Mathematics - 2014-10-03 (page 1 of 4)

I want to draw the Riemann surface of $w = \sqrt[4]{z^2}$. It seems complicated, looping around the origin might or might not pull the image out of a sheet, so $z =0$ might or might not be a branch point.

Is that some disjoint union of two copies of RSs for $w = \sqrt{z}$ or some such?

Committing to a name

Are you doing Lee's Riemannian manifolds?

Balarka Sen

@Committingtoaname Me?

Nope. I haven't read about manifolds yet.

~~I doubt I'd ever be anytime soon. I don't like that part of mathematics.~~

EEK. Wrong comment in wrong time.

ducks

Hello @TedShifrin

Semiclassical

2:09 AM

This question has had such a nice response. I'm going to have a hard time deciding which one I like best for the bounty.

Balarka Sen

@Semiclassical Essentially everything points out to the C-R equations, so I guess it is hard to choose.

The first answer seems sufficient.

Semiclassical

@BalarkaSen: right. comes down more to how well they present it

Balarka Sen

erm. did you see my question, @Semiclassical?

Committing to a name

I meant the line you dashed out, where was that directed?

Semiclassical

@BalarkaSen: yeah, thinking about it now

Balarka Sen

2:13 AM

thanks.

@Committingtoaname I never studied. How am I supposed to know?

Semiclassical

things get weird when you get stuff like $w^4=z^2$ b/c of that common factor of two

Balarka Sen

I don't like analytic (smooth, eg.) structures in particular.

Committing to a name

I meant the line you dashed out, where was it directed?

Balarka Sen

@Committingtoaname Directed?

Semiclassical

it's in those cases where my having learned Riemann surfaces fairly informally and piecemeal is problematic

Balarka Sen

2:15 AM

@Semiclassical Yeah. As the RS of $w^2 = z^2$ are just two disjoint sheets sitting one over another, I guessed that it might be two disjoint $w^2 = z$ sitting one over another.

Semiclassical

i think it might not be smooth everywhere

probably the best way to study it is to take w^4=z^2+z0^2

and then watch what happens to the surface as z0->0

Balarka Sen

urk. i don't want $\Bbb R^6$ interfering my imaginations :/

Semiclassical

R^6? i'm not envisioning z0 as an extra dimension here, just a parameter

(albeit a complex one)

Balarka Sen

ah ah

that might just work.

GBeau

How might I show $f'(x)=y'(x)\wedge y'(x)=g'(x)\implies f'(x)=g'(x)\text{ for all }f,g,y\in S$ where $S$ is the set of all differentiable functions on $\mathbb{R}$. This is for an introductory modern algebra course

it seems obviously true, but I'm not used to working with the formalisms of functions like this, so I'm not sure what needs to be done here?

Semiclassical

2:26 AM

@GBeau: it does just seem to be a transitive property

GBeau

@Semiclassical well, the issue is the question is more or less 'prove this is transitive'

Semiclassical

yeah

GBeau

and saying 'it's transitive because it's transitive' doesn't seem to reflect well, but ??

(the actual question is to prove or disprove this statement as an equivalence class: $S$ is the set of all differentiable functions on $\mathbb{R}$, and $f\sim g\iff f'(x)=g'(x)$ for all $f,g\in S$

Balarka Sen

oh, whoops, $w^2 = z^2$ is not two disjoint sheets, but one-point union (ramified, not branched at $z = 0$) of two copies of the complex plane

GBeau

where we defined equivalence classes as including the transitive property

Semiclassical

2:30 AM

@BalarkaSen: the way i try to construct it leads to some issues, hrm

Balarka Sen

bah it look complicated. can't we look at the arg?

