relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics...

34
1 Supplemental Lecture 13 Differential Forms for Physics Students III. Classical and Riemannian Differential Geometry Abstract Differential forms are employed to develop the theory of surfaces and manifolds. The method of moving frames of E. Cartan is introduced and the fundamental formulas of differential geometry are obtained in a coordinate-free fashion. This perspective leads to remarkably simple derivations of Gauss’ Theorem Egregium and the Gauss-Bonnet Theorem. The Gauss-Bonnet Theorem is developed as a structural equation of differential geometry without the need for coordinate systems or detailed formulas for the Gaussian curvature of surfaces or the geodesic curvature of curves. These ideas are generalized to higher dimensional manifolds. Topics in Riemannian and non-Euclidean geometry are introduced and studied. This lecture supplements material in the textbook: Special Relativity, Electrodynamics and General Relativity: From Newton to Einstein (ISBN: 978-0-12-813720-8) by John B. Kogut. The term “textbook” in these Supplemental Lectures will refer to that work. Keywords: Differential Forms, Classical Differential Geometry, Riemannian Geometry, E.Cartan, Orthonormal Frames, Gaussian Curvature, Theorem Egregium, Gauss-Bonnet Theorem, Hypersurfaces, Manifolds, Non-Euclidean Geometry Contents Strategy and Perspective. ................................................................................................................ 2 Moving Frames ............................................................................................................................... 3 Surfaces in R 3 .................................................................................................................................. 6 Gaussian and Mean Curvatures ...................................................................................................... 7 Theorem Egregium ......................................................................................................................... 9 Harmonic Functions ...................................................................................................................... 11 Gauss-Bonnet Theorem: Global Version ...................................................................................... 13

Transcript of relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics...

Page 1: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

1

Supplemental Lecture 13

Differential Forms for Physics Students

III. Classical and Riemannian Differential Geometry

Abstract

Differential forms are employed to develop the theory of surfaces and manifolds. The method of

moving frames of E. Cartan is introduced and the fundamental formulas of differential geometry

are obtained in a coordinate-free fashion. This perspective leads to remarkably simple

derivations of Gauss’ Theorem Egregium and the Gauss-Bonnet Theorem. The Gauss-Bonnet

Theorem is developed as a structural equation of differential geometry without the need for

coordinate systems or detailed formulas for the Gaussian curvature of surfaces or the geodesic

curvature of curves. These ideas are generalized to higher dimensional manifolds. Topics in

Riemannian and non-Euclidean geometry are introduced and studied.

This lecture supplements material in the textbook: Special Relativity, Electrodynamics and

General Relativity: From Newton to Einstein (ISBN: 978-0-12-813720-8) by John B. Kogut. The

term “textbook” in these Supplemental Lectures will refer to that work.

Keywords: Differential Forms, Classical Differential Geometry, Riemannian Geometry,

E.Cartan, Orthonormal Frames, Gaussian Curvature, Theorem Egregium, Gauss-Bonnet

Theorem, Hypersurfaces, Manifolds, Non-Euclidean Geometry

Contents Strategy and Perspective. ................................................................................................................ 2

Moving Frames ............................................................................................................................... 3

Surfaces in R3 .................................................................................................................................. 6

Gaussian and Mean Curvatures ...................................................................................................... 7

Theorem Egregium ......................................................................................................................... 9

Harmonic Functions ...................................................................................................................... 11

Gauss-Bonnet Theorem: Global Version ...................................................................................... 13

Page 2: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

2

Gauss-Bonnet Theorem: Local Version ........................................................................................ 14

Hypersurfaces in Rn+1 .................................................................................................................... 16

Intrinsic Geometry of Manifolds .................................................................................................. 20

Non-Euclidean Geometry ............................................................................................................. 26

Appendix: Making Contact with Classical Differential and Riemannian Geometry ................... 28

References ..................................................................................................................................... 34

Strategy and Perspective. In our earlier discussions of classical differential geometry, we began by setting up a coordinate

mesh on a smooth surface S embedded in three dimensional Euclidean space R3. We then used

the coordinate mesh to calculate properties of the surface and argued that those properties did not

actually depend on the particular mesh chosen but were properties of the surface itself, some

intrinsic and others dependent on the surface’s embedding in R3. In several cases we found that

this approach obscured the underlying geometry under a flood of coordinate-dependent

derivatives and algebraic detail. The approach was computational oriented, rather than concept

oriented.

By contrast, using differential forms we aim to manipulate geometric objects directly,

find identities between them and only use coordinate systems to quantify those identities when

necessary. A critical element which makes this approach successful and productive is the moving

frames approach introduced by E. Cartan [1]. Specializing to frames traveling on surfaces gives

us differential expressions for the geometry of the surface. The curvatures of curves and surfaces

are formulated from this perspective. No coordinates are necessary and only geometrical

relations result. The Gaussian and mean curvatures are introduced. The Theorem Egregium

follows easily. Minimal surfaces are introduced. The Gauss-Bonnet Theorem is easily derived in

both its global and local forms [2]. It is particularly interesting that the local form of the theorem,

which took considerable detailed work to derive in classical differential geometry drops out here

with little effort as a structural, fundamental result, without the need for detailed formulas for

Page 3: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

3

either the Gaussian or geodesic curvatures. (This success illustrates one of the tenets of M.

Spivak [3] which states that important theorems are those which are supported by formulations

and definitions which make them almost trivial!) This single result illustrates the incisiveness of

using differential forms and frames to study differential geometry. These results are generalized

to hypersurfaces in higher dimensional Euclidean spaces. Finally, using the lessons learned from

these exercises, the intrinsic geometry of manifolds is discussed without the need for an

embedding space and the covariant derivative and Riemann curvature tensor are derived in the

language of differential forms and many of their properties are derived. The non-Euclidean

geometry of the Poincare Upper Half Plane model is illustrated from this point of view.

Moving Frames The use of moving frames in R3 was pioneered by E. Cartan. To begin, consider a point �⃗�𝑥 =

(𝑥𝑥,𝑦𝑦, 𝑧𝑧) in R3 and let there be an orthonormal frame at �⃗�𝑥, {𝑒𝑒1, 𝑒𝑒2, 𝑒𝑒3} [1]. As �⃗�𝑥 moves through R3

we suppose that the frame {𝑒𝑒1, 𝑒𝑒2, 𝑒𝑒3} moves smoothly. Consider an infinitesimal change in �⃗�𝑥:

𝑑𝑑�⃗�𝑥 is a vector whose coefficients are one-forms, as introduced in Supplementary Lecture 12,

𝑑𝑑�⃗�𝑥 = 𝜎𝜎1𝑒𝑒1 + 𝜎𝜎2𝑒𝑒2 + 𝜎𝜎3𝑒𝑒3 (1)

We make the same observations for the triad,

𝑑𝑑𝑒𝑒𝑖𝑖 = 𝜔𝜔𝑖𝑖1𝑒𝑒1 + 𝜔𝜔𝑖𝑖2𝑒𝑒2 + 𝜔𝜔𝑖𝑖3𝑒𝑒3 (2)

The one-forms 𝜔𝜔𝑖𝑖𝑖𝑖 are anti-symmetric in the indices i and j. To show this, differentiate 𝑒𝑒𝑖𝑖 ∙ 𝑒𝑒𝑖𝑖 =

𝛿𝛿𝑖𝑖𝑖𝑖,

𝑑𝑑𝑒𝑒𝑖𝑖 ∙ 𝑒𝑒𝑖𝑖 + 𝑒𝑒𝑖𝑖 ∙ 𝑑𝑑𝑒𝑒𝑖𝑖 = 𝜔𝜔𝑖𝑖𝑖𝑖 + 𝜔𝜔𝑖𝑖𝑖𝑖 = 0 (3)

It is convenient to use matrix notation to write these expressions more efficiently,

𝑒𝑒 = �𝑒𝑒1𝑒𝑒2𝑒𝑒3� , 𝜎𝜎 = (𝜎𝜎1,𝜎𝜎2,𝜎𝜎3) , Ω = �

0 𝜔𝜔12 𝜔𝜔13−𝜔𝜔12 0 𝜔𝜔13−𝜔𝜔13 −𝜔𝜔13 0

� (4)

So, we have,

𝑑𝑑�⃗�𝑥 = 𝜎𝜎𝑒𝑒 , 𝑑𝑑𝑒𝑒 = Ω𝑒𝑒 , Ω = −Ω𝑇𝑇 (5)

Page 4: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

4

where “T” indicates “transpose”. The anti-symmetric character of Ω is expected: it is the

generator of a rotation, an Orthogonal transformation, whose generators have anti-symmetric

representations. This point was discussed in the context of the Frenet-Serret equations in earlier

lectures.

We are clearly not done. In order to integrate the differential equations, 𝑑𝑑�⃗�𝑥 = 𝜎𝜎𝑒𝑒 , 𝑑𝑑𝑒𝑒 =

Ω𝑒𝑒, we need to know how 𝜎𝜎 and Ω vary with �⃗�𝑥. These relations, call “Integrability Conditions”,

follow from 𝑑𝑑2 = 0 for the exterior derivative. First, from 𝑑𝑑(𝑑𝑑�⃗�𝑥) = 0, we learn,

𝑑𝑑(𝑑𝑑�⃗�𝑥) = 𝑑𝑑(𝜎𝜎𝑖𝑖𝑒𝑒𝑖𝑖) = 𝑑𝑑(𝜎𝜎𝑒𝑒) = (𝑑𝑑𝜎𝜎)𝑒𝑒 − 𝜎𝜎 ∧ 𝑑𝑑𝑒𝑒 = 0 (6)

Note that we will be using the Einstein index convention in this lecture, 𝜎𝜎𝑖𝑖𝑒𝑒𝑖𝑖 is short-hand for

∑ 𝜎𝜎𝑖𝑖𝑒𝑒𝑖𝑖𝑖𝑖 . In addition, we have been careful with signs in Eq. 6: for the exterior derivative,

𝑑𝑑(𝜆𝜆 ∧ 𝜇𝜇) = 𝑑𝑑𝜆𝜆 ∧ 𝜇𝜇 + (−1)deg𝜆𝜆𝜆𝜆 ∧ 𝑑𝑑𝜇𝜇. Since 𝑑𝑑𝑒𝑒 = Ω𝑒𝑒, Eq. 6 can be written,

𝑑𝑑(𝑑𝑑�⃗�𝑥) = (𝑑𝑑𝜎𝜎)𝑒𝑒 − 𝜎𝜎 ∧ Ω 𝑒𝑒 = (𝑑𝑑𝜎𝜎 − 𝜎𝜎 ∧ Ω) 𝑒𝑒 = 0 (7)

which implies,

𝑑𝑑𝜎𝜎 = 𝜎𝜎 ∧ Ω (8)

Similarly, from 𝑑𝑑(𝑑𝑑𝑒𝑒) = 0, we learn

𝑑𝑑Ω = Ω ∧ Ω (9)

The first two equalities in Eq. 5 are called “structural equations” and Eq.’s 8 and 9 are

“Integrability Conditions”. A past example of integrability conditions were the Codazzi-

Mainardi equations of classical differential geometry, discussed in Supplementary Lecture 9.

They guaranteed the smoothness of surfaces in R3.

Let’s illustrate the formalism of moving frames with a familiar example: spherical

coordinates, (𝑟𝑟,𝜃𝜃,𝜑𝜑). The position vector reads,

�⃗�𝑥 = 𝑟𝑟(sin𝜃𝜃 cos𝜑𝜑 , sin𝜃𝜃 sin𝜑𝜑 , cos 𝜃𝜃)

And we can construct the moving orthonormal frame,

�̂�𝑒1 =𝜕𝜕�⃗�𝑥𝜕𝜕𝑟𝑟

= (sin𝜃𝜃 cos𝜑𝜑 , sin𝜃𝜃 sin𝜑𝜑 , cos 𝜃𝜃)

Page 5: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

5

�̂�𝑒2 =1𝑟𝑟𝜕𝜕�⃗�𝑥𝜕𝜕𝜃𝜃

= (cos𝜃𝜃 cos𝜑𝜑 , cos 𝜃𝜃 sin𝜑𝜑 ,−sin𝜃𝜃)

�̂�𝑒3 =1

𝑟𝑟 sin𝜃𝜃𝜕𝜕�⃗�𝑥𝜕𝜕𝜑𝜑

= (− sin𝜑𝜑 , cos𝜑𝜑 , 0)

Taking the differential of �⃗�𝑥,

𝑑𝑑�⃗�𝑥 =𝜕𝜕�⃗�𝑥𝜕𝜕𝑟𝑟

𝑑𝑑𝑟𝑟 +𝜕𝜕�⃗�𝑥𝜕𝜕𝜃𝜃

𝑑𝑑𝜃𝜃 +𝜕𝜕�⃗�𝑥𝜕𝜕𝜑𝜑

𝑑𝑑𝜑𝜑 = (𝑑𝑑𝑟𝑟)�̂�𝑒1 + (𝑟𝑟𝑑𝑑𝜃𝜃)�̂�𝑒2 + (𝑟𝑟 sin𝜃𝜃 𝑑𝑑𝜑𝜑)�̂�𝑒3

And we identify,

𝜎𝜎1 = 𝑑𝑑𝑟𝑟 𝜎𝜎2 = 𝑟𝑟 𝑑𝑑𝜃𝜃 𝜎𝜎3 = 𝑟𝑟 sin𝜃𝜃 𝑑𝑑𝜑𝜑

Next, we can differentiate the moving frame and identify the skew-symmetric one-form 𝜔𝜔𝑖𝑖𝑖𝑖,

𝑑𝑑𝑒𝑒𝑖𝑖 = 𝜔𝜔𝑖𝑖𝑖𝑖𝑒𝑒𝑖𝑖,

𝑑𝑑𝑒𝑒1 =𝜕𝜕𝑒𝑒1𝜕𝜕𝑟𝑟

𝑑𝑑𝑟𝑟 +𝜕𝜕𝑒𝑒1𝜕𝜕𝜃𝜃

𝑑𝑑𝜃𝜃 +𝜕𝜕𝑒𝑒1𝜕𝜕𝜑𝜑

𝑑𝑑𝜑𝜑

𝑑𝑑𝑒𝑒1 = 0 + (cos𝜃𝜃 cos𝜑𝜑 , cos 𝜃𝜃 sin𝜑𝜑 ,− sin𝜃𝜃) 𝑑𝑑𝜃𝜃 + (−sin𝜑𝜑 sin 𝜃𝜃 , cos𝜑𝜑 sin𝜃𝜃 , 0) 𝑑𝑑𝜑𝜑

𝑑𝑑𝑒𝑒1 = (𝑑𝑑𝜃𝜃) 𝑒𝑒2 + (sin𝜃𝜃 𝑑𝑑𝜑𝜑) 𝑒𝑒3

And similarly,

𝑑𝑑𝑒𝑒2 = (−𝑑𝑑𝜃𝜃) 𝑒𝑒1 + (cos𝜃𝜃 𝑑𝑑𝜑𝜑) 𝑒𝑒3

𝑑𝑑𝑒𝑒3 = (− sin 𝜃𝜃 𝑑𝑑𝜑𝜑) 𝑒𝑒1 + (− cos 𝜃𝜃 𝑑𝑑𝜑𝜑) 𝑒𝑒2

And we identify from 𝑑𝑑𝑒𝑒 = Ω𝑒𝑒, Eq. 5,

Ω = �𝜔𝜔𝑖𝑖𝑖𝑖� = �0 𝑑𝑑𝜃𝜃 sin𝜃𝜃 𝑑𝑑𝜑𝜑

−𝑑𝑑𝜃𝜃 0 cos 𝜃𝜃 𝑑𝑑𝜑𝜑−sin𝜃𝜃 𝑑𝑑𝜑𝜑 −cos𝜃𝜃 𝑑𝑑𝜑𝜑 0

And finally, the familiar volume element in spherical coordinates is,

𝜎𝜎1 ∧ 𝜎𝜎2 ∧ 𝜎𝜎3 = 𝑟𝑟2 sin𝜃𝜃 𝑑𝑑𝑟𝑟 𝑑𝑑𝜃𝜃 𝑑𝑑𝜑𝜑

Page 6: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

6

Surfaces in R3

Now let’s use moving frames to describe surfaces S in R3 [1]. This is the set-up we

choose: let the frame 𝑒𝑒 moves along the surface with 𝑒𝑒3 always normal to the surface. Then 𝑒𝑒1

and 𝑒𝑒2 span the tangent plane. As �⃗�𝑥 varies across the surface, the orientation of 𝑒𝑒3 over a unit

sphere S2 varies, defining the Gauss map discussed in Supplementary Lecture 9. The Gauss map

and its derivative, the Weingarten map, contain much of the geometry of the surface. Let’s see

how the geometry of the surface is exposed using frames and forms. Note that we will be doing

all of this without setting up a coordinate mesh!

