Mathematics - 2023-02-18

Koro

00:04

here is another solution: math.stackexchange.com/a/3583697/266435

But I don't understand how they created H.

What is the intuition behind it? Is there any at all?

Thorgott

the intuition is in understanding what the picture means

it is, as has been pointed out, not a category-theoretic diagram

it's an actual square ($I\times I$), which you should imagine as domain of $F$

Koro

I understood that. But why is 3st coming in the definition of H?

Why is s-st coming in the definition of H?

Thorgott

I don't care to look at the formula. Do you have a visual understanding of the homotopy? If no, worry about that. If yes, it's just a matter of converting that intuition into formulae.

Koro

I don't understand what you mean by visual understanding of that.

Thorgott

pretend that $F$ is the identity on $I\times I$ (then argue this suffices). can you imagine a homotopy from $\alpha$ to $\gamma\circ\beta\circ\delta^{-1}$ inside $I\times I$ that keeps the two endpoints fixed?

there's a classic analogy for how to visualize this. imagine $I\times I$ as a solid floor or something. take a string of rubber and fix it into the floor at the two endpoints with some nails. now, you can drag around the string of rubber across the surface (we imagine it is infinitely stretchable and such), but of course you can't move its ends as they are fixed. this is the relation of homotopy rel end points.

Xander Henderson

00:19

Koro

@Thorgott $H(x,t)=\alpha (x) (1-t)+ \gamma\circ \beta\circ \delta^{-1} (x) t$

convex combination

Silly Goose

00:38

Is the set of all polynomials of the form $x^n$ finite? Or is it countable :P

I feel like it should be finite… since a polynomial of uncountable degree doesnt really make sense to me but i dont have an actual reason to believe one way or the other

I guess if $n$ is a natural number then it is uncountable

shintuku

@SillyGoose you mean polynomials like $x^{1.5}$, $x^{3}$, $x^{\sqrt{2}}$?

that set is uncountable

if the exponent is a natural number or a rational number, then the set is countable

but why would you think it is finite?

robjohn

02:47

@SillyGoose Each polynomial is of finite degree, but there are an infinite number of polynomials. For example, the set $\left\{x^n:n\in\mathbb{N}\right\}$ is infinite and each element has finite degree.

leslie townes

03:37

the map sending $n$ to $x^n$ is probably a bijection. it would be under usual definitions of 'polynomials' (e.g. as formal expressions in the "variable" x, or as functions on a set like $\mathbb{R}$). you might run into finiteness issues if a "polynomial" is defined, extremely unusually, to be a function on a finite field or ring whose rule is given by a polynomial formula.

Koro

05:51

0

Q: Constructing a homotopy in a square

Let F:$I\times I\to X$ be continuous, $\alpha, \beta,\gamma, \delta$ be paths in $X$ as below Thus, $\alpha(t)=F(t,0), \beta(t)=F(t,1),\gamma(t)=F(0,t), \delta(t)=F(1,t)$. Then it is to be proven that $\alpha\simeq \gamma\circ\beta\circ \delta^{-1}$ rel {0,1}. Here $\circ$ represents concatenatio...

Can anyone please help me with this? Thanks.

Koro

07:13

@Jakobian @AlessandroCodenotti: can you please advice how to construct the homotopy here?

I've been at it almost whole night but still no progress :(.

robjohn

08:00

@Koro Did you look at Xander's image?

Koro

08:12

@robjohn I did. But I don't understand how to transform it into a homotopy.

I don't understand how the picture helps.

@Berci: Can you please help me understand how the picture will help? I have posted the same question with picture here but still I don't understand how to write the homotopy. Can you please help me with that? Thanks. — Koro 1 hour ago

I looked for videos/websites which talk about homotopy but pretty much every one of them is following Hatcher's, and this exercise is not from Hatcher so it's not discussed there. Usually, just a path homotopy is defined and it is shown that if $f\simeq f'$ rel {0,1}, $g\simeq g'$ rel {0,1} and if the product fog is defined then $fog\simeq f'og'$ rel {0,1}.

There are no examples/illustrations how to create homotopy for example in my case.

I did not find the way to create the homotopy in Munkres' book also.

In my post, I have also linked to two solutions of the exercise but none of them explains how they created the homotopy. It's like: 'Define H(x,t) like this and verify that it is the desired homotopy.'

So that makes me feel like the creation of homotopy is done basis hit and trial, and there is no actual way to construct it. If it is so, I just want a confirmation of the same so that I can memorize the solution and move on.

pourjour

09:13

Hi everyone

I need some help to make an kind of an algorithm

can someone help me

I can show a graph with what Iwant

after that I want to write an algorithm in python

ILikeMathematics

10:32

Some people say the best way to prepare for a course is to use the textbook used in that course, but does using a different textbook on the same topic make a noticeable difference?
Say the course used Rudin's Principles of Mathematical Analysis and you prepared for it using Tao's Analysis, would it really be worse than if you used Rudin's Analysis from the beginning?

