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Notes to Barrett O’Neill’s Semi-Riemannian
Geometry With Applications to Relativity
Andrew Goetz
March 18, 2015
Contents
1
Manifold Theory
1
2
Tensors
2
3
Semi-Riemannian Manifolds
3
4
Semi-Riemannian Submanifolds
4
5
Riemannian and Lorentz Geometry
5
6
Special Relativity
7
i
Chapter 1
Manifold Theory
Notes
Page 30, Definition 55: The case B = ∞ might look strange at first glance. By
definition, α : [0, ∞) → M has a continuous extension α˜ : [0, ∞] → M if and only
if there is a point p ∈ M such that for every sequence {tn } approaching infinity, the
sequence {α(tn )} converges to p. Equivalently, α˜ is continuous when [0, ∞] is given
the topology of the one-point compactification of [0, ∞). This is the topology such
that the map from [0, ∞] to [0, 1] given by x 7→ x/(1 + x) for x < ∞ and ∞ 7→ 1 is a
homeomorphism.
1
Chapter 2
Tensors
Notes
2
Chapter 3
Semi-Riemannian Manifolds
Notes
Page 66, Proposition 19: Parallel translation does not depend on the parametrization
of the curve. Let α : I → M be a curve and h : J → I a reparametrization function. We
can parallel translate a vector z ∈ T α(t0 ) (M) along α giving a parallel vector field Z on α.
Alternatively, we can parallel translate z ∈ T α(h(s0 )) (M) along β = α ◦ h giving a parallel
vector field Y on β. Then an easy computation shows that Z(h(s)) satisfies the same
differential equation as Y(s), with the same initial value. By uniqueness, Y(s) = Z(h(s))
for all s ∈ J.
Page 73, line 3: As O’Neill notes just before Example 34, we also have
(3)
gi j,k (o) = 0.
Page 80, Corollary 43: This corollary is also true at a single point with constant
curvature.
Page 89, line -5: The more general result referred to in Chapter 7 is Proposition 7.38.
Page 91, Proposition 62: We can now characterize all isometries of Rnν . Let ψ :
Rnν → Rnν be an isometry which fixes the origin. The linear map dψ0 is an isometry of
T 0 (Rnν ) viewed as a vector space (it preserves the scalar product). By Lemma 7 it is
an isometry of T 0 (Rnν ) viewed as a semi-Riemannian manifold. Then by Example 34,
n
exp0 ◦ dψ0 ◦ exp−1
0 is an isometry of Rν . Its differential at the origin is equal to dψ0 . By
the proposition, it is equal to ψ. Since it is linear, ψ is linear, and the matrix of ψ in the
standard basis is the same as that of dψ0 . Thus ψ preserves the scalar product.
A general isometry of Rnν can be made to fix the origin just by composing it with a
translation. Since translations are linear isometries, we have shown that every isometry
of Rnν is affine linear, a composition of a linear isometry which fixes the origin and a
translation.
3
CHAPTER 3. SEMI-RIEMANNIAN MANIFOLDS
Page 95, Problem 18: For the definition of dθ, see Exercise 2.2.
4
Chapter 4
Semi-Riemannian Submanifolds
Notes
Pages 105-106, Definition 15 and following: If dim M = 1, then every point of M is
umbilic, and hence M is totally umbilic.
Page 107, Definition 18: Perhaps a more transparent way to define S is to define the
symmetric (0, 2) tensor B first: set B(V, W) = hII(V, W), Ui and raise an index of B to
get the (1, 1) tensor S .
Page 109, Proposition 22: Note that Λ ⊂ Rn+1
is diffeomorphic to Λ ⊂ Rn+1
ν
n−ν+1 . So, for
2
0
instance, (R − 0) × S is diffeomorphic to (R − 0) × S 1 .
Page 109, Figure 2: If 1 < ν < n, Rn+1
− (Λ ∪ 0) has two connected components
ν
consisting of spacelike and timelike vectors, respectively. If ν = 1, there are three
connected components: spacelike vectors, timelike vectors with x1 > 0, and timelike
vectors with x1 < 0. If ν = n, there are three connected components: timelike vectors,
spacelike vectors with xn+1 < 0, and spacelike vectors with xn+1 > 0. Exercise 5.4 is a
special case of this more general fact.
n
Page 110, Lemma 24: The mapping σ−1 carries Hn−ν
(r) anti-isometrically onto S νn (r).
n
n+1
The mapping σ does not. (Note that Hn−ν (r) < Rν .)