Semiclassical

writing it as z^2 = w^4-w0^4 for some complex w0, you've got a two-sheeted riemann surface with four square-root branch points on each sheet

so that'd be a torus. but the two cuts on each sheet collapse into a single in the limit of w0->0

Balarka Sen

moving once around the non uniqueness point $z = 0$ jumps the argument by $2\pi$. and doing this again the function comes back. the other sheets aren't touched at all!

oh oh taking the other branch gets it to the second sheet.

ok pretty sure it's one-point union of two $(w^2 = z)$-surfaces.

let's see if this is given at the back of the book

found it. uh.

Semiclassical

...that could be right, but i'm really bad at reading those

Balarka Sen

oh ok it is what i said then. the white cuts (not visible in R^4) confused me.

Cool.

Semiclassical

2:42 AM

yeah. though they show that rather oddly

Balarka Sen

show what @Semiclassical?

Semiclassical

eh, i just mean that i'm more used to the 'draw it like a torus' construction than that sheet presentation.

Balarka Sen

oh

yeah it's just that i am more comfortable with RSs over $\Bbb C$ than $\Bbb P^1$ ;)

well, partial credits for that goes to the book

Semiclassical

heh, fair enough

Committing to a name

@GBeau Suppose $(f,g)\in\mathbb{R}$ and $(g,h)\in\mathbb{R}$. Then $f'(x)=g'(x)$ and $g'(x)=h'(x)$. Hence $f'(x)=h'(x)$ and therefore we can see that $(f,h)\in\mathbb{R}$ and thus our relation is transitive.

Balarka Sen

2:47 AM

perspectives are not hard to change however. one just have to "flip some sheets" and "identify some boundaries".

GBeau

@Committingtoaname Oh! Hello again, I see you saw my question! Give me a second to make sure I understand you

Committing to a name

:)

Semiclassical

one thing i really like about the 'gluing riemann spheres together' is that it makes the (co)homology aspects quite apparent, which was important for what i used it for (e.g. hyperelliptic integrals)

Balarka Sen

never studied it =P i only know what a group (co)homology is

Semiclassical

heh, different strokes!

GBeau

2:53 AM

@Committingtoaname I believe I understand the formalization now

Committing to a name

Yep so you will have $(a,b)\in S$ and $(b,c)\in S$ for your Set $S$ and you just need to show that this set contains $(a,c)$

GBeau

yes I see now

I was happy with my formalization of the prime problem: i.imgur.com/gfcQdCW.png

(solved with modular arithmetic)

Committing to a name

Looks good!

GBeau

which leaves my last problem as a likely "easy" one i.imgur.com/7w7kgB9.png

Committing to a name

Probably could have left it as short as: meta.math.stackexchange.com/questions/4666/…

GBeau

3:02 AM

I understand the idea of these proofs with Euclid, I just wasn't comfortable with every step for this yet (I'll be attempting it again in a bit, not given up on it)

Committing to a name

Euclid didn't actually use contradiction

GBeau

I mean that in the sense of what our provided material claims

Committing to a name

Yeah it later on became a proof by contradiction in many texts xD

GBeau

i.imgur.com/ZGwN6hA.png (sorry for all the images!)

Committing to a name

I'll be back later on

GBeau

3:10 AM

all right

my professor included a sizable bonus on this homework for proving things related to the axiom of choice

so I'll probably still be working on it

:P

Committing to a name

xD sounds good, see you later

GBeau

(we have not been introduced to said axiom nor are we otherwise expected to know it (for this course, at least), hence the bonus)

Committing to a name

Actually before I leave, can you put up that image so I can think about it until then

GBeau

the axiom of choice one?

Committing to a name

Yep

GBeau

3:12 AM

it's a long section, it will be several, but yes

first part i.imgur.com/IM1hwmv.png

second (&final): i.imgur.com/RXfTfJW.png

Semiclassical

an xkcd link as citation. nice.

GBeau

for easy clicking xkcd.com/804

3

Committing to a name

Thank you, I'll be back later now

GBeau

all right

Balarka Sen

@GBeau LOL

"I carved and carved and the next thing I knew I had two pumpkins" - that one is a gem

Ice Boy

3:21 AM

Another "proof" that 2=1.

