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/sci/ - Science & Math


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4340592 No.4340592 [Reply] [Original]

Hey sci

If I've got the set, <span class="math">U(X)[/spoiler] of all holomorphic functions from a Riemann surface <span class="math"> X [/spoiler] to the Riemann sphere.

Let <span class="math"> G [/spoiler] be the group of all Mobius transforms with domain and codomain being the Riemann sphere (group under function composition). (Möbius group).

Consider the <span class="math"> G [/spoiler] acting on <span class="math"> U(X) [/spoiler] by left function composition. Is this transitive? Free? Faithful?

>> No.4340617
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4340617

bump

>> No.4340627

>>4340617
would fuck

dem titties

>> No.4340650
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4340650

>>4340627
Amen to that brother.

>> No.4340654

Yes, it's transitive; don't know about the others though.

>> No.4340659

>>4340654
I could of guessed that, can you give a outline for the proof?

>> No.4340675

>>4340650
Well that's not beautiful but it's certainly very very interesting

>> No.4340713
File: 273 KB, 639x862, 1328203320046.png [View same] [iqdb] [saucenao] [google]
4340713

>>4340675
>Confirmed for never doing any mathematics ever
You think
<span class="math"> 10^{k}\sum _{n=1}^k \frac{(10-n)}{10^{n}} = k+8*10^k\sum _{n=1}^k \frac{n}{10^n} [/spoiler] is interesting....

bump

>> No.4340728

>>4340713
Or in general if we're not in base 10:
<div class="math">b^k\sum _{n=1}^k \frac{b-n}{b^{n}} = k+(b-2)b^k\sum _{n=1}^k \frac{n}{b^n}</div>
Also getting <span class="math">b^k[/spoiler] in the sums and maybe switching <span class="math">k[/spoiler] for <span class="math">n-k[/spoiler] would make this look better.

>> No.4340766
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4340766

bump

>> No.4340786

Doesn't it depend on X a bit. I can't see it being the same for compact and non-compact X for example, as compact reduces the problem to polynomials, whilst non-compact would make transitivity a bit harder to get.

>> No.4340831

>>4340786
Only compact <span class="math">X[/spoiler] is fine too.

>> No.4340839

>>4340592
I severely lol'd at your image

>> No.4340865
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4340865

>>4340839
Me too. Couldn't stop for a good 20 seconds when I first saw it.

bump

>> No.4340868

>>4340654
>>4340659

How can it be transitive?

Consider <span class="math">f_1(z) = 1[/spoiler]. MTs are functions of the form <span class="math">\phi(z) = (az+b)/(cz+d)[/spoiler].
Thus, <span class="math">Gf= \{z\mapsto (a+b)/(c+d)\,:\,a,b,c,d\in\mathbb{C}\} = [/spoiler] constant functions.
Thus, <span class="math">Gf[/spoiler] cannot equal <span class="math">U(X)[/spoiler].

What am I missing?

>> No.4340873

>>4340713
>summing over n instead of over k
what the fuck is wrong with you

>> No.4340877
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4340877

>>4340868
Sorry, <span class="math">U(X)[/spoiler] is supposed to contain no constant functions.

>> No.4340883

>>4340877
The replace <span class="math">f[/spoiler] with <span class="math">f(z)=z[/spoiler]. <span class="math">Gf[/spoiler] clearly doesn't contain exp().

?

>> No.4340887

>>4340883
It doesn't need to for compact X. In the compact case all meromorphic functions are rational functions.

>> No.4340888

>>4340883
<span class="math"> exp() [/spoiler] isn't holomorphic when codomain is the Riemann sphere.

>> No.4340913

>>4340888
?exp() isn't holomorphic when codomain is the Riemann sphere.

It isn't *bi*holomorphic, but unless I'm totally confused, it is holomorphic. [Here I can take the Riemann surface to just equal the complex plane.]

Did you mean to define <span class="math">U(X)[/spoiler] to be the set of all biholomorphic functions?

>> No.4340933

>>4340913
Sorry, my brain is full of fuck. (3 am where I live). It's obvious that it isn't transitive. Abandon thread, thanks for the help.