@bolbteppa!

Hey @BalarkaSen

so, what've you been thinking about lately?

multivariable complex analysis, non-linear partial differential equations, conformal quantum field theory, it's a tough life lol

How about you?

@bolbteppa hell...

@bolbteppa been studying homology. interesting stuff.

Hi @bolbteppa

U have to study a lot @bolbteppa

@bolbteppa always wondered why there's so much big names in modern physics.

My professor told me something interesting, that all of homology, cohomology etc... was historically motivated by complex analysis, that it all finds it's natural place there, e.g. etale spaces mathoverflow.net/a/8864/38721 started in multivariable complex analysis (I think you mentioned these to me)

topological conformal quantum electrodynamics. sheesh.

@bolbteppa u know about Pauli's exclusion principle

Well theoretical physics motivates just about every area of math you can think of tbh, that's why I studied math so I could appreciate it properly and not get scared lol

@bolbteppa yeah. cohomology really originated from sheaf cohomology.

i plan to study that stuff.

@SayanChattopadhyay sure it is a consequence of the Heisenberg uncertainty principle when you apply it to many-particle wave functions

haha

lol

What books are you using Balark?

everyone will start unignoring only if you start using "you" instead of "u"

@bolbteppa Hatcher. Classic.

@BalarkaSen you finished munkres

Soory

Sorry @BalarkaSen

@SayanChattopadhyay how could massless radiation from an electromagnetic field cause the gravitational field to curve?

interesting, @bolbteppa.

That's the doubt I had because that's what I saw in a documentary

funny how one can sound scary by just translating "how the hell would that happen?" into physics.

@SayanChattopadhyay which bit of Munkres? I studied fundamental groups and covering spaces from Munkres, it doesn't have homology.

@BalarkaSen I mean algebraic topology

Yes, I finished Munkres almost 2 months ago.

Otherwise I wouldn't even have touched Hatcher.

@BalarkaSen where are you thinking to do your bsc

Haven't thought about it.

@BalarkaSen Can we find a function $f(x)$ such that, say, $$\int_0^{\infty} \frac{\sin(\sin(\sin(x)))}{x} f(x) \ dx=\int_0^{\infty} f(x) \ dx$$? I recommend you to make use of some complex analysis.

Not interested, @Chris'ssis

@BalarkaSen I haven't sat down with Hatcher properly but all I know is that when I do I will study that subject using Stokes' theorem and ask how every theorem and definition is just formalizing something that is intuitive from the P.o.V. of Stokes' theorem, note the link to complex analysis, e.g. I think the Mayer-Vietoris sequence has a really intuitive explanation from that point of view

And I don't know complex analysis either.

@bolbteppa Ugh.

You're nuts.

The general case is $$\int_0^{\infty} \frac{\overbrace{\sin(\sin(\cdots\sin(x)))}^{\displaystyle n \text{times}}}{x} f(x) \ dx=\int_0^{\infty} f(x) \ dx$$

The trivial case:

$$\int_0^{\infty} \left(\frac{\sin(x)}{x}\right)^2 \ dx = \int_0^{\infty} \frac{\sin(x)}{x} \ dx=\frac{\pi}{2}$$

@bolbteppa I understand Mayer-Vietoris as a long version of the Van Kampen theorem.

Much of the similarities between fundamental/homotopy groups and homology groups ca be made clear by looking at the mysterious chain homotopy idea I guess, but I haven't given it a thought.

Well @BalarkaSen why are the most difficult questions in mathematics in number theory.....why not algebraic topology or something else

That's not true, @Sayan.

Algebraic topology has many very difficult questions.

Take homotopy theory of spheres for example.

But no one does talk about it right ?

Many talks about it.

It's one of the greatest open problems of mathematics.

Just because something is not a millennium prize problem doesn't mean it's not great :P

@BalarkaSen I recommend Goursat's Complex Analysis + Needham's Visual Complex Analysis + the problem books by Knopp, they are old books but nothing's changed and these are the only good books I've ever found on it, you could fly through Goursat, honestly I think you should read his 5 volumes on mathematics, you will know more than your professors on some things if you do tbh

Anyway, Poincare problem was a millennium prize problem and it wasn't from number theory.

It's geometric topology.

