Mathematics - 2016-11-03 (page 3 of 3)

Ted Shifrin

21:00

Oh ...

Danu

The final thing I got stuck on was ehh

Semiclassical

@TedShifrin Yeah, should be $\partial_u(dx/y)=(x^2-x)(dx/2y^3)$

woops.

Danu

$F^T_\nabla v^*(w)=h(w,\bar F^T_\nabla v)$

Ted Shifrin

Substitute for $y^2$ and simplify. But remember $g=1$ in this case, too.

Semiclassical

oh, I know how to simplify it computationally. I figured out a method for that.

what I don't understand is the terminology

Danu

21:02

Anyways @Ted, I'm now putting in the last tiny details of my talk tomorrow

the one on Stiefel-Whitney classes (axiomatic, formal treatment in section 4 of MS)

Mostly catching typos in my notes :P

Ted Shifrin

But there's a subtlety here. This comes from varying the curve, so it's not a holo differential on the curve, @Semiclassic.

saturatedexpo

$h:Q^2->Q$ $(x,y)\to x-y$ surjective? Let h(x,y)=a. x-y=a. x=a-y. Choose y=0. Then h(a,0)=a. And a and 0 are in Q. (otherwise the a is illchoosen, sorry for unrigorousity) is this sketch ok?

Ted Shifrin

I have forgotten my variation of Hodge structure knowledge ...

Semiclassical

Ah. That'd actually help, then---my confusion was why we had to use 2 basis differentials, when for g=1 the cohomology is 1D

so if that pushes us into the full de Rham cohomology H^1 (not just H^{1,0}) then I'm a bit happier

but yeah, this is hodge structure stuff. i might not be talking about it in those terms (i really don't know how) but I know it's all of this variation business

Ted Shifrin

It seems to me we get a singularity. When I get home I'll have to look up Griffiths stuff.

Semiclassical

21:06

Could be, yeah. There's a lot where I'm flying by the seat of my pants

Danu

Are you on your trip already, Ted?

Ted Shifrin

Yup. Up at Stanford, moving to location #2 tomorrow.

Danu

Cool :)

Semiclassical

The method I worked out for getting the decomposition into basis 1-forms is something I'm reasonably happy with, but I don't know how/if it can be done if my curve is a smooth plane quartic (i.e. not hyperelliptic)

There's apparently a reduction method that owes to Griffiths and Dwork, but it kicks my butt whenever I try to read it :/

Ted Shifrin

I used to know VHS, but it's been 40 yrs.

Danu

21:10

Blu-Ray* is all the hype ;D

(just kidding)

Semiclassical

Something along the lines of: For a given holomorphic 1-form, I can consider it as the residue of a holomorphic 2-form. It's apparently easier to do the reduction of stuff up there, because of Jacobean stuff

Ted Shifrin

Not holo 2-form. Mero.

Semiclassical

Probably right

Ted Shifrin

That's the Poincaré residue I mentioned.

Semiclassical

Yeah, I figured.

Doesn't mean I understood it :/

I mean, I've seen stuff like $\text{Res }(dx\wedge dy/f) = dx/f_y$

but I don't really understand it

Ted Shifrin

21:14

But your calculation gives $(dx/2y)\cdot 1/(x-u)$, so we have a pole.

Semiclassical

Hrm. That doesn't sound right.

Let me check the details.

Danu

For any (real) vector bundle $E$, the identity map generates a 1-dimensional trivial subbundle of $\operatorname{Hom}(E,E)$, right?

saturatedexpo

@Danu Isn't this then also an Automorphism?

Semiclassical

We've got $f=y^2-(1-x^2)(1-u x^2)=0$, so $dx/f_y=dx/(2y)$

Ted Shifrin

I'm talking the u partial.

Semiclassical

21:18

getting there

Tobias Kildetoft

@saturatedexpo Isn't what an automorphism?

saturatedexpo

@TobiasKildetoft let me rephrase: what is Hom(E,E). Is it NOT an homeomorphism?

