6:06 AM
Let $A$ be the set of all positive rationals $p$ such that $p^2<2$ and let $B$ consist of all positive rationals $p$ such that $p^2>2$. Consider $A$ and $B$ as subsets of the ordered set $\Bbb{Q}$. The set $A$ is bounded above. In fact, the upper bounds of set $A$ are exactly the members of $B$. Since $B$ contains no smallest number , $A$ has no least upper bound in $\Bbb{Q}$. And for $B$ it's kinda same.
My book says this proves $\Bbb{Q}$ doesn't have least upper bound property.
But in proof they say $B$ is exactly upper bound of $A$. I think it is not right since negate of > is $\ge$. So $B$ isn't exactly upper bound.
It is wierd they proved $\Bbb{Q}$ has no LUB property. It must be wrong they should replace $B$ with $\Bbb{Q}$ and $\sqrt{2}$ should be lub.

2 hours later…
8:44 AM
@Stupidquestioninc Does $A$ have a least upper bound?

9:03 AM
@AlessandroCodenotti if $Bbb{Q}=A\cup B$ then A doesn't have
lub

@Stupidquestioninc I think you need to reread the sentence "the upper bounds of set $A$ are exactly the members of $B$"

@LeakyNun yes I have but I don't think $B$ is exactly upper bound I think a set with positive rational $p$ such that $p^2\ge 2$ can be conted exactly

what does "$B$ is exactly upper bound" mean

fully ,totally, completely

yes, and if you want to play the dictionary game, you can see that it's an adverb
2
so it can't modify a noun phrase, which "upper bound" is

9:12 AM
this means 2 needs to be member of set $B$ which is not included and $B$ has dense rational gap then it means $A$ has no lub

$2$ does belong to $B$, since $2^2> 2$

oops I means square 2

that's not a rational number

ahhh now i know where I was missing
:P

9:14 AM
You mean primary or first language I used?

native language / mother tongue / the language you're most fluent in

most fluent is English mother tongue is also English native is trad Chinese which I don't really know how to speak
I have forgotten most of the grammar which I learned in school
so apologies if it has caused some misunderstandings

but if English is the language you're born with, you don't need to learn grammar...?

@LeakyNun well that's long and complicated story I was illiterate for 3 to 4 yr due to disaster. Found some random math book and started to learn it.

I see

9:23 AM
ahh I missed right in the definition of exactly oops =P
that was two embarrassing mistake thanks for fixing it sir!

3 hours later…
12:08 PM
help help

Do I need to plot using color maps?

@AnindyaPrithvi No, you are just drawing a subset of the complex plane

I see no mention of color.

Then how do i sketch?
I know about the rotations on squaring
and the shift of real axis by 2
and the shoot up/down of magnitude
but how do i show it

12:12 PM
You just plot in the given complex numbers

then?
the given is an infinite rectangle

Then you are done, because that was what the exercise asked for

?this is a complex plane...how do i show which plots to which

You don't

Can you share an illustration...I cannot find without colours

12:15 PM
You only plot the values, not the input

@TobiasKildetoft I am not getting what you are trying to say...what i know is f(x,y)=g(x,y)+ i h(x,y) on the subset evaluates to a point....marking just that point would do the job?

Well, you need to mark all of the points, but yes

Gotcha, I was thinking this but was stuck at doing f(f(x,y) iteratively

But the exercise did not mention anything about applying f twice

I know..just that I was twisting everything

12:23 PM
Slow down, and take it one step at a time :-)

12:39 PM
I couldnt plot something meaningful
only a region that too not well defined
@skullpatrol That would take infinite time to plot the graph xD

What would you do if you were asked to sketch the original set instead of the image of it?

@TobiasKildetoft I would have used x in R and y in -1,1

And what would you do with them?

@TobiasKildetoft nothing? it asked me to plot the original set so i did

But you just mentioned some infinite set of points. Why was that easier to plot than the one you get now?

12:46 PM
hmm sounds fair
But i am unable to verify my answer now
the unit circle at origin plots as it is just shifted by 2
and the rest region's boundary looks like a near parabola
The holding parabola is y^2=4(x-1)
should it look like the area traced by red line?
NVM solved it

1 hour later…
2:02 PM
Can we have two completely different vector space having same dimension?

@prakeshkumardora Depends on what you mean by different. There is no way to tell them apart just as vector spaces
(asuming they are over the same field of course)

@TobiasKildetoft I am saying in the context of linear transformation. Where a linear operator acts on a vector belong to ( say vector space V) and transforms to other vector belonging to ( say vector space V' ). Now say dimension of both vector space V ,V' is same. Then I want to under stand how V' is different than V . Since they both have same dimension.

