Mathematics - 2017-06-22 (page 3 of 7)

8:00 AM

Hi! If I tell you : "Environments: interface, interaction, homogeneity, break". What does that inspire you in mathematics? For a project, I have to find a subject in maths related to these words, but I did not find nothing.

Daminark

Yeah it's a lot of fun, I did the apprentice version last year, which had a 5 week class on linear algebra and a bit of graph theory/combinatorics. Paper was on Sylow

Category theory is one thing that comes to mind, I guess @MiKiDe

MiKiDe

8:13 AM

Hm... I'll look for it. It's a project alike what you call "REU". I hope it isn't too hard for a second year student in university, is it?

Thank you

Daminark

I have also just finished second year and the vibe I get with category theory is that it's a new way to think about things, which isn't necessarily hard to grasp but whose advantages over doing things normally aren't apparent unless you have more background

I haven't studied any of it myself, but the vibe I get is that it wants to not think of functions via how they're acting on elements so much as how they interact with each other

For example, you can think about an invertible linear transformation in terms of just being bijective and linear, which is more elements based

Alternatively, there's the fact that a linear map $T$ has the property that $rk(T\circ S) = rk(S)$ for any linear map $S$ if and only if it's an invertible linear map

The vibe I get is that stuff along the latter lines is how you want to be thinking in categories (please someone do correct me if I'm wrong, perhaps @Steamy?)

But it's not necessarily as insightful unless you see it in the right context, usually algebra and (mostly) algebraic topology/geometry

Liad

hi. we learned about the fundamental group $\Pi_1$ . and i know that $\Pi_1(S \ ^ 1 ) = \Bbb Z$ (iso' to).

Balarka Sen

@SohamChowdhury Good, good

Liad

i need to find $f : X \to Y$ continuous s.t $X,Y$ both path connected and $f_*$ is not the trivial homomorphism / the identity.

i thought taking $X = S \ ^ 1$ and $Y = S \ ^ 1 \times S \ ^ 2$ $f(x) =(-x,x)$

Daminark

Hey there @Liad!

Liad

8:22 AM

@Daminark hey ! how are you?

Daminark

Hmm, I'd actually be interested in talking about this

Balarka Sen

hi chat

Daminark

(Also I'm alright, you?)
(And hey @Balarka)

Liad

im fine :P

Daminark

But yeah so, I think I have an idea in mind about how to show that $\pi_1(S^1) = \mathbb{Z}$

Balarka Sen

8:23 AM

(I will keep my mouth shut)

Liad

i can use this fact

Daminark

So, do you know what homotopy equivalence is?

Liad

yea

MiKiDe

@Daminark And, why the words I told you make you think about Category theory?

Daminark

OK, so what I want to show is that if 2 spaces are homotopy equivalent, they share the same fundamental group

The idea being, I've proven once before that 2 curves in the punctured plane are homotopy equivalent if and only if they have the same winding number

Liad

8:25 AM

my question is : how can i "see" the $f_*$ above. it is defined as $f_*([g] ) = [f (g )] $ (composition)

Daminark

@MiKiDe interaction and I guess to a lesser extent, interface, do the most, because of how broad the subject is

Fargle

lol damn it @Dami, you're nuking flies again

Liad

i know that this homomorphis is like between $\Bbb Z $ and $\Bbb Z \times \Bbb Z$ but im not sure how it is defined

Daminark

In proving the fundamental group of the circle?

Balarka Sen

He's not trying to prove $\pi_1(S^1) \cong \Bbb Z$.

Liad

8:28 AM

no. i can use this fact . i defined a mapping $f(x) = (-x,x) $

i want to show that $f_*$ is not trivial nor the identity. but im not sure how.

maybe it is trivial / the identity but i have a feeling it is not :P

Daminark

@Balarka I was actually replying to @Fargle

Balarka Sen

Ah ok

@Liad Well, I don't see how your map is defined. $x$ is a point on $S^1$. How are you thinking of it as a point in $S^2$?

