03.15.E1 · modern-geometry / morse-homology

Morse homology exercise pack (Schwarz Morse Homology supplement)

shippedIntermediate-onlyLean: nonepending prereqs

Anchor (Master):

Formal definition of the pack Intermediate

Schwarz's Morse Homology builds the homology of a closed manifold $M^{n}$ out of a Morse function $f$ and a Morse-Smale gradient flow. The data are the critical points, graded by the Morse index $μ (x) = dim W^{u} (x)$ ; the trajectory moduli $M (x, y)$ of unparametrised negative-gradient flow lines, of dimension $μ (x) - μ (y) - 1$ under transversality; coherent orientations supplying a sign $n (x, y) \in Z$ to each isolated flow line; and the boundary operator $\partial x = \sum_{μ (y) = μ (x) - 1} n (x, y) y$ . The central facts are $\partial^{2} = 0$ and the Morse Homology Theorem, $H M_{*} (f, g) ≅ H_{*} (M; Z)$ .

This pack collects ten problems — three easy, four medium, three hard — each with a hint and a full solution. The exercises test the layers in order: reading off indices from a Morse function, the Fredholm-index dimension count $μ (x) - μ (y)$ , the $\partial^{2} = 0$ cancellation, the Morse inequalities, and the explicit computation of $H M_{*}$ for $S^{n}$ , $T^{2}$ , and $C P^{n}$ with standard Morse functions. It is read alongside its prerequisite units rather than as a standalone development.

Conventions follow Schwarz: $f$ is Morse on a closed Riemannian $(M^{n}, g)$ ; the flow is $\overset{γ}{˙} = - \nabla f (γ)$ , so $f$ decreases along trajectories; the unstable manifold $W^{u} (x)$ has dimension equal to the index $μ (x)$ ; and $M (x, y)$ runs from the higher-index $x$ to the lower-index $y$ , with $dim M (x, y) = μ (x) - μ (y)$ and $dim M (x, y) = μ (x) - μ (y) - 1$ .

Key theorem with full solution Intermediate

We work one computation in full as an exemplar of the format. The remaining problems use the same problem/hint/answer structure.

Lead exercise. Compute the Morse homology of $S^{2}$ from the standard height function, and identify the boundary operator.

Solution. Embed $S^{2} \subset R^{3}$ as the unit sphere and let $f$ be the height $f (x, y, z) = z$ . The critical points are the north pole $N$ (a maximum, index $2$ ) and the south pole $S$ (a minimum, index $0$ ): the Hessian of $z$ in stereographic-style local coordinates is negative-definite at $N$ and positive-definite at $S$ , with no other critical points. So the Morse complex has generators in degrees $0$ and $2$ only:

C_{2} = Z ⟨ N ⟩, C_{1} = 0, C_{0} = Z ⟨ S ⟩ .

The boundary operator $\partial : C_{2} \to C_{1}$ and $\partial : C_{1} \to C_{0}$ both have zero domain or codomain, so $\partial = 0$ identically. There is no index- $1$ generator to receive $\partial N$ or to map onto $S$ . Hence

H M_{k} (S^{2}) = ⎩ ⎨ ⎧ Z 0 Z k = 0, k = 1, k = 2,

matching $H_{*} (S^{2}; Z)$ , as the Morse Homology Theorem requires.

A subtlety worth noting: this height function is not Morse-Smale-generic in the strong sense relevant to $S^{2}$ only because there are no index- $1$ critical points to break the count — but it is a legitimate Morse function and the complex is forced by the gradings. To see the boundary operator do real work one needs an index- $1$ generator, which is exactly what the torus computation (Exercise 8) supplies, where two flow lines from a saddle to the minimum cancel mod sign. $□$

This is the template for every standard computation: list critical points, read off indices, and the differential is pinned down by the index gradings together with the signed flow-line counts. When consecutive indices are present, the $\pm 1$ counts of $M (x, y)$ determine $\partial$ .

Exercises Intermediate

Exercise 1 (easy, numeric). Index of a quadratic critical point.

Let $f (x_{1}, \dots, x_{n}) = - x_{1}^{2} - \dots - x_{k}^{2} + x_{k + 1}^{2} + \dots + x_{n}^{2}$ near the origin. Verify the origin is a non-degenerate critical point and compute its Morse index.

Hint

The Hessian is diagonal. The Morse index is the number of negative eigenvalues.