Since �⃗�𝑥 is constrained to the surface S, 𝑑𝑑�⃗�𝑥 lies in the tangent plane at each location, so

𝜎𝜎3 = 0 and,

𝑑𝑑�⃗�𝑥 = 𝜎𝜎1𝑒𝑒1 + 𝜎𝜎2𝑒𝑒2 (10)

and 𝜎𝜎1 ∧ 𝜎𝜎2 is the surface area element on S.

Now consider the equations for the changes in the moving frame 𝑒𝑒 as �⃗�𝑥 varies. We tailor

the notation to surfaces in R3 by writing,

𝑑𝑑𝑒𝑒3 = 𝜔𝜔1𝑒𝑒1 + 𝜔𝜔2𝑒𝑒2 (11)

This notation mirrors Eq. 10 and brings out the geometry in the Gauss map: 𝑆𝑆 → 𝑆𝑆2 provided by

𝑒𝑒3(�⃗�𝑥) and the derivative of the Gauss map, the Weingarten map, which is realized by the linear

transformation between the tangent plane of S at �⃗�𝑥 and the tangent plane of S2 at 𝑒𝑒3(�⃗�𝑥).

Eq. 11 is accompanied by,

𝑑𝑑𝑒𝑒1 = 𝜔𝜔�𝑒𝑒2 − 𝜔𝜔1𝑒𝑒3 , 𝑑𝑑𝑒𝑒2 = −𝜔𝜔�𝑒𝑒1 − 𝜔𝜔2𝑒𝑒3 (12)

So, 𝑑𝑑𝑒𝑒 = Ω𝑒𝑒 with,

Ω = �−0 𝜔𝜔� −𝜔𝜔1𝜔𝜔� 0 −𝜔𝜔2𝜔𝜔1 𝜔𝜔2 0

� (13)

Now consider the integrability conditions, Eq. 8 and 9, 𝑑𝑑𝜎𝜎 = 𝜎𝜎 ∧ Ω and 𝑑𝑑Ω = Ω ∧ Ω. So,

Page 7: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

7

𝑑𝑑𝜎𝜎 = (𝑑𝑑𝜎𝜎1,𝑑𝑑𝜎𝜎2, 0) = (𝜎𝜎1,𝜎𝜎2, 0) ∧ �−0 𝜔𝜔� −𝜔𝜔1𝜔𝜔� 0 −𝜔𝜔2𝜔𝜔1 𝜔𝜔2 0

� =

(−𝜎𝜎2 ∧ 𝜔𝜔�,𝜎𝜎1 ∧ 𝜔𝜔�,𝜔𝜔1 ∧ 𝜎𝜎1 + 𝜔𝜔2 ∧ 𝜎𝜎2)

Writing these results out,

𝑑𝑑𝜎𝜎1 = −𝜎𝜎2 ∧ 𝜔𝜔� , 𝑑𝑑𝜎𝜎2 = 𝜎𝜎1 ∧ 𝜔𝜔� , 𝜔𝜔1 ∧ 𝜎𝜎1 + 𝜔𝜔2 ∧ 𝜎𝜎2 = 0 (14)

Next, 𝑑𝑑Ω = Ω ∧ Ω becomes,

�−0 𝑑𝑑𝜔𝜔� −𝑑𝑑𝜔𝜔1𝑑𝑑𝜔𝜔� 0 −𝑑𝑑𝜔𝜔2𝑑𝑑𝜔𝜔1 𝑑𝑑𝜔𝜔2 0

� = �−0 𝜔𝜔� −𝜔𝜔1𝜔𝜔� 0 −𝜔𝜔2𝜔𝜔1 𝜔𝜔2 0

� ∧ �−0 𝜔𝜔� −𝜔𝜔1𝜔𝜔� 0 −𝜔𝜔2𝜔𝜔1 𝜔𝜔2 0

which implies the equations,

𝑑𝑑𝜔𝜔� + 𝜔𝜔1 ∧ 𝜔𝜔2 = 0 , 𝑑𝑑𝜔𝜔1 = 𝜔𝜔� ∧ 𝜔𝜔2 , 𝑑𝑑𝜔𝜔2 = −𝜔𝜔� ∧ 𝜔𝜔1 (15)

We will see that these equations formulate differential geometry very efficiently.

Gaussian and Mean Curvatures Now let’s introduce and discuss the Gaussian and Mean curvatures on a surface S in the context

of differential forms [1-2]. Gauss introduced the intrinsic curvature K by relating the surface

element of S at �⃗�𝑥, 𝜎𝜎1 ∧ 𝜎𝜎2, to the same quantity on S2, 𝜔𝜔1 ∧ 𝜔𝜔2. Since S is a two dimensional

space, there is only one independent two-form on it. Therefore, 𝜎𝜎1 ∧ 𝜎𝜎2 and 𝜔𝜔1 ∧ 𝜔𝜔2 must be

proportional at each �⃗�𝑥. The proportionality factor, which is also a function of �⃗�𝑥, is the Gaussian

curvature K,

𝜔𝜔1 ∧ 𝜔𝜔2 = 𝐾𝐾 𝜎𝜎1 ∧ 𝜎𝜎2 (16)

This is the fundamental definition of K that was, in fact, the starting point for Gauss himself.

Recall the corresponding equation and discussion in classical differential geometry,

Supplementary Lecture 9, Eq. 55 on page 39,

𝑑𝑑𝑁𝑁��⃗ 𝑢𝑢 × 𝑑𝑑𝑁𝑁��⃗ 𝑣𝑣 = 𝐾𝐾𝑑𝑑𝑟𝑟𝑢𝑢 × 𝑑𝑑𝑟𝑟𝑣𝑣

Page 8: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

8

This relation expresses the same idea as Eq. 16, it relates the surface area element on the surface

to that on the unit sphere, the target space of the Gauss map 𝑁𝑁��⃗ (�⃗�𝑥):𝑆𝑆 → 𝑆𝑆2, but it relies on a

coordinate mesh and vector cross products in R3 for its formulation.

Let’s comment on the dimensions of the quantities in Eq. 16. Each 𝜎𝜎𝑖𝑖 has the dimension

of length as we illustrated in the discussion of spherical coordinates and is apparent in Eq. 10.

But the orthonormal frame 𝑒𝑒 is dimensionless, so Ω = �𝜔𝜔𝑖𝑖𝑖𝑖� is also. Therefore, K has dimension

𝐿𝐿−2. We learned in classical differential geometry that 1 √𝐾𝐾⁄ sets the scale of lengths: for

distances small compare to 1 √𝐾𝐾⁄ , the geometry is essentially Euclidean, but for lengths

comparable to and larger than 1 √𝐾𝐾⁄ , the geometry is curved. We learned in General Relativity

that its space-time manifold is essentially Minkowskian for nearby events in space-time. This

aspect of Riemannian manifolds is called the Equivalence Principle in the language of physicists:

freely falling frames are locally inertial and the rules of special relativity apply.

Let’s return to the discussion of the Gaussian curvature K and the mean curvature H in

R3. There are other two-forms on S. We saw that one candidate 𝜔𝜔1 ∧ 𝜎𝜎1 + 𝜔𝜔2 ∧ 𝜎𝜎2 is in fact zero.

However, 𝜎𝜎1 ∧ 𝜔𝜔2 − 𝜎𝜎2 ∧ 𝜔𝜔1 is non-trivial so it too must be proportional to 𝜎𝜎1 ∧ 𝜎𝜎2. The

proportionality is defined to be twice the mean curvature H, which has the dimension 𝐿𝐿−1,

𝜎𝜎1 ∧ 𝜔𝜔2 − 𝜎𝜎2 ∧ 𝜔𝜔1 = 𝐻𝐻 𝜎𝜎1 ∧ 𝜎𝜎2 (17)

To see that K and H are the familiar curvatures introduced in classical differential

geometry, let’s make the map between the tangent planes of S and S2 (Weingarten map) explicit.

In particular, 𝜔𝜔1 and 𝜔𝜔2 can be written as linear superpositions of 𝜎𝜎1 and 𝜎𝜎2 because the set

{𝜎𝜎1,𝜎𝜎2} is a complete set of one forms on S. So, we can find coefficients, which depend on �⃗�𝑥, so

that,

�𝜔𝜔1𝜔𝜔2� = �

𝑝𝑝 𝑞𝑞𝑞𝑞 𝑟𝑟� �

𝜎𝜎1𝜎𝜎2� (18)

The linear transformation here is symmetric because of the constraint 𝜔𝜔1 ∧ 𝜎𝜎1 + 𝜔𝜔2 ∧ 𝜎𝜎2 = 0

which reads,

𝜎𝜎1 ∧ (𝑝𝑝𝜎𝜎1 + 𝑞𝑞𝜎𝜎2) + 𝜎𝜎2 ∧ (𝑞𝑞𝜎𝜎1 + 𝑟𝑟𝜎𝜎2) = 𝑞𝑞(𝜎𝜎1 ∧ 𝜎𝜎2 + 𝜎𝜎2 ∧ 𝜎𝜎1) = 0 (19)

Page 9: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

9

The symmetric character of the linear transformation guarantees that its eigenvalues are real and

its eigenvectors are perpendicular. Inserting the transformation Eq. 18 into Eq. 16 and 17, the

definitions of K and H, we find,

𝐻𝐻 = 12

(𝑝𝑝 + 𝑟𝑟) 𝐾𝐾 = 𝑝𝑝𝑟𝑟 − 𝑞𝑞2 (20)

So, H is half the trace of the transformation and K is its determinant. Both of these quantities are

independent of the basis leading to this representation. In fact, we can diagonalize the

transformation and label the eigenvalues 𝜅𝜅1 and 𝜅𝜅2. Then,

𝐻𝐻 = 12

(𝜅𝜅1 + 𝜅𝜅2) 𝐾𝐾 = 𝜅𝜅1𝜅𝜅2 (21)

These relations were derived in the textbook and various Supplementary Lectures. They are

essential in differential geometry.

Theorem Egregium We now have the ingredients to show that K is an intrinsic property of the surface S and does not

depend on how it is embedded in R3. In other words, K is determined just by measuring lengths

and angles on the surface S itself. The key relations are 𝑑𝑑𝜔𝜔� + 𝜔𝜔1 ∧ 𝜔𝜔2 = 0 and 𝜔𝜔1 ∧ 𝜔𝜔2 =

𝐾𝐾 𝜎𝜎1 ∧ 𝜎𝜎2, which can be combined,

𝑑𝑑𝜔𝜔� + 𝐾𝐾 𝜎𝜎1 ∧ 𝜎𝜎2 = 0 (22)

So, once we know 𝜔𝜔�, 𝜎𝜎1 and 𝜎𝜎2, we can calculate K. But we know that,

𝑑𝑑𝜎𝜎1 = −𝜎𝜎2 ∧ 𝜔𝜔� , 𝑑𝑑𝜎𝜎2 = 𝜎𝜎1 ∧ 𝜔𝜔�

from Eq. 14, so 𝜔𝜔� can be found given 𝜎𝜎1 and 𝜎𝜎2. In particular 𝜔𝜔� can be written as a linear

superposition of 𝜎𝜎1 and 𝜎𝜎2, so

𝜔𝜔� = 𝑎𝑎𝜎𝜎1 + 𝑏𝑏𝜎𝜎2 (23)

implying,

𝑑𝑑𝜎𝜎1 = 𝑎𝑎𝜎𝜎1 ∧ 𝜎𝜎2 𝑑𝑑𝜎𝜎2 = 𝑏𝑏𝜎𝜎1 ∧ 𝜎𝜎2 (24)

which can be solved for a and b, thereby producing 𝜔𝜔�.

Page 10: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

10

This establishes that K is an intrinsic property of the surface. Explicit formulas for K can

be derived by placing a coordinate mesh on the surface. Let’s carry out this exercise and derive a

famous formula for the curvature K with much(!) less effort than it took us in the context of

classical differential geometry. Recall that if we use an orthogonal mesh on S so that its metric

reads 𝑑𝑑𝑑𝑑2 = 𝐸𝐸𝑑𝑑𝑑𝑑2 + 𝐺𝐺𝑑𝑑𝑑𝑑2, then we derived,

𝐾𝐾 = − 1√𝐸𝐸𝐸𝐸

� 𝜕𝜕𝜕𝜕𝑣𝑣� 1√𝐸𝐸

𝜕𝜕√𝐸𝐸𝜕𝜕𝑣𝑣� + 𝜕𝜕

𝜕𝜕𝑢𝑢� 1√𝐸𝐸

𝜕𝜕√𝐸𝐸𝜕𝜕𝑢𝑢�� (25)

Now let’s use differential forms to obtain the same result in three steps [2]! The surface

is described by a position function �⃗�𝑥(𝑑𝑑, 𝑑𝑑). We make the orthonormal frame 𝑒𝑒1 = �⃗�𝑥𝑢𝑢 √𝐸𝐸⁄ and

𝑒𝑒2 = �⃗�𝑥𝑣𝑣 √𝐺𝐺⁄ with 𝐸𝐸 = �⃗�𝑥𝑢𝑢 ∙ �⃗�𝑥𝑢𝑢 and 𝐺𝐺 = �⃗�𝑥𝑣𝑣 ∙ �⃗�𝑥𝑣𝑣. (In all these formulas the subscript u or v means

differentiation with respect to that coordinate variable: standard vector calculus notation.). Now

we can calculate the associated co-frame, (𝜎𝜎1,𝜎𝜎2), using the definition, 𝑑𝑑�⃗�𝑥 = �⃗�𝑥𝑢𝑢𝑑𝑑𝑑𝑑 + �⃗�𝑥𝑣𝑣𝑑𝑑𝑑𝑑 =

𝜎𝜎1𝑒𝑒1 + 𝜎𝜎2𝑒𝑒2,

𝜎𝜎1 = √𝐸𝐸 𝑑𝑑𝑑𝑑 𝜎𝜎2 = √𝐺𝐺 𝑑𝑑𝑑𝑑 (26)

Next we calculate 𝜔𝜔� from the relations 𝑑𝑑𝜎𝜎1 = −𝜎𝜎2 ∧ 𝜔𝜔� , 𝑑𝑑𝜎𝜎2 = 𝜎𝜎1 ∧ 𝜔𝜔�,

𝑑𝑑𝜎𝜎1 = �√𝐸𝐸�𝑣𝑣𝑑𝑑𝑑𝑑 ∧ 𝑑𝑑𝑑𝑑 = −𝜎𝜎2 ∧ 𝜔𝜔� = −√𝐺𝐺 𝑑𝑑𝑑𝑑 ∧ 𝜔𝜔�

𝑑𝑑𝜎𝜎2 = �√𝐺𝐺�𝑢𝑢𝑑𝑑𝑑𝑑 ∧ 𝑑𝑑𝑑𝑑 = 𝜎𝜎1 ∧ 𝜔𝜔� = √𝐸𝐸 𝑑𝑑𝑑𝑑 ∧ 𝜔𝜔�

which implies,

𝜔𝜔� = −�√𝐸𝐸�𝑣𝑣√𝐸𝐸

𝑑𝑑𝑑𝑑 +�√𝐸𝐸�𝑢𝑢√𝐸𝐸

𝑑𝑑𝑑𝑑 (27)

But we have 𝑑𝑑𝜔𝜔� + 𝐾𝐾 𝜎𝜎1 ∧ 𝜎𝜎2 = 0, and substituting Eq. 26 and 27 into this relation,

𝑑𝑑𝜔𝜔� = −𝜕𝜕𝜕𝜕𝑑𝑑

��√𝐸𝐸�𝑣𝑣√𝐺𝐺

�𝑑𝑑𝑑𝑑 ∧ 𝑑𝑑𝑑𝑑 +𝜕𝜕𝜕𝜕𝑑𝑑

��√𝐺𝐺�𝑢𝑢√𝐸𝐸

�𝑑𝑑𝑑𝑑 ∧ 𝑑𝑑𝑑𝑑

𝑑𝑑𝜔𝜔� = � 𝜕𝜕𝜕𝜕𝑣𝑣��√𝐸𝐸�𝑣𝑣√𝐸𝐸

� + 𝜕𝜕𝜕𝜕𝑢𝑢��√𝐸𝐸�𝑢𝑢√𝐸𝐸

�� 𝑑𝑑𝑑𝑑 ∧ 𝑑𝑑𝑑𝑑 = −𝐾𝐾 𝜎𝜎1 ∧ 𝜎𝜎2 = −√𝐸𝐸𝐺𝐺 𝐾𝐾𝑑𝑑𝑑𝑑 ∧ 𝑑𝑑𝑑𝑑

and Eq. 25 falls out! Done.

