Mathematics - 2021-01-31 (page 1 of 3)

BigSocks

00:00

and is it uninteresting? if we want the Chinese Remainder theorem to work, we probably need some stuff to be relatively prime, right?

Fargle

@BigSocks This is actually what motivates a more general definition of coprime ideals: two ideals $I,J$ in a ring $R$ are said to be coprime if $I + J = R$. (This may need to be in a commutative ring, I don't remember.)

Just as a fun little side fact.

BigSocks

@Fargle oh that's right, I think Lorenzini actually talks about this. It's been a while since I was there wow

Ted Shifrin

Yes, and this will lead to a more general CRT.

BigSocks

well if you don't like $a$ and $b$ being relatively prime, maybe we look at the factors of $n$ or something

Ted Shifrin

We were working with very specific numbers a minute ago.

They were not relatively prime. Nor was 9 relatively prime to 120.

BigSocks

00:04

$120 = 2^3 *3*5$

Ted Shifrin

Almost.

Yes, that's how I made up the example. :)

Balarka's half hour was a bit long.

Balarka Sen

@ShaVuklia Stronger: Nigel Short also famously thinks this. He's vice president of FIDE!

@TedShifrin Let's say I prepared for more than half an hour so that the talk could be half an hour instead ;)

Ted Shifrin

Talks always need more time than we think they will. :)

Balarka Sen

Yeah.

BigSocks

@BalarkaSen good luck with your talk

Balarka Sen

00:06

Also probability sucks, there's too much happening

Fargle

@BalarkaSen We will all have the last laugh, though---Nigel Short (the tall one) may be a GM, but it was Mikhail Tal (the short one) who was a world champion.

BigSocks

Ok so actually $(a) \cap (b) = (ab)$ right?

Balarka Sen

Lol

That's a good joke

Ted Shifrin

No, @BigSocks, not without hypotheses.

BigSocks

hmm ok they have to be special...

Balarka Sen

00:09

Gotta sleep, night all

Astyx

'night

Fargle

'Night!

Thorgott

night

Ted Shifrin

Night, a @Balarka.

BigSocks

cya

@TedShifrin I looked it up in Lorenzini- they have to be coprime like what @Fargle was saying

Ted Shifrin

00:11

@Thor: I just found a simpler ring reading a post on main. $K[x,y]/(x-xy^2)$.

@BigSocks: $(a)\cap (b) = (\text{lcm}(a,b))$.

BigSocks

so $(a) + (b) = (1) \Rightarrow (a) \cap (b) = (ab)$

Astyx

What's the non trivial algebra fact that is discussed?

Fargle

@Astyx In quotients of a PID, two elements that generate the same ideal are associate.

Thorgott

yeah, I considered such a thing too, but this isn't really universal in a practical manner anymore

Ted Shifrin

We're trying to prove that in $\Bbb Z/n$ two elements generate the same ideal iff they are associates.

BigSocks

00:13

right and if $gcd(a,b) = 1$, $lcm(a,b) = ab$

Ted Shifrin

@Astyx: I think it's true in any quotient of any PID, in fact.

Astyx

In Z/nZ specifically or in general?

Fargle

This discussion was specifically about $\Bbb Z$ mod $n$, sorry.

BigSocks

that is our toy example but it would be great to prove it there atm

Ted Shifrin

I'm lost.

BigSocks

00:14

what, why

Ted Shifrin

There's a completely different approach from the one I've been taking that seems to be much more straightforward. But I'm going to go do my neck/back/hip exercises.

BigSocks

hmmm ok

Mike Miller

@BalarkaSen I imagine he doesn't like that new Netflix show

BigSocks

So $(8) \cap (3) \cap (5) = (120)$ fwiw

Mike Miller

About a cent and a half, I'd estimate

Less, given how cheap computation is these days, actually.

BigSocks

00:18

depends on the computer- for a quantum computer to factor $120$ that's probably a lot of $

Mike Miller

Sure, but you're not a quantum computer.

AFAIK.

BigSocks

all according to plan

they suspect nothing

Sha Vuklia

@BalarkaSen My argument was based on age, and Short is only 2 years younger. Anyhow, so far I'm only a fan of Finegold and Nakamura, and I think they are solid, so I'm fine x)

BigSocks

I would not be surprised if I was a quantum computer. I sometimes get things right, inexplicably, but very often it's all wrong or, at best, bears a very close semblance of something you'd expect

Astyx

Naroditsky and Rosen are the best ones

Thorgott

00:22

good taste

Sha Vuklia

@Astyx Haven't heard of them yet (honestly, I only started paying attention to chess culture, if I can call it that, pretty recently)

NoName

How do some people $\LaTeX$ so fast?

