Mathematics - 2021-02-22 (page 1 of 4)

And alhough french also flock together, their speaking french - while I was the only non-speaker in the group - never felt exclusive. Like, there's two sides. People bothering about being excluded, and people anxious about excluding someone or about someone not fitting in. And it always felt casual in my experience (limited to a few occasions in different groups)

That seems like a massive generalization but man in exchanges the french, italian and spanish speakers really flock together often

partly due to their english proficiency level often not being that high, in comparison to e.g. scandinavian countries

copper.hat

i always felt bad that my linguistic abilities were so abysmal compared to my continential friends.

i can't even speak my 'native' language.

Astyx

Yes that's a very reasonable explanation. In my experience brazilian and chinese students stayed in groups

copper.hat

i think most groups do...

geocalc33

12:13 AM

0

Q: Rotation of curve in specific coordinate system

I’d like to rotate $\log(x) \log(y)=1$ in an $e^x - e^y$ coordinate system. I’d like to create a surface of revolution that sits in 3-space. I tried using a standard Cartesian rotation matrix and then I parametrised the surface. But I think I need to change the basis to get a rotation matrix comp...

copper.hat

comfort zone & efficiency.

user2103480

@copper.hat in permanent residence, definitely

this is more a statement in the context of exchange semesters and travels

copper.hat

@user2103480 i think being from a small country one is forced to interact more, so one has more practice maybe?

i like to interact with random strangers which makes it easier

and have no problem with embarrassment :-).

user2103480

@copper.hat hmm not necessarily. Dutch & scandinavian people have most movies in english with subtitles

Astyx

@copper.hat I wish I didn't

user2103480

12:15 AM

that naturally increases the language skills

copper.hat

@Astyx why?

i think i can communicate reasonably, just not in a particular language :-).

user2103480

but language definitely isn't all, I was accidentally participating in the budapest semester in mathematics program

copper.hat

@user2103480 my first real foreign experience was in the netherlands, so many things were just brilliant for me, movies in their original langauge was just one.

Astyx

My fear of embarrassment hinders my social aptitude to some extent

geocalc33

0

Q: Rotation of curve in specific coordinate system

I’d like to rotate $\log(x) \log(y)=1$ in an $e^x - e^y$ coordinate system. I’d like to create a surface of revolution that sits in 3-space. I tried using a standard Cartesian rotation matrix and then I parametrised the surface. But I think I need to change the basis to get a rotation matrix comp...

matrices geometry coordinate-systems

copper.hat

12:17 AM

@Astyx i was very shy/introverted as a child. but i found overcoming the activation energy bump of embarrassment was more than rewarded.

someone is trying to send a message, i think...

geocalc33

@anakhro any ideas?

user2103480

which was exclusively US students, and they really just stayed in their groups. There were several occasions where I made more lasting friendships in one evening than short-term friends in 3 months of uni with the US people

copper.hat

definitely a bottle of wine convo :-)

Astyx

@copper.hat Hopefully it'll click at some point :-)

copper.hat

if there was one stereotype, i find it easier to meet people in the usa, but much longer to form deep friendships.

user2103480

12:20 AM

The only one I actually bonded with was someone with a ... historically justified... german nationality. (His last name is Levy, to give some context)

geocalc33

*harder?

user2103480

@copper.hat yes their small talk is great

copper.hat

@Astyx i guess what i mean is that it is a choice, the price is embarrassment.

user2103480

(and again, I am aware I'm overgeneralizing)

copper.hat

i think there is just a huge variance, so one's experience is very path dependent.

user2103480

12:24 AM

you're probably right

robjohn

@geocalc33 How are you defining rotation in an $e^x-e^y$ coordinate system?

or do you want to move to normal coordinates rotate and then move back?

copper.hat

@user2103480 for example, when i finished my graduate work, i looked at various places, including an electronics research lab in my old university in ireland (people who knew me fairly well). a company in the usa was willing to take a risk, engage and hired me in a heartbeat whereas my old institution were dithering indefinitely (i had the support of the university president at that time too). so, for me, the fact that they were willing to take a chance on me was a huge positive.

user2103480

@copper.hat maybe that is partly due to workers' rights protection laws

copper.hat

@user2103480 absolutely. the flip side is that setting up a company, hiring, getting funding is sooooo much easier in the usa.

geocalc33

@robjohn I don't know how to define rotation in an $e^x - e^y$ coordinate system. Do you have any suggestions?

robjohn

12:30 AM

@geocalc33 not really, that's why I asked.

user2103480

@copper.hat that's why they bounced back immediately from the covid economic dent in summer

copper.hat

@user2103480 so i think a lot depends on one's risk profile and when one is young that profile looks a lot different that when you are older :-).

