Mathematics - 2025-02-11

nickbros123

09:51

@Jakobian you should continue posting your findings, I believe. Keep posting in the general topology chatroom atleast

Or like soumik said, a blog

flowian

11:43

I have hard time understanding why loop based space $\Omega S^1$ is not path-connected.

Every element of the space starts and ends at $x_0$, and is determined by the winding number. But since the end-, and the start point are the same, how can it NOT be path connected?

Thorgott

what does a path in $\Omega S^1$ look like?

flowian

@Thorgott I am not entirely sure. Letting my imagination go wild, I want to say it is a continuous constant that starts and ends at the basepoint.

Thorgott

12:05

I don't know what a "continuous constant" is supposed to be

you have to unwind the definition

cause how can you form an opinion on whether a space is path-connected if you don't know the paths in it

flowian

Fair! By definition, a path in X, between any two points a,b is a continuous map f:[0,1] -> X s.t.
f(0)=a, f(1)=b.

What I was thinking earlier: Suppose a,b are two elements in $\Omega S^1$ with distinct winding numbers, and both a,b start and end at the same point $x_0$.

Define constant map $f:[0,1]\to \Omega S^1$, $f(t)=x_0$. Then there is path between a,b ?

Thorgott

13:00

$x_0$ is not a point in $\Omega S^1$

flowian

14:36

@Thorgott Well isn't $\Omega S^1$ a based space? So can't I choose it to be based at $x_0$?

Ben Steffan

You can do that, at the risk of confusing yourself, which you already seem to be.

With the understanding that $x_0$ is some point of $\Omega S^1$, not some point of $S^1$.

You still have not unraveled what a path in $\Omega S^1$ is; do this.

flowian

14:57

@BenSteffan Thanks. So $\Omega S^1$ consists of loops based at some point $x_0$ - these are the points in $\Omega S^1$... I see now. So the path in the space must consist of series of loops of $\Omega S^1$, which are points in this space. There is no way to have a continuous path with such 'points'.

Ben Steffan

No, that's not correct. Try again.

flowian

@BenSteffan edited*

Ben Steffan

Better, but still imprecise. A path in $\Omega S^1$ is the same thing as a homotopy of based loops $I \times I \to S^1$.

flowian

@BenSteffan $I\times I$?

Ben Steffan

Do you prefer $S^1 \times I \to S^1$?

flowian

15:02

@BenSteffan Is it the same as $[0,1]\times I \to S^1$?

Well I'm still confused...

Ben Steffan

@flowian If you do not know that $I$ is standard AT notation for $[0, 1]$, then perhaps your time would be better served picking up a textbook

Thorgott

I would say that is standard notation even beyond AT

flowian

"A path is a homotopy between loops". So suppose $a,b$ are two loops with the same basepoint $x_0$ and distinct windings.
You are saying there is no homotopy from $a$ to $b$? Why wouldn't something like $a(1-t)+bt$, for $t\in[0,1]$.

Thorgott

think about what you're trying to do geometrically

you're drawing straight lines between points on the circle

flowian

yep

Thorgott

15:07

yeah, so what is the codomain of your proposed homotopy

Ben Steffan

Now we're really straying into the territory of the absolute basics of AT.

flowian

Oh shoot you are absolutely right

the co-domain is $S^1$, so I need to think in terms of circles, or arcs.

@BenSteffan It's been over a year since I did AT. I'm aware of this notation just a bit rusty. Being an ass about it serves no purpose.

Thorgott

anyway, the point is the winding number is a homotopy invariant (this may be more or less obvious depending on how you prefer to define the winding number)

flowian

So, I can characterize $S^1$ with $e^{2\pi i t}$. Suppose I want to make path from element $\alpha$ with $a$ loops to element $\beta$ with $b$ loops, $a,b\in \Bbb N, \, a<b$, both based at $x_0\in S^1.$

I am thinking of homotopy like $e^{2i\pi a(1-t) + 2i\pi bt}$.

It is probably not precise. But is the idea drifting in correct direction?

