Notes on Tangent Space
Let M be a C ∞ manifold and let p ∈ M . We let C ∞ (p) denote the
R-algebra of germs of C ∞ functions at p.
Definition 1. A tangent vector v at p is a map:
v : C ∞ (p) → R,
satisfying the following two properties:
i) v is R linear.
ii) v satisfies the Leibnitz (product) rule:
v(f.g) = v(f ) · g(p) + f (p) · v(g).
We denote the set of all tangent vectors at p by Tp (M ).
Remark 1. For f ∈ C ∞ (p), the value f (p) is well-defined.
Remark 2. If A is an R-algebra, a map D : A → R satisfying (i) and
(ii) in Definition 1 is called a derivation of A. Therefore: a tangent
vector at p is a derivation of the algebra C ∞ (p).
Remark 3. If v is as above and we let 1 ∈ C ∞ (p) denote the function
identically equal to 1 in M , then we have
v(1.1) = v(1).1 + 1.v(1) = 2v(1)
which implies v(1) = 0. For any other constant function c we also have
v(c) = c · v(1) = 0.
Example 2. This a very important example. In fact, it will turn out
that all tangent vectors arise in this way: Let
γ : (−ε, ε) ⊂ R → M
be a C ∞ map such that γ(0) = p. We call γ a (C ∞ ) curve through p.
Then we define:
d
vγ (f ) := |t=0 f (γ(t)).
dt
It is easy to check that vγ satisfies (i) and (ii) in Definition 1 and so it
is a tangent vector.
Example 3. Let M = Rn and p ∈ Rn . Then the map C ∞ (p) → R
that sends
∂f
f ∈ C ∞ (p) 7→
|p
∂xi
is a derivation and defines an element ∂/∂xi |p ∈ Tp (Rn ).
1
2
Given v1 , v2 ∈ Tp (M ) and a1 , a2 ∈ R and f ∈ C ∞ (p) we can define
(a1 v1 + a2 v2 )(f ) := a1 (v1 (f )) + a2 (v2 (f ))
and it is easy to check that a1 v1 + a2 v2 ∈ Tp (M ), i.e. satisfies (i) and
(ii) in Definition 1. Hence, we have:
Proposition 4. Tp (M ) is naturally a vector space over R.
Example 5. Let us return to the case where M = Rn . Given p ∈ Rn
we can define a linear map
(1)
n
n
θ : R → Tp (R ) by θ((λ1 , . . . , λn )) :=
n
X
i=1
λi
∂
|p
∂xi
In other words,
θ((λ1 , . . . , λn ))(f ) :=
n
X
λi
i=1
∂f
(p)
∂xi
is the directional derivative of f at p in the direction of the vector
(λ1 , . . . , λn ) ∈ Rn .
The following is the key result:
Theorem 6. The map θ is a linear isomorphism. In particular, Tp (Rn )
is an n-dimensional real vector space and the partial derivatives ∂/∂xi |p
are a basis of Tp (M ).
Proof. We begin by showing that θ is injective. Suppose θ(λ) = 0, for
λ = (λ1 , . . . , λn ) ∈ Rn . This means that
θ(λ)(f ) = 0
for all f ∈ C ∞ (p). But taking f = xj we have:
0 = θ((λ1 , . . . , λn ))(xj ) :=
n
X
i=1
λi
∂xj
(p) = λj ,
∂xi
and therefore λ = 0.
Next we want to show that θ is surjective. Let v ∈ Tp (Rn ) and let
f ∈ C ∞ (p). Pick a representative C ∞ function f : U → R, p ∈ U , and
recall the following result proved in class:
Given a C ∞ function f : U → R, p ∈ U , there exists an open set W ,
p ∈ W ⊂ U and C ∞ functions g1 , . . . , gn : W → R such that
f (x) = f (p) +
n
X
i=1
gi (x)(xi − pi ) for all x ∈ W.
3
and gi (p) = ∂f /∂xi (p). This means that as germs, i.e. in C ∞ (p), we
have
n
X
f = f (p) +
gi · (xi − pi ),
i=1
where now f (p) denotes the constant function with value f (p), xi denotes the funcion: “projection on the i-th coordinate” and pi denotes
the constant function with value pi . Then, using properties (i) and (ii),
in Definition 1, that characterize tangent vectors we have:
!
n
X
v(f ) = v(f (p)) + v
gi · (xi − pi )
i=1
=
=
n
X
i=1
n
X
(v(gi ) · (xi − pi )(p) + gi (p) · v((xi − pi )))
gi (p) · v(xi ) =
i=1
n
X
v(xi ) ·
i=1
∂f
(p)
∂xi
Since this is true for every f ∈ C ∞ (p), we have:
(2)
v =
n
X
v(xi ) ·
i=1
∂
|p .
∂xi
Remark 4. Note that v(xi ) means the value of v applied to the coordinate function xi ∈ C ∞ (p) and that these are exactly the coefficients of
v written on the basis ∂/∂xi |p .
Example 7. Again let M = Rn , p ∈ Rn and let γ : (−ε, ε) ⊂ R → Rn
be a curve through p, i.e. γ(t) = (γ1 (t), . . . , γn (t)). Let us write the
tangent vector vγ of Example 2 in terms of the basis ∂/∂xi |p of Tp (Rn ).
By (2) we have:
n
X
∂
|p .
vγ =
vγ (xi ) ·
∂xi
i=1
But, by the definition of vγ in Example 2 we have:
vγ (xi ) =
d
d
|t=0 (xi (γ(t))) =
|t=0 (γi (t)) = γi0 (0).
dt
dt
Therefore
vγ =
n
X
i=1
γi0 (0) ·
∂
|p .
∂xi
4
and, via the identification of Tp (Rn ) with Rn given by the basis ∂/∂xi |p
we have
vγ = γ 0 (0) = (γ10 (0), . . . , γn0 (0)).
That is, via these identifications, vγ is the usual tangent vector to γ at
p.
Note that we can now verify for Rn the assertion in Example 2 that
every tangent vector is a vγ for some curve γ through p. Indeed, given
v ∈ Tp (Rn ) we can write
v =
n
X
∂
|p .