How does $x$ in the second coordinate of $(-x, x) \in S^1 \times S^2$ make sense?

Fargle

I would say so but that's because the second result (about winding number) doesn't strike me as any more obvious

Liad

it is a point in $S \ ^ 1 \times S \ ^ 1$

did i wrote $S \ ^ 2$ ? oops .

Balarka Sen

Ah, alright.

I'll let Dami answer.

Daminark

8:33 AM

@Fargle so, I think I know how to do the winding number stuff

So let $\omega_0 = \frac{-ydx + xdy}{x^2 + y^2}$

Crap my internet is being dumb

I had to switch to my phone briefly and cannot really type it out, gimme a bit

Fargle

iss ok

Liad

maybe there is a simpler $f:X \to Y$ s.t $X,Y$ are path connected and $f_*$ is not the identity/trivial ?

Balarka Sen

Yes.

There is a much simpler one. I am not sure if I should answer or @Daminark wants to.

Daminark

I haven't learned enough about this sort of thing in particular so I shall defer

Balarka Sen

@Liad Ok, so first, your example works. To see this, consider your map $f: S^1 \to S^1 \times S^1$ given by $f(x) = (x, -x)$. Let $p_1, p_2 : S^1 \times S^1 \to S^1$ be the projection maps to the two circle factors. Then $p_1 \circ f$ and $p_2 \circ f$ are the identity and the antipodal map respectively.

That means $(p_1 \circ f)_*$ and $(p_2 \circ f)_*$ are the identity $\Bbb Z \to \Bbb Z$ homomorphism and the multiplication by -1 homomorphism respectively.

Since $(p_i \circ f)_\star = (p_i)_\star \circ f_\star$, that implies $f_\star : \Bbb Z \to \Bbb Z \times \Bbb Z$ is actually the homomorphism $f(n) = (n, -n)$, which is not identity nor trivial.

SteamyRoot

8:44 AM

@Daminark Well, I've only had a basic intro in Category Theory and that was a few years ago by now. My feeling was that, if you're only interested in a specific mathematical structure, there's little point to studying it on the category level.
On the other hand, category theory was amazing in showing how (sometimes very different) structures can be abstracted and suddenly are very similar. I really liked the whole initial and terminal objects thing

Balarka Sen

@Liad In any case, it does not particularly make sense to say whether or not $f_*$ is identity if $\pi_1(X)$ and $\pi_1(Y)$ are not isomorphic.

For the simpler example(s): how many maps $f : S^1 \to S^1$ do you know, upto homotopy?

MiKiDe

@Daminark Finally, the category theory is to difficult for the time I have for my subject and the things I learnt. But thank you!

Liad

wait can we go a bit back? @BalarkaSen im not sure about your conclusion that $f(n) = (n,-n)$ this is what i wanted to say too, but what is your justification for that?

Balarka Sen

I just gave a justification...

Daminark

Okay I hope for the love of God that this works out

The internet in my apartment is garbage, I'm connecting to the hospital guest wifi nearby

Balarka Sen

8:49 AM

If the induced map is $f_* : \Bbb Z \to \Bbb Z \times \Bbb Z$, then composing with the projection homomorphisms to each of the $\Bbb Z$ factors you get (1) the identity map (2) the multiplication by -1 homomorphism.

Daminark

@Fargle

user8469759

hi guys

I have a math question (although related to a different field)

Fargle

@Daminark fer shur

Daminark

The first thing to note about the angle form I defined above is that its integral over the circle traversed $n$ times is that it has integral $2\pi n$

Thus, $\frac{1}{2\pi}\int_{nS^1} \omega_0 = n$, where I'm saying $nS^1$ to mean traversing it $n$ times

user8469759

say you have $p$ integer numbers of $k$ bit each, namely x_1,...,x_p and you sum this numbers together you will obtain a sum s and a carry c

s + 2^k c

what's a lower bound on the max value for s

?

