give an independent proof that [V, W ]M = [V, W ], then we would have an
alternative proof that S is self-adjoint. To this end note that
[V, W ] = V W
= V W W V = [V, W ]M .
Thus to prove that [V, W ]M = [V, W ] it is enough to show that [V, W ] =
The operation is also known as the standard Levi-Civita connection of
R . If W is a tangent vectoreld of A, i.e., a mapping W : A Rn such
that Wp is a tangent vector of A for all p A, then we set
W f (p) := Wp f
and (W V )p := Wp V.
Note that W f : A R
Exercise 1. What is det(li ) equal to?.
Exercise 2. Show that Ni = dn(Xi ) = S (Xi ).
Next we compute the Christoel symbols. To this end note that
Xij , Xk =
l glk ,
Xl , Xk =
which in matrix notation reads
Xij , X1
Xij , X2
1 g11 +
Then, since is a curve on M ,
distM (p, pi ) Length[ ]
x2 + yi + h(txi , tyi ), (xi , yi ) 2 dt
x2 + yi + (x2 + yi )2 dt
x2 + yi
(1 + ) p pi
So, for any > 0 we have
distM (p, pi )
provided that pi is suciently close
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
Exercise 4. Dene :
Exercise 4. Check that (V, n) is indeed a local gauss map.
Exercise 5. Show that n : S2 S2 , given by n(p) := p is a Gauss map
(Hint: Dene f : R3 R by f (p) := p 2 and compute its gradient. Note
that S2 is a level set of f . Thus the gradient of f at p mu
from the south pole gives a homeomorphism between R2 and S2 (0, 0, 1);
+ (x, y, z ) := ( 1z , 1z , 0), and (x, y, z ) := ( z1 , z1 , 0).
Exercise 3. (Surfaces as graphs) Let U R2 be an open subset and
f : U R be a continuous map. Then
graph(f ) :=
Exercise 1. Show that if the torsion of a curve : I R3 is zero everywhere
then it lies in a plane. (Hint : We need to check that there exist a point p
and a (xed) vector v in R3 such that (t) p, v = 0. Let v = B , and p
be any point of the curve.)
Exercise 5. Verify the last sentence, i.e., show that if : I R2 is a simple
closed convex planar curve, and is any line in the plane which intersects
(I ), then either intersects in exactly two points, or (I ) lies on one side
.(Hint : Show that if inters
(0) = |f (0)|.
Next note that the center of the circle which is tangent to at (0, 0) must
lie on the y -axis at some point (0, r), and for this circle to also pass through
the point (s, f (s) we must have:
r2 = s2 + (r f (s)2 .
Solving the above equa
N in terms of this frame. The denition of N already yields that, when is
parametrized by arclength,
T (t) = (t)N (t).
To get the corresponding formula for N , rst observe that
N (t) = aT (t) + bN (t).
for some a and b. To nd a note that, since T , N = 0,
denote the n-dimensional unit sphere in Rn+1 . Dene a mapping from Sn Sn
to R by
distSn (p, q ) := angle(p, q ).
Exercise 2. Show that (Sn , distSn ) is a metric space.
The above metric has a simple geometric interpretation described as follows. By a grea
Next note that, since f preserves the inner product, if ei , i = 1 . . . n, is an
orthonormal basis for Rn , then so is f (ei ). Further,
f (p + q ), f (ei )
p + q, ei = p, ei + q , ei
f (p), f (ei ) + f (q ), f (ei )
f (p) + f (q ), f (ei ) .
t t, t2 ) R2 ,
and the cubic curve
t t, t2 , t3 ) R3 .
Exercise 1. Sketch the cubic curve (Hint: First draw each of the projections
into the xy , yz , and zx planes).
Exercise 2. Find a formula for the curve which is traced by the motion
Since this is such a remarkable and far reaching result we will include here
three dierent proofs. The rst proof is quite short and slick, but also highly
nontransparent, i.e., it is not easy to see how someone could come up with
that. The second proof is