∂xi
λi ·
i=1
and it suffices to take γ(t) = (p1 + λ1 t, . . . , pn + λn t).
In order to extend all these results to an arbitrary manifold M we
will make use of the “functorial” properties of the definition of tangent
vectors.
Definition 8. Let M, N be C ∞ manifolds and F : M → N a C ∞ map.
Let p ∈ M and q = F (p) ∈ N . Then we define:
F∗,p : Tp (M ) → Tq (N )
by
[F∗,p (v)] (g) := v(g ◦ F ) ;
v ∈ Tp (M ), g ∈ C ∞ (q)
In order for Definition 8 to make sense we must check that F∗,p (v) ∈
Tq (N ), that is that it satisfies the two conditions that define tangent
vectors. In fact, we have:
Theorem 9. Let F∗,p be as above. Then:
i) For every v ∈ Tp (M ), F∗,p (v) ∈ Tq (N ).
ii) The map F∗,p : Tp (M ) → Tq (N ) is linear.
iii) Let id : M → M denote the identity map in M . Then, for every
p ∈ M , id∗,p is the identity map in Tp (M ).
iv) Let F : M → N and G : N → X be C ∞ maps between C ∞
manifolds. Let p ∈ M and q = F (p) ∈ N . Then
(G ◦ F )∗,p = G∗,q ◦ F∗,p
(Chain Rule)
v) Let F : M → N be a diffeomorphism. Then, for every p ∈ M ,
F∗,p : Tp (M ) → Tq (N )
is a linear isomorphism.
5
Proof. The first two statements are straightforward verifications and
are left to the reader. The third statement is also clear from the definition: given v ∈ Tp (M ) and f ∈ C ∞ (p),
[id∗,p (v)] (f ) = v(f ◦ id) = v(f ).
Hence id∗,p (v) = v as asserted.
We now prove the Chain rule: Given v ∈ Tp (M ) and f ∈ C ∞ (p) we
have:
[(G ◦ F )∗,p (v)] (f )
=
=
v(f ◦ (G ◦ F )) = v((f ◦ G) ◦ F )
[F∗,p (v)] (f ◦ G) = [G∗,q (F∗,p (v))] (f ).
Since this holds for all f ∈ C ∞ (p), we have:
(G ◦ F )∗,p (v) = G∗,q (F∗,p (v)) ;
for all v ∈ Tp (M ).
Hence (G ◦ F )∗,p = G∗,q ◦ F∗,p .
Finally, suppose F : M → N is a diffeomorphism and let G : N → M
be the inverse map. Then G ◦ F = idM and F ◦ G = idN . Hence, if
F (p) = q it follows from the Chain rule and from (3) that:
G∗,q ◦ F∗,p = (G ◦ F )∗,p = (idM )∗,p = idTp (M ) ,
which implies that F∗,p is a linear isomorphism.
Remark 5. Let U be an open subset of a manifold M and suppose
∞
(p) be a germ of a C ∞ function at p
p ∈ U ⊂ M . Then let ϕ ∈ CM
considered as a point in M . Then for any representative f : W → R of
ϕ, the restriction
f |U ∩W : U ∩ W → R
defines a germ of a C ∞ function at p considered as a point in U . This
∞
map identifies CM
(p) with CU∞ (p) and, consequently identifies the space
of derivations, i.e. Tp (U ) is identified with Tp (M ).
Example 10. We return to Example 2. Let
γ : (−ε, ε) ⊂ R → M
be a curve through p = γ(0) ∈ M . We get a linear map
γ∗,0 : T0 (R) → Tp (M ).
On the other hand, it follows from Theorem 6 that T0 (R) is a onedimensional vector space with basis
d
|t=0 .
dt
(We write d/dt instead of ∂/∂t in the one-variable case.)
6
We now compute γ∗,0 (d/dt|t=0 ). Let f ∈ C ∞ (p), then
d
d
γ∗,0
|t=0 (f ) =
|t=0 (f ◦ γ) = vγ (f ).
dt
dt
Hence:
d
γ∗,0
|t=0 = vγ .
dt
Definition 11. Given a curve γ : (−ε, ε) → M through p ∈ M we call
the vector
d
γ∗,0
|t=0
dt
the tangent vector to γ at p. We will abuse notation and denote the
tangent vector of γ at p by γ 0 (0) as we would in Rn .
Theorem 12. Let F = (f1 , . . . , fm ) : Rn → Rm be a C ∞ map. Let
p ∈ Rn and q = F (p) ∈ Rm . Let ∂/∂xj |p , j = 1, . . . , n and ∂/∂yi |q ,
i = 1, . . . , m be the bases of Tp (Rn ) and Tq (Rm ), respectively, given by
Theorem 6. Then,
m
X
∂
∂
∂fi
F∗,p
|p =
(p)
|q .
∂xj
∂xj
∂yi
i=1
In particular, the matrix of F∗,p with respect to these bases agrees with
the matrix of partial derivatives
∂fi
(p) ; 1 ≤ i ≤ m, 1 ≤ j ≤ n,
∂xj
which means that if we use these bases to identify Tp (Rn ) ∼
= Rn , Tq (Rm ) ∼
=
m
R , the linear map F∗,p agrees with the usual differential DF (p); i.e.
we have a commutative diagram:
F*,p
Tp (Rn ) −−−→ Tq (Rm )



∼
∼
=y
y=
(3)
DF(p)
Rn −−−→ Rm
Proof. Applying (2) we can write:
m
X
∂
∂
∂
|p =
F∗,p
|p (yi )
|q .
F∗,p
∂xj
∂x
∂y
j
i
i=1
But
F∗,p
as asserted.
∂
∂
∂
∂fi
|p (yi ) =
|p (yi ◦ F ) =
|p (fi ) =
(p).
∂xj
∂xj
∂xj
∂xj
7
We return now to the case of an n-dimensional manifold M .