Answer

The gradient is $\nabla f = (- 2 x_{1}, \dots, - 2 x_{k}, 2 x_{k + 1}, \dots, 2 x_{n})$ , which vanishes only at the origin, so the origin is the unique critical point near $0$ . The Hessian is the constant diagonal matrix $diag (- 2, \dots, - 2, 2, \dots, 2)$ with $k$ entries $- 2$ and $n - k$ entries $+ 2$ . It is non-singular, so the critical point is non-degenerate.

The Morse index is the number of negative eigenvalues of the Hessian, which is $k$ . This is the content of the Morse lemma: every non-degenerate critical point of index $k$ looks, in suitable coordinates, exactly like this normal form, and $dim W^{u} = k$ .

Exercise 2 (easy, numeric). Dimension of a trajectory moduli space.

Let $x, y$ be critical points of a Morse-Smale pair with $μ (x) = 4$ and $μ (y) = 1$ . State the dimensions of $M (x, y)$ and of the unparametrised quotient $M (x, y)$ .

Hint

The Fredholm index of the flow section is $μ (x) - μ (y)$ , and quotienting by the $R$ -action of time-translation drops dimension by one.

Answer

Under the Morse-Smale transversality condition, $M (x, y) = W^{u} (x) \cap W^{s} (y)$ is a smooth manifold of dimension equal to the Fredholm index of the linearised flow section,

dim M (x, y) = μ (x) - μ (y) = 4 - 1 = 3.

The flow carries a free $R$ -action by time-reparametrisation (for $x \neq = y$ ), and the unparametrised moduli space is the quotient:

dim M (x, y) = μ (x) - μ (y) - 1 = 2.

The boundary operator counts the zero-dimensional $M$ , i.e. index-difference exactly $1$ ; the $\partial^{2} = 0$ identity reads off the one-dimensional $M$ , index-difference $2$ (compare Exercise 5).

Exercise 3 (easy, numeric). Critical points of the height function on $S^{n}$ .

For the unit sphere $S^{n} \subset R^{n + 1}$ with height function $f (x) = x_{n + 1}$ , list the critical points and their indices, and write down the Morse complex.

Hint

There are exactly two critical points; one is a maximum and one a minimum.

Answer

The height function $f (x) = x_{n + 1}$ on $S^{n}$ has critical points where the tangent plane is horizontal: the north pole $N = (0, \dots, 0, 1)$ and the south pole $S = (0, \dots, 0, - 1)$ . At $N$ the function is maximal and the Hessian is negative-definite, so $μ (N) = n$ ; at $S$ it is minimal, so $μ (S) = 0$ .

The Morse complex has one generator in degree $n$ and one in degree $0$ :

C_{n} = Z ⟨ N ⟩, C_{0} = Z ⟨ S ⟩, C_{k} = 0 (k \neq = 0, n) .

For $n \geq 2$ there is no consecutive index pair, so $\partial = 0$ and $H M_{k} (S^{n}) = Z$ for $k \in {0, n}$ , zero otherwise — the homology of $S^{n}$ .

Exercise 4 (medium, proof). Negative gradient decreases $f$ .

Show that along any non-constant negative-gradient trajectory $\overset{γ}{˙} = - \nabla f (γ)$ , the value $f (γ (t))$ is strictly decreasing, and that the limits $lim_{t \to \pm \infty} γ (t)$ are critical points.

Hint

Differentiate $f \circ γ$ . For the limits, use that $f$ is bounded on the compact $M$ and that $∣\nabla f ∣^{2}$ is integrable along the trajectory.

Answer

Differentiate: $\frac{d}{d t} f (γ (t)) = ⟨ \nabla f (γ), \overset{γ}{˙} ⟩ = ⟨ \nabla f, - \nabla f ⟩ = - ∣\nabla f (γ) ∣^{2} \leq 0$ . The derivative is strictly negative wherever $\nabla f \neq = 0$ , so $f \circ γ$ is strictly decreasing on any interval avoiding critical points; if $γ$ ever met a critical point it would be constant there by uniqueness of solutions, so a non-constant trajectory avoids critical points and $f \circ γ$ is strictly decreasing.

For the limits: $M$ is compact so $f$ is bounded below, hence $\int_{- \infty}^{\infty} ∣\nabla f (γ) ∣^{2} d t = f (γ (- \infty)) - f (γ (+ \infty)) < \infty$ . By compactness $γ (t)$ has limit points as $t \to + \infty$ ; any such limit point $p$ has $\nabla f (p) = 0$ (otherwise $f$ would keep decreasing past $f (p)$ ). Since the critical points are isolated (non-degeneracy) and $f$ is monotone, the limit set is a single critical point. The same argument with $t \to - \infty$ gives a critical point of higher $f$ -value. So every trajectory runs from a critical point to a lower one.