Page 111, line -12: Of course, U is the outward unit normal P/r.
Page 115, Corollary 34: The statements in the corollary are also true at a single point
which has constant curvature.
5
Chapter 5
Riemannian and Lorentz
Geometry
Notes
Page 131, Lemma 9: Here is a proof of the lemma: It is clear that E ◦ ∆ = idC ×C .
Let W = ∆(C × C ), and take any w ∈ W (say w = ∆(p, q) where p, q ∈ C ). Let
ψ : D˜ → T M be the flow of the vector field G on T M in Proposition 3.28. The domain
D˜ of the flow is an open subset of T M × R, and the mapping is ψ(v, s) = γv0 (s). Integral
curves of G correspond under the projection π : T M → M to geodesics of M. The map
˜ Furthermore, (π ◦ ψ)−1 (C )
π ◦ ψ : D˜ → M is smooth, so (π ◦ ψ)−1 (C ) is open in D.
contains (w, t) for 0 ≤ t ≤ 1 because (π ◦ ψ)({w} × [0, 1]) = γw ([0, 1]) ⊂ C . Since [0, 1] is
compact, we can find an open neighborhood V of w such that V × [0, 1] ⊂ (π ◦ ψ)−1 (C ).
Then V ⊂ W because if we take any v ∈ V and set p0 = γv (0) and q0 = γv (1), then
γv |[0, 1] is a geodesic in C running from p0 to q0 . So v = ∆(p0 , q0 ). This shows that W
is open.
Next, since C is a normal neighborhood of p, exp p is nonsingular at w ∈ T p (M). By
Lemma 6, E is nonsingular at w. Hence there is a neighborhood U ⊂ W such that E|U
is a diffeomorphism onto a neighborhood of (p, q) in C × C . Since w was arbitrary, this
shows that E|W is a local diffeomorphism. Since it is also injective (E ◦ ∆ = idC ×C ), it
is a diffeomorphism. Its inverse is the diffeomorphism ∆ : C × C → W .
Page 131, Lemma 10: The second sentence of the proof requires explanation. Suppose
M is a metric space and C∗ is an open cover of M. Then there is an open cover B of M
such that if two of its elements meet then their union is contained in some element of
C∗ . To produce B, around every point p in M let Br (p) be an open ball centered at p
contained in some element of C∗ . Let B consist of the sets Br/2 (p) for p ∈ M. It is easy
to see that B has the required property.
The second sentence of the proof now follows from the fact that M is metrizable
(this is a well-known fact about second-countable manifolds).
6
CHAPTER 5. RIEMANNIAN AND LORENTZ GEOMETRY
7
Page 132, lines 6-9: Note that if α : [a, b] → M is piecewise smooth and h : [c, d] →
[a, b] is a piecewise smooth reparameterization function, then α◦ h need not be piecewise
smooth. For example, take M = R, α : [−e−1 sin(1), e−1 sin(1)] → R defined by



t t ≤ 0
α(t) = 

0 t ≥ 0,
and h : [−1, 1] → [−e−1 sin(1), e−1 sin(1)] defined by
 −2


e−s sin(1/s) s , 0
h(s) = 

0
s = 0.
Then α is piecewise smooth and h is smooth, but because α has a break at t = 0 and the
zeros of h are not isolated, α ◦ h is not piecewise smooth. Indeed, α ◦ h is the function
h− (s) obtained from h(s) by setting all the positive function values of h to zero.
However, if h is a monotone function, then α ◦ h is guaranteed to be piecewise
continuous. Proof: We may assume h is nondecreasing. The preimages of the break
points of α form a partition [c, d]. Adding in the break points of h, we obtain a refined
partition. Restricting α ◦ h to any subinterval of the partition gives a smooth function,
so α ◦ h is piecewise continuous.
