Mathematics - 2018-09-13 (page 1 of 2)

anakhro

02:38

math.uwaterloo.ca/~cgodsil/html_notes/exlinalg.html

Under "Fun", question 1 about the Lie bracket.

Is this not wrong?

Basically says that $[[A,B]^2,C] = 0$ for all $2\times 2$ matrices.

I am trying to figure what it wanted to say.

Mike Miller

What's a counterexample?

anakhro

I used [1,2;3,4] and [5,6;7,8]

Mike Miller

There are three variables, what's your input?

I think the statement is true.

anakhro

For A, B

Mike Miller

I think [A,B]^2 is a multiple of the identity.

anakhro

02:47

For $[A,B]^2$ you then get $\begin{bmatrix}16&144\\144&16\end{bmatrix}$

Mike Miller

Hmm.

anakhro

Let me double check

Mike Miller

Wolfram gives me that too.

But I don't understand. I seem to be able to also prove that is impossible.

Aha!

[A,B] is [-4,12;12,4]. Now, when we compute that squared on Wolfram, it seems to be squaring the coefficients, not doing matrix multiplication.

anakhro

Why does wolfram do that

WHY

Mike Miller

No idea.

Spite?

anakhro

02:51

Spite it is.

Faust

MORNING

Mike Miller

Here was my proof. 1) Commutators are traceless. Therefore I can write the matrix [A,B] as [a,b;c,-a]. I squared that by hand and got $(a^2+bc) * I$.

Therefore $[A,B]^2$ commutes with everything.

Faust

i got a question

WTF IS AN INVERSE LIMIT

Mike Miller

Flip that limit upside down.

anakhro

Yeah mike, that was what I had and then I thought it was wrong because when I tested it in wolfram, it gives me nonsense.

Lesson is to never trust a computer

So goodbye proof assistants, wolfram put me off of anything remotely mechanical.

NO MORE SLIDE RULE

Mike Miller

02:56

Actually here is a less bashy proof.

First put it in Jordan form. It's still traceless and squares to a multiple of the identity.

Easier computation. One variable.

anakhro

Heh. That's a good one.

Ah, well thanks for finding out the issue, @MikeMiller

Talk to you later!

Faust

@MikeMiller no that definatly wasnt it but the arrow did go the wrong way

@MatheinBoulomenos get to do my C* algebras this semester going to die

Mithlesh Upadhyay

05:29

Please vote to undelete this post:

math.stackexchange.com/questions/1656686/…

This post has 10,920 times views and 2 years, 7 months old.

Mithlesh Upadhyay

05:45

Accidentally, I clicked on 'Mobile' button given in lower bar. Now, its showing display like in Mobile, how can I return to Desktop view ??

Solved, thanks.

Faust

06:26

cant see the post

Mithlesh Upadhyay

@Faust , you need 10,000 reputations to view and vote for a deleted post.

37

Q: How do you view deleted questions?

10k rep members are allowed to see deleted questions, is there a way to find questions that were deleted or to see a list of them somewhere? Thanks.

Pritt Balagopal

07:02

Hi, I am trying to learn Z-transforms, could you guys suggest any good reference for learning it? Textbooks or websites would be great

Manolis Lyviakis

07:31

differential geometry or measure is a must on ones knowlegde

measurre theory*

Tobias Kildetoft

@ManolisLyviakis Depends on what you want to do

Manolis Lyviakis

not very interested in either

i just have to pick one of those 2 for the semester

i tink i wanna ready algebra :p

Tobias Kildetoft

either one could be useful for algebra depending on the branch of algebra

Manolis Lyviakis

where does measure theory come up?

Tobias Kildetoft

In stuff like automorphism representations, where a lot of things are done by integrating over a group with respect to a suitable measure

Manolis Lyviakis

07:41

well i already know some differential geometry

i could just be better at that :P

Tobias Kildetoft

I would say that of the two, differential geometry is the one I most often wish I knew more about.