Balarka Sen

@IceBoy Banach-Tarski is well-known

Ice Boy

Not as well-known as "proofs" that 2=1.

Balarka Sen

2=1 is just BS. it depends on what you're working with. For example, every interval is homeomorphic to a point so that'd imply a set with uncountable cardinality is null if you wrongly interpret that way.

besides, give higher mathematical jokes a respect. not everything is 2 = 1.

=P

Ice Boy

Come on pal, I respect higher mathematical jokes, I'm just saying that they are built upon lower mathematical jokes :D

Balarka Sen

not really. Banach-Tarski most definitely isn't a variation of 2 = 1

in fact, this made me realize that it's not even related to homeomorphisms

Ice Boy

3:27 AM

I have 2 pumpkins now?

Balarka Sen

gah this is just trolling

zips lip

Mike Miller

4:24 AM

hello @Karl

Karl Kronenfeld

sup @MikeMiller

Mike Miller

how's it

Karl Kronenfeld

considering i don't have much to do with it, i'm not sure why you'd ask me that

Mike Miller

I have more questions than sense

GBeau

$4n-1\subseteq4n+3$ for $n\in\mathbb{N}$ ?

Karl Kronenfeld

4:30 AM

hm

Mike Miller

as stated and without further context that string of symbols is nonsense

it's deep stuff

yeah I mean the set

or a triviality

I'm being lazy with notation here

because I'm a bad person

Karl Kronenfeld

4:33 AM

we don't get it.

be a better person

GBeau

ok

I'm being asked to prove 4n-1 has inf primes

it is my belief 4n+3 having infinite primes would prove the above

Mike Miller

that's correct

GBeau

how would you state the modular relation

formally

The Substitute

off topic: If taking the integral (with respect to $\theta$) from $0$ to $2\pi$ of a function $i*f(\theta)$, then can I treat $i$ like a constant or is there a definition for evaluation of "imaginary" integrals?

Mike Miller

but unfortunately restating it that way is unlikely to prove anything

yikes

Anthony

4:34 AM

@MikeMiller !!!!

Mike Miller

@Anthony is here

Anthony

I know

GBeau

yes I restate it only in as I could more easily construct a proof of one

Anthony

And I bet you know why I'm here

GBeau

but I cannot even restate it if I can't justify my restatement

in which case what I'm asking is

why is what I stated just now true

(why can I rewrite it like that)

Mike Miller

4:35 AM

I don't, @Anthony

Anthony

Wow

Mike Miller

I bet it's to chat about Philz coffee

GBeau

As in, how do I formally go through modular arithmetic to get this (as opposed to hand waving)

Karl Kronenfeld

hi @pedro

Anthony

@MikeMiller I still haven't been there.

I actually need some homework help, if you are willing.

Pedro Tamaroff

4:38 AM

@KarlKronenfeld Heya!

Mike Miller

@Anthony I will help until Smash bros finishes downloading, and no longer.

Anthony

oh dear

alright

Pedro Tamaroff

@KarlKronenfeld I have a stupid question.

Anthony

@MikeMiller We have our set $A$ in $X$, and $B$ in $Y$. Taking our open subset $\mathcal{O}$ in $X \times Y$ containing $A \times B$, we seek to show there is an open subset $U$ in $X$, and $V$ in $Y$, containing $A$ and $B$ respectively, with $U \times V$ also in $\mathcal{O}$.

Pedro Tamaroff

Consider the curve $(3t,2t^2,2t^3)$.

This should have constant angle with the line $y=0,x=z$.

Karl Kronenfeld

4:40 AM

mkay

Pedro Tamaroff

"show that the tangent lines to the regular parametrized curve a(t)=(3t, 2t^2, 2t^3) make a constant angle with the line y=0, x=z"

Mike Miller

I'm not sure I really get what you're asking, @Anthony.

