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Unformatted text preview: Mar 3, 20041 Math 497C
Curves and Surfaces
Fall 2004, PSU Lecture Notes 10
2.3 Meaning of Gaussian Curvature In the previous lecture we gave a formal deﬁnition for Gaussian curvature
K in terms of the diﬀerential of the gauss map, and also derived explicit
formulas for K in local coordinates. In this lecture we explore the geometric
meaning of K .
2.3.1 A measure for local convexity Let M ⊂ R3 be a regular embedded surface, p ∈ M , and Hp be hyperplane
passing through p which is parallel to Tp M . We say that M is locally convex
at p if there exists an open neighborhood V of p in M such that V lies on
one side of Hp . In this section we prove:
Theorem 1. If K (p) > 0 then M is locally convex at p, and if k (p) < 0
then M is not locally convex at p.
When K (p) = 0, we cannot in general draw a conclusion with regard to
the local convexity of M at p as the following two exercises demonstrate:
Exercise 2. Show that there exists a surface M and a point p ∈ M such
that M is strictly locally convex at p; however, K (p) = 0 (Hint: Let M be
the graph of the equation z = (x2 + y 2 )2 . Then may be covered by the Monge
patch X (u1 , u2 ) := (u1 , u2 , ((u1 )2 + (u2 ))2 ). Use the Monge Ampere equation
derived in the previous lecture to compute the curvature at X (0, 0).).
Exercise 3. Let M be the Monkey saddle, i.e., the graph of the equation
z = y 3 − 3yx2 , and p := (0, 0, 0). Show that K (p) = 0, but M is not locally
convex at p.
1 Last revised: October 31, 2004 1 After a rigid motion we may assume that p = (0, 0, 0) and Tp M is the
xy plane. Then, using the inverse function theorem , it is easy to show that
there exists a Monge Patch (U, X ) centered at p, as the follwing exercise
demonstrates:
Exercise 4. Deﬁne π : M → R2 by π (q ) := (q 1 , q 2 , 0). Show that dπp is
locally onetoone. Then, by the inverse function theorem, it follows that π
is a local diﬀeomorphism. So there exists a neighborhood U of (0, 0) such
that π −1 : U → M is onetoone and smooth. Let f (u1 , y 2 ) denote the z coordinate of π −1 (u1 , u2 ), and set X (u1 , u2 ) := (u1 , u2 , f (u1 , u2 )). Show that
(U, X ) is a proper regular patch.
The previous exercisle shows that local convexity of M at p depends on
whether or not f changes sign in a neighborhood of the origin. To examine
this we need to recall the Taylor’s formula for functions of two variables:
2 f (u1 , u2 ) = f (0, 0) + 2 Di f (0, 0)ui +
i=1 1
Dij (ξ1 , ξ2 )ui uj ,
2 i,j =1 where (ξ1 , ξ2 ) is a point on the line connecting (u1 , u2 ) to (0, 0).
Exercise 5. Prove the Taylor’s formula given above. (Hints : First recall Taylor’s formula for functions of one variable: g (t) = g (0) + g (0)t +
(1/2)g (s)t2 , where s ∈ [0, t]. Then deﬁne γ (t) := (tu1 , tu2 ), set g (t) :=
f (γ (t)), and apply Taylor’s formula to g . Then chain rule will yield the
desired result.)
Next note that, by construction, f (0, 0) = 0. Further D1 f (0, 0) = 0 =
D2 f (0, 0) as well. Thus
2 1
Dij (ξ1 , ξ2 )ui uj .
f (u1 , u2 ) =
2 i,j =1
Hence to complete the proof of Theorem 1, it remains to show how the
quanitity on the right hand side of the above euation is inﬂuence by K (p).
To this end, recall the MongeAmpere equation for curvature:
det(Hess f ξ1 , ξ2 ) = K (f (ξ1 , ξ2 )) 1 + grad f (ξ1 , ξ2 ) 2 22 . Now note that K (f (0, 0)) = K (p). Thus, by continuity, if U is a suﬃciently
small neighborhood of (0, 0), the sign of det(Hess f ) agrees with the sign of
K (p) throughout U .
Finally, we need some basic facts about quadratic forms. A quadratic
form is a function of two variables Q : R2 → R given by
Q(x, y ) = ax2 + 2bxy + cy 2 ,
where a, b, and c are constants. Q is said to be deﬁnite if Q(x, x) = 0 whenver
x = 0.
Exercise 6. Show that if ac − b2 > 0, then Q is deﬁnite, and if ac − b2 < 0,
then Q is not deﬁnite. (Hints : For the ﬁrst part, suppose that x = 0, but
Q(x, y ) = 0. Then ax2 +2bxy +cy 2 = 0, which yields a+2b(x/y )+c(x/y )2 = 0.
Thus the discriminant of this equation must be positive, which will yield a
contradiction. The proof of the second part is similar).
Theorem 1 follows from the above exercise.