Theorem 13. If M is an n-dimensional manifold and p ∈ M , then
Tp (M ) is an n-dimensional vector space. Moreover, a choice of local coordinates (U, ϕ) around p determines a distinguished basis E1ϕ , . . . , Enϕ
of Tp (M ), where
∂
ϕ
−1
Ei := (ϕ )∗,ϕ(p)
|ϕ(p) .
∂xi
Proof. Let (U, ϕ) be local coordinates around p. Then ϕ : U → ϕ(U )
is a diffeomorphism. By (iv) in Theorem 9,
ϕ∗,p : Tp (U ) → Tϕ(p) (ϕ(U ))
is an isomorphism. But we have already remarked that we can identify
Tp (U ) with Tp (M ) and, since ϕ(U ) is open in Rn , Tϕ(p) (ϕ(U )) may
be identified with Tϕ(p) (Rn ). But by Theorem 6, Tϕ(p) (Rn ) is an ndimensional vector space and, therefore, so is Tp (M ).
The second assertion follows from the fact that ∂x∂ i |ϕ(p) , i = 1, . . . , n
is a basis of Tϕ(p) (Rn ) and that (ϕ−1 )∗,ϕ(p) is an isomorphism.
Remark 6. It is useful to note that if v ∈ Tp (M ) and (U, ϕ) are local
coordinates around p as above. Then we can write v in terms of the
basis Eiϕ as follows:
(4)
v =
n
X
v(ϕi ) Eiϕ .
i=1
Indeed, ϕ∗,p (v) ∈ Tϕ(p) (Rn ) and hence, by (2) we can write:
ϕ∗,p (v) =
n
X
i=1
n
X
∂
∂
[ϕ∗,p (v)] (xi )
|ϕ(p) =
v(ϕi )
|ϕ(p)
∂xi
∂xi
i=1
and, applying (ϕ−1 )∗,ϕ(p) to both sides of the above identity we get:
X
n
n
X
∂
−1
|ϕ(p) =
v =
v(ϕi ) (ϕ )∗,ϕ(p)
v(ϕi ) Eiϕ .
∂x
i
i=1
i=1
Theorem 14. Let M and N be manifolds of dimension m and n,
respectively, and let F : N → M be a C ∞ map. Suppose p ∈ N , q =
F (p) ∈ M , and (U, ϕ), (V, ψ) are local coordinates around p and q,
respectively, such that F (U ) ⊂ V . Set
F̃ = (f˜1 , . . . , f˜m ) := ψ ◦ F ◦ ϕ−1 : ϕ(U ) ⊂ Rn → ψ(V ) ⊂ Rm .
8
Let E1ϕ , . . . , Enϕ be the basis of Tp (N ) defined by the local coordinates
(U, ϕ) and let Gψ1 , . . . , Gψm be the basis of Tq (M ) defined by the local
coordinates (V, ψ). Then
F∗,p
Ejϕ
m
X
∂ f˜i
=
(p) Gψi .
∂x
j
i=1
Consequently, the matrix of F∗,p with respect to the bases Ejϕ and Gi ψ
is the matrix of partial derivatives:
!
∂ f˜i
(p) .
∂xj
Proof. We proceed as in the proof of Theorem 12 but using the expression (6). Indeed, we have:
F∗,p
Ejϕ
=
n
X
F∗,p Ejϕ (ψi ) · Gψi
=
n
X
ϕ
Ej (ψi ◦ F ) · Gψi
i=1
=
=
i=1
n X
i=1
n
X
i=1
=
−1
(ϕ )∗,ϕ(p)
∂
|ϕ(p)
∂xj
(ψi ◦ F ) · Gψi
∂(ψi ◦ F ◦ ϕ−1 )
(p) · Gψi
∂xj
n
X
∂ f˜i
(p) · Gψi
∂x
j
i=1
as asserted.
For completeness sake we give a second proof of this Theorem: Because of the Chain rule, the commutative diagram of maps:
U


ϕy
F
−−−→
F̃
V

ψ
y
ϕ(U ) ⊂ Rn −−−→ ψ(V ) ⊂ Rm
9
gives rise to a commutative diagram of vector spaces and linear maps:
Tp (N )

ϕ*,p 
y
F*,p
−−−→
Tq (M )

ψ
y *,q
F̃*,ϕ(p)
Tϕ(p) (Rn ) ∼
= Rm
= Rn −−−−→ Tψ(q) (Rm ) ∼
Since the vertical maps are isomorphisms and the bases {Ejϕ } and {Gψi }
correspond under these isomorphisms to the standard bases {∂/∂xj }
and {∂/∂yi }, respectively, then the matrix of F∗,p agrees with the matrix of F̃∗,ϕ(p) and the result now follows from Theorem 12.
Remark 7. Let v ∈ Tp (M ) and let (U, ϕ) be a local coordinate system
around p. Suppose
n
n
X
X
∂
ϕ
−1
|ϕ(p) .
v =
vi · Ei =
vi · (ϕ )∗,ϕ(p)
∂x
i
i=1
i=1
Then, if (U, ψ) is another system of coordinates around p, we can write:
n
n
X
X
∂
ψ
−1
v =
wj · Gj =
wj · (ψ )∗,ψ(p)
|ψ(p)
∂y
j
j=1
j=1
and it follows from Theorem 14 by taking F = id that:
n
X
∂(ψj ◦ ϕ−1 )
(ϕ(p)) · vi
wj =
∂x
i
i=1
Thus, the matrix
(5)
−1
D(ψ ◦ ϕ )(ϕ(p)) =
∂(ψj ◦ ϕ−1 )
(ϕ(p))
∂xi
is the change of basis matrix.
Corollary 15. Let M be a C ∞ manifold and p ∈ M . Then for any
v ∈ Tp (M ) there exists a curve γ : (−ε, ε) → M through p (i.e. γ(0) =
p) such that:
v = γ∗,0 (d/dt|0 ) = vγ .
Proof. Let (U, ϕ) be local coordinates around p and let
n
X
v =
vi · Eiϕ
i=1
be the local coordinate expression of v. Then, it is easy to check that
γ(t) := ϕ−1 (ϕ(p) + t · ν) ,
10
where ν = (v1 , . . . , vn ) ∈ Rn and ε is small enough so that ϕ(p) + t · ν ∈
ϕ(U ) for |t| < ε, is such a curve.