@user2103480 education for kids & health care are dismal in the usa if you do not have money.

that does not mean it is not bad elsewhere, nowhere is perfect.

user2103480

hahaha let's not get into that, I'm very pro welfare state and I can't comprehend the US system

copper.hat

i have had European friends who have had medical nightmares with things as simple as routine appendectomies.

And my 90 year old aunt in ireland still has not been offered a vaccination.

user2103480

vaccination is so slow here. people somehow think germany is efficient but that's far from the truth

copper.hat

12:33 AM

well, that is what i do not understand. even from a purely economic perspective it makes sense to have a good educational & health support system, regardless of one's social perspective.

the word welfare is interpreted very differently in the usa.

user2103480

@copper.hat emphatic yes

welfare = socialism to many lol

copper.hat

i am not in a vaccination group yet, but i can drive down to a local racetrack (horse) and get a 'spare' vaccination if i am willing to wait around.

i so need to get back to being able to interact with people.

i am sure there will be a sigh of relief in this chat room when that happens :-)

user2103480

I must admit my main sigh of relief will occur when my mother gets her first jab :P

My father already got two jabs, he's a healthcare worker close to retirement

copper.hat

my concern is not getting it, but possibly being a spreader.

my brother is a dr, he got two jabs, but tested positive afterwards.

user2103480

@copper.hat oh yeah that's bad too. But at least the most vulnerable groups will most likely be safe by then

copper.hat

12:38 AM

that's a much longer story that we have time for here.

a relative of my wife is a doctor in san jose who picked up covid from some much ballyhooed party a few months ago, she had the two jabs too.

geocalc33

@robjohn I have a definition of rotation actually

copper.hat

@user2103480 anyway, i should do something useful with the remainder of the day :-). nice chatting!

user2103480

@copper.hat and I guess she also tested positive?

@copper.hat me as well, enjoy!

copper.hat

@user2103480 yes, she did.

Clarinetist

Question on the Borel-Cantelli lemma: let's say I want to show $\limsup_{n \to \infty}X_n = c$ almost surely.

Do I then aim to show that for any $\epsilon > 0$, $\sum_{n=1}^{\infty}P(X_n > c - \epsilon) < \infty$?

geocalc33

12:49 AM

@robjohn thanks for helping I've figured out my real question now

user2103480

I always forget what lim sup and lim inf exactly say as sets. You're trying to prove that for almost all omega, the lim sup of the sequence is c, right?

Clarinetist

Sorry, the $X_n$ are random variables, not sets. In this case, the answer is yes.

user2103480

This is the thing you use right:

uhh

wrong case

robjohn

@geocalc33 okay. Sorry, I slipped out for dinner. Now, the dogs are asking to go for a walk.

user2103480

gimme a minute

So $$
\mathbb{P}\left(\lim \sup X_{n} \leq a\right)=1 \Longleftrightarrow \mathbb{P}\left(\limsup \left\{X_{n} \geq a+\varepsilon\right\}\right)=0
$$
(for all epsilon > 0)

but you want to prove equality so you want to prove this is zero for all a < c? Borel cantelli is definitely the right approach, I'm only trying to follow how exactly one phrases this, you probably have it right already

robjohn

12:56 AM

@user2103480 The thing that irks me is that the pot workers are getting ahead of the teachers. They have a richer lobby, evidently.

user2103480

Let's start with the a = c case. Then you wrote down the right formula, assuming the distribution is not atomic

Clarinetist

@user2103480 Great, it's not atomic, so I'm good there. Thanks.

user2103480

@robjohn uhh pot = weed?

robjohn

@user2103480 marijuana, cannabis sativa

user2103480

@Clarinetist but for equality you also need to prove it's not smaller I'd think

ah that is ridiculous

robjohn

12:58 AM

I think so, but as I said, they have a more powerful lobby

user2103480

very pro legalization, but that is excessive

shouldnt then all retail workers get the vaccine before teachers?

