Find the mechanism before the formula
Ordinary Morse theory asks topology to change at isolated nondegenerate critical points. Symmetry often produces whole circles, tori or other manifolds of critical points instead. Morse–Bott theory keeps the useful local quadratic picture by demanding nondegeneracy only transverse to that critical manifold.
We will identify the correct normal Hessian, build a circle-valued worked model, translate the index into a negative normal disk bundle, and mark the precise boundary between a critical submanifold and a merely degenerate critical set.
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Start with the question closest to the decision you need to make. Every route returns to the same worked evidence and transfer checks.
| Route | Chapters | Use it when |
|---|---|---|
| 1 | 1–3 | You need definitions, the normal Hessian and index. |
| 2 | 4–5 | You want the local normal form and sublevel attachment. |
| 3 | 6–8 | You need a worked circle example and changed-context transfer. |
| 4 | 9–10 | You must control orientations, perturbations and misconceptions. |
| 5 | 11–12 | You want FAQs and a proof checklist. |
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Build the model
Test the mechanism
1. A critical set can have dimension
For a smooth f:M→ℝ, the critical set Crit(f) consists of points where df=0. In Morse theory, each critical point is isolated and the Hessian is nonsingular. In the Morse–Bott setting, a connected component C of Crit(f) may itself be a smooth submanifold.
Being a submanifold is necessary but not sufficient. The Hessian must have no additional zero directions transverse to C. Tangent directions to C are zero because f is locally constant along the critical component; those zeros are expected.
- First prove C is an embedded submanifold.
- Then identify TpC inside TpM.
- Test the Hessian on the normal quotient or a chosen normal bundle.
2. Nondegeneracy lives in the normal directions
At p∈C, the Hessian of f is a symmetric bilinear form on TpM. For a critical submanifold, TpC lies in its kernel. The Morse–Bott condition is ker Hesspf=TpC.
Equivalently, the induced quadratic form on the normal space νpC=TpM/TpC is nonsingular. This formulation does not depend on a metric. A metric is convenient because it identifies a complementary normal subspace.
If the kernel is larger than TpC, the function is degenerate in a transverse direction and the Morse–Bott lemma does not apply.
3. The index becomes a vector bundle rank
The normal Hessian splits the normal bundle into negative and positive subbundles ν⁻ and ν⁺ when the critical component is compact and the index is locally constant. The Morse–Bott index λ(C) is rank ν⁻.
For an isolated critical point, C is a point and this reduces to the ordinary number of negative Hessian eigenvalues. For a higher-dimensional C, the fibres can twist as one moves around C, so ν⁻ contains more information than its rank alone.
| Object | Ordinary Morse | Morse–Bott |
|---|---|---|
| Critical locus | Point p | Submanifold C |
| Zero Hessian directions | None | Exactly TC |
| Negative data | λ-dimensional vector space | Rank-λ bundle ν⁻→C |
| Attachment | One λ-handle | Disk bundle D(ν⁻) |
4. The Morse–Bott local normal form
Near a point of C, suitable coordinates separate tangent coordinates u along C from negative normal coordinates x and positive normal coordinates y. The function has the form f=f(C)−||x||²+||y||².
No quadratic term appears in u because every point of C is critical and f is constant on each connected component. The normal form shows exactly why transverse nondegeneracy is enough: all topology-changing directions remain controlled by a genuine quadratic form.
This is a local statement. A global product C×ℝr need not exist because the normal bundles can be nontrivial.
5. Crossing a critical value attaches a negative disk bundle
Assume a compact interval [a,b] contains one critical value c and the corresponding Morse–Bott critical component C, with no boundary complications. Up to the appropriate homotopy or diffeomorphism formulation, the upper sublevel Mb is obtained from Ma by attaching the disk bundle D(ν⁻) along its sphere bundle S(ν⁻).
Positive normal directions thicken the local model but do not form the attaching core. When C is a point, D(ν⁻) is an ordinary λ-disk and the statement becomes the familiar λ-handle picture.
The theorem needs hypotheses: compactness or properness for flow control, regular values around c, and a clean treatment of boundaries. The slogan alone is not a proof.
6. Worked model: a critical circle with mixed normal signs
Let M=S¹×ℝ² with coordinates (θ,x,y) and f(θ,x,y)=−x²+y². The differential is df=−2x dx+2y dy, so Crit(f)=S¹×{(0,0)}. This is a critical circle C.