Semiclassical

and then $\partial_u(dx/y)=(x^4-x^2)(dx/2y^3)$

NaCl

@saturatedexpo However, I didn't need the inverse. I was able to prove surjectivity by induction

Tobias Kildetoft

@saturatedexpo No, that set is not a homeomorphism.

Semiclassical

21:21

@TedShifrin Where would that have a pole? The only place where one could be is when $y=0$, and those are the branch points

saturatedexpo

@NaCl i'm interested, maybe post it as a topic or here?

Semiclassical

you're right that it simplifies, though. becomes $(dx/2y)/(1-u x^2)$

but would that really be a pole, if there's already a branch point there?

Ted Shifrin

$

Semiclassical

oh, wait. bollocks

I've got two parametrizations of elliptic curves in my head, and I've switched between them

should've been $f=y^2-x(x-1)(x-u)$, ugh

Ted Shifrin

You can always switch $dx/y$ to $dy/?$. But I'm not sure what's going on.

Semiclassical

21:28

No, I mean I went from $y^2=x(x-1)(x-u)$ to $y^2=(1-x^2)(1-u x^2)$

thought I was using the latter earlier when I was actually using the former

that still gives $dx/f_y=dx/(2y)$, but now $\partial_u(dx/y)=(x^2-x)(dx/2y^3)=(dx/2y)\cdot 1/(x-u)$

which is what you had above

saturatedexpo

@NaCl what do you study?

Semiclassical

My response is still the same, though. there's a branch point at $x=u$, so I don't see how $(dx/2y)\cdot 1/(x-u)$ would have a pole @TedShifrin

NaCl

@saturatedexpo Computer Science. Are you online tomorrow? I don't really have the time to convert my handwritten stuff to TeX right now

Semiclassical

So for now I'd stand behind that being holomorphic on the Riemann surface

saturatedexpo

yes i am. (ich freu mich drauf^^)

Ted Shifrin

21:33

I'm not sure, but since the holo structure is varying, pretty sure this can't be holo anyhow.

NaCl

@saturatedexpo Ok, thats good, as I need to submit my problem sheet tomorrow and there is still way too much left unsolved, lol

Semiclassical

Wouldn't shock me

saturatedexpo

@NaCl which city?

Danu

How do I induce a Hermitian structure on a tensor product of Hermitian bundles? Simply by taking the product, i.e. $h_{1\otimes 2}=h_1\cdot h_2$ (in schematic notation)?

NaCl

@saturatedexpo Saarbrücken. :D

Semiclassical

21:35

@TedShifrin I guess this is what my puzzlement all comes down to. I know that the right phrase here is "Gauss-Manin connection" but that's an empty gesture

saturatedexpo

@NaCl no Düsseldorf. Cuz im too lazy to move haha

Ted Shifrin

On basic tensors, @Danu. But then use bilinearity.

Danu

I see.

NaCl

@saturatedexpo You study computer science as well?

Ted Shifrin

But @Semiclassic, you have

$dx/y = 2 dy/(....)$ is holo, nonzero, and now you divide by $x-u$.

saturatedexpo

21:37

@NaCl no, i study math (und anwendungsgebiete). don't know which other field i will study. (i guess IT, but im not sure)

Semiclassical

...huh. that's true.

saturatedexpo

@NaCl if there is any chance to study french as applied mathematics, i do that :) (hipsterlevel>9000)

NaCl

@saturatedexpo aah, okay. Pure math is a bit too abstract for my taste

lol

Ted Shifrin

Ah, but pole of order 2, it appears.

Semiclassical

Yeah. So no residue.

Ted Shifrin

21:39

But still not holo.

Semiclassical

Right.

saturatedexpo

@NaCl only computerscience with no passion seems dry to me too ;) (and i dont have passion for that, altho i like the stuff in the newspaper)

NaCl

@saturatedexpo haha, good point

Semiclassical

@TedShifrin That might be enough, though. I can break that up into a polar part and a holomorphic part

hm, but

I know that there exist $A(u),B(u)$ such that $\partial_u(dx/y)$ and $(Ax+B)(dx/y)$ are cohomologous

I've calculated them before, so I'm confident on that

saturatedexpo

@NaCl my knowledge of computerscience is how to install linux. And that is probably even a littlebit offensive to computerscientists.