Page 11: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

11

Eq. 25 was central to many of the illustrations in Supplementary Lecture 9. The simple

derivation of Eq. 25 using forms shows that manipulating these geometric objects has real

advantages. We will discuss an even better(!) example when we come to the Gauss-Bonnet

Theorem.

Let’s end this section with a few more calculations that will be useful in applications [1].

Consider the vector cross product 𝑑𝑑�⃗�𝑥 × 𝑑𝑑�⃗�𝑥 for 𝑑𝑑�⃗�𝑥 in the tangent plane of a surface. Since 𝑑𝑑�⃗�𝑥 =

𝜎𝜎1𝑒𝑒1 + 𝜎𝜎2𝑒𝑒2 consists of one-forms and vectors, we must be careful with the order of operations

when doing calculations,

𝑑𝑑�⃗�𝑥 × 𝑑𝑑�⃗�𝑥 = (𝜎𝜎1𝑒𝑒1 + 𝜎𝜎2𝑒𝑒2) × (𝜎𝜎1𝑒𝑒1 + 𝜎𝜎2𝑒𝑒2) = 𝜎𝜎1 ∧ 𝜎𝜎2(𝑒𝑒1 × 𝑒𝑒2) + 𝜎𝜎2 ∧ 𝜎𝜎1(𝑒𝑒2 × 𝑒𝑒1)

𝑑𝑑�⃗�𝑥 × 𝑑𝑑�⃗�𝑥 = 2(𝜎𝜎1 ∧ 𝜎𝜎2) 𝑒𝑒3 (28)

where we note the appearance of the vectorial surface area. We can write this expression in terms

of coordinates,

𝑑𝑑�⃗�𝑥 × 𝑑𝑑�⃗�𝑥 = (𝑑𝑑𝑥𝑥,𝑑𝑑𝑦𝑦,𝑑𝑑𝑧𝑧) × (𝑑𝑑𝑥𝑥,𝑑𝑑𝑦𝑦,𝑑𝑑𝑧𝑧) = �𝑖𝑖 𝑗𝑗 𝑘𝑘𝑑𝑑𝑥𝑥 𝑑𝑑𝑦𝑦 𝑑𝑑𝑧𝑧𝑑𝑑𝑥𝑥 𝑑𝑑𝑦𝑦 𝑑𝑑𝑧𝑧

𝑑𝑑�⃗�𝑥 × 𝑑𝑑�⃗�𝑥 = 2(𝑑𝑑𝑦𝑦 ∧ 𝑑𝑑𝑧𝑧,𝑑𝑑𝑧𝑧 ∧ 𝑑𝑑𝑥𝑥,𝑑𝑑𝑥𝑥 ∧ 𝑑𝑑𝑦𝑦) = 2(𝜎𝜎1 ∧ 𝜎𝜎2) 𝑒𝑒3 (29)

Similarly,

𝑑𝑑�⃗�𝑥 × 𝑑𝑑𝑒𝑒3 = 2𝐻𝐻(𝜎𝜎1 ∧ 𝜎𝜎2) 𝑒𝑒3

And,

𝑑𝑑𝑒𝑒3 × 𝑑𝑑𝑒𝑒3 = 2𝐾𝐾(𝜎𝜎1 ∧ 𝜎𝜎2) 𝑒𝑒3

These expressions and ones like them will be useful in our next discussion on harmonic

functions.

Harmonic Functions Let’s do a simple application of these ideas [1]. We want to find the shape of minimal surfaces.

These are surfaces which have vanishing mean curvatures. Soap bubbles are examples of

Page 12: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

12

minimal surfaces. Soap bubbles can be supported by loops, as every child knows, and can also be

created without a boundary.

First let’s recall the expression of the Laplacian, ∇2= 𝜕𝜕2 𝜕𝜕𝑥𝑥2⁄ + 𝜕𝜕2 𝜕𝜕𝑦𝑦2⁄ + 𝜕𝜕2 𝜕𝜕𝑧𝑧2⁄ , in

the language of differential forms. Given a function in R3, we have its differential,

𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑥𝑥𝑑𝑑𝑥𝑥 + 𝑑𝑑𝑦𝑦𝑑𝑑𝑦𝑦 + 𝑑𝑑𝑧𝑧𝑑𝑑𝑧𝑧

where 𝑑𝑑𝑥𝑥 = 𝜕𝜕𝑑𝑑 𝜕𝜕𝑥𝑥⁄ , etc. Take the Hodge star of this one form in R3,

∗ 𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑥𝑥𝑑𝑑𝑦𝑦 ∧ 𝑑𝑑𝑧𝑧 + 𝑑𝑑𝑦𝑦𝑑𝑑𝑧𝑧 ∧ 𝑑𝑑𝑥𝑥 + 𝑑𝑑𝑧𝑧𝑑𝑑𝑥𝑥 ∧ 𝑑𝑑𝑦𝑦

And finally, applying the exterior derivative to this result,

𝑑𝑑 ∗ 𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑥𝑥𝑥𝑥 𝑑𝑑𝑥𝑥 ∧ 𝑑𝑑𝑦𝑦 ∧ 𝑑𝑑𝑧𝑧 + 𝑑𝑑𝑦𝑦𝑦𝑦 𝑑𝑑𝑦𝑦 ∧ 𝑑𝑑𝑧𝑧 ∧ 𝑑𝑑𝑥𝑥 + 𝑑𝑑𝑧𝑧𝑧𝑧 𝑑𝑑𝑧𝑧 ∧ 𝑑𝑑𝑥𝑥 ∧ 𝑑𝑑𝑦𝑦

𝑑𝑑 ∗ 𝑑𝑑𝑑𝑑 = (𝑑𝑑𝑥𝑥𝑥𝑥 + 𝑑𝑑𝑥𝑥𝑥𝑥 + 𝑑𝑑𝑥𝑥𝑥𝑥) 𝑑𝑑𝑥𝑥 ∧ 𝑑𝑑𝑦𝑦 ∧ 𝑑𝑑𝑧𝑧 = (∇2𝑑𝑑) 𝜎𝜎

Let’s apply this result to functions defined on the surface S. Then its differential is, using

a coordinate mesh {𝑑𝑑, 𝑑𝑑},

𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑢𝑢𝑑𝑑𝑑𝑑 + 𝑑𝑑𝑣𝑣𝑑𝑑𝑑𝑑

Then applying the Hodge operator in R2,

∗ 𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑢𝑢𝑑𝑑𝑑𝑑 − 𝑑𝑑𝑣𝑣𝑑𝑑𝑑𝑑

(Be careful with signs: In R2, ∗ 𝑑𝑑𝑑𝑑 = −𝑑𝑑𝑑𝑑 and, ∗ 𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑑𝑑. Consult the previous Supplementary

Lecture 12 for an introduction to the Hodge duality * operator.). And finally, apply the exterior

derivative,

𝑑𝑑 ∗ 𝑑𝑑𝑑𝑑 = 𝑑𝑑(𝑑𝑑𝑢𝑢𝑑𝑑𝑑𝑑 − 𝑑𝑑𝑣𝑣𝑑𝑑𝑑𝑑) = (𝑑𝑑𝑢𝑢𝑢𝑢 + 𝑑𝑑𝑣𝑣𝑣𝑣) 𝑑𝑑𝑑𝑑 ∧ 𝑑𝑑𝑑𝑑 = (∇2𝑑𝑑)𝑑𝑑𝑑𝑑 ∧ 𝑑𝑑𝑑𝑑

The same ideas apply to vectors on S, 𝑑𝑑�⃗�𝑥 = 𝜎𝜎1𝑒𝑒1 + 𝜎𝜎2𝑒𝑒2, so ∗ 𝑑𝑑�⃗�𝑥 = 𝜎𝜎2𝑒𝑒1 − 𝜎𝜎1𝑒𝑒2. Notice

that,

𝑑𝑑�⃗�𝑥 × 𝑒𝑒3 = (𝜎𝜎1𝑒𝑒1 + 𝜎𝜎2𝑒𝑒2) × 𝑒𝑒3 = 𝜎𝜎2𝑒𝑒1 − 𝜎𝜎1𝑒𝑒2

So, we have the identity,

∗ 𝑑𝑑�⃗�𝑥 = 𝑑𝑑�⃗�𝑥 × 𝑒𝑒3

Page 13: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

13

And, using the remark after Eq. 29 above, 𝑑𝑑�⃗�𝑥 × 𝑑𝑑𝑒𝑒3 = 2𝐻𝐻(𝜎𝜎1 ∧ 𝜎𝜎2) 𝑒𝑒3,

𝑑𝑑 ∗ 𝑑𝑑�⃗�𝑥 = −𝑑𝑑�⃗�𝑥 × 𝑑𝑑𝑒𝑒3 = −2𝐻𝐻(𝜎𝜎1 ∧ 𝜎𝜎2) 𝑒𝑒3

Collecting,

∇2�⃗�𝑥 = (∇2𝑥𝑥,∇2𝑦𝑦,∇2𝑧𝑧) = −2𝐻𝐻𝑒𝑒3

This is the result we wanted: A minimal surface (𝐻𝐻 = 0, locally) is a surface whose coordinate

fuctions are harmonic. So, each point on the surface is a saddle point.

There is much more to be said about minimal surfaces from the perspectives of geometry

and the calculus of variations (minimax principles). The reader should consult the references in

[2] to get started.

Gauss-Bonnet Theorem: Global Version Let’s return to the Gauss map, �⃗�𝑥 → 𝑒𝑒3, taking S to S2. The Gaussian curvature K is defined as the

ratio of the area elements on S (𝜎𝜎1 ∧ 𝜎𝜎2) and S2 (𝜔𝜔1 ∧ 𝜔𝜔2) [1-2],

𝜔𝜔1 ∧ 𝜔𝜔2 = 𝐾𝐾 𝜎𝜎1 ∧ 𝜎𝜎2 (16)

As �⃗�𝑥 varies over S, 𝑒𝑒3 varies over S2. By elementary theorems in topology, we know that if �⃗�𝑥

varies over S, then 𝑒𝑒3 varies over S2 an integral number of times, called the degree of the map, n.

So, integrating Eq. 16,

∫ 𝐾𝐾 𝜎𝜎1 ∧ 𝜎𝜎2 = 4𝜋𝜋 𝑛𝑛𝑆𝑆 (30)

which is a form of the Global Gauss-Bonnet Theorem derived in Supplementary Lecture 3 and

11 in the context of traditional classical differential geometry. In particular, if S is a closed,

convex surface, then 𝑒𝑒3 varies over S2 exactly once, so then

∫ 𝐾𝐾 𝜎𝜎1 ∧ 𝜎𝜎2 = 4𝜋𝜋 𝑆𝑆

The generalization to a smooth surface with g holes (genus) is,

∫ 𝐾𝐾 𝜎𝜎1 ∧ 𝜎𝜎2 = 4𝜋𝜋 𝑆𝑆 (1 − 𝑔𝑔) (31)

Page 14: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

14

In particular, for a torus, ∫ 𝐾𝐾𝜎𝜎1 ∧ 𝜎𝜎2𝑆𝑆 = 0 which we showed explicitly in Supplementary Lecture

11 where K was calculated from the principle curvatures on the surface, 𝐾𝐾 = 𝜅𝜅1𝜅𝜅2, and the

integral was done by elementary means.

The power of Eq. 31, the global form of the Gauss-Bonnet Theorem, is that it holds even

as S is distorted and K changes smoothly. The theorem expresses a global topological invariant

which is the Euler characteristic of the surface. See previous Supplementary Lectures for more!

Gauss-Bonnet Theorem: Local Version There is another very useful form of the Gauss-Bonnet Theorem, pioneered the French

mathematician by Pierre Ossian Bonnet, which applies to regions on a surface that are bounded

by a curve C. If the curve is smooth, continuous and differentiable, then it reads [2],

∫ 𝐾𝐾 𝑑𝑑𝑎𝑎 + ∫ 𝜅𝜅𝑔𝑔 𝑑𝑑𝑑𝑑 = 2𝜋𝜋𝜕𝜕𝜕𝜕𝜕𝜕 (32)

where M labels a connected region on the surface, 𝜕𝜕𝜕𝜕 is its boundary, the curve C, and 𝜅𝜅𝑔𝑔 is its

geodesic curvature. We used this form of the Gauss-Bonnet Theorem and its generalization to

piece-wise smooth curves (curves with a finite number of “kinks”, abrupt changes in direction:

think of a triangle) to discuss the non-Euclidean geometry on curved surfaces. The traditional

derivations of Eq. 32 are very detailed and require the formulas for K and 𝜅𝜅𝑔𝑔 on a coordinate

mesh describing the surface. By contrast, differential forms provide a different perspective of the

theorem and allow a short, incisive derivation which does not require such detail. The crucial

element in the proof is simply the definition of K and the integrability condition,

𝑑𝑑𝜔𝜔� + 𝜔𝜔1 ∧ 𝜔𝜔2 = 𝑑𝑑𝜔𝜔� + 𝐾𝐾𝜎𝜎1 ∧ 𝜎𝜎2 = 0 (33)

To begin, let’s describe the curve C that bounds the region M. Along the curve there is a

unit tangent 𝑡𝑡(𝑑𝑑), parametrized by arc-length s, and there is a second unit vector 𝑘𝑘�⃗ which lies in

the tangent plane to the surface at that location and is perpendicular to 𝑡𝑡(𝑑𝑑). The rate of change

of 𝑡𝑡(𝑑𝑑), 𝑑𝑑𝑡𝑡 𝑑𝑑𝑑𝑑⁄ , has a component in the tangent plane which points in the direction 𝑘𝑘�⃗ and has a

magnitude which defines the geodesic curvature 𝜅𝜅𝑔𝑔. We can write the rate of change of both 𝑡𝑡(𝑑𝑑)

and 𝑘𝑘�⃗ (𝑑𝑑), projected onto the tangent plane,

Page 15: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

15

𝐷𝐷𝑑𝑑𝑑𝑑�𝑡𝑡𝑘𝑘�⃗� = �

0 𝜅𝜅𝑔𝑔−𝜅𝜅𝑔𝑔 0 � �

𝑡𝑡𝑘𝑘�⃗� (34)

where we use the notation 𝐷𝐷 𝑑𝑑𝑑𝑑⁄ to indicate the projection of 𝑑𝑑 𝑑𝑑𝑑𝑑⁄ onto the tangent plane. (This

is the covariant derivative, as introduced in the lectures on classical differential geometry and

Riemannian manifolds.) The 2 × 2 matrix in Eq. 34 is anti-symmetric to guarantee that the pair

�𝑡𝑡(𝑑𝑑),𝑘𝑘�⃗ (𝑑𝑑) � remain orthonormal as s varies. We also can relate �𝑡𝑡(𝑑𝑑),𝑘𝑘�⃗ (𝑑𝑑) � to the pair of

orthonormal vectors that define the coordinate mesh, {𝑒𝑒1, 𝑒𝑒2}: they are related by a rotation,