More generally, if the preimages of the break points of α are a finite set in [c, d],
then α ◦ h is piecewise smooth by the same argument as above. In our counterexample
above, 0 was a breakpoint of α and h−1 (0) was infinite. One condition on h that ensures
this is true is piecewise analyticity, because then h can take on any value only a finite
number of times.
Page 132, Lemma 12: Here is a proof of (1): We may assume h is nondecreasing. As
noted above, α ◦ h is piecewise smooth. Let c = s0 , s1 , . . . , sn−1 , sn = d be a partition of
[c, d] such that (α ◦ h)|[si−1 , si ] is smooth. Let ti = h(si ). Then a = t0 , t1 , . . . , tn−1 , tn = b
is a partition of [a, b] (some points may be repeated) and α|[ti−1 , ti ] is smooth.
L(α ◦ h) =
n Z
X
=
|(α ◦ h)0 (s)| ds
si−1
i=1
=
si
n Z
X
si
|α0 (h(s))| · |h0 (s)| ds
i=1 si−1
n Z ti
X
|α0 (t)| dt
i=1
ti−1
= L(α).
The third equality above follows from the change of variables theorem in one dimension.
Here is a proof of (2): Let a = t0 , t1 , . . . , tn−1 , tn = b be a partition of [a, b] such that
α|[ti−1 , ti ] is smooth. Define
Z
t
g(t) =
a
|α0 (t0 )| dt0 .
CHAPTER 5. RIEMANNIAN AND LORENTZ GEOMETRY
8
Then g is a strictly increasing function from [a, b] to [0, d] for some d > 0. Furthermore,
g|[ti−1 , ti ] is smooth so g is piecewise smooth. Letting si = g(ti ), we get a partition
0 = s0 , s1 , . . . , sn−1 , sn = d of [0, d]. Letting h : [0, d] → [a, b] be the inverse of g, it
is easy to see that h is strictly increasing and h|[si−1 , si ] is smooth, so h is piecewise
smooth. The claim is that β = α ◦ h has |β0 | = 1. This is an easy computation: for any
s ∈ [0, d], let t = h(s). Then
|β0 (s)| = |α0 (h(s))| · |h0 (s)| = |α0 (t)| · |g0 (t)|−1 = 1
from the definition of g and the fundamental theorem of calculus. If s = si for some
1 ≤ i ≤ n − 1, then the above computation needs to be interpreted as being done either
from the left or from the right side of s. In either case, |β0 (s)| = 1.
Page 133, Lemma 14: We prove the following statement which encompasses the
lemma and both refinements mentioned at the top of page 134:
Let σ : [0, 1] → M be a geodesic from o to p. Let τ : [0, b] → T o (M) be a piecewise
smooth curve such that τ(0) = 0, τ(b) = v where expo (v) = p, and d expo is singular for
at most a finite number of points along τ. The curve α = expo ◦ τ is a piecewise smooth
curve from o to p and
(a) L(α) ≥ L(σ)
(b) If L(α) = L(σ), then α is a monotone reparameterization of σ.
Proof: Assume first that τ(t) , 0 for t > 0. Let 0 = t0 , t1 , . . . , tn−1 , tn = b be
a partition of [0, b] such that τ|[ti−1 , ti ] is smooth and for all t not in the partition,
(d expo )τ(t) is nonsingular. Let U˜ be the vector field along τ obtained by restricting the
˜
radial unit vector field on T o (M) to τ. (The value of U(0)
is not defined but this will not
matter.) Using d expo we obtain a vector field U along α. It is a unit vector field by the
Gauss lemma. The vector fields U˜ and U are smooth on each subinterval [ti−1 , ti ] where
τ and α are smooth.