Proposition 16. Let M and N be manifolds let F : N → M be a C ∞
map. Let p ∈ N and q = F (p) ∈ M . Let γ(t) be a curve through p and
define:
µ(t) := F (γ(t)).
Then µ(t) is a curve through q and:
µ0 (0) = F∗,p (γ 0 (0)) .
Proof. This is a direct consequence of the definition of the tangent
vector to a curve and of the chain rule. Indeed:
µ0 (0)
=
=
=
=
µ∗,0 (d/dt|t=0 )
(F ◦ γ)∗,0 (d/dt|t=0 )
(F∗,γ(0) ◦ γ∗,0 ) (d/dt|t=0 )
F∗,p (γ 0 (0)).
Combining Corollary 15 with Proposition 16 gives a useful method
for computing the differential of a map.
Example 17. Let M = GL(n, R) and N = Sn (R) ⊂ Mn (R), where
Mn (R) denotes the vector space of n × n real matrices and Sn (R) the
subspace of symmetric matrices. Consider the map
F: M →M ;
F (X) = X T · X.
We will compute the differential F∗,I , where I denotes the identity
2
matrix. Since M is an open set in Rn , the tangent space TI (M ) is
2
naturally identified with Rn which is identified, in turn, with Mn (R).
Thus we may think of the tangent vectors V ∈ TI (M ) as n×n matrices.
As we saw above, we can consider the curve
γ(t) = I + tV
and we will have γ 0 (0) = V . In order to compute F∗,I (V ) we use
Proposition 16 to obtain that F∗,I (V ) is the tangent vector to the curve
F (γ(t)), i.e. to the curve
µ(t) = F (γ(t)) = (I + tV )T · (I + tV ) = I + t(V T + V ) + t2 (V T · V )
Therefore
(6)
F∗,I (V ) = µ0 (0) = V T + V.
11
Example 18. We use a similar argument to compute the differential
at the identity of the map
F : M → R ; F (X) = det(X).
2
Given V ∈ TI (M ) ∼
= Mn (R), we consider γ(t) = I + tV and
= Rn ∼
then
d
(7)
F∗,I (V ) = |t=0 (det(I + tV )) = tr(V ).
dt
To understand the last equality recall the expansion of the determinant
and note that the only terms with a single power of t would be those
containing n − 1 diagonal “ones”, but then the remaining term must
be a t-term from the diagonal.
Rank of a map. Tangent space of a submanifold.
Given a C ∞ map F : M → N , and p ∈ M , we have defined the rank:
rankp (F ) := rankp (D(ψ ◦ F ◦ ϕ−1 ))(ϕ(p))
for any choice of local coordinates (U, ϕ) around p and (V, ψ) around
q = F (p) and such that F (U ) ⊂ V . But it follows from Theorem 14
that the matrix D(ψ ◦ F ◦ ϕ−1 )(ϕ(p)) is the matrix of F∗,p relative to
the coordinate bases: {Eiϕ } and Gψj }. Hence we have:
(8)
rankp (F ) = rankF∗,p .
In particular, suppose N ⊂ M is an (immersed or embedded) submanifold. Then the inclusion map i : N → M has rank equal to dim(N )
at every point and this means that the differential
(9)
i∗,p : Tp (N ) → Tp (M )
is injective for all p ∈ N . From now on we will use i∗,p to identify
Tp (N ) with a subspace of Tp (M ).
Suppose now that f : M → X is a C ∞ map and let N ⊂ M be an
(immersed or embedded) submanifold. Then it is easy to check that
(10)
(f |N )∗,p = (f∗,p ) |Tp (N ) .
We recall that if N is an (immersed or embedded) submanifold of M
and p ∈ N , then there exist local coordinates (U, ϕ) so that
V := {q ∈ U : ϕn+1 (q) = · · · = ϕm (q) = 0}
is an open set in N (if N is embedded then we can choose U so that
V = U ∩ N ) and ψ = (ϕ1 , . . . , ϕn ) defines local coordinates for N on
12
V . It is easy to check that in the corresponding coordinate bases, the
map i∗,p is given by:
i∗,p (Gψi ) = Eiϕ ;
i = 1, . . . , n.
That is, under the identification (9), Tp (N ) is identified with the subspace of Tp (M ) generated by the first n standard basis vectors.
More generally, we have the following classical construction:
Example 19. Let S ⊂ Rn be an (immersed/embedded) k-dimensional
submanifold, U ⊂ Rk an open set, a C ∞ map φ : U → Rn is said to be
a local parametrization of S if and only if:
i) φ(U ) ⊂ S
ii) φ : U → S is continuous (this condition is automatically satisfied if S is embedded).
iii) rank(φ) = k everywhere in U .
Note, that such φ is locally 1:1 so we can make such an assumption,
locally.
By HW Problem 19 we know that φ : U → S is C ∞ and, by the rank
condition, we have that for all p ∈ U , the linear map
φ∗,p : Tp (Rk ) → Tφ(p) (S) ⊂ Rn
is an isomorphism and therefore the tangent vectors:
∂
φ∗,p
|p ; j = 1, . . . , k,
∂uj
are a basis of Tφ(p) (S). Note that according to Theorem 14 we have
φ∗,p
∂
|p
∂uj
=
n
X
∂φi
i=1
∂uj
(p)
∂
|φ(p) .
∂xi
As a particular example consider the case where S ⊂ R3 is the graph
of a C ∞ function f : U → R, where U is an open subset of R2 . Then
φ : U → R3 ;
φ(u, v) := (u, v, f (u, v))
is a (global) parametrization of S and for each p = (u0 , v0 , f (u0 , v0 )) ∈
S we have that:
∂
∂f
∂
∂
∂f
∂
+
(u0 , v0 ) ·
;
+
(u0 , v0 ) ·
∂x ∂u
∂z
∂y ∂v
∂z
are a basis of Tp (S) ⊂ R3 .
