Mathematics - 2020-07-16 (page 1 of 2)

Balarka Sen

00:28

@user10478 I wrote down a list of important distributions, their pdf/pmf, mean, variance and moment generating functions in my first course in probability

You should not just memorize this list you should be able to derive these things by yourself quickly. For example, compute variance of the hypergeometric distribution; it's honestly a good exercise and you'll end up noticing something (hypergeometric is a sum of weakly dependent bernoulli distributions)

its important to understand some continuous distributions are scaled limits of discrete ones, so you can derive stuff about the latter from stuff about the former (eg characteristic function)

it should be a healthy mix of examples and theory. too much of the former will make you an insane statistician, too much of the latter will make you an insane measure theorist. there's a middle ground, called being a probabilist

just play with lots of examples. sit down and see if you can prove the central limit theorem for the bernoulli distribution one day.

Konformist Liberal

01:19

Hi! Do you think $\langle f_i(A)\rangle^{n-1}_{i=0}$ is the correct way to write the group that's generated by the images of $f_i$?
I mean the group that's generated with all of them together

ronak jain

02:09

Please tell me how to use math jax in mobile

I am having problems in understanding maths problems and answers just because of symbols used in chat

ronak jain

02:29

@KonformistLiberal can you please tell me how to use math jax in mobile

Konformist Liberal

math.ucla.edu/~robjohn/math/mathjax.html here at the end of the page there's some explanation

ronak jain

@KonformistLiberal can you suggest me a PDF containing question for the topic "Function" . So that I can strengthen this. I am having problems in this while preparing for jee

Yuvraj

03:34

@robjohn hi sir

i need your help

robjohn

@Yuvraj yes?

Yuvraj

@robjohn this one sir

robjohn

@Yuvraj is $\frac{\sin(x)}{x}$ an even function, an odd function, or neither?

Yuvraj

03:51

it is a even function

@robjohn

robjohn

@Yuvraj so what would be the coefficient of $x$ in an even function?

Yuvraj

it should be even

robjohn

@Yuvraj is $x$ an even function, an odd function, or neither?

Yuvraj

odd

@robjohn

robjohn

@Yuvraj so what would be the coefficient of $x$ in an even function?

Yuvraj

04:06

got it sir @robjohn

robjohn

so?

Yuvraj

it should be odd

Edward Evans

07:18

0

A: Equality of fields being deduced from isomorphism of Galois groups

Two proofs (Following TokenToucan's comment): Conceptual proof: We have $E_\sigma \subset K^{\text{ur}}E_\sigma \subset K^{\text{ab}}.$ By Galois theory we have $$K^{\text{ur}}E_\sigma = (K^{\text{ab}})^{\operatorname{Gal}(K^{\text{ab}}/K^{\text{ur}}E_\sigma)}$$ and $$\operatorname{Gal}(K^{\text{...

rofl

Knight

I want to space two images within a line.

Can somebody suggest something

??

Calvin Khor

What is the context, @Knight? In a picture? In a word document? In an arts and crafts project? In a question on MSE?

Knight

Hello Calvin, How are you?

Calvin Khor

so-so, sleepy, busy

Knight

I'm writing an answer and I need two blue squares and then a proper space and then three blue squares

How can I do it?

Calvin Khor

07:25

$$ {\color{blue}\blacksquare} {\color{blue}\blacksquare} \qquad {\color{blue}\blacksquare} {\color{blue}\blacksquare} {\color{blue}\blacksquare}?$$

Knight

Yes

Yes, can you please make them a little bigger?

Calvin Khor

$$ {\Huge \color{blue}{\blacksquare\blacksquare \qquad \blacksquare \blacksquare \blacksquare}} ?$$

Knight

Wow! Thats more than best

Calvin Khor

horizontal spacing can be added with \, \ \quad \qquad and if you need finer control \hspace{x em} where x can be negative also and have idk 3 decimal places. also \! removes one \ space

Also \phantom{} adds spacing equal to whats in the {}, so for instance if you want the empty space to fit one of the cubes exactly, you can do $${\Huge \color{blue}{\blacksquare\blacksquare \phantom{\blacksquare} \blacksquare \blacksquare \blacksquare}} $$

Knight

Oh

Taufi

07:35

@Ted

Shifrin: I see the connection now to the binomial theorem, but what to do about the missing binomial coefficients that come up in the expansion?

