Imagine that there is a road that runs along the Earth’s equator and that you are driving along this road at constant speed. The road would appear perfectly straight – you would not have to turn to the right or left to continue along the road. Thus, you would perceive your velocity (and not just your speed) as being constant. On the other hand, an observer in space would see that your velocity is not constant as you are travelling in a circle rather than in a straight line. The resolution of this apparent paradox is that an observer restricted to the surface of the Earth perceives only the component of the acceleration tangential to the surface. From the point of view of the observer in space, the acceleration vector points towards the centre of the Earth, and so has zero tangential component.
In general, suppose thatγis a curve on a surfaceS and letvbe atangent vector fieldalongγ, i.e., a smooth map from an open interval (α, β) toR3such
thatv(t)∈Tγ(t)S for allt∈(α, β). To an observer moving along the curveγ, the perceived rate of change of v is the tangential component of ˙v, i.e., the orthogonal projection of ˙v = dv/dt onto Tγ(t)S. If N is a unit normal to σ, the component of ˙vperpendicular to the surface is ( ˙vãN)N, so the tangential component is
∇γv= ˙v−( ˙vãN)N. (7.11) Note that this is unchanged ifNis replaced by−N, so∇γvis well defined on any surface, orientable or not.
. N
v v
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v
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Definition 7.4.1
Letγbe a curve on a surfaceSand letvbe a tangent vector field alongγ. The covariant derivative of v along γ is the orthogonal projection ∇γv of dv/dt onto the tangent planeTγ(t)S at a pointγ(t).
In particular, an inhabitant ofS would perceivevas being constant along γ if∇γv=0. In this case, v is sometimes said to becovariant constant, but the usual terminology is contained in
Definition 7.4.2
With the notation in Definition7.4.1,vis said to beparallel alongγif∇γv=0 at every point ofγ.
Proposition 7.4.3
A tangent vector fieldvis parallel along a curveγ on a surface S if and only if ˙vis perpendicular to the tangent plane of S at all points ofγ.
Proof
This is clear from the definitions: if the right-hand side of Eq.7.11is zero, then
˙
vis obviously parallel toN. Conversely, if ˙v=λNfor some scalarλ, then
∇γv= ˙v−(λNãN)N= ˙v−λN=0.
To establish the existence of parallel tangent vector fields, we shall express the covariant derivative in terms of the parametersu, vof a parametrizationσ of the surface. To do this, we shall need the following calculation, which will also be used later.
Proposition 7.4.4 (Gauss Equations)
Letσ(u, v) be a surface patch with first and second fundamental forms Edu2+ 2F dudv+Gdv2 and Ldu2+ 2M dudv+N dv2. Then,
σuu= Γ111σu+ Γ211σv+LN, σuv= Γ112σu+ Γ212σv+MN, σvv= Γ122σu+ Γ222σv+NN, where
Γ111= GEu−2F Fu+F Ev
2(EG−F2) , Γ211= 2EFu−EEv−F Eu 2(EG−F2) , Γ112= GEv−F Gu
2(EG−F2), Γ212= EGu−F Ev
2(EG−F2), Γ122= 2GFv−GGu−F Gv
2(EG−F2) , Γ222= EGv−2F Fv+F Gu
2(EG−F2) .
The six Γ coefficients in these formulas are calledChristoffel symbols. Note that they depend only on thefirstfundamental form ofσ.
Proof
Since{σu,σv,N}is a basis ofR3, scalar functionsα1, . . . , γ3 satisfying σuu=α1σu+α2σv+α3N,
σuv=β1σu+β2σv+β3N, (7.12) σvv=γ1σu+γ2σv+γ3N,
certainly exist. Taking the dot product of each equation withNgives α3=L, β3=M, γ3=N.
Now we take the dot product of each equation in (7.12) with σu and σv. This gives six scalar equations from which we determine the remaining six coefficients. For example, taking the dot product of the first equation in (7.12) withσuandσv gives the two equations
Eα1+F α2=σuuãσu=1 2Eu,
F α1+Gα2=σuuãσv= (σuãσv)u−σuãσuv=Fu−1 2Ev. Solving these equations givesα1 = Γ111, α2= Γ211; similarly for the other four coefficients in Eq.7.12.