Manolis Lyviakis

yeah i think measure theory is too specific :p

i mean i can learn a thing or 2 when im in need

but im not really interested about its foundation. i may use it as a tool though

also what i mean by differential geometry it is not in the broad sense of manifolds. Its surfaces/curves and geodesics gaussian curvature fundamental forms etc.

Ash catcham

@LeakyNun how to show that every element in field is diffrence of two squares?

Leaky Nun

interesting question

if the field is not characteristic two, then $a = a \times 1 = \left(\frac{a+1}2\right)^2 - \left(\frac{a-1}2\right)^2$

I don't think it's true in $\Bbb F_2(t)$

is $t$ a difference of two squares in $\Bbb F_2(t)$?

Ash catcham

07:59

right thank you

Leaky Nun

what???

you asked me how to show

and when I doubted it you said right thank you?

so you were tricking me?

Ash catcham

for your argument for field of characterstic not 2

Leaky Nun

but that's not a complete answer

Ash catcham

actually I am interseted in field with characterstic not equal to two because I am dealing with bilinear form and quadratic form correspondance

Leaky Nun

...

whatever

Ash catcham

08:03

for characterstic 2 a^2+b^2=(a+b)^2 so a^2=(a+b)^2-b^2 so this may not seems true in characterstic 2

@LeakyNun sorry for mistake

Leaky Nun

I'm confused now, why does that mean that it's not true?

oh ok

right

we know that $t$ is not a square in $\Bbb F_2(t)$

Mary Star

We have the rings $R_1, R_2$, the elements $a,b\in R_1$, the ideal $I=(a)$ of $R_1$ and a ring homomorphism $f:R_1\rightarrow R_2$. It is given that $\ker f=I$. It holds in $R_1$ that $b\mid a$ but not that $a\mid b$.

If $b$ is not a prime in $R_1$ then is $R_2$ an integral domain?

For that we have to know if the image of a prime element is also prime or not?

Leaky Nun

so if $t=a^2-b^2$ then $t=(a+b)(a-b) = (a+b)^2$

thanks @Ashcatcham

that was interesting

Ash catcham

all right nice talking to you

ssk

08:38

I am learning about gaussian processes and stationary processes. Unfortunately, I am the only one learning this subject in my place. Is there anybody who is also currently studying these subjects and would like to discuss regularly.

usukidoll

10:47

Quick question. Can the zero vector be in the form of <0,0>

Tobias Kildetoft

@usukidoll What do the brackets mean there?

usukidoll

11:04

I was in an inner product space

<x, x> is $\mid x \mid^{2}$

@TobiasKildetoft

Tobias Kildetoft

@usukidoll then no, because the zero vector lives in a different place that the value of an inner product

Or do you mean whether the zero vector is determined by the fact that it has $0$ as inner product with itself?

usukidoll

0 like nothing

Tobias Kildetoft

I have no idea what that last sentence should mean

mick

11:22

0

Q: What is the value for $ 2 + 3 + 5 + 7 + 11 + 13 + 17 + ... $?

Consider the sum of all primes $$ A = 2 + 3 + 5 + 7 + 11 + 13 + 17 + ... $$ This is divergent , so we must use a summability method. What is the best value for $A$ ? I suggest using zeta regularization $$ \zeta(z) = \sum_n \space p_n \space n^{-z} $$ $$ A = \zeta(0) $$ Notice that the pri...

prime-numbers divergent-series zeta-functions analytic-continuation summation-method

Any ideas ?

Tobias Kildetoft

@mick What is the definition of a value being "best" for such a sum?

user 2646

11:35

best

Adjective: best

superlative form of good: most good.
(Can we date this quote?) William Shakespeare
When he is best, he is a little worse than a man.
(Can we date this quote?) John Milton
Heaven's last, best gift
Most; largest.

Adverb: best

superlative form of well: most well
(Can we date this quote?) John Milton
Thou serpent! That name best befits thee.
(Can we date this quote?) Samuel Taylor Coleridge
He prayeth best, who loveth best / All things both great and small.
At her invitation he outlined for her the succeeding chapters with terse military accuracy ; and what she liked best and best understood was avoidance of that...

