Mukhi S., Mukunda N. Lectures on Advanced Mathematical Methods for Physicists
Подождите немного. Документ загружается.
12
Chapter
1.
Topology
topology.
The
reader
should
check
this
statement.
Exercise:
In
the
usual
topology
on
IR
2
,
give some examples
of
open
sets
that
are
not
open
discs. Also find a basis for
the
same
topology consisting
of
open
sets
that
are
different from
open
discs.
1.5
Closure
In
many
familiar cases, we have seen
that
the
distinction
between
open
and
closed
sets
is
that
the
former do
not
contain
their
"end
points" while
the
latter
do.
In
trying
to
make
this
more
precise, we
are
led
to
the
concept of
closure
of
a set.
In
IR
2
,
for example,
the
set
Xl
= {
x
c
IR
2
I
i!2
<
1 } defines
an
open
set,
the
open
unit
disc.
It
is
open
because every
point
in
it
can
be
enclosed
in
a small
open
disc which lies inside
the
bigger
one
(see Fig.
l.1).
Points
on
the
unit
circle i!2
=
1
are
not
in
the
unit
open
disc, so we
don't
have
to
worry
about
enclosing
them.
.
.
.
~
~"'
...
'"
(~"'.
,
.......
,
,
,
,
.........
,
Figure
l.1:
The
open
unit
disc.
Every
point
in
it
can
be
enclosed
in
an
open
disc.
On
the
other
hand
consider
the
set
X
2
= {
xC
IR2
I
i!2
:s
1 }.
In
addition
to
points
within
the
unit
circle,
this
set
contains
all
points
on
the
unit
circle.
But
clearly
X
2
is
not
an
open
set, since
points
on
the
boundary
circle
cannot
be
enclosed
by
open
discs
in
X
2
.
In
fact
X
2
is a
closed
set, as one
can
check
by
going
back
to
the
axioms.
So
it
seems
that
we
can
add
some
points
to
an
open
set
and
make
it
a
closed set.
Let
us
make
this
precise
via
some definitions.
Definition:
Let
(S,
U)
be
a topological space.
Let
A
C
S.
A
point
s
E
S
is
called a
limit point
of
A
if, whenever
s
is
contained
in
an
open
set
u
E
U,
we
have
(u-{s})nA#¢
1.6.
Connected and Compact Spaces 13
In
other
words,
S
is a limit
point
of
A
if every
open
neighbourhood
of
S
has
a
non-empty
intersection
with
A.
Exercise:
Show
that
all
points
on
the
boundary
of
an
open
disc
are
limit
points
of
the
open
disc.
They
do
not,
however, belong
to
the
open
disc.
This
shows
that
in
general, a limit
point
of
a
set
A
need
not
be
contained
in
A.
Definition:
The
closure
of
a
set
A
c
Sis:
A
=
A
u {
limit
points
of
A }
In
other
words, if we
add
to
a set all
its
limit points, we get
the
closure
of
the
set.
This
is so
named
because
of
the
following result:
Theorem:
The
closure
of
any
set
A
c
S
is a closed set.
Exercise:
Prove
this
theorem.
As always,
it
helps
to
go
back
to
the
definition.
What
you
have
to
show is
that
the
complement
of
the
closure
A
is
an
open
set.
Given a topological space
(S,
U),
we
can
define a topology
on
any
subset
A
c
S.
Simply choose
U
A
to
be
the
collection
of
sets
ui
n
A,
Ui
C
U.
This
topology is called
the
relative topology
on
A.
Note
that
sets
which
are
open
in
A
C
S
in
the
relative topology need
not
themselves
be
open
in
S.
Exercise:
Consider
subsets
of
IR
and
find
an
example
to
illustrate
this
point.
1.6
Connected
and
Compact
Spaces
Consider
the
real
line
IR
with
the
usual
topology.
If
we
delete
one
point,
say
{O},
then
IR
-
{O}
falls
into
two disjoint pieces.
