No of printed pages: 3

Sardar Patel UniversityMathematics

M.Sc. Semester IIMonday, 29 April 201311.00 a.m. to 2.00 p.m.

PS02CMTH03 Differential Geometry

Maximum Marks: 70

Q.1 Choose the correct option for each of the following. [8]

(1) (a)(2) (a)

(3) (b)(4) (a)

(5) (b)(6) (c)

(7) (d)(8) (d)

Q.2 Attempt any Seven. [14](a) Let γ be a regular curve in R3 with constant curvature and zero torsion. Show that

γ is part of a circle.Sol: We may assume that γ is unit-speed. Now d

dt

(γ + 1

κn)

= t + 1κ(τb − κt) = 0.

This gives γ + 1κn = c for some constant vector c. Therefore ‖γ − c‖ = ‖ 1

κn‖ = 1

κ.

Hence γ is lying on a sphere. Since τ = 0, γ is lying on a plane. Hence γ is lying onboth plane and sphere, i.e., γ is part of a circle.

(b) Find a unit-speed reparametrization of γ(t) = (et cos t, et sin t).

Sol: We have s(t) =∫ t0‖γ(u)‖du =

∫ t0eu√

2du =√

2et. This gives t = ln(s/√

2).Therefore unit-speed reparametrization of γ may be given byγ(t) = 1√

2(t cos(ln(t/

√2)), t sin(ln(t/

√2))).

(c) State the Isoperimetric Inequality and the Four Vertex Theorem.Sol: The Isoperimetric inequality: Let γ be a simple closed curve in R2. Let ` bethe length of γ and A be the are of the interior of γ. Then `2 ≥ 4πA, with equalityholding if and only if γ is a circle.

The Four Vertex theorem: Let γ be a simple closed convex curve in R2. Thenγ has at least four vertices.

(d) Let S1,S2,S3 be smooth surfaces. If f : S1 → S2 and g : S2 → S3 are smooth maps,then show that Dp(g ◦ f) = Df(p)g ◦Dpf for every p ∈ S1.Sol: Let p be a point on S1, and let w ∈ TpS1. Let γ be a curve in S1 passingthrough p whose tangent at p is w. Let γ1 = f ◦ γ. Then γ1 is a curve in S2passing through f(p). Let w1 be the tangent to γ1 at f(p). Then Dpf(w) = w1. Letγ2 = g ◦ γ1 = g ◦ f ◦ γ. Then γ2 is curve in S3 passing through g(f(p)). Let w2 bethe tangent to γ2 at g(f(p)). Then Df(p)g(w1) = w2, i.e, Df(p)g(Dpf(w)) = w2, i.e.,(Df(p)g ◦Dpf)(w) = w2. Also, w2 is the tangent to (g ◦ f) ◦ γ at g ◦ f(p). ThereforeDp(g ◦ f)(w) = w2. This shows that Dp(g ◦ f)(w) = (Df(p)g ◦Dpf)(w). Since w is anarbitrary point of TpS1, we get Dp(g ◦ f) = Df(p)g ◦Dpf .

(e) Show that composition of local isometries is a local isometry.Sol: Let f : S1 → S2 and g : S2 → S3 be local isometries. We need to proveg ◦ f : S1 → S3 is a local isometry. Let γ be a curve in S1. Since f is a local isometry,the curves f ◦ γ in S2 and γ in S1 have the same length. Since g is a local isometry,

1

the curves g ◦ f ◦ γ is S3 and f ◦ γ in S2 have the same length. Hence the curves γin S1 and g ◦ f ◦ γ in S3 have the same length. Therefore g ◦ f is a local isometry.

(f) Let σ be a surface patch with the unit normal N. Then show that Nuσu = −L,Nuσv = −M = Nvσu and Nvσv = −N .Sol: Since σu and σv generates the tangent space and N is orthogonal to the tangentspace, we have Nσu = Nσv = 0. Differentiating Nσu = 0 partially with respect to u,we obtain Nuσu + Nσuu = 0. But Nσuu = L. Hence Nuσu = −L. By differentiatingNσu = 0 partially with respect to v, we get Nvσu = −M . Similarly, by differentiatingNσv = 0 partially with respect to u and v, we get the relations Nvσu = −M andNvσv = −N respectively.

