@semiC How does that work?

the second method

and then life is easy, because that can be factored as $s_m e^{i(k x+\omega t+\phi/2)} (e^{i\phi/2}+e^{-i\phi/2})=s_m e^{i(k x+\omega t+\phi/2)}\cdot 2 \cos (\phi/2)$

which evidently has imaginary part $s_m \sin (kx+\omega t+\phi/2)\cdot 2 \cos(\phi/2)$

@Sophie Because it's not true for random things. $\Bbb Z/6[x]$ doesn't have unique factorization.

Hi @Adeek

this is basically just using $\text{Im }e^{i k x}=\sin k x$ and $\text{Im}(z+w)=(\text{Im } z)+(\text{Im }w)$

Actually swap $\sin$ and $\cos$ in the right hand side of my equation

It is, however, true that poly rings over UFD's are also UFD's

anyways. this all amounts to saying a superposition of two waves of identical wavenumber and frequency is another wave of that same wavenumber and frequency.

But things start getting complicated there, eg those start becoming not PID's. One just wants to prove so much as he wants to use.

@TedShifrin So, I took a look at your conormal bundle proof for $N_V=[V]\big|_V$. I can get most of it, but I've got one weird problem. If $V$ is locally defined by $f_\alpha$, then we have $f_\alpha=\varphi_{\alpha\beta}f_\beta$. It looks like you're getting that the transformation rule for $N^*_V$ is $df_\alpha=\varphi_{\alpha\beta} df_\beta$. But then it's the same as that for $[V]\big|_V$, not as $[-V]\big|_V$...

Does there exist a grouphomomorphism that maps 2 to 1 between (Z,+) and (Z,+)? no because 1 has to be mapped to one?

There doesn't exist one but that's not the reason

@semiC Using the methods you provided inevitably leads to having a cos and a sin part, correct?

sure.

@saturatedexpo Why don't you just check the properties?

@BalarkaSen what is a vector bundle intuitively.

I am required to have this become one wave because I need a single quantity for $\omega$ and $s_m$

It's because then $f(n) = f(n\times1) = n\times f(1) = 2n$, thus $f(\Bbb Z) = 2\Bbb Z \ne \Bbb Z$ thus f is not surjective

@Adeek A continuous association of vector spaces to points in a manifold, sort of.

draw a bunch of arrows on a piece of paper, then cut them out and stack them together. there, you've bundled a bunch of vectors together :P

context: two identical sound waves described above, differing in phase, travel through a tube of air. I am to find the avg. power given the phase difference, amplitude, and angular frequency of the wave

@Astyx A group homomorphism can fail to be surjective

I think the definition is simple enough to understand.

yeah the definition that I have is not hard to understand

(by that logic, i guess a spinor bundle would be a merry-go-round and a tensor bundle would be a final exam :P)

Oh it's just a homomorphism, then what I said does not hold

@Astyx You're still correct but it's really because 1/2 isn't an integer

Are you familiar with the tangent bundle of a smooth manifold?

Of course

yeah

@saturatedexpo No homomorphism exists mapping $2$ to $1$ but one exists mapping $1$ to $2$, namely $f(n) = 2n$. (Check that this is a homomorphism.)

It is the union of the all tangent spaces at point p.

disjoint union.

Well, you gotta topologize that and actually make it a smooth manifold.

@Fargle is a "homomorphism"="grouphomomorphism"?

if it preserves the group structure, sure.

@saturatedexpo As I mean it, yes

I need help with this

yeah

and $2n+2m=2(m+n)$, so yeah

Ok. Then I don't know why you're lacking intuition - you seem to have both the definition and the interesting examples.

Be more specific?

@KasmirKhaan can't bounty it

@saturatedexpo The reason one doesn't exist mapping $2$ to $1$: if $f(2) = 1$, then $f(1 + 1) = f(1) + f(1) = 1$, and there is no integer $a$ such that $a + a = 1$

I have been waiting almost 2 days now

i want to work that myself out haha

I cant bounty it yet :(

@saturatedexpo can you help me bounty it in 40 mins?

ping me then ok @KasmirKhaan

@semiC I take that back the description you gave works fine. lol

@saturatedexpo will do thanks sir

why do we need the commutative diagram definition with the regard to the picture you gave me @BalarkaSen ?