Thorgott

09:00

Topology fact of the day: For each $k\in\mathbb{N}$, a $4k$-dimensional connected, orientable, compact smooth manifold with Betti numbers $b_0=b_{4k}=1$, $b_1=...=b_{2k-1}=b_{2k+1}=...=b_{4k-1}=0$ and $b_{2k}$ odd exists iff $k=1$ or $k=2^a+2^b$ for some $a,b\in\mathbb{N}$.

Balarka Sen

That follows from CGB?

Thorgott

no

I don't think that has anything to do with CGB

Balarka Sen

Why is it true

Thorgott

do I look like I know research-level mathematics

I can't tell you man

Balarka Sen

why is that research level math

seems like a cool exercise unless its very hard

Thorgott

09:04

it's very hard

Balarka Sen

ref?

Thorgott

the reason this is true is because the existence of such a manifold with specified $b_{2k}$ is equivalent to the solubility of a certain equation involving Bernoulli numbers and that equation can be solved mod $2$

Balarka Sen

that doesnt tell me anything tho

what kind of tools are used?

Thorgott

for all multiples of $4$ smaller than $2^{25}$, $b_{2k}=5$ is realizable, but for dimension $2^{25}$, the smallest odd $b_{2k}$ is $34511$

Balarka Sen

Just give me a reference instead of random numbers man

lol

Thorgott

09:06

idk, I'm just here for the fun facts

I don't know the specific reference, these are joint results from Kreck and Zagier

Balarka Sen

did Kreck tell you this stuff

Thorgott

yes

Balarka Sen

weird guy

Thorgott

he decided to make the last lecture about which Betti numbers are realizable

it has sth to do with the Hirzebruch signature theorem and Pontryagin classes, I think

Balarka Sen

Ah OK

That makes sense

The signature theorem is a ramped up CGB that's what I had in mind

Thorgott

09:11

orientable and of dimension 4k is the hardest case for some reason

for non-orientable manifolds, the only obstruction is that the Euler characteristic has to vanish in odd dimensions

Balarka Sen

I think there's some general theory here. If you have a bunch of candidate Betti numbers for a potential orientable manifold it has to satisfy Poincare duality

Then you make a complex with those Betti numbers

That's a rational PD space; then you surger to modify it to be a manifold

So here's the question to be precise. If $X$ is a, say, an integral PD space, i.e., has cohomological dimension $n$ and $H_k(X; \Bbb Z) \cong H^{n-k}(X; \Bbb Z)$, when is $X$ a manifold?

Can you modify $X$ in some canonical way without changing homology so that it's a manifold?

I don't know how to do this but I think this has good answers

Thorgott

sounds like a hard question

Balarka Sen

yeah

Ask Kreck man

Threnody

I'm sorry to invade - I have a somewhat subjective question so I don't think it's a fit for the main site

Why do we necessarily tie the concept of counting with $|\mathbb{N}|$ instead of ordinal numbers?

I'm a beginner in this idea of ordinals but they seem so much more expressive, so why limit counting to the naturals?

Thorgott

09:29

Conceptually we don't, that's precisely why ordinals exist. Terminologically, it's probably for historical reasons, since counting with natural numbers preceded the conceptualization thereof.

Threnody

The reason I ask is because we also do this limitation in computer science when it comes to what is computable and what isn't.

Balarka Sen

@Thorgott How do you formulate Poincare duality

Thorgott

I don't know any CompSci, but I assume when you actually try to get results from a computer, you want it to finish doing it's computation in finite time/a finite amount of steps, no?

Threnody

Yes, we want it to "halt" in real life

But, I care more about the theory

Thorgott

@Balarka $H^k(M)\cong(H_c^{n-k}(M))^{\ast}$

Threnody

09:33

https://cs.stackexchange.com/questions/128356/whats-wrong-with-this-proof-that-mathbbr-is-enumerable

For example, I asked this ^. The proof, I think, works if we accept counting beyond $\omega_0$. So my issue is whether or not a theoretical computer can actually go through $\mathbb{R}$ - since by the axiom of choice we can well-order any set hence we can ask what's next - am I wrong?