2:19 PM
As I said, there is no way to distinguish them as vector spaces

Ok thanks.

of course, they can be disjoint as sets

2:59 PM
Consider a symplectic manifold $(M, \omega)$. We have the following isomorphism $i_{\omega} : TM \to T*M$ which induces an isomorphism between $\theta : \wedge^{2}(TM) \to \wedge^{2}(T*M)$, this induces a map on the global sections, $\tilde{\theta} : \Gamma(M, \wedge^2(TM) ) \to \Gamma(M, \wedge^2(T*M))$
I have to show that for $F,G \in C^{\infty}(M)$, $\{ F, G\} := (\Lambda, dF \wedge dG) = \omega(X_F,X_G)$ where $\Lambda$ is the pullback of the symplectic form by $\tilde{\theta}$ and $(\cdot ,\cdot)$ is just the contraction of an element of $\Gamma(M, \wedge^2TM)$ and $\Gamma(M, \wedge^2T Does this straight up follow from the fact that$(\Lambda, dF \wedge dG) = (X_H \wedge X_G, \omega) = \omega(X_H, X_G)$? 3:53 PM It is also stated that$\{H, \cdot \} = X_H$. I fail to see this, by the above hypothesis,$\{H, \cdot \} = \omega(X_H, \cdot) = dH$, so how is this$X_H$? You made a mistake.$\{H, f\} = \omega(X_H, X_f) = dH(X_f)$. You should be able to prove$df(X_g) = X_f(g)$, I believe. Ehhh horrible Horrible error 4:08 PM I mean, write$df(X_H) = X_H(f)$, that's all really. Hm, you get a sign$\omega(X_f, X_g) = df(X_g) = -dg(X_f)$Depending on convention it's either$X_H$or$-X_H$. I think they define Poisson bracket as$\{f, g\} = \omega(X_g, X_f)$Maybe not According to the given definition, there shouldn't be a sign Just figure it out, something I mean I looked up and they say$\{\cdot, H\} = X_H$Sure sure, thanks So there is a sign You wrote the identity wrong Maybe that's a typo in the problem set It says the opposite in it, but anyway, the idea remains the same 4:20 PM What is the point of the Poisson structure I don't have any understanding beyond pure formalism @BalarkaSen As of now I only have a physics motivation and that too has to be made formal. So in physics, these functions$f \in C^{\infty}(M)$are called classical observables. Noether's theorem states that such conservation of such observables have a "symmetry law" associated with them. To formalise this, one can use Poisson brackets. So consider the vector field$\{K, \cdot \}$on a symplectic manifold$M$with a hamiltonian, then this vector field actually preserves the flow generated by$H$. But I remember that there is a theorem which says that a poisson manifold can be partitioned into immersed symplectic submanifolds but I do not know how one would prove this A friend of mine uses them a lot in geometric quantization and Kontseivich stuff but I have no clue what she talks about Hm OK I don't understand all of this very well but I can buy this I don't actually know what classically symplectic geometry people used to do Nowadays it's just hard analysis Oh, I thought people will be doing more geometry, with all that J-holomorphic curves business That seems like bullshit analysis lmao So hard lol I don't know, maybe I will end up doing some derived algebraic geometry stuff coming from symplectic geometry, that stuff sounds interesting (if all this has too much analysis lol) 4:35 PM lolok If you have an injective holomorphic map$f$from$U$to$V$(both are open and in$\mathbb{C}^n$, and$n > 1$), how do you exclude the possibility that for$h(z):= \det(Jac(f(z)))$, there's a point$z_0 \in U$such that$h(z_0) = 0$then$h'(z_0) = 0$as well ? This is for some reason assumed to be true in GH's proof that$f$would infact be bijective to it's image. I didn't understand the theorem.$f$is an injective holomorphic map... injective maps are in fact bijective to their images :P OK I mean you get a holomorphic inverse basically you need to show that$Jac(f(z))$is full rank everywehre he assumes in the proof that$rank(Jac(f(z))) > 0$everywhere, if I'm understanding it correctly, in the proof But I'm not finding it entirely trivial to rule out the case$Jac(f(z)) = 0$. It's trivial if$z$is a regular point of$h(z)$(we get a contradiction using implicit function theorem) Taylor expand Huh, can you elaborate ? 4:50 PM If the Jacobian was zero you should be able to argue by Taylor expanding$f$, and noting that the first order terms just die. Are you sure you want to rule out "the case Jac f(z) = 0" yes I am fixing a point$z_0$, how do you rule out$Jac(f(z_0)) $is not the zero matrix Ok, then it should follow exactly like the version for$\Bbb C$Try Taylor expanding and arguing Well for$n = 1$, I get a contradictio to injectivity easily using Rouche's theorem if the first order term vanishes. How does this generalizes to higher dimension ? No that's a terrible way to do it for$n = 1$lol (or, if$f(z) = z^n g(z)$with$g(0) \neq 0$, then taking$z \sqrt[n]{g(z)}$as a coordinate) 5:01 PM How many cubic numbers that can divide$2^{38}\times 3^{17}\times 5^7 \times 7^4$are there? My answer is 468, is it correct? sounds about right @Thorgott Are you responding my question? yes @Thorgott "Sounds" about right? My answer is correct? 