@Fargle just want to know, is that the only reason such a function doesnt exist (0.5 no integer)

I'd split up D into D_1 with y>yc and D_2 with y<yc, with yc being the solution to $y=1/x=x-1$

@saturatedexpo I'm sure there are other reasons it fails, but they're all pretty much equivalent

I'm going, good day to all of you !

@Adeek The commutative diagram just says that locally the bundle looks like the standard product $X \times \Bbb R^n$ aka trivial bundle.

@saturatedexpo My explanation: If it's going to be 2-to-1, it needs to do so everywhere. In particular, you need $f(x)=f(y)=1$ for some distinct $x,y$

@Danu 'sfine

@BalarkaSen Disguuuuusting :P

The arrows are so wrong

ok I see @BalarkaSen so we are trying to generalize products right ?

I mean vector bundles tries to generalize products right ?

@Adeek In some sense.

Vector bundles are "twisted products", is the point.

...hm. my explanation goes nowhere.

I guess we can explain this using some kind of univerisal property.

not sure how you get to $f(2)=1$ without loss of generality.

@Danu There are more disgusting things in the words. Like the results of the US presidential elections this year.

vector bundles I guess form a category

@BalarkaSen Take a cartesian product and make it weird :P

actually our assignments are really easy, because everytime something is right, there just stands "show it", and if not "give an example or disprove it"

with objects E

@BalarkaSen Ehhh, there's a difference. those pictures only offend the eye

and morphisms being the commutative diagrams.

@Adeek Sure, why not?

I don't think that the result of the presidential election is that disgusting, well.

@Adeek Well, morphisms of vector bundles are just (continuous, smooth, whatever you are considering) maps of total spaces that restrict to morphisms of fibers on each fiber.

So for vector bundles you need the restriction to a fiber to be linear.

what is a total spaces ?

In this case, you need to fix the base space.

@FrankScience You don't need to, why?

@Adeek The space that locally looks like a product.

ok I see.

@Semiclassical i read your explanation. so a isomorphism is bijective, but a normal morphism is only ...?

The definition he has in mind is equivalent to what you're saying, Danu.

But I guess yours give more intuition.

@BalarkaSen Hmm?

i'm not good with the morphism language, so i dunno

Morphism = natural type of map in given setting

@Danu Maps $E \to E'$ which commutes with the projections.

I am fixing the base here, but feel free not to.

(of course you can invent categories where this intuition is a bit mangled but it's essentially right)

@BalarkaSen But why would one fix the base?

ok sure I understand now the definition. That is cool. So it is something that looks locally like a products. I already have a specific picture in my mind.

For example. We can consider a line and above each point in the line a fiber

then this will be a vector bundle.

Yeah, so that's just a product.

The classic not-as-trivial example is the Moebius strip.

@Danu Usually one looks at a slice category because one can take limit and colimits there.

(maybe you should prove it's not a product!)

@KasmirKhaan I think there's an easy solution. First off, that integrand is positive everywhere on the domain of integration (x can't be negative, so neither can y>=1/x)

But the definition I gave works fine without fixing any base.

@BalarkaSen What's a slice category?

It's just a pair of maps $f : X \to Y$, $f' : E \to E'$ which commute with the projections.

But it looks like that integrand diverges along the line $x=1/y$ as a function of $y\geq \frac{1+\sqrt{5}}{2}$

yeah

@Semiclassical yes it should be easy solution because we are doing multivariable calculus not real analysis

I have different definition for slice category @BalarkaSen.