Fargle

For me, years of experience. I was the dweeb who was doing freshman math homework in LaTeX just to show off or something, I think.

Sha Vuklia

But Finegold and Nakamura are defo bigger memers than Naroditsky and Rosen from what I can tell given a 5 sec impression of their YT channels, so I'm still good.

Fargle

(Not that I'm above that now, but I'm in public a lot less.)

Astyx

00:27

"Oh no, my queen"

NoName

I'm sure the tutors marking the homework appreciated it.

Fargle

@ShaVuklia Finegold is where I first learned of the Evans Gambit. Named, of course, for Mr. Gambit.

Mike Miller

@NoName It depends what you mean. Most TeX is very uncomplicated --- stick the stuff you were already writing in dollar signs. The more complicated things are things you don't really do when liveTeXing.

Fargle

Yeah I always need to re-Google, for example, how to do tikz commutative diagrams. Never can remember exactly how it goes.

Also heya Mike.

Mike Miller

Hi

Oh yeah forget commutative diagrams

I can now do that live in AMScd but only with practice

There are now websites you can just draw the diagrams in though

Thorgott

00:30

Daniel "I know I'm taking a while for each game, but that's the point of a speedrun" Naroditsky

Astyx

I may be saying complete garbage, but isn't asking that A has no non-associate elements that give the same ideal the same as asking that the Picard group of Spec A is trivial?

Fargle

There was that one site that let you draw it that I knew of, but it always breaks for me.

Thorgott

@Astyx algebraist spotted

Astyx

I'm trying my best

NoName

@MikeMiller Do you think using a local editor makes a difference? I use Overleaf now.

Mike Miller

00:32

I use Overleaf for everything because I am used to it and I am a creature of habit.

BigSocks

@Astyx apparently the same as asking $Pic(A)$ be trivial, if the above is true?

Mike Miller

Oh, I guess you're recompiling and fixing as you go?

I just fix errors later.

Astyx

I don't know what Pic A is, but that's probably the same object

BigSocks

isomorphic yea

Astyx

(if it isn't I'm stopping maths for ever)

Mike Miller

00:32

@Fargle It's pretty new, the good one

BigSocks

(you are safe)

Mike Miller

q.uiver.app

Fargle

@NoName I use TeXstudio personally, but I don't feel that my workflow is any better or worse there than it was on Overleaf for years.

Sha Vuklia

https://www.youtube.com/watch?v=M0MWQ8nWyF0
> Nakamura (16 sec)

NoName

Yeah, I'm gonna try the fix errors later approach

Fargle

00:33

@MikeMiller Bless you, kind soul.

Astyx

OH NO MY QUEEN...OH NO MY QUEEN

►

Mike Miller

Thank github user varkor, not me

Astyx

It's a colleague of my brother apparently

I was using it to show my brother some stuff

And he recognized it and said he'd tell his collegue he would be happy to know some people actually use it to do serious maths

Small world

Mike Miller

@NoName In my experience it's about 4:1 time spent typing to correcting code. I never tried to correct code while TeXing live, but I can definitely imagine it slowing me down pretty drastically.

I only ever took one class where I simultaneously wanted to liveTeX and do commutative diagrams --- in that one I did the diagrams on paper and wrote in the document {DIAGRAM} to add in later. When I want to draw a picture, again, I do it on paper and write {PICTURE} in the document.

That way I can ctrl+f.

Sha Vuklia

@Astyx That's a mild meme

Astyx

00:40

Oh no my meme

Thorgott

god, this algebra fact is way too bizarre

BigSocks

pretty nuts huh

cranks the imposter syndrome up

Astyx

I think the C(R,R) examples gives the intuition

Fargle

The moral, of course, is that algebra's fake.

BigSocks

@Astyx continuous functions from $R$ to itself?

Astyx

00:47

Yes, the one Thorgott gave some time ago

BigSocks

oh I must have not been paying attention. what was it about then?