�𝑒𝑒1𝑒𝑒2� = �cos 𝜃𝜃 − sin𝜃𝜃

sin𝜃𝜃 cos𝜃𝜃 � �𝑡𝑡𝑘𝑘�⃗�

So, the rate of change of {𝑒𝑒1, 𝑒𝑒2} along C, projected onto the tangent plane is,

𝐷𝐷𝑒𝑒1𝑑𝑑𝑑𝑑

= �−𝑡𝑡 sin𝜃𝜃 − 𝑘𝑘�⃗ cos 𝜃𝜃� 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

+ 𝐷𝐷𝑡𝑡𝑑𝑑𝑑𝑑

cos 𝜃𝜃 − 𝐷𝐷𝑘𝑘�⃗

𝑑𝑑𝑑𝑑sin 𝜃𝜃

𝐷𝐷𝑒𝑒1𝑑𝑑𝑑𝑑

= − 𝑒𝑒2𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

+ �𝑘𝑘�⃗ 𝜅𝜅𝑔𝑔� cos 𝜃𝜃 − �−𝑡𝑡 𝜅𝜅𝑔𝑔� sin𝜃𝜃

𝐷𝐷𝑒𝑒1𝑑𝑑𝑑𝑑

= − 𝑒𝑒2𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

+ �𝑡𝑡 sin𝜃𝜃 + 𝑘𝑘�⃗ cos 𝜃𝜃�𝜅𝜅𝑔𝑔 = − 𝑒𝑒2𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

+ 𝑒𝑒2𝜅𝜅𝑔𝑔 = 𝑒𝑒2 �𝜅𝜅𝑔𝑔 −𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑�

But we had an expression, Eq. 12, for the rate of change of 𝑒𝑒1 in terms of the one-forms on the

surface,

𝑑𝑑𝑒𝑒1 = 𝜔𝜔�𝑒𝑒2 − 𝜔𝜔1𝑒𝑒3

So, we identify,

𝜔𝜔� = �𝜅𝜅𝑔𝑔 −𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑� 𝑑𝑑𝑑𝑑 (35)

The Gauss-Bonnet Theorem Eq. 32 now follows from Stoke’s Theorem,

�𝑑𝑑𝜔𝜔� = �𝜔𝜔�𝜕𝜕𝜕𝜕𝜕𝜕

Using Eq. 33,

− �𝐾𝐾𝜎𝜎1 ∧ 𝜎𝜎2 = ��𝜅𝜅𝑔𝑔𝑑𝑑𝑑𝑑 − 𝑑𝑑𝜃𝜃�𝜕𝜕𝜕𝜕𝜕𝜕

Page 16: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

16

which we collect,

�𝐾𝐾 𝑑𝑑𝑎𝑎𝜕𝜕

+ �𝜅𝜅𝑔𝑔𝑑𝑑𝑑𝑑 = �𝑑𝑑𝜃𝜃𝜕𝜕𝜕𝜕𝜕𝜕𝜕𝜕

= 2𝜋𝜋

and we are done! The simplicity of this derivation indicates that the Gauss-Bonnet Theorem is a

structural statement about surfaces and is fundamental. It is discussed at length in other lectures

in this series.

Hypersurfaces in Rn+1

Now let’s turn to hypersurfaces in Rn+1 before we discuss Riemannian geometry [1]. A

hypersurface means a n-dimensional manifold M. At the point �⃗�𝑥 on M, the orthonormal set

{𝑒𝑒1, 𝑒𝑒2, … , 𝑒𝑒𝑛𝑛} spans its tangent space. If 𝑛𝑛� is the unit normal to the tangent plane at �⃗�𝑥, then

{𝑒𝑒1, 𝑒𝑒2, … , 𝑒𝑒𝑛𝑛,𝑛𝑛�⃗ } make up an orthonormal basis of Rn+1. Since 𝑑𝑑�⃗�𝑥 lies in the tangent plane, one

can write,

𝑑𝑑�⃗�𝑥 = 𝜎𝜎1𝑒𝑒1 + 𝜎𝜎2𝑒𝑒2 + ⋯+ 𝜎𝜎𝑛𝑛𝑒𝑒𝑛𝑛 (36)

where 𝜎𝜎𝑖𝑖 are one-forms on M.

Now let’s turn to the variation of members of the basis, {𝑒𝑒1, 𝑒𝑒2, … , 𝑒𝑒𝑛𝑛,𝑛𝑛�⃗ }. Since,

𝑒𝑒𝑖𝑖 ∙ 𝑒𝑒𝑘𝑘 = 𝛿𝛿𝑖𝑖𝑘𝑘 𝑒𝑒𝑖𝑖 ∙ 𝑛𝑛�⃗ = 0 𝑛𝑛�⃗ ∙ 𝑛𝑛�⃗ = 1

we have the differentials,

𝑑𝑑𝑒𝑒𝑖𝑖 ∙ 𝑒𝑒𝑘𝑘 + 𝑒𝑒𝑖𝑖 ∙ 𝑑𝑑𝑒𝑒𝑘𝑘 = 0 𝑑𝑑𝑒𝑒𝑖𝑖 ∙ 𝑛𝑛�⃗ + 𝑒𝑒𝑖𝑖 ∙ 𝑑𝑑𝑛𝑛�⃗ = 0 𝑛𝑛�⃗ ∙ 𝑑𝑑𝑛𝑛�⃗ = 0

So, if we write,

𝑑𝑑𝑒𝑒𝑖𝑖 = 𝜔𝜔𝑖𝑖𝑖𝑖𝑒𝑒𝑖𝑖 − 𝜔𝜔𝑖𝑖𝑛𝑛�⃗ (37)

where 𝜔𝜔𝑖𝑖𝑖𝑖 are one-forms on M and 𝜔𝜔𝑖𝑖 are one-forms on Rn+1. But 𝑑𝑑𝑒𝑒𝑖𝑖 ∙ 𝑒𝑒𝑘𝑘 + 𝑒𝑒𝑖𝑖 ∙ 𝑑𝑑𝑒𝑒𝑘𝑘 = 0 implies

that

𝜔𝜔𝑖𝑖𝑖𝑖 + 𝜔𝜔𝑖𝑖𝑖𝑖 = 0

And 𝑑𝑑𝑒𝑒𝑖𝑖 ∙ 𝑛𝑛�⃗ + 𝑒𝑒𝑖𝑖 ∙ 𝑑𝑑𝑛𝑛�⃗ = 0 and 𝑛𝑛�⃗ ∙ 𝑑𝑑𝑛𝑛�⃗ = 0, imply,

Page 17: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

17

𝑑𝑑𝑛𝑛�⃗ = 𝜔𝜔𝑖𝑖𝑒𝑒𝑖𝑖

As in our discussion on surfaces in R3, it is convenient to define matrices,

𝑒𝑒 =

⎜⎜⎜⎛

𝑒𝑒1𝑒𝑒2....𝑒𝑒𝑛𝑛⎠

⎟⎟⎟⎞

, �⃗�𝜎 = (𝜎𝜎1,𝜎𝜎2, … ,𝜎𝜎𝑛𝑛) , 𝜔𝜔��⃗ = (𝜔𝜔1,𝜔𝜔2, … ,𝜔𝜔𝑛𝑛)

And,

Ω = �𝜔𝜔𝑖𝑖𝑖𝑖�

Then we can write,

𝑑𝑑�⃗�𝑥 = �⃗�𝜎𝑒𝑒 , 𝑑𝑑 �𝑒𝑒𝑛𝑛�⃗� = �Ω −𝜔𝜔��⃗ 𝑇𝑇

𝜔𝜔��⃗ 0� �𝑒𝑒𝑛𝑛�⃗� , Ω𝑇𝑇 = −Ω (38)

Next we can write out the integrability condition,

0 = 𝑑𝑑(𝑑𝑑�⃗�𝑥) = (𝑑𝑑�⃗�𝜎)𝑒𝑒 − �⃗�𝜎(𝑑𝑑𝑒𝑒) = (𝑑𝑑�⃗�𝜎)𝑒𝑒 − �⃗�𝜎 ∧ (Ω𝑒𝑒 − 𝜔𝜔��⃗ 𝑇𝑇𝑛𝑛�⃗ ) = (𝑑𝑑�⃗�𝜎 − �⃗�𝜎 ∧ Ω)𝑒𝑒 + (�⃗�𝜎 ∧ 𝜔𝜔��⃗ 𝑇𝑇)𝑛𝑛�⃗

So,

𝑑𝑑�⃗�𝜎 = �⃗�𝜎 ∧ Ω , �⃗�𝜎 ∧ 𝜔𝜔��⃗ 𝑇𝑇 = 0 (39)

Next,

0 = 𝑑𝑑 �𝑑𝑑 �𝑒𝑒𝑛𝑛�⃗�� = �dΩ −𝑑𝑑𝜔𝜔��⃗ 𝑇𝑇

𝑑𝑑𝜔𝜔��⃗ 0� �𝑒𝑒𝑛𝑛�⃗� − �Ω −𝜔𝜔��⃗ 𝑇𝑇

𝜔𝜔��⃗ 0�𝑑𝑑 �𝑒𝑒

𝑛𝑛�⃗�

= �dΩ −𝑑𝑑𝜔𝜔��⃗ 𝑇𝑇𝑑𝑑𝜔𝜔��⃗ 0

� �𝑒𝑒𝑛𝑛�⃗� − �Ω −𝜔𝜔��⃗ 𝑇𝑇

𝜔𝜔��⃗ 0�2�𝑒𝑒𝑛𝑛�⃗�

= �dΩ − Ω ∧ Ω + 𝜔𝜔��⃗ 𝑇𝑇 ∧ 𝜔𝜔 −𝑑𝑑𝜔𝜔��⃗ 𝑇𝑇 + Ω ∧ 𝜔𝜔��⃗ 𝑇𝑇𝑑𝑑𝜔𝜔 − 𝜔𝜔 ∧ Ω 𝜔𝜔 ∧ 𝜔𝜔��⃗ 𝑇𝑇

� �𝑒𝑒𝑛𝑛�⃗�

Collecting,

dΩ − Ω ∧ Ω + 𝜔𝜔��⃗ 𝑇𝑇 ∧ 𝜔𝜔 = 0 , 𝑑𝑑𝜔𝜔 = 𝜔𝜔 ∧ Ω (40)

Page 18: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

18

The combination dΩ − Ω ∧ Ω occurs frequently and is significant geometrically, as we will see,

so define,

Θ = �𝜃𝜃𝑖𝑖𝑖𝑖� = dΩ − Ω ∧ Ω (41)

which satisfies,

Θ + 𝜔𝜔��⃗ 𝑇𝑇 ∧ 𝜔𝜔 = 0 (42)

Collecting everything in terms of entries in the matrix,

𝑑𝑑𝜎𝜎𝑖𝑖 = 𝜎𝜎𝑖𝑖 ∧ 𝜔𝜔𝑖𝑖𝑖𝑖 , 𝜔𝜔𝑖𝑖𝑖𝑖 = −𝜔𝜔𝑖𝑖𝑖𝑖 , 𝜎𝜎𝑖𝑖 ∧ 𝜔𝜔𝑖𝑖 = 0 (43)

And,

𝑑𝑑𝜔𝜔𝑖𝑖 = 𝜔𝜔𝑖𝑖 ∧ 𝜔𝜔𝑖𝑖𝑖𝑖 , 𝜃𝜃𝑖𝑖𝑖𝑖 + 𝜔𝜔𝑖𝑖 ∧ 𝜔𝜔𝑖𝑖 = 0 (44)

Since the set {𝜎𝜎𝑖𝑖} forms a basis of one-forms on M, we can write,

𝜔𝜔𝑖𝑖 = 𝑏𝑏𝑖𝑖𝑖𝑖𝜎𝜎𝑖𝑖 (45)

Since �⃗�𝜎 ∧ 𝜔𝜔��⃗ 𝑇𝑇 = 𝜎𝜎𝑖𝑖 ∧ 𝜔𝜔𝑖𝑖 = 0, we learn that 𝑏𝑏𝑖𝑖𝑖𝑖 is symmetric, 𝑏𝑏𝑖𝑖𝑖𝑖 = 𝑏𝑏𝑖𝑖𝑖𝑖.

Generalizing from our earlier discussion of surfaces S in R3, we define the mean

curvature H and the Gaussian curvature K,

𝐻𝐻 =1𝑛𝑛𝑏𝑏𝑖𝑖𝑖𝑖 𝐾𝐾 = 𝑑𝑑𝑒𝑒𝑡𝑡�𝑏𝑏𝑖𝑖𝑖𝑖�

It follows from Eq. 45 that the volume element,

𝜔𝜔1 ∧ 𝜔𝜔2 ∧ …∧ 𝜔𝜔𝑛𝑛 = 𝑏𝑏1𝑖𝑖𝜎𝜎𝑖𝑖 ∧ 𝑏𝑏2𝑘𝑘𝜎𝜎𝑘𝑘 ∧ …∧ 𝑏𝑏𝑛𝑛𝑛𝑛𝜎𝜎𝑛𝑛 = 𝑑𝑑𝑒𝑒𝑡𝑡�𝑏𝑏𝑖𝑖𝑖𝑖�𝜎𝜎1 ∧ 𝜎𝜎2 ∧ …∧ 𝜎𝜎𝑛𝑛

= 𝐾𝐾𝜎𝜎1 ∧ 𝜎𝜎2 ∧ …∧ 𝜎𝜎𝑛𝑛

and K represents the ratio of areas on M and Sn.

Now consider a vector field �⃗�𝑑 on M. �⃗�𝑑 is tangent to M, so,

�⃗�𝑑 = 𝑐𝑐𝑖𝑖𝑒𝑒𝑖𝑖

which we can differentiate,

Page 19: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

19

𝑑𝑑�⃗�𝑑 = 𝑑𝑑𝑐𝑐𝑖𝑖𝑒𝑒𝑖𝑖 + 𝑐𝑐𝑖𝑖𝑑𝑑𝑒𝑒𝑖𝑖 = 𝑑𝑑𝑐𝑐𝑖𝑖𝑒𝑒𝑖𝑖 + 𝑐𝑐𝑖𝑖�𝜔𝜔𝑖𝑖𝑖𝑖𝑒𝑒𝑖𝑖 − 𝜔𝜔𝑖𝑖𝑛𝑛�⃗ � = �𝑑𝑑𝑐𝑐𝑖𝑖 + 𝑐𝑐𝑖𝑖𝜔𝜔𝑖𝑖𝑖𝑖�𝑒𝑒𝑖𝑖 − 𝑐𝑐𝑖𝑖𝜔𝜔𝑖𝑖𝑛𝑛�⃗

where we used Eq. 37. We learn that if �⃗�𝑑 does not change on the surface, then

𝑑𝑑𝑐𝑐𝑖𝑖 + 𝑐𝑐𝑖𝑖𝜔𝜔𝑖𝑖𝑖𝑖 = 0 (46)

If this is true for the vector field �⃗�𝑑, we say that �⃗�𝑑 moves by “parallel transport” along the surface.

This is the same concept discussed at length in the textbook and in several Supplementary

Lectures, especially #9. Using the same analysis as in our discussion of surfaces in R3, we see

that two vector fields, �⃗�𝑑 and 𝑤𝑤��⃗ , which move by parallel transport along M, have a constant inner

product, �⃗�𝑑 ∙ 𝑤𝑤��⃗ .

In addition, suppose that 𝑃𝑃�⃗ (𝑑𝑑) is a curve on M, where s is the curve’s arc-length, and

𝑡𝑡(𝑑𝑑) = 𝑑𝑑𝑃𝑃�⃗ 𝑑𝑑𝑑𝑑⁄ is the curve’s unit tangent, then the curve is a geodesic if 𝑡𝑡(𝑑𝑑) moves by parallel

transport. We discussed the geometry underlying this terminology in Supplementary Lecture 9: if

the tangent moves by parallel transport, then the curve is as straight as possible on the

hypersurface M.