@Danu ncatlab.org/nlab/show/overcategory

@Semiclassical can you pls post solution ? because i cantsee your answer , well i do but in messed up way

By the way @Balarka, wikipedia gives a bundle morphism as requiring both properties (restricting to fiber morphism, and commuting with projection)

all i'm saying is that if you look the behavior along the curve x=1/y, the integrand doesn't go to zero

The terminologies are sometimes confusing.

and it really should, if you're going to get a finite result.

not sure I can make that precise, though.

bundle map, bundle morphism, etc

@BalarkaSen given a category C. Suppose we consider fixed objects A,B. We define a new category $C_{A,B}$ with objects being morphisms from some objects to both A and B and morphisms are commutative diagrams between them.

This is how allufi defines slice category.

@Semiclassical am very new on the topic , this is my first improper double integral and it would be really helpfull to get good solution so i can do the next 7 similiar problems on my own

@Danu Commuting with projection automatically gives restriction to fibers. What you mean is preserving the linear structure, for which you need to say that the fiberwise restriction you get is linear.

If you can recommend a book or pdf that i can use it would be great !

eh, it's been too long since i've done multivariable to have any reccs

@BalarkaSen That's what I said.

(restrict to a morphism on each fiber)

@Adeek I don't know what you're referring to. I linked Danu the article on slice category in reply, instead of giving him any definition.

@KasmirKhaan Actually, though.

when I check it by another route, I get something which should be finite :/

soooo i dunno

haha they are tricky !

if you could help me set up a bounty it would be great , im gonna to set a bounty as well but dont have enough energy points

@Danu Yes; I just said your definition is the same as the one Adeek has in mind.

eh, you've got enough rep that you can do it yourself. you just have to wait for it to be open for bounty

Commuting with the projection is the same as saying "it sends fiber to fiber"

if you look below your question, you'll see a little thing telling you how long until you can post a bounty.

@BalarkaSen Yeah, of course :)

so the fact that 2 gets mapped to one, does not imply that 1 gets mapped to something else then 1. Because injectivity is not nessesary?

right :) didn't mean to imply you were wrong or anything.

@Semiclassical it sais 20 min

then wait 20 minutes :)

@Semiclassical will do sir !

@Danu So, how's your new math-life?

mmkay

@BalarkaSen Scary and exciting :)

its very annoying there are no videos about double improper integrals online

but i want to bounty for badge lol

I'm getting close to settling on a precise thesis topic now.

@saturatedexpo you can bounty me in few mins :D

I presume it's going to be complex geometry?

@saturatedexpo It doesn't imply immediately that $1$ must be mapped to something other than $1$, but it does imply that it must be mapped so that $f(1) + f(1) = 1$, and certainly $f(1) = 1$ doesn't work for that

This year I'll also write a thesis.

@BalarkaSen Yeah, sure. Characteristic classes on complex manifolds.

and in fact no value for $f(1)$ works for that

But not clear what exactly.

@danu You could help me figure out my stuff :P

@Danu Cool stuff.

@Fargle thanks, now i understand your argument at last ;)

I'm trying to read up on cohomology of non-hyperelliptic curves

@saturatedexpo Huzzah!

@Semiclassical Meh :P

We;ll see.

specifically with an eye towards stuff Griffiths did re: residue maps

lol

Ah, well

I have a question about characteristic classes

it should be easy

I'm looking for stuff like: Suppose I have a family of smooth plane quartics $f(x,y;t)=0$, indexed by $t\in \Bbb C$

I don't know much about them yet @Frank.

But you can always ask.

Suppose that $B$ is a compact manifold and $p\colon E\to B$ constitutes a vector bundle of real dimension $n$.

in the case of interest I'd have $f(x,y)=p(x,y)-t q(x,y)$ with $p$ quartic and $q$ at least cubic.

We note $u\in H^n(E,E_0)$ the Thom class, where $E_0$ is the set of nonzero vectors.

You've lost me. :)

^

I'm interested in period maps, e.g. what integrals like $\oint R(x,y)\,dx$ look like for $R(x,y)$ some rational function

What I want to show is very simple:

Ain't nobody got time for $\{\text{anything}\}\setminus \{\text{Chern classes}\}$ :P

since $R(x,y)\,dx$ is an element of the first de Rham cohomology, I should be able to write it in terms of the basis of such.

