Mathematics - 2022-11-25 (page 1 of 2)

shintuku

00:05

hot take

studying for law is combinatorics

leslie townes

i agree

shintuku

only way to get good grades is to find strange patterns in the content

there's just too much info

at most you'll have time to review it once

so, given a power point slide and a concept and it's characteristics

you only got once shot

(do not miss your chance to blow)

cough sorry

you only got once shot to memorize it for a long amount of time

so a concept and it's characteristics is a puzzle: what's the way to memorize the whole thing without individually memorizing the elements

leslie townes

01:57

i agree

shintuku

this is why the concept of the "administrative function of the public administration understood as an entity" is solved by a concrete bridge sustained on a tripod-like pillars, stabbed into a teat-like terrain elevation with spanish bullfighters running on it

it is permanent, unmediated and material, and we notice the conjunction of "ma" and "me" which are alike in these characteristics, which remind of the french word "mamelle", thus the tripod structure on the terrain of that shape

generally spontaneous, where "pont" is a bridge in french, and bullfighters because... they're spanish, so that explains the "s" in front of "pont"

and well it is a concrete bridge because... the administrative function is concrete

i feel like i'm using controverted mnemonics like Michael's in the Office, but in principle the image is way faster to manifest than this whole explanatory text

let no one doubt that this is hard science

shintuku

03:02

my room is full of obscene scenes with stick figures

leslie townes

04:05

i think that's enough peyote for one day

PM 2Ring

04:38

@Ajay If $n|m$ then $F_n|F_m$. $F_3=2$, so every 3rd Fibonacci number is even. $2+\frac{F_{3k}/2}{F_{3k+3}/2}$ is a good approximation to $\sqrt5$

Dyslexia is fun...

Eg, $(2+\frac{72}{305})^2 + \frac1{305^2} = 5$

@shintuku Reminds me of administratium

PNDas

04:58

A path connected space is 0-connected. Right?

Not Tfue

This is the only nobel prize I can hope for lol

I don't want to ruin the joke so removed

(removed)

Silly Goose

are zero divisors not allowed in the denominator of elements in fraction rings because they cause distinct elements to be mapped to 0?

Ted Shifrin

05:15

Can a zero divisor have a multiplicative inverse?

Not Tfue

samperson.itch.io/desktop-goose

Ted Shifrin

For example, @SillyGoose, can there be $x\in \Bbb Z_6$ with $\bar 2x= \bar 1$?

copper.hat

05:39

Happy thanksgiving

I think tonight was our first really traditional Thanksgiving in may years, as in turkeym stuffing, etc.

leslie townes

i asked my daughter what she was thankful for and she said she wasn't thankful for having to go to day care. so i changed the subject to complaints about the previous year. it was a lot of fun actually

Ted Shifrin

Did you win the most complaints?

leslie townes

no, munchkin did.

Ted Shifrin

06:07

I meant — hers about others.

leslie townes

oh, the cat won. in fact, it turned into a collective airing of grievances about the cat, who happened to be sitting on the dinner table at the time.

Ajay

06:36

@copper.hat my mom cooked a full thanksgiving lunch. I got to stuff the turkey, hehe.

Ajay

06:48

jokes aside, it was a lovely meal.

shintuku

07:14

while ye celebrate and Thank, I have become Pain

student of Law. I have prostituted my well-being, forsaken my sanity, all for what?

about three fiddy credits? These salacious drawings of stick figures engaging in the most indecent acts of debauchery?

no, this will not do, I shall not Thank

although stuffing turkey probably has some sweet mnemonic potential

one potato two potato

07:34

Any holomorphic map preserves the orientation?

Lukas Heger

@onepotatotwopotato yes

one potato two potato

@LukasHeger How can I show that?

Lukas Heger

Complex matrices have positive determinant when considered as real matrices

More precisely, det over R is | det over C|^2

If you want an abstract argument: GL_n(C) is connected

And the real determinant maps into R \0

So the image is connected

Thus it can't contain negative numbers

Calvin Khor

07:51

im not sure if this should surprise me but north korea has a math department. first author here has listed address "Institute of Mathematics, State Academy of Sciences, Pyongyang, Democratic People's Republic of Korea"

Not Tfue

they need icbm they need math

google.com/amp/s/www.upi.com/amp/Top_News/World-News/2019/07/22/…

Koro

Hi @CalvinKhor!!