Secret

12:56

I wonder if user 2646 and user 2236 are both skullpatrol

15 hours ago, by Alessandro Codenotti

Not having a geometric approach: a geometric approach

Rudi_Birnbaum

Are there any $\Bbb R^n \to \Bbb R$ versions of the Rolle theorem?

Secret

5

Q: Prove an analog of Rolle's theorem for several variables

On p. 135 of Buck's Advanced Calculus, he asks the reader to prove an analog of Rolle's theorem for functions of two variables (I suspect the number two is arbitrary). The hint is to assume that $f=0$ on the boundary of a bounded open set, and I belive what he's looking for is that if $f\in $ $C...

multivariable-calculus derivatives

You need more conditions on the boundary to ensure that

GENESECT

13:48

Can someone tell me how to see which of sinx or arcsin(sinx) is bigger from 0to π.... Any hint

The graph of arcsin(sinx) is your=x from 0 you π/2 and y=-x+2π but I can't seem to compare them

Semiclassical

@GENESECT first, you should note how both functions behave for x<pi/2 compared with x<pi/2. For instance, how are sin(x) and sin(pi-x) related?

GENESECT

sinx and sin(pi-x) are identical functions... right

Semiclassical

right.

that in turns means that arcsin(sin(pi-x)) and arcsin(sin(x)) are identical as well

GENESECT

Yea

Semiclassical

which means that if arcsin(sin(x)) > sin(x) for 0<x<pi/2, then that's also true for pi/2<x<pi

hence it's enough to look at 0<x<pi/2

in which case you're trying to show that x>sin(x) for 0<x<pi/2

GENESECT

13:58

Okay

Semiclassical

(In fact, x>sin(x) is true for all x>0. But you're only interested in 0<x<pi/2 for the present purpose.)

GENESECT

In 0<x<π\2, the graph of arc sinx is coming greater then because it's just y=x in that interval

Semiclassical

Sure. If you already have an argument that x>sin(x) on that interval, then you're done.

GENESECT

Got it. Thanks

Semiclassical

np

@AkivaWeinberger amusingly, I notice that you've got an answer regarding how to prove that inequality

I got a chuckle out of the last line :P

Akiva Weinberger

14:03

Can confirm

I was

Semiclassical

lol

most amusing typo I saw recently was somebody doing $\epsilon_{jam}$ instead of $\epsilon_{jkm}$

jam on

Akiva Weinberger

Oh wait

Is $\pi_4(S^2)=\Bbb Z_2$ and $\pi_2(SO(3))=\Bbb Z_2$ not a coincidence??

Or maybe the first one should be $\pi_4(S^3)$

Semiclassical

of course not. how could it be? (I don't actually know.)

Akiva Weinberger

(which is also $\Bbb Z_2$)

Ugh, stupid high dimensional visualizations

Erm, yeah it should be $\pi_4(S^3)$ I think

Semiclassical

I'm not sure what kind of mechanism one would look for to relate $\pi_4$ of one space to $\pi_2$ of another

Akiva Weinberger

14:10

but now my head hurts

Semiclassical

$\pi_4$ and $\pi_3$, I could understand

Akiva Weinberger

@Semiclassical It's an annoying sequence of pictures

Semiclassical

ew

Akiva Weinberger

Or, um

Let $x$ be the generator of $\pi_3(S^3)$

Semiclassical

if Balarka was around I'd bug him about it

Akiva Weinberger

14:11

Then an element of $\pi_4(S^3)$ is a weird sequence that goes from $0$ to stuffs and back to $0$

So morph $0$ into $x+(-x)$

the latter of which kinda looks like two spheres

er, hyperspheres

joined at a basepoint

Now rotate one of the hyperspheres 360 deg

(aka rotate its equator, which is a sphere, 360 deg

and the rest of the hypersphere rotates along with it keeping the basepoint stationary)

and then morph $x+(-x)$ back into $0$

Semiclassical

I'm going to just smile and nod

Nick

@Semiclassical welcome to the club..