@Clarinetist wait it should be a plus epsilon, no? Or am I not focused enough

Clarinetist

I'm honestly not sure, but I'll try the $+ \epsilon$ anyway. I haven't found a single book that explains clearly how to translate a statement you want to prove about $\limsup$s of sequence of random variables to events that we want to find the probability of so as to apply Borel-Cantelli.

user2103480

"omega in lim sup of sets" means "for all n there exists m>=n with omega in E_m"

So if the lim sup of the sets you wrote is 0, by borel cantelli, this means that for almost all omega, there is an n such that they are in none of the E_n above that

geocalc33

@robjohn my question is: "How do you derive the Lorentz boost in an $e^x - e^y - e^z$ coordinate system?

I've done it for $e^x - e^y$ coordinates

robjohn

@geocalc33 not sure what a "Lorentz boost" is

geocalc33

1:07 AM

instead of circular rotation

it's hyperbolic rotation

so imagine a point zooming around a hyperbola instead of a circle

user2103480

@Clarinetist So in math: If $\sum_{n=1}^{\infty}P(X_n > c - \epsilon) < \infty$, then $$\Bbb P\left(\bigcap_n \bigcup_{m \geq n} \{X_m \geq c-\varepsilon\}\right) = 0$$

robjohn

Not sure what that looks like mathematically. I need to walk dogs. BBL

Akiva Weinberger

Does the universal cover of the loop space correspond to the loop space of something

user2103480

thus for almost all omega, they are only finitely many times bigger than $c-\epsilon$, so the suprema must be less than $c-\varepsilon$. This does not work out

If you put a plus there, this yields that the RV are finitely many times bigger than $c+\varepsilon$, so the lim sup must be at most $c+\varepsilon$

geocalc33

@34Pancakes hi

can I have some pancakes? maybe 17?

user2103480

1:19 AM

Now you also want to prove that for every $\varepsilon$, the the RV are infinitely often bigger than $c - \varepsilon$. So we want $$\Bbb P\left( \forall n \, \exists m \geq n \colon \{X_m \geq c - \varepsilon \} \right) = \Bbb P\left(\bigcap_n \bigcup_{m \geq n} \{X_m \geq c-\varepsilon\}\right) = 1$$

34Pancakes

@geocalc33 Hi ! Well for now I have only 10

Mike Miller

@AkivaWeinberger No, nothing you'd like. I can give a kind of geometric description of the universal cover but it's not a loop space in a natural or canonical way.

user2103480

If the RV are independent and the sum of the probabilities of the events diverges, then you get this by borel-cantellis other case

There are more refined versions though

geocalc33

@34Pancakes okay I'll take 5 pancakes then

34Pancakes

@geocalc33 Okay ! Should I send it to you ?

dc3rd

2:17 AM

I was curious as to what is flawed (or am I missing) from the procedure I did for this question:

The question asks: Find the equation of the circle whose diameter extends from point $A = (-3,-2)$ to $B = (1,4)$.

The following is a diagram and solution from the text:

So what I was going to do was get the distance of $AB$, take half of it which would then give me a radius and from there use the radius to get my center. Well as can be seen the radius is $\sqrt{13}$. Here is where I got stuck. How could I use the radius to get the right coordinates for the center ?

There way makes sense to me as well, but I'm just attempting to reconcile things with what my first "instinct" was to do. Surely I can find the coordinate just from my radius.

robjohn

@geocalc33 this is for rotation. I remember about 50 years ago I talked to Dr William Kaufman and told him that I had found the linear formula for adding relativistic speeds: $c\tanh(\operatorname{arctanh}(v_1/c)+\operatorname{arctanh}(v_2/c))$. I was in junior high at the time, so he blew me off, but I know the formula is correct. However, it is linear and not rotational.

Ted Shifrin

@dc3rd The hell with the radius. Find the midpoint of the diameter.

Thorgott

sanity check: I already have a SES $0\rightarrow VE\rightarrow TE\rightarrow\pi^{\ast}TM$ whenever $\pi\colon E\rightarrow M$ is a submersion, right?

robjohn

@TedShifrin that's a very centrist view.