Ben Steffan

@flowian Frankly they way you're writing suggests that you've never studied these things to any depth at all. I'm telling you that your time would be better served actually sitting down with a textbook and studying.

flowian

15:20

@BenSteffan Thank you

Ben Steffan

@flowian Case in point, the message that what you're trying to do here is impossible seems to have missed you, still.

flowian

No it has not missed me. I'm trying to understand why it is not possible by creating counterexamples that do not work...well my above counterexample may not be precise but I still do not have the intuition why it doesn't work.

You said the path in $\Omega S^1$ is the homotopy of based loops and that's what I tried to create, a path on $S^1$ between two loops

leslie townes

try examples like a = 0 and b = 1 or a = 1 and b = 2

Ben Steffan

@flowian You're asking why two loops on $S^1$ with different winding number are not homotopic, a question that texts computing $\pi_1 S^1$ usually discuss...

flowian

@leslietownes So I want to do something like $e^{2i\pi( t+ 2(1-t))}=e^{2i\pi(2-t)}$...can I get some hint?

Thorgott

15:35

it's hard to give a hint as to carrying out a procedure that provably cannot work

but I can take a guess at to what your intuition is missing

flowian

@Thorgott what it could be, please?

Thorgott

if you wanna write $a(t)=e^{2\pi i\alpha(t)}$, then $\alpha$ is not necessarily continuous

flowian

@Thorgott I don't quite follow. Any suggestions where I could read about this stuff?

I mean, if $a$ has two windings and $b$ has three, then I would want to do something like $e^{2i\pi(2(1-t)+3t)}$ to move from $a$ to $b$. Not sure why it wouldn't work?

Thorgott

that's a path, not a homotopy of paths

flowian

Ahh makes sense!

it's been a while

Thank you

@BenSteffan I'm sure it does, many questions that are asked here can be found in textbooks.

Also maybe you should consider getting teaching qualifications if you like giving out advice how others should spend their time studying maths

Ben Steffan

15:55

@flowian Oh, I will, I will.

leslie townes

yeah haha a lot of folks here have them (or probably would have them if they worked in places that required them)

Ben Steffan

I mean, teaching qualifications don't exist here, so to speak

getting into an academic career automatically qualifies you for teaching

leslie townes

same in the US at the university level

you usually need some kind of certification to mold young minds but the damage has been done by the time they get to college

Ben Steffan

...but I'm honored that you think I'd make a good teacher, despite you spending several hours this afternoon asking about something here you could have read up on in about ~10min in any standard text, against my advice.

flowian

@BenSteffan Yea with PhD...

@BenSteffan You being a good teacher is a questionable thing😅 I've never said this. I've spend maybe 30mins at most on this problem. Your comments just came out toxic.

But I'm sure you can be a good teacher!

Ben Steffan

16:08

@leslietownes it's a whole separate category of degree here, for school teaching. funnily enough they make them take some of the first-year math courses here, the math-ed students

and by "funnily enough" I mean "not funny at all:" that's just excessively cruel :^)

and mostly pointless

leslie townes

ben it's kind of interesting in the US school teaching is mostly regulated at the state level, with varying requirements depending on what state you get your job in (also, there are often exemptions and other loopholes). and private schools generally have less regulation than public

Thorgott

@BenSteffan Î had to do like a one-day course to get a qualification for being a Tutor the first time

leslie townes

its pretty common (outside of like MIT or wherever) for most of the math majors at a university to intend to be math teachers, and very ironically they are often the most hostile to what a mathematician would regard as 'actual' math instruction

Thorgott

@BenSteffan lol they have separate courses here

Ben Steffan

@Thorgott we have a course but it's optional

leslie townes

16:13

and usually the success of a program can be measured by how much they do not require prospective teachers to take "regular" math courses

like i am not actually that sure of the merits of the argument for making every potential middle school teacher learn real analysis, although that was definitely part of the math ed track at my undergrad

Thorgott

I think it's supposed to be mandatory here, but I know some people who avoided it

also due to bureaucratic covid complications

Ben Steffan

we generally treat tutoring as a bit of a "whatever" job

will it be a lot of work? will it be very little work? will the lecturer provide you with solutions to the exercises? will you have to do all the exercises on your own?