@AkivaWeinberger Is that even close to a full rotation in 4D space?

Akiva Weinberger

Why wouldn't it be?

Nick

Rotations in hyperspace, now that's something I;ve never tried imagining.

Akiva Weinberger

Oh yeah they can be complicated

like you can rotated along two planes at the same time

so I don't know how you'd measure the degrees there

but here I'm not doing that

Nick

14:16

Rotations are inherently planar.

Like figure skaters on ice stay on the ice.

Akiva Weinberger

If your four axes are w, x, y, and z, try rotating on the wx-plane at the same time as rotating on the yz-plane

Nick

At most I'd say you can jump in the air and rotate along the plane which contains the line $z = sqrt(x^2+y^2)$

Akiva Weinberger

I recommend purchasing a 4D ball and playing with it

(/s)

Nick

In 4D, I can't imagine the figure skater "curving/slanting" that much.

Akiva Weinberger

In 4D, you can't imagine the figure skater

Nick

14:19

@AkivaWeinberger How so? It'll look like a normal 3D ball when its stopped at a phase passing through normal space.

@AkivaWeinberger You can imagine this 2D text in a 3D world, right? 3D objects in a 4D world is not much of a long shot.

I'm talking about applying 4th dimensional rotation on 3D objects, not just 4D.

Rithaniel

Quick question, but is $\{\o , \{a\}, S\} \subseteq \{\o , \{a, b\}, S\}$ ?

Hmmm, does the empty set symbol not work in chat? Or am I using the wrong code?

Abdullah UYU

Try \emptyset

Rithaniel

$\{\emptyset , \{a\}, S\} \subseteq \{\emptyset , \{a, b\}, S\}$

That works

user131753

@AkivaWeinberger: I think the following lemma is also true:

Akiva Weinberger

Also \varnothing $\varnothing$

$\emptyset\varnothing$

Rithaniel

14:26

Now, is this true? Or are $\{a\}$ and $\{a,b\}$ not comparable?

I think that they should be, but I want to be sure.

Akiva Weinberger

@Rithaniel It is not true.

user131753

"Let $(X,\tau_X)$ and $(Y,τ_Y)$ be topological spaces. Let $f,g:I→Y$ be two continuous functions. Then $f$ and $g$ are homotopic iff there is a path between $f(x)$ and $g(x)$ for all $x∈I$."

Akiva Weinberger

There exists an element of the first set that is not an element of the second

Rithaniel

Okay, thank you @Akiva

Abdullah UYU

Is $\{a\}\in \{\emptyset,\{a,b\},S\}$? @Rithaniel

Akiva Weinberger

14:27

$\{a\}$ is an element of the first set, but the second set only has the three elements $\emptyset$, $\{a,b\}$, and $S$

user131753

The lemma that we will need to prove this is as follows:

Akiva Weinberger

None of which equal $\{a\}$

unless $S=\{a\}$ I guess

@user170039 Does $f(0)=g(0)$ and $f(1)=g(1)$? Are they held constant throughout the homotopy?

Rithaniel

Nah, S is the whole set, so $S = \{a\}$ would imply that $\{a,b\}$ could not exist.

user131753

@AkivaWeinberger Nope.

user131753

Well let me first elaborate the lemma which I think will most likely be true (but I am lacking a proof of it).

Akiva Weinberger

14:30

Wait, so $S=\{\emptyset,\{a,b\},S\}$?

$S\in S$ isn't actually allowed in ZFC set theory

(Well, there exist other set theories that allow it but they're weird)

user131753

"Let $(X,τ_X)$ and $(Y,τ_Y)$ be topological spaces. Let $f,g:X→Y$ be two continuous functions. Then $f$ and $g$ are homotopic iff there is a path between $f(x)$ and $g(x)$ for all $x∈X$."

user131753

Sorry @AkivaWeinberger, see the above corrected version of the theorem that I want to prove.

user131753

The lemma is,

Rithaniel

Well, nah, I misspoke. These are topologies on a set S.