NaCl

21:44

@saturatedexpo it is indeed offensive

saturatedexpo

(and the film terminator, good Lebenslauf)

NaCl

I'm triggered

Semiclassical

The basic tool I've used is to take a form like $dx/y$, and integrate it by parts to get (up to an exact form) $x dy/y^2 =-x f_x (dx/2y^3)$

saturatedexpo

i find it also funny if someone sais they have A grades in Abitur. because Abitur is like nothing compared to university stuff.

(kinda have to laugh at myself)

Semiclassical

that and $dx/y= y^2 dx/(y^3)=p(x)(dx/y^3)$

i.e. expressing all of those in terms of $x^k(dx/y^3)$ and using that to work out those in terms of $dx/y$, $x(dx/y)$

NaCl

21:48

@saturatedexpo That is as true as my bad grades in school

Ted Shifrin

Remember, $g=1$, so one-dimensional holo coho.

Semiclassical

hmm

I guess that confuses me as well. Does that mean $x(dx/y)$ also has a pole?

saturatedexpo

@NaCl Abitur is like: you know a little bit of everything. here you go.

Ted Shifrin

Yup.

NaCl

yep

Semiclassical

21:49

huh.

Danu

@TedShifrin I guess sesquilinearity

Semiclassical

I hadn't expected that.

bbl, gotta catch transit

Ted Shifrin

Sure, Danu, my sloppiness.

Danu

The complex conjugations will drive us all mad

Semiclassical

Oh, I can use my phone for this. Right

saturatedexpo

21:53

@NaCl its dumb because in the early years children can learn the most. But you get bored with silly stuff you never use again. And then in university you got to learn a shit ton, but you cant because you are old.

And i still wait to see the man who buys 50 melons

Semiclassical

@ted How do I see that it has a pole?

Ted Shifrin

Look at $\infty$.

Semiclassical

And, uh, where? Has to be at zero or infinity

Ah

Think I see it, yeah, upon doing z=1/x (I'm on a light rail train, so can't do pencil & paper)

Ted Shifrin

Have to change y too, of course.

saturatedexpo

What is a nice way to highlight notes in Latex (which are also available in the PDF). Like ("Here we got to solve for x" or something like that, which to the experienced is redundand)

A better word might be "downlight". because the opposite of highlighting is meant.

Danu

22:03

@TedShifrin Showing that the curvature of the tensor product of two positively-curved bundles is again positive has devolved in a terrible mess of indices. Any nicer way?

I can probably just do it on a basis and avoid summing?

Ted Shifrin

 Can't you use $\Omega_E\otimes 1 + 1\otimes\Omega_F$ and plug in vectors?

Danu

So the positivity condition I'm using is the one that asserts $h(\Omega s,s)(v,\bar v)>0$

Griffiths positivity

So I have to deal with Hermitian structures throughout

Semiclassical

$(x,y)\mapsto(1/x,y/x^2)?$

I'm doing my algebra in the mobile prompt, so no guarantee

Danu

Meh, the indices aren't actually that terrible. I think I'm going to be okay.

Semiclassical

That gives $x(dx/y)\mapsto -1/x(dx/y)$, so yeah. Pole.

Sophie

22:14

Do you know who Doron Zeilberger is?

math.rutgers.edu/~zeilberg/Opinion113.html

Ted Shifrin

@Semiclassic, I had to think it through. Your COV seems right.

Semiclassical

Kk

Could I then break that into a holomorphic form and a polar form?

My guess is no, since the other example was evidently meromorphic despite having no residue

saturatedexpo

22:29

$f(x)=1$ if x<1

$f(x)=2$ if x>0

Normally it's done with a bick squirly bracket.

How is such a list called?