@bolbteppa Thanks! I really plan to study complex analysis after algebraic topology.

See coz when I go to some institution in mathematics in India I bet you noone will tell you anything about homotopy spheres you see

@zed111 it doesn't mention closure because it's thought to be obvious, technically one shouldn't define maps unless closure is satisfied, those 'failure of closure' exercises are just to test you to make sure you understand what's going on

@SayanChattopadhyay You're mistaken.

I might be but not in Bangalore for sure

There is so much more to mathematics than number theory.

I know

@SayanChattopadhyay You're still more mistaken.

IISc is a fort of geometers.

I doubt you'll hear anything about number theory, on the other hand.

^agree

That's why I am so interested in these topics.....

These topics are something u can visualise right @BalarkaSen

Yeah.

Ji @G-man

Kind of.

hello to all

Thats why I want to study these......

@BalarkaSen u did start topology after calculus and all right?

Yes. I started topology after studying algebra, in fact.

algebra......group theory

@bolbteppa I hear some physicists are being overenthusiastic about symplectic geometry. What's this stuff about?

The guy who sits next to me in my prof's office draws weird spaces all day and he says he studies symplectic topology.

Weird guy.

@SayanChattopadhyay Group theory is only a part of algebra.

Hey I had a question. Can any human actually, truly visualize the fourth dimension of space?

@bolbteppa everything is finite in physics right

U mean abstract algebra right @BalarkaSen

Yes.

And for heavens sake use "you". :P

I am sorry again

@BalarkaSen well the motivation for symplectic comes from physics. Classical mechanics is defined via Lagrangian mechanics which says that the true path $y$ that a particle follows is the path that minimizes the action $S(x,y,y') = \int \mathcal{L}dt$ in Lagrangian form, where the Lagrangian is a function constructed from physical principles (e.g. symmetries) that the system your modelling satisfies

If anyone missed my previous post, I post it again, the integrals are simply too nice. math.stackexchange.com/questions/1164183/…

@bolbteppa Interesting.

@bolbteppa if a theory in physics results that universe is infinitely expanding then is that theory wrong?

Here $(y,y')$ are the position and velocity of the particle, and I am constructing the surface $S$ using two points $y$ & $y'$. But intuitively I should also be able to generate the surface $S$ using points $y$ and tangent lines to the points $y$ right? (So long as some convexity condition is satisfied). Something called a Legendre transform allows you to equivalently generate the surface $S$ by points and tangent lines.

Let me see if Cleo or M.N.C.E. came up with something new ...

No, nothing new in terms of answers.

@BalarkaSen how is hungeford for algebra

@bolbteppa Makes sense. So you claim that if the tangent space is given the manifold can be drawn, is that right?

The tangent line $p = \frac{\partial \mathcal{L}}{\partial y'}$ is given the name momentum, shocking, so we have expressed our mechanics in terms of $(y,p)$ coordinates, this is what is known as Hamiltonian mechanics. So in Lagrangian mechanics we use position and velocity coordinates, and in Hamiltonian mechanics we use position and momentum coordinates (where the momentum coordinate in Hamiltonian form is a tangent line viewed in Lagrangian form).

@SayanChattopadhyay Try Dummit-Foote.

@SayanChattopadhyay no an expanding universe is just an example of Einstein's general relativity, it's like a model and can easily be changed

@BalarkaSen the tangent space is always assumed to be drawable, but there are some issues with being able to generate the surface if some convexity issue fails, I never looked into it because in physics you always assume it's possible

Drawing the manifold given the tangent space seems suspiciously like a lot of work using differential equations and whatnots...

And for an expanding universe the cosmological constant has to be equal to 1

So anyway, when your mechanics is in Hamiltonian form, you end up with a 1-form $\omega = pdq - Hdt$ and because mechanics assumes $\omega = d \eta$, i.e. that the system is conservative, we have $d \omega = 0$ or $dH \wedge dt = dp \wedge dq$ which gives you the Hamiltonian equations of motion and so one must assume $dp \wedge dq$ is non-zero for this to hold, when this defined and non-zero on a manifold we have a symplectic structure

I think it's expanding just for a while. I personally doubt Big Bang ever existed, and my theories about that can be confirmed by mathematicians I talked to many years ago (when the new theories about the Big Bang wasn't out there).