Now consider the curvature matrix Θ. Its matrix elements 𝜃𝜃𝑖𝑖𝑖𝑖 are two-forms so they can

be written as linear superpositions of �𝜎𝜎𝑖𝑖,𝜎𝜎𝑖𝑖�. The coefficients make up the famous Riemann

curvature tensor,

𝜃𝜃𝑖𝑖𝑖𝑖 = 12𝑅𝑅𝑖𝑖𝑖𝑖𝑘𝑘𝑛𝑛𝜎𝜎𝑘𝑘 ∧ 𝜎𝜎𝑛𝑛 , 𝑅𝑅𝑖𝑖𝑖𝑖𝑘𝑘𝑛𝑛 + 𝑅𝑅𝑖𝑖𝑖𝑖𝑛𝑛𝑘𝑘 = 0 (47)

We will check later that 𝑅𝑅𝑖𝑖𝑖𝑖𝑘𝑘𝑛𝑛 is the Riemann curvature tensor introduced in the textbook and

Supplementary Lectures in terms of parallel transport and holonomy. This will be done after

introducing Christoffel connections.

Back to the business at hand. We had the defining relations, Eq. 44,

𝜃𝜃𝑖𝑖𝑖𝑖 + 𝜔𝜔𝑖𝑖 ∧ 𝜔𝜔𝑖𝑖 = 0

So, we can relate this to Eq. 47 using the fact, Eq. 45, that 𝜔𝜔𝑖𝑖 = 𝑏𝑏𝑖𝑖𝑖𝑖𝜎𝜎𝑖𝑖,

Page 20: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

20

𝜔𝜔𝑖𝑖 ∧ 𝜔𝜔𝑖𝑖 = 𝑏𝑏𝑖𝑖𝑘𝑘𝑏𝑏𝑖𝑖𝑛𝑛 𝜎𝜎𝑘𝑘 ∧ 𝜎𝜎𝑛𝑛 =12�𝑏𝑏𝑖𝑖𝑘𝑘𝑏𝑏𝑖𝑖𝑛𝑛 − 𝑏𝑏𝑖𝑖𝑛𝑛𝑏𝑏𝑖𝑖𝑘𝑘�𝜎𝜎𝑘𝑘 ∧ 𝜎𝜎𝑛𝑛

So the Riemann curvature tensor satisfies,

𝑅𝑅𝑖𝑖𝑖𝑖𝑘𝑘𝑛𝑛 + 𝑑𝑑𝑒𝑒𝑡𝑡 �𝑏𝑏𝑖𝑖𝑘𝑘 𝑏𝑏𝑖𝑖𝑛𝑛𝑏𝑏𝑖𝑖𝑘𝑘 𝑏𝑏𝑖𝑖𝑛𝑛

� = 0 (48)

This result gives the symmetry relations of 𝑅𝑅𝑖𝑖𝑖𝑖𝑘𝑘𝑛𝑛,

𝑅𝑅𝑖𝑖𝑖𝑖𝑘𝑘𝑛𝑛 + 𝑅𝑅𝑖𝑖𝑖𝑖𝑛𝑛𝑘𝑘 = 0 , 𝑅𝑅𝑖𝑖𝑖𝑖𝑘𝑘𝑛𝑛 + 𝑅𝑅𝑖𝑖𝑖𝑖𝑘𝑘𝑛𝑛 = 0

𝑅𝑅𝑖𝑖𝑖𝑖𝑘𝑘𝑛𝑛 + 𝑅𝑅𝑖𝑖𝑘𝑘𝑛𝑛𝑖𝑖 + 𝑅𝑅𝑖𝑖𝑛𝑛𝑖𝑖𝑘𝑘 = 0

𝑅𝑅𝑖𝑖𝑖𝑖𝑘𝑘𝑛𝑛 = 𝑅𝑅𝑘𝑘𝑛𝑛𝑖𝑖𝑖𝑖

Now let’s briefly discuss the equations,

𝑑𝑑�⃗�𝜎 = �⃗�𝜎 ∧ Ω , Ω + Ω𝑇𝑇 = 0

We will see in the next section that they determine Ω. This means that Ω is determined by �⃗�𝜎

alone. Given this result, it will follow that,

Θ = dΩ − Ω ∧ Ω

is also completely determined by �⃗�𝜎. In other words, the Riemann curvature tensor is an intrinsic

property of the surface S. More below.

Intrinsic Geometry of Manifolds Now let’s ask what aspects of the geometry of the manifold M exist independent of any

Euclidean space it might be embedded in [1]. In this case we must postulate the existence of a

local geometry, we cannot inherit it from Rn+1. Of course, the reason we discussed manifolds

embedded in Rn+1 was to motivate this step and to make the best definitions!

So, let M be a n-dimensional manifold. The manifold is endowed with an inner product,

so if �⃗�𝑑 and 𝑤𝑤��⃗ are two tangent vectors, then �⃗�𝑑 ∙ 𝑤𝑤��⃗ is a given, smooth real function. Suppose that

there is a basis {𝑒𝑒1, 𝑒𝑒2, … , 𝑒𝑒𝑛𝑛} which is orthonormal, 𝑒𝑒𝑖𝑖 ∙ 𝑒𝑒𝑖𝑖 = 𝛿𝛿𝑖𝑖𝑖𝑖. If P is a point on M, then we

postulate,

Page 21: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

21

𝒅𝒅𝑃𝑃 = 𝜎𝜎𝑖𝑖𝑒𝑒𝑖𝑖 (49)

where the differential operator 𝒅𝒅 will prove to be the covariant derivative introduced earlier in

the context of classical differential geometry (Some authors use the notation D here, but we will

follow the conventions of [1] and use a bold face 𝒅𝒅. See the appendix to this lecture for more.) If

there is a coordinate system {𝑑𝑑1,𝑑𝑑2, … ,𝑑𝑑𝑛𝑛} on M, then,

𝒅𝒅𝑃𝑃 = 𝑑𝑑𝑑𝑑𝑖𝑖𝜕𝜕𝜕𝜕𝑑𝑑𝑖𝑖

and � 𝜕𝜕𝜕𝜕𝑢𝑢1

, 𝜕𝜕𝜕𝜕𝑢𝑢2

, … , 𝜕𝜕𝜕𝜕𝑢𝑢𝑛𝑛

� is a natural frame associated with the coordinate system. A standard

application of the chain rule shows that 𝒅𝒅𝑃𝑃 is independent of the choice of a particular

coordinate system.

Now we have to potulate how 𝑒𝑒𝑖𝑖 changes under a displacement 𝒅𝒅. In this case there is no

“normal direction”, so we postulate,

𝒅𝒅𝑒𝑒𝑖𝑖 = 𝜔𝜔𝑖𝑖𝑖𝑖𝑒𝑒𝑖𝑖 (50)

The inspiration for this postulate, the defining condition for 𝒅𝒅, is Eq. 37. If we were in Rn+1,we

would describe 𝒅𝒅 as the projection of the differential in Rn+1 onto the tangent plane, which is, in

fact, the definition of the covariant derivative. With this understanding, much of the

developments here can be borrowed from our earlier discussions of hypersurfaces in Rn+1.

Now we need to find the one-forms 𝜔𝜔𝑖𝑖𝑖𝑖 which are consistent with the integrability

conditions,

𝒅𝒅𝑒𝑒𝑖𝑖 ∙ 𝑒𝑒𝑘𝑘 + 𝑒𝑒𝑖𝑖 ∙ 𝒅𝒅𝑒𝑒𝑘𝑘 = 0 , 𝒅𝒅(𝒅𝒅𝑃𝑃) = 0

The first condition implies 𝜔𝜔𝑖𝑖𝑖𝑖 + 𝜔𝜔𝑖𝑖𝑖𝑖 = 0, as before. The second condition reads,

𝒅𝒅(𝜎𝜎𝑖𝑖𝑒𝑒𝑖𝑖) = 0

So,

𝒅𝒅𝜎𝜎𝑖𝑖𝑒𝑒𝑖𝑖 − 𝜎𝜎𝑖𝑖𝒅𝒅𝑒𝑒𝑖𝑖 = �𝒅𝒅𝜎𝜎𝑖𝑖 − 𝜎𝜎𝑖𝑖 ∧ 𝜔𝜔𝑖𝑖𝑖𝑖�𝑒𝑒𝑖𝑖 = 0

which implies,

Page 22: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

22

𝒅𝒅𝜎𝜎𝑖𝑖 = 𝜎𝜎𝑖𝑖 ∧ 𝜔𝜔𝑖𝑖𝑖𝑖 (51)

So, our problem at hand is to solve 𝒅𝒅𝜎𝜎𝑖𝑖 = 𝜎𝜎𝑖𝑖 ∧ 𝜔𝜔𝑖𝑖𝑖𝑖 subject to the condition 𝜔𝜔𝑖𝑖𝑖𝑖 + 𝜔𝜔𝑖𝑖𝑖𝑖 = 0. First,

{𝜎𝜎𝑘𝑘} is a complete set of one-forms, so there must be coefficients, called connection coefficients

Γ𝑖𝑖𝑖𝑖𝑘𝑘, or Christoffel symbols,

𝜔𝜔𝑖𝑖𝑖𝑖 = Γ𝑖𝑖𝑖𝑖𝑘𝑘𝜎𝜎𝑘𝑘

We will see later that Γ𝑖𝑖𝑖𝑖𝑘𝑘 coincide with the Christoffel symbols introduced earlier in the

textbook and Supplementary Lectures. The anti-symmetry of 𝜔𝜔𝑖𝑖𝑖𝑖 implies,

Γ𝑖𝑖𝑖𝑖𝑘𝑘 + Γ𝑖𝑖𝑖𝑖𝑘𝑘 = 0

But the 𝒅𝒅𝜎𝜎𝑖𝑖 are known because the 𝜎𝜎𝑖𝑖 are known. We can write out completeness again,

𝒅𝒅𝜎𝜎𝑖𝑖 =12𝑐𝑐𝑖𝑖𝑖𝑖𝑘𝑘𝜎𝜎𝑖𝑖 ∧ 𝜎𝜎𝑘𝑘 , 𝑐𝑐𝑖𝑖𝑖𝑖𝑘𝑘 + 𝑐𝑐𝑖𝑖𝑘𝑘𝑖𝑖 = 0

Then,

𝒅𝒅𝜎𝜎𝑖𝑖 =12𝑐𝑐𝑖𝑖𝑖𝑖𝑘𝑘𝜎𝜎𝑖𝑖 ∧ 𝜎𝜎𝑘𝑘 = 𝜎𝜎𝑖𝑖 ∧ 𝜔𝜔𝑖𝑖𝑖𝑖 = Γ𝑖𝑖𝑖𝑖𝑘𝑘 𝜎𝜎𝑖𝑖 ∧ 𝜎𝜎𝑘𝑘 =

12�Γ𝑖𝑖𝑖𝑖𝑘𝑘 − Γ𝑘𝑘𝑖𝑖𝑖𝑖�𝜎𝜎𝑖𝑖 ∧ 𝜎𝜎𝑘𝑘

So,

Γ𝑖𝑖𝑖𝑖𝑘𝑘 − Γ𝑘𝑘𝑖𝑖𝑖𝑖 = 𝑐𝑐𝑖𝑖𝑖𝑖𝑘𝑘

We can solve this system of equations for Γ𝑖𝑖𝑖𝑖𝑘𝑘 by writing down the two other cyclic permutations

of the indices (𝑗𝑗𝑖𝑖𝑘𝑘 → 𝑘𝑘𝑗𝑗𝑖𝑖 → 𝑖𝑖𝑘𝑘𝑗𝑗), adding the first two and subtracting the third to isolate Γ𝑘𝑘𝑖𝑖𝑖𝑖.

The result is,

Γ𝑖𝑖𝑖𝑖𝑘𝑘 =12�𝑐𝑐𝑖𝑖𝑘𝑘𝑖𝑖 + 𝑐𝑐𝑖𝑖𝑖𝑖𝑘𝑘 − 𝑐𝑐𝑘𝑘𝑖𝑖𝑖𝑖�

All of this will become more familiar when we turn to metric spaces below.

Let’s derive a further integrability condition by calculating the exterior derivative of

𝑑𝑑�⃗�𝜎 = �⃗�𝜎 ∧ Ω. Now,

0 = 𝑑𝑑(𝑑𝑑�⃗�𝜎) = 𝑑𝑑�⃗�𝜎 ∧ Ω − �⃗�𝜎 ∧ 𝑑𝑑Ω = (�⃗�𝜎 ∧ Ω) ∧ Ω − �⃗�𝜎 ∧ 𝑑𝑑Ω = �⃗�𝜎 ∧ (Ω ∧ Ω − dΩ)

Page 23: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

23

Therefore,

�⃗�𝜎 ∧ Θ = 0

where Θ is the curvature form introduced earlier,

Θ = dΩ − Ω ∧ Ω

The exterior derivative of Θ is informative,

𝑑𝑑Θ = d(dΩ) − 𝑑𝑑(Ω ∧ Ω) = 0 − (dΩ) ∧ Ω + Ω ∧ (dΩ)

= (−Θ + Ω ∧ Ω) ∧ Ω + Ω ∧ (Θ + Ω ∧ Ω) = Ω ∧ Θ − Θ ∧ Ω

which comprise the “Bianchi identities” which were introduced briefly in the textbook.

Finally, we should consider the rate of change of 𝒅𝒅𝑒𝑒𝑖𝑖,

𝒅𝒅𝟐𝟐𝑒𝑒𝑖𝑖 = 𝒅𝒅(𝒅𝒅𝑒𝑒𝑖𝑖) = 𝒅𝒅(Ω ∧ 𝑒𝑒𝑖𝑖) = 𝑑𝑑Ω ∧ 𝑒𝑒𝑖𝑖 − Ω ∧ 𝒅𝒅𝑒𝑒𝑖𝑖 = (dΩ − Ω ∧ Ω)𝑒𝑒𝑖𝑖 = Θ𝑒𝑒𝑖𝑖

where we have identified the curvature form again. So, the “acceleration” of the frame is

proportional to the curvature.

As discussed in Eq. 47, the two-forms 𝜃𝜃𝑖𝑖𝑖𝑖 which make up the curvature matrix, Θ =

�𝜃𝜃𝑖𝑖𝑖𝑖�, can be expanded in terms of the basis �𝜎𝜎𝑖𝑖 ∧ 𝜎𝜎𝑖𝑖�. The coefficients of the completeness

statement will be identified as the components of the Riemann tensor, 𝑅𝑅𝑖𝑖𝑖𝑖𝑘𝑘𝑛𝑛,

𝜃𝜃𝑖𝑖𝑖𝑖 = 12𝑅𝑅𝑖𝑖𝑖𝑖𝑘𝑘𝑛𝑛𝜎𝜎𝑘𝑘 ∧ 𝜎𝜎𝑛𝑛 (47)

Several of the important symmetries of 𝑅𝑅𝑖𝑖𝑖𝑖𝑘𝑘𝑛𝑛 follow directly from this equation,

𝑅𝑅𝑖𝑖𝑖𝑖𝑘𝑘𝑛𝑛 + 𝑅𝑅𝑖𝑖𝑖𝑖𝑛𝑛𝑘𝑘 = 0 , 𝑅𝑅𝑖𝑖𝑖𝑖𝑘𝑘𝑛𝑛 + 𝑅𝑅𝑖𝑖𝑖𝑖𝑘𝑘𝑛𝑛 = 0 (52.a)

In addition, the integrability condition �⃗�𝜎 ∧ Θ = 0 becomes,

𝑅𝑅𝑖𝑖𝑖𝑖𝑘𝑘𝑛𝑛𝜎𝜎𝑖𝑖 ∧ 𝜎𝜎𝑖𝑖 ∧ 𝜎𝜎𝑘𝑘 = 0

which implies that,

𝑅𝑅𝑖𝑖𝑖𝑖𝑘𝑘𝑛𝑛 + 𝑅𝑅𝑖𝑖𝑘𝑘𝑛𝑛𝑖𝑖 + 𝑅𝑅𝑖𝑖𝑛𝑛𝑖𝑖𝑘𝑘 = 0 (52.b)

The final symmetry,

Page 24: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

24

𝑅𝑅𝑖𝑖𝑖𝑖𝑘𝑘𝑛𝑛 = 𝑅𝑅𝑘𝑘𝑛𝑛𝑖𝑖𝑖𝑖

follows from combining Eq. 52.a and 52.b.