It
is
natural
to
say
that
IR
is
connected
but
IR
-
{O}
is disconnected. To
make
this
precise
and
general,
we
need
to
give a definition in
terms
of
a given topology
on
a set.
Definition:
A topological
space
(S, U)
is
connected
if
it
cannot
be
expressed
as
the
union
of
two disjoint
open
sets
in
its
topology.
If
on
the
other
hand
we
can
express
it
as
the
union
of
disjoint
open
sets, in
other
words if we
can
find
open
sets
Ul,
U2
E
U
such
that
Ul
n
U2
=
¢,
Ul
U
U2
=
S,
then
the
space is
said
to
be
disconnected.
Theorem:
In
a
connected
topological
space
(S,
U),
the
only
sets
which
are
both
closed
and
open
are
Sand
¢.
Exercise:
Prove
this
theorem.
This
definition tells us in
particular
that
with
the
usual
topology, IR
n
is
connected for all
n.
IR
-
{O}
is disconnected,
but
IR
n
-
{O}
is
connect~d
for
n
2':
2.
On
the
other
hand,
the
space
]R.2
- {
(x,
0)
I x
E
IR}, which is
IR2
with
a line removed, is disconnected. Similarly,
IR2
- {
(x,
y)
I
x
2
+
y2
=
I},
14
Chapter
1.
Topology
which is
IR
2
with
the
unit
circle removed, is disconnected.
IR3
minus a
plane
is
disconnected,
and
so on.
In
these
examples
it
is sufficient
to
rely
on
our
intuition
about
connectedness,
but
in
more
abstract
cases one needs
to
carefully follow
the
definition above.
Note
that
connectivity
depends
in
an
essential way
on
not
just
the
set,
but
also
the
collection
of
open
sets
U,
in
other
words
on
the
topology.
For
example,
IR
with
the
discrete
topology is disconnected:
IR=Ua{a},
aEIR
(1.10)
Recall
that
each
{a}
is
an
open
set, disjoint from every
other,
in
the
discrete
topology. We
can
also conclude
that
IR
is disconnected in
the
discrete topology
from a
theorem
stated
earlier.
In
this
topology,
IR
and
¢
are
not
the
only
sets
which
are
both
closed
and
open, since
each
{a}
is also
both
closed
and
open.
We now
turn
to
the
study
of
closed, bounded
3ubsets
of
IR
which will
turn
out
to
be
rather
special.
Definition:
A
cover
of
a set
X
is a family
of
sets
{Fa}
=
F
such
that
their
union
contains
X,
thus
Xc
UaFa
If,
(S,
U)
is a topological space
and
XeS,
then
a cover {
Fa}
is
said
to
be
an
open cover
if
Fa
E
U
for all
n,
namely, if
Fa
are
all
open
sets.
Now
there
is a famous
theorem:
Heine-Borel
Theorem:
Any
open
cover
of
a closed
bounded
subset
of
IR
n
(in
the
usual topology)
admits
a finite subcover.
Let
us see
what
this
means
for
IR.
An
example
of
a closed
bounded
sub-
set
is
an
open
interval
[a,
bj.
The
theorem
says
that
if
we
have
any
(possibly
infinite) collection
of
open
sets
{Fa}
which cover
[a,
bj
then
a finite
subset
of
this
collection also exists which covers
[a,
bj.
The
reader
should
try
to
convince
herself
of
this
with
an
example.
An
open
interval
(a,
b)
is
bounded
but
not
closed, while
the
semi-infinite
interval
[0,00)
= {
x
E
IR
I
x
>
0 }
is closed
but
not
bounded.
So
the
Heine-
Borel
theorem
does
not
apply
in
these
cases.
And
indeed, here is
an
example
of
an
open
cover
of
(a,
b)
with
no finite
sub
cover. Take
(a,
b)
=
(-1,
1).
Let
Fn=
(-1+~,1-~)
n=2,3,4,
...
n n
(1.11)
With
a
little
thought,
one sees
that
U~=l
Fn
=
(-1,1).