Balarka Sen

Ah, sucks, so the isomorphism comes from the pairing $(\omega, \eta) \mapsto \int_{[M]} \omega \wedge \eta$

Hm

Thorgott

ye

I don't know what a "theoretical computer" can or cannot do.

Threnody

A Turing machine, essentially. Although this might be more cs than math, in the sense that I don't think mathematicians cover Turing stuff?

Edward Evans

$\text{CS} \subset \text{Math}$

Threnody

Idk, I definitely would ask my professors about this if I could actually go to them :D

Yeah I agree that theoretical cs is a branch of math

Balarka Sen

09:39

I don't think any of us know the theory well enough to respond to your question properly.

Thorgott

the Turing machines I know are finite objects

Edward Evans

Truth

Threnody

Yeah neither do I :D

Edward Evans

I might take theoretical cs next semester

Threnody

Turing machines, are in theory described as having infinite memory

The crux is that the literature takes that to mean "countably infinite"

Although I'm just an undergrad, I have no authority on what people think things mean...

It's just how it was always taught to me

ofc in reality computers have finite memory, but they're, in theory, as capable as finite memory turing machines

Edward Evans

09:41

@Balarka old Dimmu Borgir is cool af

Thorgott

yeah, what I mean is that they have finitely many states, use finite alphabets and halt in finitely many steps when they do so

Threnody

@Thorgott yeah, when they do so :D

Balarka Sen

Turing-recognizability is by definition a countable phenomenon but what do I know.

You only call subsets of $\Bbb N$ to be Turing-recognizable

Drathora

Do you understand what's wrong with the proof you wrote now Threnody? The one that you linked

Threnody

yes, my question is what's up with that limitation

Balarka Sen

09:43

Anyway we don't know this stuff I am sure. if anyone does it's one of @Alessandro, @LeakyNun or @AkivaWeinberger

Thorgott

I think the point is that a Turing machine's reader does things step by step and only moves left or right on the tape each step, so it can't go beyond anything countably infinite

Threnody

@Drathora I really think I do but I'm not "convinced" in a sense

@Thorgott perhaps a better abstraction is a type-0 grammar

@Thorgott they're equivalent, but i think grammars are more ...digestible?

Thorgott

while I do have a limited understanding of what a Turing machine is, I don't know what that is anymore, I'm afraid

Drathora

So your issue is that if you can't "enumerate" over the entirety of $S_\infty$ the way you've done

Thorgott

my point is that you can't go beyond anything countable by just repeatedly doing $+1$

Drathora

09:45

Because there are $2^{\omega_0}$ Dedekind cuts, not $\omega_0$

Threnody

@Thorgott but isn't that what ordinal numbers do?

Drathora

So you're trying to do a countable enumeration of something that isn't countable

Balarka Sen

No, the point is ordinal numbers need not have immediate predecessors

Threnody

@Drathora Yes i understand that, but that's why I said "keep going"

Balarka Sen

The Turing machine only turns left and right, and that's all it does. You have both immediate successor and predecessor

Threnody

09:46

@Drathora as in, going from $\omega_0$ to $2 \omega_0$ doesn't change much when it comes to describing $\mathbb{R}$

Thorgott

no, ordinal numbers go further, but not by just doing $+1$

Drathora

Going from $\omega_0$ to $2\omega_0$ doesn't change anything, that's correct

But going from $\omega_0$ to $2^{\omega_0}$ changes a whole bunch

Thorgott

but I'll leave this in the hands of the actual computer scientist here

Threnody

@Drathora yes - that's what I'm trying to convey

@Drathora and assumingly, you could go beyond $\mathbb{R}$ if you were to accept that somehow you can "keep going" even further than $2^{\omega_0}$

Drathora

You can keep going indefinitely with this process and never even leave the interval $[0,1]$

Threnody

09:49

@Drathora well yes, if you look at it as a step by step process

@Drathora hmm, I see your point

Balarka Sen

@Thorgott Anyway I wanted to say a PD space is slightly more general than how I described it because it specified what map $H^k(X) \to H_{n-k}(X)$ should be an isomorphism, but maybe this is a little tricky for me to explain with your formulation of PD

Because it's a cohomology-to-cohomology duality; you can do it but I'm too lazy to spell it out

Anyway if $X$ is a simply connected PD space of cohomological dimension at least $5$ and odd, it's homotopy equivalent to a manifold.