2.3.2 Ratio of areas In the previous subsection we gave a geometric interpretation for the sign
of Gaussian curvature. Here we describe the geometric signiﬁcance of the
magnitude of K .
If V is a suﬃciently small neighborhood of p in M (where M , as always,
denotes a regular embedded surface in R3 ), then it is easy to show that there
exist a patch (U, X ) centered at p such that X (U ) = V . Area of V is then
deﬁned as follows:
D1 X × D2 X du1 du2 . Area(V ) :=
U Using the chain rule, one can show that the above deﬁnition is independent
of the the patch.
Exercise 7. Let V ⊂ S2 be a region bounded in between a pair of great
circles meeting each other at an angle of α. Show that Area(V ) = 2α(Hints :
Let U := [0, α] × [0, π ] and X (θ, φ) := (cos θ sin φ, sin θ sin φ, cos φ). Show
that D1 X × D2 X =  sin φ. Further, note that, after a rotation we may
assume that X (U ) = V . Then an integration will yield the desired result).
3 Β Α β
γ C α Τ Exercise 8. Use the previous exercise to show that the area of a geodesic
triangle T ⊂ S2 (a region bounded by three great circles) is equal to sum of
its angles minus π (Hints: Use the picture below: A + B + C + T = 2π , and
A = 2α − T , B = 2β − T , and C = 2γ − T ).
Let Vr := Br (p) ∩ M . Then, if r is suﬃciently small, V (r) ⊂ X (U ), and,
consequently, Ur := X −1 (Vr ) is well deﬁned. In particular, we may compute
the area of Vr using the patch (Ur , X ). In this section we show that
Area(n(Vr ))
.
r →0 Area(Vr ) K (p) = lim Exercise 9. Recall that the mean value theorem states that
f du1 du2 =
U
1
2
1
2
f (¯ , u ) Area(U ), for some (¯ , u ) ∈ U . Use this theorem to show that
u¯
u¯
D1 N (0, 0) × D2 N (0, 0)
Area(n(Vr ))
=
r →0 Area(Vr )
D1 X (0, 0) × D2 X (0, 0)
lim (Recall that N := n ◦ X .)
Exercise 10. Prove Lagrange’s identity: for every pair of vectors v, w ∈ R3 ,
v×w 2 v, v
w, v = det v, w
w, w . Now set g (u1 , u2 ) := det[gij (u1 , u2 )]. Then, by the previous exercise it
follows that D1 X (0, 0) × D2 X (0, 0) = g (0, 0). Hence, to complete the
proof of the main result of this section it remains to show that
D1 N (0, 0) × D2 N (0, 0) = K (p) g (0, 0).
4 We prove the above formula using two diﬀerent methods:
METHOD 1. Recall that K (p) := det(Sp ), where Sp := −dnp : Tp M → Tp M
is the shape operator of M at p. Also recall that Di X (0, 0), i = 1, 2, form a
basis for Tp M . Let Sij be the coeﬃcients of the matrix representation of Sp
with respect to this basis, then
2 Sp (Di X ) = Sij Dj X.
j =1 Further, recall that N := n ◦ X . Thus the chain rule yields:
Sp (Di X ) = −dn(Di X ) = −Di (n ◦ X ) = −Di N.
Exercise 11. Verify the middle step in the above formula, i.e., show that
dn(Di X ) = Di (n ◦ X ).
From the previous two lines of formulas, it now follows that
2 −Di N = Sij Dj X.
j =1 Taking the inner product of both sides with Dk N , k = 1, 2, we get
2 −Di N, Dk N = Sij Dj X, Dk N .
j =1 Exercise 12. Let F , G : U ⊂ R2 → R3 be a pair of mappings such that
F, G = 0. Prove that Di F, G = − F, Di G .
Now recall that Dj X, N = 0. Hence the previous exercise yields:
Dj X, Dk N = − Dkj X, N = −lij .
Combining the previous two lines of formulas, we get: Di N, Dk N =
which in matrix notation is equivalent to 2
k=1 Sij ljk ; [ Di N, Dj N ] = [Sij ][lij ].
Finally, recall that det[ Di N, Dk N ] = D1 N × D2 N 2 , det[Sij ] = K , and
det[lij ] = Kg . Hence taking the determinant of both sides in the above
equation, and then taking the square root yields the desired result.
5 Next, we discuss the second method for proving that D1 N × D2 N =
√
K g.
METHOD 2. Here we work with a special patch which makes the computations easier:
Exercise 13. Show that there exist a patch (U, X ) centered at p such that
[gij (0, 0)] is the identity matrix. (Hint: Start with a Monge patch with
respect to Tp M )
Thus, if we are working with the coordinate patch referred to in the
above exercise, g (0, 0) = 1, and, consequently, all we need is to prove that
D1 N (0, 0) × D2 N (0, 0) = K (p).
Exercise 14. Let f : U ⊂ R2 → S2 be a diﬀerentiable mapping. Show that
Di f (u1 , u2 ), f (u1 , u2 ) = 0 (Hints: note that f , f = 1 and diﬀerentiate).