Let J = I − {t0 , . . . , tn }. For t ∈ J we can write
α0 = hα0 , UiU + N
where N is a vector field along α orthogonal to U. There are two possible values for
N at t1 , . . . , tn−1 but this does not matter; we leave N undefined at these points. The
definition of N at t0 = 0 and tn = b also does not matter. For t ∈ J, N is smooth and we
have
˜ = (˜r ◦ τ)0 = |τ|0 .
|α0 | = [hα0 , Ui2 + hN, Ni]1/2 ≥ |hα0 , Ui| ≥ hα0 , Ui = hτ0 , Ui
We used the Gauss lemma and the fact that U˜ = grad r˜. By the fundamental theorem of
calculus,
Z
Z
L(α|[ti−1 , ti ]) =
ti
ti−1
|α0 | dt ≥
ti
|τ|0 dt = |τ(ti )| − |τ(ti−1 )|
ti−1
since |τ| is continuous [ti−1 , ti ] and differentiable on (ti−1 , ti ). Summing over i, we obtain
L(α) ≥ |τ(b)| − |τ(0)| = |v| = L(σ),
CHAPTER 5. RIEMANNIAN AND LORENTZ GEOMETRY
9
which proves (a).
For (b), if L(α) = L(σ), then the inequalities above all become equalities. We
conclude that
N=0
and
|τ|0 = hα0 , Ui = |hα0 , Ui| ≥ 0
for t ∈ J. Thus α0 = |τ|0 U for t ∈ J. Applying d(expo )−1 (our assumption about the
singularites of the exponential map is used here, and only here), we get that
τ0 = |τ|0 U˜ = |τ|0 ·
τ
|τ|
for t ∈ J. We compute for such t that
d τ
|τ|τ0 − τ|τ|0
=
= 0.
dt |τ|
|τ|2
Thus on each open interval (ti−1 , ti ), τ/|τ| is a constant vector. By continuity, τ/|τ| is
constant on (0, b]. Since τ(b) = v = exp−1
o (p), we get that
τ(t) = |τ(t)| ·
v
(˜r ◦ τ)(t)
=
v = h(t)v
|v| (˜r ◦ τ)(b)
for all t ∈ [0, b], where h : [0, b] → [0, 1] is piecewise smooth and nondecreasing.
Applying expo , we get
α(t) = (σ ◦ h)(t),
which shows that α is a monotone reparameterization of σ.
Now we remove the assumption that τ(t) , 0 for t > 0. Assuming this is false, since
τ−1 (0) is closed there is some t0 ∈ (0, b) such that τ(t0 ) = 0 and τ(t) , 0 for t ∈ (t0 , b].
Applying the previous result to τ|[t0 , b] and α|[t0 , b], we get that
L(α) ≥ L(α|[t0 , b]) ≥ L(σ)
so (a) holds. For (b), it is clear that if τ leaves 0 in the interval [0, t0 ], then L(α|[0, t0 ]) > 0
so that
L(α) = L(α|[0, t0 ]) + L(α|[t0 , b]) > L(σ).
Hence L(α) = L(σ) implies τ|[0, t0 ] ≡ 0. Define H : [0, b] → [0, 1] by



t ∈ [0, t0 ]
0
H(t) = 

h(t) t ∈ [t0 , b]
where h is the reparameterization function produced by applying our previous work to
τ|[t0 , b] and α|[t0 , b]. Then H is a monotone piecewise smooth function and α = σ ◦ H.
Page 139, line -9: At this point it’s not hard to show that if t ∈ T and t0 < t, then t0 ∈ T .
Proof: Note that for any t0 ∈ [0, d],
d = d(p, q) ≤ d(p, γ(t0 )) + d(γ(t0 ), q) ≤ t0 + d(γ(t0 ), q).
CHAPTER 5. RIEMANNIAN AND LORENTZ GEOMETRY
10
Thus to show t0 ∈ T we need to show the reverse inequality d ≥ t0 + d(γ(t0 ), q). Assume
that t ∈ T and t0 < t. As noted in the text, γ|[0, t] is minimizing, so γ|[t0 , t] is minimizing.
Hence d(γ(t0 ), γ(t)) = t − t0 . Then we have that
d(γ(t0 ), q) ≤ d(γ(t0 ), γ(t)) + d(γ(t), q) = t − t0 + d − t = d − t0
which is what we wanted to show.
Page 139, line -8: More simply, T is nonempty since 0 ∈ T . But also by (∗), δ ∈ T . By
the above remark, we know at this point at least that [0, δ] ⊂ T .