Ted Shifrin

But if you insist, parametric equations for the line segment and go halfway.

dc3rd

2:26 AM

@TedShifrin, makes sense. Out of curiosity how could I use the radius?

Ted Shifrin

@robjohn I can't always be leftist!

dc3rd

Ok, so I would have to set up parametric equations. But midpoint is simpler. Good to know an alternative. @robjohn....funny....lol

Ted Shifrin

How do you “know” the radius?

robjohn

This discussion is probably over my head. When you are talking about radius, it is not simply half the diameter of a circle, is it?

Ted Shifrin

Oh, half the diameter. So we're back to ratio and proportion again.

Time to review vectors.

robjohn

2:31 AM

@TedShifrin I like ratios and proportions. That's why I figured it did not apply here.

dc3rd

Well the radius is half the distance of the diameter.

Another question I had which is a bit more abstract. So today in my readings the trig ratios were covered and I went off to thinking about how Spivak had defined$\cos$ in his text. Looking at the wikipedia page they didn't do that either. ..I' guess I'm wondering how the "simple" form of $\cos$ taught in high school courses reconciles with Spivak's

my radius answer is dated but I did have the response ready before you lot got to conversing

Ted Shifrin

@robjohn dc3rd is filling gaps in his education.

robjohn

@TedShifrin Ah, chasmic education

Ted Shifrin

Don't worry about Spivak's definition for now. He's basing things on area because arc length of an arc of a circle is more subtle.

dc3rd

Nah it was all my fault @robjohn, when I was in high school years ago I avoided some of these essential courses and then when I came back to university I initially thought "what will I need geometry and trig for?...shapes aren't used in finance..........."

and here we are................😭😭

Ted Shifrin

2:35 AM

You could go back to finance? 😻

robjohn

@TedShifrin I used areas in an answer to show that $\lim\limits_{x\to0}\frac{\sin(x)}x=1$.

user2103480

@Thorgott since when are you going full topology

dc3rd

I've seen too much...........the lie of shapes not being in finance won't hold over anymore @TedShifrin

Ted Shifrin

@robjohn Totally standard.

dc3rd

Pandora's box got opened....and now "truth" has become a very abstract notion.....ah how the days of naievete are sometimes desired for...🤣

Thorgott

2:39 AM

I've been going full topology for most of this semester

Ted Shifrin

Especially deep at night!

Thorgott

currently fleshing out the last details for my seminar talk on tuesday

on time

Ted Shifrin

Allegedly.

robjohn

@dc3rd "Ignorance is bliss" is perhaps the phrase you're looking for.

Ted Shifrin

@robjohn dc3rd's biggest mistake was picking up my textbook.

Thorgott

2:43 AM

after that, I will go full physics for a week and hope to not fail my exam on march 1st lol

robjohn

@TedShifrin Oh, no! He didn't...

dc3rd

@robjohn, exactly that, but I wanted to be esoteric and come up with my own superfluous "quote".............

Ted Shifrin

Thor, you won't fail.

Thorgott

you underestimate how inept I am at physics

but a week of preparation should hopefully do the job

Ted Shifrin

You might be right, mr category.

dc3rd

2:45 AM

You say it in jest @TedShifrin, but it was probably the best decision I made because I was going to use Munkres's Analysis on Manifolds, but my gut told me I was missing some foundational understanding. You actually explain things without all the verbose math talk....very much a relief.

Thorgott

yeah, precisely

user2103480

I can get any abelian group as a fundamental group right

Thorgott

I'll be bereft of the beauty of elegant axiomatization and consistent abstraction

@user2103480 any group, even

user2103480

Of a 1-dimensional CW complex as well?

Ted Shifrin

He assumes basic analysis and a knowledge of linear algebra and multivariable. I assume none of that.

Thorgott

2:48 AM

no, 2-dimensional CW-complex is best possible

1-dimensional CW-complexes have free fundamental group

user2103480

Ah yes true

Ted Shifrin

Relations!

Thorgott

but 2-dimensional will work

Ted Shifrin

We can't all be unrelated!

user2103480

Hmm how do I ensure that the second homology of that thing is trivial

Ted Shifrin

2:49 AM

Huh?