Now let’s introduce the metric. We have the natural frame ��⃗�𝑑1 = 𝜕𝜕𝜕𝜕𝑢𝑢1

, … , �⃗�𝑑𝑛𝑛 = 𝜕𝜕𝜕𝜕𝑢𝑢𝑛𝑛

�.

Then the entries in the metric tensor are defined to be the inner products,

𝑔𝑔𝑖𝑖𝑖𝑖 = �⃗�𝑑𝑖𝑖 ∙ �⃗�𝑑𝑖𝑖

so the invariant distance element is,

𝑑𝑑𝑑𝑑2 = 𝒅𝒅𝑃𝑃 ∙ 𝒅𝒅𝑃𝑃 = ��⃗�𝑑𝑖𝑖𝑑𝑑𝑑𝑑𝑖𝑖� ∙ ��⃗�𝑑𝑖𝑖𝑑𝑑𝑑𝑑𝑖𝑖� = �⃗�𝑑𝑖𝑖 ∙ �⃗�𝑑𝑖𝑖 𝑑𝑑𝑑𝑑𝑖𝑖𝑑𝑑𝑑𝑑𝑖𝑖 = 𝑔𝑔𝑖𝑖𝑖𝑖𝑑𝑑𝑑𝑑𝑖𝑖𝑑𝑑𝑑𝑑𝑖𝑖

Next we write 𝒅𝒅�⃗�𝑑𝑖𝑖 in terms of the complete set {�⃗�𝑑𝑖𝑖},

𝒅𝒅�⃗�𝑑𝑖𝑖 = 𝜂𝜂𝑖𝑖 𝑖𝑖�⃗�𝑑𝑖𝑖

where �𝜂𝜂𝑖𝑖 𝑖𝑖� are one-forms which can be written as linear superpositions of �𝑑𝑑𝑑𝑑𝑖𝑖�

𝜂𝜂𝑖𝑖 𝑖𝑖 = Γ 𝑖𝑖𝑘𝑘

𝑖𝑖 𝑑𝑑𝑑𝑑𝑘𝑘

Now let’s find Γ𝑖𝑖𝑘𝑘𝑖𝑖 in terms of the metric and its derivatives to show that Γ𝑖𝑖𝑘𝑘

𝑖𝑖 is indeed the

Christoffel symbols introduced earlier. First, differentiating 𝑔𝑔𝑖𝑖𝑖𝑖 = �⃗�𝑑𝑖𝑖 ∙ �⃗�𝑑𝑖𝑖,

𝑑𝑑𝑔𝑔𝑖𝑖𝑖𝑖 = 𝒅𝒅�⃗�𝑑𝑖𝑖 ∙ �⃗�𝑑𝑖𝑖 + �⃗�𝑑𝑖𝑖 ∙ 𝒅𝒅�⃗�𝑑𝑖𝑖

We have,

𝑑𝑑𝑔𝑔𝑖𝑖𝑖𝑖 = 𝜂𝜂𝑖𝑖 𝑘𝑘�⃗�𝑑𝑘𝑘 ∙ �⃗�𝑑𝑖𝑖 + �⃗�𝑑𝑖𝑖 ∙ 𝜂𝜂𝑖𝑖 𝑘𝑘�⃗�𝑑𝑘𝑘 = 𝜂𝜂𝑖𝑖 𝑘𝑘𝑔𝑔𝑘𝑘𝑖𝑖 + 𝜂𝜂𝑖𝑖 𝑘𝑘𝑔𝑔𝑖𝑖𝑘𝑘 = Γ 𝑖𝑖𝑛𝑛𝑘𝑘 𝑑𝑑𝑑𝑑𝑛𝑛𝑔𝑔𝑘𝑘𝑖𝑖 + Γ 𝑖𝑖𝑛𝑛

𝑘𝑘 𝑑𝑑𝑑𝑑𝑛𝑛𝑔𝑔𝑖𝑖𝑘𝑘

𝑑𝑑𝑔𝑔𝑖𝑖𝑖𝑖 = �Γ 𝑖𝑖𝑛𝑛𝑘𝑘 𝑔𝑔𝑘𝑘𝑖𝑖 + Γ 𝑖𝑖𝑛𝑛

𝑘𝑘 𝑔𝑔𝑖𝑖𝑘𝑘�𝑑𝑑𝑑𝑑𝑛𝑛

So, we learn,

𝜕𝜕𝑔𝑔𝑖𝑖𝑖𝑖𝜕𝜕𝑢𝑢𝑙𝑙

= Γ 𝑖𝑖𝑛𝑛𝑘𝑘 𝑔𝑔𝑘𝑘𝑖𝑖 + Γ 𝑖𝑖𝑛𝑛

𝑘𝑘 𝑔𝑔𝑖𝑖𝑘𝑘 (53)

But requiring,

0 = 𝒅𝒅(𝒅𝒅𝑃𝑃) = −𝑑𝑑𝑑𝑑𝑖𝑖 𝒅𝒅�⃗�𝑑𝑖𝑖

Page 25: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

25

we learn a symmetry property of the Γ𝑖𝑖𝑛𝑛𝑘𝑘,

0 = −𝑑𝑑𝑑𝑑𝑖𝑖 𝒅𝒅�⃗�𝑑𝑖𝑖 = −𝑑𝑑𝑑𝑑𝑖𝑖𝜂𝜂𝑖𝑖 𝑖𝑖�⃗�𝑑𝑖𝑖 = Γ 𝑖𝑖𝑘𝑘

𝑖𝑖 𝑑𝑑𝑑𝑑𝑖𝑖 ∧ 𝑑𝑑𝑑𝑑𝑘𝑘�⃗�𝑑𝑖𝑖

which implies,

Γ 𝑖𝑖𝑘𝑘𝑖𝑖 = Γ 𝑘𝑘𝑖𝑖

𝑖𝑖

We lower indices with

Γ𝑖𝑖𝑖𝑖𝑘𝑘 = 𝑔𝑔𝑖𝑖𝑛𝑛Γ 𝑖𝑖𝑘𝑘𝑛𝑛

So Eq. 53 becomes,

𝜕𝜕𝑔𝑔𝑖𝑖𝑖𝑖𝜕𝜕𝑢𝑢𝑙𝑙

= Γ𝑖𝑖𝑖𝑖𝑛𝑛 + Γ𝑖𝑖𝑖𝑖𝑛𝑛 (54)

We have met this set of equations before and know how to solve it for a particular Γ𝑖𝑖𝑖𝑖𝑘𝑘: We write

the equation and then obtain two more by cyclically permuting 𝑙𝑙𝑖𝑖𝑗𝑗 twice. Taking the sum of the

first two and subtracting the third isolates a single Γ𝑖𝑖𝑖𝑖𝑛𝑛 and we find,

Γ𝑖𝑖𝑖𝑖𝑘𝑘 =12�𝜕𝜕𝑘𝑘𝑔𝑔𝑖𝑖𝑖𝑖 + 𝜕𝜕𝑖𝑖𝑔𝑔𝑖𝑖𝑘𝑘 − 𝜕𝜕𝑖𝑖𝑔𝑔𝑖𝑖𝑘𝑘�

where 𝜕𝜕𝑘𝑘 = 𝜕𝜕 𝜕𝜕𝑑𝑑𝑘𝑘⁄ . This is precisely the formula for the Christoffel symbols that we derived in

the tensor analysis section in the textbook.

Next, let’s discuss parallel transport in this scenario. If �⃗�𝑑 is a tangent vector in M, then

�⃗�𝑑 = 𝑐𝑐𝑖𝑖𝑒𝑒𝑖𝑖

where 𝑐𝑐𝑖𝑖 will depend on the position P. The derivative of �⃗�𝑑 is,

𝒅𝒅𝒗𝒗��⃗ = 𝑑𝑑𝑐𝑐𝑖𝑖𝑒𝑒𝑖𝑖 + 𝒅𝒅𝑒𝑒𝑖𝑖 = 𝑑𝑑𝑐𝑐𝑖𝑖𝑒𝑒𝑖𝑖 + 𝑐𝑐𝑖𝑖𝜔𝜔𝑖𝑖𝑖𝑖𝑒𝑒𝑖𝑖 = �𝑑𝑑𝑐𝑐𝑖𝑖 + 𝑐𝑐𝑖𝑖𝜔𝜔𝑖𝑖𝑖𝑖�𝑒𝑒𝑖𝑖

So, if �⃗�𝑑 moves by parallel transport, meaning, 𝒅𝒅𝒗𝒗��⃗ = 0, then

𝑑𝑑𝑐𝑐𝑖𝑖 + 𝑐𝑐𝑖𝑖𝜔𝜔𝑖𝑖𝑖𝑖 = 0 , 𝑜𝑜𝑟𝑟 𝜕𝜕𝑐𝑐𝑖𝑖𝜕𝜕𝑢𝑢𝑘𝑘

+ Γ𝑖𝑖𝑖𝑖𝑘𝑘𝑐𝑐𝑖𝑖 = 0 (55)

which is in fact the formula for parallel transport derived in the textbook.

Page 26: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

26

Non-Euclidean Geometry Let’s apply this formalism [1] to the geometry of the Poincare Upper-Half Plane model of a

space with a constant negative Gaussian curvature, 𝐾𝐾 = −1. This topic was discussed in

Supplementary Lecture 4 using complex analysis.

Recall the metric for the upper half plane, (𝑦𝑦 ≥ 0, ∞ ≥ 𝑥𝑥 ≥ −∞), model

𝑑𝑑𝑑𝑑2 = 𝑑𝑑𝑥𝑥2+𝑑𝑑𝑦𝑦2

𝑦𝑦2 (56.a)

So,

𝑔𝑔11 = 𝑦𝑦−2 , 𝑔𝑔12 = 0 , 𝑔𝑔22 = 𝑦𝑦−2 (56.b)

An orthonormal basis for the space will be,

𝑒𝑒1 = (𝑦𝑦, 0) and 𝑒𝑒2 = (0,𝑦𝑦) (57)

Let’s check that 𝑒𝑒𝑖𝑖 ∙ 𝑒𝑒𝑖𝑖 = 𝛿𝛿𝑖𝑖𝑖𝑖 .First,

𝑒𝑒1 ∙ 𝑒𝑒1 = 𝑦𝑦2(1,0) ∙ (1,0) = 𝑦𝑦2𝑔𝑔11 = 1

And similarly 𝑒𝑒1 ∙ 𝑒𝑒2 = 0 and 𝑒𝑒2 ∙ 𝑒𝑒2 = 1.

If we write,

𝒅𝒅𝑃𝑃 = (𝑑𝑑𝑥𝑥,𝑑𝑑𝑦𝑦) = 𝜎𝜎𝑖𝑖𝑒𝑒𝑖𝑖

then,

𝜎𝜎1 = 𝑑𝑑𝑥𝑥𝑦𝑦

and 𝜎𝜎2 = 𝑑𝑑𝑥𝑥𝑦𝑦

(58)

Now we can start investigating the geometry of this space. The integrability conditions

read,

𝑑𝑑𝜎𝜎1 = −1𝑦𝑦2𝑑𝑑𝑦𝑦 ∧ 𝑑𝑑𝑥𝑥 =

1𝑦𝑦2𝑑𝑑𝑥𝑥 ∧ 𝑑𝑑𝑦𝑦 = 𝜎𝜎1 ∧ 𝜎𝜎2 , 𝑑𝑑𝜎𝜎2 = 0

We can identify the matric Ω from its definition,

𝑑𝑑�⃗�𝜎 = �⃗�𝜎 ∧ Ω

Page 27: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

27

Explicitly,

(𝑑𝑑𝜎𝜎1,𝑑𝑑𝜎𝜎2) = (𝜎𝜎1,𝜎𝜎2) ∧ Ω

Or,

(𝜎𝜎1 ∧ 𝜎𝜎2, 0) = (𝜎𝜎1,𝜎𝜎2) ∧ Ω

So, we infer,

Ω = �𝜔𝜔𝑖𝑖𝑖𝑖� = � 0 𝜎𝜎1−𝜎𝜎1 0 �

using the fact that Ω is anti-symmetric to infer ω12 = 𝜎𝜎1.

Now we can calculate the curvature matrix,

Θ = dΩ − Ω ∧ Ω = � 0 𝑑𝑑𝜎𝜎1−𝑑𝑑𝜎𝜎1 0 � − � 0 𝜎𝜎1

−𝜎𝜎1 0 � ∧ �0 𝜎𝜎1−𝜎𝜎1 0 � = � 0 1

−1 0� 𝑑𝑑𝜎𝜎1

= � 0 1−1 0� 𝜎𝜎1 ∧ 𝜎𝜎2

But recall the definition 𝜃𝜃𝑖𝑖𝑖𝑖 = 12𝑅𝑅𝑖𝑖𝑖𝑖𝑘𝑘𝑛𝑛 𝜎𝜎𝑘𝑘 ∧ 𝜎𝜎𝑛𝑛. So, we read off

𝜃𝜃12 =12𝑅𝑅1212 𝜎𝜎1 ∧ 𝜎𝜎2 +

12𝑅𝑅1221 𝜎𝜎2 ∧ 𝜎𝜎1 = 𝑅𝑅1212 𝜎𝜎1 ∧ 𝜎𝜎2

So, 𝑅𝑅1212 = 1. Now we can obtain the Gaussian intrinsic curvature K using the integrability

condition 𝑑𝑑𝜔𝜔� + 𝐾𝐾𝜎𝜎1 ∧ 𝜎𝜎2 = 0 and 𝜔𝜔� = 𝜔𝜔12 = 𝜎𝜎1. So, 𝑑𝑑𝜔𝜔� = 𝑑𝑑𝜎𝜎1 = 𝜎𝜎1 ∧ 𝜎𝜎2 which implies 𝐾𝐾 =

−1. We learn that 𝐾𝐾 = −1 = −𝑅𝑅1212 and the space has a constant negative curvature, as

claimed.

Finally let’s check that semi-circles with their centers on the real axis are geodesics in the

upper-half plane. We discussed this in Supplementary Lecture 4 using complex analysis and it

would be instructive to see how it works out here. In (𝑥𝑥,𝑦𝑦) coordinates, the geodesics read,

𝑥𝑥 = 𝑎𝑎 + 𝑟𝑟 cos𝜑𝜑, 𝑦𝑦 = 𝑟𝑟 sin𝜑𝜑

for 0 ≤ 𝜑𝜑 ≤ 𝜋𝜋. We need the tangent to 𝑃𝑃 = �𝑥𝑥(𝜑𝜑),𝑦𝑦(𝜑𝜑)�,

𝒅𝒅𝑃𝑃𝑑𝑑𝜑𝜑

= 𝑟𝑟(− sin𝜑𝜑, cos𝜑𝜑) =𝑟𝑟𝑦𝑦

[(− sin𝜑𝜑)𝑒𝑒1 + (cos𝜑𝜑)𝑒𝑒2] =1

sin𝜑𝜑[(− sin𝜑𝜑)𝑒𝑒1 + (cos𝜑𝜑)𝑒𝑒2]

Page 28: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

28

But along the curve,

𝑑𝑑𝑑𝑑2 =𝑑𝑑𝑥𝑥2 + 𝑑𝑑𝑦𝑦2

𝑦𝑦2=

(𝑟𝑟2 sin2 𝜑𝜑 + 𝑟𝑟2 cos2 𝜑𝜑)(𝑑𝑑𝜑𝜑)2

𝑟𝑟2 sin2 𝜑𝜑=

(𝑑𝑑𝜑𝜑)2

sin2 𝜑𝜑

So, the unit tangent vector is,

𝑡𝑡 =𝒅𝒅𝑃𝑃𝑑𝑑𝑑𝑑

=𝒅𝒅𝑃𝑃𝑑𝑑𝜑𝜑

𝑑𝑑𝜑𝜑𝑑𝑑𝑑𝑑

= (− sin𝜑𝜑)𝑒𝑒1 + (cos𝜑𝜑)𝑒𝑒2

But along the curve,

𝜔𝜔12 = 𝜎𝜎1 =𝑑𝑑𝑥𝑥𝑦𝑦

=−𝑟𝑟 sin𝜑𝜑 𝑑𝑑𝜑𝜑𝑟𝑟 sin 𝜑𝜑

= −𝑑𝑑𝜑𝜑 = − sin 𝜑𝜑𝑑𝑑𝑑𝑑

And,

𝒅𝒅𝑒𝑒 = Ω𝑒𝑒 = � 0 𝜔𝜔12−𝜔𝜔12 0 � �𝑒𝑒1

𝑒𝑒2�

So,

𝒅𝒅𝑒𝑒1 = −𝑑𝑑𝜑𝜑 𝑒𝑒2 𝒅𝒅𝑒𝑒2 = 𝑑𝑑𝜑𝜑 𝑒𝑒1

And finally,

𝒅𝒅�⃗�𝒕 = 𝒅𝒅[(− sin𝜑𝜑)𝑒𝑒1 + (cos𝜑𝜑)𝑒𝑒2]

= − cos 𝜑𝜑 𝑑𝑑𝜑𝜑 𝑒𝑒1 − sin𝜑𝜑 𝑑𝑑𝜑𝜑 𝑒𝑒2 − sin𝜑𝜑(−𝑑𝑑𝜑𝜑𝑒𝑒2) + cos𝜑𝜑(𝑑𝑑𝜑𝜑 𝑒𝑒1) = 0

So, the unit tangent �⃗�𝒕 moves by parallel transport and the curve, the semi-circle with center on

the real axis, is indeed a geodesic.