@saturatedexpo Right.

furthermore, I can differentiate this form w/r/t $t$ to get other such forms.

$u\cap$ sends the fundamental class of $(E,E_0)$ to the fundamental class of $B$, where $H_m(E)\cong H_m(B)$, and $m=\dim B$.

And since the first cohomology has finite dimension, there should be some linear combination of $\partial_t^n R(x,y)\,dx$ that vanishes up to an exact form.

The question is then, how do I do that practically?

And that's where my head hurts. :/

Practical calculations

I know how to do it if it's a hyperelliptic curve, at least in principle

The worst kind

:D

But doing it for a non-hyperelliptic curve is nooooot so fun

Yeah.

Where in physics is this relevant?

Well, in the hyperelliptic case, it's easy

I have a quick question. @BalarkaSen if we consider $E_p$ (which is $\mathbb{R}$ vector space) as before then $h_j^{-1}(p,e_i)$ for $i \in \{1,...,r\}$ is basis for this ?

Like, you can't do that on a torus? You can compute the intersections number of the dual of that form with the basis circles by integrating, can't you?

Whenever you want to do semiclassical stuff, you usually end up computing action integrals $\int p(x)\,dx$ where $p^2+V(x)=E$ is just conservation of energy

Sorry

I mean why have that $h_j|_{E_p}$ is a diffeomorphism onto $p \times \mathbb{R}^r$

You mean

and if $V(x)$ is a polynomial potential, that amounts to integrating $p(x)\,dx$ on the Riemann surface defined by $p^2+V(x)=E$

explicitly calculate the $H^1$?

@Adeek Yep.

but why does it carry basis to basis ?

it is not linear no ?

That's the easy version.

It's a linear isomorphism, @Adeek.

That hard version emerges in the context of parametrically driven oscillators, and their semiclassical theory.

In that case you start with a polynomial potential, then add on a resonant driving oscillation and some damping

@balarka In the definition we didn't assume it is linear isomorphism ?

we just assumed it was a diffeomorphism.

To study that, you usually do what's known as a rotating wave approximation. Basically you go to the interaction picture and drop certain higher order terms.

Look again carefully.

In that case, you get left with a conservation of 'pseudo-energy'.

oh ok yes I see @BalarkaSen.

(like pseudomomentum in the context of periodic potentials, except in time not space)

and the examples I know of those look something like $G=(p^2+q^2)^2+$ lower-order stuff

with $G$ being the conserved quantity.

I forgot these physics.

I remember there was something called Hamiltonian.

@danu so you still end up being interested in integrals $\int p(q)\,dq$, but the curve you're on is no longer hyperelliptic.

@FrankScience Yeah, this would be a pseudo-Hamiltonian in the language of this stuff

it's not a standard thing

How do I write this inequality corrrectly algebraically? I have that p+j=100 meaning j=100-p. Then we're told that p<50, therefore j=100-p<100-50=50 or j<50... but actually, j>= 50 since we must maintain p+j=100.

Basically, you have a curve then you want to calculate the periods?

right.

and i'd like to have it set up in a way that doesn't depend overly-much on the specific (closed) path of integration on the curve.

well, actually 50 <=j <=100.... but the important part for this question is that j should be 50 or more, but following the algebra i get less than 50.

@Semiclassical I see.

since i'd like to consider the periods of various cycles.

I don't understand why you think that hyperelliptic case is easy. Even for elliptic curves, the elliptic functions are not easy.

I guess most physicists would've defaulted to numerical methods long ago :D

@FrankScience It's not elliptic functions here, though, but elliptic integrals.

and those satisfy differential equations. the idea here is to reduce the calculation of the periods to the solution of some (complicated) ODE.

@Jeff j = 100 - p > 100 - 50 is the step where you erred. If p < 50, -p > -50

But as far as I've understood your formulation, you need to find a base for $H^1$ and then try to compute the periods of them?

Not quite.

@Fargle oh.... duh (sheepish grin). TY :D

@Jeff No problem!

Given a closed form he wants to write it down in terms of some canonical choice of basis.