Astyx

It's not about Z/nZ

But the idea is that an invertible function in that space has constant sign

So you can look at the subspace of functions that cancel at 0

And $x^2$ and $x|x|$ give the same ideal, but are not associate

Wait no I oversimplified it

BigSocks

there needs to be some "quotient of PID" aspect I think. might be there and I'm not seeing it

Astyx

That wouldn't work because you need some room to switch signs

Thorgott

it's super weird, because the property apparently is true for all PIRs, not just PIDs

because PIRs are products of quotients of PIDs

Astyx

00:52

R = rings?

Thorgott

yes

BigSocks

weird...

Fargle

@Thorgott That's...deeply disconcerting. That would mean I could prove a ring isn't a PIR by finding two elements that are non-associate but generate the same principal ideal, I think, unless I'm being a doof.

Thorgott

yeah, it would mean that

Fargle

Check out this one weird trick; constructivists hate it!

NoName

00:57

@MikeMiller Yeah, I think that's what's been slowing me down. Should have realised it tbh because I've a decent typing speed otherwise. Thanks.

Mike Miller

Sure, let me know if it works well.

Ted Shifrin

@NoName Part of the art is having a plentiful list of macros ... and knowing them well enough that you recall what they are :P

@Fargle @Thor: I wouldn't blame you if no one wants to associate with me any longer.

Fargle

01:21

@TedShifrin groan

robjohn

@TedShifrin I'd never remember what they were, so I just buckle down and type.

zacts

01:45

salvēte

Ted Shifrin

@robjohn If you’re writing books with tons of matrices (with differing alignments), systems of equations, even partial derivatives .

Fargle

@zacts Salve.

user2103480

02:33

@BalarkaSen that is a very good question damn

I think this should work

BigSocks

03:42

hey @TedShifrin so how did we ever reconcile the statement we wanted to prove with $9$ not being a unit?

(in $\Bbb Z/120$?)

I just had some food, took a little break. Might sleep soon, but it's good to keep some math in mind for the subconscious to chew on

Ted Shifrin

Could change to something equivalent to 9 mod 20 that will beca unit mod 120.

I’m thinking the right proof is to factor 120 into prime powers and use CRT.

BigSocks

yeah I looked at it and I learned about the version for rings. I guess you work with a map $\Bbb Z \rightarrow \Pi_{p \vert n}(\Bbb Z / p^k \Bbb Z)$ with kernel the product (also the intersection) of all the $(p^k)$

Ted Shifrin

04:01

If you have relatively prime factors, there is no kernel. So you get an isomorphism to the product.

Mike Miller

Note the domain, @TedShifrin.

BigSocks

isn't the kernel the product of all these $(p^k)$? so it's actually $(n)$

Right

so if I was saying $\Bbb Z/(n) \to \Pi_{d \vert n} (\Bbb Z/ p^k \Bbb Z)$

that's an iso

123

Hello World..

copper.hat

print("hello world!")

Ted Shifrin

Yeah, you’re right. I didn't have typesetting on.

Sorry.

BigSocks

04:16

oh ok no worries

well I had a lot of fun thinking about this today- maybe tomorrow we'll crack it :) see ya @TedShifrin

Ted Shifrin

Night, Big

user 85795

04:33

cya

Amin Idelhaj

04:48

Hey!

Ted Shifrin

Howdy, Demonark.

Amin Idelhaj

How's everything going?

Ted Shifrin

Nothing exciting to report.

Fargle

Heya @Amin

Amin Idelhaj

Fargle what's up? And yeah on my end the semester has started which is nice

Ted Shifrin

04:59

What zoom classes are you taking?

Fargle

Just vibing, mostly. School's light so far, little to do.

Ted Shifrin

I thought I was asking Demonark :)

Amin Idelhaj

I'm doing automorphic forms, Lie theory, and "Topics in Algebraic Topology", which is doing intersection homology and perverse sheaves

I think he responded to my "Fargle what's up?"

Ted Shifrin

Oh, intersection homology will be interesting for you.

Oh, he responded to you. This is too confuzling.

Fargle

@TedShifrin Yeah, I was talking to him, jeez. >:c

Ted Shifrin

05:00

I'll go back to watching tennis. Don't mind me.

Fargle

:)

user 85795

who's playing?

