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Geometry of curved spacetime, part 4

This post has been migrated from my old blog, the math-physics learning seminar.

Today I had to try to explain connections and curvature in local frames (as opposed to coordinates), and I really feel that Wald’s treatment of this is just awful (this is one of the few complaints I have with an otherwise classic textbook). It is particularly baffling since the treatment in Misner, Thorne, and Wheeler is just perfect. What follows is the modern math (as opposed to physics) point of view. This is more abstract than any introductory GR (or even Riemannian geometry) text I’ve seen, but in this case the abstraction absolutely clarifies and simplifies things.

Let MM be a smooth manifold and suppose EE is a smooth vector bundle over MM. A connection on EE is a map ∇\nabla taking sections of EE to sections of T∗M⊗ET^\ast M \otimes E, R\mathbb{R}-linear and satisfying the Leibniz rule

∇(fσ)=df⊗σ+f∇σ. \nabla(f\sigma) = df \otimes \sigma + f \nabla \sigma.

Now consider the sheaf of EE-valued pp-forms on MM. Call it Ωp(E)\Omega^p(E). Then we can extend the connection to a map

∇:Ωp(E)→Ωp+1(E) \nabla: \Omega^p(E) \to \Omega^{p+1}(E)

via the Leibniz rule:

∇(η⊗σ)=dη⊗σ+(−1)pη∧∇σ. \nabla(\eta \otimes \sigma) = d\eta \otimes \sigma + (-1)^p \eta \wedge \nabla \sigma.

Let us define the curvature FF associated to a connection ∇\nabla by the composition

F=∇2:Ωp(E)→Ωp+2(E). F = \nabla^2: \Omega^p(E) \to \Omega^{p+2}(E).

Claim FF is C∞C^\infty-linear, i.e. it is tensorial.

Proof

∇(∇(fσ))=∇(df⊗σ+f∇σ)=d2f⊗σ−df∧∇σ+df∧∇σ+f∇2σ=f∇2σ. \begin{aligned} \nabla(\nabla(f \sigma)) &= \nabla( df \otimes \sigma + f \nabla \sigma) \cr &= d^2 f \otimes \sigma - df \wedge \nabla \sigma + df \wedge \nabla \sigma + f \nabla^2 \sigma \cr &= f \nabla^2 \sigma. \end{aligned}

So far we have not made any additional choices (beyond ∇\nabla). In order to actually compute something locally, we have to make some choices. Let e^a\hat{e}_a be a frame, i.e. a local basis of sections of EE. Then ∇e^a\nabla \hat{e}_a is an EE-valued 1-form, hence it can be expressed as a sum

∇e^a=∑bωab⊗e^b \nabla \hat{e}_a = \sum_b \omega_a^b \otimes \hat{e}_b

where the coefficients ωab\omega_a^b are 1-forms, often called the connection 1-forms. Let Ω\Omega denote the matrix of 1-forms whose entries are exactly ωab\omega_a^b.

Claim Let σ=σae^a\sigma = \sigma^a \hat{e}_a. Then we have

∇σ=dσ+Ωσ. \nabla \sigma = d\sigma + \Omega \sigma.

Proof The coefficients σa\sigma^a are functions (i.e. scalars), so ∇σa=dσa\nabla \sigma^a = d\sigma^a. Using the Leibniz rule we have

∇(σae^a)=(∇σa)e^a+σa∇e^a=dσae^a+σaωabe^b=dσae^a+ωcaσce^a=(dσ+Ωσ)ae^a. \begin{aligned} \nabla(\sigma^a \hat{e}_a) &= (\nabla \sigma^a) \hat{e}_a + \sigma^a \nabla \hat{e}_a \cr &= d\sigma^a \hat{e}_a + \sigma^a \omega_a^b \hat{e}_b \cr &= d\sigma^a \hat{e}_a + \omega_c^a \sigma^c \hat{e}_a \cr &= (d\sigma + \Omega \sigma)^a \hat{e}_a. \end{aligned}

Claim The curvature satisfies F=dΩ−Ω∧ΩF = d\Omega - \Omega \wedge \Omega.

Proof Just apply the above formula twice using Leibniz.

Connection 1-forms from Christoffel symbols. Suppose now that we are in the Riemannian setting and we already know the Christoffel symbols in some coordinates. Then we can express our frame e^a\hat{e}_a in terms of coordinate vector fields, i.e. e^a=e^ai∂∂xi \hat{e}_a = \hat{e}_a^i \frac{\partial}{\partial x^i} Then we have that ∇je^ai=∂e^ai∂xj+Γjkie^ak \nabla_j \hat{e}_a^i = \frac{\partial \hat{e}_a^i }{\partial x^j} + \Gamma^i_{jk} \hat{e}_a^k

So, as a vector-valued 1-form, we have ∇e^=∂e^ai∂xjdxj⊗∂∂xi+Γjkie^akdxj⊗∂∂xi. \nabla \hat{e} = \frac{\partial \hat{e}_a^i}{\partial x^j} dx^j \otimes \frac{\partial}{\partial x^i} + \Gamma^i_{jk} \hat{e}_a^k dx^j \otimes \frac{\partial}{\partial x^i}.

Juggling things a bit using the metric, we find ∇e^a=∂e^ai∂xje^ibdxj⊗e^b+Γjkie^ake^ibdxj⊗e^b. \nabla \hat{e}_a = \frac{\partial \hat{e}_a^i}{\partial x^j} \hat{e}^b_i dx^j \otimes \hat{e}_b + \Gamma^i_{jk} \hat{e}_a^k \hat{e}_i^b dx^j \otimes \hat{e}_b.

So the connection 1-forms are given by ωab=∂e^ai∂xje^ibdxj+Γjkie^ake^ibdxj. \omega_a^b = \frac{\partial \hat{e}_a^i}{\partial x^j} \hat{e}^b_i dx^j + \Gamma^i_{jk} \hat{e}_a^k \hat{e}_i^b dx^j.

To come later (if I ever get around to it): some explicit computations.