Liad

8:52 AM

@BalarkaSen alright.

user8469759

I have a reference that says 2^k - 1

but I'd like to understand why

Daminark

Now, given any curve in the punctured plane $\gamma$, you can project it to $S^1$ via normalization

So now, you've homotoped it to some $e^{i\alpha(t)}$

By smoothness of the exponential, homotoping $\alpha(t)$ is a homotopy of the whole curve

MiKiDe

I'm asking again if someone else as an idea: If I tell you : "Environments: interface, interaction, homogeneity, break". What does that inspire you in mathematics (or in computer science maybe)?

Daminark

Then you homotope $\alpha(t)$ to be straight, this gives you that it's homotopic to some traversal of the circle

SteamyRoot

@user8469759 Well, $2^k - 1$ is the maximum value of a $k$ bit number, no?

Daminark

8:56 AM

So you know any curve can be homotoped to some $n$-times traversal of the circle. But since $\omega_0$ is closed, the only possibility for $n$ is $\frac{1}{2\pi}\int_{\gamma} \omega_0$

So thus, we know that the fundamental group of the punctured plane is $\mathbb{Z}$

(It's path connected, so this is true for any base point)

Now, the idea behind proving that homotopy equivalence preserves the fundamental group seems to fall out of using the thing Liad was talking about

I looked it up just now, the idea is that given a function $f$ from $X$ to $Y$, you can take loops based at $x_0$ and run them through $f$ to get loops based at $y_0$

SteamyRoot

@Daminark te.xel.io/img/posts/…

Daminark

This plays nicely with the group operation and everything, so in fact is an induced homomorphism between their based fundamental groups

(KEK @Steamy, that was beautiful)

But yeah, homotopic maps have the same induced homomorphism, which I think checks out because $f\circ \gamma$ is homotopic to $g\circ \gamma$ if $f$ is to $g$

So, if you have two homotopy equivalent spaces, say with maps $f$ and $g$, you know the induced homomorphisms must compose to the identity

This implies that those induced homomorphisms are isomorphisms

Perhaps you might have to worry about basepoints to make that jump, but at least if they're path connected this definitely checks out

So, if two path connected spaces are homotopy equivalent, they have isomorphic fundamental groups

Now, applying that here, we know that the punctured plane has a fundamental group of $\mathbb{Z}$

So, consider the normalization map and embedding of $S^1$ in $\mathbb{R}^2\setminus 0$

Those two maps compose to ones which are homotopy equivalent to the identity in each case. Hence, $\pi_1(S^1) =\mathbb{Z}$

Sounds good @Fargle?

Fargle

Oh I thought you were proving in the other direction, lol.

Daminark

Oh lmao

I don't actually know how to compute $\pi_1(S^1)$ directly yet

Just this roundabout way

Hmm, I think an analogue of this argument could likely be used for $\pi_n(S^n)$, yeah?

Like, perhaps $\pi_n(\mathbb{R}^{n+1}\setminus 0) = \mathbb{Z}$

Balarka Sen

9:12 AM

Not directly. You need a crucial theorem which says $H_n(S^n) = \pi_n(S^n)$ where $H_*$ is the homology group.

This comes from Hurewicz theorem, which is an entirely algerbo-topologic machinery

Once you know that, you can compute $H_n(\Bbb R^{n+1} - 0)$ from deRham stuff

The point is integration only captures winding numbers upto homology, not homotopy.

Daminark

Is this just for $S^n$ or more generally?

Oh wait a second there's an easier way to do this perhaps

Hopf degree theorem

Balarka Sen

If $H_i(X) = 0$ for all $0 < i < n$ then $H_n(X) = \pi_n(X)$ for any CW complex $X$.

