00:00 - 15:0015:00 - 23:00

12:28 AM
@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.

1:19 AM
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

2:09 AM
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

2:29 AM
@KonformistLiberal can you please tell me how to use math jax in mobile

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

@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

1 hour later…
3:34 AM
@robjohn hi sir

@Yuvraj yes?

@robjohn this one sir

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

3:51 AM
it is a even function
@robjohn

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

it should be even

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

odd
@robjohn

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

4:06 AM
got it sir @robjohn

so?

it should be odd

3 hours later…
7:18 AM
0

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 $$\operatorname{Gal}(K^{\text{... rofl I want to space two images within a line. Can somebody suggest something ?? What is the context, @Knight? In a picture? In a word document? In an arts and crafts project? In a question on MSE? Hello Calvin, How are you? so-so, sleepy, busy I'm writing an answer and I need two blue squares and then a proper space and then three blue squares How can I do it? 7:25 AM$$ {\color{blue}\blacksquare} {\color{blue}\blacksquare} \qquad {\color{blue}\blacksquare} {\color{blue}\blacksquare} {\color{blue}\blacksquare}?$$Yes Yes, can you please make them a little bigger?$$ {\Huge \color{blue}{\blacksquare\blacksquare \qquad \blacksquare \blacksquare \blacksquare}} ?$$Wow! Thats more than best horizontal spacing can be added with \, \ \quad \qquad and if you need finer control \hspace{x em} where x can be negative also and have idk 3 decimal places. also \! removes one \ space Also \phantom{} adds spacing equal to whats in the {}, so for instance if you want the empty space to fit one of the cubes exactly, you can do$${\Huge \color{blue}{\blacksquare\blacksquare \phantom{\blacksquare} \blacksquare \blacksquare \blacksquare}} 

Oh

7:35 AM
@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?

1 hour later…
9:00 AM
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 $a,b\in\mathbb{N}$.

That follows from CGB?

no
I don't think that has anything to do with CGB

Why is it true

do I look like I know research-level mathematics
I can't tell you man

why is that research level math
seems like a cool exercise unless its very hard

9:04 AM
it's very hard

ref?

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 $2$

that doesnt tell me anything tho
what kind of tools are used?

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 $34511$

Just give me a reference instead of random numbers man
lol

9:06 AM
idk, I'm just here for the fun facts
I don't know the specific reference, these are joint results from Kreck and Zagier

did Kreck tell you this stuff

yes

weird guy

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

Ah OK
That makes sense
The signature theorem is a ramped up CGB that's what I had in mind

9:11 AM
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

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

sounds like a hard question

yeah

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?

9:29 AM
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.

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.

@Thorgott How do you formulate Poincare duality

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?

Yes, we want it to "halt" in real life
But, I care more about the theory

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

9:33 AM
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?

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

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

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

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

Yeah I agree that theoretical cs is a branch of math

9:39 AM
I don't think any of us know the theory well enough to respond to your question properly.

the Turing machines I know are finite objects

Truth

Yeah neither do I :D

I might take theoretical cs next semester

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

9:41 AM
@Balarka old Dimmu Borgir is cool af

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

@Thorgott yeah, when they do so :D

Turing-recognizability is by definition a countable phenomenon but what do I know.
You only call subsets of $\Bbb N$ to be Turing-recognizable

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

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

9:43 AM
Anyway we don't know this stuff I am sure. if anyone does it's one of @Alessandro, @LeakyNun or @AkivaWeinberger

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

@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?

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

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

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

9:45 AM
Because there are $2^{\omega_0}$ Dedekind cuts, not $\omega_0$

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

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

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

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

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

9:46 AM
@Drathora as in, going from $\omega_0$ to $2 \omega_0$ doesn't change much when it comes to describing $\mathbb{R}$

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

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

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

@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 $2^{\omega_0}$

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

9:49 AM
@Drathora well yes, if you look at it as a step by step process
@Drathora hmm, I see your point

@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

idek what a PD space is

"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

@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.

In even dimension 2k there's only signature restriction which vanishes when k is odd
So 4k is the only troubling dimension

9:58 AM
@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.

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

can I understand curvature without understanding Riemannian geometry?

Just learn some altopo so we can read this shit man

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

Seems classical

10:00 AM
@Drathora I'm actually intrigued to see that...
@Drathora the single number is fine I can extrapolate the partitioning, somewhat

@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)

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

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

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

@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 $|\mathbb{N}|$

10:02 AM
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

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

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

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"

@Drathora yeah precisely

10:06 AM
@Thorgott Can you tell me how you define the connection 1-form?

