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Jul
18
comment How does Ahlfors define derivative on a Riemann Surface?
I assume $\theta$ may be a chart map which you pre- and post- compose with f or g, to get a map from $\mathbb C$ to itself; that is the only way I know of dealing with differentiability of maps between surfaces/manifolds.
Jul
18
comment How does Ahlfors define derivative on a Riemann Surface?
What is the def. of $\pi$?
Jul
18
comment Engineer: Unable to appreciate importance of L2 space
@Illegal Immigrant: Don't mean to nitpick , but one usually writes $L^(..)$, where ... is the region on which f satisfies the square-integrability condition.
Jul
18
comment Engineer: Unable to appreciate importance of L2 space
@TylerHG: Don't you mean $L^2$?
Jul
15
comment Homology groups equal when $\cdots \rightarrow 0 \rightarrow H_n(\mathbb{S}^m) \rightarrow H_{n-1}(\mathbb{S}^m) \rightarrow 0 \rightarrow \cdots$
thanks for the input, downvoting coward.
Jul
15
comment the mapping class group of the disk is trivial proof
Thanks for your input; it was helpful.
Jul
15
revised the mapping class group of the disk is trivial proof
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Jul
15
comment the mapping class group of the disk is trivial proof
I perceived you determined to show me I am wrong --I do not believe I am-- and lecturing me, instead of examining my argument fairly.
Jul
15
revised the mapping class group of the disk is trivial proof
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Jul
15
comment the mapping class group of the disk is trivial proof
I didn't mean to come off as offensive, sorry; I wish you had let me explain myself before concluding I was wrong, which I was upset about . My map is actually z->ze^{i\pit} ;t in [0,1], which is a homeo. for each t, and has the continuous inverse e^{i(-t)}.
Jul
14
comment the mapping class group of the disk is trivial proof
So, I guess you can guarantee that all proofs that $MCG (\mathbb R^n)$ necessarily use that $MCG (D^n)=Id$, and that rotation by $\pi $ for $n=2$ is not an isotopy, in order to counter my proof. You have not done either.
Jul
14
revised the mapping class group of the disk is trivial proof
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Jul
14
comment the mapping class group of the disk is trivial proof
Well, I know of a proof that does not use this fact, so there is no circularity within my layout. And a rotation by $\pi$ is an isotopy (at least for n=2).
Jul
14
comment the mapping class group of the disk is trivial proof
@MartianInvader:Please see my edit.
Jul
14
revised the mapping class group of the disk is trivial proof
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Jul
14
revised the mapping class group of the disk is trivial proof
added 8 characters in body
Jul
14
revised the mapping class group of the disk is trivial proof
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Jul
14
comment Homology groups equal when $\cdots \rightarrow 0 \rightarrow H_n(\mathbb{S}^m) \rightarrow H_{n-1}(\mathbb{S}^m) \rightarrow 0 \rightarrow \cdots$
no problem, downvoter, I will explain the answer to you one of these days.
Jul
14
comment the mapping class group of the disk is trivial proof
pretty sure if you knew, you would have written something.
Jul
14
comment continuous and COVERING MAP
@Claire yes, sorry for the delay.Let me continue.