Therefore
Fn
provides
an
open
cover
of
(-1,1).
But
no
finite
subset
of
the
collection
{Fn}
is a cover
of
(-1,1).
Exercise:
Find
an
open
cover
of
[0,
00)
which
has
no
finite subcover.
Closed
bounded
subsets
of
IR
n
have several
good
properties.
They
are
1.7.
Continuous
Functions
15
known as
compact
sets
on
lRn.
Now we need
to
generalize
this
notion
to
arbitrary
topological spaces.
In
that
case
we
cannot
always give a
meaning
to
"bounded"
,
so we proceed using
the
equivalent notions
provided
by
the
Heine-Borel
theorem.
Definition:
Given a topological
space
(3,
U),
a
set
X
c
3 is
said
to
be
compact
if every
open
cover
admits
a finite
sub
cover.
Theorem:
Let
3
be
a
compact
topological space.
Then
every infinite
subset
of
3
has
a limit point.
This
is
one
example
of
the
special
properties
of
compact
sets.
Exercise:
Show
by
an
example
that
this
is
not
true
for
non-compact
sets.
It
is
worth
looking
up
the
proof
of
the
above
theorem.
Theorem:
Every
closed
subset
ofa
compact
space is
compact
in
the
relative
topology.
Thus,
compactness
is preserved
on
passing
to
a topological subspace.
1.7
Continuous
Functions
Using
the
general
concept
of
topological spaces developed above, we now
turn
to
the
definition
of
continuity. Suppose (3,
U)
and
(T,
V)
are
topological spaces,
and
f :
3
----t
T
is a function. Since
f
is
not
in general injective (one
to
one),
it
does
not
have
an
inverse in
the
sense
of
a function
f-
1
:
T
----t
3.
But
we
can
define a
more
general
notion
of
inverse, which
takes
any
subset
of
T
to
some
subset
of
3.
This
will provide
the
key
to
understanding
continuity.
Definition:
If
T'
c
T,
then
the
inverse
f-
1
(T')
c
3 is defined by:
f-l(V)
= {
s
E
3
I
f(s)
E
T'
}
Note
that
the
inverse is defined
on
all
subsets
T'
of
T,
including
the
indi-
vidual
elements
{t}
C
T
treated
as special, single-element subsets. However
it
does
not
necessarily
map
the
latter
to
individual
elements s
E
3,
but
rather
to
subsets
of 3
as
above.
The
inverse
evaluated
on
a
particular
subset
of
T
may
of
course
be
the
empty
set
¢
C
3.
For
the
special case of bijective (one-to-one
and
onto)
functions, single-
element
sets
{t}
c
T
will
be
mapped
to
single-element
sets
f-
1
(
{t})
c
3.
In
this
case we
can
treat
the
inverse as a function
on
T,
and
the
definition above
coincides
with
the
usual
inverse function.
Consider
an
example
in
lR
with
the
usual
topology:
Example:
f :
lR
----t
lR+ is defined by
f(x)
=
x
2
.
Then,
f-
1
:
{y}
C
lR+
----t
{Vfj,
-Vfj}
c
lR (Fig.
1.2).
Now
take
f-
1
on
an
open set
of
lR+, say
(1,4).
Clearly
rl:
(1,4) C
lR+
----t
{(I,
2),
(-1,
-2)}
C
lR.
(1.12)
16
Chapter
1.
Topology
x
-2
-I
2
Figure
1.2:
f(x)
=
x
2
,
an
example
of
a
continuous
function.
Thus
th
e inverse
maps
open
sets
of
lR.+
to
open
sets
of
lR..
One
can
convince oneself
that
this
is
true
for
any
continuous
function
lR.
---+
lR.
+.
Moreover,
it
is false for
discontinuous
functions,
as
the
following
example
shows (Fig. 1.3):
f(x)
=
x
+ 1,
x
<
0
=
x
+
2,
x
~
0
f(x)
-\
112
x
Figure
1.3:
An
example
of
a
discontinuous
function.