Akiva Weinberger

Ahh

So $S=\{a,b,\text{possibly }c\}$ or something?

user131753

14:34

"Let $(X,τ_X)$ and $(Y,τ_Y)$ be topological spaces. Let $f,g:X→Y$ be two continuous functions. Then $f$ and $g$ are homotopic iff there is family $\mathscr{F}:=\{\varphi_x:I\to Y\mid x\in X\}$ such that $\varphi_x$ is a path connecting $f(x)$ and $g(x)$ for all $x\in X$."

Akiva Weinberger

@user170039 That seems wrong. Consider $X=S^1$ (a loop) and $Y$ an annulus.

mercio

^

Rithaniel

Yeah, the task I'm assigned with is finding all topologies on $\{a,b,c\}$ and showing how they relate in coarseness/fineness.

mercio

if $Y$ is path connected there always is a path between $f(x)$ and $g(x)$

which would make every function homotopic to each other according to your lemma

user131753

Yes. Ok.

user131753

14:36

Very silly mistake.

Akiva Weinberger

2 mins ago, by user 170039

"Let $(X,τ_X)$ and $(Y,τ_Y)$ be topological spaces. Let $f,g:X→Y$ be two continuous functions. Then $f$ and $g$ are homotopic iff there is family $\mathscr{F}:=\{\varphi_x:I\to Y\mid x\in X\}$ such that $\varphi_x$ is a path connecting $f(x)$ and $g(x)$ for all $x\in X$."

Basically, you need it so that, if $x$ is close to $y$, then $\varphi_x$ is close to $\varphi_y$, in a sense.

And that's what the definition of homotopy makes rigorous.

Rithaniel

I found nine types, which I think is all of them, with a total of 26 total topologies among these types.

Akiva Weinberger

$\varphi_x(t)$ is essentially $h_t(x)$.

(or $h(t,x)$, where $h$ is the function from $I\times X$ to $Y$)

user131753

@AkivaWeinberger Is it? Here when you are defining $\varphi_x$, your $x$ is fixed but when you are defining $h_t$, your $t$ is fixed.

user131753

@AkivaWeinberger Sorry, but I don't understand. Can you elaborate?

Akiva Weinberger

14:40

$\varphi_x(t)$ and $h_t(x)$ are points, not functions

user131753

@AkivaWeinberger: I meant this comment.

Akiva Weinberger

@user170039 Perhaps it's useful to think about a specific counterexample

Here, $Y$ is an annulus, and $X$ is a circle

I've drawn the images of two maps from the circle to the annulus

These maps are not homotopic

However, every point is path-connected to every other point.

What went wrong?

Think about what it would look like if you drew a path from $f(x)$ to $g(x)$ for every $x$

There would be two points on $S^1$, call them $x$ and $y$, such that $x$ and $y$ are really close to each other (and thus $f(x)$ and $f(y)$ are close to each other), but $\varphi_x$ is a path that goes one way around the hole and $\varphi_y$ is a path that goes another way around the hole

Manolis Lyviakis

nice example

Akiva Weinberger

Wish drawing and posting images was more efficient

I think I have tons of random math drawings on my Imgur account

Manolis Lyviakis

and its kinda mandatory when explaining algebraic topology

Akiva Weinberger

14:50

I wonder if that's public

Manolis Lyviakis

the question is why the first curve is not homotopic to S1?

Akiva Weinberger

Also, technically those both should have orientations but whatever

Like I could have made it so that $f$ is a loop that goes clockwise around the hole and $g$ is a loop that goes counterclockwise

Manolis Lyviakis

isnt the problem that the homotopy wont be able to map the self intersection?