I think.

what I ultimately want is to find a linear combination of $\partial_t^k p(x,y)\,dy$ which vanishes up to an exact form

because then that automatically gives a differential operator $\mathcal{L}$ which annihilates any integral $\oint p(x,y)\,dx$

i.e. a differential equation in the parameter

Main trouble is how to make this practical. I know a way (not necessarily the most elegant) how to do it in the hyperelliptic case.

In that case $dx/y$ is a holomorphic form and I can simplify things a lot.

not going to be the case on a non-hyperelliptic curve, and that's the problem.

Griffiths had done some stuff on this, which these notes discuss: math.stonybrook.edu/~cschnell/pdf/notes/picardfuchs.pdf

And as I understand it, the method there really doesn't require anything about hyperellipticity.

problem is, reading that is hard because I don't have all the background.

I don't even know what primitive cohomology is :/

@BalarkaSen just want to make sure I understand this correctly.

so here we are considering $(p,v) \in (U_i \cap U_j)\times \mathbb{R}^r$

then we are defining $(p,g_{ij}(p)v)$ to be $(h_i \circ h_j^{-1})(p,v)$ ?

I mean what is $g_{ij}$ we haven't even defined it yet ?

@KasmirKhaan I pinged your question with this, but I get a final answer using Mathematica which is far nicer than I'd have expected

namely, I just get $1/e$ for that entire integral :/

$h_i \circ h_j^{-1}$ is a map $(U_i \cap U_j) \times \Bbb R^r \to (U_i \cap U_j) \times \Bbb R^r$. $g_{ij}$ is the matrix of the linear map on the second component.

Hi

There's presumably a reason for it, but I'm nonplussed.

I have a pretty simple question regarding calculus

I have a point where the derivtive of the function is equal to zero and i want to check if a local extremum is there. In order to do so I can calculate the second derivative if the second derivative is equal to zero i can calculate the third derivative and so on, to finally see which one is non zero

That is to say, you get a $r \times r$ matrix for each point in $U_i \cap U_j$. That's precisely the associations $U_i \cap U_j \to \text{GL}(r, \Bbb R)$.

oh ok.. I see we get this matrix from the isomorphism.

Right.

oh ok

At each different point we get a matrix that we can work with and we define the map using this matrix.

Ok that is quite some fancy stuff cool.

Nah, it just tells you how the $U_i \times \Bbb R^r$ and $U_j \times \Bbb R^r$ are "glued" inside the vector bundle.

now, lets say that the point for which i am cehcking the extremum is $x_0$. and $f''(x_0)=0$. Now i check wheter for $x_0<0$ $f''(x_0)<0$ and see the sign of $f''(x_0)$ for $x_0>0$

@Semiclassical hello mate how did you do that?

Mathematica and a lot more good fortune than I expected

if in those both intervals the function is concave (or non-cave) then the function has a local extremum

@Semiclassical it cant be convergent

is this reasoning proper?

btw you can set up a bounty now if u want :)

oh I see @BalarkaSen that is pretty cool.

I will try to add a bounty on my own too but i dotn have alot

first, I rewrote the domain of integration as $1/x\leq y\leq x-1$ for $x>\frac{1}{2}(1+\sqrt{5})=1.618...$

@Semiclassical you have been supportive since yesterday thanks alot

np

I will ask my teacher if he made a mistake

maybe he wrote diverge insteed of converge , wont be first time he makes mistakes :D

I asked Mathematica to integrate that w/r/t $y$ for generic $x$, and it succeeded in giving an answer (in terms of special functions)

@Adeek Try explicitly computing the transition functions on (1) the trivial bundle (2) the moebius line bundle over $S^1$.

and then I integrated the result from $x=1.618...$ to $\infty$

@KasmirKhaan i made a bounty :)

after that, I did a full simplify call and got 1/e :/

@saturatedexpo thanks alot bro <3

yeah that is good example @BalarkaSen i will do it.

thanks i ll have to ask my teacher!

@ping me if an answer fullfills your needs ;)

will do sir !