Page 145, line -9: At a point (u0 , . . . , un ) ∈ S 1n ,
tan ∂0 = (1 + (u0 )2 )∂0 + (u0 u1 )∂1 + · · · + (u0 un )∂n
nor ∂0 = −(u0 )2 ∂0 − u0 u1 ∂1 − · · · u0 un ∂n = −u0 P
where P = u0 ∂0 + · · · + un ∂n is the unit normal. Then htan ∂0 , tan ∂0 i = −(1 + (u0 )2 ) so
tan ∂0 is indeed a timelike vector field on S 1n .
Page 145, line -4: Let I = (−1, 1). On R × I, let
g1 = −dt2 + dx2
be the standard Lorentz metric inherited from R21 and let
g2 = − cos(x) dt2 − 2 sin(x) dt dx + cos(x) dx2
Let π : R × I → S 1 × I be the quotient map (x, t) 7→ (eix , t). In fact π is a smooth
covering map. Pushing g1 down to S 1 × I via π gives us a time-orientable Lorentz
metric on S 1 × I. A timelike vector field on S 1 × I is obtained by pushing down ∂t via
π. On the other hand, g2 can also be pushed down via π to give a Lorentz metric on
S 1 × I. But now S 1 × I is not time-orientable. Proof: If Y is a smooth timelike vector
field on S 1 × I, pull it back via π to give a smooth timelike vector field (which we also
denote by Y) on (R × I, g2 ). Let Z = cos(x/2)∂t + sin(x/2)∂ x . Then g2 (Z, Z) = −1 so Z
is a smooth timelike vector field on (R × I, g2 ). Hence g2 (Y, Z), being smooth, and the
scalar product of two timelike vectors, cannot change sign. This is a contradiction since
Y(0, 0) = Y(2π, 0) but Z(0, 0) = −Z(2π, 0).
Similarly g1 can be pushed down to the M¨obius band to give a not time-orientable
metric and g2 to give a time-orientable metric (a smooth timelike vector field being
given by the pushdown of Z).
Page 146, line 6: After doing Exercise 3 and constructing a few more examples, one
can say the following. Let V be a Lorentz vector space with dim V ≥ 3. Denote one
causal cone by “+” and the other by “−”. Let v and w be vectors in V. The following
table gives all the information it is possible to know about the sign of hv, wi if we know
only the causal characters of v and w.
CHAPTER 5. RIEMANNIAN AND LORENTZ GEOMETRY
sign hv, wi
v t.l. +
v t.l. −
v null +
v null −
v s.l.
w t.l. +
−
+
−
+
−, 0, +
w t.l. −
+
−
+
−
−, 0, +
w null +
−
+
−, 0
+, 0
−, 0, +
w null −
+
−
+, 0
−, 0
−, 0, +
11
w s.l.
−, 0, +
−, 0, +
−, 0, +
−, 0, +
−, 0, +
Remember that if v and w are null, then hv, wi = 0 if and only if they are colinear. If
dim V = 2, then the only change we need to make in the table above is to delete the
possibility that the scalar product of a null vector and a spacelike vector is 0.
Page 146, Lemma 33: In the proof we must interpret the vector field P as being defined
only along the curve α. It might not be defined in a neighborhood of a point of α since
no assumption is made that d expo is nonsingular at each point of β.
The proof of the lemma where timelike and timecone are replaced by causal and
causal cone requires more than just “minor changes”. There are several points where
there is no obvious modification to make; for example, we cannot straightforwardly
conclude that initially β is in a single causal cone. This is because β0 (0) could be null,
and β might never even enter a causal cone. To prove the desired result, we use a
variational argument.
Let 0 = t0 , t1 . . . , tn−1 , tn = b be a partition of [0, b] such that α|[ti−1 , ti ] is smooth.
Define a vector field V(t) along α(t) by picking a timelike vector in the same causal
cone as α0 (0) and parallel translating it along α. The vector field V(t) is timelike for all
t. On the interval [t0 , t1 ], hα0 (t), V(t)i is a continuous function which starts off negative
and cannot pass through zero, so it is negative everywhere. Hence V(t) lies in the same
causal cone as α0 (t) for t ∈ [t0 , t1 ]. At the breakpoint t1 , α0 (t1+ ) lies in the same causal
cone as α0 (t1− ), hence also V(t1 ). We can then repeat the argument to show that V(t) lies
in the same causal cone as α0 (t) for t ∈ [t1 , t2 ], etc.