Consider now the case of product manifolds. Remember that if M, N
are C ∞ manifolds, the product M × N is also a C ∞ manifold and
M × N has a covering by local coordinate charts which are products
13
of coordinate charts in M and N . Recall also Homework Problem 17,
where you showed that given (p, q) ∈ M × N , the maps
ιq : M → M × N ;
ιp : N → M × N
defined by
ιq (x) := (x, q) ;
ιp (y) := (p, y)
are embeddings. This means, in particular, that the differential (ιq )∗,p
(respectively, (ιp )∗,q ) identifies Tp (M ) (respectively, Tq (N )) with a subspace of T(p,q) (M × N ).
On the other hand we also have projection maps
πM : M × N → M ;
πN : M × N → N.
The following identities are easy to check:
i) πM ◦ ιq = idM for all q ∈ N and therefore (πM )∗,(p,q) ◦ (ιq )∗,p =
idTp (M ) .
ii) πN ◦ ιp = idN for all p ∈ M and therefore (πN )∗,(p,q) ◦ (ιp )∗,q =
idTq (N ) .
iii) πM ◦ ιp : N → M is a constant map and therefore (πM )∗,(p,q) ◦
(ιp )∗,q = 0.
iv) πN ◦ ιq : M → N is a constant map and therefore (πN )∗,(p,q) ◦
(ιq )∗,p = 0.
It is now an easy exercise to check that these identities imply that if we
identify Tp (M ) and Tq (M ) with subspaces of T(p,q) (M × N ) as above,
then
(11)
T(p,q) (M × N ) = Tp (M ) ⊕ Tq (N ).
Next, we want to study the tangent space of a manifold defined
“implicitly”.
Recall the following Theorem proved in class:
Theorem 20. Let F : M → N be a C ∞ map, q ∈ N , and suppose that
F has constant rank k in an open set containing the set Z = F −1 (q) ⊂
M . Then Z is an embedded submanifold of M and
(12)
dim Z = dim M − k.
We can now extend this theorem
Theorem 21. Let F : M → N be a C ∞ map, q ∈ N , and suppose that
F has constant rank k in an open set containing the set Z = F −1 (q) ⊂
M . Then, for every p ∈ Z,
Tp (Z) = ker{F∗,p : Tp (M ) → Tq (N )}
14
Proof. Since F∗,p has rank k,
dim (ker{F∗,p : Tp (M ) → Tq (N )}) = dim M − k.
Therefore, it suffices to prove that
Tp (Z) ⊂ ker{F∗,p : Tp (M ) → Tq (N )}
Now, for any v ∈ Tp (Z) there exists a smooth curve γ : (−ε, ε) → Z
through p and such that γ 0 (0) = v. We then have that F∗,p (v) = µ0 (0),
where µ(t) = F (γ(t)). But since γ(t) ∈ Z, µ(t) = q for all t and
µ0 (0) = 0. Hence,
F∗,p (v) = 0 ;
for all v ∈ Tp (Z).
Corollary 22. Let F : M → N be a smooth map and let q ∈ N be a
regular value such that X = F −1 (q) 6= ∅. Then, for every p ∈ X, the
following sequence is exact:
i*,p
F*,p
0 −−−→ Tp (X) −−−→ Tp (M ) −−−→ Tq (N ) −−−→ 0
Example 23. Let f : Rn+1 → R and suppose rankp f = 1 for all
p ∈ Z = f −1 (0), i.e., assume that 0 is a regular value of f . We
assume, moreover, that Z 6= ∅. Then, Z is an embedded submanifold
of dimension n in Rn+1 . We call Z a hypersurface of Rn+1 . We now
have:
Tp (Z) = ker{f∗,p : Tp (Rn+1 ) → T0 (R)}.
Via the usual identifications Tp (Rn+1 ) ∼
= Rn and T0 (R) ∼
= R, we know
that the matrix of f∗,p is given by:
∂f
∂f
(p), . . . ,
(p)
∂x0
∂xn
which as a row-vector may be identified with the gradient ∇f (p). Hence
we may identify:
Tp (Z) = (∇f (p))⊥ .
In particular, if
f (x0 , . . . , xn ) = x20 + · · · + x2n − 1
then for p = (p0 , . . . , pn ) ∈ Z ∼
= S n,
Tp (S n ) = (p0 , . . . , pn )⊥ .
15
Example 24. We continue with the notation of Example 23. Suppose
g : Z → R is a C ∞ map and suppose we want to find the critical points
of g, that is, we want to find the points p ∈ Z where rankp (g) = 0.
This means that
Tp (Z) ⊂ ker{g∗,p : Tp (Rn+1 ) → Tg(p) (R)}
but since both these spaces are n-dimensional they must be equal.
Hence
(∇f (p))⊥ = (∇g(p))⊥
and consequently, there exists λ ∈ R∗ such that
∇f (p) = λ · ∇g(p)
This is the so-called Lagrange multiplier equation.
Example 25. Let F : GL(n, R) → GL(n, R) be as in Example 17.
Then, since F∗,I (V ) = V T + V , we see that the image of F∗,I is the
subspace in Mn (R) consisting of all symmetric matrices. Hence
n(n + 1)
= dim TI (Sn (R)).
2
We claim that this is, in fact, the rank at every X0 ∈ F −1 (I): Consider
the commutative diagram:
rankF∗,I =
F
GL(n, R) −−−→ Sn (R)




LX0 y
yid
F
GL(n, R) −−−→ Sn (R)
where LX0 denotes left-multiplication by X0 . The fact that this diagram commutes follows from the fact that
F ((LX0 )(X)) = (X0 · X)T · (X0 · X) = X T · (X0T · X0 ) · X = F (X)
since F (X0 ) = X0T ·X0 = I. By the Chain rule, the above commutative
diagram gives rise to a commutative diagram of vector spaces and linear
maps:
F*,I
TI (GL(n, R)) −−−→ TI (Sn (R))




(LX0 )*,I y
yid
F*,X
0
TX0 (GL(n, R)) −−−→
TI (Sn (R))
Since the vertical maps are isomorphisms, the horizontal maps have
the same rank.