Exercise 5 (medium, proof). Why $\partial^{2} = 0$ .

Explain, from the structure of the compactified one-dimensional moduli space, why the Morse boundary operator satisfies $\partial^{2} = 0$ .

Hint

Fix $x, z$ with $μ (x) - μ (z) = 2$ . The coefficient of $z$ in $\partial^{2} x$ counts boundary points of the compact 1-manifold $\overline{M} (x, z)$ .

Answer

Fix critical points $x, z$ with $μ (x) - μ (z) = 2$ . The coefficient of $z$ in $\partial^{2} x = \partial (\partial x)$ is

μ (y) = μ (x) - 1 \sum n (x, y) n (y, z),

a signed count of broken trajectories $x \to y \to z$ passing through an intermediate critical point $y$ of index $μ (x) - 1$ .

The unparametrised moduli $M (x, z)$ has dimension $μ (x) - μ (z) - 1 = 1$ . Its compactification $\overline{M} (x, z)$ is a compact 1-manifold with boundary (compactness theorem: sequences subconverge to broken trajectories; gluing theorem: each once-broken trajectory $x \to y \to z$ is the limit of a unique one-parameter family, so it is a genuine boundary endpoint). Therefore $\partial \overline{M} (x, z)$ is exactly the set of once-broken trajectories.

A compact 1-manifold with boundary has an even number of boundary points, and the coherent orientations make the two endpoints of each arc carry opposite signs. Summing the signs of all boundary points gives $0$ . But that signed sum is precisely $\sum_{y} n (x, y) n (y, z)$ . Hence the coefficient of $z$ in $\partial^{2} x$ vanishes for every $z$ , so $\partial^{2} = 0$ . The coherence of orientations is what makes the cancellation occur over $Z$ rather than only mod $2$ .

Exercise 6 (medium, numeric). Morse inequalities for a genus- $g$ surface.

Let $Σ_{g}$ be the closed orientable surface of genus $g$ . State the weak and strong Morse inequalities, and find the minimum possible number of critical points of a Morse function on $Σ_{g}$ .

Hint

The Betti numbers are $b_{0} = 1$ , $b_{1} = 2 g$ , $b_{2} = 1$ . The weak inequality is $c_{k} \geq b_{k}$ ; the alternating sum gives the Euler characteristic.

Answer

Let $c_{k}$ be the number of index- $k$ critical points and $b_{k} = dim H_{k} (Σ_{g}; Q)$ the Betti numbers: $b_{0} = 1$ , $b_{1} = 2 g$ , $b_{2} = 1$ .

Weak Morse inequalities: $c_{k} \geq b_{k}$ for each $k$ . So $c_{0} \geq 1$ , $c_{1} \geq 2 g$ , $c_{2} \geq 1$ .

Strong Morse inequalities: $c_{k} - c_{k - 1} + \dots \pm c_{0} \geq b_{k} - b_{k - 1} + \dots \pm b_{0}$ for each $k$ , with equality (Euler-Poincaré) at the top: $\sum_{k} (- 1)^{k} c_{k} = \sum_{k} (- 1)^{k} b_{k} = χ (Σ_{g}) = 2 - 2 g$ .

Minimum total count: the weak inequalities force $c_{0} + c_{1} + c_{2} \geq 1 + 2 g + 1 = 2 g + 2$ . This bound is achieved by the standard "upright torus / pretzel" height function with one minimum, $2 g$ saddles, and one maximum. So a Morse function on $Σ_{g}$ has at least $2 g + 2$ critical points, with $2 g + 2$ attainable.

For $g = 1$ (the torus) this gives the minimum $4$ : one minimum, two saddles, one maximum — recovered explicitly in Exercise 8.

Exercise 7 (medium, numeric). Morse homology of $C P^{n}$ .

The Morse function $f ([z_{0} : \dots : z_{n}]) = \frac{\sum _{j} j ∣ z _{j} ∣ ^{2}}{\sum _{j} ∣ z _{j} ∣ ^{2}}$ on $C P^{n}$ has $n + 1$ critical points, the coordinate points $p_{j} = [0 : \dots : 1 : \dots : 0]$ , with $μ (p_{j}) = 2 j$ . Compute $H M_{*} (C P^{n})$ .

Hint

All critical points have even index, so no two indices are consecutive. What does that force on $\partial$ ?

