Two overlapping charts (U, φ) and (V, ψ) on a manifold M have the transition map ψ ∘ φ⁻¹ : φ(U ∩ V) → ψ(U ∩ V). For M to be a smooth manifold, what must be true about this transition map?
AIt must be a homeomorphism (continuous with continuous inverse)
BIt must be infinitely differentiable (C∞) as a map between open subsets of ℝⁿ
CIt must be an isometry (preserving distances between coordinate representations)
DIt must be a linear map between the coordinate domains
A smooth manifold requires all transition maps to be C∞ (infinitely differentiable). Since transition maps go between open subsets of ℝⁿ, the standard definition of differentiability from multivariable calculus applies directly. Merely being a homeomorphism (option A) gives a topological manifold, not a smooth one. Isometry (option C) is far too restrictive — it would force the manifold to be flat. Linearity (option D) would make all charts affinely related, which is also far too restrictive.
Question 2 True / False
Every topological manifold admits a unique smooth structure.
TTrue
FFalse
Answer: False
This is false in general. Some topological manifolds admit multiple non-diffeomorphic smooth structures. The most famous example is ℝ⁴, which admits uncountably many distinct smooth structures (exotic ℝ⁴s). The 7-sphere S⁷ admits exactly 28 distinct smooth structures (Milnor's exotic spheres). However, in dimensions 1, 2, and 3, every topological manifold admits a unique smooth structure up to diffeomorphism. The relationship between topological and smooth structures is one of the deepest questions in differential topology.
Question 3 Multiple Choice
Let f : M → ℝ be a function on a smooth manifold M with atlas {(Uα, φα)}. What does it mean for f to be smooth?
Af is continuous as a map between topological spaces
BFor every chart (Uα, φα), the composition f ∘ φα⁻¹ : φα(Uα) → ℝ is C∞
Cf has a Taylor expansion at every point of M
DThe graph of f is a smooth submanifold of M × ℝ
Smoothness on a manifold is defined by pulling back to coordinate charts. The function f : M → ℝ is smooth if for every chart (Uα, φα), the coordinate representation f ∘ φα⁻¹ is a C∞ function from an open subset of ℝⁿ to ℝ — where standard multivariable calculus defines smoothness. The compatibility of the smooth structure (C∞ transition maps) guarantees that if f ∘ φα⁻¹ is smooth in one chart, it is smooth in every overlapping chart. Option D is actually equivalent but is not the definition — it is a consequence.
Question 4 Short Answer
Why is the requirement that transition maps be smooth (rather than merely continuous) essential for doing calculus on manifolds?
Think about your answer, then reveal below.
Model answer: Without smooth transition maps, the notion of a differentiable function on the manifold would depend on which chart you use to compute the derivative. A function could appear differentiable in one chart but not in another. Smooth transition maps ensure that the chain rule transfers differentiability between charts: if f ∘ φ⁻¹ is differentiable and ψ ∘ φ⁻¹ is smooth, then f ∘ ψ⁻¹ = (f ∘ φ⁻¹) ∘ (φ ∘ ψ⁻¹) is also differentiable. This makes 'f is smooth on M' a chart-independent statement.
The chain rule is the mechanism that makes smooth structures work. When you change coordinates from φ to ψ, a function's coordinate representation transforms by composition with the transition map. If the transition map is only continuous (not differentiable), differentiability of the composite cannot be guaranteed. The smooth atlas ensures all notions of calculus — derivatives, tangent vectors, differential forms — are consistently defined across chart boundaries.