Not Tfue

just curious

one potato two potato

To steal bitcoin they anyway need to study math so there should be a math dept.

Calvin Khor

@Koro yo, long time no see, how you doing? :)

@onepotatotwopotato yes lol but still

I was stunned for a moment

Koro

08:03

@CalvinKhor I'm doing alright. How about you? Indeed, it's been long since you were here.

Your room is also not visible in the list of rooms anymore.

Calvin Khor

p alright :) yeah the room is frozen lol

Not Tfue

Steal bitcoin?

Koro

I have one question: Suppose that (F,f) is a free module on set S, then f is 1-1.
Proof: $f:S\to F$. Suppose f(x)=f(y). I'll show that x=y. Let (S) be the module generated by S, and let in:$S\to (S)$ be the identity. So by definition of free module, there exists a morphism $h:F\to (S)$ such that hof= in. This implies that in(x)=h(f(x))=h(f(y))=in(y), whence x=y. QED.

Calvin Khor

@NotTfue whaaaaaaaaat

Koro

Can anyone please let me know in case there is anything missing in this? I ask because the book gives a very different proof.

Not Tfue

08:07

@CalvinKhor you saw it right

Calvin Khor

08:19

I don't believe it, its @NotTfue

@Koro whaddaheckisa free module

Koro

I have posted my question on MSE here and the definition is included there.

@CalvinKhor

Calvin Khor

I think it would be good to have "I ask because the book gives a very different proof." as the last sentence in your Q

@Koro

Koro

I wanted to ask this in DaRT room too but it seems that it is only for ring theory.

I removed that part from the post.

To show that f is injective, let x, y in S be such that $x\ne y$; we have to
show that $f (x) \ne f ( y)$. For this purpose, let M be an R-module having more than
one element (e.g. R itself will do) and choose any mapping g : S --> M such that
$g(x) \ne g( y)$. Let $h : F \to M$ be the unique R-morphism such that $h\circ f = g$. Then
since $h[ f (x)] = g(x) \ne g( y) = h[ f ( y)]$ we must have $f (x) \ne f ( y)$.

Here is the the proof in the book.

I think that my proof is also correct.

Ajay

09:59

@CalvinKhor Hello! Remember me?

Koro

11:02

Given than (F,f) is a free R module on S. If there is a unique R morphism j: F--> F' s.t. jof=f', then how to show that (F',f') is also a free R-module?

The problem is that: j is not given to be invertible.

someoneinexistence

11:21

Hello!

one potato two potato

11:37

0

A: Stein complex analysis exercise 4.12

I looked up this problem again today and I found that it's not a very hard question. I think I'm misunderstanding something when I asked this question. From the post, I concluded that $$|F(z)| = |g(z)e^{\gamma z}| = |g(z)|e^{\operatorname{Re}\gamma z} = ce^{\pi|z|(1-\epsilon_{\theta})},\quad \eps...

Not a very hard question. I don't know why I asked at that time.

someoneinexistence

Ello

Thorgott

@Koro not true

you very much want to assume j invertible

Koro

@Thorgott thanks a lot :-).

I guess that's a typo in the book then.

It is to be noted however that the book does assume invertibility of j in the proof.

Thorgott

note that, by definition, there is always a unique R morphism j s.t. jf=f' for any f', so if that claim was true, youd be saying any pair (F',f') is a free module and that's absurd

one potato two potato

11:57

1000

Koro

@onepotatotwopotato mine is visited 1105 days, 62 consecutive.

my consecutive streak broke and I don't know why. Else it would be more than 1 years consecutive!!

Aug 20 at 22:18, by Koro

At of now, I visited mse 1009 days, 547 consecutive.

@Thorgott yes :-).

one potato two potato

12:56

If $f$ is holomorphic on the strip $0<Im(z)<1$ continuous up to the boundary and $f$ is bounded on the boundary $Im(z) = 0,1$ then $f$ is bounded on the entire strip?