Amin Idelhaj

Yeah I'm hyped for it, first week has been a motivational one. Mentioned the Kahler package for complex projective manifolds, why it fails for singular varieties, and hinted at how intersection homology recovers things

Ted Shifrin

It tries to replicate Poincaré duality on singular spaces.

user 85795

@copper.hat hello world!

copper.hat

05:14

@user85795 hi there :-)

user 85795

how's life, my friend

copper.hat

covid times :-)

user 85795

yeah, they're going a bit overboard with the regionalized names of the different variants

copper.hat

part of some international blame game

user 85795

05:30

realistically, in the long run if the most populated countries in the world don't get it under control we're all finished

user 85795

05:41

but, that's just my $0.02 worth

copper.hat

i don't think so. it is certainly more deadly than flu, but most people are fine.

i think the economic & educational deficit will have a far deeper impact on the population.

i am not saying it is ok for some people to die.

a brother got the pfizer vaccine but still tested positive.

user 85795

wow

copper.hat

unfortunately i think the only realistic way of managing it is testing & tracking.

and people resist tracking.

user 85795

google is trying their best

they tracked a lot of the rioters at the Capitol

copper.hat

05:56

you mean from a covid perspective or just location tracking?

user 85795

location

copper.hat

i suspect the rioters would not be amenable to covid tracking.

i think it will end up being like a more deadly flu. people will learn to accept the new reality, those at risk will get vaccinated.

people didn't care so much when it was the flu even though it was still fairly deadly.

user 85795

but the possible mutations are completely unpredictable

copper.hat

06:19

yes, but to some extent that was true of flu as well.

Balarka Sen

06:31

@AminIdelhaj Let's read IH together

@ShaVuklia Yeah that's fair. But I imagine it doesn't help if establishment promotes bigotry. The Twitch crowd is excellent, yeah.

They know it's just a game.

@MikeMiller Yeah lol all of this came up in response to that show

I haven't seen it though and I probably won't, too far from my interests. I heard it's good.

Fargle

It's pretty alright

robjohn

06:54

@TedShifrin Ah, I see, that would help. I use a lot of copy-paste for things like that, but I guess macros might be better.

123

In isoceles triangle ABC, where AB = AC, and D, E, F are midpoints of sides AB, BC, AC respectively.

How to proof Midsegment DE and median BF intersection is the midpoint of BF.

Pls help me to figure this. I don't understand by triangle theorems.

Semiclassical

among a physicist's greatest nightmares: trying to find a missing factor of 2

(another is trying to find a minus sign error)

user 85795

the sign error is universal

copper.hat

once had to completely redo someone's paper because they had used an inequality in the wrong direction. at least i became a co author then.

Semiclassical

of course, sometimes that sign "mistake" turns out to be correct. the discovery of asymptotic freedom in QCD is an instance of that

in the case of the answer i wrote above, it's particularly irritating because mathematica includes the factor of 2 in some ways of writing it

but if i try to simplify said calculation in what i think is the right way, then I lose the factor of 2

123

07:06

1drv.ms/u/s!AozWlUoG8z4tigjfSUb3eXV66nSl

Pls see the link diagram..

zacts

I'm working on this problem: mathb.in/49583.

123

@copper.hat pls see

user 85795

@Semiclassical have you tried comparing the different ways you're writing it?

Semiclassical

i obtained the latter by simplifying in what i think are valid ways, in context

123

You can share any hint or theorem which help to understand/solve this.

user 85795

07:08

without simplifying

Semiclassical

hmm

user 85795

try to look at it through Mathematica's eyes :-)

Semiclassical

lol

hmm, that may have helped at least in terms of pinning down where the disagreement arises

the irritating bit of this is that it deals with the function f(x)=x/|x| which in general isn't well-defined at x=0

but within the context of how it arises (from a certain integral representation) I -thought- it was evident that one had to treat f(0)=0

copper.hat

@123 perhaps i am missing something, but $BCF$ and $BEG$ are similar and $|BC|=2|BE|$ so $|FB|=2|BG|$.

Semiclassical

specifically, it's the representation $\displaystyle |p|^{-1}=\int_0^\infty \frac{\sin(p \lambda)}{p \lambda}\,d\lambda$ for real $p$

then $\displaystyle p/|p|=\int_0^\infty \frac{\sin (p\lambda)}{\lambda}d\lambda$

copper.hat

07:17

it bugs me a bit when people ask a question on mse, you help them and then they delete the question.

6

Semiclassical

that's gross

user 85795

yeah, flag it

Semiclassical

lately i've had two instances where i stared at a problem for a very long time and while i was trying to put together an answer it got deleted

i mean, it's not like I had an answer ready

but yeah, if you've actually answered it then flag the question

@copper.hat i'm not sure what i find more annoying, tho

copper.hat

yeah, not a real biggie, just bugs me :-.

