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1. Mathematical Theory and Modeling
ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)
Vol.3, No.13, 2013
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Local Properties and Differential Forms of Smooth Map and
Tangent Bundle
Md. Shafiul Alam 1 , Khondokar M. Ahmed 2 & Bijan Krishna Saha 3
1
Department of Mathematics, University of Barisal, Barisal, Bangladesh
2
3
Department of Mathematics, University of Dhaka, Dhaka, Bangladesh
Department of Mathematics, University of Barisal, Barisal, Bangladesh
1
Corresponding author: Email: shafiulmt@gmail.com, Tel: +88-01819357123
Abstract
In this paper, some basic properties of differential forms of smooth map and tangent bundle are
d e fi n ed b y
is t he d eri va tive of at a,
developed. The li n ear m ap
where
is a smooth map and
. If
is a smooth bijective ma p and if the maps
are all injective, then is a diffeomorphism. Finally, it is shown that if X is a
vector field on a manifold M then there is a radial neighbourhood
gh
of
in
and a
smooth map
such that
and
, where
is given
by
.
Keywords: Smooth map, manifold, tangent bundle, isomorphism, diffeomorphism, vector field.
1. Introduction
S. Kobayashi1 and K. Nomizu1 first worked on differential forms of smooth map and tangent bundle. Later H.
Flanders2, F. Warner3, M. Spivak4 and J. W. Milnor5 also worked on differential forms of smooth map and
tangent bundle. A number of significant results were obtained by S. Cairns 6, W. Greub9, E. Stamm9, R. G.
Swan10, R. L. Bishop12, R. J. Crittenden12 and others.
Let M be a smooth manifold and let S (M) be the ring of smooth functions on M. A tangent vector of
M at a point
is a linear map S (M)
such that
S (M).
under the linear operations
The tangent vectors form a real vector space
S (M).
is called the tangent space of M at a.
Let E be an n-dimensional real vector space. Let
be an open subset of E and let
define a linear isomorphism
. If
is a smooth map of
sp ace F, then the classic al d erivative o f at a is the linear map
Moreover, in the special case
. We shall
into a second vector
given by
we have the product formula
S (O).
S (O)
given by
This shows that the linear map
of O at a. Hence, we have a canonical linear map
is a tangent vector
given by
.
2. The Derivative of a Smooth Map
Let
be a smooth map and let
. The li nea r map
is ca ll ed t h e d er i va tive of at a. It will be denoted by
,
101
d efi ned b y

2. Mathematical Theory and Modeling
ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)
Vol.3, No.13, 2013
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S (N),
If
map
is a second smooth map, then
we have
and
Lemma 1 . Let
to
map from
Proof. Let
.
. Moreover, for the identity
. In particular, if
is a diffeomorphism, then
are inverse linear isomorphisms.
, then
.
and the corre spondence
be a smooth map.
S (N),
.
defines a linear
S (N)
induces a homomorphism
S (M) given by
is a linear map from S (N) to . Moreover,
S (N)
and so
. Clearly
is linear. Therefore, the lemma is proved.
be a basis for
Proposition 1. Let
(a)
the functions
the functions
S ( ).
Then
S ( ) are given by
(b)
and let
S ( ) satisfy
Proof. By the fundamental theorem of calculus we have
Thus the theorem follows, with
Proposition 2. The canonical linear map
.
is an isomorphism.
Proof. First we consider the case
. We show first that
is surjective. Let
and let
S
where the
satisfy conditions (a) and (b) of Proposition 1.
( ). Write
Since maps the constant function
into zero, so
Since the functions are independent of f, we can write
surjective. To show that
is injective, let be any linear function in . Then for
Now suppose
. Then
. Hence
and so
is injective.
Finally, let be any open subset of E and let
identity map and we have the commutative diagram
be the inclusion map. Then
102
. Thus
,
is
.
is the