Akiva Weinberger

No, the self-intersection is fine

Manolis Lyviakis

ohh

Akiva Weinberger

14:56

For example, the maps represented by the bottom two images there are homotopic

Manolis Lyviakis

ohh i get it

Akiva Weinberger

Let me share another drawing I made yesterday but with no context

(Completely unrelated)

(Uh, those are supposed to be surfaces in 3D space but I can't really draw them)

(Surfaces of revolution across the y-axis)

Abdullah UYU

A drawing tablet may be useful for you @AkivaWeinberger

With an appropriate software that automatically uploads to a image hosting service like imgur.

That can be interesting as a workflow.

Akiva Weinberger

As opposed to the iPhone I'm currently using

That's a thing I might get just 'cause it sounds cool

(and then almost never use)

Abdullah UYU

:) Actually, I'm saying because I have the dream of such a workflow.

Requires a bit of coding though.

But yeah, don't need to be whimsical for no reason :)

Akiva Weinberger

15:09

Or you can draw on a whiteboard and photograph it with an iPhone scanner app

Ex:

scribbles

(scanner app converts anything dark enough to pure black and anything light enough to pure white)

(If this were something like Discord, I wouldn't even have to upload to Imgur, I could just post from my camera roll directly)

user131753

@AkivaWeinberger I can feel it but can't rigorously formulate it.

Abdullah UYU

That's also a good solution.

user131753

It's clear that the notion of such a family as I have mentioned earlier is strictly weaker than that of homotopy.

user131753

But I can't understand what conditions I will need to make it equivalent to homotopy.

user131753

I am particularly interested in this kind of definition because of it's nice intuitive appeal.

Akiva Weinberger

15:26

You need $(t,x)\mapsto\phi_x(t)$ to be a continuous function of two variables

user131753

@AkivaWeinberger Sorry, but can you be more precise?

user131753

I mean, I need a function $f:I\times X\to Y$ such that...

Akiva Weinberger

The function $h:I\times X\to Y$ defined by $h(t,x)=\varphi_x(t)$ (with $\varphi$ being defined as you did earlier) must be a continuous function

($h$, $f$, same thing)

Incidentally, a note on functions on two variables

Say we have a function $f:A\times B\to C$

eg $f(a,b)=c$

We can also write this as a family of functions

$f_{a,-}$ defined by $f_{a,-}(b)=f(a,b)$

and as another family of functions

$f_{-,b}$ defined by $f_{-,b}(a)=f(a,b)$

(That's not standard notation but it's the best thing I can think of)

The former is functions you get by taking $a$ constant, the latter is functions you get by taking $b$ constant

$f$ is continuous iff $f_{a,-}$ are continuous for all $a$ and $f_{-,b}$ are continuous for all $b$

---So the upshot is, for your case

your thing failed 'cause you only required $\varphi_x$ to be a continuous function for each $x$ (that maps $t$ to $\varphi_x(t)$)

but you also need the functions that map $x$ to $\varphi_x(t)$ to be continuous for each $t$

user131753

@AkivaWeinberger So I also need a function $f_t:X\to Y$ defined as $f_t(x)=\varphi_x(t)$ to be continuous. Am I right?

Akiva Weinberger

Yeah

user131753

15:39

I see.

user131753

But how can then we prove the modified lemma?

Akiva Weinberger

@user170039 Oh, no, I made a silly mistake

A very silly mistake actually

user131753

No problem.

user131753

Just think of my earlier mistake.

Akiva Weinberger

$f:A\times B\to C$ is not necessarily continuous, even if its continuous when we keep one variable constant

So it's not actually enough that $f_t$ and $\varphi_x$ are continuous

You really do need to define a map $f:I\times X\to Y$ by $f(t,x)=\varphi_x(t)$ and say that it's continuous

(I hope you remember how the topological space $I\times X$ is defined, incidentally)

user131753

15:45

But when we give the equivalent definition of homotopy using the parameter $t$ why there is no such dependence?

Akiva Weinberger

What equivalent definition do you mean?

user131753

"A homotopy between $f,g:X\to Y$ is a family of continuous maps $\{\varphi_t:X\to Y\}$ such that $\varphi_0\equiv f$ and $\varphi_1\equiv g$."

Akiva Weinberger

Yeah, that isn't enough.