16
Proposition 26. The space
(13)
O(n, R) := {X ∈ GL(n, R) : X T · X = I}
is an embedded submanifold of GL(n, R) of dimension n(n−1)/2, called
the orthogonal group. Moreover, the tangent space TI (O(n, R)) may
be identified with the subspace:
{A ∈ Mn (R) : AT = −A},
i.e. with the subspace of skew-symmetric matrices, which will be denoted so(n, R).
Remark 8. O(n, R) has two connected components, the one containing
the origin:
(14)
SO(n, R) := {X ∈ GL(n, R) : X T · X = I ; det(X) = 1}
is called the special orthogonal group.
Proof. This is a direct consequence of the fact that the map
F : GL(n, R) → S(n, R)
has maximal rank at every point in O(n, R) and hence in a neighborhood of O(n, R). The result then follows from Theorems 20 and
21.
Transversality.
Suppose F : M → N is a C ∞ map and Z ⊂ N is an embedded
submanifold. Suppose the dimensions of M , N , and Z are m, n, and
k respectively. We would like to find conditions that guarantee that
X = F −1 (Z) is an embedded submanifold of M . This may be seen as
a generalization of “implicitly” defined submanifolds, in which case Z
is just a point.
Let q ∈ Z. Then, since Z is an embedded submanifold of N , there
exists a local coordinate system (V, ψ) of q in N such that:
V ∩ Z = {y ∈ V : ψk+1 (y) = · · · = ψn (y) = 0}.
Let η denote the projection
η : V → Rn−k ;
η = (ψk+1 , . . . , ψn ).
Then, setting U = F −1 (V ) ⊂ M we have
(15)
U ∩ X = (η ◦ F )−1 (0).
Therefore it would suffice to have 0 be a regular value of
G := η ◦ F : U → Rn−k
17
to be able to conclude that U ∩ X is an embedded submanifold of U
which, in turn, would imply that X is an embedded submanifold of M .
We will now translate the condition “0 is a regular value of G” in
terms of the original data. We have the following diagram, where i
denotes the inclusion Z ⊂ N and the horizontal sequence is exact by
Corollary 22.
Tp (N )

F
y *,p
i*,q
η*,q
0 −−−→ Tq (Z) −−−→ Tq (N ) −−−→ Rn−k −−−→ 0
Now we have:
Proposition 27. With the notation as above, η∗,q ◦ F∗,p is surjective if
and only if
(16)
Tq (N ) = F∗,p (Tp (N )) + Tq (Z),
where, as usual, we use i∗,q to identify Tq (Z) with a subspace of Tq (N ).
Proof. We begin by showing the if direction. Suppose w ∈ Rn−k , then
since η∗,q is surjective, there exists v ∈ Tq (N ) such that w = η∗,q (v).
Now, by asumption we can write
v = F∗,p (u) + v1 ;
u ∈ Tp (M ), vi ∈ Tq (Z).
But then
w = η∗,q (v) = η∗,q (F∗,p (u)) + η∗,q (v1 ) = η∗,q (F∗,p (u)) ,
where the last identity follow from the fact that the composition of the
horizontal maps is zero. Hence, η∗,q ◦ F∗,p is surjective.
Suppose now that η∗,q ◦ F∗,p is surjective and let v ∈ Tq (N ), then
there exists u ∈ Tp (N ) such that
η∗,q (F∗,p (u)) = η∗,q (v).
But this implies that
v − F∗,p (u) ∈ ker(η∗,q ) = Tq (Z).
Definition 28. Let F : M → N be a smooth map and Z ⊂ N an
embedded submanifold. Then we say that F is transversal to Z and
write F >
∩ Z if (16) holds for every p ∈ M and q = F (p). Given
18
embedded submanifolds X, Y in M we say that X is transversal to Y ,
and write X >
∩ Y if and only if for every p ∈ X ∩ Y :
(17)
Tp (M ) = Tp (X) + Tp (Y ).
Example 29. If F : M → N is a submersion then F >
∩ Z for all
embedded submanifolds Z ⊂ N . Indeed, since for all p ∈ M , q =
F (p) ∈ N , F∗,p (Tp (M )) = Tq (N ), condition (16) holds independently
of Z.
Example 30. If X, Y are curves in R2 (or any two-dimensional manifold), we have X >
∩ Y if and only if the tangent lines Tp (X) and Tp (Y )
are different for all p ∈ X ∩ Y , i.e. if X and Y have no tangential
intersection.
Theorem 31. Suppose F : M → N be a smooth map and let Z ⊂ N
be an embedded submanifold. Then, if F >
∩ Z and X = F −1 (Z) 6= ∅, X
is an embedded submanifold of M and
dim M − dim X = dim N − dim Z.
(18)
Moreover, given p ∈ X and q = F (p) ∈ Z, we have:
Tp (X) = (F∗,p )−1 (Tq (Z)).
(19)
Proof. It follows from (15) that if F >
∩ Z and p ∈ X, then there exists
an open neighborhood U of p in M such that
X ∩ U = (η ◦ F )−1 (0),
where η takes values in Rn−k , n = dim N , k = dim Z, and 0 ∈ Rn−k is
a regular value of η ◦ F . Hence by Theorem 20, it follows that X ∩ U
is an embedded submanifold of U and, therefore, X is an embedded
submanifold of M . We have moreover:
dim X = dim M − (n − k),
which implies (18). The second statement follows from the fact that:
Tp (X) = Tp (X ∩ U ) = ker{(η ◦ F )∗,p : Tp (M ) → Rn−k }.
But, since Tq (Z) = ker{η∗,q : Tq (N ) → Rn−k }, we have that
v ∈ ker(η ◦ F )∗,p if and only if F∗,p (v) ∈ ker η∗,q
that is, if and only if
F∗,p (v) ∈ Tq (Z)
as asserted.
19
Remark 9. It is useful to define for a submanifold X ⊂ M :
codim(X) := dim M − dim X.