So it's basically saying the volume between the coordinates and momentum in phase space should be well-defined so that you can solve your equations of motion, it's just a mathematical trick to ensure the EOM exist and are well-defined

Anyway.

@BalarkaSen generating a manifold given information in the tangent space is basically going from a Lie algebra structure to it's corresponding Lie group and is done by setting up flows on a manifold and solving the systems of differential equations you get

I need to write some proofs. BBL (when you want my creations @BalarkaSen that require complex analysis, especially done for that, just let me know)

@Chris'ssis there is a lot of evidence for the big bang

earthsky.org/space/what-if-the-universe-had-no-beginning

@Chris'ssis that article is based on the news from two weeks ago about the 'Quantum Raychaudhuri equation' paper where they used a logically inconsistent 'theory' of quantum mechanics, known as Bohmian mechanics which basically assumes classical mechanics is valid, to make their claims, it's really bad stuff

Also discussed here phys.org/news/2015-02-big-quantum-equation-universe.html

Also discussed here motls.blogspot.ie/2015/02/…

I have the set $\{(-2)^n : n \in \mathbb{N}\} \cup \{0\}$ and I need to determine its closure. I don't really know where to begin with regards to doing this.

We've defined the closure as the set of points $x$ whose every neighbourhood meets both H and its complement.

@bolbteppa Is this a way of approaching things? "The first question that a TRF reader may be asking now is: Haven't I already seen the name of Ahmed Farag Ali somewhere?"

For instance, no one heared of me and still I wanna publish the best book in the area of integrals, series and limits. Besides that, I wanna see how many mathematicians I'll meet in this life that have an original solution to the Basel problem, say, not ideas taken from other papers, and how many have more results found than Ramanujan found.

Their original work, not things taken from others.

I read sometime an answer of this guy and then I said to myself I won't do it again. It was a mathematical question posted somewhere, maybe on MSE.

And then "he second thing that a laymen may want to notice is that this paper has, after 10 months, just one non-selfie citation. That's not too many, you know. Papers that are really transforming physics may get close to 10,000 citations, like Maldacena's AdS/CFT."

Is there a rule about that?

I personally if I could I'd cite nothing because I need no citation at all (referring to my book). However, it's not bad citing things, it'a also recommended, it's that way it is said the community works.

Anyway.

I think I have to finish some proofs. BBL

Oh yeah, and this is viewed as a subset of $\mathbb{R}$ under the euclidean topology. Would the closure be $\mathbb{Z}$ as this includes the limit points of $(-2)^n$?

Or would the closure be $\mathbb{R}$ as we are viewing it as a subset of $\mathbb{R}$ rather than $\mathbb{Z}$?

@DanielFischer Would you be able to help with this?

@user112495 That would be the boundary. The closure of $H$ is the set of points such that every neighbourhood intersects $H$, nothing forbids that some neighbourhood would be contained in $H$.

@Chris'ssis you misunderstand what he means about having already seen those guys names before. You are not a TRF reader so you shouldn't be asking yourself that question, he's talked about them before, that's why he says this...

@user112495 $\mathbb{Z}$ is a closed subset of $\mathbb{R}$, so the closure is a subset of $\mathbb{Z}$. But it's not all of $\mathbb{Z}$, for example there's no number of the form $(-2)^n$ in the interval $(5,7)$, so $6$ does not belong to the closure.

@DanielFischer So is the closure the set itself then?

@user112495 Can you give an argument for or against that?

@DanielFischer Not really. Thinking about it, I think the interior of that set would be the empty set as there is no r>0 s.t. the ball of radius r about a point is a subset of the set. I think that this would then mean that we can't find any neighbourhoods for any points.

@user112495 Take a look at $(-2)^n$ for the first handful of $n$. Can you see how things proceed? Then it's visually clear that and why the set is closed, and that suggests a way of proving it.

And yes, the interior is empty in this case.

@DanielFischer So if you consider $\mathbb{R}-H$, then you can draw an open ball of some positive radius around any point, where the open ball is a subset of the set, and so it must be open. Thus H is closed and so it is equal to its closure?

@user112495 Yes. Of course you need to prove it still.

@bolbteppa OK. Question (since I am an heck of an algebraist) : Given informations about the Zariski tangent space, can you always construct the algebraic variety?