No the question statement is stronger

one potato two potato

13:12

Oh, the question is assuming $f$ is bounded. Ignore the question.

PNDas

13:27

I want to use this result (planetmath.org/everymapintospherewhichisnotontoisnullhomotopic) to prove that $\pi_1(S^n)=0$ for $n\ge2$.

But to do that I need to show that $[0,1]\to S^n$ loops are not surjective. How to argue that?

Lukas Heger

there are surjective loops $[0,1] \to S^n$

one can prove that every surjective loop is homotopic to a non-surjective one, but that is difficult

one could proceed e.g. by the smooth approximation theorem in differential topology

if you want to show that $S^n$ is simply connected for $n \geq 2$, it's easier to use van Kampen

PNDas

okay that's why they used that partition lemma

@LukasHeger hmm but I was trying to use the result given in the link.

@LukasHeger in algebraic topology do we assume loops to be not self intersecting? Also the surjective loops you are saying, are they self-intersecting?

anyway I'll just use Van Kampen theorem.

Lukas Heger

loops are allowed to self-intersect, the maps $S^1 \to X$ need not be injective

the surjective loops are non-injective yeah

PNDas

okay

Lukas Heger

maybe this is just due to my ignorance in topology, but I know of no easy way to prove directly that every map $S^1 \to S^n$ is homotopic to a non-surjective map

maybe @Balarka knows this

this follows from some big theorems, such as smooth approximation, cellular approximation or of course if you just compute $\pi^1$ via van Kampen

PNDas

13:39

staff.ustc.edu.cn/~wangzuoq/Courses/21F-Manifolds/Notes/…

see corollary 1.7

Lukas Heger

this is exactly what I said!

I said use smooth approximation lol

PNDas

hmm I was just giving a reference.

Lukas Heger

okay

Koro

I didn't think that it would be so complicated to prove S^n simply connected.

😮

Path connected follows from pathconnectedness of $R^{n+1}-\{0\}$.

PNDas

@Koro after you prove Van Kampen then it'll be easy.

Lukas Heger

13:49

iirc the existence of surjective continuous maps $S^n \to S^m$ for $n<m$ is a major annoyance when you try to prove the cellular approximation theorem

so this problem is a general technicality that can't always be solved with Van Kampen

one potato two potato

0

Q: $\sup_{x\in\Bbb R}|F(x+iy)|\leq M_0^{1-y}M_1^y$ on the strip $0<Im(z)<1$ for bounded holomorphic function $F$

$\newcommand{\ep}{\epsilon}$ The following is Problem 4.3 in Stein complex analysis Suppose $F(z)$ is holomorphic and bounded in the strip $S = \{z:0<Im(z)<1\}$ and continuous on its closure. If $|F(z)|\leq 1$ on the boundary lines, then $|F(z)|\leq 1$ throughout the strip. Question. For the mor...

complex-analysis

Hmm...

Koro

@onepotatotwopotato: I think that such F can be extended continuously to a map F on all of C and having absolute values in [0,1].

This is basically by Tietze extension theorem.

one potato two potato

How does that help?

Koro

Your problem would be solved if the extension were unique but the uniqueness is not guaranteed by Tietze extension theorem.

Ohh. Please ignore the message. I was thinking about the part before 'Question' in your post as your question.

Lukas Heger

14:08

@onepotatotwopotato can you apply Phragmèn-Lindelöf to G?

one potato two potato

@LukasHeger Maybe. That's what I'm trying to.

@Koro The bound is only on the boundary lines. Extension does not show the prescribed $F$ is bounded by $1$.

If it's analytic then true by the identity theorem but what maniac would prove that extension is analytic?

Raffallo

hello guys!
https://math.stackexchange.com/questions/4513761/find-distance-between-a-point-and-a-catenary-curve?noredirect=1#comment9478217_4513761
I have formula like this from question. My `q` will be 0.0.
Now i need to find `p` value where `y` difference between x=0 and x=my value will be equal to detination difference. So basically i have two values of `x` and want to find `p` that will give me destination `y` diff between both `x`

Koro

14:37

Does anyone know what symbol this is?