Semiclassical

people who ask a question and then disappear when a response is given, or people who pepper with responses that make clear they have no idea what they're doing

one of those "is it better to appear the fool or open one's mouth to remove all doubt" problems

123

07:25

@copper.hat thanks. I think this way can I prove the conclusion what I need. Let me check

Semiclassical

hmm. i think that my above reasoning may run into an issue of exchanging limits

if you compute $\int_0^\infty \frac{\sin(p\lambda}{\lambda}d\lambda$ for $p>0$, and then consider $p\to 0^+$, then it goes to $1$.

but if you take $p\to 0$ first, then the integral is trivially zero

that's...vexatious

Lelouch

08:02

If $X$ is a topological vector space, and $N$ is a closed subspace of $X$, and if $N^*$ is a complementary subspace of $N$, then is $X/N \simeq N^*$, where the former is equipped with the quotient topology and the latter with the subspace topology?

This shouldn't be true (since it otherwise renders two theorems in Functional Analysis, Rudin to be trivial) but I'm unable to cook up some counterexample

maths student

08:20

We have a coin with probability
$$
p
$$
of Head, box1 with 6 white and 4 black balls and box2 with 4 white and 6 black balls.
Toss the coin. If you get a $\mathrm{H}$ in the toss, draw the first and second ball from box $1,$ with replacement and if you get a tail do the same from box $2 .$
Then
$$
W_{1}, W_{2}
$$
drawing white in the first and drawing white in the second are independent

This is true or false anybody explain the reason

Leonhard Euler

10:46

Let n be a positive integer

And v(n) be the number of groups of order n

I derived a (seemingly bad) bound for v(n)

$$v(n)\leq n^{n^2}$$

Any nicer bounds?

Leaky Nun

11:15

@LeonhardEuler oeis.org/A000001

a(p^e) ~ p^((2/27)e^3 + O(e^(8/3)))

polite proofs

Is this valid notation? $$\lim_0 \frac{\sin \bigg|_{\mathbb{R} \setminus \{0\}}}{\mathrm{id}_{\mathbb{R} \setminus \{0\}}} = 1$$

math.stackexchange.com/questions/4006767/… Made a thread about it, since not too many people here right now.

Alessandro Codenotti

@Lelouch closed disk on the plane

Astyx

@politeproofs Yes, but you never want to write/read that

And your vertical line is weirdly big

polite proofs

@Astyx I absolutely do! ;)

Astyx

I'm not going to debate this, but the goal when writing math is to allow people to understand you as quickly as possible, not encrypt the information in the crudest way possible

polite proofs

11:27

It seems as if the community agrees with you, since even my question got downvoted.

Mike Miller

12:01

@BalarkaSen I found it very compelling but agree it is not really up your alley. Ridiculous that FIDE president decided to speak out about a TV show lol

Lelouch

12:54

@AlessandroCodenotti Uh how is that a closed subspace?

user15072279

[![enter image description here][1]][1]

Here, theta equals d theta . Therefore , we can say s is almost like a line . Then , that makes it a triangle .

I have read that to find theta , we say it is theta = S / H .

Now , let us say theta is a final velocity at of particle P at A and initial velocity of particle P at B . [![enter image description here][2]][2]

[1]: https://i.sstatic.net/f7k94.jpg
[2]: https://i.sstatic.net/wA1sv.jpg

Then , we also say that theta = delta v / Vf. I did it understand how did we solve this and why did we do it like this here as well ? Is it possib…

Mike Miller

He parsed subspace as topological subspace, not vector subspace.

@Lelouch Take X = ell^1 and N = 0.

Then you're just saying that X is not continuously isomorphic to its dual ell^inf.

Mike Miller

13:11

Oh, you're not talking about duals. That's very irritating. Don't use the asterisk!

There is a natural map $N^* \to X/N$ which is a continuous map which is an isomorphism of vector spaces. You want to now appeal to the closed graph theorem to conclude. If X is a Banach space, then this is true. The closed graph theorem doesn't hold for fully general topological vector spaces though. So I'd look at counterexamples to the closed graph theorem.

Note that not every closed subspace is complemented.

Lelouch

@MikeMiller So probably taking something like $C^\infty([a,b]) \subset C([a,b])$ would work, but I'm not sure. Thanks!