It follows from the previous exercise that Di N, N = 0. Now recall that
N (0, 0) = n ◦ X (0, 0) = n(p). Hence, we may conclude that N (0, 0) ∈ Tp M .
Further recall that {D1 X (0, 0), D2 X (0, 0)} is now an orthonormal basis for
Tp M (because we have chosen (U, X ) so that [gij (0, 0)] is the identity matrix).
Consequently,
2 Di N = Di N, Dk X Dk X,
k=1 where we have omitted the explicit reference to the point (0, 0) in the above
formula in order to make the notation less cumbersome (it is important to
keep in mind, however, that the above is valid only at (0, 0)). Taking the
inner product of both sides of this equation with Dj N (0, 0) yields:
2 Di N, Dj N = Di N, Dk X Dk X, Dj N .
k=1 Now recall that Di N, Dk X = − N, Dij X = −lij . Similarly, Dk X, Dj N =
−lkj . Thus, in matrix notation, the above formula is equivalent to the following:
[ Di N, Dj N ] = [lij ]2
Finally, recall that K (p) = det[lij (0, 0)]/ det[gij (0, 0)] = det[lij (0, 0)]. Hence,
taking the determinant of both sides of the above equation yields the desired
result.
6 2.3.3 Product of principal curvatures For every v ∈ Tp M with v = 1 we deﬁne the normal curvature of M at p
in the direction of v by
kv (p) := γ (0), n(p) ,
where γ : (− , ) → M is a curve with γ (0) = p and γ (0) = v .
Exercise 15. Show that kv (p) does not depend on γ .
In particular, by the above exercise, we may take γ to be a curve which
lies in the intersection of M with a plane which passes through p and is
normal to n(p) × v . So, intuitively, kv (p) is a measure of the curvature of an
orthogonal cross section of M at p.
Let U Tp M := {v ∈ Tp M  v = 1} denote the unit tangent space of M
at p. The principal curvatures of M at p are deﬁned as
k1 (p) := min kv (p),
v and k2 (p) := max kv (p),
v where v ranges over U Tp M . Our main aim in this subsection is to show that
K (p) = k1 (p)k2 (p).
Since K (p) is the determinant of the shape operator Sp , to prove the above
it suﬃces to show that k1 (p) and k2 (p) are the eigenvalues of Sp .
First, we need to deﬁne the second fundamental form of M at p. This is
a bilinear map IIp : Tp M × Tp M → R deﬁned by
IIp (v, w) := Sp (v ), w .
We claim that, for all v ∈ U Tp M ,
kv (p) = IIp (v, v ).
The above follows from the following computation
Sp (v ), v = − dnp (v ), v
= − (n ◦ γ ) (0), γ (0)
= (n ◦ γ )(0), γ (0)
= n(p), γ (0)
7 Exercise 16. Verify the passage from the second to the third line in the
above computation, i.e., show that − (n ◦ γ ) (0), γ (0) = (n ◦ γ )(0), γ (0)
(Hint: Set f (t) := n(γ (t)), γ (t) , note that f (t) = 0, and diﬀerentiate.)
So we conclude that ki (p) are the minimum and maximum of IIp (v ) over
U Tp M . Hence, all we need is to show that the extrema of IIp over U Tp M
coincide with the eigenvalues of Sp .
Exercise 17. Show that IIp is symmetric, i.e., IIp (v, w) = IIp (w, v ) for all v ,
w ∈ Tp M .
By the above exercise, Sp is a selfadjoint operator, i.e, Sp (v ), w =
v , Sp (w) . Hence Sp is orthogonally diagonalizable, i.e., there exist orthonormal vectors ei ∈ Tp M , i = 1, 2, such that
Sp (ei ) = λi ei .
By convention, we suppose that λ1 ≤ λ2 . Now note that each v ∈ U Tp M
may be represented uniquely as v = v 1 e1 + v 2 e2 where (v 1 )2 + (v 2 )2 = 1. So
for each v ∈ U Tp M there exists a unique angle θ ∈ [0, 2π ) such that
v (θ) := cos θe1 + sin θe2 ;
Consequently, bilinearity of IIp yields
IIp (v (θ), v (θ)) = λ1 cos2 θ + λ2 sin2 θ.
Exercise 18. Verify the above claim, and show that minimum and maximum values of IIp are λ1 and λ2 respectively. Thus k1 (p) = λ1 , and
k2 (p) = λ2 .
The previous exercise completes the proof that K (p) = k1 (p)k2 (p), and
also yields the following formula which was discovered by Euler:
kv (p) = k1 (p) cos2 θ + k2 (p) sin2 θ.
In particular, note that by the above formula there exists always a pair
of orthogonal directions where kv (p) achieves its maximum and minimum
values. These are known as the principal directions of M at p. 8 ...
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This note was uploaded on 08/25/2011 for the course MATH 6456 taught by Professor Staff during the Spring '08 term at Georgia Institute of Technology.
 Spring '08
 Staff
 Math, Geometry

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