@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

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 $i,j=1,...,n$.

Ah ok great.

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

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

10:09 AM
Nah locally is good enough

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

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

@LeakyNun but presumably, why not?

what exactly is the question?
I haven't followed the entirety of the conversation

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

10:12 AM
@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.

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 $1 + \omega = \omega$

@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?

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

@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.

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 $\omega^i_q(e_p) - \omega^i_p(e_q) = \theta^i([e_p, e_q])$

10:21 AM
Computability is tied to enumberability for the reason I said before. Computation happens in steps that are isomorphic to the natural numbers.

Wow dude so much CS happening

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

@Thorgott come to garbology

Sorry haha

I'm sorry honestly

10:22 AM
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

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

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 $n$

woo, seminar on elliptic curves was just announced

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)

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

10:35 AM
@AlessandroCodenotti Yeah that's what I suggested earlier

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

Surely some nutcases do ordinal length Turing machines
some weird computation happens at the limit ordinals

I guess there are all sorts of generalizations

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

10:42 AM
Of course he works in automatic proof verifications

Hales does to, but he is an honest geometer pfff

i learnt yesterday that Mike Freedman does quantum computing now
topological quantum computing
in microsoft research
lmao

I have a feeling that pays better than geometry

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

life goals

10:47 AM
Grisha reacts only

@BalarkaSen or run for mayor in Paris

Oh yeah that

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

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

Who's that?

10:51 AM
lol I was about to reference that one

Daniel Biss

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

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

@Thorgott lmfao really
i know he made some stupid error i didnt know what it was
just 1 error man
scary stuff

yeah, I think that was it

10:52 AM
LMFAO

Kreck likes telling that story

I love Kreck

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

Kreck is awesome

Which is hilarious to me, because the main person who's bringing all this money is this man:
Most bizarre lecture series I've ever attended

10:55 AM
@Drathora Hello

hey ^^

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

@Drathora what the hell is that outfit

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

10:56 AM
lol

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

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

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?

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

lmao that's wild

10:58 AM
fucking nut

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

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

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

no but this man
this man...

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

10:59 AM
scares me
My god
$(\infty, 1)$-masculinity
No soy for this guy

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

I understand completely

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

oh my

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?

11:14 AM
"My very unfortunate duty is.."
spoken like a true assassin

lmfao yeah

1 hour later…
12:26 PM
Why the unit tangent bundle is a sphere bundle (when for example the underlying mainfold is not locally flat) ?

May be of interest to some people:
13

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 ...

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

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

12:37 PM
Just proved $[E_\sigma : E_\sigma \cap K^{ur}] = [E_\sigma : E_\sigma \cap K^{ur}]$
the dream

I think I have a very short proof of that result

go on
rofl

Well they are equal, therefore they are equal. QED

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

@Lelouch Um uh, what is local flatness?

12:40 PM
Like flatness, but locally

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

"moves hands in a plane, but only locally"
@EdwardEvans lmao

alessandro on drugs
what are you smoking man i want some of that

General topology

Nevermind I don't want it

12:42 PM
I've been thinking about the pseudoarc the whole afternoon

You can have it

It'll do that to you

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

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

I have done 0 work today

12:44 PM
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"

Yeah measure theory wasn't introduced until the 2nd year of my undergrad, and wasn't mandatory
As a formal concept at least

@Alessandro same

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

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

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

12:46 PM
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

His last paper was percolation on hyperbolic groups

Ok now that sounds pretty cool

Lmao

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

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

12:49 PM
any help is appreciated

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

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

It's alright man the analysis is actually bearable

@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.

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

1:00 PM
What is a locally flat manifold

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?

shit, I meant it's flat metric
@BalarkaSen
somehow in my head "locally isometric to Euclidean metric" became "locally flat"

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.

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

@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

1:02 PM
@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

@Thorgott perfect, thanks

@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?

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...

tensor commutes with direct sum (naturally)

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

1:07 PM
So in general $B\otimes_A (C\oplus D)\cong (B\otimes_A C)\oplus (B\otimes_A D)$?

yes
even for infinite sums

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

@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

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

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

1:17 PM
yeah

Makes sense, thanks

@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

2:14 PM
is any one online
need help understanding a solution

2:41 PM
@BalarkaSen karlin.mff.cuni.cz/~pyrih/e/e2001v1/c/ect/node93.html look at the beautiful pictures at the end of this page

tf man

They're just chains, but very wiggly

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