Thorgott

I think if your group is countable, you can even always realize it as fundamental group of some manifold in dimension 4 or 5 or sth like that

dc3rd

I thought doing Spivak combined with Insel would've given me what I need. Plus I have "done" a multivariable course.....didn't get much from it, simply cause I was not ready for it....but that is changing with doing your text and............that linear algebra book...

Thorgott

but I don't know how that works exactly, though I think Balarka has written about it here before

Ted Shifrin

Haven’t seen balarka in a while!

Thorgott

@user2103480 why do you want that

user2103480

2:52 AM

I'd like to get any sequence of abelian groups as reduced homology groups of a connected CW complex, and I'd like to reduce this to finding a space that has only nontrivial first homology and then suspending + wedging

(it was one of our exercises)

robjohn

@TedShifrin Having a Balarka Yen?

Thorgott

too complicated

you already know an easy space that has non-trivial reduced homology only in dimension n

nvm, that doesn't cut it

user2103480

I don't know how to get any abelian group with the spheres though

also, that is also just suspending smh

Thorgott

yeah, right

Mike Miller

@Thorgott Yes. four suffices

Thorgott

2:55 AM

ah no wait, the obvious thing works, doesn't it

at least for Z cofeficients

user2103480

covfefecients

Mike Miller

@user2103480 Think about what the simplest possible cellular complex of something with H_1 = Z/n is

user2103480

@MikeMiller will do, that's something I already wondered about

Thorgott

yeah, that will work, I was confusing myself

Mike Miller

Just think chain complexes first

user2103480

3:04 AM

I have to get a map thats multiplication by three, with one cell in each dimension. The space is probably the same as the disk with a circle glued to the boundary by the map that winds three times. gotta think about how to do this formally now

okay, I mean, I already have the attaching map

sorry

been thinking bout Z/3Z

Thorgott

this is correct

user2103480

And local degree yields this quickly

ok thanks, then trivial second homology follows immediately

can I get any abelian group as a direct sum of Zs and Z_ps

sounds wrong

Thorgott

any finitely generated one

this might get a bit ugly if you want arbitrary abelian groups

user2103480

Let $G_{1}, G_{2}, \ldots$ be a (possible infinite) sequence of (not necessarily finitely presented) abelian groups. Construct a $C W$ -complex $X$ with reduced homology groups
$$
\widetilde{H}_{k}(X) \cong\left\{\begin{array}{l}
G_{k} ; \text { if } k \in \mathbb{N}, \\
0 ; \text { else. }
\end{array}\right.
$$

That's the problem statement so it might get ugly

Thorgott

ugly

actually not sure how do this immediately

user2103480

3:14 AM

@Thorgott There's an alternate reality where you intentionally put a comma after "sure" to make this a grammatically correct sentence

do this immediately

Thorgott

you mean after "how"?

user2103480

there's an alternate reality where I am smart

yes

Thorgott

it's definitely possible, but dunno if it's possible in 2 dimensions in general

user2103480

okay, gonna head to sleep now, I'm pleased enough with the wedges of circles and weird disks and then suspending + wedging

Thorgott

good night

Mike Miller

4:16 AM

@Thorgott Every Abelian group has a presentation --- AKA a 2-step resolution by free groups --- because subgroups of free Abelian groups are free.

Take A, pick a presentation, make a presentation complex out of it.

Thorgott

4:32 AM

Can this always be realized as cellular chain complex?

user8469759

9:12 AM

Hi guys, does anyone know if there's a way to use the Conjugate gradient to perform the inverse of a symmetric positive definite matrix?

Mike Miller

10:31 AM

@Thorgott What's stopping you? They're free Abelian groups, so you make a wedge or circles on the number of generators in degree 1 and choose a loop for each generator in degree 2 to attach a disc along

Monty

10:44 AM

Hi guys, can anyone say much about the closure of the set of gradients of $C^\infty ( \mathbb{R}^d $ (or say $C^1 ( \mathbb{R}^d $ ) functions ?

Mike Miller

11:25 AM

That's a closed set.

In the first case, at least. The second case is not as obvious.