By a very explicit homotopy equivalence

Thorgott

idek what a PD space is

Balarka Sen

"Satisfies Poincare duality"

It makes sense to say a PD space is a space $X$ such that $H^k(X) \cong H_{n-k}(X)$. Just take that as a definition; it's a little more than that but this is ok for now

Drathora

@Threnody one final remark about this. A way to think about this is "is there an algorithm I could theoretically write that will "eventually" cover everything I'm going to enumerate. When enumerating the rationals there are well-known ways of doing this via diagonalization. If you think about how you'd enumerate S_i though, no algorithm you can ever give me wouldn't get "stuck" in a state where there's something it's clearly never going to reach.

Balarka Sen

In even dimension 2k there's only signature restriction which vanishes when k is odd

So 4k is the only troubling dimension

Threnody

09:58

@Drathora Hmm, well of course such an algorithm would never halt even for just $\mathbb{N}$ but that's "ok" since I'm staying theoretical. So the problem then is the word enumerating.

Drathora

It's not that it would never halt. It's that I can give you an actual example of a Dedekind cut that your algorithm would never reach

Thorgott

can I understand curvature without understanding Riemannian geometry?

Balarka Sen

Just learn some altopo so we can read this shit man

Drathora

Whereas in the natural or rational case, there is no such example for some algorithms

Balarka Sen

Seems classical

Threnody

10:00

@Drathora I'm actually intrigued to see that...

@Drathora the single number is fine I can extrapolate the partitioning, somewhat

Balarka Sen

@Thorgott Yeah for sure. Curvature to you is just the curvature 2-form, right? Do you define a Levi-Civita connection or use connection 1-forms?

I can tell you the geometry

(I think)

Thorgott

yeah, the latter, and I would love to hear it

Drathora

Well, the problem is that I can give you $2^{\omega_0}$ many Dedekind cuts that you'll never reach with any algorithm you give me :P

Thorgott

I wanna amp my understanding rate of CGB to like 30% at least

Threnody

@Drathora I do know of the uncomputable numbers - but they're so called "uncomputable" because of the drawn line of enumerability being "no more than $|\mathbb{N}|$

Balarka Sen

10:02

Ok let me think how to phrase this, I am not super versed with the moving frames language. I'll do this in exchange of you explaining CGB to me btw

Threnody

@Drathora When I asked about this on the CS site I was led to "hypercomputation" and subsequently told that it is... a "crank-adjacent" subject :D

Thorgott

I can tell you about it once I fill my 5 or 6 large understanding gaps

Threnody

@Drathora perhaps because some people think it's physically realisable... but again i'm fine with it not being so

Drathora

Yeah I'm unfamiliar with the topic. Although the quote I found did give me a good laugh

"if non-computable inputs are permitted, then non-computable outputs are attainable"

Threnody

@Drathora yeah precisely

Balarka Sen

10:06

@Thorgott Can you tell me how you define the connection 1-form?

Threnody

@Drathora so my dilemma is why do we draw the line to what is computable and what is not if it's just a matter of ... "just stop at the cardinality of N it's enough"

@Drathora if anything i'm trying to find where the rigor is, if it is a matter of rigor and not just a convention/definition

Thorgott

Take a frame $X_1,...,X_n$ and pass to the dual frame $\theta^1,...,\theta^n$. The connection forms are the unique forms $\omega_j^i$, $i,j=1,...,n$ satisfying $\omega_j^i=-\omega_i^j$ and $d\theta^i=\sum_{k=1}^n\theta^k\wedge\omega_k^i$ for all $i,j=1,...,n$.

Balarka Sen

Ah ok great.

Thorgott

apparently the "better" framework is to construct these on the frame bundle instead, but we didn't prove that and only got a reference in the 5th volume of Spivak for it that I haven't looked up yet

Leaky Nun

@Threnody would your issue be resolved if instead of saying "infinite memory" we say "arbitrarily large memory"?

Balarka Sen

10:09

Nah locally is good enough

Threnody

@LeakyNun eh, ... if we allow such memory... I'm not too well versed in ZFC to make statements like these to be honest :D

Thorgott

yeah, it's probably better to understand it in terms of frames first before passing to the frame bundle

Threnody

@LeakyNun but presumably, why not?