The θ direction is tangent to C and lies in the Hessian kernel. On the normal (x,y)-plane, the Hessian is diag(−2,2), which is nonsingular. Therefore f is Morse–Bott with index one.
Here ν⁻ is the x-line bundle and ν⁺ is the y-line bundle. In this product example both are trivial. The local crossing attaches D¹×S¹ along S⁰×S¹: a family of one-dimensional handles parameterised by the critical circle.
| Direction | Hessian value | Role |
|---|---|---|
| θ | 0 | Tangent to critical circle |
| x | −2 | Negative normal direction |
| y | +2 | Positive normal direction |
7. A small perturbation can resolve the family
Choose a Morse function h:C→ℝ and extend it into a tubular neighbourhood. For small ε, f+εh often breaks the critical submanifold into isolated critical points. Their ordinary indices are λ(C)+indexh(p).
For C=S¹, a height-like h has one minimum and one maximum. In the worked index-one model, the perturbed function produces isolated critical points of indices one and two. Their ordinary handles collectively recover the disk-bundle attachment.
This perturbation explains the Morse–Bott polynomial contribution tλ(C)Pt(C), but it must be constructed with control so no unintended critical points appear.
8. Transfer check: add a flat transverse direction
Change the model to g(θ,x,y)=−x²+y⁴. The critical set is still S¹×{(0,0)}, and the θ direction is still tangent. However, the normal Hessian is diag(−2,0). The y direction is transverse and also lies in the kernel.
The equality ker Hess=T C now fails, so g is not Morse–Bott even though its critical set is a smooth circle. The quartic term cannot be replaced by a positive quadratic term through the Morse–Bott lemma.
This is the decisive transfer test: preserve the critical set, change only the transverse curvature, and recheck the kernel. A smooth critical manifold does not guarantee Morse–Bott nondegeneracy.
9. Relative homology records the attachment
Under standard orientation hypotheses, the relative homology across the critical level is related by a Thom isomorphism to the homology of C shifted by λ: H*(Mb,Ma)≈H*−λ(C), with coefficients adjusted for the orientation local system of ν⁻ when necessary.
This is why a whole critical component contributes its topology, not merely one generator. If ν⁻ is nonorientable, using integer coefficients without the local system can lose information; field coefficients may simplify but do not make the twist disappear geometrically.
Morse–Bott inequalities compare the sum of these shifted critical-component contributions with the homology of M, just as ordinary Morse inequalities compare critical points with Betti numbers.
10. Limits and misconceptions
- A critical submanifold is not automatically Morse–Bott.
- The index is the rank of the negative normal bundle, not the dimension of C.
- A disk-bundle attachment need not be a product C×Dλ.
- Perturbing to Morse points is a method, not proof that the original function was Morse–Bott.
- Distinct critical components can share a critical value; the crossing must account for all of them.
- Noncompact manifolds require properness or separate control of escaping gradient trajectories.
Symmetry often creates Morse–Bott families, but symmetry can also produce worse degeneracy. Compute the normal Hessian rather than relying on the source of the family.
11. Frequently asked questions
Is f constant on a critical component?
Yes on each connected component, because df vanishes along every path in it. Different components may have the same or different critical values.
Why use the quotient normal space?
It removes the tangent directions that must lie in the Hessian kernel and makes the nondegeneracy condition independent of a chosen metric complement.
Does every Morse–Bott function perturb to a Morse function?
Under standard compactness and local-construction hypotheses, one can perturb using Morse functions on the critical components. The global statement still requires care near overlaps and boundaries.
12. A proof and computation checklist
| Stage | Question |
|---|---|
| Critical set | Is it exactly computed and an embedded submanifold? |
| Tangent space | Is TC identified at every component? |
| Hessian | Is its kernel exactly TC? |
| Index | What are ν⁻, ν⁺ and their ranks? |
| Crossing | Which disk bundle attaches at each critical value? |
| Coefficients | Does ν⁻ require an orientation local system? |
The gold standard is the transverse test: compute the critical manifold, quotient out its tangent directions, prove the remaining Hessian is nonsingular, and let the negative normal bundle—not a guessed handle count—describe the topology change.
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