Solution to Differential Geometry question paper April 2013

1. No of printed pages: 3
Sardar Patel University
Mathematics
M.Sc. Semester II
Monday, 29 April 2013
11.00 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
κ
n = t + 1
κ
(τb − κt) = 0.
This gives γ + 1
κ
n = c for some constant vector c. Therefore γ − c = 1
κ
n = 1
κ
.
Hence γ is lying on a sphere. Since τ = 0, γ is lying on a plane. Hence γ is lying on
both 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) =
t
0
˙γ(u) du =
t
0
eu
√
2du =
√
2et
. This gives t = ln(s/
√
2).
Therefore unit-speed reparametrization of γ may be given by
γ(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 be
the length of γ and A be the are of the interior of γ. Then 2
≥ 4πA, with equality
holding if and only if γ is a circle.
The Four Vertex theorem: Let γ be a simple closed convex curve in R2
. Then
γ 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 passing
through p whose tangent at p is w. Let γ1 = f ◦ γ. Then γ1 is a curve in S2
passing through f(p). Let w1 be the tangent to γ1 at f(p). Then Dpf(w) = w1. Let
γ2 = g ◦ γ1 = g ◦ f ◦ γ. Then γ2 is curve in S3 passing through g(f(p)). Let w2 be
the 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). Therefore
Dp(g ◦ f)(w) = w2. This shows that Dp(g ◦ f)(w) = (Df(p)g ◦ Dpf)(w). Since w is an
arbitrary 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 prove
g ◦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

2. the curves g ◦ f ◦ γ is S3 and f ◦ γ in S2 have the same length. Hence the curves γ
in 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 tangent
space, 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 differentiating
Nσu = 0 partially with respect to v, we get Nvσu = −M. Similarly, by differentiating
Nσv = 0 partially with respect to u and v, we get the relations Nvσu = −M and
Nvσv = −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 Weingarten
map Wp are called the principal curvatures of S at p and the corresponding eigen
vectors 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 linearly
independent, there exist function Γk
ij such that
σuu = Γ1
11σu + Γ2
11σv + LN
σuv = Γ1
12σu + Γ2
12σv + MN
σvv = Γ1
22σu + Γ2
22σv + NN



The functions Γk
ij 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 if
the geodesic curvature of γ is zero everywhere.
Sol: Let γ be a unit-speed curve on a surface S. We know that a geodesic curvature
of 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-speed
curve 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) =
s
s0
k(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
cos ϕ(u)du,
s
s0
sin ϕ(u)du .
Then ˙γ(s) = (cos ϕ(s), sin ϕ(s)) for all s. Therefore γ is unit-speed curve. It follows
that ϕ is the turning angle of γ satisfying ϕ(s0) = 0 and hence the signed curvature
of γ is ˙ϕ. But ˙ϕ = k. Hence the signed curvature of γ is k.
2

3. 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) =
s
s0
k(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) = (cos ψ(s), sin ψ(s)) for all s. Let a = γ(s0). Integrating the above equation,
we have
γ(s) =
s
s0
cos ψdu,
s
s0
sin ψdu + a
= Ta
s
s0
cos ψdu,
s
s0
sin ψdu
= Ta
s
s0
cos(ϕ + θ)du,
s
s0
sin(ϕ + θ)du
= Ta
s
s0
(cos ϕ cos θ − sin ϕ sin θ)du,
s
s0
(sin ϕ sin θ + cos ϕ cos θ)du
= Ta cos θ
s
s0
cos ϕdu − sin θ
s
s0
sin ϕdu, cos θ
s
s0
cos ϕdu + sin θ
s
s0
sin ϕdu
= Ta ◦ Rθ
s
s0
cos ϕdu,
s
s0
sin ϕdu
= Ta ◦ Rθ(γ(s)).
Therefore γ(s) = M(γ(s)) for all s, where M = Ta ◦ Rθ is a direct isometry. Hence
γ = 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
˙κ
κ2τ
.
Sol: We may assume that γ is unit-speed. Assume that γ is spherical. Then there
exist 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 we
have (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
κ
n +
˙κ
τκ2
b.
But then R2
= γ − a 2
= 1
κ2 + ˙κ2
τ2κ4 . Differentiating it we obtain
τ2κ42 ˙κ¨κ− ˙κ2(2τ ˙τκ4+τ24κ2 ˙κ)
τ4κ8 − 2 ˙κ
κ3 = 0, i.e., κ2¨κτ−2κ ˙κ2τ−κ2 ˙κ ˙τ
κ4τ2 − τ
κ
= 0, i.e., τ
κ
= d
dt
˙κ
κ2τ
.
(OR again differentiating (γ−a)b = ˙κ
τκ2 we get t b−(γ−a)τn = d
dt
( ˙κ
τκ2 ). Therefore
τ
κ
= d
dt
˙κ
κ2τ
.)
3