Leaky Nun

what exactly is the question?

I haven't followed the entirety of the conversation

Drathora

To me, it's that computations flow in discrete steps. We can label the steps of a computation using the natural numbers. To compute something of cardinality higher than that of $\mathbb{N}$ would give us an injection from this "something" to the naturals

Threnody

10:12

@Drathora effectively the ordinals, from what I understand?

@LeakyNun I can phrase the question as:

@LeakyNun "Why do we limit the definition of computable to the idea of enumeration up to $|\mathbb{N}|$ when ordinals are much more expressive"

@LeakyNun strictly staying in theory, not reality

@LeakyNun Don't get me wrong I'm absolutely fine if it's a definition or just a convention we chose in the past that stuck - What I'm worried about is if I'm lacking rigor if I talk about this.

Drathora

So is what you have in mind to describe a program that has to enumerate the rationals and then do one final step using the ordinal $\omega + 1$?

Whereas a program that does one step and then enumerates the rationals is described by $1 + \omega = \omega$

Balarka Sen

@Thorgott I'll call the moving frame $e_1, \dots, e_n$, and everything else will be notationally same. The connection equation is (matrix form) $d\theta = \theta \wedge \omega$.

What's $d\theta^i(e_p, e_q)$ from this? Because $\theta^i(e_j) = \delta_{ij}$, $d\theta^i(e_p, e_q) = (\theta^p \wedge \omega^i_p + \theta^q \wedge \omega^i_q)(e_p, e_q)$

Which is what? $\omega^i_p(e_q) - \omega^i_q(e_p)$, yeah?

Drathora

The point here is that as soon as we go beyond $\omega$, even to the successor $\omega + 1$, nothing is computable as I can pinpoint a step that will never be reached

Threnody

@Drathora Well yes, I understand that but the definition of what is computable is tied to enumerability which is tied to the natural numbers not the ordinals.

Balarka Sen

On the other hand, remember for a $1$-form $\omega$, $d\omega(X, Y) = X \omega(Y) - Y \omega(X) - \omega([X, Y])$, so $d\theta^i(e_p, e_q) = - \theta^i([e_p, e_q])$. So we get $\omega^i_q(e_p) - \omega^i_p(e_q) = \theta^i([e_p, e_q])$

Drathora

10:21

Computability is tied to enumberability for the reason I said before. Computation happens in steps that are isomorphic to the natural numbers.

Balarka Sen

Wow dude so much CS happening

Threnody

@Drathora Yep. So it is after all just a matter of definition?

Balarka Sen

@Thorgott come to garbology

Drathora

Sorry haha

Threnody

I'm sorry honestly

Drathora

10:22

Well, sort of. Of course you could say it in terms of ordinals

But as soon as you exceed the ordinal $\omega$, everything is non-computable

So it's the exact same theory, just with a relabelling, as far as I'm concerned

Not accounting for any of this strange hypercomputation

Threnody

Mhm. I think I understand

Thank you, sorry for the CS spam but I think you guys would've been more likely to know what ordinals are :D

Drathora

This all kind of ties back to the explanation I used for ordinals using queues the other day

If I give you an infinite length queue, and you walk down it, if I give you a person $n$, you'll eventually reach them, no matter which $n$ it is

On the other hand, if I give you an $\omega + 1$ queue, then you'll never reach that $\omega + 1$th person

If you think of walking past position $n$ as being computation step $n$

Edward Evans

woo, seminar on elliptic curves was just announced

Alessandro Codenotti

Turing machines have a tape which is $\Bbb N$ because you want to be able to move both ways

Longer well orders have elements with no immediate predecessor

(I don't really know much about them, but this seems like a reasonable guess to me)

Thorgott

yeah, I think this is essentially the "computation happens in discrete steps" idea that Drathora brought up

Balarka Sen

10:35

@AlessandroCodenotti Yeah that's what I suggested earlier

Alessandro Codenotti

Ah I didn't see that, I only looked at the messages around the one where you pinged me :P