Semiclassical

A piecewise function?

saturatedexpo

Mmh, i don't mean the function itself, i mean the list defining the function.

if it has no name it's ok too to me ;)

Semiclassical

Though the definition you give is inconsistent for 0<x<1

saturatedexpo

yes, but the function(or its consistensy) is not what i ask, its just an example. I mean the list defining the function.

Semiclassical

Fair enough. But I don't really get what you're after. That list is the function

Take a look at the wiki page on such functions, maybe? en.m.wikipedia.org/wiki/Piecewise

saturatedexpo

22:35

Well, let me explain why i need a name for that. Does there exist a function with infinite entries on that list, that are not redundant. (my final question goes in such a direction)

Tobias Kildetoft

@saturatedexpo Sure, just list every single value of the function separately

Semiclassical

For a practical example, you could write an infinite piecewise definition of $\lfloor 1/x\rfloor$

There's a subdomain mapped to zero, another to one, etc

Brody

Hi chat

Suppose $C$ is a curve in the ordinary $xyz$-space defined parametrically by $x=\cos t,\, y=\sin t,\, z=\frac{1}{t};\; t\in\mathbb{R}$.

Does adding in the unit circle centered at the origin in the $xy$-plane "fill in" $C$ and make it "continuous" in some sense?

saturatedexpo

@Semiclassical then f(0)= undefined or infinity, f(0.1)=10, f(0.11)=9. Do you mean it that way?

Ted Shifrin

Hi @Brody

Brody

22:49

Hi there @TedShifrin. How's it do?

Ted Shifrin

Alive and kicking. Of course you mean $t\ne 0$.

Brody

Yes, I realized too late and couldn't edit in the change :P

$t\in\mathbb{R}\setminus \{0\}$

Ted Shifrin

So you're thinking about the curve spiralling down/up to $z=0$ as $|t|\to \infty$?

Brody

Uhuh, how it "coils" very very tightly near $z=0$ without ever touching it, like a removable discontinuity (but a curve rather than a point)

I guess the question is more about how to fix in the weirdness of an essential singularity (??)

DHMO

@Brody but it is already continuous, or you mean t can be negative?

Ted Shifrin

22:54

So putting in the curve makes a connected set but it's still not path-connected.

DHMO

@Brody is it an essential singularity?

Brody

@TedShifrin Those terms I don't understand

@DHMO Maybe not. I recklessly borrowed the word

Ted Shifrin

Yeah, you'll learn those words eventually. It just spirals, so you can never get to $z=0$.

Essential sing is totally the wrong word.

DHMO

@TedShifrin why is it connected?

Brody

@TedShefrin lol, I was taking the rough concept from $\cos(x^{-1})$ around $x=0$, though it's not really the same phenomenon

DHMO

22:58

isnt essential singularity for complex functions?

Ted Shifrin

Yup.

@DHMO, because you're throwing in limit points of the set.

DHMO

@TedShifrin but the limit points can never be reached

Brody

@DHMO @TedShifrin Oh. I read a forum post a while ago where someone said the region around $x=0$ for $y=\cos(x^{-1})$ is an essential singularity, or something like that

Semiclassical

@saturatedexpo right, though I should've restricted to the domain $x>0$.

DHMO

yes, when cosine is the complex cosine

Ted Shifrin

23:00

For complex cos, yes, @Brody.

Brody

What is complex cosine, taking a complex argument?

Ted Shifrin

Not removable, not a pole.

DHMO

yes

Brody

Ok, thanks

Ted Shifrin

@DHMO, have you learned limit points in analysis or topology?

DHMO

23:02

i know what they are

Ted Shifrin

Well, they're limits of points in the set.

DHMO

sure

Ted Shifrin

@Brody yes.

Brody

Speaking of, what can be said about the union of $y=\cos(x^{-1}),\, x\ne 0$ and the vertical line segment between $(0,-1)$ and $(0,1)$?