We can now establish the conditions for a tangent vector field v to be parallel along a curve γ(t) = σ(u(t), v(t)) on a surface patchσ(u, v). Since the tangent plane ofσis spanned by the vectorsσu andσv, there are smooth scalar functionsαandβ such that
v(t) =α(t)σu+β(t)σv, the derivatives ofσ being evaluated atσ(u(t), v(t)).
Proposition 7.4.5
Letγ(t) =σ(u(t), v(t)) be a curve on a surface patchσ, and letv(t) =α(t)σu+ β(t)σvbe a tangent vector field alongγ, whereαandβare smooth functions of t. Then,vis parallel alongγ if and only if the following equations are satisfied:
˙
α+ (Γ111u˙+ Γ112v)α˙ + (Γ112u˙+ Γ122v)β˙ = 0
β˙+ (Γ211u˙ + Γ212v)α˙ + (Γ212u˙+ Γ222v)β˙ = 0. (7.13) Note that these equations involve only the firstfundamental form ofσ.
Proof
Using the Gauss equations, we have
˙
v= ˙ασu+ ˙βσv+αu(Γ˙ 111σu+ Γ211σv+LN) (7.14) + (αv˙+βu)(Γ˙ 112σu+ Γ212σv+MN) +βv(Γ˙ 122σu+ Γ222σv+NN).
By Proposition7.4.3,vis parallel alongγif and only if ˙vis parallel toN, which means that the coefficients of σu and σv on the right-hand side of Eq.7.14 must both be zero. But these coefficients are the left-hand sides of the two equations in (7.13).
Eqs. (7.13) are of the form
˙
α=f(α, β, t), β˙ =g(α, β, t), (7.15) wheref andgare smooth functions of three variables. It is proved in the theory of ordinary differential equations that such equations have a unique solution for any given set ofinitial conditions, i.e., if t0 is some particular value of t, andα0, β0 ∈ R, there are unique smooth functions α(t) andβ(t), defined on an open interval containingt0, that satisfy Eq.7.16and are such that
α(t0) =α0, β(t0) =β0. (7.16) In the situation considered in Proposition 7.4.5, the initial conditions (7.16) are equivalent to
v(t0) =α0σu+β0σv. So we obtain
Corollary 7.4.6
Letγbe a curve on a surfaceSand letv0be a tangent vector ofS at the point γ(t0). Then, there is exactly one tangent vector fieldvthat is parallel alongγ and is such thatv(t0) =v0.
Example 7.4.7
Take γ to be a circle of latitude θ = θ0 (−π/2 < θ0 < π/2) on the unit sphere with the latitude–longitude parametrization σ(θ, ϕ) (Example 4.1.4);
thus, γ(ϕ) = σ(θ0, ϕ). The first fundamental form of σ is dθ2 + cos2θdϕ2 (Example6.1.3) from which we find
Γ111= Γ211= Γ222= Γ112= 0, Γ212=−tanθ, Γ122= sinθcosθ.
The differential equations (7.13) become
˙
α=−βsinθ0cosθ0, β˙=αtanθ0. (7.17) Ifθ0= 0, thenαandβ are constant. Ifθ0= 0, eliminatingβ gives
¨
α+αsin2θ0= 0,
which has the general solution
α(ϕ) =Acos(ϕsinθ0) +Bsin(ϕsinθ0),
whereAandB are constants; the second equation in (7.17) now gives β =Asin(ϕsinθ0)
cosθ0 −Bcos(ϕsinθ0) sinθ0 .
Let us consider the special case in which v = σϕ is tangent to γ when ϕ= 0. Then, α= 0,β = 1 whenϕ= 0, which givesA= 0, B=−sinθ0 and hence
v(ϕ) =−sinθ0sin(ϕsinθ0)σθ+ cos(ϕsinθ0)σϕ
(this solution is also correct whenθ0= 0). Note thatv(ϕ) is not tangent toγ atγ(ϕ) in general. However, ifθ0 = 0 thenv is tangent to γ for all ϕ. Thus, the tangent vector ofγ is parallel alongγ if and only ifγ is a great circle. We shall see the reason for this in Section9.1.