Now, there's a chance I messed up the algebra at some point and Mathematica's doing the wrong integral.

good evening

As far as I know they cant handle such integration limits

i mean they can only do rektangular

[a,b] x [ c,d]

It can be done, but it requires care.

I have matematica installed but never had the need to use it

i use walframalfa to check my primitive and thats it

Hrm.

I'm doing it in another way right now, and Mathematica no longer agrees about it being convergent :/

Hi can some one help explain what i am misunderstanding with my simplification. I have e^2*ln(c) = 2c but the answer is c^2

i don't quite get why its squared

e^(2 ln c)=(e^(ln c))^2 = c^2

@saturatedexpo If you have a group with $e$, $x = x^{-1}$ and $y = y^{-1}$, then $xy$ is also in that group.

alternatively, e^(2 ln c) = e^(ln(c^2)) = c^2

Your subgroup must have $3$ elements. There's only one group (up to isomorphism) with $3$ elements!

shouldn't ln c be multipled by 2 since its 2(ln c)

@SteamyRoot so for 3 elements, e has to be in it and x and $x^{-1}$?

it is being multiplied by 2.

yup

2 ln c = ln c + ln c = ln(c*c)=ln(c^2)

but how do i notate the elements of $S_5$? (to specify x)

Usually elements of $S_n$ are notated as cycles

Like, $(1 3 2)$

thats kinda confusing to me

well, what's e^(ln c)?

so (1,2) and (2,1) would be ok?

That means "the permutation that maps $1$ to $3$; $3$ to $2$ and $2$ to $1$

c

No, because $(1 2) = (2 1)$

right. and does it make sense that e^(a+b)=(e^a)(e^b)?

yes

okay. then note that 2 (ln c) = ln c + ln c

@saturatedexpo Are you familiar with cycle notation?

so therefore e^(2 ln c) = e^(ln c + ln c) = e^(ln c)*e^(ln c)

@Fargle yes, i just dont have figured it out yet ;)

ahh

now i understand :)

thank you !

mmkay

np

The best part about cycle notation is the "disjoint cycle notation"

You can write any permutation as a series of cycles with no common elements

and the order of a $k$-cycle is $k$. So for your subgroup of order $3$, you want a subgroup generated by a $3$-cycle.

@KasmirKhaan Okay, I managed to prove that it works using the substitution $u=xy,$ $v=x-y$

that parametrizes the integrand as $(u,v)\in [1,\infty)^2$

and taking into account the Jacobian, the integrand is simply $e^{-v}/u^2$

Which integrates to $1/e$.

Okay, back later

@SteamyRoot so (12) can't be in it because it's selfinverse, neither can any 2-cycle?

every 2-cycle is a self-inverse :)

a^2=1 <-> a=a^-1 after all

in the group with $3$ elements no elements apart from the identity is its own inverse though

mmh here lies my problem: (321)=(31) 2cylce :/

@Semiclassical thanks alot ill work on it as well

Uh, (21)(21)=(1)

S_3 certainly has 2-cycles

Oh, did you mean the cyclic group of order 3?

the subgroup with $3$ elements of $S_5$ doesn't

Yeah, thought you meant S_3 altogether

@Semiclassical thank you sir , the substitution did work exactly

@Semiclassical but for someone who dont have such good intuition how would he find such good substitution ?

$(1+\sqrt{3})^{n+1}+(1-\sqrt{3})^{n+1}\in^?\mathbb{N}$

@NaCl for all n?

yep

induction

or is this already the step? ;)

thats the step

@NaCl Open the binomial. The terms where the exponent on $\sqrt{3}$ is odd are going to cancel out

idk what to do with $(1-\sqrt{3})^{n+1}$

i think i solved a very similar excercise using the e (euler) notation

using log and all that jizz

The expression is "all that jazz". What you said is probably inappropriate for any other situation besides a secret conversation in an Access Hollywood bus.

@PVAL-inactive i tend to agree

@MikeMiller How's the formation of the Republic of California going?

oh gawd

nevermind, that was so easy

@Sophie thanks for the hint with the binomial

did you use the assumtion @NaCl

you don't need to