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- #1

- Jan 17, 2013

- 1,667

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

- Thread starter ZaidAlyafey
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- Thread starter
- #1

- Jan 17, 2013

- 1,667

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

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- #2

- Jan 26, 2012

- 4,193

I think you must mean that the two loops smoothly deform to the

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- #3

- Jan 17, 2013

- 1,667

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 .

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- #4

- Jan 26, 2012

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True, but you need more than that, right? Is the wording of the original problem such that both contours can be smoothly deformed to theAssume 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 .

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- #5

- Jan 17, 2013

- 1,667

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.True, but you need more than that, right? Is the wording of the original problem such that both contours can be smoothly deformed to thesamepoint? If so, how can you leverage that information to get the deformation you need?

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- #6

- Jan 26, 2012

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- #7

- Jan 17, 2013

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\(\displaystyle z_1(s,t) = (1-s)z_1(t) +s z_1 \) 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 \(\displaystyle \Gamma_2\)?

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- #8

- Jan 26, 2012

- 4,193

Right. And there's one more deformation you can write down. How about $z_{1}$ to $z_{2}$? Why can you do that one?

\(\displaystyle z_1(s,t) = (1-s)z_1(t) +s z_1 \) 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 \(\displaystyle \Gamma_2\)?

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- #9

- Jan 17, 2013

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- #10

- Jan 26, 2012

- 4,193

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?

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- #11

- Jan 17, 2013

- 1,667

Done, I see what you mean . Thanks for your time .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?

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- #12

- Jan 26, 2012

- 4,193

You're very welcome, as always!Done, I see what you mean . Thanks for your time .