Mike Miller

That's not a closed subspace

Anyway I have never in my life found a need for things at this level of generality, so I don't have an example offhand.

user 85795

13:26

Aha! I found it:

Jan 11 '11 at 22:17, by Asaf Karagila

Real numbers

His first message in this room.

not too long after that the first vid was posted

chat.stackexchange.com/transcript/36/2011/1/15

Leonhard Euler

14:19

Why is the kernel notation $Ker\,f$ but not $Ker(f)$

When writing with hand I sometimes write it as $Kerf$

Although I can recognize it, it looks bad

Astyx

You can write it however you want

$\ker f$ is nice because there are less lines

But ofc that depends on your handwriting

Leonhard Euler

Algebra notation is sometimes annoying

Every time I see $xx^{-1}$ I want to write it as $1$ but I can't(unless the identity is denoted by 1 but I think it is a bad notation)

Sayan Chattopadhyay

4 hours, 4 godamn hours, I was teaching for 4 hours, I have lost my voice

They better pay me good bucks

user 85795

lol

Leonhard Euler

Hehe

Thorgott

14:28

writing $Ker$ instead of $\ker$ is a crime against humanity

user 85795

why?

Sayan Chattopadhyay

I did that a hole lot of times today, I don't care, Linear algebra is amazing, but 4 hours of it, and 4 hours of explaining row reduction is not

user 85795

All 4 hours on zoom?

Sayan Chattopadhyay

Meet but who cares, everything is fuzzy now

user 85795

Imagine if you did it in person -_-

Thorgott

14:33

poor sayan

mentally scarred

Leonhard Euler

Can someone teach differential geometry and algebraic geometry for 4 hours

user 85795

nope, only linear algebra

maybe arithmetic to primary kids

satan 29

hey everyone, i had a multiple integrals question

Leonhard Euler

What is better: saying that $f$ is one one onto or saying that it is bijective?

satan 29

we need to find the volume of the solid obtained when the cylinder x^2 + y^2=9is bounded by the planes z=1 and x+z=5

Astyx

14:40

@LeonhardEuler My profs did AG and ANT this semester

satan 29

$$\int_{-3}^{3} \int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}} \int_{1}^{5-x} dzdxdy$$

is what I came up with

but it gives the answer as 36(pi -1), and the answer is 36(pi)

oh no wait, the integral does give 36 pi

My bad

Leonhard Euler

Is anyone here vaccinated

Against covid 19

Mike Miller

15:26

@SayanChattopadhyay It is very tiring, yes.

IMO the trickiest part of row reduction is explaining that it is an algorithm, with pre-defined steps, and that one should not simply use their cleverness to solve it. One should run the algorithm. (After all, we do not care so much about me knowing how to invert a 5 x 5 matrix by row reduction, but rather to program something that can do that. So I must know the algorithm.)

BigSocks

@MikeMiller you mean "tricky" as in getting students to accept/understand it is tricky?

Thorgott

15:46

is a fiber bundle over a compact space with compact fibers itself compact?

the answer is a clear yes if the base is additionally locally compact

Mike Miller

@BigSocks It is the most conceptually difficult part

The algorithm itself can be clearly explained with catchphrases

But the idea of an algorithm is not trained, despite the fact that we train people to do algorithms

@Thorgott So this is only interesting in the non Hausdorff case? Lol no thanks

polite proofs

math.stackexchange.com/questions/4006767/… Could someone clarify about the vertical bar differences here?

BigSocks

hmm I see. I have never taught/TA'd linear algebra so I would not know. I would have thought just giving them the list of steps and saying "here it is, do this until it's in rref" would work

Mike Miller

@politeproofs I have no idea what that person is talking about. I understand your notation.

And I think theirs is uglier notation for the same thing.

polite proofs

Okay

Thorgott

15:52

@MikeMiller fair

Mike Miller

@BigSocks But the Gauss Jordan algorithm is not always the optimal route to rref. So a clever student naturally wants to save themselves energy by finding a more optimal route.

This misses the point, as I am not trying to train someone to solve Ax = b for explicit by hand matrices, really, but rather an algorithm.

BigSocks

Ah ok, I see how intuitively someone might wanna find a shortcut instead

Mike Miller

Which is a great thing to do, right? You want to be writing efficient algorithms. Notice enough shortcuts for a certain kind of matrix and you've written a better algorithm.