(g) Define principal curvatures and principal directions at a point on a smooth surface.Sol: Let p be a point on a smooth surface S. The eigen values of the Weingartenmap Wp are called the principal curvatures of S at p and the corresponding eigenvectors of Wp are called the principal directions of S at p.

(h) Define the Christoffel’s symbols of second kind for a surface patch σ.Sol: Let σ be a surface patch of the surface σ. Since σu, σv and N are linearlyindependent, there exist function Γkij such that

σuu = Γ111σu + Γ2

11σv + LNσuv = Γ1

12σu + Γ212σv + MN

σvv = Γ122σu + Γ2

22σv + NN

The functions Γkij are called the Christoffel symbols of second kind.

(i) Let γ be a unit-speed curve on a surface S. Show that γ is a geodesic if and only ifthe geodesic curvature of γ is zero everywhere.Sol: Let γ be a unit-speed curve on a surface S. We know that a geodesic curvatureof a curve γ is given by κg = γ(N× γ).

Assume that γ is a geodesic. Then γ ‖ N. Hence κg = γ(N× γ) = (γ ×N)γ = 0.Conversely, assume that κg = 0. Then γ is perpendicular to N × γ. Since γ, N

and γ ×N are mutually perpendicular and γ is perpendicular to γ, we have γ ‖ N.Therefore γ is a geodesic.

Q.3(a) Let k : (a, b)→ R be a smooth map. Show that there is a unit-speed curve γ : (a, b)→ [6]

R2 whose signed curvature is k. Moreover, if γ : (a, b) → R2 is any other unit-speedcurve whose signed curvature is k, then show that there is a direct isometry M of R2

such that γ = M ◦ γ.Sol: Let s0 ∈ (a, b). Define ϕ : (a, b) → R by ϕ(s) =

∫ ss0k(u)du. Since k is smooth,

ϕ is a smooth map. Note that ϕ(s0) = 0 Define γ : (a, b)→ R2 by

γ(s) =

(∫ s

s0

cosϕ(u)du,

∫ s

s0

sinϕ(u)du

).

Then γ(s) = (cosϕ(s), sinϕ(s)) for all s. Therefore γ is unit-speed curve. It followsthat ϕ is the turning angle of γ satisfying ϕ(s0) = 0 and hence the signed curvatureof γ is ϕ. But ϕ = k. Hence the signed curvature of γ is k.

2

Let γ : (a, b) → R2 be a unit-speed curve with signed curvature k. Let ψ be the

turning angle of γ. Then ψ = k. But then ψ(s) =∫ ss0k(u)du + ψ(s0) = ϕ(s) + θ,

where θ = ψ(s0) is a constant. So we have ψ = ϕ+ θ. Since ψ is the turning angle of

γ, ˙γ(s) = (cosψ(s), sinψ(s)) for all s. Let a = γ(s0). Integrating the above equation,we have

γ(s) =

(∫ s

s0

cosψdu,

∫ s

s0

sinψdu

)+ a

= Ta

(∫ s

s0

cosψdu,

∫ s

s0

sinψdu

)= Ta

(∫ s

s0

cos(ϕ+ θ)du,

∫ s

s0

sin(ϕ+ θ)du

)= Ta

(∫ s

s0

(cosϕ cos θ − sinϕ sin θ)du,

∫ s

s0

(sinϕ sin θ + cosϕ cos θ)du

)= Ta

(cos θ

∫ s

s0

cosϕdu− sin θ

∫ s

s0

sinϕdu, cos θ

∫ s

s0

cosϕdu+ sin θ

∫ s

s0

sinϕdu

)= Ta ◦Rθ

(∫ s

s0

cosϕdu,

∫ s

s0

sinϕdu

)= Ta ◦Rθ(γ(s)).