3. Mathematical Theory and Modeling
ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)
Vol.3, No.13, 2013
www.iiste.org
is a linear isomorphism, the canonical linear map
Since
completes the proof.
. Then dim
Corollary 1. Let M be a smooth manifold and let
Proof. Let
we find dim
is an isomorphism, which
be a chart for M such that
= dim
= dim = dim .
= dim
m
. Using the result of Lemma 1 and Proposition 2
be a smooth map
Proposition 3. The derivative of a constant map is zero. Conversely, let
Then, if M is connected, is a constant map.
such that
Proof. Assume that
function given by
follows that each
is the constant map
where
. Hence, for
.
S ( ) is the constant
. It
. Then, for
,
is a smooth map satisfying
and let
be
Conversely, assume that
connected. Then, given two points
and , there exists a smooth curve
such
and
. Consider the map
.
We have
that
.
Now using an atlas for N we see that
must be constant. In particular
. Thus is a constant map and the proposition is proved.
and so
3. Local Properties of Smooth Maps
be a smooth map. Then
is called a local diffeomorphism at a point
is a linear isomorphism. If is a local diffeomorphism for all points
called a local diffeomorphism of M into N.
if the map
, it is
Theorem 1. Let
point. Then
be a given
Let
be a smoo th map where d im M = n and dim N = r. Let
(a)
If is a local diffeomorphism at a, there are neighbourhoods U of a and V of b such that
maps U diffeomorphically onto V.
(b)
If
is injective, there are neighbourhoods U of a, V of b, and
such that
a diffeomorphism
(c)
If
of 0 in
.
is surjective, there are neighbourhoods U of a, V of b, and W of 0 in
a diffeomorphism
the projection.
and
such that
where
, and
is
. In part (a), then, we are
Proof. By using charts we may reduce to the case
assuming that
is an isomorphism, and the conclusion is the inverse function theorem.
For part (b), we choose a subspace E of
given by
such that
, and consider the map
. Then
103

4. Mathematical Theory and Modeling
ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)
Vol.3, No.13, 2013
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.
It follows that
i s i n j e c t i v e a n d t h u s a n i s o m o r p h i s m (r = dim Im
+ dim E = n +
dim E ). Thus part (a) implies the existence of neighborhoods U of a, V of b, and
of 0 in E such that
is a diffeomorphism. Clearly,
such that
Finally, for part (c), we choose a subspace E of
projection induced by this decomposition and define
. Let
. It follows easily that
Then
Hence there are neighborhoods U of a, V of b and W of
diffeomorphism. Hence the proof of the theorem is complete.
be the
by
is a linear isomorphism.
such that
is a smooth bijective map and if the maps
Proposition 4. If
all injective, then is a diffeomorphism.
is a
are
is inj ective, we have
. Now we show that r
Proof. Let d im M = n , dim N = r. Since
= n. In fact, according to Theorem 1, part (b), for every
there are neighborhoods
of a, V of
and
of
together with a diffeomorphism
such that the diagram
commutes (i denotes the inclusion map opposite 0).
of
such that each
is compact and contained
Choose a countable open covering
in some
. Since is surjective, it follows that
. Now assume that r > n. Then the
diagram implies that no
contains an open set. Thus, N could not be Hausdorff. This contradiction
shows that
. Since
, is a local diffeomorphism. On the other hand, is bijective. Since it
is a local diffeomorphism, Theorem 1 implies that its inverse is smooth. Thus is a diffeomorphism and the
theorem is proved.
Definition 1. A q uotient manifold o f a manifold M is a manifold N together with a smooth map
such that and each linear map
is surjective and thus dim M dim
N.
Lemma 2. Let
make N into a quotient manifold of M. Then the map
is open.
, there is a neighbourhood U of a such that the restriction
Proof. It is sufficient to show that, for each
of to U is open. This follows at once from part (c) of Theorem 1.
4. Submanifolds
Let M be a manifold. An embedded manifold is a pair
, where N is a second manifold and
is a smooth map such that the derivative
is injective. In particular, since the maps
are injective, it follows that dim N dim M. Given an embedded manifold
, the subset
may be considered as a bijective map
. A submanifold of a
manifold M is an embedded manifold
such that
is a homeomorphism, when
is
given the topology induced b y the topology of M. If N is a subset o f M and is the inclusion map,
we say simply that N is a submanifold of M.
104