But it's not where you should start from. Need to understand the basic algorithm before you can improve it

@politeproofs I will say I'm not big on lim_0, that might throw people off. And I do agree with them that it's unnecessary to do the restriction of domain here, since the definition of a limit to 0 indeed only uses the restriction to R - 0.

But the bar thing was a weird quibble.

polite proofs

Good point about R - 0

@MikeMiller Yes, I am trying to obfuscate my proofs.

Thorgott

@Mike So what's the conceptual way of thinking about the $D(E)/S(E)\cong\mathbb{CP}^{n+1}$ fact from yesterday. I've managed to cook up a proof since, but not a lot of insight.

Mike Miller

16:00

What's your proof

I'm writing lectures so may be slow to respond

Thorgott

I wrote down a homeo by trial and error. Take an element in $D(E)$, represent it as $([x],\lambda x)$ with $x\in S^{2n+1}$, map it to $[((1-|\lambda|)x,\overline{\lambda})]$. This is well-defined and with a little bit of work, one sees that this induces a continuous bijection $D(E)/S(E)\rightarrow\mathbb{CP}^{n+1}$ which is a homeo since the domain is compact and codomain Hausdorff.

Mike Miller

Oof

Can you prove that the total space of the tautological bundle is homeomorphic to CP^{n+1}, minus a point?

This one should be pretty clear

Thorgott

similar idea: map $([x],\lambda x)$ with $x\in S^{2n+1}$ to $[(x,\overline{\lambda})]$, only misses the point at infinity

Mike Miller

Lame

You should be able to write this without ever saying anything about spheres

Think in terms of linear algebra

Perhaps doing this for RP^2 - pt instead will make it clearer?

Once you have a conceptual understanding of this step I'll tell you the quick trick for the rest

BigSocks

16:27

So for a sequence of $A$-modules, $0 \rightarrow M' \rightarrow_f M \rightarrow_g M'' \rightarrow 0$, the sequence $0 \rightarrow im(f) \rightarrow_i ker(g) \rightarrow _{\pi} ker(g)/im(f) \rightarrow 0$ is exact. That means $ker(g)/im(f) = (0)$ right?

yeah ok nvm I think that is pretty obvious

Thorgott

obviously wrong, that is

BigSocks

wut

but I thought they were equal

wait no it can't be. they didn't tell me the first one was exact

it would be getting exactness for free... so it is impossible

yeah this sequence being exact just tells me that $ker(\pi) / im(i)$ should be zero

which is just $im(f)/im(i) = im(f)/im(f)$

so not very surprising

Thorgott

Hmm, I'm not seeing it. The zero section of the tautological bundle surely should correspond to the canonical inclusion $\mathbb{CP}^n\rightarrow\mathbb{CP}^{n+1}$ and then the fibers of the bundle should parametrize the last homogenous coordinate in $\mathbb{CP}^{n+1}$ (so the one point missed would be the point $[0\colon...\colon0\colon1]$), but I can't figure a good way of doing this other than the stupid map I've already written down.

I mean, I know that the punctured projective plane is a Möbius strip, but I only think about that in terms of spheres, not projectively.

Curious youth

16:43

Soon I will be pursuing PhD in antiderivative

good night peepeeople

user3559014

Dear experts, i have a geometric problem to solve: Is there a known algorithm to figure out, whether a number of points are distributed "equally" along a given polyline. Equally means there are points along the path of the polyline. The background is to figure out whether a gps track has fully covered a specific street and can be marked as "processed". I have no idea what to google so i thought maybe one of you has some hints for finding a proper problem description or solution... :-)

Curious youth

why i see two parallel polyline with path 1 meet after a vodku

bezier curve is moving like a worm

BigSocks

@user3559014 don't really know about this stuff, but googling polyline lead me to this wiki ref en.wikipedia.org/wiki/Polygonal_chain#cite_note-11

Curious youth

I also see Riemann staring me in my soul

oh I got it he was staring me because it meets when you project it on blue balls

Mike Miller

@Thorgott So the condition that our lines are not [0: ... : 0 : 1] is equivalent to saying that our lines in C^{n+1} project to lines in C^n. That gives a map CP^n - pt -> CP^{n-1}. Do you see why the fibers of this map are complex lines, and how to identify this with the tautological bundle?

user3559014

16:56

@BigSocks thanks, i will look into this

Mike Miller

I'm glad you brought this up again, IMO this is a good picture to understand.

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