Therefore γ(s) = M(γ(s)) for all s, where M = Ta ◦ Rθ is a direct isometry. Henceγ = M ◦ γ.

(b) Let γ be a regular curve in R3 with nowhere vanishing curvature. Show that γ is [6]spherical if and only if τ

κ= d

dt

(κκ2τ

).

Sol: We may assume that γ is unit-speed. Assume that γ is spherical. Then thereexist a constant vector a ∈ R3 and a constant R ∈ R such that (γ − a)(γ − a) =‖γ − a‖2 = R2. Differentiating it we get (γ − a)t = 0. Differentiating it again wehave (γ − a)n = − 1

κ. Again differentiation gives (γ − a)b = κ

τκ2. Since {t,n,b} is an

orthonormal basis of R3, γ − a = ((γ − a)t)t + ((γ − a)n)n + ((γ − a)b)b. Therefore

γ − a = −1

κn +

κ

τκ2b.

But then R2 = ‖γ − a‖2 = 1κ2

+ κ2

τ2κ4. Differentiating it we obtain

τ2κ42κκ−κ2(2τ τκ4+τ24κ2κ)τ4κ8

− 2κκ3

= 0, i.e., κ2κτ−2κκ2τ−κ2κτκ4τ2

− τκ

= 0, i.e., τκ

= ddt

(κκ2τ

).

(OR again differentiating (γ−a)b = κτκ2

we get t b−(γ−a)τn = ddt

( κτκ2

). Thereforeτκ

= ddt

(κκ2τ

).)

3

Conversely, assume that τκ

= ddt

(κκ2τ

). Let a = γ + 1

κn− κ

τκ2b. Then

a = t− κ

κ2n +

1

κn− d

dt

(κ

κ2τ

)b− κ

τκ2b

= t− κ

κ2n +

1

κ(τb− κt)− d

dt

(κ

κ2τ

)b− κ

τκ2(−τ)n

=

(τ

κ− d

dt

(κ

κ2τ

))b = 0.

Hence a is a constant vector. Now ‖γ − a‖2 = 1κ2

+ κ2

τ2κ4. Again using τ

κ= d

dt

(κκ2τ

), it

follows that 1κ2

+ κ2

τ2κ4is a positive constant, say, R2. Therefore ‖γ − a‖2 = R2, and

hence γ is spherical.(OR Let ρ = 1

κand σ = 1

τ. Since τ

κ= d

dt

(κκ2τ

), we have ρ

σ= − d

dt(ρσ) = −(ρσ)�, i.e.,

ρσ

+(ρσ)� = 0. Let a = γ+ρn+ ρσb. Then a = t+ ρn+ρ( 1σb− 1

ρt)+(ρσ)�b− ρσ 1

σn =

( ρσ

+ (ρσ)�)b = 0. Hence a is a constant vector. Now ‖γ − a‖2 = ρ2 + (ρσ)2. Againusing ρ

σ+ (ρσ)� = 0, we have ρ2 + (ρσ)2 is constant. Hence γ is spherical.)

OR

(b) Characterize all curves in R3 with constant curvature and constant torsion. State the [6]result you use.

Let α > 0 and β be constant. Let a = αα2+β2 and b = β

α2+β2 . Consider the curve

γ(t) =(a cos

(t√

a2+b2

), a sin

(t√

a2+b2

), bt√

a2+b2

). The curve γ is a helix. Note that γ

is a unit-speed curve. Computations show that the curvature κ of γ is k = aa2+b2

= α

and the torsion τ of γ is τ = ba2+b2

= β. Hence given α > 0 and β constant we havea curve whose curvature is α and torsion is β. If γ is any other (unit-speed) curvewith curvature α and torsion β, then by the fundamental theorem of space curves itfollows that there is a direct isometry M of R3 such that M(γ) = γ. Since γ is ahelix, M(γ) is a helix. Hence it follows that all the curves having constant curvatureand constant torsion are helices only.