Felix.C

What is the meaning of an equal sign with three lines? Is it like defined as?
Edit: Not in the context of modulo operations of course

Daminark

Hmm, well, hopefully we'll do that stuff soon in atop

Balarka Sen

Yes, this is a consequence of the Hopf degree theorem.

Daminark

9:17 AM

Lol I remember May was telling us after mentioning the weak homotopy equivalence thingy (which I think I now get after this shtick) that any decent atop class jumps into CW complexes, so hopefully he'll get into that soon enough

In the meantime I should probably go to bed because he's doing the extra atop talk at noon where he gets into homology/cohomology

Balarka Sen

Yeah good night

Daminark

(Which is exactly what I need to understand that stuff you said :P)

Good night!

Aditya Kumar

Hey, @Balarka. Textme when you are online

Balarka Sen

I am online.

Aditya Kumar

Hi, I saw your name in the ATM workshop on diff geo

I am a junior UG at IIT-K

Balarka Sen

9:29 AM

Ah yes I am supposed to go there

Are you going too?

Aditya Kumar

No, my semester will start

So I didn't apply

Balarka Sen

Ok, I see

Aditya Kumar

Actually I started using MathSE only a couple of days back

So, was pleasantly surprised so see a name I recognized in the chatroom :3

Balarka Sen

Got it. Nice to know you, welcome to MSE.

I don't use the main site frequently but I do roam the chat rather frequently.

Aditya Kumar

I know that this may not be a very polite question, but I was just wondering that how were you initiated into math?

Is someone in your family a math prof, or something like that

Balarka Sen

9:34 AM

That's a fair question. It's hard to answer that I guess; no, none of my family members studied mathematics extensively, but I guess I looked at a few Russian pop-math books.

Aditya Kumar

Oh those green ones <3

Anyway

Have you heard of the caratheodory conjecture?

Balarka Sen

I don't think so.

What does it say?

Aditya Kumar

It basically states that a closed and convex surface has atleast 2 umbilic points

It is open for about 80 years

Balarka Sen

I have forgotten what an umbillic point is. The two principal curvatures are equal?

Aditya Kumar

Showing the existence of one umbilic is trivial because of the poincare hopf index theorem

Yeah

I was reading about it a few minutes back and thought of going at it via the borsuk ulham theorem

As in we look at the continuous function $k_1 - k_2$on the surface

(since closed convex surface is homeomorphic to sphere so we can do all sorts of things)

Balarka Sen

9:41 AM

Let me try to understand it a bit better, I am sort of rusty at differential geometry of surfaces. So $k_1 = k_2$ means the shape map stretches both the principle directions by the same factor.

Why's it clear there is at least one umbilic point?

Aditya Kumar

Yeah, $k_1 = k_2$ means that both the principle curvatures are equal, that is to say that the surface is locally spherical.

That is a consequnce of the poincare hpf index theorem

*hopf

Balarka Sen

Ok, but I want to understand how Poincare-Hopf implies that.

Aditya Kumar

sorry, my mother had just called me

Yeah

Okay so assume that there is no umbilic

Balarka Sen

OK

Aditya Kumar

But the sphere has a non zero euler characteristic

Here 2

Balarka Sen

9:51 AM

I agree.

Aditya Kumar

So now, cosider the vector field of the principal curvature line

Now if you had no umbilic points it can't lead to a non-zero euler characterisitic

I am sorry I missed one point

The surfaces are assumed to be atleast twice differentiable

Balarka Sen

Surfaces usually mean smooth surface to me so that's fine.

Aditya Kumar

Okay

Balarka Sen

I am not sure what you mean by the vector field of the principal curvature line though. The principal directions are defined locally, how do you extend that to a vector field on all of the surface?

Aditya Kumar

But they are contious

*continous

Balarka Sen

9:57 AM

Hm, I guess. You are simply taking the eigenspace of the IInd fundamental form at each tangent space.

That does give a 1-dimensional vector field on the surface.

What next?