A C^k vector field (k>0) is a gradient of a C^{k+1} function iff all its mixed partials are equal. That means that the set of gradients of C^{k+1} functions is a closed set in the space of C^k vector fields.

love_sodam

In the drawing in the problem, is there a disk on the intersected hole?

i.e. inside and outside are distinguished

Mike Miller

Yes

This is what "A Klein bottle intersecting itself" means; I'd there was no disc there the original object (before self-intersection) wouldn't be a Klein bottle, which is a topological surface

It'd be a Klein bottle with a hole, intersecting itself along the boundary of that removed disc :)

Balarka Sen

11:56 AM

It is interesting that even though C^infty gradient vector fields are closed in C^infty, they're far from closed in C^0

They can even have non-conservative limits in C^0

Here is a precise statement: If A is a polyhedral subset of R^n, given a C^infty vector field near A, one can always find a C^0-small ambient isotopy of A after applying which the initial vector field is C^0-close to a C^infty-gradient vector field on its domain of definition.

The C^0-small pertubation breaks topology (you cannot argue the limit will have 0 integral over a loop contained in A, so it has to be conservative, say) and the C^0 approximation breaks analysis (you cannot argue that grad o curl = 0 so the limiting vector field has to be irrotational)

This phenomenon is known as an h-principle

love_sodam

12:30 PM

@MikeMiller Thanks!

Edward Evans

12:56 PM

Let $R$ be a noetherian ring and $I \subseteq R[X]$ an ideal. Set $I_n := \lbrace f \in I : \deg f \leq n\rbrace$ (which is an $R$-module) and let $b_n : I_n \to R$ send $f$ to its leading coefficient. Call $B_n := \operatorname{im} b_n$, which is an ideal in $R$. For any $f \in I_n$ you have $Xf \in I_{n+1}$ and $b_n(f) = b_{n+1}(Xf) \in B_{n+1}$, giving an ascending chain of ideals. Since $R$ is noetherian, this chain stabilises, so there's an $m \in \Bbb N$ with $B_m = B_n$ for all $m \geq n$

Monty

@MikeMiller @BalarkaSen many thanks :) ! I may hit you with a follow up, but i'll have to think for a bit.

Balarka Sen

You can ignore my comment, it's an example of a wild phenomenon.

@EdwardEvans You're proving Hilbert basis? :)

Edward Evans

ye

Balarka Sen

Coolz

Edward Evans

but

I'm having trouble interpreting the stabilisation of this chain

Balarka Sen

12:58 PM

So the ideal generated by all the leading coefficients of $I$ is finitely generated.

That's what this is saying

The last step IIRC is a trick, take a finite generating set of the leading coefficients ideals, suppose $f_1, \cdots, f_n$ are polynomials which have those precise leading coefficients. Then take any polynomial $f$ in $I$; it's leading coefficient must be a linear combo of the ones of $f_1, \cdots, f_n$

Then you take an appropriate linear combo so that $f - \sum_{i = 1}^n c_i x^{\deg(f) - \deg(f_i)} f_i$ is a polynomial of degree less than $f$

Edward Evans

Yeah right

Balarka Sen

You can write the proof constructively like this but induction on degree is aesthetically pleasing I suppose. This is why they like to grade it by degree.

Edward Evans

the proof I'm going through shows $I = R[X]I_n$ and shows $I_n$ is an f.g. $R$-module, so that $I = \sum_{i=0}^r R[X]f_i$ for the generators of $I_n$

the content of which is what you wrote lol

Balarka Sen

Yeah, it's a little mentally easier for me to say it in terms of polynomials

Symbols confuse my small brain

Edward Evans

with this $n$ being the point at which the chain of the $B_i$ stabilises lol

Balarka Sen

1:04 PM

Yeah

Edward Evans

@BalarkaSen this helped tho

seemed kinda surprising

for a reason I can't articulate

Balarka Sen

It's interesting to upgrade finite generation from leading coefficients to the full ideal of polynomials

Not actually totally obvious

Edward Evans

guess that's the meat

Balarka Sen

I mean not obvious at all lol what about $k[x]$ where there is no leading coefficients

I guess the point is then everything is divisible by anything

Edward Evans

lmao

rip nt

Balarka Sen

1:07 PM

You can do polynomial long division

Astyx

no leading coefficient?