Q.4(c) Show that a local diffeomorphism f : S1 → S2 is a local isometry if and only if, for [6]

any surface patch σ of S1, the patches σ and f ◦ σ of S1 and S2, respectively, havethe same first fundamental form.Sol: Let E,F and G be the first order magnitudes of σ and E1, F1 and G1 be those off ◦σ. Assume that f is a local isometry. Then to prove both σ and f ◦σ have the samefundamental form, we need to prove E = E1, F = F1 and G = G1. Since f is a localisometry f ∗〈v, w〉p = 〈v, w〉p for all p ∈ S1. In particular, f ∗〈σu, σu〉p = 〈σu, σu〉p,f ∗〈σu, σv〉p = 〈σu, σv〉p and f ∗〈σv, σv〉p = 〈σv, σv〉p. Now,

E = 〈σu, σu〉p = f ∗〈σu, σu〉p = 〈Dpf(σu), Dpf(σv)〉f(p) = 〈σu, σu〉f(p) = E1.

Similar arguments shows that F = F1 and G = G1. Hence the first fundamentalforms of σ and f ◦ σ are same.

Conversely, assume that the first fundamental forms of σ and f ◦ σ are same.Then E = E1, F = F1 and G = G1. Fix p = σ(u, v). Let w = λ1σu + µ1σv, z =

4

λ2σu + µ2σv ∈ TpS. Then

f ∗〈w, z〉p = 〈Dpf(w), Dpf(z)〉f(p) = 〈Dpf(λ1σu + µ1σv), Dpf(λ2σu + µ2σv)〉f(p)= λ1λ2〈Dpf(σu), Dpf(σu)〉f(p) + (λ1µ2 + λ2µ1)〈Dpf(σu), Dpf(σv)〉f(p)

+µ1µ2〈Dpf(σv), Dpf(σv)〉f(p)= λ1λ2〈σu, σu〉p + (λ1µ2 + λ2µ1)〈σu, σv〉p + µ1µ2〈σv, σv〉p= 〈λ1σu + µ1σv, λ2σu + µ2σv〉p = 〈w, z〉p.

Hence f ∗〈·, ·〉p = 〈·, ·〉p. Since p ∈ S1 is arbitrary, f ∗〈·, ·〉p = 〈·, ·〉p for every p ∈ S1,i.e., f is a local isometry.

(d) (i) Show that z = y2 + x2 is a smooth surface. [3]Sol: We may parametrize z = y2 + x2 by σ(u, v) = (u, v, u2 + v2), u, v ∈ R.Clearly the map σ is smooth. Now σu = (1, 0, 2u) and σv = (0, 1, 2v). Therefore‖σu × σv‖ = ‖(−2u,−2v, 1)‖ =

√1 + 4u2 + 4v2 > 0. Hence the vectors σu and

σv are linearly independent. This proves that z = y2 + x2 is a smooth surface.(ii) Let S1 and S2 be smooth surfaces. When is a map f : S1 → S2 called smooth? [3]

Define the derivative of f at p ∈ S1.Sol: Let σ1 : U1 → R3 and σ2 : U2 → R3 be parametrizations (surface patches) ofS1 and S2. A map f : S1 → S2 is called smooth if the map σ−12 ◦f ◦σ1 : U1 → U2

is a smooth map.Let f be smooth and let p ∈ S1. Let w ∈ TpS1. Let γ be a curve in S1 passingthrough p whose tangent at p is w. Let w be the tangent to f ◦ γ at f(p). Thenthe derivative Dpf of f at w ∈ TpS1 is defined by Dpf(w) = w.