@DanielFischer How would I go about proving it formally? I can see that as we are only looking at singular point holes, for any point that isn't one of the $(-2)^n$, I can draw a ball of some radius around it which doesn't touch one of the (-2)^n (for example, I could calculate the distance from any point to the closest of the (-2)^n, and then draw a ball of radius half this distance).

@user112495 Right. You calculate the distance to the closest point (or find a positive lower bound for that distance, but in this case that's no easier than determining the exact distance), and that's basically it.

@DanielFischer And then as this set isn't bounded, that means it isn't compact doesn't it?

@user112495 Right.

Isn't this answer an utter BS?

Surely summing the edges of the polygons will give me half the number of edges in the 1-skeleton?

@DanielFischer And it isn't connected because it isn't an interval (as if a subset of $\mathbb{R} is connected, then it is an interval)?

I'd be surprised to know that a tetrahedron has 12 edges, otherwise :P

@user112495 Yes.

@BalarkaSen If it's a very edgy tetrahedron maybe.

:P

A truncated tetrahedron has 12 edges, yes.

Actually it still works.

In case of K_5, it says the number of edges must be at least 21/2, i.e., more than 11. Contradiction.

K_{3, 3} is a bit harder to deal with but I guess I can do it.

@BalarkaSen I wont pretend to have studied Zariski tangent spaces but this en.wikipedia.org/wiki/Restricted_Lie_algebra#Examples says that a Zariski tangent space is a lie algebra on an algebraic group (with some extra information), and it talks about how each element of the lie algebra defines a left-invariant vector field, which means that each vector can be interpreted as the tangent vector to a curve on the algebraic variety (this curve is called the flow)

You should be able to exponentiate out of the Lie algebra to get the lie group which gives you the algebraic variety, since the curves of the flow generate the surface

@BalarkaSen waiving my hands here, this math.stackexchange.com/a/676494/82615 seems to claim that you can interpret the Zariski tangent space as the differential $dG$ of a homogeneous map $G$, this looks very similar to interpreting "homogeneous spaces" as lie groups and the corresponding lie algebra as the differential of the lie group

that just went above my head

Hmmm okay, an algebraic variety is the set of solutions of a polynomial equation right? (ignore technicalities unless imperative lol)

i know what an algebraic variety is :P

Can you generate those equations from symmetric polynomials?

sure.

What about homogeneous symmetric polynomials?

yes.

Still waiving my hands, if you view the space generated by a homogeneous polynomial as a homogeneous space, we can then notice homogeneous spaces are related to Lie groups, so any polynomial is a bunch of homogeneous spaces stuck together somehow (serious hand-waiving, but stay with me!). If we think of a lie group as related to e^(tX) for some operator X then the lie algebra is the first order approximation 1 + tX (tangent vector is first order approximation, hence we get tangent spaces)

so in the polynomial example in this math.stackexchange.com/a/676494/82615 link we see the 1st degree homogeneous polynomials form the tangent space of the variety, and we should be able to exponentiate and re-create the lie group, the way you recreate a polynomial by combining symmetric polynomials. So all the scary big manifolds language is literally nothing but the 17'th century intuition with big words attached.

lol

i can see what mike meant.

About what?

anyway, yeah, i believe you. it makes sense.

If $C, D$ are connected subsets of a topological space T such that $\bar C \cap \bar D \neq \phi$, is it true that $C \cup D$ is necessarily connected?

Well try to justify what I've said if you're interested I'm sure I was wrong about something lol

@bolbteppa this.

I'm leaning towards no, as otherwise this would have made a better statement of the theorem. But I can't come up with a counterexample.

@user112495 take $C$ to be the ball $|z| < 1$ and $D$ to be $|z| > 1$. why wouldn't this work?

$C, D$ subsets of $\Bbb C$

Or $(0,1)$ & $(1,2)$ on the real line?

yeah :P

The difference between a Physicist's and a Mathematician's hand waving argument is that the Physicist uses his hands :D

I think yours is the complex version of mine Balark lol

more generally, two balls close enough in any metric space, separated by their closures.

and a mathematical physicist abuses math with his hands...

yeah @bolbteppa

that was not a reply to the mathematical physicist comment :P

@bolbteppa @BalarkaSen Thanks.

haha

no problem, @user112495. try visualizing often when you're doing topology, you'll see coming up with counterexamples is not so hard :)

Ok later guys

that's why i love topology, really. you need the least amount of rigor to prove something.

byes.

later

@evinda Is there anything that you're unsure about in that?