Ted Shifrin

Connected, not path-connected :)

Brody

23:05

Ah, same stick. Thanks :)

Semiclassical

@TedShifrin not to belabor it, but is there a way to write $1/x\cdot (dx/y)$ as a holomorphic 1-form plus a polar part?

saturatedexpo

@Semiclassical if if you did that. f(0.00000000...01) would still make no sense.

Ted Shifrin

Write the Laurent series, as usual, near infinity, @Semiclassic.

Semiclassical

@saturatedexpo What?

PVAL-inactive

@Ted Apparently the 5 minute talk I'm giving has to be a beamer talk.

Semiclassical

23:06

$f(10^{-n})=10^n$ is perfectly well-defined

Brody

my calc IV prof (last spring) mentioned "path-connected" in passing, can't remember the exact context though

saturatedexpo

@Semiclassical ah ok thanks :).

Ted Shifrin

In 5 minutes that's all you can do (or slides), PVAL.

DHMO

@TedShifrin I do not see how exp(1/(1-z)) being expressible with Taylor series can have anything funny happening when z=1...

Ted Shifrin

Laurent series, not Taylor

Infinitely many terms with negative exponents

DHMO

23:08

I don't see how there is negative exponents

Ted Shifrin

You expand in powers of $z-1$.

DHMO

right, i forgot that 1+z+z^2+... is not always convergent

what is its radius of convergence actually?

Ted Shifrin

1

saturatedexpo

is there a MUST-Read on Operations?

DHMO

@TedShifrin and why?

Ted Shifrin

23:12

Plug in 1 and what happens?

DHMO

But complex number?

Ted Shifrin

Gotta go. Sorry.

saturatedexpo

|z|=1

DHMO

@saturatedexpo yes but why?

Brody

always wondered why it's called a "radius" but I might've just caught up as to why

DHMO

23:15

because it is a circle lol

saturatedexpo

actually a spiral, no?

DHMO

i'm talking about z

saturatedexpo

ah well xd

Brody

@DHMO, as I just realized from @saturatedexpo's earlier comment :P

I took AP calculus in high school and haven't encountered series since, so all I know is how they work with real numbers

DHMO

complex numbers complete real numbers

saturatedexpo

23:18

mmh, how is this thing called again?

1+z+z^2...

DHMO

a series

Brody

a lot of ppl seem to hype up complex analysis as this wonderful tidy-up, so I definitely get the impression

DHMO

@Brody look, powers isn't even closed on the reals

Brody

@DHMO, you're talking about $x^y,\; x,y\in\mathbb{R}$?

DHMO

yes

saturatedexpo

23:21

geometric series

Brody

@DHMO yeah, that does seem unfortunate...lol

DHMO

a geometric series with starting term 1 and complex ratio

@Brody also, can you imagine a continuous real function with image R\{0}? it is simply impossible on the reals

Brody

@DHMO Right

I highly anticipate getting to complex analysis. maybe then I'll finally get the hype around residues and fractals and such

Forever Mozart

yes mmhmm

saturatedexpo

@DHMO can you take the absolute value of complex numbers? (i mean you, not in math ;)) Then it's understandable why that is. (think first about real numbers like -1)

Forever Mozart

23:27

@topology

Brody

hmm, $\mathbb{C}$ doesn't seem to have "order" like the real number line does

perhaps unsurprisingly since neither does $\mathbb{R}^2$

Benjamin R

Hi guys

is there a symbol for vertex adjacency in Graph theory?

DHMO

@saturatedexpo no, I don't understand?

Benjamin R

for example, we have a symbol for incidence

(lemme find it)

DHMO

@Brody correct

Benjamin R

23:32

$\Phi$ is for incidence

saturatedexpo

@DHMO if you plugin -1. The series converges by the lemma of absolute convergency. This Lemma is what you need to understand then. applieing it to complex numbers is rather easy

PVAL-inactive

@Brody $\Bbb C$ has none that play well with both multiplication and addition. Certainly it has a nice ordering which behaves very well under addition.

DHMO

so complex integral is actually line integral

arctic tern

@DHMO the continuous image of a connected space is connected

DHMO

@saturatedexpo but that is for real numbers

saturatedexpo

23:34

"applieing it to complex numbers is rather easy" @DHMO Or do you have trouble at exactly that point?

