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Continuous deformation

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Prove that If two loops \(\displaystyle \Gamma_1 , \Gamma_2\) in a domain \(\displaystyle D\) can be continuously deformed to a point then they can be continuously deformed to each other .

I have no experience in topology , I found this note in a complex analysis book .
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
Re: continuous deformation

I think you must mean that the two loops smoothly deform to the same point. Is that correct? If so, can you think of a smooth deformation that takes one loop to the other?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: continuous deformation

Assume that we have two closed contours \(\displaystyle \Gamma_1 , \Gamma_2 \) then we can continuously deform \(\displaystyle \Gamma_1 \) to \(\displaystyle \Gamma_2\) if we can find a deformation function \(\displaystyle z(s,t)=(1-s)z_1(t)+s z_2(t) \) where \(\displaystyle s \in [0,1] , t \in [0,1]\) and \(\displaystyle z_1(t),z_2(t) \) are parametrization for \(\displaystyle \Gamma_1,\Gamma_2 \) respectively .
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
Re: continuous deformation

Assume that we have two closed contours \(\displaystyle \Gamma_1 , \Gamma_2 \) then we can continuously deform \(\displaystyle \Gamma_1 \) to \(\displaystyle \Gamma_2\) if we can find a deformation function \(\displaystyle z(s,t)=(1-s)z_1(t)+s z_2(t) \) where \(\displaystyle s \in [0,1] , t \in [0,1]\) and \(\displaystyle z_1(t),z_2(t) \) are parametrization for \(\displaystyle \Gamma_1,\Gamma_2 \) respectively .
True, but you need more than that, right? Is the wording of the original problem such that both contours can be smoothly deformed to the same point? If so, how can you leverage that information to get the deformation you need?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: continuous deformation

True, but you need more than that, right? Is the wording of the original problem such that both contours can be smoothly deformed to the same point? If so, how can you leverage that information to get the deformation you need?
Clearly we don't need the two contours to deform to the same point . I never heard of smooth deformation , it is never mentioned in the context . But, by definition these contours are finite sequence of smooth curves.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
You're in a domain, so going from one point to the other is no issue. This proof is one of those "what do you know" kind of proofs. By the time you finish writing down all the stuff you've been given, the answer kind of pops out at you. Can you write down all the deformation functions you know you have?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Since we are given that both loops deform to a point in $D$ , let \(\displaystyle \Gamma_1 \) deforms to the point $z_1 \in D$ and \(\displaystyle \Gamma_2 \) deforms to the point $z_2 \in D$ so we can describe these deformations as follows

\(\displaystyle z_1(s,t) = (1-s)z_1(t) +s z_1 \) for \(\displaystyle \Gamma_1\)

\(\displaystyle z_2(s,t) = (1-s)z_2(t) +s z_2 \) for \(\displaystyle \Gamma_2\)

But I cannot see how to deduce the deformation from \(\displaystyle \Gamma_1 \) to \(\displaystyle \Gamma_2\)?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
Since we are given that both loops deform to a point in $D$ , let \(\displaystyle \Gamma_1 \) deforms to the point $z_1 \in D$ and \(\displaystyle \Gamma_2 \) deforms to the point $z_2 \in D$ so we can describe these deformations as follows

\(\displaystyle z_1(s,t) = (1-s)z_1(t) +s z_1 \) for \(\displaystyle \Gamma_1\)

\(\displaystyle z_2(s,t) = (1-s)z_2(t) +s z_2 \) for \(\displaystyle \Gamma_2\)

But I cannot see how to deduce the deformation from \(\displaystyle \Gamma_1 \) to \(\displaystyle \Gamma_2\)?
Right. And there's one more deformation you can write down. How about $z_{1}$ to $z_{2}$? Why can you do that one?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Since any domain is connected then any two points in $D$ can be transformed to each other because they can be connected through a finite set of straight lines , is that right ?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
Since any domain is connected then any two points in $D$ can be transformed to each other because they can be connected through a finite set of straight lines , is that right ?
Right. So you have a deformation from $\Gamma_{1}$ to $z_{1}$, from $z_{1}$ to $z_{2}$, and from $\Gamma_{2}$ to $z_{2}$. Any ideas from here?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Right. So you have a deformation from $\Gamma_{1}$ to $z_{1}$, from $z_{1}$ to $z_{2}$, and from $\Gamma_{2}$ to $z_{2}$. Any ideas from here?
Done, I see what you mean . Thanks for your time .
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193