-1=b_1-b_0$. They therefore cannot be the counts of a Morse function on $S^3$; the strong inequalities detect that additional obstruction.

If all the strong inequalities are equalities, all $r_i$ vanish and $c_i=b_i$. An integer boundary matrix with rational rank zero is the zero matrix. Thus the integral complex itself has zero differential, not merely its rationalization. Consequentlyand higher homology vanishes. In particular there is no homology torsion. This is an integral consequence of a Perfect Morse function, rather than just an assertion about ranks.

For ${\Sigma }_g$, the integral homology is free with Betti numbers $(1,2g,1)$. The Künneth theorem therefore has no torsion terms, and the product's Poincaré polynomial isSumming the weak inequalities yields the lower boundChoose a surface Morse function $h$ with one minimum, $2g$ saddles and one maximum. It can be built from one closed disk, the standard $2g$ one-handles, and a closing closed disk, using the local forms $u^2+v^2$, $-u^2+v^2$, and $-u^2-v^2$ at the handle centres. Regular collars interpolate between the handles without new critical points. Now set $F(x,y)=h(x)+h(y)$. Its critical points are the ordered pairs of those of $h$, and its Hessian matrix is a block direct sum; the indices add. This product of Morse functions has exactly $(2g+2{)}^2$ critical points and realizes every displayed Betti count. Small bumps constant near the critical pairs can separate their values without altering the Hessians or creating new critical points.

The same minimum does not apply to arbitrary smooth functions. Here is an explicit finite-critical-point counterexample, avoiding any assumption that degeneracy leaves the Morse lower bound intact. On $T^2=(\mathbb{R}/\pi \mathbb{Z}{)}^2$, takeIt is well defined because shifting either coordinate by $\pi$ reverses two of its factors. Its derivatives are $\sin v\sin (2u-v)$ and $\sin u\sin (u-2v)$. If either $\sin u$ or $\sin v$ vanishes, both derivatives vanish only at $(0,0)$. Otherwise $2u-v$ and $u-2v$ must be multiples of $\pi$, giving exactly $(\pi /3,2\pi /3)$ and $(2\pi /3,\pi /3)$ in addition to the origin. At the first point the Hessian is $\sqrt{3}\left(\begin{array}{cc}1& -1/2\\ -1/2& 1\end{array}\right)$, positive definite; at the second it is its negative. They are a minimum and maximum. The origin is an isolated degenerate monkey saddle, with leading term $uv(u-v)$. Thus this three-critical-point function on a torus has exactly three critical points.

For $g>1$, perform connected-sum gluing of functions between this torus and a perfect Morse surface of genus $g-1$: delete a small neighbourhood of the torus maximum and of the other surface's minimum, and join their regular boundary circles by a monotone cylinder. Collar coordinates ensure nonzero differential throughout the neck, so precisely those two critical points are lost and none are added. The result on ${\Sigma }_g$ has $3+[2(g-1)+2]-2=2g+1$ critical points, retaining the degenerate point. For $g=1$, simply use $h_0$. Calling this smooth function $\tilde{h}$ and the earlier perfect Morse function $h$, the smooth product $\tilde{h}(x)+h(y)$ hascritical points. This explicitly proves the negative answer to the unrestricted smooth-minimum question; it does not assert that this smaller illustrative count is sharp.