On a smooth manifold M, a 1-form ω assigns to each point p a linear map ωp : TpM → ℝ. If f : M → ℝ is a smooth function, what is the 1-form df?
Adf_p(v) = f(p) · v for each tangent vector v
Bdf_p(v) = v(f) — the tangent vector v applied to f as a derivation
Cdf_p(v) = ∇f · v — the gradient dot product, which requires an inner product
Ddf_p(v) = the directional derivative of v in the direction of f
The differential df of a smooth function f is defined by df_p(v) = v(f) for any tangent vector v ∈ TpM. This is purely algebraic — the tangent vector, being a derivation, already knows how to act on f. No metric or inner product is needed. In coordinates, df = (∂f/∂xⁱ)dxⁱ, where dxⁱ are the coordinate 1-forms (the duals of ∂/∂xⁱ). Option C confuses df with the gradient ∇f — the gradient requires a Riemannian metric to convert the 1-form df into a vector field.
Question 2 True / False
The coordinate 1-forms dx¹, ..., dxⁿ at a point p form the dual basis to the coordinate tangent vectors ∂/∂x¹, ..., ∂/∂xⁿ at p.
TTrue
FFalse
Answer: True
By definition, dxⁱ(∂/∂xʲ) = δⁱⱼ (the Kronecker delta — 1 if i=j, 0 otherwise). This is the duality pairing between the cotangent space T*pM and the tangent space TpM. The coordinate 1-forms form a basis for the cotangent space, just as the coordinate vector fields form a basis for the tangent space. A general 1-form at p is ω_p = ωᵢdxⁱ, where the components ωᵢ transform by the inverse Jacobian (covariant transformation law) — dual to the contravariant transformation of vector components.
Question 3 Multiple Choice
A 2-form on a 3-manifold with coordinates (x, y, z) can be written as ω = f dy∧dz + g dz∧dx + h dx∧dy. How many independent components does a k-form have on an n-manifold?
Anᵏ
Bn!/(k!(n-k)!) — the binomial coefficient C(n,k)
Cn·k
D2ⁿ
A k-form at a point is an alternating multilinear map on k tangent vectors. The space of k-forms at a point has dimension C(n,k) = n!/(k!(n-k)!). For the example: 2-forms on a 3-manifold have C(3,2) = 3 independent components — exactly the three coefficients f, g, h. The alternating property (ω(v,w) = -ω(w,v)) and multilinearity reduce the nᵏ possible components down to C(n,k). Note that C(n,k) = 0 for k > n, so there are no (n+1)-forms on an n-manifold.
Question 4 Short Answer
Why are differential forms (rather than vector fields) the natural objects to integrate on manifolds?
Think about your answer, then reveal below.
Model answer: Integration requires an object that transforms correctly under coordinate changes to give a well-defined number. An n-form on an n-manifold transforms by the determinant of the Jacobian under coordinate changes, which is exactly what is needed to make the change-of-variables formula work. Vector fields transform contravariantly and do not have this property. Additionally, forms of degree k naturally integrate over k-dimensional submanifolds without requiring a metric — they are intrinsically adapted to measuring oriented k-dimensional volumes.
This is the deep reason differential forms exist. In multivariable calculus, the change-of-variables formula involves a Jacobian determinant. Differential forms package this transformation law into their definition: an n-form automatically picks up the Jacobian determinant when you change coordinates. This makes integration coordinate-independent. By contrast, integrating a vector field requires converting it to a form using a metric (like g(X, ·)), which adds extra structure.