Appendix: Making Contact with Classical Differential and Riemannian

Geometry The presentation of differential forms in this lecture has been very brief because many of the

concepts and formulas are already familiar from past discussions of classical differential

geometry and tensor analysis. Let’s make the relations more explicit.

Page 29: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

29

Consider a hypersurface in Rn+1. Then points on the n-dimensional surface S are located

with a vector 𝑅𝑅�⃗ (𝑑𝑑1, … ,𝑑𝑑𝑛𝑛), using �𝑑𝑑𝑖𝑖� for a smooth, singularity-free n-dimensional mesh. The

tangent space at 𝑅𝑅�⃗ is spanned by vectors 𝑒𝑒𝑖𝑖 = 𝜕𝜕𝑅𝑅�⃗ 𝜕𝜕𝑑𝑑𝑖𝑖⁄ . The partial derivatives with respect to the

mesh coordinates produce vectors spanned by {𝑒𝑒𝑖𝑖} and 𝑛𝑛�⃗ , the unit normal to S at 𝑅𝑅�⃗ . Using

notation that follows the conventions in the textbook,

𝜕𝜕𝑖𝑖𝑒𝑒𝑖𝑖 = 𝑒𝑒𝑘𝑘Γ𝑖𝑖𝑖𝑖𝑘𝑘 + 𝑛𝑛�⃗ 𝑏𝑏𝑖𝑖𝑖𝑖 (A.1)

where 𝜕𝜕𝑖𝑖 = 𝜕𝜕 𝜕𝜕𝑑𝑑𝑖𝑖⁄ , Γ𝑖𝑖𝑖𝑖𝑘𝑘 are the Christoffel connections and 𝑏𝑏𝑖𝑖𝑖𝑖 will be identified as the second

fundamental form of the hyperspace S. We can easily see that Γ𝑖𝑖𝑖𝑖𝑘𝑘 and 𝑏𝑏𝑖𝑖𝑖𝑖 are symmetric in their

indices i and j. This follows from the identity, an integrability condition,

𝜕𝜕𝑖𝑖𝑒𝑒𝑖𝑖 =𝜕𝜕2𝑅𝑅�⃗

𝜕𝜕𝑑𝑑𝑖𝑖𝜕𝜕𝑑𝑑𝑖𝑖=

𝜕𝜕2𝑅𝑅�⃗𝜕𝜕𝑑𝑑𝑖𝑖𝜕𝜕𝑑𝑑𝑖𝑖

= 𝜕𝜕𝑖𝑖𝑒𝑒𝑖𝑖

So,

𝜕𝜕𝑖𝑖𝑒𝑒𝑖𝑖 = 𝜕𝜕𝑖𝑖𝑒𝑒𝑖𝑖 = 𝑒𝑒𝑘𝑘Γ𝑖𝑖𝑖𝑖𝑘𝑘 + 𝑛𝑛�⃗ 𝑏𝑏𝑖𝑖𝑖𝑖 = 𝑒𝑒𝑘𝑘Γ𝑖𝑖𝑖𝑖𝑘𝑘 + 𝑛𝑛�⃗ 𝑏𝑏𝑖𝑖𝑖𝑖

which proves the point.

Next we need to make definitions to help identify intrinsic properties of S. An observer

on S would measure the 𝑒𝑒𝑘𝑘Γ𝑖𝑖𝑖𝑖𝑘𝑘 terms in 𝜕𝜕𝑖𝑖𝑒𝑒𝑖𝑖 but the variation normal to the surface 𝑛𝑛�⃗ 𝑏𝑏𝑖𝑖𝑖𝑖 would be

out of his sight. So, we introduce the Covariant Derivative to pick out the terms in Eq. A.1 lying

in the tangent space,

𝐷𝐷𝑒𝑒𝑖𝑖 = 𝐷𝐷𝑒𝑒𝑖𝑖𝜕𝜕𝑢𝑢𝑖𝑖

𝑑𝑑𝑑𝑑𝑖𝑖 = 𝑒𝑒𝑘𝑘Γ𝑖𝑖𝑖𝑖𝑘𝑘𝑑𝑑𝑑𝑑𝑖𝑖 (A.2)

The result is a differential form and 𝑑𝑑 is the exterior derivative. We define 𝐷𝐷 applied to a scalar

function to be the ordinary differential, 𝐷𝐷𝑑𝑑 = 𝑑𝑑𝑑𝑑. And when 𝐷𝐷 applies to a vector, we define,

𝐷𝐷�⃗�𝑑 = 𝐷𝐷�𝑒𝑒𝑖𝑖𝑑𝑑𝑖𝑖� = 𝑒𝑒𝑖𝑖 𝑑𝑑𝑑𝑑𝑖𝑖 + (𝐷𝐷𝑒𝑒𝑖𝑖)𝑑𝑑𝑖𝑖 = 𝑒𝑒𝑖𝑖 𝜕𝜕𝑣𝑣𝑖𝑖

𝜕𝜕𝑢𝑢𝑖𝑖𝑑𝑑𝑑𝑑𝑖𝑖 + 𝑒𝑒𝑘𝑘Γ𝑖𝑖𝑖𝑖𝑘𝑘𝑑𝑑𝑖𝑖𝑑𝑑𝑑𝑑𝑖𝑖 = 𝑒𝑒𝑘𝑘 �

𝜕𝜕𝑣𝑣𝑘𝑘

𝜕𝜕𝑢𝑢𝑖𝑖+ Γ𝑖𝑖𝑖𝑖𝑘𝑘𝑑𝑑𝑖𝑖� 𝑑𝑑𝑑𝑑𝑖𝑖 (A.3)

which is a vector-valued differential form. We learn that the components of the Covariant

Derivative are,

Page 30: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

30

𝐷𝐷𝑣𝑣𝑘𝑘

𝜕𝜕𝑢𝑢𝑖𝑖= 𝜕𝜕𝑣𝑣𝑘𝑘

𝜕𝜕𝑢𝑢𝑖𝑖+ Γ𝑖𝑖𝑖𝑖𝑘𝑘𝑑𝑑𝑖𝑖 (A.4)

which indeed is the covariant derivative introduced in the textbook discussion of tensor analysis.

To see that we can use D to formulate the curvature of S, consider the second covariant

derivative of 𝑒𝑒𝑖𝑖. Here we must consistently use the definition of the exterior derivative 𝑑𝑑,

𝐷𝐷2𝑒𝑒𝑖𝑖 = 𝐷𝐷(𝐷𝐷𝑒𝑒𝑖𝑖) = 𝐷𝐷�𝑒𝑒𝑘𝑘Γ𝑖𝑖𝑖𝑖𝑘𝑘𝑑𝑑𝑑𝑑𝑖𝑖� = 𝑒𝑒𝑘𝑘𝑑𝑑�Γ𝑖𝑖𝑖𝑖𝑘𝑘𝑑𝑑𝑑𝑑𝑖𝑖� + (𝐷𝐷𝑒𝑒𝑘𝑘)Γ𝑖𝑖𝑖𝑖𝑘𝑘𝑑𝑑𝑑𝑑𝑖𝑖

= 𝑒𝑒𝑛𝑛𝜕𝜕Γ𝑖𝑖𝑖𝑖𝑛𝑛

𝜕𝜕𝑑𝑑𝑚𝑚𝑑𝑑𝑑𝑑𝑚𝑚 ∧ 𝑑𝑑𝑑𝑑𝑖𝑖 + 𝑒𝑒𝑛𝑛Γ𝑘𝑘𝑚𝑚𝑛𝑛 𝑑𝑑𝑑𝑑𝑚𝑚 ∧ Γ𝑖𝑖𝑖𝑖𝑘𝑘𝑑𝑑𝑑𝑑𝑖𝑖

Collecting terms,

𝐷𝐷2𝑒𝑒𝑖𝑖 = 12𝑒𝑒𝑛𝑛 �

𝜕𝜕Γ𝑖𝑖𝑖𝑖𝑙𝑙

𝜕𝜕𝑢𝑢𝑚𝑚− 𝜕𝜕Γ𝑖𝑖𝑚𝑚

𝑙𝑙

𝜕𝜕𝑢𝑢𝑖𝑖+ Γ𝑘𝑘𝑚𝑚𝑛𝑛 Γ𝑖𝑖𝑖𝑖𝑘𝑘 − Γ𝑘𝑘𝑖𝑖𝑛𝑛 Γ𝑖𝑖𝑚𝑚𝑘𝑘 � 𝑑𝑑𝑑𝑑𝑚𝑚 ∧ 𝑑𝑑𝑑𝑑𝑖𝑖 = 𝑒𝑒𝑛𝑛θ𝑖𝑖𝑛𝑛 (A.5)

where we identified the curvature two-form and the explicit formula for the Riemann curvature

tensor,

θ𝑖𝑖𝑛𝑛 = 𝑅𝑅𝑖𝑖 𝑚𝑚𝑖𝑖 𝑛𝑛 𝑑𝑑𝑑𝑑𝑚𝑚 ∧ 𝑑𝑑𝑑𝑑𝑖𝑖 (A.6a)

with

𝑅𝑅𝑖𝑖 𝑚𝑚𝑖𝑖 𝑛𝑛 =𝜕𝜕Γ𝑖𝑖𝑖𝑖

𝑙𝑙

𝜕𝜕𝑢𝑢𝑚𝑚− 𝜕𝜕Γ𝑖𝑖𝑚𝑚

𝑙𝑙

𝜕𝜕𝑢𝑢𝑖𝑖+ Γ𝑘𝑘𝑚𝑚𝑛𝑛 Γ𝑖𝑖𝑖𝑖𝑘𝑘 − Γ𝑘𝑘𝑖𝑖𝑛𝑛 Γ𝑖𝑖𝑚𝑚𝑘𝑘 (A.6b)

Let’s make closer contact with the notation in the body of this lecture. We have written

Eq. 50,

𝐷𝐷𝑒𝑒𝑖𝑖 = 𝑒𝑒𝑘𝑘𝜔𝜔𝑖𝑖 𝑘𝑘

So we identify, in the notation of standard tensor analysis in this appendix,

𝜔𝜔𝑖𝑖 𝑘𝑘 = Γ𝑖𝑖𝑖𝑖𝑘𝑘𝑑𝑑𝑑𝑑𝑖𝑖 (A.7)

So,

𝐷𝐷2𝑒𝑒𝑖𝑖 = 𝐷𝐷�𝑒𝑒𝑘𝑘𝜔𝜔𝑖𝑖 𝑘𝑘� = 𝑒𝑒𝑘𝑘𝑑𝑑𝜔𝜔𝑖𝑖

𝑘𝑘 + (𝐷𝐷𝑒𝑒𝑘𝑘) ∧ 𝜔𝜔𝑖𝑖 𝑘𝑘 = 𝑒𝑒𝑘𝑘𝑑𝑑𝜔𝜔𝑖𝑖

𝑘𝑘 + 𝑒𝑒𝑛𝑛𝜔𝜔𝑘𝑘 𝑛𝑛 ∧ 𝜔𝜔𝑖𝑖

𝑘𝑘 = 𝑒𝑒𝑛𝑛�𝑑𝑑𝜔𝜔𝑖𝑖 𝑛𝑛 + 𝜔𝜔𝑘𝑘

𝑛𝑛 ∧ 𝜔𝜔𝑖𝑖 𝑘𝑘�

So we collect results,

Page 31: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

31

𝐷𝐷2𝑒𝑒𝑖𝑖 = 𝑒𝑒𝑛𝑛𝜃𝜃𝑖𝑖𝑛𝑛 , 𝜃𝜃𝑖𝑖𝑛𝑛 = 𝑑𝑑𝜔𝜔𝑖𝑖 𝑛𝑛 + 𝜔𝜔𝑘𝑘

𝑛𝑛 ∧ 𝜔𝜔𝑖𝑖 𝑘𝑘 (A.8)

where 𝜃𝜃𝑖𝑖𝑛𝑛 are the elements of the two-form curvature Θ. We can obtain our formula Eq. A.6a for

𝜃𝜃𝑖𝑖𝑛𝑛,

𝜃𝜃𝑖𝑖𝑛𝑛 = 𝑑𝑑�Γ𝑖𝑖𝑖𝑖𝑛𝑛 𝑑𝑑𝑑𝑑𝑖𝑖� + �Γ𝑘𝑘𝑚𝑚𝑛𝑛 𝑑𝑑𝑑𝑑𝑚𝑚� ∧ �Γ𝑖𝑖𝑖𝑖𝑘𝑘𝑑𝑑𝑑𝑑𝑖𝑖� = �𝜕𝜕Γ𝑖𝑖𝑖𝑖

𝑙𝑙

𝜕𝜕𝑢𝑢𝑚𝑚+ Γ𝑘𝑘𝑚𝑚𝑛𝑛 Γ𝑖𝑖𝑖𝑖𝑘𝑘� 𝑑𝑑𝑑𝑑𝑚𝑚 ∧ 𝑑𝑑𝑑𝑑𝑖𝑖 (A.9)

And we can extract the Riemann curvature tensor 𝑅𝑅𝑖𝑖 𝑚𝑚𝑖𝑖 𝑛𝑛 as above, being careful to explicitly anti-

symmetrize in the indices m and j.

Recall from the discussion of parallel transport and holonomy in the textbook and

Supplementary Lectures 9 and 11, that one can also obtain the Riemann curvature tensor from

the commutator of components of the covariant derivative. Let’s sketch this calculation. We

begin with Eq. A.4 for the covariant derivative of the components of a vector,

𝐷𝐷𝑘𝑘𝑑𝑑𝑖𝑖 =𝐷𝐷𝑑𝑑𝑖𝑖

𝜕𝜕𝑑𝑑𝑘𝑘=𝜕𝜕𝑑𝑑𝑖𝑖

𝜕𝜕𝑑𝑑𝑘𝑘+ Γ𝑖𝑖𝑘𝑘𝑖𝑖 𝑑𝑑𝑖𝑖

Then a careful but straightforward calculation produces,

[𝐷𝐷𝑘𝑘,𝐷𝐷𝑛𝑛]𝑑𝑑𝑚𝑚 = 𝑅𝑅𝑖𝑖 𝑛𝑛𝑘𝑘 𝑚𝑚 𝑑𝑑𝑖𝑖 (A.10)

This formula was used extensively in the textbook. Its fundamental feature is that the result of

the commutator of the components of the covariant derivative applied to a vector is proportional

to the vector itself. There are no derivatives of 𝑑𝑑𝑚𝑚 on the right-hand-side of Eq. A.10: the result

is “ultra-local”. And the coefficient of the right-hand-side is the Riemann curvature tensor which

is a property of the surface S, independent of 𝑑𝑑𝑚𝑚. This equation was responsible for many of the

fundamental properties of Riemann manifolds presented in Supplementary Lectures 9 and 11.