Balarka Sen

Surely some nutcases do ordinal length Turing machines

some weird computation happens at the limit ordinals

Alessandro Codenotti

I guess there are all sorts of generalizations

Balarka Sen

math.uni-bonn.de/people/koepke/Talks/Ordinal_computability.pdf

said nutcase from your uni

Alessandro Codenotti

Oh I know him

He's retired now officially, but he mostly works on automated proof verification with Hales nowadays

Balarka Sen

10:42

Of course he works in automatic proof verifications

NERD ALERT

Alessandro Codenotti

Hales does to, but he is an honest geometer pfff

Balarka Sen

i learnt yesterday that Mike Freedman does quantum computing now

topological quantum computing

in microsoft research

lmao

Alessandro Codenotti

I have a feeling that pays better than geometry

Balarka Sen

he has millions of those multi-authored papers in arxiv in quantum CS

it pays better than bing topology at least

standard career tricks once you get a fields medal

leave math and troll applied people or live on your moms pension

Thorgott

life goals

Balarka Sen

10:47

Grisha reacts only

Alessandro Codenotti

@BalarkaSen or run for mayor in Paris

Balarka Sen

Oh yeah that

Alessandro Codenotti

Or you can go the Grothendieck way and retire in a cave in the mountains or something like that

Balarka Sen

you can also almost get the Fields medal and chill as a Senator in the US only to learn your results are all wrong

@AlessandroCodenotti you mean becoming a pickup artist

Alessandro Codenotti

Who's that?

Thorgott

10:51

lol I was about to reference that one

Balarka Sen

Daniel Biss

Alessandro Codenotti

@BalarkaSen lol I have a classmate who calls Grothendieck "the Rasputin of math"

Thorgott

forgot to check simply connected for Seifert-van-Kampen, I think

Balarka Sen

@Thorgott lmfao really

i know he made some stupid error i didnt know what it was

just 1 error man

scary stuff

Thorgott

yeah, I think that was it

Balarka Sen

10:52

LMFAO

Thorgott

Kreck likes telling that story

Balarka Sen

I love Kreck

Drathora

Can confirm that quantum CS does pay well

Our quantum department is raking in money, I'm pretty sure the US Navy or something is involved somehow

Thorgott

Kreck is awesome

Drathora

Which is hilarious to me, because the main person who's bringing all this money is this man:

Bob Coecke: "A survey of categorical quantum mechanics"

►

Most bizarre lecture series I've ever attended

Knight

10:55

@Drathora Hello

Drathora

hey ^^

Knight

How was your meeting?

Drathora

Good, my undergraduate student is ready to begin implementing my results, while I sit back and never program again

Balarka Sen

@Drathora what the hell is that outfit

Drathora

I know lmao. He's a peculiar guy, although quite brilliant I guess

Knight

10:56

lol

Balarka Sen

i'd be running in the opposite direction if a man in that outfit started writing category theory on the board

Drathora

The course he teaches is really strange. It's quantum mechanics via box/string diagram, all via some symmetric monoidal category or something

Knight

I have written a long answer to this question, but I'm still not sure what OP wanted. Can you please have a look at it?

Drathora

I forget the details, it's one of those things I learned and then never thought about again

Thorgott

lmao that's wild

Balarka Sen

10:58

fucking nut

Drathora

But my experience is that it teaches you all of this quantum stuff, but in a language that nobody from any other institution knows, since they use traditional notation

Thorgott

@Balarka is there any man you wouldn't run from when he starts writing category theory on the board except Gromov

Drathora

So you know a lot of the same things as them, but you can't actually communicate with them

Balarka Sen

no but this man

this man...

Drathora

But it does make a lot of the results in quantum computing etc fall out very easily

Balarka Sen

10:59

scares me

My god

$(\infty, 1)$-masculinity

No soy for this guy

Drathora

Now imagine the startled 4th year Computer Science students at Oxford when he walks into the room, with a cap on that looks like it's been made out of 10 other recycled caps glued together, a string vest and about 5 chains hanging from his belt

Quite a far cry from what they imagined when they applied I bet haha

Balarka Sen

I understand completely

Thorgott

@Balarka I misremembered the SvK thing, it seems. Check this: arxiv.org/pdf/0709.1291.pdf

Balarka Sen

oh my

Drathora

It's hard to assess your answer @Knight, since I don't really know exactly what the question-asker needs. I wonder if they wanted things like distributivity, associativity and commutativity to also be expressed

It's not really made clear, so I think the person asking the question needs to give a better picture of what's required

Not sure if I'd be mortified or proud if something like that got released about my work

On one hand, damn, all that time was for nothing

On the other... at least someone read it?