In
this
example,
the
open
set
(~,
~)
c
lR.
is
mapped
by
f-
1
to
the
set
[O,~)
which is
not
open.
In
fact
it
can
be
shown
that
on
lR.
with
the
usual
topology,
the
continuous
functions
are
those
for
whom
the
inverse always
takes
open
sets
to
open
sets. For
discontinuous
functions (i.e. functions
having
a
"break"
in
their
graph)
there
will
be
at
least
one
open
set
which is
mapped
by
the
inverse
1.8.
Homeomorphisms
17
function
to
a
non-open
set.
Exercise:
Try
to
formulate a
general
proof
of
the
above
statement.
As before, we use
this
property
of
th
e real line
as
a way
of
defining contin-
uous functions
on
arbitrary
topological spaces.
Definition:
For general topological spaces
(S,
U)
and
(T,
V),
a function
I :
S
---7
T
is called
continuous
if
its
inverse
takes
open
sets
of
T
to
open
sets
of
S.
Exercise:
Show
that
if we
put
the
discrete topology
on
Sand
T
then
every
function
I:
S
---7
T
is
continuous. Clearly
this
topology is
too
crude
to
capture
any
interesting
information
about
continuity. Also if we
put
the
indiscrete
topology
on
both
Sand
T
then
only very few functions
are
continuous
(which
are
those?). So
this
topology is also
rather
uninteresting
from
the
point
of
view
of
continuity
.
Exercise:
Consider 2 x 2
real
matrices
M
=
[~
~]
(1.13)
Think
of
the
four
entries
a,
b,
c,
d as
points
in
IR4
with
the
metric
topology. Show
that
the
"determinant
map",
det
:
IR4
---7
IR
is continuous, where
det(a
,
b,
c,
d)
=
ad
-
be.
1.8
Homeomorphisms
One
may
define
many
apparently
different topological spaces,
but
some
of
these
may
be
"equivalent" for all purposes. To
make
this
notion
precise we need
to
define a kind
of
map
betw
een topological spaces, such
that
whenever
such
a
map
exists,
the
spaces will
be
considered
to
be
the
same. Basically,
what
such
a
map
should
do
is
to
establish
a 1-1 correspondence
between
elements
of
the
two spaces viewed as sets,
in
such a way
that
the
open
sets
of
one
are
mapped
onto
open
sets
of
the
other.
Definition:
f :
S
---7
T
is a
homeomorphism
if
f
is bijective
and
both
f
and
I-I
are
continuous.
Since
1
is bijective,
we
can
think
of
1-1
as
mapping
points
of
T
to
unique
points
of
S.
The
bijective
property
of
I
establishes
an
equivalence
between
points
of
S
and
points
of
T,
while
both-ways
continuity
ensures
that
open
sets
go
to
open
sets in each direction.
If
a
homeomorphism
ex
ists
between
two topological spaces
Sand
T
then
they
are
said
to
be
homeomorphic
and
we
think
of
them
thereafter
as
being
the
same.
Theorem:
If
I:
S
---7
T
is a
homeomorphism
then
S
is
connected
if
and
only
18
Chapter
1.
Topology
if
T
is connected,
and
S
is
compact if
and
only if
T
is
compact.
Exercise:
Prove
this
theorem.
This
is
merely a
matter
of carefully working
through
the
definitions.
1.9
Separability
We close
this
chapter
with
a
few
definitions
and
comments concluding
with
the
notion of separability.
Let
(S,
U)
be
a topological space.
Definition:
The
interior
of a subset
XeS,
written
xo,
is
the
union of all
open sets
Ui
contained in
X:
xo
=
UUiCX,UiEU
Ui
Definition:
The
boundary
of
XeS
is
the
difference between
the
closure
X
of
X
and
the
interior
XO
of
X:
b(X)
=
X
-Xo
Examples:
(i) Let
X
=
(a,
b)
C
lR in
the
usual topology.