DHMO

@arctictern yes?

@saturatedexpo is rather intuitive, but not easy

arctic tern

@DHMO I thought you were asking if R\0 could be the image of R, sorry

Brody

I don't fully understand @PVAL-inactive. In what way should an ordering be compatible with a given operation? (mind that I'm murky on the formal definition of a total ordering)

DHMO

@Brody what do you mean?

Brody

@DHMO I get the intuition behind a linearly ordered set. But when talking about fields, how do addition and multiplication come into play?

saturatedexpo

23:41

are links allowed to videos if they are ontopic?

@DHMO well here you go: khanacademy.org/math/calculus-home/series-calc/… can't explain it better

Brody

looking up the definition on Wikipedia, I see the gist now @DHMO @PVAL-inactive

Forever Mozart

you have to modulate

Devilius

Hey there, do you guys provide wolfram alpha support?

DHMO

@Brody how does addition and multiplication have anything to do with ordering?

@saturatedexpo any links are allowed

saturatedexpo

in 99% of cases wolfram alpha doesn't get you where you want to be. @Devilius

Benjamin R

23:49

more info: Consider the following statement from "Pearls in Graph Theory" (Harstfield, Ringel):

"Two graphs $\text{G}_1, \text{G}_2$ with $p$ vertices are said to be $isomorphic$ if the vertices of $\text{G}_1$ and $\text{G}_2$ can be labeled with the numbers from 1 to $p$ such that whenever vertex $i$ is adjacent to vertex $j$ in $\text{G}_1$, then vertex $i$ is adjacent to vertex $j$ in $\text{G}_2$ and conversely.

Such a labeliing is the same as a one-to-one correspondence between $\text{V(G}_1)$ and $\text{V(G}_2)$ that preserves adjancency."

saturatedexpo

@Devilius but feel free to ask what you want to know

Benjamin R

For example I could state:

Let $v_i \sim v_j$ denote that vertex $i$ is adjancent to vertex $j$ in some graph $\text{G}$.

Brody

@DHMO considering whether or not something can have a linear ordering, as apparently this relates to ordered fields

saturatedexpo

@DHMO my prof did make a nice graph, showing that the radius of convergence makes sense for complex numbers too. the problem: i cant remember it haha.

(it was a spiral)

Devilius

So, I am trying to use the ForAll and Exists quantifiers together, but am having trouble. The following is my attempt, after reading through their docs, to enter the following: for all x, there exists a y, there exists a z, such that (for x, y, z all positive integers) x^2+y^2 = z

forAll[x, Element[x, Integers > 0],Exists[y,Element[y, Integers > 0],Exists[z,Element[z, Integers > 0],(x^2+y^2=z]]]

Anyway, I'm assuming I am not stacking the forAll and Exists correctly, but can't find any examples on how use them together.

saturatedexpo

23:54

That is abuse of math language. Just my opinion.

also that looks like one of fermats problems no?

Brody

I think if one knows Mathematica language (I don't), they would know this

saturatedexpo

yes

Devilius

may I also ask why you find that wolfram is not the way to go ?

saturatedexpo

because you try to proof something via computational power. and then pat yourself on the back?

(or verify)

DHMO

@saturatedexpo I just wasted 5 minutes

Benjamin R

23:57

Hi Devillus, you need to remember that $\forall$ and $\exists$ are not layerable.

saturatedexpo

@DHMO then it's not clear to me where exactly your problem is.

Benjamin R

In fact we learnt that you never use them in the same statement

One is a symbol for universality, the other for existence

DHMO

@Brody of course you can always define an ordering in C

Benjamin R

lemme find you what my Discrete Math prof said about it

DHMO

@BenjaminR no, e.g. the epsilon-delta definition of limit

Devilius

23:59

ah. My discrete mathematics book "stacks" them

calls it "nested quantifiers"

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$Mathematics$ Mathematics