Questions: Connections and Covariant Derivative

Question 1

Why can't you simply differentiate a vector field Y in the direction of X using the ordinary directional derivative X(Yⁱ)∂/∂xⁱ?

• A. Because vector fields are not functions and cannot be differentiated
• B. Because the expression X(Yⁱ)∂/∂xⁱ is not coordinate-independent — it transforms incorrectly under coordinate changes
• C. Because the directional derivative of a vector field is always zero
• D. Because X and Y live in different tangent spaces and cannot interact

Answer: (B)

The expression X(Yⁱ)∂/∂xⁱ differentiates the components of Y, but the basis vectors ∂/∂xⁱ also change from point to point. The 'naive' derivative misses the change in the basis vectors, and the result is not a tensor — it does not transform correctly under coordinate changes. A connection adds the correction term Γᵏᵢⱼ Xⁱ Yʲ that accounts for how the basis vectors twist as you move. The full covariant derivative is (∇_X Y)ᵏ = X(Yᵏ) + Γᵏᵢⱼ Xⁱ Yʲ, which transforms as a vector.

Question 2

An affine connection ∇ on a manifold is uniquely determined by the Riemannian metric.

• T. True
• F. False

Answer: False

A smooth manifold admits infinitely many connections. What IS unique is the Levi-Civita connection — the one that is both torsion-free (∇_X Y - ∇_Y X = [X,Y]) and metric-compatible (∇g = 0). The Fundamental Theorem of Riemannian Geometry states that given a Riemannian metric, there exists a unique such connection. But other connections exist: connections with torsion appear in Einstein-Cartan theory, and non-metric connections arise in affine differential geometry.

Question 3

The Christoffel symbols Γᵏᵢⱼ are the components of a tensor.

• T. True
• F. False

Answer: False

The Christoffel symbols do NOT transform as tensor components. Under a coordinate change x → x', they transform as Γ'ᵏᵢⱼ = (∂x'ᵏ/∂xˡ)(∂xᵐ/∂x'ⁱ)(∂xⁿ/∂x'ʲ)Γˡₘₙ + (∂x'ᵏ/∂xˡ)(∂²xˡ/∂x'ⁱ∂x'ʲ). The second term — involving second derivatives of the coordinate transformation — is the non-tensorial part. It is precisely this non-tensorial term that cancels the non-tensorial part of the naive derivative X(Yⁱ), making the full covariant derivative ∇_X Y a well-defined tensor. The difference of two connections IS a tensor, which is why the space of connections is an affine space.

Question 4

A connection ∇ on a manifold must satisfy three algebraic properties. Which property distinguishes it from the Lie bracket?

Model answer: C∞(M)-linearity in the first argument: ∇_{fX} Y = f∇_X Y for any smooth function f. The Lie bracket fails this — [fX, Y] = f[X,Y] - Y(f)X has an extra term. Both operations are ℝ-linear in both arguments and satisfy a Leibniz rule in the second argument (∇_X(fY) = X(f)Y + f∇_X Y). But the C∞(M)-linearity in X makes ∇_X Y depend only on the value of X at a point, not its derivatives — this is tensoriality in the first slot.

This tensoriality means (∇_X Y)_p depends only on X_p (the value of X at p) and on Y along a curve tangent to X at p. The Lie bracket [X,Y]_p depends on the derivatives of both X and Y. This makes the covariant derivative the right tool for defining parallel transport (which should depend on the direction of transport, not on how the direction field extends away from the curve).