DIFFERENTIAL GEOMETRY ATTACKS THE TORUS William Schulz Department of Mathematics and Statistics Northern Arizona University, Flagstaff, AZ 86011

1. INTRODUCTION

This is a pdf showing computations of Differential Geometry quantities using the Torus as example. No advantage is taken of the particular qualities of the torus; the calcu- lations are done as they would be for any surface, but of course have simpler results because the surface is simple. We make use of the embedding at the beginning to get the metric coefficients (gij ) and then proceed in a Riemannian manner. For the torus enthusiast I have added a section at the end that treats the normal vector and uses it to find the Gaussian Curvature.

2. NOTATION AND METRIC COEFFICIENTS

We begin by parametrizing the torus by longitude and latitude as usual. The torus is parametrized by θ which is the angle going round the big sweep of the torus from 0 to 2π and by φ which is the angle going around the little waist of the torus, also from 0 to 2π. The parametrization is

~x(θ, φ)=< (R + r cos φ)cos θ, (R + r cos φ) sin θ, r sin φ> Remember that for systematic computation purposes we set the parameters u1 = θ and u2 = φ. This order guarantees that the normal vector will point outward. Next we compute the unit tangent vectors:

∂~x ~e1 = ∂u1 = < −(R + r cos φ) sin θ, (R + r cos φ)cos θ, 0 >

∂~x ~e2 = ∂u2 = < −r sin φ cos θ, −r sin φ sin θ, r cos φ >

Next is to find the matrix of metric coefficients gij = ei · ej . These are easily found to be

2 2 2 2 g11 = e1 · e1 = (R + r cos φ) sin θ + (R + r cos φ) cos θ +0 = (R + r cos φ)2

g12 = e1 · e2 = (R + r cos φ)r(sin θ sin φ cos θ − cos θ sin φ sin θ) = 0 2 2 2 2 2 2 g22 = e2 · e2 = r (sin φ cos θ + sin φ sin θ + cos φ) = r2

1 In matrix form this is (R + r cos φ)2 0 (g )= ij 0 r2   We will also need the inverse matrix 1 ij −1 (R+r cos φ)2 0 (g ) = (gij ) = 1 0 2  r 

3 CONNECTION AND CURVATURE FORMS

We first want to compute the Christoffel symbols for which we need the basic formulas 1 ∂g ∂g ∂g Γ = ik + jk − ij ij|k 2 ∂uj ∂ui ∂uk and   i im Γjk = g Γjk|m i i From these we get, remembering that Γjk =Γkj , 1 ∂g Γ = 11 =0 11|1 2 ∂u1 1 ∂g ∂g ∂g Γ = 12 + 12 − 11 = (R + r cos φ)r sin φ 11|2 2 ∂u1 ∂u1 ∂u2 1∂g ∂g ∂g  Γ = 11 + 21 − 12 = −(R + r cos φ)r sin φ 12|1 2 ∂u2 ∂u1 ∂u1 1∂g ∂g ∂g  Γ = 12 + 22 − 12 =0 12|2 2 ∂u2 ∂u1 ∂u2 1∂g ∂g ∂g  Γ = 21 + 21 − 22 =0 22|1 2 ∂u2 ∂u2 ∂u1 1∂g  Γ = 22 =0 22|2 2 ∂u2 Then 1 1m 11 Γ11 = g Γ11|m = g Γ11|1 =0 2 2m 22 Γ11 = g Γ11|m = g Γ11|2 = 1 (R + r cos φ) sin φ = ((R + r cos φ)r sin φ)= r2 r 1 1m 11 Γ12 = g Γ12|m = g Γ12|1 = 1 r sin φ = − (R + r cos φ)r sin φ = − (R + r cos φ)2 (R + r cos φ) 2 2m 22  Γ12 = g Γ12|m = g Γ12|2 =0 1 1m 11 Γ22 = g Γ22|m = g Γ22|1 =0 2 2m 22 Γ22 = g Γ22|m = g Γ22|2 =0

