Moving on from * first part*.

Suppose ∇ is a derivative operator, and g_{ab} is a metric, on the manifold M. Then ∇ is compatible with g_{ab} iff ∇_{a} g_{bc} = 0.

Suppose γ is an arbitrary smooth curve with tangent field ξ^{a} and λ^{a} is an arbitrary smooth field on γ satisfying ξ^{n}∇_{n}λ^{a} = 0. Then

ξ^{n}∇_{n}(g_{ab}λ^{a}λ^{b}) = g_{ab}λ^{a}ξ^{n} ∇_{n}λ^{b} + g_{ab}λ^{b}ξ^{n} ∇_{n}λ^{a} + λ^{a}λ^{b}ξ^{n} ∇_{n}g_{ab}

= λ^{a}λ^{b}ξ^{n}∇_{n}g_{ab}

Suppose first that ∇_{n}g_{ab} = 0. Then it follows immediately that ξ^{n}∇_{n}g_{ab}λ^{a}λ^{b} = 0. So ∇ is compatible with g_{ab}. Suppose next that ∇ is compatible with g_{ab}. Then ∀ choices of γ and λ^{a} (satisfying ξ^{n}∇_{n}λ^{a} =0), we have λ^{a}λ^{b}ξ^{n}∇_{n}g_{ab} = 0. Since the choice of λ^{a} (at any particular point) is arbitrary and g_{ab} is symmetric, it follows that ξ^{n}∇_{n}g_{ab} = 0. But this must be true for arbitrary ξ_{a} (at any particular point), and so we have ∇_{n}g_{ab} = 0.

Note that the condition of compatibility is also equivalent to ∇_{a}g^{bc} = 0. Hence,

0 = g^{bn}∇_{a}δ^{c}_{n} = g^{bn}∇_{a}(g_{nr}g^{rc}) = g^{bn}g_{nr}∇_{a}g^{rc} + g^{bn}g^{rc}∇_{a}g_{nr}

= δ^{b}_{r}∇_{a}g^{rc} + g^{bn}g^{rc}∇_{a}g_{nr} = ∇_{a}g^{bc} + g^{bn}g^{rc}∇_{a}g_{nr}.

So if ∇_{a}g_{bc} = 0,it follows immediately that ∇_{a}g^{bc} = 0. Conversely, if ∇_{a}g^{bc} =0, then g^{bn}g^{rc}∇_{a}g_{nr} = 0. And therefore,

0 = g_{pb}g_{sc}g^{bn}g^{rc}∇_{a}g_{nr} = δ^{n}_{p}δ^{r}_{s}∇_{a}g_{nr} = ∇_{a}g_{ps}

The basic fact about compatible derivative operators is the following.

Suppose g_{ab} is a metric on the manifold M. Then there is a unique derivative operator on M that is compatible with g_{ab}.

It turns out that if a manifold admits a metric, then it necessarily satisfies the countable cover condition. And then it guarantees the existence of a derivative operator.) We do prove that if M admits a derivative operator ∇, then it admits exactly one ∇′ that is compatible with g_{ab}.

Every derivative operator ∇′ on M can be realized as ∇′ = (∇, C^{a}_{bc}), where C^{a}_{bc} is a smooth, symmetric field on M. Now

∇′_{a}g_{bc} = ∇_{a}g_{bc} + g_{nc} C^{n}_{ab} + g_{bn} C^{n}_{ac} = ∇_{a}g_{bc} + C_{cab} + C_{bac}. So ∇′ will be compatible with g_{ab} (i.e., ∇′_{a}g_{bc} = 0) iff

∇_{a}g_{bc} = −C_{cab} − C_{bac} —– (1)

Thus it suffices for us to prove that there exists a unique smooth, symmetric field C^{a}_{bc} on M satisfying equation (1). To do so, we write equation (1) twice more after permuting the indices:

∇_{c}g_{ab} = −C_{bca} − C_{acb},

∇_{b}g_{ac} = −C_{cba} − C_{abc}

If we subtract these two from the first equation, and use the fact that C_{abc} is symmetric in (b, c), we get

C_{abc} = 1/2 (∇_{a}g_{bc} − ∇_{b}g_{ac} − ∇_{c}g_{ab}) —– (2)

and, therefore,

C^{a}_{bc} = 1/2 g^{an} (∇_{n}g_{bc} − ∇_{b}g_{nc} − ∇_{c}g_{nb}) —– (3)

This establishes uniqueness. But clearly the field Cabc defined by equation (3) is smooth, symmetric, and satisfies equation (1). So we have existence as well.

In the case of positive definite metrics, there is another way to capture the significance of compatibility of derivative operators with metrics. Suppose the metric g_{ab} on M is positive definite and γ : [s_{1}, s_{2}] → M is a smooth curve on M. We associate with γ a length

|γ| = ∫_{s1}^{s2} g_{ab}ξ^{a}ξ^{b} ds,

where ξ^{a} is the tangent field to γ. This assigned length is invariant under reparametrization. For suppose σ : [t_{1}, t_{2}] → [s_{1}, s_{2}] is a diffeomorphism we shall write s = σ(t) and ξ′^{a} is the tangent field of γ′ = γ ◦ σ : [t_{1}, t_{2}] → M. Then

ξ′^{a} = ξ^{a}ds/dt

We may as well require that the reparametrization preserve the orientation of the original curve – i.e., require that σ (t_{1}) = s_{1} and σ (t_{2}) = s_{2}. In this case, ds/dt > 0 everywhere. (Only small changes are needed if we allow the reparametrization to reverse the orientation of the curve. In that case, ds/dt < 0 everywhere.) It

follows that

|γ’| = ∫_{t1}^{t2} (g_{ab}ξ′^{a}ξ′^{b})^{1/2} dt = ∫_{t1}^{t2} (g_{ab}ξ^{a}ξ^{b})^{1/2} ds/dt

= ∫_{s1}^{s2} (g_{ab}ξ^{a}ξ^{b})^{1/2} ds = |γ|

Let us say that γ : I → M is a curve from p to q if I is of the form [s_{1}, s_{2}], p = γ(s_{1}), and q = γ(s_{2}). In this (positive definite) case, we take the distance from p to q to be

d(p,q)=g.l.b. |γ|:γ is a smooth curve from p to q.

Further, we say that a curve γ : I → M is minimal if, for all s ∈ I, ∃ an ε > 0 such that, for all s_{1}, s_{2} ∈ I with s_{1} ≤ s ≤ s_{2}, if s_{2} − s_{1} < ε and if γ′ = γ|[s_{1}, s_{2}] (the restriction of γ to [s_{1}, s_{2}]), then |γ′| = d(γ(s_{1}), γ(s_{2})) . Intuitively, minimal curves are “locally shortest curves.” Certainly they need not be “shortest curves” outright. (Consider, for example, two points on the “equator” of a two-sphere that are not antipodal to one another. An equatorial curve running from one to the other the “long way” qualifies as a minimal curve.)

One can characterize the unique derivative operator compatible with a positive definite metric g_{ab} in terms of the latter’s associated minimal curves. But in doing so, one has to pay attention to parametrization.

Let us say that a smooth curve γ : I → M with tangent field ξ^{a} is parametrized by arc length if ∀ ξ^{a}, g_{ab}ξ^{a}ξ^{b} = 1. In this case, if I = [s_{1}, s_{2}], then

|γ| = ∫_{s1}^{s2} (g_{ab}ξ^{a}ξ^{b})^{1/2} ds = ∫_{s1}^{s2} 1.ds = s_{2} – s_{1}

Any non-trivial smooth curve can always be reparametrized by arc length.

[…] Part 1 and Part 2. […]