Anthony

Cool, I'm really good writing.

Mike Miller

:P

Anthony

Can I just copy the question statement, I wonder.

Karl Kronenfeld

4:42 AM

@PedroTamaroff So you don't feel like taking dot products or something? :P

Anthony

Let X and Y be topological spaces, let A be a compact subset of X, and let B be a
compact subset of Y . Let O be a subset of X × Y that is open for the product topology.
Prove that if A × B ⊆ O then there exist an open subset U of X and an open subset V of
Y such that A ⊆ U and B ⊆ V and U × V ⊆ O. (Hint: consider first the case in which A
contains only one point.)

Yay!

Pedro Tamaroff

@KarlKronenfeld I am not getting constant stuff.

Karl Kronenfeld

Oh, lemme check

Mike Miller

LOL THAT'S NOT WHAT YOU ASKED ME AT ALL

Anthony

lol

Pedro Tamaroff

4:42 AM

Maybe the problem has a typo.

Anthony

So bad

Mike Miller

This seems vaguely fun

Anthony

I mean besides the compact part, isn't it what I wrote? Oh well.

I don't know how to go about this.

I drew a pretty picture.

Mike Miller

Someone should back me up that what you wrote was incomprehensible. :)

Anthony

I agree with you.

Mike Miller

4:43 AM

You drew apretty picture of two... arbitrary topological spaces?

Anthony

Yeah.

It's beautiful.

The next big thing.

Mike Miller

This is very cute but probably above my pay grade, since Smash is almost here.

Anthony

NOOOO PLEAAASSEE NNOOOOOO

Mike Miller

Can you do it given the hint?

Karl Kronenfeld

@PedroTamaroff Yeah, seems like it.

Pedro Tamaroff

4:45 AM

@KarlKronenfeld There's an effing typo.

Anthony

@MikeMiller I mean I read the hint, and I didn't see how it helped.

Mike Miller

Can you do the case $A = *$?

Karl Kronenfeld

9+16t^2+36t^4 sure aint (3+6t^2)^2

Anthony

I guess I'll just stare longer if you don't think it's too bad.

*?

Mike Miller

One-point space.

Anthony

4:46 AM

I'm not sure. Let me try.

That just means O is an open set in B right?

Karl Kronenfeld

@PedroTamaroff Perhaps they meant (3t,3t^2,2t^3)

Anthony

Basically?

Mike Miller

Why's it mean that?

OK I'm out

Sorry

Anthony

It's cool.

Thanks @MikeMiller.

Pedro Tamaroff

@KarlKronenfeld Yes.

That's what I got.

Karl Kronenfeld

4:50 AM

@Anthony O is still fairly arbitrary even in the case A=*

Anthony

@KarlKronenfeld Really?

Karl Kronenfeld

yeah, draw a line segment in the plane and a salamander around it

Anthony

Whhhhhhhhhat. I thought if A was a point that would just be a line on the plane.

Karl Kronenfeld

AxB is a subset not a superset of O

Anthony

Oh so it's all sets containing a line segment. I see.

Then I have no idea how to show this, especially considering I'm going to need to use compactness, right?

Karl Kronenfeld

4:54 AM

Yeah, the compactness is definitely necessary

although I haven't thought of a proof of the easy case yet either

Balarka Sen

@Karl! @Pedro!

Oh noes I missed @Mike

Anthony

@KarlKronenfeld Hmmm

Karl Kronenfeld

@Anthony Write O as a union of products of open sets.

Balarka Sen

throws table at @Karl for a lack of hello

Karl Kronenfeld

is it a table of integrals?

Anthony

4:59 AM

I mean, just some arbitrary one?

Balarka Sen

@KarlKronenfeld It's a desk.

Karl Kronenfeld

@Anthony yeah, like $\bigcup_{i\in I}U_i\times V_i$

Anthony

Indeed.

Karl Kronenfeld

then use compactness of $B$

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