The advantage of using Rudin from the start is that the problems you work through will be the same as in the course, so you will probably be able to solve them quickly in the course, but is there another advantage? Does the ma…

pourjour

10:45

Here's an example of what I want

uhoh

11:15

2

Q: What are "transcritical, pitchfork, period-doubling, torus, and subharmonic bifurcations" in the context of three-body orbits? (e.g. butterflies)

MattB's excellent and well-sourced answer to Where do the butterflies land on this bifurcation plot? (Earth-Moon three-body butterfly orbits) begins: I'm fairly certain that the butterfly family is not present in this diagram, sadly. The butterfly family originates from a period-doubling bifurca...

orbital-mechanics mathematics halo-orbit three-body

added +100 bounty

mick

12:16

0

Q: When does this commutative non-associative algebra have nilpotent elements?

Consider a non-associative commutative unital algebra of finite dimension where the product is defined by a Cayley table such that elements are generated with real number coefficients $(a_0, \dotsc, a_n) $ for a basis $\{1, i_1, \dotsc, i_n \}$. If $a,b,c,d$ are elements of the algebra then $ab =...

abstract-algebra commutative-algebra nilpotence universal-property nonassociative-algebras

Koro

14:21

@Jakobian the same question that we were discussing yesterday :-).

0

Q: Homeomorphism between sets of $\Bbb R^3$

Let $\Bbb R^3$ be a topological space with usual topoloy and $\Bbb Q$ denotes the set of all rational numbers. Define the subspaces $X$, $Y$, $Z$ and $W$ as follows: $$X=\{(x,y,z)\in \Bbb R^3 :|x|+|y|+|z|\in \Bbb Q\}$$ $$Y=\{(x,y,z)\in \Bbb R^3 :xyz=1\}$$ $$Z=\{(x,y,z)\in \Bbb R^3 :x^2+y^2+z^2=1\...

real-analysis general-topology analysis

The Real Masochist

15:36

Is the following question clear enough for most people?

When I read this question, my solution is $26 \choose 3$.

Koro

15:57

Koro

16:07

Here $c_p$ represents the constant map at $p$ (that is, it takes I to p in X). It is to be shown here that $f\simeq c_p\circ f$. Using this picture, they defined $H(s,t)=\begin{cases}p; t\ge 2s\\ f(\frac{2t-s}{2-t}); t\le 2s\end{cases}$. I have no idea how they defined $H(s,t)$ for the case $t\le 2s$ using the picture. How does the picture help here in constructing H?

Thorgott

for starters, the variables in your denominator are switched

think about it slice-wise, the path $H_t$ at time $t$ should be constant with value $p$ for $s\le t/2$ and then a rescaled version of $f$ for $s\ge t/2$. note $s=t/2$ is precisely the bold line in the picture. the fraction is just the result of making that rescaling explicit.

Koro

That was a typo. It should be $f(\frac{2s-t}{2-t})$. But still, the question stays. :(

Thorgott

16:23

which part of my explanation is unclear?

Koro

the part for $s\ge t/2$.

Why is it specifically $f(\frac{2s-t}{2-t})$?

Thorgott

no highly particular reason, piecewise-linear maps are just very easy to work with (both theoretically and algorithmically), so we tend to use them

I mean, we just want a family of continuous maps $[t/2,1]\rightarrow[0,1]$ mapping $t/2\mapsto0$, $1\mapsto1$ and that is continuous in the parameter $t$. There isn't really anything easier you can do than just use the unique piecewise-linear maps satisfying this.

In a practical sense, too. Once you understand that first sentence is what you wanna do, you can just write down a system of two linear equations and explicitly solve for the coefficients of the piecewise linear map, which yields $\frac{2s-t}{2-t}$.

But the explicit form really is not important at all relative to the idea.

soupless

18:15

Hello. I want to ask if there's a formula for $\tau(xy)$ when $\gcd(x,y) \neq 1$.

Koro

I don't understand how you get [t/2,1]-->[0,1]. :(

I can't bounty the question until tomorrow.

Thorgott

leslie townes

all of that piecewise linear stuff is more or less dictated by, lets carve this parameter space into pieces that do __ continuously but have __ result at various endpoints or edges. there may be arbitrary choices within that.