Luigi M

quick question, what is a representative of 0 in Ext^2(A,B)? where A,B are modules (say over R). The sum should be the Baer sum, but I found difficult to write down explicitly an exact sequence 0->B->X_1->X_2->A->0 representing 0

Aditya Kumar

@balarka

so isn't the proof obvious now

Balarka Sen

@AdityaKumar I don't see how so, though. The vector fields you constructed on the surface might be zero, say, if the surface is flat somewhere.

You want a nonzero vector field to contradict $\chi(S^2) = 2$ via Poincare-Hopf.

Aditya Kumar

But umbilic points can be a flat region, there are plenty of such cases

Just to set everythingstraight

The field that we have, it's zeros correspond to the umbilic points

BAYMAX

what is a great way to write latex online ?

like easy way

Aditya Kumar

10:01 AM

Overleaf

Best way

BAYMAX

oh nice!

Aditya Kumar

tremensous way

*tremendous

Best, the best, the very very best

*it

Even the president uses t

user8469759

@SteamyRoot yes, the max, but how come that's the min?

BAYMAX

oh

which president?

:)

Aditya Kumar

The greatest of them all

TheDonald

BAYMAX

10:02 AM

oh

Aditya Kumar

@bala

do you use hangouts

?

Balarka Sen

@AdityaKumar The field is just in the direction of the one of the eigenspaces of $\Bbb{II}$, right? Doesn't the zeroes simply correspond to when $k_1 = 0$? Or am I misunderstanding your vector field?

I use email but not as frequently as this chat. I'd rather talk here.

Aditya Kumar

um, oh I see your point now

No

I was incorrect. I goofed up somewhere

Balarka Sen

I think it's an interesting idea to use Poincare-Hopf, but I think some work is needed.

Aditya Kumar

First of all, it is a well known proof. I have read it bedore. Having problems in reconstructing it :3

(for one umbilic point)

Balarka Sen

10:07 AM

I understand, no problem, I have hard time recalling a lot of things I have learnt. I have personally never seen the proof, myself.

Aditya Kumar

Btw, do apply to the Analytic Number theory ATM at ISI

You know Ritabrata Munshi right?>

Balarka Sen

Oh, interesting, I didn't know about that ATM. I have met him.

Aditya Kumar

It is in december

Balarka Sen

I am not much of a number theory person, however, myself :P I wouldn't have gone to the geometry conference either if not for the peer pressure.

Aditya Kumar

Oh, when, I mean how did you meet him

Lol even I am not a number theory or algebraic geometry person

Balarka Sen

10:11 AM

Ah, so I was there in ISICal at a small talk on number theory & dynamics. We had a small chat over stuff, but that's it.

Aditya Kumar

Hmm

What are you into these days?

Balarka Sen

Majorly, preparing for various exams to get into a good university. But on the math side of things, mainly topology.

Foliations, in particular, I suppose.

Aditya Kumar

You are in 11th class right?

Balarka Sen

12th right now, but yes.

Aditya Kumar

Oh, you SHOULD apply to Princeton and MIT and Harvard

user8469759

10:14 AM

0

Q: Lower bound on the residual of a sum of $p$ numbers of $k$ bits each

Say we have $x_1,\ldots,x_p$ numbers represented in $k$ bits. The sum of such numbers will be generally something of the form $$ \sum_{j=1}^{p} x_j = s + 2^k c $$ There's a subtle point I can't understand from a reference I have, which states something like Since $s$ is the residual after s...

Aditya Kumar

Also don't miss out on UCB

Balarka Sen

I don't plan to go outside of the country any time soon.

Aditya Kumar

Oh then CMI is the place to be

However

Balarka Sen

yeah that's one of the options I have in mind.

Aditya Kumar

If you want to have a slightly broad education(which I in my finite wisdom will recommend)

Go to IISc

There is math dept has more breadth

Balarka Sen

10:15 AM

As in, broader into science than just mathematical sciences?