2 hours later… 6:47 PM Is which argument is linear in an inner product just a matter of convention? 7:01 PM You’re talking about a hermitian inner product? Yes. @BalarkaSen ok, I found after some googling you need to do some stuff with Weierstrass division theorem and weak nullstellensatz to exclude this case, this is not entirely trivial (at least, not to me). I still don't see why Taylor's expansion is relevant here. ok ty @BalarkaSen nevermind lol, it's trivial idk I was probably overthinking 7:37 PM @BalarkaSen no wait, how does taylor expansion help exactly? Hey @Ted! 2 hours later… 9:37 PM @Lelouch I think he was thinking of the 1-variable case, where you can Taylor expand and then show that$f^{-1}(\epsilon)$is$n$points for small nonzero$\epsilon$, or find a compositional inverse. In higher dimensions I think the appropriate version of this is weiestrass preparation as you asy. Weierstrass preparation is still a Taylor expansion result :P There's some work but this is definitely easier than Weierstrass. I tried to see it without Weierstrass but I didn't see. I was thinking about how to say it but it's not worth the effort. You need to understand the analytic variety$\det(Df) = 0$You can always find points on this guy around which you can write it as a graph of some holomorphic function I don't care anyway Yeah that's right, you just want to write$\det(Df) = 0$in coordinates as$z_1 = 0$, and then by GH's stuff$Df$vanishes everywhere on$z_1 = 0$which forces$f(0, z_2, \cdot, z_n)$to be constant It's easier than Weierstrass preperation to find points in analytic hypersurfaces which are given by coordinate function = 0 I believe 10:06 PM Monologue: Let$\eta$be a point process on a space$X$(recall that's a random counting measure on$X$), then define Laplace transform of$\eta$to one which eats a nonnegative measurable function$f : X \to \Bbb R$and spits$\mathscr{L}_\eta(f) = \Bbb E e^{-\int f d\eta}\mathscr{L}_{\eta}$determines$\eta$What do you mean with a random counting measure? (sorry for messing up the monologue part) and what do you mean by$\mathbb E$? A counting measure to me is the space of countable linear combinations of$\Bbb N_0$valued measures on$X$Such measures in fact make a measurable space with sigma algebra given by cylinder sets Fix a measurabe subset of$X$, fix a number in$\Bbb N_0 \cup \{\infty\}$, say your measure takes that value on that measurable subset So a random counting measure is a map from some probability space$\Omega$to this measurable space of counting measures @LeakyNun$e^{-\int_X f d\eta}$is a random variable on$\Omega$, because$\eta$(the random counting measure) is parametrized over$\Omega$So you can take expectation expectation wrt$\eta$? Ya Here's a cool fact. Suppose you're throwing countably many darts at a measured space$(X, \mathcal{A}, \lambda)$in a way that for a measurable set$A \subset X$, the number of darts that fell in$A$is a Poisson random variable with parameter$\lambda(A)$, i.e., probability that the number of darts within$A$is$k$is equal to$e^{-\lambda(A)} \lambda(A)^k/k!$This is actually a random counting measure on$X$, because these "dart counting" random variables makes up a counting measure on$X$as your vary$A \in \mathcal{A}$, except that they're... random. Then you can always write these guys as a random sum of finitely many random Dirac deltas Hm, I need$\lambda$to be a probability measure on$X$for this That is, pick random variables$P_1, P_2, \cdots$valued in$X$, with law$\lambda$. The experiment is "pick countably many random points on$X$following the measure$\lambda$". Let$Z$be a plain vanilla Poi(1) random variable Then$\sum_{i = 1}^Z \delta_{P_i}\$ is the random counting measure described above
Somehow, even though you throw countably many darts, you can write the dart counting measure as counting finitely many points -- given you get to choose finitely many random points and you get to choose randomly many such points
Pretty bizarre
(random finite)*(random finite) = (random infinite)
I am echoing @Secret at this point