OR

(d) (i) Define first fundamental form on a surface S. Compute it for the surface [3]σ(u, v) = (u, v, f(u, v)).Sol: Let S be a smooth surface, and let p ∈ S. Then the first fundamentalform on S at p is a mapping Ip : TpS × TpS → R given by Ip(v, w) = 〈v, w〉p =v · w (v, w ∈ TpS).Here σ(u, v) = (u, v, f(u, v)). Therefore σu = (1, 0, fu) and σv = (0, 1, fv). Thisgives E = σuσu = 1 + f 2

u , F = σuσv = fufv and G = σvσv = 1 + f 2v . Hence the

first fundamental form of the given surface is (1+f 2u)du2+2fufvdudv+(1+f 2

v )dv2.(ii) Let f : S1 → S2 be a local diffeomorphism, and let γ be a regular curve in S1. [3]

Show that f ◦ γ is a regular curve in S2.Sol: We note that (f ◦ γ)•(t) is a tangent to the curve f ◦ γ at f ◦ γ(t). Henceby the definition of the derivative Dγ(t)f(γ(t)) = (f ◦ γ)•(t) for all t. Since fis a local diffeomorphism, Dpf is invertible for all t. In particular, Dγ(t)f is

invertible for all t. Fix t. Since γ(t) 6= 0, and the linear map Dγ(t)f is invertible,

we have (f ◦ γ)•(t) = Dγ(t)f(γ(t)) 6= 0. Since t is arbitrary, it follows that,‖(f ◦ γ)•(t)‖ > 0 for all t, i.e., f ◦ γ is regular.

Q.5(e) Define normal curvature of a surface at a point p in the unit direction v in TpS.

Compute the normal curvature of σ(u, v) = (u, v, v2 − u2) at (0, 0, 0) in the direction( 1√

2, 1√

2, 0).

5

Sol: Let p be a point on a smooth surface S, and let v be a unit vector in a givendirection. Then the normal curvature of S at p in direction v is given by 〈Wp(v), v〉p,where Wp is the Weingarten map at p.

Here σ(u, v) = (u, v, v2 − u2). Since σ(0, 0) = (0, 0, 0), we find all the quantitiesat (0, 0). We have σu = (1, 0, 0), σv = (0, 1, 0), σuu = (0, 0,−2), σuv = (0, 0, 0) andσvv = (0, 0, 2). Therefore E = σuσu = 1, F = σuσv = 0, G = σvσv = 1, N = (0, 0, 1),L = σuuN = −2, M = σuvN = 0 and N = σvvN = −2. The principal curvatures are

the roots of

∣∣∣∣L− λE M − λFM − λF N − λG

∣∣∣∣ = 0. Hence we get (−2−λ)(2−λ) = 0, i.e., λ = ±2.

Now the principal direction x = ζσu + ησv = (ζ, η, 0) corresponding to 2 is given by(L− 2E M − 2FM − 2F N − 2G

)(ζη

)=

(00

), i.e.,

(−4 00 0

)(ζη

)=

(00

). This gives ζ = 0

and η ∈ R is arbitrary. Take η = 1. Hence the principal direction corresponding toκ2 = 2 is t2 = (0, 1, 0). Similarly, the principal direction corresponding to κ1 = −2 is

t1 = (1, 0, 0). Since the oriented angle t1t2 = π2

and v = ( 1√2, 1√

2, 0) makes an oriented

angle π4

with t1, by Euler’s theorem the normal curvature in the direction ( 1√2, 1√

2, 0)

at (0, 0, 0) is κn = κ1 cos2 θ + κ2 sin2 θ = −2 cos2(π4

)+ 2 sin2

(π4

)= 0.

(f) Define the Weingarten map at a point on a surface. Show that it is a symmetric [6]linear map.Sol: Let p be a point on a surface S. Then map Wp : TpS → TpS defined byWp = −DpG is called the Weingarten map.

Since the derivative DpG of the Gauss map is linear, the map Wp is linear. Nowwe show that Wp is symmetric, i.e., 〈Wp(x), w〉p = 〈x,Wp(w)〉p (x,w ∈ TpS). Firstwe prove the following.

