Differential Topology - Computer Graphics and Geometric Modeling - page 477

8.4.3. Theorem. Let M be a k-dimensional submanifold of R n and let p ΠM . Let V
be an open neighborhood of p in M and let F : U Æ V be any parameterization of V ,
where U is an open set in R k . Let q ΠU and F( q ) = p .

(1) If M = R k = R n , then T p ( R k ) = R k .
(2) T p ( M ) is a k-dimensional vector space. In fact, T p ( M ) = DF( q )( R k ).
(3) The tangent plane of M at p is a k-dimensional plane.

∂
∂

F
u i

(4) The vectors

()

, i = 1,2,..., k, are a basis for T p ( M ).

q

Proof. Figure 8.13 should help the reader follow the proof. To prove (1) note that
the lines through q parallel to the coordinate axes in R k correspond to some very

special curves g i defined by

() =+ ( )

g i t qe t = q ,..., q , q + t q , ,..., q .

i 1 i - 1 i i + 1 k

Clearly, g i (0) = q . Furthermore, it is also easy to check that g i ¢(0) = e i . This gives us a
clue as to how to define a curve g(t) through q that has as a tangent vector an arbi-
trary vector v ΠR k . Simply let

k

Â

() =+ ( )

g t q tv e = q

11 ,...,

+ v t q + v t .

ii k k

i = 1

Then g¢(0) = v , from which (1) follows.
To prove (2) note that if g(t) is a parametric curve in R k

through q , then m(t) =

F(g(t)) is a curve in M through p . The chain rule implies that

R n

T p (M)

V m¢ 2 (0)

M

m 1 (t)

p

m 2 (t)

m¢ 1 (0)

F

R k

g 2 (t)

q

g 1 (t)

U

Figure 8.13. Defining the

standard basis
for the tangent
space.