4. Conversely, assume that τ
κ
= d
dt
˙κ
κ2τ
. Let a = γ + 1
κ
n − ˙κ
τκ2 b. Then
˙a = t −
˙κ
κ2
n +
1
κ
˙n −
d
dt
˙κ
κ2τ
b −
˙κ
τκ2
˙b
= t −
˙κ
κ2
n +
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
= 1
κ2 + ˙κ2
τ2κ4 . Again using τ
κ
= d
dt
˙κ
κ2τ
, it
follows that 1
κ2 + ˙κ2
τ2κ4 is a positive constant, say, R2
. Therefore γ − a 2
= R2
, and
hence γ is spherical.
(OR Let ρ = 1
κ
and σ = 1
τ
. Since τ
κ
= d
dt
˙κ
κ2τ
, we have ρ
σ
= − d
dt
( ˙ρσ) = −( ˙ρσ) , i.e.,
ρ
σ
+( ˙ρσ) = 0. Let a = γ +ρn+ ˙ρσb. Then ˙a = t+ ˙ρn+ρ(1
σ
b− 1
ρ
t)+( ˙ρσ) b− ˙ρσ 1
σ
n =
(ρ
σ
+ ( ˙ρσ) )b = 0. Hence a is a constant vector. Now γ − a 2
= ρ2
+ ( ˙ρσ)2
. Again
using 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 = α
α2+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 = a
a2+b2 = α
and the torsion τ of γ is τ = b
a2+b2 = β. Hence given α > 0 and β constant we have
a curve whose curvature is α and torsion is β. If γ is any other (unit-speed) curve
with curvature α and torsion β, then by the fundamental theorem of space curves it
follows that there is a direct isometry M of R3
such that M(γ) = γ. Since γ is a
helix, M(γ) is a helix. Hence it follows that all the curves having constant curvature
and 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, have
the same first fundamental form.
Sol: Let E, F and G be the first order magnitudes of σ and E1, F1 and G1 be those of
f ◦σ. Assume that f is a local isometry. Then to prove both σ and f ◦σ have the same
fundamental form, we need to prove E = E1, F = F1 and G = G1. Since f is a local
isometry 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 fundamental
forms 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

5. 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) of
S1 and S2. A map f : S1 → S2 is called smooth if the map σ−1
2 ◦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 passing
through p whose tangent at p is w. Let w be the tangent to f ◦ γ at f(p). Then
the 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 fundamental
form 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). This
gives E = σuσu = 1 + f2
u, F = σuσv = fufv and G = σvσv = 1 + f2
v . Hence the
first fundamental form of the given surface is (1+f2
u)du2
+2fufvdudv+(1+f2
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). Hence
by the definition of the derivative Dγ(t)f(˙γ(t)) = (f ◦ γ)•
(t) for all t. Since f
is a local diffeomorphism, Dpf is invertible for all t. In particular, Dγ(t)f is
invertible for all t. Fix t. Since ˙γ(t) = 0, and the linear map Dγ(t)f is invertible,
we have (f ◦ γ)•
(t) = Dγ(t)f(˙γ(t)) = 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

6. Sol: Let p be a point on a smooth surface S, and let v be a unit vector in a given
direction. 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 quantities
at (0, 0). We have σu = (1, 0, 0), σv = (0, 1, 0), σuu = (0, 0, −2), σuv = (0, 0, 0) and
σvv = (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 − λF
M − 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 − 2F
M − 2F N − 2G
ζ
η
=
0
0
, i.e.,
−4 0
0 0
ζ
η
=
0
0
. This gives ζ = 0
and η ∈ R is arbitrary. Take η = 1. Hence the principal direction corresponding to
κ2 = 2 is t2 = (0, 1, 0). Similarly, the principal direction corresponding to κ1 = −2 is
t1 = (1, 0, 0). Since the oriented angle t1t2 = π
2
and v = ( 1√
2
, 1√
2
, 0) makes an oriented
angle π
4
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 π
4
+ 2 sin2 π
4
= 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 by
Wp = −DpG is called the Weingarten map.
Since the derivative DpG of the Gauss map is linear, the map Wp is linear. Now
we show that Wp is symmetric, i.e., Wp(x), w p = x, Wp(w) p (x, w ∈ TpS). First
we 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 and
Wp(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