Thorgott

11:14

"My very unfortunate duty is.."
spoken like a true assassin

Balarka Sen

lmfao yeah

Lelouch

12:26

Why the unit tangent bundle is a sphere bundle (when for example the underlying mainfold is not locally flat) ?

Nike Dattani

May be of interest to some people:

13

Q: What are the types of DFT?

Similar to: What are the different types of charge analysis?, What are the types of bond orders?, and What are some recent developments in density functional theory?, I would like to ask: What are the different variations/flavors of DFT (density functional theory)? I ask users to stick to one of ...

Lelouch

The only condition I've trouble verifying is the local trivialization property. Don't we need local flatness for that ?

Nike Dattani

Very easy answers, just pick one "type of DFT" in the list and explain it in an answer.

Edward Evans

12:37

Just proved $[E_\sigma : E_\sigma \cap K^{ur}] = [E_\sigma : E_\sigma \cap K^{ur}]$

the dream

Alessandro Codenotti

I think I have a very short proof of that result

Edward Evans

go on

rofl

Alessandro Codenotti

Well they are equal, therefore they are equal. QED

Edward Evans

jesus

Corollary: $n = n$ for all $n n n$

Balarka Sen

@Lelouch Um uh, what is local flatness?

Alessandro Codenotti

12:40

Like flatness, but locally

Edward Evans

Alessandro throwing out the proofs like a farmer throwing out cow shit today

Alessandro Codenotti

"moves hands in a plane, but only locally"

@EdwardEvans lmao

Balarka Sen

alessandro on drugs

what are you smoking man i want some of that

Alessandro Codenotti

General topology

Balarka Sen

Nevermind I don't want it

Alessandro Codenotti

12:42

I've been thinking about the pseudoarc the whole afternoon

Balarka Sen

You can have it

Alessandro Codenotti

It'll do that to you

Edward Evans

"Love to mate, love to, but this is mine and I want it all.. so.." - Super Hans

Alessandro Codenotti

I'm procrastinating because I should read the proof of Borel determinacy, but I skimmed it in the book and three pages into the proof he starts splitting cases into subcases

Balarka Sen

I have done 0 work today

I'm supposed to read probability

Alessandro Codenotti

12:44

Why are you doing so much probability? How crazy are the probability courses at your uni?

My undergrad probability course was like "this is a gaussian and this is a Poisson and don't ever dare mention measure theory"

Drathora

Yeah measure theory wasn't introduced until the 2nd year of my undergrad, and wasn't mandatory

As a formal concept at least

Edward Evans

@Alessandro same

Alessandro Codenotti

We had both measure theory and the probability course I was talking about in the 2nd year as mandatory courses, but then we also had a non mandatory 3rd year course on measure theoretic probability

Edward Evans

Lukas hadn't graduated from his bachelor's at Heidelberg because he couldn't bring himself to take intro probability hahahaha

Balarka Sen

@AlessandroCodenotti Mahan is thinking a lot about percolations so he's forcing me to

Drathora

12:46

I forget what the first year probability course entailed. I remember being overwhelmed with notation because I'd never really written formal mathematics before my undergraduate

Balarka Sen

His last paper was percolation on hyperbolic groups

Alessandro Codenotti

Ok now that sounds pretty cool

Balarka Sen

Lmao

Charlie Shuffler

Hi all, I have a question about the error term coming up when approximating an integral with the trapezoidal rule

the rule states that we can make the approximation: $\int^b_a f(x) dx \approx (b-a)*(f(a)+f(b))/2$

is there a way to bound the error term that comes up here with some sort of big O term?

like $O(b^3)$ or whatever

Balarka Sen

We don't have too much probability either despite being Indian Statistical Institute

Charlie Shuffler

12:49

any help is appreciated

Balarka Sen

Last year (so next for me) has a Markov chains course

or well, stochastic processes

I like prob its intuitive; I need to get a hang of these inequalities though

Alessandro Codenotti

You're the first person ever to tell me probability is intuitive

Balarka Sen

It's alright man the analysis is actually bearable

Thorgott

@CharlieShuffler You can bound it explicitly in terms of the second derivative of $f$ (this should make sense intuitively if you draw a picture). I don't see why you want to bound this in terms of $b$. The way both the integral and approximation will change when $b$ changes will depend drastically on how $f$ changes therewith and you usually want to keep $a,b$ fixed either way.