Let σ be a surface patch of a surface S containing a point p of S, say, p = σ(u, v).Then for any v, w ∈ TpS,

IIp〈w, x〉 = Ldu(w)du(x) +M(du(w)dv(x) + du(x)dv(w)) +Ndv(w)dv(x),

where du, dv : TpS → R are defined by du(w) = λ and dv(w) = µ when w =λσu + µσv ∈ TpS.

Proof. Since DpG(σu) = Nu and DpG(σv) = Nv, we have Wp(σu) = −Nu andWp(σv) = −Nv. Let w = λ1σu + µ1σv and x = λ2σu + µ2σv be in TpS. Then

IIp〈w, x〉p = 〈Wp(w), x〉p= 〈Wp(λ1σu + µ1σv), λ2σu + µ2σv〉p= 〈λ1Wp(σu) + µ1Wp(σv), λ2σu + µ2σv〉p= 〈−λ1Nu − µ1Nv, λ2σu + µ2σv〉p= −λ1λ2Nuσu − λ1µ2Nuσv − λ2µ1Nvσu − µ1µ2Nvσv

= Ldu(w)du(x) +M(du(w)dv(x) + du(x)dv(w)) +Ndv(w)dv(x).

This proves the result. �6

By definition of the second fundamental form IIp〈w, x〉p = 〈Wp(w), x〉p. The aboveresult shows that IIp〈w, x〉p = IIp〈x,w〉p. Hence 〈Wp(x), w〉p = 〈Wp(w), x〉p or〈Wp(x), w〉p = 〈x,Wp(w)〉p. Therefore Wp is symmetric.

OR

(f) (i) Let p be a point on a surface S, and let κ1 and κ2 be the principal curvatures [3]at p. Show that the mean of the normal curvatures at p in the orthogonal unitdirections in TpS is the mean curvature at p.Sol: Let κ1 and κ2 be principal curvatures of S at p, and let t1 and t2 be

corresponding directions. We may assume that t1 t2 = π2. Let v and w be

orthogonal directions at p. Let θ be the oriented angle between t1 and v, i.e.,

t1v = θ. Then t1w is either θ + π2

or θ + +3π2

. By Euler’s theorem the normal

curvature (κn)1 in the direction v is (κn)1 = κ1 cos2 θ + κ2 sin2 θ and the normalcurvature (κn)2 in the direction w is (κn)2 = κ1 cos2

(θ + π

2

)+ κ2 sin2

(θ + π

2

)or (κn)2 = κ1 cos2

(θ + 3π

2

)+ κ2 sin2

(θ + 3π

2

). In any case (κn)2 = κ1 sin2 θ +

κ2 cos2 θ. Hence (κn)1+(κn)22

= κ1+κ22

, which is the mean curvature of S at p.Hence the mean of the normal curvatures at p in the orthogonal unit directionsin TpS is the mean curvature at p.

(ii) Compute the second fundamental form of the surface patch [3]σ(u, v) = (sinhu sinh v, sinhu cosh v, sinhu).Sol: We have σu = (coshu sinh v, coshu cosh v, coshu), σv = (sinhu cosh v, sinhu sinh v, 0),σuu = (sinhu sinh v, sinhu cosh v, sinhu), σuv = (coshu cosh v, coshu sinh v, 0)and σvv = (sinhu sinh v, sinhu cosh v, 0). Now N = σu×σv

‖σu × σv‖. Therefore

N = 1√2(− sinh v

cosh v, 1,− 1

cosh v). Now L = σuuN = 1√

2(− sinhu cosh v+ coshu sinh v),

M = σuvN = 0 and N = σvvN = sinhu√2 cosh v

.

Q.6(g) Consider the surface of revolution σ(u, v) = (f(u) cos v, f(u) sin v, g(u)), where f > 0 [6]

and(dfdu

)2+(dgdu

)2= 1. Prove the following statements.

(I) Every meridian is a geodesic, i.e., the curves α(t) = σ(u(t), v0) is a geodesic.(II) A parallel u = u0 (say) is a geodesic if and only if df

du(u0) = 0.