7. By definition of the second fundamental form IIp w, x p = Wp(w), x p. The above
result 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 unit
directions 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 = π
2
. 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 + π
2
or θ + +3π
2
. By Euler’s theorem the normal
curvature (κn)1 in the direction v is (κn)1 = κ1 cos2
θ + κ2 sin2
θ and the normal
curvature (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)2
2
= κ1+κ2
2
, which is the mean curvature of S at p.
Hence the mean of the normal curvatures at p in the orthogonal unit directions
in TpS is the mean curvature at p.
(ii) Compute the second fundamental form of the surface patch [3]
σ(u, v) = (sinh u sinh v, sinh u cosh v, sinh u).
Sol: We have σu = (cosh u sinh v, cosh u cosh v, cosh u), σv = (sinh u cosh v, sinh u sinh v, 0),
σuu = (sinh u sinh v, sinh u cosh v, sinh u), σuv = (cosh u cosh v, cosh u sinh v, 0)
and σvv = (sinh u sinh v, sinh u 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
(− sinh u cosh v + cosh u sinh v),
M = σuvN = 0 and N = σvvN = sinh u√
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 df
du
2
+ dg
du
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 = ( df
du
cos v, df
du
sin v, dg
du
) and σv = (−f sin v, f cos v, 0). Therefore E = 1,
F = 0 and G = f2
, and so Eu = Ev = Fu = Fv = Gv = 0 and Gu = 2f df
du
.
Let γ(t) = σ(u(t), v(t)) be a unit-speed curve on S. Since γ is unit-speed,
1 = σu ˙u + σv ˙v 2
= ˙u2
+ f2
˙v2
. (1)
Then γ is a geodesic iff
d
dt
( ˙u) = f
df
du
˙v2
and
d
dt
(f2
˙v) = 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 is
a geodesic.
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8. (II) Let β(t) = σ(u0, v(t)) be a parallel. Since u = u0, ˙u = 0. Since β is
unit-speed, ˙v = ± 1
f(u0)
[see (1)]. Clearly the second equation in (2) holds
and the first equation holds iff 0 = d
dt
( ˙u) = f(u0) df
du
(u0), i.e., iff df
du
(u0) = 0
as f(u0) > 0. Hence β is a geodesic iff df
du
(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 the
results you use.
Sol: Let γ be the triangle and that it is positively oriented. Since γ is made up
of geodesics of the surface, κg = 0. We also note that the Gaussian curvature K of
given surface is −1. Let α1, α2, α3 be the exterior angles of γ, and let β1, β2, β3 be the
exterior 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
3
i=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 S
with 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 preserved
under 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 Gauss
equations that the Gaussian curvature K1 of S1 can be expressed by the first
order magnitudes E1, F1 and G1 and their partial derivatives. Since the first
order magnitudes E2, F2 and G2 are same as E1, F1 and G1 respectively, the
Gaussian curvature K2 of S2 is same as the Gaussian curvature K1 of S1. Hence
the Gaussian curvature of a surface is preserved under local isometries.
(ii) Let σ be a surface patch of a surface S. Prove that Nu = 1
EG−F2 ((MF −LG)σu + [3]
(LF − ME)σv) and Nv = 1
EG−F2 ((NF − MG)σu + (MF − NE)σv).
Sol: We know that the matrix of Wp with respect to the basis {σu, σv} is
E F
F G
−1
L M
M N
= 1
EG−F2
LG − MF MG − NF
−LF + ME −MF + NE
. Therefore Nu =
−Wp(σu) = − LG−MF
EG−F2 σu + −LF+ME
EG−F2 σv , i.e., Nu = 1
EG−F2 ((MF − LG)σu +
(LF − ME)σv). Similarly one gets the desired expression for Nv.
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