Homotopy an its impact on integration

Sessions

 

In this session all paths are in an open subset D of C and parametrized in [0,1].

Definitions.  Let  and  be two paths in D.

1)      If (0)=(0)=a, say and (1)=  (1)=b, we say that  is homotopic to  in D

if  there is a continuous mapping

D such that

 and f

and

and .

The mapping is called a homotopy from  to .

The above identities define a family of paths in D which connects  and .  Intuitively, this means that  can be continuously deformed to .

2)  In particular, closed paths  and  are homotopic in D if there is a continuous mapping

D such that

 and

and

More particularly, a closed path  is homotopic to a point a in D if  is homotopic to the path  defined by .

Thus we can now rephrase the definition of simple connected set D to be a set in which every closed curve in D is homotopic to a point in D.

Theorem.  Let  and let S denote the set of all paths in D with initial point a and terminal point b.  Consider the binary relation H defined on S by

H are homotopic.

Then H is an equivalence relation.

Proof. 1) H is  reflexive:  If  S, define

by

.

Since  is continuous, so is .  Moreover,

and

.

So is a homotopy from  to .

2)      H is symmetric:  Suppose such that H.  Let be a homotopy from  to . Define  by

                                  .

Clearly  is continuous.  Moreover,

.

So is a homotopy from  to .

3)      H is transitive:  Let such that  and .  Let  and  be two homotopies from  to  and from  to  respectively.  Define

 by

if  and  if .  Note that

 is continuous on  and [0,1]x( and at we get

1)=.  So  is continuous on (t,.  Hence is continuous on [0,1]x[0,1].  Moreover,

and

if  and if .

Similarly, .  Therefore is a homotopy from  to .

We now begin to explore the impact of homotopy on integration

Definition.  Let C   and let   be a closed path in C -{a}.  The index (or winding number) of  with respect to a, denoted by , is defined by

.

We state without proof some properties of the index in the following

Theorem. 

a)  Let  be a closed path and  be its complement.  Then is an integer-valued function on  which is constant in each component of  and which is 0 in the unbounded component of .

b)  Let  and  be homotopic closed paths in C -{0}.  Then

.

c)      Let  and  be homotopic closed paths in C such that  does not take the value 0 and such that , then  does not take the value 0 and

.

Lemma. If  and  are closed paths and if a is a complex number such that

, then =.

Proof.  Note that the inequality guarantees that  and  are in C -{a}.  Let .  Then  and again using the inequality .  Thus the graph of  is contained in the disk  and hence outside  and therefore  by a) of the previous theorem. Integrating  with respect to t on [0,1]  gives the result.

As we’ll see now, this lemma leads to a very important theorem where if we add homotopy to the hypothesis of the lemma, then we can dispose of the inequality (this shows the impact of homotopy on the index integral and as we shall see in later sessions on more general integrals):

Theorem.   Let  and  be homotopic closed paths in an open subset D of C.  Suppose .  Then =

      Proof.  Since   and  be homotopic in D, there is a continuous mapping

D such that

 and

and

and .

Since  is compact and the continuous image of a compact set is compact,

 is compact .  Hence there exists  such that 

.  Moreover, by compactness, is uniformly continuous.  Thus there exists a positive integer n such that

 whenever .

Define polygonal closed paths  by

 if and .

Thus .  In particular, taking the extreme cases  and , we get .Moreover, .

On the other hand, .

It follows, from the last three inequalities and  applications of the previous lemma, that a has the same index with respect to each of the paths  which completes the proof. 

Example.  Let C . Let   be the positively oriented circle  such that , then using the last theorem we have

=.

Thus , in this case, =1 which explains the terminology of winding number.