Lelouch

@BalarkaSen I meant I thought I need to assume that it's a locally flat manifold

Balarka Sen

13:00

What is a locally flat manifold

Alessandro Codenotti

Sanity check, for an $A$-module $B$, to say $B\otimes_A A\cong B$ it's enough to note that $b\otimes a=ba\otimes 1$, right?

Lelouch

shit, I meant it's flat metric

@BalarkaSen

somehow in my head "locally isometric to Euclidean metric" became "locally flat"

Balarka Sen

Ah I see. No, the unit tangent bundle is a bundle regardless, although I see your point that locally isometric to R^n makes the proof immediate.

Lelouch

Hmm @BalarkaSen so how do you show the trivialization condition of the bundle ?

Thorgott

@Alessandro yeah, the iso is $b\otimes a\mapsto ab$ in one direction and $b\mapsto b\otimes1$ in the other direction; inverseness one way is obvious and inverseness the other way is what you're noting

Charlie Shuffler

13:02

@Thorgott I am indeed familiar with such a bounding using the second derivative of $f$, but I don't see how that holds up when using only 2 points in my trapezoidal rule application

Alessandro Codenotti

@Thorgott perfect, thanks

Balarka Sen

@Lelouch You just need to show the fiberwise unit bundle of $U \times \Bbb R^n$ with a pointwise smoothly varying metric $g_p$ on the fibers $\{p\} \times \Bbb R^n$ is diffeomorphic to $U \times S^{n-1}$, yeah?

Alessandro Codenotti

And why is $B\otimes_A A^n\cong (B\otimes_A A)^n$ true? I don't understand anything about modules or their tensor products...

Thorgott

tensor commutes with direct sum (naturally)

Balarka Sen

@AlessandroCodenotti u can write down the same argument u were writing to see it's $B^n$

just componentwise

Alessandro Codenotti

13:07

So in general $B\otimes_A (C\oplus D)\cong (B\otimes_A C)\oplus (B\otimes_A D)$?

Thorgott

yes

even for infinite sums

Alessandro Codenotti

With the iso being $b\otimes (c,d)\mapsto (b\otimes c,b\otimes d)$ then I suppose

Thorgott

@CharlieShuffler there's a corresponding error bound that works for any amount of nodes; it's on Wikipedia for example

@Alessandro yeah, and the inverse comes from the inclusions componentswise

Charlie Shuffler

Yeah I've seen it, but I actually just realized I'm trying to apply the trapezoidal rule to an improper integral

So i've got new problems to worry about

Alessandro Codenotti

@Thorgott So if I write it out explicitely it looks like $(b_1\otimes c,b_2\otimes d)\mapsto b_1\otimes(c,0)+b_2\otimes(0,d)$

Thorgott

13:17

yeah

Alessandro Codenotti

Makes sense, thanks

Lelouch

@BalarkaSen Yes, exactly. The main difficulty is that if $(u, v)$ goes to $(u, \Psi_u(v))$, then defining $\Psi_u$ in a smoothly varying way. This is probably outright trivial (being the first exercise in Lee), but I don't see how ?

Doesn't the fact $(U, g)$ being isometric to $(O, \cdot_{\mathbb{R}^n})$ imply that the manifold must be flat ?

Shit, nevermind. It's $T^1(U)$ should be diffeomorphic to $U \times S^{n-1}$, not isometric.

So we can probably do it like this: Consider the map $h(p, T) = TG_PT^{-1}$, where $g_p(v, w) = v^T G_P w$. Now we can do some bullshit on the preimage on identity

beta_me me_beta

14:14

is any one online

need help understanding a solution

Alessandro Codenotti

14:41

@BalarkaSen karlin.mff.cuni.cz/~pyrih/e/e2001v1/c/ect/node93.html look at the beautiful pictures at the end of this page

Edward Evans

tf man

Alessandro Codenotti

They're just chains, but very wiggly

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$Mathematics$ Mathematics