2 For future purposes we place these in matrices:

1 1 r sin φ i Γ11 Γ21 0 − R+r cos φ γ1 = (Γj1)= 2 2 = (R+r cos φ) sin φ Γ11 Γ21 0   r ! and Γ1 Γ1 − r sin φ 0 γ = (Γi )= 12 22 = R+r cos φ 2 j2 Γ2 Γ2 0 0  12 22    Now we want the connection one forms. These are defined by

i i k ωj =Γjk du or in matrix form by Γ1 Γ1 Γ1 Γ1 (ω i)= 11 21 du1 + 12 22 du2 j Γ2 Γ2 Γ2 Γ2  11 21   12 22  which explicitly for the sphere gives

0 − r sin φ − r sin φ 0 (ω i)= R+r cos φ dθ + R+r cos φ dφ j (R+r cos φ) sin φ 0 0 0 r !   or r sin φ r sin φ i − R+r cos φ dφ − R+r cos φ dθ (ωj )= (R+r cos φ) sin φ r dθ 0 ! Next we want to compute the Riemann Curvature Tensor Form which is given by Ω= dω + ω ∧ ω As the calculations are gross and the intermediate results of almost no interest, they will be relegated to an Appendix to this section. The final result is

r cos φ 0 R+r cos φ Ω= dω + ω ∧ ω = cos φ dθ ∧ dφ − r (R + r cos φ) 0 ! Recalling that here the matrix entries of Ω is R 1 R 1 Ω= 1 12 2 12 dθ ∧ dφ R 2 R 2  1 12 2 12  we can read off the values of the Riemann Curvature Tensor as

1 1 r cos φ R1 12 =0 R2 12 = R+r cos φ 2 cos φ 2 R1 12 = − r (R + r cos φ) R2 12 =0 We can now get the Gaussian Curvature from the good old standard formula of Gauss g R m K = − 2m 1 12 det(gij )

3 which here gives us g R 2 K = − 22 1 12 det(gij ) r2 (− cos φ) (R + r cos φ) = − r r2(R + r cos φ)2 cos φ = r(R + r cos φ) It is amusing to compute the curvatura integra at this point. 2π 2π cos φ K dS = det(gij)dθdφ T 2 0 0 r(R + r cos φ) Z Z Z q 2π 2π cos φ = r(R + r cos φ)dθdφ r(R + r cos φ) Z0 Z0 2π 2π 2π = cos φ dθdφ =2π cos φ dφ =0 Z0 Z0 Z0 which is the result we expect from the Gauss-Bonnet theorem. APPENDIX: Calculation of the Riemann Curvature Tensor Recall that the curvature form ω is given by

ω = γ1dθ + γ2dφ where

1 1 r sin φ i Γ11 Γ21 0 − R+r cos φ γ1 = (Γj1)= 2 2 = (R+r cos φ) sin φ Γ11 Γ21 0   r ! and Γ1 Γ1 − r sin φ 0 γ = (Γi )= 12 22 = R+r cos φ 2 j2 Γ2 Γ2 0 0  12 22    To compute dω we need ∂ r sin φ (R + r cos φ)(r cos φ) − (r sin φ)(−r sin φ) − = − ∂φ R + r cos φ (R + r cos φ)2   r(R cos φ + r) = − (R + r cos φ)2 and ∂ (R + r cos φ) sin φ 1 = (R cos φ + r(cos2 φ − sin2 φ)) ∂φ r r   so we have

r(R cos φ+r) 0 2 dω (R+r cos φ) dθ dφ = 1 2 2 ∧ − r (R cos φ + r(cos φ − sin φ))) 0 !