"who knows lol"

Thorgott

yeah, same here

we were supposed to track our hours as Tutors, but I think everybody just made their hour charts up

cause no way you spend like 6 hours in a week grading hw towards the end of the semester

leslie townes

that kind of stuff is often barely a job in the US

Ben Steffan

@Thorgott ah yes, that

leslie townes

16:21

i knew someone who listed what were nominally teaching jobs as part of his financial support (i.e. grants etc) in a CV, which seems like a faux pas at a minimum, but pretty spot on in terms of what the "job" was

at my last academic job i had a TA assigned to me who was worse than useless, he just confused students and told them wrong things that i had to correct, and when i complained the department told me that the "job" was part of his aid package and was not negotiable

so i told him to just stay home and not talk to my students for the rest of the semester

Ben Steffan

@Thorgott I'm technically getting a sweet deal this term: because they had so many students and still decided to do oral exams they ended the course a week early so they start with examinations, plus since its oral exams I won't be involved with exams at all

in particular I will be paid a regular 7h a week this month when I've worked a literal 0h

leslie townes

ben were you my TA

haha

Ben Steffan

lol

Joe

16:45

Given a topological space $X$, we can define the category $\operatorname{Ouv}_X$ of open sets of $X$, with the morphisms simply being inclusions $U\hookrightarrow V$. We can then define a presheaf $\mathcal F$ on $X$ with values in $\mathcal C$ as simply being a contravariant functor from $\operatorname{Ouv}_X$ to $\mathcal C$.

My question is: is there a particular reason why we want this functor to be contravariant? Couldn't we just define a presheaf to be a (covariant) functor $\operatorname{Ouv}_X\to\mathcal C$?

Soumik Mukherjee

16:56

@Joe Consider the (pre)sheaf F that assigns to each open set U, the ring of continuous functions from U to R. Now if you have a inclusion of open subsets U \subseteq V then you have a morphism in the opposite direction F(V)->F(U) that is just the restriction homomorphism. Taking values in an arbitrary category instead of rings generalizes this idea.

psie

17:06

Consider $[a,b]\subset\mathbb R$. Don't you think it is strange that if $f^3$ is Riemann integrable, then so is $f$, whereas if $f^2$ is Riemann integrable, it doesn't follow that $f$ is? I think it is strange, because if I recall correctly, $L^p(\mu)\subset L^1(\mu)$ for a finite measure $\mu$ and $p>1$ and the Lebesgue integral coincides with the Riemann integral on $[a,b]$.

leslie townes

psie i am not sure it is helpful to think too long about this stuff but squaring destroys sign information, and riemann integration is very sensitive to sign stuff even on sets of measure zero. think of e.g. the indicator function of Q, minus 1/2.

at a super vague level, it is not that weird to me that something that destroys sign stuff might change integrability more than something that does not.

Thorgott

@Joe it's the variance that you see in most of the natural examples, both geometrically (as in Soumik's comment) and algebraically (see Yoneda lemma) speaking

leslie townes

also as a side note, the riemann integral being sensitive to stuff happening even on a set of measure zero is a strong signal that riemann integrability might not be that useful of a threshold for thinking about stuff.

psie

ok, I will not dwell on this too long :)

leslie townes

the riemann integral is certainly helpful when it makes sense because it links integration to something extremely intuitive, and it is rewarding when our intuitive thing coincides with the thing we are doing. but maybe also the riemann integral is not what anybody should be thinking of as 'the' integral

psie at a really vague level squaring on R is not one-to-one but cubing is, one thing 'destroys' 'information' and the other doesn't

thats the model in my head anyway

Ben Steffan

17:16

"What's geometry? It's when the functors are contravariant." :)

Thorgott

except when we have duality

duality is something that translates contravariant geometry into covariant geometry

psie

@leslietownes Ok 👍 did you ever do some numerical integration, out of curiosity? I think the Lebesgue integral has very little meaning there, but I may be mistaken.