Lukas Heger

flat

it's a musical symbol

ah not flat

but's it's a musical symbol

Natural (music)

In music theory, a natural (♮) is an accidental which cancels previous accidentals and represents the unaltered pitch of a note. A note is natural when it is neither flat (♭) nor sharp (♯) (nor double-flat nor double-sharp ) (nor triple-flat nor triple-sharp). Natural notes are the notes A, B, C, D, E, F, and G represented by the white keys on the keyboard of a piano or organ. On a modern concert harp, the middle position of the seven pedals that alter the tuning of the strings gives the natural pitch for each string. The scale of C major is sometimes regarded as the central, natural or basic...

ILikeMathematics

14:51

A tangent on the graph of $f(x) = 4 - 0.25x^2$ goes through the point $P(0|5)$.
At what point does this tangent touch $f$?

Can we solve this using $t(x) = mx + k$ or do we need $t(x) = f'(a) * (x - a) + f(a)$?

Using $t(x) = f'(a) * (x - a) + f(a)$, we can plug in $x = 0$ and calculate $a$ which will be the point's x-coordinate

Raffallo

I have formula: $a\cosh\left(\dfrac{x-p}{a}\right)$
My example data will be equal a = 1138 and two points P1(0, 1138) and P2(77.10, 1145.05).
I need to fit my curve to this two points, so I need to convert this formula to formula that I could minimize using Newton'` method.
Can someone help me in creating function to minimize?

basically it don't need to fit to the points, but i need to have "part" of the curve, where on distance 77.10 i will change y value by 7.05

Koro

15:29

@LukasHeger thanks :)

Not Tfue

15:44

@CalvinKhor Shark doesn't know that cow exists

The best thing about North korea is it's low crime rate. Nothing impressive about the math olympaid.

And it's the amount of light pollution is so low that it's the best place to observe the cosmos.

schn

@TedShifrin yes, that's the correct definition of $|x|$. However, it is also true that $\sqrt{x^2}=|x|$ and my question was; can that equality be proven? Anyway, I already found an answer on the main site. $\sqrt{x^2}=|x|$ holds since this makes the square root a function :)

ILikeMathematics

16:04

There in LaTeX

Thorgott

16:21

@LukasHeger I mean, that's what the van Kampen argument does and I'd say that qualifies as direct

it's the easier half of SvK, to boot

Lukas Heger

fair enough

but I meant directly from the definitions

Van Kampen becomes unavailable when you ask about why $\pi_n(S^m)=0$ for $n<m$

I did tell PNDas that one should just use Van Kampen, so I do agree with you

Van Kampen seems by far the easiest for $n=1$ though

Thorgott

there's a M-V/SvK-esque principle you can use to argue $\pi_n(S^m)=0$ for $n<m$

namely, if $X$ is covered by the interiors of two subspaces $X_1$ and $X_2$, $X_1$ and $X_2$ are $n$-connected and $X_1\cap X_2$ is $n-1$-connected, then $X$ is $n$-connected

Lukas Heger

oh okay

Thorgott

the argument is essentially from definition

but it involves some nasty subdivision business

and I do want to use the term "cofibration" when proving it

Lukas Heger

in my mind I always use cellular approximation for $\pi_n(S^m)$ for $n <m$

Thorgott

16:29

can you prove cellular approximation without already proving this fact, tho?

Lukas Heger

not quite sure

Thorgott

16:43

@Thorgott I should add that this is proven in "essentially the same way" as the van Kampen $n=1$ case, though a couple of more complicated techniques are invisible in the $n=1$ case cause there's an explicit alternative way to do it there

PNDas

Hi how to calculate the fundamental group of three circles where the middle circle touches left and right circles (like two infinity symbol joined and middle two become one).

Lukas Heger

SvK twice

really not substiantially different from $S^1 \vee S^1 \vee S^1$

PNDas

but there three circles join at a common point right?

Thorgott

the two spaces are homotopy equivalent

Lukas Heger

that's correct

PNDas

16:52

how to see that?