Then
XO
=
(a, b),
X
=
[a,
b],
and
b(X)
=
{a,b}.
The
same
is
true
for each of
the
choices
X
=
(a,b], X
=
[a,b),
X
=
[a,b].
(ii) Let
X
= {
(x, y)
E
lR
2
I
x
2
+
y2
<
1 },
the
unit
open
disc.
Then
XO
=
X,
and
X
= {
(x,
y)
E
lR
2
I
x
2
+
y2
:::;
1 }, which
is
the
unit
closed disc.
The
boundary
is
b(X)
= {
(x,
y)
E
lR
2
I
x
2
+
y2
=
1 }, which
is
the
unit
circle.
(iii)
On
lR
with
the
usual topology, consider
the
subset Q
= {
x
E
lR
I
x
=
~
}
where
p
E
71..,
q
E
71..
-
{o}.
These
are
the
rational
numbers. One
can
see
that
every
point of lR
is
a limit
point
of
Q,
because
any
open
interval contains infinitely
many
rational
numbers.
Thus
Q
=
lR.
We say
that
the
rationals are
dense
in
the
reals. Clearly
the
interior
QO
of
Q
is
<p,
since no
open
set
fits inside
the
rationals, so
that
we also have
b(Q)
=
lR.
An
important
concept in topological spaces is
that
of
separability.
Given
two
distinct
points, it
may
be
possible
to
enclose one in
an
open
set
which does
not
contain
the
other.
One
may be able
to
do
better,
and
enclose
both
of
them
in two disjoint
open
sets.
There
are
various degrees of separability which a
topological space
may
have.
The
one which
is
most relevant in
the
study
of
manifolds
and
hence of physics is
the
following:
Definition:
A topological space
S
is
Hausdorff
if, whenever
S1,82
are
two
distinct
elements of
S,
there
exist disjoint
open
sets
U1,U2
such
that
81
E
U1,
82
E
U2.
1.9.
Separability
19
Exercises:
(i) Show
that
lR
n
with
the
usual
topology is Hausdorff.
(ii) Show
that
lR
with
the
indiscrete topology is
not
Hausdorff.
(iii) Consider
the
following
type
of space.
S
is
any
infinite set.
u
C
S
is
an
open
set
if
u'
is a finite set. Check
that
this
defines a topology
on
S.
Is
this
a
Hausdorff topology?
(iv) Show
that
every
metric
space is Hausdorff.
In
fact,
metric
spaces
are
normal,
a
stronger
condition
implying
that
disjoint closed
sets
can
be
enclosed
in disjoint
open
sets.
Exercise:
Show
that
every
compact
subset
of
a Hausdorff
space
is closed.
This page intentionally left blankThis page intentionally left blank
Chapter
2
Homotopy
2.1
Loops
and
Homotopies
In
this
chapter
we discuss ways
to
understand
the
connectivity of a topological
space.
These
will consist largely of
the
study
of
"closed loops"
on
a topological
space,
and
the
possibility
of
deforming these into each other. Many,
though
not
all, essential properties of a topological space emerge
on
studying
connectivity.
We have
already
defined a connected topological space: one which
cannot
be
expressed as
the
union
of
two disjoint
open
sets.
There
is
another
kind of
"connectivity"
property
of
topological spaces which will prove very
important.
Consider as
an
example
the
plane
m?
with
the
unit
disc
cut
out
(Fig. 2.1).
Figure
2.1:
m?
with
a disc
cut
out.
On
this
space, a loop like
h
(we will give a precise definition of "loops"
later),
which does
not
encircle
the
. disc, has
the
property
that:
(i)
h
can
be
continuously
shrunk
to
a point.
(ii)
h
can
be
continuously deformed
to
any
other
loop
not
encircling
the
disc.
(iii)
h
cannot
be
continuously deformed
to
a loop like
l2
which encircles
the
disc
once.
The study
of
whether
loops in a topological space
can
be
defonned into
others
is
part
of
homotopy theory,
and
is
an
important
tool
in characterizing