Ben Steffan

"covariant geometry" you mean algebra? :^)

leslie townes

psie i do not think anyone would be doing numerical integration on something that is not riemann integrable. i do not think the outer limits of lebesgue integrability are at all relevant to anything. which sometimes makes me question the questions you ask on here if i am being honest :)

in my own mind, the point of the lebesgue integral is that it's a well behaved mathematical object that lets you identify something that you can later prove is riemann integrable

Thorgott

sheaf cohomology vs cosheaf homology, the algebra always wins

leslie townes

17:20

but i'm a crank, nobody should listen to me

Ben Steffan

but I'm a crank, I'm a weirdo
what the hell am I doing here, I don't belong here :(

4

leslie townes

to be clear i'm not saying that the riemann integral or its intuitive interpretations are useless. i'm saying that it's not useful to focus on the technical criteria for riemann integrability

Alessandro Codenotti

@Thorgott Chain complexes and cochain mplexes

Governor

17:38

Any ideas here? I want to show for $X$ a real-valued random variable with right-continuous CDF $F_X(x) = P(X \le x)$ satisfies:
$$P(F_X(X) \le \alpha) \le \alpha$$ I am trying to prove it using the quantile function but I'm getting a bit confused as it is bounded below by $\alpha$

My attempt was to show the event $\{F_X(X) \le \alpha\} \subseteq $ an appropriate set including a quantile function ideally $\{X \le Q^+(\alpha)\}$ where $Q^+(\alpha) = \sup\{x : P(X < x) \le \alpha\}$ but I can't justify it...

Thorgott

@AlessandroCodenotti lol

Governor

This seems quite helpful: math.stackexchange.com/questions/3763458/… but I still struggle to justify an appropriate bound

pie

18:18

I noticed that any inequity that involve $ab+bc+ac=1$ and the question is to prove that $\sum\limits_{cyc}\frac{\sqrt{f(a,b,c)}}{g(a,b,c)}\ge \sum\limits_{cyc}\frac{\sqrt{f(0,1,1)}}{g(1,1,1)} $ has never been answered on MSE (without using mathematica ) these question seem to be difficult but I wonder if there is some unified approach that can solve all of these problems since $(0,1,1)$ is always the minimum

[Minimizing $ \sum\limits_{cyc}\frac{\sqrt{5a+8bc}}{8a+5bc}$ with $ \sum\limits_{cyc}ab=1$](https://math.stackexchange.com/questions/4802066/minimizing-sum-limits-cyc-frac-sqrt5a8bc8a5bc-with-sum-limits)

[If $ab+bc+ca=1,$ prove that $\frac{\sqrt{5a+4}}{a+bc}+\frac{\sqrt{5b+4}}{b+ca}+\frac{\sqrt{5c+4}}{c+ab}\ge 8.$](https://math.stackexchange.com/questions/4790340/if-abbcca-1-prove-that-frac-sqrt5a4abc-frac-sqrt5b4bca?noredirect=1&lq=1)

[If $ab+bc+ca=1,$ prove $\frac{\sqrt{a+1}}{a+bc}+\frac{\sqrt{b+1}}{b+ca}+\frac{\sqrt{c+1}}{c+ab}\ge 1+2\sqrt{2}.$](https://math.stackexchange.com/questio…

I think there must be some general theorem that grantees the minimum is always at (0,1,1) when $f (a,b,c)=x_1a+y_1bc +z_1$ for all $x,y,z>0$ and $g(a,b,c)=x_2a+y_2bc +z_2$

Xander Henderson

18:33

@psie I don't find it strange at all. Cubing is injective and continuous, squaring is not. You kind of expect injective functions to preserve structures better than non-injective functions.

leslie townes

even if you do like an abstract sigma algebra setup for your integration, if E is not in that sigma algebra, (1_E - 1/2)^2 will be measurable although 1_E - 1/2 will not

User198

21:46

Is there a connection between a Lie derivative and a Material derivative? The second one being: ${\displaystyle {\frac {\mathrm {D} y}{\mathrm {D} t}}\equiv {\frac {\partial y}{\partial t}}+\mathbf {u} \cdot \nabla y,}$ ?

As I understand the two coincide when $y$ is a scalar function. Is that true?
What about if $y$ becomes a vector field?

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