Thorgott

The concatenation of the red and blue part is the path $H_t$. On the red part, it is supposed to be constant with value $p$. The blue part is defined on the interval from $s=t/2$ (the slanted diagonal in green) to $s=1$ (the right edge in yellow) at height $t$. There, the path is supposed to do what $f$ does, just rescaled, so we have to rescale $[t/2,1]$ to $[0,1]$.

D.C. the III

una buena yerba mate para comenzar el dia de estudios.

leslie townes

18:33

do you have the whole setup? the metal straw?

D.C. the III

leslie townes

that's legit

D.C. the III

What I'm truly missing is one of the gourds made out of a calabosh

leslie townes

i wasn't going to point that out, but yes

D.C. the III

but I chose not to get one because it is wood so it cna build mildew

and I'm not into the extra level of service that would require

leslie townes

18:39

when i lived in iowa city one of the local cafes made a pretty decent mate (they put honey in but you could tell them not to), i was basically always in there fueling up

D.C. the III

Iowa city would not be where I'd imagine you'd get your mate introduction. But it does wonders for focus and as you know isn't as jittery as coffee but gives you the "zoned in" buzz

Next step is to get the gourd and one of those mate carriers. Then I've reached peak Argentine

I used to make them with some sweetener, but I toughned up and now drink it straight. ALso I learned how to prepare it better too. The difference between putting boiling water on the leaves vs putting on water that is around 90 degrees celsius is night and day

Sine of the Time

chat is boring without ted

Koro

18:55

I think that Ted hasn't been here today.

scrolling back, I see only my comments from last night.

Sine of the Time

@Koro are you enjoying algebraic topology?

Koro

Yes, I am.

Sine of the Time

I prefer general topology

Koro

I have been stuck at understanding creating homotopy using pictures for about 1 and half day now.

and still no progress

@SineoftheTime I like that too :-).

Sine of the Time

@Koro It's not immediate, but it's important to understand because it the core part of the fundamental group

Koro

18:59

I'm inclined towards analysis , topology. I am not so much into abstract algebra -modules, rings etc.

Sine of the Time

@Koro 🤝. I love analysis

Koro

And I don't care about Sylow theorems or finding if a group of order so and so exists or not. 😅

4

Sine of the Time

@Koro sure, I like in general theoretical subjects like algebra, geometry and analysis, but I definitely prefer the analysis

Koro

I like linear algebra too so I am finding functional analysis to be amazing as well. :-)

Sine of the Time

yes, linear algebra is good. I think my favourite course will be functional analysis, but I'm still in my second year so I've to wait 2 years

Koro

19:05

the fun part in functional analysis is: often we have to show one space to be isometric with some other, and we define a map. Now, to show that the norms are preserved, one shows inequalities from both sides. One can often show one inequality easily but the reverse inequality is often tricky.

Proving the reverse inequality sometimes requires Hahn Banach or sometimes open mapping theorem etc.

Sine of the Time

I've had a taste of functional analysis when we proved Picard Lindelof theorem

Koro

when I complete my masters, I'll study inequalities. I like them so I especially like the part of functional analysis where we show isometries by showing inequalities from both sides.

Sine of the Time

I still don't know what to study, but maybe armonic analysis and EDP

Koro

@SineoftheTime is it the one that one usually meets in ODEs?

Sine of the Time

@Koro yes, but we did in calc 2

The subjects are different

Next semester I've the course of ODE

Koro

19:31

@Thorgott thanks a lot. I think I understand it now. :-)

When I saw $H_t$ there, I understood it :-).

Koro

19:57

still no idea about this one though

14 hours ago, by Koro

0

Q: Constructing a homotopy in a square

Let F:$I\times I\to X$ be continuous, $\alpha, \beta,\gamma, \delta$ be paths in $X$ as below Thus, $\alpha(t)=F(t,0), \beta(t)=F(t,1),\gamma(t)=F(0,t), \delta(t)=F(1,t)$. Then it is to be proven that $\alpha\simeq \gamma\circ\beta\circ \delta^{-1}$ rel {0,1}. Here $\circ$ represents concatenatio...

general-topology algebraic-topology homotopy-theory

PM 2Ring

20:38

@D.C.theIII Here's something sweet from Argentina

Nos Siguen Pegando Abajo - Charly Garcia (cover de Pacifica)

►

Koro

0

Q: Understanding how to obtain homotopy from a picture.

Suppose that we are given two paths $f:I\to X$ and $f^{-1}(t):=f(1-t)$. Let $c_p$ be the constant map at $p\in X$. It is to be proven that $c_p\simeq f\circ f^{-1}$, where $\circ$ represents concatenation of paths. The book gives the following diagram. Then somehow concludes from it that the des...

general-topology algebraic-topology homotopy-theory

D.C. the III

21:13

@PM2Ring This is more my speed 😃