I have the set $S=\{(x, y) \in \mathbb{R}^2 : \text{x or y is rational}\}$. I want to determine whether or not this set is connected. I can see that this is just a countable union of lines in $\mathbb{R}^2$. As the intersection of these is non-empty, and lines are closed, that means that S is connected, doesn't it?

Only the formulation @DanielFischer

and not compact as it isn't bounded?

@evinda You can say it that way. But, you wrote in the text that the while-loop is executed $m$ times, but your computation says it's executed $m+1$ times. Which of the two does the code agree with?

Oh, right... It is executed m+1 times, right? @DanielFischer

@user112495 It's not compact, since it's not bounded. And it's not closed, which is another thing showing that it's not compact. $S$ is indeed connected, but your argument for that is incorrect.

@evinda Yes.

@DanielFischer Am I supposed to use 'the continuous image of a connected space is connected'?

@user112495 I suppose the "and lines are closed" was a mistyping and you meant "lines are connected"? Then your main problem is that "the intersection of these is non-empty" isn't correct, since e.g. the two lines $\{x = 0\}$ and $\{x = 1\}$ have empty intersection. You need to qualify, which intersections aren't empty, and explain how that implies that the whole thing is connected.

@user112495: Might I suggest you think about paths?

hi @DanielF :)

Hi @Ted.

Quiet weekend in mod land?

@TedShifrin o/

@TedShifrin Any idea on this one ? :D

Salut @Hippa ... ça fait bien longtemps !

@TedShifrin Eh oui

@hippa you didn't come for weeks...

Salut @Ramanewb

@user112495 I think you should show the space is path-connected, which is stronger than you need.

hi @ted

Good morning @Ted

@bolbteppa OK

@Chris'ssis o/

@evinda I would have to guess whether they expect the use of pow or not. Your guess is as good as mine.

@Hippalectryon How are you doing? :-)

@Hippa: You're on the right track. If the $0$-eigenspace is $2$-dimensional, then in fact the solutions lie on a line. If the $0$-eigenspace is $1$-dimensional, your argument is correct. Because the matrix is real, if there are complex eigenvalues, they come in conjugate pairs, which corresponds to a real $2$-dimensional subspace. Best thing to do is write $P^{-1}AP=\Delta$, where $\Delta$ is either diagonal or close to diagonal.

morning, @teadawg

@Chris'ssis Fine, what about you ?

@Hippalectryon Great! Extremely creative these days. :-)

@Hippa: Exercise for you. If you have a real $2\times 2$ matrix with eigenvalues $\lambda = a\pm bi$, then there is a basis with respect to which the matrix becomes $\begin{bmatrix} a & -b \\ b & a\end{bmatrix}$.

@TedShifrin So if you start at the point (x0, y0) and want to get to (x1, y1) (suppose x0 and x1 are both rational, while y0 and y1 are both irrational) then you could travel up the line x=x0 to some other rational number r. Then travel along the line y=r to the point x1. And then travel along the line x=x1 until you reach the point (x1, y1). Could I just use this kind of argument for all the other cases to show it is path connected?

Why not? @user112495 You don't really have to write out four cases. Probably two should do it (and then a "similarly ...")

Can I ask a quick question?

@Hippalectryon Try this one math.stackexchange.com/questions/1164183/…

You just did, @Silenttiffy

@Silenttiffy Just ask; don't ask to ask

;) Is this true: $$ t_n \equiv j \mod k \implies t_{n+1}^{t_n} \equiv t_{n+1}^j \mod k$$ ?

NOOOOOO

At least not if $t_{n+1}$ is general.

$1\equiv 4\pmod 3$; now consider $2^1$ and $2^4$ mod $3$.

@Hippalectryon you only need a good start there.

@Chris'ssis i'll think about it :D

@Ramanewbie see the problem at the link above.

@Hippa: T'as lu ce que je t'ai écrit?