4 Note the reversal of signs because the straightforward application of d to ω results in dφ ∧ dθ which is the wrong order. Next we compute ω ∧ ω. Recall that

ω = γ1dθ + γ2 dφ so that

ω ∧ ω = (γ1dθ + γ2 dφ) ∧ (γ1dθ + γ2 dφ)

= (γ1γ2 − γ2γ1) dθ ∧ dφ

So we need γ1γ2 and γ2γ1

0 − r sin φ r sin φ R+r cos φ − R+r cos φ 0 γ1γ2 = (R+r cos φ) sin φ 0 0 0 r !   0 0 = − sin2 φ 0   r sin φ r sin φ − R+r cos φ 0 0 − R+r cos φ γ2γ1 = 0 0 (R+r cos φ) sin φ 0   r ! 2 2 r sin φ 0 2 = (R+r cos φ) 0 0 ! so we have 2 2 r sin φ 0 − (R+r cos φ)2 ω ∧ ω = 2 dθ ∧ dφ − sin φ 0 ! To put dω and ω ∧ ω together we need

r(R cos φ + r) r2 sin2 rR cos φ + r2 − r2 sin2 φ − = (R + r cos φ)2 (R + r cos φ)2 (R + r cos φ)2 rR cos φ + r2 cos2 φ = (R + r cos φ)2 r cos φ(R + r cos φ) = (R + r cos φ)2 r cos φ = (R + r cos φ) so that we finally have

r cos φ 0 R+r cos φ Ω= dω + ω ∧ ω = cos φ dθ ∧ dφ − r (R + r cos φ) 0 !

5 4. The NORMAL VECTOR AND QUANTI- TIES ASSOCIATED WITH IT

Recall that the parametrization of the surface is ~x(θ, φ)=< (R + r cos φ)cos θ, (R + r cos φ) sin θ, r sin φ> with tangent vectors ~e1 and ~e2 given by ∂~x ~e1 = ∂u1 = < −(R + r cos φ) sin θ, (R + r cos φ)cos θ, 0 >

∂~x ~e2 = ∂u2 = < −r sin φ cos θ, −r sin φ sin θ, r cos φ > The normal vector ~n is given by ∂~x ∂~x ~n = × ∂u1 ∂u2 = < (R + r cos φ)r cos θ cos φ, (R + r cos φ)r sin θ cos φ, (R + r cos φ)r sin φ> = (R + r cos φ)r < cos φ cos θ, cos φ sin θ, sin φ> and its length, which also gives the multiplier in the area element for the torus, is |~n|2 = (R + r cos φ)2r2 cos2 φ cos2 θ + cos2 φ sin2 θ + sin2 φ = (R + r cos φ)2r2 |~n| = (R + r cos φ)r This gives a unit tangent vector nˆ =< cos φ cos θ, cos φ sin θ, sin φ> Recall the standard formula ∂ ~e i = ~e Γk +ˆnb ∂uj k ij ij

We wish to compute the bij . This is easy for donuts, because it is so easy to take derivatives ofn ˆ. We obviously have ∂ ~e b = i · nˆ ij ∂uj

Since ~ei · nˆ = 0, we can rewrite this as ∂nˆ b = −~e · ij i ∂uj We need ∂nˆ = < − cos φ sin θ, cos φ cos θ, 0 > ∂θ ∂nˆ = < − sin φ cos θ, − sin φ sin θ, cos φ> ∂φ

6 Now we have ∂nˆ b = − ~e · 11 1 ∂θ = − (R + r cos φ) sin2 φ cos φ + (R + r cos φ)cos2 φ cos φ +0 = −(R + r cos φ)cos φ
∂nˆ b = b = − ~e · 21 12 2 ∂θ = − r sin φ cos θ cos φ sin θ − r sin φ sin θ cos φ cos θ +0 =0 ∂nˆ b = −~e ·
22 2 ∂φ = − r sin2 φ cos2 θ + r sin2 φ sin2 θr cos2 φ = −r and thus
−(R + r cos φ)cos φ 0 (b )= ij 0 −r   With this we can find the Gaussian Curvature which relates the infinitesmal area on the unit sphere to the infinitesmal area on the surface via the Gauss Map given byn ˆ:

det(b ) r((R + r cos φ)cos φ cos φ K ij = = 2 2 = det(gij ) r (R + r cos φ) r(R + r cos φ)

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