Next let’s derive the formula for the Christoffel symbols in terms of the metric on the

surface S. From 𝜕𝜕𝑖𝑖𝑒𝑒𝑖𝑖 = 𝑒𝑒𝑘𝑘Γ𝑖𝑖𝑖𝑖𝑘𝑘 + 𝑛𝑛�⃗ 𝑏𝑏𝑖𝑖𝑖𝑖 , we can extract the Christoffel symbols,

𝑔𝑔𝑘𝑘𝑛𝑛Γ𝑖𝑖𝑖𝑖𝑘𝑘 = 𝑒𝑒𝑛𝑛 ∙ 𝜕𝜕𝑖𝑖𝑒𝑒𝑖𝑖 , 𝑔𝑔𝑘𝑘𝑛𝑛Γ𝑖𝑖𝑖𝑖𝑘𝑘 = Γ𝑖𝑖𝑖𝑖𝑛𝑛

where 𝑔𝑔𝑘𝑘𝑛𝑛 = 𝑒𝑒𝑘𝑘 ∙ 𝑒𝑒𝑛𝑛. These expressions suggest that Γ𝑖𝑖𝑖𝑖𝑘𝑘 can be written in terms of 𝑔𝑔𝑘𝑘𝑛𝑛 and its

derivatives 𝜕𝜕𝑖𝑖𝑔𝑔𝑘𝑘𝑛𝑛. To show this, begin with,

Page 32: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

32

𝜕𝜕𝑘𝑘𝑔𝑔𝑖𝑖𝑖𝑖 = �𝜕𝜕𝑘𝑘𝑒𝑒𝑖𝑖 ∙ 𝑒𝑒𝑖𝑖� + �𝑒𝑒𝑖𝑖 ∙ 𝜕𝜕𝑘𝑘𝑒𝑒𝑖𝑖� = Γ𝑖𝑖𝑘𝑘𝑖𝑖 + Γ𝑖𝑖𝑘𝑘𝑖𝑖 (A.11)

Similarly,

𝜕𝜕𝑖𝑖𝑔𝑔𝑖𝑖𝑘𝑘 = Γ𝑖𝑖𝑖𝑖𝑘𝑘 + Γ𝑘𝑘𝑖𝑖𝑖𝑖 , 𝜕𝜕𝑖𝑖𝑔𝑔𝑘𝑘𝑖𝑖 = Γ𝑘𝑘𝑖𝑖𝑖𝑖 + Γ𝑖𝑖𝑖𝑖𝑘𝑘 (A.12)

To isolate Γ𝑖𝑖𝑖𝑖𝑘𝑘 we add the two equations in Eq. A.12 and subtract Eq. A.11. This gives, after

some algebra which uses the symmetry Γ𝑖𝑖𝑖𝑖𝑘𝑘 = Γ𝑖𝑖𝑖𝑖𝑘𝑘,

Γ𝑖𝑖𝑖𝑖𝑘𝑘 = 12�𝜕𝜕𝑖𝑖𝑔𝑔𝑖𝑖𝑘𝑘 + 𝜕𝜕𝑖𝑖𝑔𝑔𝑘𝑘𝑖𝑖 − 𝜕𝜕𝑘𝑘𝑔𝑔𝑖𝑖𝑖𝑖� (A.13a)

And raising the index k, we obtain a familiar result,

Γ𝑖𝑖𝑖𝑖𝑚𝑚 = 12𝑔𝑔𝑚𝑚𝑘𝑘�𝜕𝜕𝑖𝑖𝑔𝑔𝑖𝑖𝑘𝑘 + 𝜕𝜕𝑖𝑖𝑔𝑔𝑘𝑘𝑖𝑖 − 𝜕𝜕𝑘𝑘𝑔𝑔𝑖𝑖𝑖𝑖� (A.13b)

It is interesting in this context to consider how inner products vary around the surface.

This exercise will lead to an important insight into the covariant derivative and the related

subject of parallel transport which lies at the heart of curved manifolds and the Riemann

curvature tensor, as we have seen in the textbook and in Supplementary Lectures 9 and 11. We

calculate,

𝜕𝜕𝑘𝑘(�⃗�𝑑 ∙ 𝑤𝑤��⃗ ) = 𝜕𝜕𝑘𝑘�𝑔𝑔𝑖𝑖𝑖𝑖𝑑𝑑𝑖𝑖𝑤𝑤𝑖𝑖� = �𝜕𝜕𝑘𝑘𝑔𝑔𝑖𝑖𝑖𝑖�𝑑𝑑𝑖𝑖𝑤𝑤𝑖𝑖 + 𝑔𝑔𝑖𝑖𝑖𝑖�𝜕𝜕𝑘𝑘𝑑𝑑𝑖𝑖�𝑤𝑤𝑖𝑖 + 𝑔𝑔𝑖𝑖𝑖𝑖𝑑𝑑𝑖𝑖�𝜕𝜕𝑘𝑘𝑤𝑤𝑖𝑖�

𝜕𝜕𝑘𝑘(�⃗�𝑑 ∙ 𝑤𝑤��⃗ ) = �Γ𝑖𝑖𝑘𝑘𝑖𝑖 + Γ𝑖𝑖𝑘𝑘𝑖𝑖�𝑑𝑑𝑖𝑖𝑤𝑤𝑖𝑖 + 𝑔𝑔𝑖𝑖𝑖𝑖�𝜕𝜕𝑘𝑘𝑑𝑑𝑖𝑖�𝑤𝑤𝑖𝑖 + 𝑔𝑔𝑖𝑖𝑖𝑖𝑑𝑑𝑖𝑖�𝜕𝜕𝑘𝑘𝑤𝑤𝑖𝑖�

𝜕𝜕𝑘𝑘(�⃗�𝑑 ∙ 𝑤𝑤��⃗ ) = 𝑔𝑔𝑖𝑖𝑛𝑛�Γ𝑖𝑖𝑘𝑘𝑛𝑛 𝑑𝑑𝑖𝑖 + 𝜕𝜕𝑘𝑘𝑑𝑑𝑛𝑛�𝑤𝑤𝑖𝑖 + 𝑔𝑔𝑖𝑖𝑛𝑛�Γ𝑖𝑖𝑘𝑘𝑛𝑛 𝑤𝑤𝑖𝑖 + 𝜕𝜕𝑘𝑘𝑤𝑤𝑛𝑛�𝑑𝑑𝑖𝑖 = (𝐷𝐷𝑘𝑘�⃗�𝑑 ∙ 𝑤𝑤��⃗ ) + (�⃗�𝑑 ∙ 𝐷𝐷𝑘𝑘𝑤𝑤��⃗ )

So, finally,

𝜕𝜕𝑘𝑘(�⃗�𝑑 ∙ 𝑤𝑤��⃗ ) = (𝐷𝐷𝑘𝑘�⃗�𝑑 ∙ 𝑤𝑤��⃗ ) + (�⃗�𝑑 ∙ 𝐷𝐷𝑘𝑘𝑤𝑤��⃗ ) (A.14)

We learn the important fact that the differential of the inner product is determined by the

covariant derivatives of the individual vectors,

𝑑𝑑(�⃗�𝑑 ∙ 𝑤𝑤��⃗ ) = 𝜕𝜕𝑘𝑘�𝑔𝑔𝑖𝑖𝑖𝑖𝑑𝑑𝑖𝑖𝑤𝑤𝑖𝑖� 𝑑𝑑𝑑𝑑𝑘𝑘 = (𝐷𝐷𝑘𝑘�⃗�𝑑 ∙ 𝑤𝑤��⃗ )𝑑𝑑𝑑𝑑𝑘𝑘 + (�⃗�𝑑 ∙ 𝐷𝐷𝑘𝑘𝑤𝑤��⃗ )𝑑𝑑𝑑𝑑𝑘𝑘 (A.15)

So, if 𝐷𝐷𝑘𝑘�⃗�𝑑 and 𝐷𝐷𝑘𝑘𝑤𝑤��⃗ vanish, then �⃗�𝑑 ∙ 𝑤𝑤��⃗ is constant.

Page 33: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

33

We also recall from earlier discussions that if 𝑝𝑝𝑘𝑘𝐷𝐷𝑘𝑘�⃗�𝑑 = 0 where 𝑝𝑝𝑘𝑘 = 𝑑𝑑𝑥𝑥𝑘𝑘 𝑑𝑑𝑑𝑑⁄ is the

tangent to a curve 𝑥𝑥𝑘𝑘(𝑑𝑑), then �⃗�𝑑 moves by parallel transport along the curve. And finally, we

discussed “metric compatibility”, 𝐷𝐷𝑖𝑖𝑔𝑔𝑖𝑖𝑘𝑘 = 0, in the textbook. We recognize that Eq. A.11 and

A.12 can be written in just this form. See the textbook and Supplementary Lecture 11 for more

detail.

Let’s end this appendix with a short discussion of curves, geodesics and the second

fundamental form. Imagine a curve in R3 parametrized by its arc-length. Its path is given by

�⃗�𝑥(𝑑𝑑). Then the unit tangent to the curve is 𝑡𝑡 = 𝑑𝑑�⃗�𝑥 𝑑𝑑𝑑𝑑⁄ . The rate at which 𝑡𝑡 turns determines the

curve’s curvature 𝜅𝜅,

𝑑𝑑𝑡𝑡𝑑𝑑𝑑𝑑

= 𝜅𝜅𝑁𝑁��⃗ (A.16)

where 𝑁𝑁��⃗ is a unit normal to 𝑡𝑡. We studied curves in Supplementary Lectures 4 and 9, the

intuition behind Eq. A.16 and the geometric significance of 𝜅𝜅.

Now consider a curve traveling on the surface S. In this case it is useful to resolve 𝑑𝑑𝑡𝑡 𝑑𝑑𝑑𝑑⁄

into components in the tangent plane at �⃗�𝑥 and parallel to the normal 𝑛𝑛�⃗ to the surface there,

𝑑𝑑𝑡𝑡𝑑𝑑𝑑𝑑

= 𝜅𝜅𝑁𝑁��⃗ = 𝜅𝜅𝑔𝑔𝑘𝑘�⃗ + 𝜅𝜅𝑛𝑛𝑛𝑛�⃗ (A.17)

where 𝑘𝑘�⃗ is a unit vector in the tangent plane and 𝑛𝑛�⃗ is the familiar unit normal to the surface. We

can relate 𝜅𝜅𝑔𝑔 and 𝜅𝜅𝑛𝑛 to the curve �⃗�𝑥(𝑑𝑑) more explicitly. Begin with,

𝑡𝑡 =𝑑𝑑�⃗�𝑥𝑑𝑑𝑑𝑑

=𝜕𝜕�⃗�𝑥𝜕𝜕𝑑𝑑𝑖𝑖

𝑑𝑑𝑑𝑑𝑖𝑖

𝑑𝑑𝑑𝑑= 𝑒𝑒𝑖𝑖

𝑑𝑑𝑑𝑑𝑖𝑖

𝑑𝑑𝑑𝑑

So,

𝜅𝜅𝑁𝑁��⃗ =𝑑𝑑𝑡𝑡𝑑𝑑𝑑𝑑

=𝑑𝑑𝑑𝑑𝑑𝑑�𝑒𝑒𝑖𝑖

𝑑𝑑𝑑𝑑𝑖𝑖

𝑑𝑑𝑑𝑑� = 𝑒𝑒𝑖𝑖

𝑑𝑑2𝑑𝑑𝑖𝑖

𝑑𝑑𝑑𝑑2+𝜕𝜕𝑒𝑒𝑖𝑖𝜕𝜕𝑑𝑑𝑖𝑖

𝑑𝑑𝑑𝑑𝑖𝑖

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑖𝑖

𝑑𝑑𝑑𝑑= 𝑒𝑒𝑖𝑖

𝑑𝑑2𝑑𝑑𝑖𝑖

𝑑𝑑𝑑𝑑2+ �𝑒𝑒𝑛𝑛Γ𝑖𝑖𝑖𝑖𝑛𝑛 + 𝑛𝑛�⃗ 𝑏𝑏𝑖𝑖𝑖𝑖�

𝑑𝑑𝑑𝑑𝑖𝑖

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑖𝑖

𝑑𝑑𝑑𝑑

Collecting,

𝜅𝜅𝑔𝑔𝑘𝑘�⃗ + 𝜅𝜅𝑛𝑛𝑛𝑛�⃗ = 𝑒𝑒𝑛𝑛 �𝑑𝑑2𝑑𝑑𝑛𝑛

𝑑𝑑𝑑𝑑2+ Γ𝑖𝑖𝑖𝑖𝑛𝑛

𝑑𝑑𝑑𝑑𝑖𝑖

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑖𝑖

𝑑𝑑𝑑𝑑� + 𝑛𝑛�⃗ 𝑏𝑏𝑖𝑖𝑖𝑖

𝑑𝑑𝑑𝑑𝑖𝑖

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑖𝑖

𝑑𝑑𝑑𝑑

Page 34: relativitydoctor.com · 2020-02-04 · 1 . Supplemental Lecture 13 . Differential Forms for Physics Students . III. Classical and Riemannian Differential Geometry . Abstract . Differential

34

So,

𝜅𝜅𝑔𝑔𝑘𝑘�⃗ = 𝑒𝑒𝑛𝑛 �𝑑𝑑2𝑢𝑢𝑙𝑙

𝑑𝑑𝑑𝑑2+ Γ𝑖𝑖𝑖𝑖𝑛𝑛

𝑑𝑑𝑢𝑢𝑖𝑖

𝑑𝑑𝑑𝑑𝑑𝑑𝑢𝑢𝑖𝑖

𝑑𝑑𝑑𝑑� , 𝜅𝜅𝑛𝑛 = 𝑏𝑏𝑖𝑖𝑖𝑖

𝑑𝑑𝑢𝑢𝑖𝑖

𝑑𝑑𝑑𝑑𝑑𝑑𝑢𝑢𝑖𝑖

𝑑𝑑𝑑𝑑 (A.18)

The first expression shows us that if the curve is a geodesic on S, 𝜅𝜅𝑔𝑔 = 0, then its differential

equation is,

𝑑𝑑2𝑢𝑢𝑙𝑙

𝑑𝑑𝑑𝑑2+ Γ𝑖𝑖𝑖𝑖𝑛𝑛

𝑑𝑑𝑢𝑢𝑖𝑖

𝑑𝑑𝑑𝑑𝑑𝑑𝑢𝑢𝑖𝑖

𝑑𝑑𝑑𝑑= 0 (A.19)

This is a familiar result: since Eq. A.19 is a second order differential equation, it proves that

there is a geodesic through every point on S pointing in any initial direction indicated by 𝑑𝑑𝑑𝑑𝑖𝑖 𝑑𝑑𝑑𝑑⁄

at 𝑑𝑑 = 0. In addition, the second relation in Eq. A.18 shows that 𝜅𝜅𝑛𝑛 is the ratio of the second

fundamental form to the first fundamental form,

𝜅𝜅𝑛𝑛 =𝑏𝑏𝑖𝑖𝑖𝑖𝑑𝑑𝑑𝑑𝑖𝑖𝑑𝑑𝑑𝑑𝑖𝑖

𝑑𝑑𝑑𝑑2

another familiar result in classical differential geometry which leads to some classic results on

the Gaussian curvature K and the mean curvature H.

References

1. H. Flanders, Differential Forms with Applications to the Physical Sciences, Academic

Press, New York, 1963.

2. M. P. Do Carmo, Differential Forms and Applications, Springer-Verlag, Berlin, 1971.

3. M. Spivak, Calculus on Manifolds, Westview Press, Princeton, New Jersey, 1965.