Otherwise you could just define $\varphi_t\equiv f$ for $t\le\frac12$ and $\varphi_t\equiv g$ for $t>\frac12$.

In other words, $\varphi_t$ would "jump" suddenly from being $f$ to being $g$.

We need to prevent that from happening.

And the way we do that is to demand that the function $I\times X\to Y$ that maps $(t,x)\in I\times X$ to $\varphi_t(x)\in Y$ be a continuous function.

user131753

Here the definition is explicitly written.

user131753

I think I understand now.

Akiva Weinberger

16:05

The key phrase is "continuously depending on a parameter $t\in[0,1]$" which is kinda vague

user131753

@AkivaWeinberger Yes. That's what was the reason for my confusion.

user131753

So, how would a modified lemma (of the version which I wanted to prove initially) look like?

user131753

I guess something like the following,

user131753

Let $(X,τ_X)$ and $(Y,τ_Y)$ be topological spaces. Let $f,g:X→Y$ be two continuous functions. Then $f$ and $g$ are homotopic iff there is family $F:=\{φ_x:I→Y∣x∈X\}$ such that $φ_x$ is a path connecting $f(x)$ and $g(x)$ for all $x∈X$ and...(what would be a precise formulation of the additional property @AkivaWeinberger?)

Akiva Weinberger

$(t,x)\mapsto\varphi_x(t)$ is a continuous map from $I\times X\to Y$

You can also phrase it in terms of sequential continuity:

If $t_n$ is a sequence of points in $I$ such that $\lim_{n\to\infty}t_n=t$, and $x_n$ is a sequence of points in $X$ such that $\lim_{n\to\infty}x_n=x$, then $\lim_{n\to\infty}\varphi_{x_n}(t_n)=\varphi_x(t)$

user131753

16:17

Ok. Thank you very much @AkivaWeinberger, for having such a long discussion with me. It was really helpful.

user131753

By the way, I was wondering, does there exist any topology on $C(I,Y)$ with the help of which we can define homotopy between $f$ and $g$ as a continuous map from $X$ to $C(I,Y)$ satisfying some additional properties @AkivaWeinberger? (Here $C(I,Y)$ denoted the set of paths from $f(x)$ to $g(x)$.)

Jendrik Stelzner

I’ve just received the Taxonomist badge (creating a tag used by 50 questions) for the tag "expected-value", but have never created a tag before (as far as I remember) nor have anything to do with probabilty theory. Is this a software mistake, or am I missing something?

Akiva Weinberger

Compact-open topology

In mathematics, the compact-open topology is a topology defined on the set of continuous maps between two topological spaces. The compact-open topology is one of the commonly used topologies on function spaces, and is applied in homotopy theory and functional analysis. It was introduced by Ralph Fox in 1945.If the codomain of the functions under consideration have a uniform structure or a metric structure then the compact-open topology is the "topology of uniform convergence on compact sets." That is to say, a sequence of functions converges in the compact-open topology precisely when it converges...

^I've heard that phrase thrown around on this chat but I don't know much about it

Hopefully it does what you said

@user170039

Adarsh Kumar

16:36

can anyone please help me

please ?

many users suddenly down voting one of my question. I think a lot about that question and that I know it will help in future research also. I request you all to see it once and help me. Its my idea and I don't want it lose. help me

the question is math.stackexchange.com/questions/2888642/…

rschwieb

16:52

@AdarshKumar Help you what? A net score of -3 is a piddling negative, and people can use their downvotes as they see fit. There aren't even any close votes. I don't see what the problem is. It sounds like you're begging for sympathy upvotes?

MatheinBoulomenos

@AkivaWeinberger @user170039 the compact-open topology does indeed do the job. Let's assume everything is locally compact and Hausdorff, then one has $C(X \times Y,Z) \cong C(X,C(Y,Z))$ for three spaces $X,Y,Z$ (this is called the "exponential law"), so you get that every continuous map $X \times I \to Y$ corresponds to a continuous map $X \to C(I,Y)$

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