Aditya Kumar

Both

IISc has a broad math dept as well

FOr example

CMI is terrible in analysis

But at IISc you have people like THangavelu

Balarka Sen

Hah, I have heard of that.

Aditya Kumar

Also IISc will give you a better life

As in, CMI is a depressing place

IISc has great weather and is quite pleasant.

Balarka Sen

Indeed? Do you mean that in terms of workload and pressure?

Aditya Kumar

No

That will depend on what kind of courses you take

But IISc is better as a place

As in CMI is in the outskirts and is a very small campus

Balarka Sen

10:17 AM

I see. So you mean the atmosphere.

Aditya Kumar

Whereas IISc has a very big lush and green campus at the center f the city

And people there are generally happier

Balarka Sen

Technically, I am a well-recognized madman who prefers depression and solitude more, but I appreciate that suggestion. I might apply for IISc.

I think one has to take the KVPY's for applying there?

Aditya Kumar

Yeah

I am pretty sure you will do great at bth places

*both

It is just that I personally like to dive into multiple areas of math

Fianlly

One thing that you must not overlook is that TIFR-CAM is also in banglore

Now, I have know idea why people shun PDE's in India

But at TIFR_CAM there are some read bigshots like Adimurthy etc

You can also take courses there if you are at IISc.

Balarka Sen

TIFR is a nice place; I went to TIFR Bombay a year or so ago.

Aditya Kumar

For?

Balarka Sen

10:21 AM

Nothing much. I studied some algebraic geometry there.

Aditya Kumar

Ah, btw do you know about John Pardon
?

John Pardon

John Vincent Pardon (born June 1989) is an American mathematician who works on geometry and topology. He is primarily known for having solved Gromov's problem on distortion of knots, for which he was awarded the 2012 Morgan Prize. He is currently a full professor of mathematics at Princeton University. == Education and accomplishments == Pardon's father, William Pardon, is a mathematics professor at Duke University, and when Pardon was a high school student at the Durham Academy he also took classes at Duke. He was a three-time gold medalist at the International Olympiad in Informatics, in 2005...

Balarka Sen

Yep, I have heard of him.

Aditya Kumar

This dude

Balarka Sen

Something something Hilbert-Smith conjecture.

Aditya Kumar

Read the wikipedia page

Balarka Sen

10:23 AM

Some academic record he has.

Aditya Kumar

Yeah

And then there are lesser mortals like me :3

Balarka Sen

It's hard to be a god, man.

Aditya Kumar

Um, not like I want to be. Anyway, in math it is hard work and patience that matters more than anything else. Of course you'd know a thing or two about it

What all have you done in geometry/topology till now?

Balarka Sen

Absolutely, I don't think being a god is a good goal to have. I like what I spend my time doing, even if it's doing stupid things. At least I think.

Well, my geometry is weak, so I can't brag about it :P I like topology. I have studied a bit of algebraic topology.

Aditya Kumar

What else other than hatcher?

Balarka Sen

10:29 AM

I don't like any of the other classical books on algebraic topology, unfortunately. I have used Guillemin-Pollack for differential topology, on the other hand.

Aditya Kumar

Yeah, that is a really cute book. I read it this summer only.

Balarka Sen

I like it a lot.

Milnor's little book is good too, but there's not enough exercises.

Aditya Kumar

The one on vector bundles and characteristic classes?

Balarka Sen

Oh, no, that's Milnor-Stasheff

I meant intro to differentiable manifolds

or something, I forget the title

Aditya Kumar

Oh that. Um, yeah. It is not at all rigourous though

Balarka Sen

10:32 AM

I like the pedagogy. Usually rigor is secondary in topology; the pictures are primary.

Aditya Kumar

Anyway, I just saw this pretty nice theorem

Hadamard-Levy

Balarka Sen

Is that the global inverse function theorem?