5. Mathematical Theory and Modeling
ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)
Vol.3, No.13, 2013
Proposition 5. Let
www.iiste.org
be a submanifold of M. Assume that Q is a smooth manifold and
is a commutative diagram of maps. Then
is smooth if and only if
is smooth.
Proof. If is smooth then clearly is smooth. Conversely, assume that is smooth. Fix a point a
Q and set
. Since
is injective, there are neighbourhoods U, V of b in N and M respectively,
such that
.
and there is a smooth map
Since N is a submanifold of , the map is continuous. Hence there is a neighbourhood W of a such that
. Then
, where
denote the restrictions of
to W. It follows that
and so is smooth in W; thus is a smooth map. Hence the proposition is proved.
5. Vector Fields
A vector field X on a manifold M is a cross-section in the tangent bundle
. Thus a vector field X
a tangent vector
such that the map
so obtained is smooth.
assigns to every point
The vector fields on M form a module over the ring S ( ) which will be denoted by
.
onto the S ( )-module of derivations in the
Theorem 2. There is a canonical isomorphism of
algebra S ( ), i.e., Der S ( ).
Proof. Let
be a vector field. For each
S ( ), we define a function
.
is smooth. To see this we may assume that
is smooth. Hence every vector field X on M determines a map
S( )
Obviously,
S ( ) given by
.
is a derivation in the algebra S ( ). The assignment
Der S ( ).
We show now that
is an isomorphism. Suppose
on M by
. But then
, for some
defines a homomorphism
. Then
S ( ).
This implies that
Then, for every point
; i.e. X = 0. To prove that
, determines the vector
is surjective, let
, given by
be any derivation in S ( ).
S ( ).
. T o sho w that this map is s mo o th, fix a point
. Using a
construct vector fields
and smooth functions
on M such that
, (V is some neighbourhood of a). Then
the vectors
form a basis for
. Hence, for each
, there is a unique
system of numbers
such that
. Applying
to
we obtain
. Hence
. Since the
are smooth
functions on M, this equation shows that X is smooth in V; i.e. X is a vector field. Finally, it follows from the
definition that
. Thus is surjective, which completes the proof.
We d efine
chart, it is
easy
by
to
6. Differential Equations
Let X be a vector field on a manifold M. An orbit for X is a smooth map
105
where
is some

6. Mathematical Theory and Modeling
ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)
Vol.3, No.13, 2013
open interval such that
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.
Co n sid er t h e p ro d u ct m an i fo ld
. W e cal l a s ub se t
or
where is an open interval on
union and finite intersection of radial sets is again radial.
and
Proposition 6. Let X be a vector field on M. Fix
of X such that
. Moreover, if
an orbit
some
then
.
Proof. For the first statement of the proposition we may assume
stand ard P icard existenc e theo rem.
radial if for each
,
containing the point 0. The
. Then there is an interval and
are orbits for X which agree at
. In this case it is the
for which
is both closed and
To prove the second part we show that the set of
open, and hence all of . It is obviously closed. To show that it is open we may assume
and
then we apply the Picard uniqueness theorem, which completes the proof.
Theorem 3. Let X be a vector field on a manifold M. Then there is a radial neighbourhood
in
and a smooth map
such that
of
and
where
is given by
. Moreover,
is uniquely determined by X.
Proof. Let
be an atlas for M. The Picard existence theorem implies our theorem for each
. Hence there are radial neighbourhoods
of
in
and there are smooth maps
and
.
such that
. Then
is a radial neighbourhood of
in
. Moreover,
is a
Now set
; if
, then
is an interval
radial neighbourhood of
containing . Clearly,
are orbits of X agreeing at 0, and so by Proposition 6 they
agree in . It follows that they agree in
. Thus the
defines a smooth map
which
follows from Proposition 6. Thus the theorem is proved.
has the desired properties. The uniqueness of
7. Conclusions
Differential forms are among the fundamental analytic objects treated in this paper. The
derivative of a smooth map appears as a bundle map between the corresponding tangent
bundles. They are the cross-sections in the exterior algebra bundle of the dual of the tangent
bundle and they form a graded anticommutative algebra. Vector fields on a manifold has been
introduced as cross-sections in the tangent bundle. The module of vector fields is canonically
isomorphic to the module of derivations in the ring of smooth functions.
References
1. Kobayashi, S. and K. Nomizu, 1963. Foundations of Differential Geometry, Vol. I, Wiley,
New York, 51-57.
2. Flanders, H., 1963. Differential Forms, Academic Press, New York, 44-49.
3. Warner, F., 1971. Foundations of Differentiable Manifolds and Lie Groups, ScottForesman, Glenview, Illinois, 29-38.
4. Spivak, M., 1970. A Comprehensive Introduction to Differential Geometry, Brandeis
Univ., Waltham, Massachusetts, 37-42.
5. Milnor, J. W., 1965. Topology from the Differentiable Viewpoint, Univ. Press of Virginia,
Charlottesville, 87-95.
6. Cairns, S., 1965. Differential and Combinatory Topology, Princeton Univ. Press, Princeton,
New Jersey, 76-82.
106

7. Mathematical Theory and Modeling
ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)
Vol.3, No.13, 2013
www.iiste.org
7. Alam, M. S., 2009. Manifolds with Fibre, Vector, Tangent Bundles and De Rham
Cohomology, MS thesis, Department of Mathematics, University of Dhaka, 33-44.
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11. Narasimhan, R., 1968. Analysis on Real and Complex Manifolds, North-Holland Publ.,
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107

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