Lukas Heger

If you know the fundamental group of $S^1 \vee S^1$ via Seifert-van Kampen once, you can wedge this at a different point from original wedge point with $S^1$, this describes the space you have

and then just van Kampen again

PNDas

Yes that's what I was trying to do

Lukas Heger

even if you wedge at different points, it's still a wedge of three copies of $S^1$

@PNDas then where are you stuck?

finding $U$ and $V$?

PNDas

$X_1 = S^1\vee S^1$,$X_2 = S^1\vee S^1$, $X_1\cap X_2 = S^1$ right?

Lukas Heger

you need open sets!

and no that's too complicated

you want $X_1 = S^1 \vee S^1 + \varepsilon$ and $X_2=S^1 + \varepsilon$

Thorgott

16:58

in general, if $X$ is path-connected, then there is only one $X\vee S^1$ up to homotopy equivalence, i.e. the resulting spaces of wedging at different point are always homotopy-equivalent. but this is stronger than what you need, anyway.

Lukas Heger

then $X_1 \cap X_2$ is contractible

someoneinexistence

hey

Lukas Heger

makes the amalgamated product easier

PNDas

Let me just write what I did: $\pi_1(X_1,x_0) = \mathbb Z\ast \mathbb Z, $, $\pi_1(X_2,x_0) = \mathbb Z\ast\mathbb Z$, $\pi_1(X_1\cap X_2,x_0) = \mathbb Z$

Lukas Heger

what you did doesn't work because SvK needs open sets

you cannot ignore this

and even if you add the epsilons to make it open, it's far easier to work with a contractible intersection

I mean what you did works fine

PNDas

17:01

Like we do for $S^1\vee S^1$, I am deleting a point from left circle and right circle. Doesn't it make that open?

Lukas Heger

yes that makes it open

so you wrote your sets only up to homotopy equivalence

I didn't know that

PNDas

Yes

Lukas Heger

because I didn't know where you are stuck

then you get $\langle a,b,c,d \mid bc^{-1} \rangle= \langle a,b,c\rangle = \Bbb Z * \Bbb Z * \Bbb Z$

PNDas

After finding the fundamental groups, what to do now? In the examples I have tried the intersection was trivial

Lukas Heger

I mean, I already told you how to make the intersection trivial in this example

someoneinexistence

17:04

سلام!

Lukas Heger

but you can compute the amalgmated product in terms of generators and relations

that's the easiest

@someoneinexistence sorry for now anwering your greeting, I wasn't sure if you exist

PNDas

I don't know how to do that. Any reference example

someoneinexistence

ظهو هبسعرذع مع

PNDas

?

someoneinexistence

you observe me

Lukas Heger

17:05

well, we can take this example

someoneinexistence

my arabic keyboard was on

sorry

Lukas Heger

meine deutsche Tastatur ist auch an

PNDas

@LukasHeger what does epsilon mean here?

someoneinexistence

ach, deutsch?

Lukas Heger

just add a little bit to make it open

but if you want to work with a non-trivial amalgmation, we can take your decomposition

someoneinexistence

17:06

what are we doing right now?

Lukas Heger

@PNDas do you know what generators and relations for a group are?

PNDas

Yes

@LukasHeger Actually I made that example to understand the non trivial case.

Lukas Heger

ah I see

I didn't get what you wanted to do

okay, so take this example, let's call the three circles $C_1$ and $C_2$ and $C_3$. We have that $X_1=C_1 \vee C_2$ and $X_2=C_2 \vee C_3$. We need to work geometrically with explicit choices of generators for our groups here. so $a$ means $C^1$ say counter-clockwise, $b$ means $C_2$ counter-clockwise so that we have $\pi_1(C_1 \vee C_2)= \langle a,b\mid \rangle$

similarly, let $c$ be $C_2$ counter-clockwise, but as an element of $\pi_1(X_2)$ and $d$ be $C_3$ counter-clockwise, so that we have $\pi_1(C_2 \vee C_3) = \langle c,d |\rangle$

someoneinexistence

@LukasHeger why are you typing like that? Is that just mathjax that isn't showing up on my computer?