Aditya Kumar

not aware of that, sorry

Let f:Rn→Rn of class C2. Then f is a C1-diffeomorphism on Rn→Rn if, and only if, f is a proper map and the determinant of the Jacobian matrix is non-null everywhere.

Balarka Sen

That's what I meant. It is a global version of the inverse function theorem.

classical inverse function theorem says $f$ is locally a C2 diffeomorphism at x if Df(x) $\neq$ 0, but with the assumption that it's proper and Df $\neq$ 0 everywhere you get it's globally a C1 diffeomorphism

Aditya Kumar

Ah, yeah

Balarka Sen

10:36 AM

It's a very, very nice theorem.

Aditya Kumar

Yeah, a diffeomorphism generator :D

Balarka Sen

This can be proved topologically, in fact. Note that because $Df(x) \neq 0$ for every $x$, it's a local diffeomorphism.

Proper local diffeomorphisms are covering maps

And the only covering map $f : \Bbb R^n \to \Bbb R^n$ is... a diffeomorphism

Aditya Kumar

Yeah, I know exactly what you mean

It was an exercise in Narsimhan's book

Raghavan Narsiman

Balarka Sen

I have heard of it, actually. There was a few times I looked into it and found some insightful comments.

Aditya Kumar

Narsimhan was one of the gods actually

He gave n invited talk at ICM at age 25

Balarka Sen

10:40 AM

damn

Aditya Kumar

you have 7-8 years to catch up xP

Balarka Sen

I don't think I want to catch up :P

(Not that I would be able to)

Fargle

Yeah lol, I only got 3. I'm already older than Galois was when he was 20. Feels bad.

Aditya Kumar

@fargle LOL :p

Balarka Sen

kill yourself if you don't get fields medal yo

Aditya Kumar

10:42 AM

Well, I count till 54 :p

Ever since Yitang Zhang got his result

Fargle

@BalarkaSen It took me 13 years of concerted effort to learn how to speak to girls. I'll be lucky if I can even read a whole paper by the time I'm 80.

3

Aditya Kumar

@Fargl

you were doing things the inefficient way

Balarka Sen

what a waste of those good 13 years

Aditya Kumar

You should have watched coupling first in 3 hours

and that is more than enough preparation

Fargle

@BalarkaSen yeah I mean I'm single now, clearly I didn't learn much

Balarka Sen

10:43 AM

+1

Aditya Kumar

speaking from experience and it delivered empirically verifiable reuslts

*results :p

anyway

gtg now

@balarka See ya

Fargle

Take care.

Balarka Sen

Talk later.

@Fargle I'm currently in the process of unlearning how to talk to human beings generically.

Aditya Kumar

Andrew Bird - "Imitosis"

►

Listen to this

Both of you

Fargle

@BalarkaSen Hopefully that goes better for you than it has for me.

Aditya Kumar

10:45 AM

(with the lyrics)

Fargle

I think I have some kind of empathy deficit that I've covered for with mirror-neuron-type behaviors.

Balarka Sen

@Fargle I'm climbing such divine heights of misanthropy as never before, of course I'll be fine

read Notes from The Underground, my literary bible

that book is the closest i have to a religion

Fargle

@BalarkaSen I've never been less of a misanthrope. I've really started to like people. It's great except when people don't feel the same back at you.

Hitchhiker's Guide to the Galaxy is that book for me.

(Well, the whole five-book trilogy)

Balarka Sen

@Fargle hahah. I'm happy for you; I guess that's a bit better for getting a functioning functional life.

I have heard of those books, but never read them.

Akiva Weinberger

@Daminark So this is how you do it:

First, let's show that the sum of opposite values in a 3x3 square must be zero. (By 3x3 I mean the top side has three dots, not three edges)

Well, there are four small squares in the 3x3, so take the alternating sum. That gives you (using your notation from earlier) $f(1)-f(3)+f(7)-f(9)=0$ when it all cancels out.