Define a two-parameter map x(s, t) = exp(stV(t)). Since stV(t) is a continuous map
into T M, the preimage under this map of the domain of exp (which is open) is an open
set in R2 which contains the line segment {0} × [0, b]. This segment is compact so x
is defined on some rectangle (−ε, ε) × [0, b]. Furthermore on each interval [ti−1 , ti ] on
which α is smooth, x is smooth on the rectangle (−ε, ε) × [ti−1 , ti ]. Thus for each fixed s,
x(s, t) is a piecewise smooth curve starting at o.
We now show that for sufficiently small s > 0, x(s, t) is timelike. First, x s (0, t) =
tV(t) and hence x st (0, t) = V(t) (since V is parallel along α). Thus on each rectangle
(−ε, ε) × [ti−1 , ti ],
∂ hxt , xt i = 2 hxts , xt i| s=0 = 2 hx st , xt i| s=0 = 2hV(t), α0 (t)i < 0.
∂s s=0
Again by compactness, there is a rectangle (−ε0 , ε0 ) × [0, b] on which (∂/∂s)hxt , xt i is
negative (at breaks, (∂/∂s)hxt , xt i is double-valued). At s = 0, hxt , xt i = hα0 , α0 i ≤ 0.
Therefore for sufficiently small s > 0, hxt , xt i < 0. At a breakpoint ti , hxt (s, ti− ), xt (s, ti+ )i
is a continuous function of s which, as the scalar product of timelike vectors, is nonzero.
At s = 0, it is negative, so it is negative everywhere. Thus x(s, t) is a piecewise smooth
timelike curve for sufficiently small s > 0 (say 0 < s < δ).
CHAPTER 5. RIEMANNIAN AND LORENTZ GEOMETRY
12
˜ (0, t) = β(t). By the the timelike/timecone result
Define x˜ (s, t) = exp−1
o ◦ x. Then x
in Lemma 5.33, for each 0 < s < δ, x˜ (s, t) remains in a single timecone of T o (M). Since
x˜ is continuous and the two timecones in T o (M) are disconnected, these timecones must
all be the same, say the timecone C. Taking the limit as s approaches zero from above
shows that β remains in the closure of C. To show that β remains in the causal cone
¯ we must show that the only time t for which β(t) = 0 is t = 0. An examination of
C,
the proof the timelike/timecone case shows, by continuity, that d(q˜ ◦ β)/dt ≤ 0 on each
interval [ti−1 , ti ]. Thus if β returns to 0 (where q˜ is zero) at time T it must lie on the
nullcone (plus the origin) and d(q˜ ◦ β)/dt = hα0 , Pi = 0 for 0 ≤ t ≤ T . Since P is null
in this case and α0 is causal, α0 must be null as well, and hence P and α0 are colinear
for 0 ≤ t ≤ T . Therefore β0 and P˜ are colinear for 0 ≤ t ≤ T . Hence β follows a ray in
the nullcone and α is a reparameterization of a null geodesic. So β cannot return to the
origin, and we’re done.
Note that we have proved the following general fact: if α : [0, b] → M is a piecewise
smooth causal curve starting at o given by expo ◦ β where β : [0, b] → T o (M) is a
piecewise smooth curve starting at 0 and remaining in one component of the nullcone,
then β is a reparametrization of a ray in the nullcone and α is a reparametrization of a
null geodesic.
Page 147, Proposition 34: The refinement mentioned at the bottom of the page can be
proved using the same method that we used in these notes to prove the refinement to
Lemma 14. In fact we can prove the following statement which is a further refinement:
Let σ : [0, 1] → M be a timelike geodesic from o to p. Let α : [0, b] → M be a
piecewise smooth causal curve from o to p given by expo ◦ β where β : [0, b] → T o (M)
is a piecewise smooth curve (lying in a single causal cone, by Lemma 33) such that
β(0) = 0, β(b) = v where expo (v) = p, and d expo is nonsingular along β. Then
(a) L(α) ≤ L(σ)
(b) If L(α) = L(σ) and α is timelike, then α is a monotone reparameterization of σ.