(20)
Then (18) becomes:
(21)
codim(X) = codim(Z).
Example 32. Suppose M1 , M2 , N are C ∞ manifolds and Fi : Mi → N ,
i = 1, 2, are C ∞ maps. Assume moreover that F1 is a submersion. Then
we claim that
X := {(x1 , x2 ) ∈ M1 × M2 : F1 (x1 ) = F2 (x2 )}
is an embedded submanifold of M1 × M2 and
codim(X) = dim N.
Indeed, consider the map
G := F1 × F2 : M1 × M2 → N × N.
Then X = G−1 (∆), where ∆ ⊂ N × N is the diagonal. Thus, by
Theorem 31, it suffices to show that G >
∩ ∆. Now, we know by a HW
problem that for any (q, q) ∈ ∆,
T(q,q) (∆) = {(u, u) ∈ Tq (N ) × Tq (N )}.
But, on the other hand, if (p1 , p2 ) ∈ X and F1 (p1 ) = F2 (p2 ) = q, then
G∗,(p1 ,p2 ) (T(p1 ,p2 ) (M1 × M2 ))
=
=
(F1 )∗,p1 (Tp1 (M1 )) × (F2 )∗,p2 (Tp2 (M2 ))
Tq (N ) × (F2 )∗,p2 (Tp2 (M2 ))
since F1 is a submersion. It is now easy to check that
T(q,q) (N × N )
=
=
Tq (N ) × Tq (N )
G∗,(p1 ,p2 ) (T(p1 ,p2 ) (M1 × M2 )) + T(q,q) (∆).
Indeed, given any (v1 , v2 ) ∈ Tq (N ) × Tq (N ), we can write
(v1 , v2 ) = (v1 − v2 , 0) + (v2 , v2 )
and (v2 , v2 ) ∈ T(q,q) (∆) and
(v1 − v2 , 0) ∈ Tq (N ) × (F2 )∗,p2 (Tp2 (M2 )).
The codimension statement follows from (21).
Example 33. Consider the set
X := {(v, L) ∈ (Rn+1 \{0}) × Pn : v ∈ L},
where we are thinking of Pn as the set of lines in Rn+1 . Then we may
apply the previous example with M1 = Rn+1 \{0}, M2 = N = Pn ,
F1 = π, the natural projection, and F2 = identity, to conclude that X
is an embedded submanifold of (Rn+1 \{0}) × Pn and codim(X) = n,
20
that is dim X = n + 1. We will see later that we can extend X to a
submanifold of Rn+1 × Pn .
Theorem 34. Suppose X, Y ⊂ M are embedded submanifolds. Then,
if X >
∩ Y and X ∩ Y 6= ∅, X ∩ Y is an embedded submanifold of M and
(22)
dim(X ∩ Y ) = dim X + dim Y − dim M
Moreover, given p ∈ X ∩ Y ,
Tp (X ∩ Y ) = Tp (X) ∩ Tp (Y ).
(23)
Proof. The statement follows from Theorem 31 after observing that if
we denote by iX : X → M the inclusion of X into M then X >
∩ Y if
and only if iX >
∩ Y and that
X ∩ Y = i−1
X (Y ).
Finally, the identity (18) becomes:
dim X − dim(X ∩ Y ) = dim M − dim Y,
which easily yields (22). The statement about the tangent space is a
direct consequence of (19).
Remark 10. If X, Y ⊂ M are embedded submanifolds of complementary dimension, i.e.:
dim X + dim Y = dim M,
and X >
∩ Y , then, by (22), X ∩ Y is a zero-dimensional embedded
submanifold. In particular, X ∩ Y is a discrete set and if either X or
Y is compact we have that X ∩ Y is a finite set. This construction
plays a very important role in algebraic and differential topology. We
will return to this later in the course.
Remark 11. If X, Y, X ∩ Y ⊂ M are embedded submanifolds and
dim X + dim Y ≥ dim M,
then (22) gives what is usually called the “expected dimension” of X∩Y .
We should note, however, that (22) may fail if X and Y do not intersect
transversally.
Sard’s Theorem.
The main references for this section are “Differential Topology” by
Guillemin and Pollack ([GP]) (Chapter 2 and Appendix 1), and Chapter 10 of Lee’s “Introduction to Smooth Manifolds”.
Recall that a set A ⊂ Rn is said to have (Lebesgue) measure zero if
and only if given any ε > 0 there exists a covering of A by open cubes
Cj := {x ∈ Rn : aij < xi < bij } ,
j = 1, . . . , ∞ ,
21
such that
X
n
X Y
voln (Cj ) =
(bij − aij )
j
j
!
< ε.
i=1
Example 35. If k < n then Rk ⊂ Rn has measure zero, where we
identify Rk with the subset of points whose last n − k coordinates are
zero. Indeed, we can clearly cover Rk with a countable collection of
cubes C̃j such that volk (C̃j ) = 1, consider then
−ε1/k
ε1/k
<
y
<
}
`
2j/k
2j/k
Then, voln (Cj ) = ε/2j and the sum of the volumes is at most ε.
Cj := C̃j × {y ∈ Rn−k :
A crucial property of sets of measure zero is the following:
Theorem 36. The countable union of sets of measure zero is of measure zero.
S
Proof. Let A =
i Ai be a countable union of sets Ai each of which
is of measure zero. Given ε > 0, cover Ai with a countable collection
of cubes Cji so that
X
voln (Cji ) < ε/2i .
j
Then the cubes
{Cji }
cover A and
P
i,j
voln (Cji ) < ε.
We also note the following:
Proposition 37. An open cube:
C := {x ∈ Rn : ai < xi < bi } ,
with ai < bi does not have measure zero.
Proof. See [GP], page 203.
Corollary 38. A set of measure zero may not contain any open sets;
i.e., a set of measure zero is nowhere dense and, in particular, its
complement is a dense set.
Although the property of having measure zero does not behave well
with respect to continuous maps, it does with respect to differentiable
maps. This is so because a differentiable map is Lipschitz on a compact
set.
Theorem 39. Let F : Rn → Rn be a differentiable map. If A ⊂ Rn
has measure zero then F (A) has measure zero.