Balarka Sen

10:59 AM

@Fargle Do you watch movies?

Akiva Weinberger

Add the four corners of that square ($f(1)+f(3)+f(7)+f(9)=0$), and divide by two, getting $f(1)+f(9)=0$. Lemma proven.

Now, let's look at a 5x5 grid.

Call the four corners $f(1)$, $f(5)$, $f(21)$, and $f(25)$, and call the middle point $f(13)$

From the lemma, we know that $f(1)+f(13)=0$, $f(5)+f(13)=0$, $f(21)+f(13)=0)$, and $f(25)+f(13)=0$.

Adding them all up, and subtracting off the corners $f(1)+f(5)+f(21)+f(25)=0$, gives $4f(13)=0$, or $f(13)=0$. QED

…Incidentally, the following set of values works for a 6x6 area:

$$\begin{bmatrix}0&0&+1&-1&0&0\\0&0&-1&+1&0&0 \\-1&+1&0&0&-1&+1\\+1&-1&0&0&+1&-1\\ 0&0&+1&-1&0&0\\0&0&-1&+1&0&0 \end{bmatrix}$$

Alessandro Codenotti

hi chat

what's going on?

Secret

@LeakyNun I have no idea. While the bionomial coefficients have the symmetry $\begin{pmatrix}m \\ r \end{pmatrix} = \begin{pmatrix} m \\ m-r \end{pmatrix}$, the way the numbers combined (checking cases $(m,n)=(2,1),(3,2),(4,3),(4,2)$) always cancel out but seemed to have no dependence on where they are in the expansion. I cannot see any inherent pattern (if any)

Akiva Weinberger

However, up to constant multiples, that's the only way to fill out the 6x6.

@Semiclassical See the above proof

Applying an affine transformation to it gives us a proof for parallelograms as well.

Secret

In addition, the ratio between the r+1th and rth terms is $-\frac{(r+1)^{n-1}}{r^n(m-r-1)}$ which again does not show any useful pattern (and in face blows up for $r=0,r+1=1$)

Akiva Weinberger

11:16 AM

@AlessandroCodenotti See the array right above your comment?

If you take any four points that form a square in it, and add the values there, you get $0$

(Only ones with sides parallel to the axes)

Also, that solution is unique up to constant multiples.

Alessandro Codenotti

oh, interesting

how do you know it's unique?

Akiva Weinberger

The explanation I wrote above (which shows that it must be zero everywhere if it's on a a plane) can be used to show that as well @AlessandroCodenotti

But you could consider it to be a puzzle if you want

BAYMAX

skull patrol changed his name!

:)

Just win baby

:)

Akiva Weinberger

11:32 AM

Hi scald control

BAYMAX

scald?

hot liquid patrol

;)

Just win baby

Hi pals...

BAYMAX

was that a slap reverse :)

s.harp

If I have a cofibration $A\overset i\to X$, how is the cofiber usually defined?

let $A$ and $X$ be pointed spaces i guess

is it the mapping cylinder?

(google is super unhelpful)

Secret

11:57 AM

Ok I seemed to be getting nowhere:

[E,L]=0

Pick a basis. Under this basis, then $L=\sum_{i=1}^nl_{ij}$ and $E=\sum_{i=1}^ns_{ij}$. Let a polynomial $f(x)=\sum_{k=1}^na_kx^k$. Now if we compute, the following, the coefficients becomes:

$LEf=\sum_{i,j,k=1}^nl_{ij}s_{jk}a_k$

$ELf=\sum_{i,j,k=1}^ns_{ij}l_{jk}a_k$

Since it is given $[E,L]=0$, and $f$ is arbitrary, we get the relation $s_{ij}l_{jk}=l_{ij}s_{jk}$

Now suppose there's a linear map $C$ such that it maps $E$ to $E^{\frac{1}{2}}$. Then under this basis we have:

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$Mathematics$ Mathematics