@TedShifrin Je suis en train d'essayer ton exercice :c

@Chris'ssis This one ? math.stackexchange.com/questions/1164183/…

@Ramanewbie Aha.

bien, ça va :)

@Chris'ssis I'm not doing integrals yet...

@Ramanewbie Aha ...

@Chris'ssis Is it supposed to be very easy ?

@Ramanewbie For me both are pretty easy. Maybe because I'm a bit experienced.

Right?

not for ME @Chris'ssis... lol

hm...Good point. But if $3^{3^3} \equiv 87 \mod 100$ then $3^87 \equiv 3^{3^{3^3}} \mod 100$ seems to be true

Wait, @Silenttiffy. So how is your sequence $t_n$ defined, precisely?

And do you have a specific $k$ in terms of it?

$t_0 = 3$ and $t_{n+1} = 3^{t_n}$

OK, and is $k$ a special number (you used $100$)?

@Ramanewbie ;)

$k \in \mathbb{N}$

@evinda If pow indeed computes the power using $j-1` multiplications, that's correct. If pow is implemented more efficiently, it's better than $O(m^2)$.

but im most interested in k = 10^n n \in N

Oh, then it's definitely wrong, @Silenttiffy, if you mean $k$ can be chosen generally.

The powers of $3$ definitely have a pattern once you fix a particular $k$. You need to see what the pattern is and then see how your sequence fits it.

Ok alright, thank you very much! Then I have to find another way to show that the last digits of $t_n$ converge to a fixed sequence for large n :)

ah ...

@TedShifrin that exercise is annoying >:c it seems simple

But it's not that simple

it actually is: Write down the complex eigenvector and look at real & imaginary parts of the equation $Av = \lambda v$.

@TedShifrin for the first one that gives me a system of 4 equations, 8 unknows (If I separate real and imaginary parts)

No, no, no ... Write $v=x+iy$ where $x,y\in\Bbb R^2$. Do not write out coordinates.

Oh ok

@TedShifrin Do I develop the product with $A$ ? (if so, I more or less come back to my previous system...)

But you should see where that matrix I gave you comes from. What is $Ax$? What is $Ay$?

I'm out for 1-2 hours. @Ramanewbie I won't be here to tell me the solution, but later on.

@TedShifrin I do see the $a-b,a+b$ appearing, but....

Or should I just try a new basis $x-y,x+y$ ?

ooooooh

Oh wait that doesn't exactly work

No, double-check your work.

@Chris'ssis See you soon then...

@Ramanewbie Hope you'll provide with a great solution in the meantime. :-)

@Chris'ssis I can try ...

@Ramanewbie This is not the right attitude. You need to be conviced you're going to be successful. ;)

@TedShifrin Basically all I have is that if we take the eigenvector $X_p=X_p'+iY_p'$ then $AX_p=(aX_p'-bY_p')+i(aY_p'+bX_p')$

@Chris'ssis "I will do it" ? ;)

Right, @Hippa. So $AX_p' = ?$

@Ramanewbie Yes!:D

@TedShifrin Well, the first part of course

(aX′p−bY′p)

well, then ... so, change my sign around and you're done.

Sorry to bother you again but 3^{n} \equiv 3^{n \mod k} (\mod k) is true, isn't it (where $n,k \in N$)?

@evinda I has complexity O(exp), but what does it do?

No, @Silenttiffy, I already gave you a counterexample to that.

@DanielFischer Doesn't it raise an integer to the power of another integer?

@evinda Have you tried it?

Yes, with base=2 and exp=3 and result has the value 8, at the end of the function. @DanielFischer

@evinda That's one of the few cases where it indeed raises base to the power exp. Do more tests.

@TedShifrin I think I'm missing something pretty stupid here but... why does $A\begin{bmatrix}X_1'\\Y_1'\end{bmatrix}=\begin{bmatrix}a&-b\\b&a\end{bmatrix}\beg‌​in{bmatrix}X_1'\\Y_1'\end{bmatrix}$ give us the answer ?

@Hippa: Review how you write a matrix for a linear map with respect to a given basis.

@DanielFischer I tried it with base=5 and exp=3 and ahain I got the right result..

@TedShifrin Oh you're right, thanks

@Hippa, ça se passe de temps en temps :P

@evinda Do more tests. It's not hard to write a small testing loop.