Given vector fields X = ∂/∂x and Y = x∂/∂y on ℝ², what is [X, Y]?
A0 (the zero vector field)
B∂/∂y
Cx∂/∂x
D-∂/∂y
Computing directly: [X,Y](f) = X(Y(f)) - Y(X(f)). We have Y(f) = x·∂f/∂y, so X(Y(f)) = ∂/∂x(x·∂f/∂y) = ∂f/∂y + x·∂²f/∂x∂y. Also X(f) = ∂f/∂x, so Y(X(f)) = x·∂²f/∂x∂y (since ∂/∂y of ∂f/∂x = ∂²f/∂y∂x). The difference is ∂f/∂y. Therefore [X,Y] = ∂/∂y. The bracket is nonzero because the coefficient of Y depends on x, which X differentiates.
Question 2 True / False
The Lie bracket is C∞(M)-linear in both arguments: [fX, Y] = f[X, Y] for any smooth function f.
TTrue
FFalse
Answer: False
The Lie bracket is NOT C∞(M)-linear — it satisfies [fX, Y] = f[X, Y] - Y(f)·X instead. The extra term Y(f)·X arises because Y differentiates the function f. This is why the Lie bracket is not a tensor: tensorial operations are by definition C∞(M)-multilinear. The Lie bracket's failure of C∞(M)-linearity means it cannot be computed pointwise from the values of X and Y — it depends on their derivatives. This distinguishes it from operations like the metric pairing g(X,Y) which is tensorial.
Question 3 Short Answer
The Jacobi identity states [X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y]] = 0. What role does this identity play in the structure of vector fields?
Think about your answer, then reveal below.
Model answer: The Jacobi identity ensures that the Lie bracket makes the space of vector fields into a Lie algebra — a vector space with a bilinear, antisymmetric bracket satisfying this three-term relation. The Jacobi identity is the Lie algebra analogue of associativity for groups. It guarantees that the adjoint representation (ad_X(Y) = [X,Y]) is itself a Lie algebra homomorphism, which is essential for the theory of Lie groups and their actions on manifolds.
Without the Jacobi identity, the bracket would be an arbitrary antisymmetric bilinear operation, and much of the theory of Lie algebras, Lie groups, and symmetry in differential geometry would collapse. The identity can be verified by direct computation using the definition [X,Y] = XY - YX and the associativity of composition of derivations.
Question 4 True / False
If the flows of two vector fields X and Y commute (φs ∘ ψt = ψt ∘ φs for all s, t), then [X, Y] = 0.
TTrue
FFalse
Answer: True
This is correct, and the converse also holds: [X, Y] = 0 if and only if the flows commute. The Lie bracket [X, Y]_p can be computed as the limit lim_{t→0} (1/t²)(ψ_{-t} ∘ φ_{-t} ∘ ψ_t ∘ φ_t(p) - p), measuring the gap when you flow along X, then Y, then back along X, then back along Y. If the flows commute, this loop closes exactly and the bracket vanishes. Coordinate vector fields ∂/∂xⁱ always have vanishing brackets because the coordinate flows (translations along coordinate axes) commute.