Here σu = ( dfdu

cos v, dfdu

sin v, dgdu

) and σv = (−f sin v, f cos v, 0). Therefore E = 1,

F = 0 and G = f 2, and so Eu = Ev = Fu = Fv = Gv = 0 and Gu = 2f dfdu

.Let γ(t) = σ(u(t), v(t)) be a unit-speed curve on S. Since γ is unit-speed,

1 = ‖σuu+ σvv‖2 = u2 + f 2v2. (1)

Then γ is a geodesic iff

d

dt(u) = f

df

duv2 and

d

dt(f 2v) = 0. (2)

(I) Let α(t) = σ(u(t), v0). Note that since v = v0, v = 0. Since α is unit-speed, u = ±1 [see (1)]. Hence the equations in (2) are satisfied, i.e., α isa geodesic.

7

(II) Let β(t) = σ(u0, v(t)) be a parallel. Since u = u0, u = 0. Since β isunit-speed, v = ± 1

f(u0)[see (1)]. Clearly the second equation in (2) holds

and the first equation holds iff 0 = ddt

(u) = f(u0)dfdu

(u0), i.e., iff dfdu

(u0) = 0

as f(u0) > 0. Hence β is a geodesic iff dfdu

(u0) = 0.(h) Let S be a surface whose Gaussian curvature is −1 everywhere. Show that the sum [6]

of the interior angles of a triangle on S is strictly less than π. State carefully theresults you use.Sol: Let γ be the triangle and that it is positively oriented. Since γ is made upof geodesics of the surface, κg = 0. We also note that the Gaussian curvature K ofgiven surface is −1. Let α1, α2, α3 be the exterior angles of γ, and let β1, β2, β3 be theexterior angles of γ. Then αi+βi = π (i = 1, 2, 3). Clearly int(γ) is simply connected.Hence by the Gauss-Bonnet Theorem we have

∫∫int(γ)

(−1)+∑3

i=1 αi = 2π. Therefore∑3i=1 βi = π −

∫∫int(γ)

dS < π.

The result which we have used is:[Gauss-Bonnet Theorem] Let γ be a positively oriented piecewise smooth curve, con-sisting on n smooth arcs with exterior angles α1, α2, . . . , αn, on a smooth surface Swith Gaussian curvature K. Let int (γ) be simply connected. Then∫

γ

κg(s)ds+

∫∫int(γ)

K dS +n∑i=1

αi = 2π, (3)

where κg is the geodesic curvature of γ.

OR

(h) (i) State and prove the Gauss’ Theorema Egregium. [3][Gauss’ Theorema Egregium] The Gaussian curvature of a surface is preservedunder local isometries.

Proof. Let f : S1 → S2 be a local isometry. Then the first fundamental forms(or the first order magnitudes) of S1 and S2 are same. It follows from the Gaussequations that the Gaussian curvature K1 of S1 can be expressed by the firstorder magnitudes E1, F1 and G1 and their partial derivatives. Since the firstorder magnitudes E2, F2 and G2 are same as E1, F1 and G1 respectively, theGaussian curvature K2 of S2 is same as the Gaussian curvature K1 of S1. Hencethe Gaussian curvature of a surface is preserved under local isometries. �

(ii) Let σ be a surface patch of a surface S. Prove that Nu = 1EG−F 2 ((MF−LG)σu+ [3]

(LF −ME)σv) and Nv = 1EG−F 2 ((NF −MG)σu + (MF −NE)σv).

Sol: We know that the matrix of Wp with respect to the basis {σu, σv} is(E FF G

)−1(L MM N

)= 1

EG−F 2

(LG−MF MG−NF−LF +ME −MF +NE

). Therefore Nu =

−Wp(σu) = −(LG−MFEG−F 2 σu + −LF+ME

EG−F 2 σv), i.e., Nu = 1

EG−F 2 ((MF − LG)σu +

(LF −ME)σv). Similarly one gets the desired expression for Nv.

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