Proof. See [GP], page 204.
22
Theorems 36 and 39 allow us to define the notion of measure zero in
a manifold as follows:
Definition 40. Let M be a manifold, a subset A ⊂ M is said to be of
measure zero if there exists a countable covering of M by coordinate
charts (Uj , ϕj ) such that ϕj (A ∩ Uj ) is of measure zero in Rn for all j.
It now follows from Example 35, Theorems 36 and 39, and the Rank
Theorem that:
Theorem 41 (Easy Sard). If F : M → N is a smooth map and
dim M < dim N then the image F (M ) is of measure zero in N .
Theorem 42 (Sard’s Theorem). Let F : M → N be a smooth map.
Then the set C ⊂ N of critical values of F is of measure zero in N .
Proof. We refer to [GP].
Remark 12. Recalll that a point q ∈ N is a critical value of F if there
exists p ∈ N such that F (p) = q but F∗,p : Tp (M ) → Tq (N ) is not
surjective. In particular, the set of critical values is always contained
in the image F (M ) and we see that in the case dim M < dim N , Sard’s
theorem is equivalent to Theorem 41. Even this simple version of Sard’s
theorem is very useful.
The following theorem provides the link between Sard’s Theorem
and transversality problems:
Theorem 43. Let M, B, N be manifolds and Y ⊂ N an embedded
submanifold. Suppose
F: M ×B →N
is a C ∞ map and that F >
∩ Y . Let X = F −1 (Y ). Let p : M × B → B
be the projection and let π : X → B denote the restriction of p to X.
For each t ∈ B we denote by ft : M → N the map
ft (x) := F (x, t).
Then t is a regular value for π if and only if ft >
∩ Y . In particular,
ft >
∩ Y for all t in the complement of a set of measure zero in B.
Proof. Suppose t is a regular value of π, then for all (p, t) ∈ X,
(24)
π∗,(p,t) T(p,t) (X) = Tt (B).
We would like to prove that if (24) holds then ft >
∩ Y , that is that for
−1
all q ∈ Y and all p ∈ (ft ) (q) we have
(25)
Tq (N ) = (ft )∗,p (Tp (M )) + Tq (Y ).
23
On the other hand since, by assumption, F >
∩ Y we have that
Tq (N ) = F∗,(p,t) (Tp (M ) × Tt (B)) + Tq (Y ).
(26)
But, suppose w ∈ F∗,(p,t) (Tp (M ) × Tt (B)) then
w = (ft )∗,p (u) + (gp )∗,t (v),
where u ∈ Tp (M ), v ∈ Tt (B) and gp : B → N is the map defined by
gp (t) = F (p, t).
Now, by (24) we can write
v = π∗,(p,t) (ν)
with ν ∈ T(p,t) (X) and therefore:
0
(gp )∗,t (v) = (gp )∗,t (π∗,(p,t) (ν)) = F∗,(p,t) (ν) − (ft )∗,q (π∗,(p,t)
(ν)),
where π 0 : M × B → M is the natural projection. But, since by (19)
we have:
T(p,t) (X) = (F∗,(p,t) )−1 (Tq (Y )),
we have:
F∗,(p,t) (ν) ∈ Tq (Y )
and consequently
v ∈ (ft )∗,p (Tp (M )) + Tq (Y ),
and this yields (25).
Example 44. Let f : M → Rn be a C ∞ map and Y ⊂ Rn an embedded
submanifold. Then for any ε > 0, and almost all (i.e. except for a set
of measure zero) vectors λ ∈ Rn , k λ k< ε, the map
fλ : M → R n ;
fλ (x) := f (x) + λ ,
is transversal to Y .
Indeed, let B = {λ ∈ Rn :k λ k< ε} and define
F : M × B → Rn
by F (x, λ) = f (x) + λ. Then, since F is clearly a submersion, we
have by Example 29 that F >
∩ Y and the statement now follows from
Theorem 43.
Similarly, we can consider M = X ⊂ Rn and f = iX , then we have
that for any ε > 0, and almost all (i.e. except for a set of measure
zero) vectors λ ∈ Rn , k λ k< ε:
(27)
Xλ >
∩ Y , where Xλ := {x + λ : x ∈ X}.
24
Example 45. Suppose X ⊂ Rn is an embedded submanifold and let
0 ≤ k ≤ n. Then we claim that almost every k-dimensional subspace
of Rn intersects X transversally. By “almost every” we mean that the
set of points in the grassmannian G(k, n) representing subspaces which
do not intersect X transversally is of measure zero in G(k, n).
Indeed, let us denote by S ⊂ G(k, n) the set of k-dimensional subspaces that do not intersect X transversally. It is enough to show that
S ∩ UI , I = {1 ≤ i1 < · · · < ik ≤ n}, has measure zero for all index
sets I and where UI is the usual coordinate neighborhood in G(k, n).
Now, we may assume without loss of generality that I = {1, . . . , k}.
Then any subspace Ω ∈ UI has a unique basis of the form:
νi = ei + ωi ;
i = 1, . . . , k,
where e1 , . . . , ek is the standard basis of Rk ⊂ Rn and
ωi ∈ Rn−k ∼
= {v ∈ Rn : v1 = · · · = vk = 0}.
Consider now the map
2
F : Rk × Rk → Rn
that maps
((λ1 , . . . , λk ), (ω1 , . . . , ωk )) 7→
k
X
λj (ej + ωj ) ,
j=1
where λj ∈ R, ωj ∈ Rk . Clearly F is a submersion and therefore it is
2
transversal to X. Thus, for almost all (ω1 , . . . , ωk ) ∈ Rk the restriction
of F is also transversal to X, but that restriction has as image the
subspace with basis ej + ωj and consequently almost all suspaces in UI
intersect X transversely.
A particular interesting case is when k = n − 1 since the only way
that a hyperplane H may fail to be transversal to X is that for some
point x ∈ H ∩ X, Tp (X) ⊂ H, i.e. if H is “tangential” to X. In
projective geometry this is called the dual variety.