Ifpandqare two points on a curveγon a surfaceS, the covariant derivative enables us to associate to any vector in the tangent planeTpS a vector in the tangent planeTqS. Indeed, suppose thatpandqcorrespond to the parameter valuest0 andt1, letv0∈TpS, letv(t) be the unique parallel vector field along γsuch that v(t0) =v0 (see Corollary7.4.6), and letv1=v(t1)∈TqS.
Definition 7.4.8
With the above notation, the map Πpqγ :TpS →TqS that takesv0∈TpS to v1∈TqS is calledparallel transport frompto qalongγ.
Proposition 7.4.9
With the notation in Definition7.4.8, (i) Πpqγ :TpS →TqS is a linear map
(ii) Πpqγ is an isometry, i.e., it preserves lengths and angles.
Proof
Letv0, w0∈TpS and letλ, μ∈R. Letv(t), w(t) be the parallel vector fields alongγsuch thatv(t0) =v0,w(t0) =w0. IfV=λv+μw, then ˙V=λv˙+μw˙ is parallel to the unit normalNofS because ˙vand ˙ware parallel toN, soV is parallel alongγ. Hence,
Πpqγ (λv0+μw0) = Πpqγ (V(t0)) =V(t1) =λv1+μw1=λΠpqγ (v0)+μΠpqγ (w0),
which proves that Πpqγ is linear.
For (ii), note that d
dt(vãw) = ˙vãw+vãw˙ = (( ˙vãN)N)ãw+vã(( ˙wãN)N) = 0, sincevãN=wãN= 0 (vand ware tangent to the surface). Hence,
v0ãw0=v1ãw1.
Thus, Πpqγ preserves dot products of vectors. Since lengths and angles are expressible in terms of dot products, part (ii) is proved.
Example 7.4.10
Take γ to be the equator of S2. We saw in Example 7.4.7 that the tangent vectorσϕof γis parallel alongγ. Now, at points of the equator,σθ is a unit vector perpendicular toσϕ. By Proposition7.4.9, the parallel vector field along γequal toσθwhenϕ= 0 has the same property. It must therefore be equal to σθ; in other words,σθis also parallel alongγ(this can also be checked directly from the formulas in Example7.4.7). Since parallel transport is a linear map, it follows that, for any two pointspandqon the equator, and anyλ, μ∈R,
Πpqγ (λσθ+μσϕ) =λσθ+μσϕ. (7.18) (Note, however, that Πpqγ is not the identity map unless pand q coincide ! If p=qthe derivativesσθandσϕon the two sides of (7.18) are being evaluated at different points ofS2.)
EXERCISES
7.4.1 Let ˜γ be a reparametrization ofγ, so that ˜γ(t) =γ(ϕ(t)) for some smooth functionϕwithdϕ/dt= 0 for all values oft. Ifvis a tangent vector field alongγ, show that ˜v(t) =v(ϕ(t)) is one along ˜γ. Prove that
∇γ˜v˜= dϕ dt∇γv,
and deduce that v is parallel along γ if and only if ˜v is parallel along ˜γ.
7.4.2 Show that the parallel transport map Πpqγ :TpS →TqSis invertible.
What is its inverse?
7.4.3 Suppose that a triangle on the unit sphere whose sides are arcs of great circles has vertices p, q, r. Let v0 be a non-zero tangent vector to the arcpqthroughpandqatp. Show that, if we parallel transportv0 alongpq, then along qrand then alongrp, the result is to rotatev0 through an angle 2π− A, whereAis the area of the triangle. For an analogous result see Theorem 13.6.4.
Gaussian, mean and principal curvatures 8
In this chapter, we show how to extract geometric information from the second fundamental form of a surface or, equivalently, from its Weingarten map.