Balarka Sen

Which is exactly why $k[x]$ is a PID

Astyx

hi btw

Balarka Sen

I mean all of them are units

Edward Evans

hi @Astyx

Balarka Sen

1:07 PM

Hi

$k[x]$ to $k[x, y]$ is the nontrivial jump

OK, I am convinced now

Maybe you can also try to understand it in terms of zero sets

Astyx

What do you mean by k[x] ? Polynomials of one variable with coefficients in a field k?

Balarka Sen

The problem is leading coefficients is nonsense. It doesn't mean much, unlike the terminal coefficient

Yeah

Astyx

Oh ok I get what you meant

Edward Evans

lmao I'm just going through module basics again so I can do some homological algebra

Balarka Sen

Good shit

My homework wants me to verify $(A \otimes B) \otimes C \cong A \otimes (B \otimes C)$

I drew a commutative pentagon

Edward Evans

1:12 PM

last semester I did the same course but the lecturer did everything over general abelian categories and this lecturer just embeds it all into modules and writes everything else off

Balarka Sen

Nice

The correct approach

Edward Evans

he's the kinder lecturer

slash, they're probably both as nice as each other, the other dude is just a nutter

Balarka Sen

lol

Edward Evans

that's Lukas' supervisor

lmao

Balarka Sen

loool

of course

Edward Evans

1:14 PM

obv

he's also the guy who fell asleep 50 times during someone's elliptic curves talk

on camera

Balarka Sen

massive

Edward Evans

and then laughed as he jolted himself awake

Balarka Sen

that last thing got me

Edward Evans

lmao

Balarka Sen

The Alan Parsons Project - I Robot (1977) (2016 RM, MFSL UDSACD-2174)

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listening to good, classic prog rock

Edward Evans

1:17 PM

lool

fonky

Septic Flesh [Annubis]

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this album is my listening today

Balarka Sen

pure death metal?

Astyx

I love paperwork

I'm meant to fill in someone's email on a form

But the space given doesn't allow enough characters

Alessandro Codenotti

lol

Astyx

The form allows for people with names of length <= 4

Balarka Sen

death metal band from Athens lmao

the real Athens

Astyx

1:19 PM

(first + last name)

Alessandro Codenotti

I´ve had this problem before where the form says to write out your full name and lastname clearly in capital letters and there´s a 4cm long white space to do that

And I don´t have such a long name

Edward Evans

@BalarkaSen Greece's answer to Fleshgod Apocalypse

Balarka Sen

name checks out

Edward Evans

but about 500bpm slower

Mike Miller

@BalarkaSen Are gradients C^0 dense?

Balarka Sen

1:26 PM

Not literally but near codimension > 0 subspaces, upto small background isotopies

So you could say C^0-homeomorphism classes of germs of gradient fields are C^0-dense

Germs not only near points but near submanifolds of positive codimension

Monty

@BalarkaSen @MikeMiller My question stems from wanting to compare two general type of PDE : $ \partial_t u = \nabla f + stuff $ and $ \partial_t u = v + stuff $ ,

Balarka Sen

Oh then whatever I am saying is very relevant

ams.org/journals/bull/2012-49-03/S0273-0979-2012-01376-9/…

Monty

Your a star :) thanks ill have a look after lunch!

Balarka Sen

thats provably false but glad this might be useful

Mike Miller

Note that this is a study of low regularity cases. In high regularity, which is the case in physical applications, you get a different theory

Wish I still had my notes from Laszlo's class though

Monty

1:37 PM

I would have higher regularity

Balarka Sen

yeah then not relevant haha

these are $C^{1, \alpha}$ solutions at best

Mike Miller

I feel there is a lot hiding in the word "stuff" above. But if you truly ignore that and just go to the ODE level, notice that the dynamics of gradient vector fields look very different than the dynamics of non-gradient vector fields. The flow of a gradient vector field near an equilibrium point usually "looks like" one of the three model cases: a sink, a source, or a mix (think the flow of <2x, -2y>, whose flowlines are hyperbolae)

But you can have much more complicated dynamics of general vector fields: you can rotate around your equilibrium point, approach while spiraling around or leave while spiraling around, and in higher dimensions or when your critical point is not isolated things get very complicated.

Edward Evans

2:01 PM

Just found out that a Halmos functions as a useful little checkbox to tick off proofs you've read and understood