Proof: First suppose that α is timelike so that β remains in a single timecone of
T o (M). Let 0 = t0 , t1 , . . . , tn−1 , tn = b be a partition of [0, b] such that β|[ti−1 , ti ] is smooth
and for all t not in the partition, (d expo )β(t) is nonsingular. Let U˜ be the vector field
along β obtained by restricting the radial unit vector field on T o (M) to β. (The value of
˜
U(0)
is not defined but this will not matter.) Using d expo we obtain a vector field U
along α. It is a unit vector field by the Gauss lemma. The vector fields U˜ and U are
smooth on each subinterval [ti−1 , ti ] where β and α are smooth.
Let J = I − {t0 , . . . , tn }. For t ∈ J we can write
α0 = −hα0 , UiU + N
where N is a vector field along α orthogonal to U. There are two possible values for
N at t1 , . . . , tn−1 but this does not matter; we leave N undefined at these points. The
definition of N at t0 = 0 and tn = b also does not matter. For t ∈ J, N is smooth and we
have
˜ = (˜r ◦ β)0 = |β|0 .
|α0 | = [hα0 , Ui2 − hN, Ni]1/2 ≤ |hα0 , Ui| = −hα0 , Ui = −hβ0 , Ui
CHAPTER 5. RIEMANNIAN AND LORENTZ GEOMETRY
13
We used the Gauss lemma and the fact that U˜ = − grad r˜. By the fundamental theorem
of calculus,
Z ti
Z ti
0
L(α|[ti−1 , ti ]) =
|α | dt ≤
|β|0 dt = |β(ti )| − |β(ti−1 )|
ti−1
ti−1
since |β| is continuous [ti−1 , ti ] and differentiable on (ti−1 , ti ). Summing over i, we obtain
L(α) ≤ |β(b)| − |β(0)| = |v| = L(σ),
which proves (a).
For (b), if L(α) = L(σ), then the inequalities above all become equalities. We
conclude that
N=0
and
|β|0 = −hα0 , Ui = |hα0 , Ui| ≥ 0
for t ∈ J. Thus α0 = |β|0 U for t ∈ J. Applying d(expo )−1 , we get that
β0 = |β|0 U˜ = |β|0 ·
β
|β|
for t ∈ J. We compute for such t that
|β|β0 − β|β|0
d β
=
= 0.
dt |β|
|β|2
Thus on each open interval (ti−1 , ti ), β/|β| is a constant vector. By continuity, β/|β| is
constant on (0, b]. Since β(b) = v = exp−1
o (p), we get that
β(t) = |β(t)| ·
(˜r ◦ β)(t)
v
=
v = h(t)v
|v| (˜r ◦ β)(b)
for all t ∈ [0, b], where h : [0, b] → [0, 1] is piecewise smooth and nondecreasing.
Applying expo , we get
α(t) = (σ ◦ h)(t),
which shows that α is a monotone reparameterization of σ.
If α is causal, then as we have shown in our proof of the causal/causal cone case
of Lemma 5.33 in these notes, we can use a variation of α to produce a sequence
of piecewise smooth timelike curves αi : [0, b] → M starting at o such that for each
t ∈ [0, b], αi (t) → α(t). The construction shows that the lengths of these curves approach
the length of α. For sufficiently large i there is a timelike geodesic σi : [0, 1] → M
starting at o with the same endpoint as αi . Furthermore our assumption that d expo is
nonsingular along β implies that for sufficiently large i, αi = expo ◦ βi for piecewise
smooth curves βi : [0, b] → T o (M) and d expo is nonsingular along βi . Our previous
work shows that L(αi ) ≤ L(σi ). Taking the limit as i → ∞ gives L(α) ≤ L(σ) as desired.
Chapter 6
Special Relativity
Notes
Page 163, line -3: A particle β : I → M is regular because it is nonspacelike, so we can
never have β0 = 0. That β(I) is a one-dimensional submanifold of M is a consequence
of M being isometric to R41 ; it is not true of a general spacetime.
Page 164, Lemma 12: The lemma is not quite correct as stated. We need to add the
requirement that ξ maps p to some fixed point, the most natural point being the origin
in R41 . Otherwise the uniqueness is true only up to any translation of R41 .
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