Answer

The generators sit in degrees $0, 2, 4, \dots, 2 n$ , one in each even degree. Because every index is even, there is no pair of critical points whose indices differ by exactly $1$ . The boundary operator $\partial$ lowers degree by $1$ , so it must map each generator into a zero group: $\partial = 0$ identically.

Therefore

H M_{k} (C P^{n}) = {Z 0 k = 0, 2, 4, \dots, 2 n, k odd,

which is exactly $H_{*} (C P^{n}; Z)$ . A Morse function all of whose indices have the same parity is called a perfect Morse function: the Morse inequalities are equalities, $c_{k} = b_{k}$ , and the complex computes homology with no cancellation. The complex projective spaces are the model case.

Exercise 8 (hard, proof). The boundary operator for the torus height function.

Embed $T^{2}$ upright in $R^{3}$ (the inner tube vertical). The height function has one minimum $m$ (index $0$ ), two saddles $s_{1}, s_{2}$ (index $1$ ), and one maximum $M$ (index $2$ ). Compute $\partial$ on each generator and verify $H M_{*} (T^{2}) = H_{*} (T^{2}; Z)$ .

Hint

From the maximum there are two flow lines to each saddle; from each saddle two flow lines to the minimum. With coherent orientations the two lines in each pair cancel.

Answer

Generators: $C_{0} = Z ⟨ m ⟩$ , $C_{1} = Z ⟨ s_{1} ⟩ \oplus Z ⟨ s_{2} ⟩$ , $C_{2} = Z ⟨ M ⟩$ .

$\partial M$ . From the maximum $M$ (index $2$ ) to a saddle $s_{i}$ (index $1$ ), $M (M, s_{i})$ is zero-dimensional. Geometrically there are two gradient lines descending from $M$ to each saddle (one down each side of the tube). With coherent orientations the two lines carry opposite signs, so $n (M, s_{1}) = + 1 - 1 = 0$ and likewise for $s_{2}$ . Hence $\partial M = 0$ .

$\partial s_{i}$ . From a saddle $s_{i}$ to the minimum $m$ , there are again two descending gradient lines, oppositely oriented, so $n (s_{i}, m) = 0$ . Hence $\partial s_{1} = \partial s_{2} = 0$ .

So $\partial = 0$ on the whole complex. Then

H M_{0} = Z, H M_{1} = Z^{2}, H M_{2} = Z,

matching $H_{*} (T^{2}; Z)$ . The check on $\partial^{2} = 0$ is automatic here, but the real content is the signed cancellation: over $Z /2$ each $n (M, s_{i})$ would be $0$ too (two lines, $1 + 1 = 0 mod 2$ ), so the mod-2 complex also has zero differential and gives $H_{*} (T^{2}; Z /2) = (Z /2, (Z /2)^{2}, Z /2)$ . The coherent orientations promote this to the integral statement.

Exercise 9 (hard, proof). $R P^{2}$ over $Z$ versus $Z /2$ .

A Morse function on $R P^{2}$ has one critical point of each index $0, 1, 2$ . Compute the integral and mod-2 Morse boundary operators and the resulting homologies, explaining the discrepancy.

Hint

The single index-1-to-index-0 moduli space has two points (a double cover phenomenon); over $Z$ they are oppositely oriented, over $Z /2$ they add. The index-2-to-index-1 count is $0$ .

Answer

Generators $a$ (index $0$ ), $b$ (index $1$ ), $c$ (index $2$ ); minimal Morse function on $R P^{2}$ .

The index-1 saddle $b$ connects to the minimum $a$ by two gradient lines: $R P^{2}$ is the quotient of $S^{2}$ by the antipodal map, and the single descending separatrix of $b$ lifts to a pair of arcs whose two ends both hit $a$ . With coherent orientations these two lines carry opposite signs over $Z$ , so $n (b, a) = + 1 - 1 = 0$ ; over $Z /2$ they both count $+ 1$ , so the mod-2 coefficient is $1 + 1 = 0$ . The index-2-to-index-1 connection $c \to b$ : two lines, integrally cancelling, mod-2 also $0$ ? Here the key asymmetry is that $\partial c$ over $Z$ is $\pm 2 b$ while over $Z /2$ it is $0$ .

Concretely the integral complex is $Z \cdot 2 Z 0 Z$ (degrees $2 \to 1 \to 0$ ). Then

H M_{0} = Z, H M_{1} = Z /2, H M_{2} = ker (\cdot 2) = 0,

which is $H_{*} (R P^{2}; Z) = (Z, Z /2, 0)$ — the torsion class appears as the cokernel of multiplication by $2$ .

Over $Z /2$ the differential is identically $0$ (multiplication by $2$ is the zero map mod $2$ ), giving $H_{*} (R P^{2}; Z /2) = (Z /2, Z /2, Z /2)$ . The discrepancy — three $Z /2$ 's versus $Z \oplus Z /2$ — is exactly the universal-coefficient contribution of the $Z /2$ -torsion, and it is the cleanest illustration of why coherent orientations matter: only the signed, integral count detects the $2$ -torsion.

Exercise 10 (hard, proof). Invariance under change of Morse function.

Sketch why $H M_{*} (f_{0}, g_{0}) ≅ H M_{*} (f_{1}, g_{1})$ for any two Morse-Smale pairs, so that Morse homology is an invariant of $M$ alone.

Hint

Interpolate by a homotopy $(f_{s}, g_{s})$ and count trajectories of the non-autonomous flow to build a continuation chain map. Two such maps are chain-homotopic.

Answer

Choose a smooth homotopy $(f_{s}, g_{s})_{s \in [0, 1]}$ between the two pairs (generic, so the interpolating data are admissible). The continuation map $Φ : C_{*} (f_{0}) \to C_{*} (f_{1})$ counts trajectories of the non-autonomous gradient flow $\overset{γ}{˙} = - \nabla_{g_{s}} f_{s} (γ)$ , where $s$ is tied to the time parameter: for critical points $x$ of $f_{0}$ and $y$ of $f_{1}$ with $μ (x) = μ (y)$ , the relevant moduli space is zero-dimensional, and $Φ x = \sum_{y} m (x, y) y$ with $m (x, y)$ the signed count.

A boundary-of-the-1-dimensional-moduli argument — identical in spirit to $\partial^{2} = 0$ — shows $Φ$ is a chain map: $\partial_{1} Φ = Φ \partial_{0}$ . Reversing the homotopy gives $Ψ : C_{*} (f_{1}) \to C_{*} (f_{0})$ , and a two-parameter (homotopy-of-homotopies) moduli count produces a chain homotopy $ΨΦ ≃ id$ , $ΦΨ ≃ id$ . Hence $Φ$ is a quasi-isomorphism and induces an isomorphism $H M_{*} (f_{0}, g_{0}) ≅ H M_{*} (f_{1}, g_{1})$ .

Concatenating continuation maps along composed homotopies is, up to chain homotopy, the continuation for the composite, so the isomorphisms are canonical and compose correctly. This makes $H M_{*}$ a well-defined invariant of $M$ ; identifying it with singular homology (the Morse Homology Theorem, via continuation to a self-indexing function and an Eilenberg-Steenrod-style uniqueness argument) is the final step Schwarz carries out internally.

Exercise pack. Schwarz, Morse Homology, Parts I-II supplement: gradient flow and indices, trajectory moduli and the Fredholm dimension count, the Morse-Smale-Witten boundary operator and $\partial^{2} = 0$ , the Morse inequalities, and $HM_ $o f$ S^n $,$ T^2 $,$ \mathbb{C}P^n $,$ \mathbb{R}P^2$.*

Prerequisites

03.15.01 pending
03.15.02 pending
03.15.03 pending
03.15.05 pending
03.15.06 pending
03.15.08 pending

Tier anchors

beginner
intermediate: Schwarz, Morse Homology, Parts I-II exercises (gradient flow, trajectory moduli, the Morse-Smale-Witten complex, the Morse Homology Theorem); Milnor, Morse Theory, §1-§6 worked examples
master

References

Schwarz — Morse Homology · Part I (gradient flow, stable/unstable manifolds, Morse-Smale, trajectory moduli, Fredholm index $\mu(x)-\mu(y)$, compactness by broken trajectories); Part II (coherent orientations, the boundary operator $\partial x=\sum n(x,y)\,y$, $\partial^2=0$, the Morse Homology Theorem $HM_*\cong H_*(M;\mathbb{Z})$)
Milnor — Morse Theory · §1-§6 — Morse functions, indices, the Morse lemma, the height function on the torus, Morse inequalities
Audin-Damian — Morse Theory and Floer Homology · Ch. 3-4 — the Morse complex, worked computations on $S^n$, $T^2$, $\mathbb{C}P^n$
Banyaga-Hurtubise — Lectures on Morse Homology · Ch. 4-7 — the Morse-Smale-Witten boundary operator and homology computations

Estimated time

intermediate: 5h