Lukas Heger

and we have $e$ for again $C_2$ counter-clockwise, but this time as an element of $X_1 \cap X_2$, so $\pi_1(X_1 \cap X_2) = \langle e | \rangle$

@someoneinexistence everyone here just renders mathjax in their head

someoneinexistence

17:12

oh god

I'm scared

Lukas Heger

no, jk

someoneinexistence

phew

I thought you all were legendary

Lukas Heger

you can find at the top right a link which instructs you to enable mathjax in chat

someoneinexistence

like all of you were arceuses- why did I not notice that

PNDas

@LukasHeger just to be clear what is your base point?

Lukas Heger

17:13

uhm

I mean here's it's really irrelevant

note that we have an inclusion map $X_1 \cap X_2 \to X_1$, this induces a map on $\pi_1$s that sends $e$ to $b$

likewise, there's an inclusion map $X_1 \cap X_2 \to X_2$ that sends $e \to c$

PNDas

then b and c are identified?

Lukas Heger

exactly

that's what the amalgmation does here

in general for an algmated product of groups $G \star_H K$, you need five things: three groups $G,H,K$ and two homomorphisms $H \to G, H\to K$. Once you have all these assembeled, there's an explicit formula in terms of generators and relations to compute the amalgmated product

PNDas

Okay thanks

Lukas Heger

I think this example is not very instructive

PNDas

let me do more examples to understand this.

Lukas Heger

17:17

try computing $\pi_1$ of a torus in a weird view

or $\pi_1$ of a Klein bottle

but the intuition for amalgmated products is like pushouts for spaces

you glue $G$ and $K$ together along $H$ somehow

well at least, you take the free product and then identify elements according to the homomorphisms $H \to G$ and $H \to K$

Balarka Sen

hi all

Lukas Heger

hey @Balarka!

I'm trying to give @PNDas examples for van Kampen computations where the amalgmation is nontrivial

probably you as our resident topologist have better ones

Balarka Sen

Try a cylinder S^1 x [0, 1] attached to a circle on one end by z^2 and on another end to another circle by z^3

ILikeMathematics

https://chat.stackexchange.com/transcript/message/62440508#62440508
if you only had $t(x) = mx + k$, then you wouldnt make much progress here, right?
$t(x) = mx + k$ and it passes through (0|5), so
$t(0) = 5$, so $k = 5$
So: $t(x) = mx + 5$
$m = f'(a)$, so $m = -0.5a$ => $t(x) = -0.5ax + 5$
You would want to find a which isn't really possible here, is it?

Balarka Sen

Or try two copies of RP^2 glued along a common RP^1

Lukas Heger

17:24

that's a picture of a torus (think of the edges identified accordingly). The blue set has hole in the middle, compute $\pi_1$ of a torus using this decomposition

that's an example where the amalgmation isn't trivial, but one of the factors is

very similar to the Klein bottle computation

Balarka Sen

I like the two Mobius strips decomposition of Klein bottle as well

Lukas Heger

oh yeah that's a good one, too

it's important to be able to "see" the generators and relations in your $\pi_1$ and not just think of them as abstract groups

Balarka Sen

shocking statements, coming from you

Lukas Heger

:P

I learnt topology from Banagl

he's very visual

Balarka Sen

oh hes great

I spent a lot of time reading his stratified spaces text

Lukas Heger

17:30

ah I see

one of my professors made a great analogy between homological algebra and multivariable calculus today

PNDas

In my notes $S^2$ is presented with a two sided polygon the sides are identified and the two points are identified

how to see this?

similarly for $\mathbb RP^2$ but sides have reverse diection

Lukas Heger

he said something along the lines of "deriving a functor is not unlike deriving a function. If you take total derivatives, then the derivative of the composition is simply the composition of derivatives (so he means using derived categories), but if you work with the Jacobian matrix of partial derivatives, you need to do some complicated matrix multiplication. That's the Grothendieck spectral sequence in homological algebra"

Balarka Sen

yeah its a lot like taking derivatives

Lukas Heger

So total derivatives is $D(F \circ G)= D(F) \circ D(G)$ and taking the matrix product of Jacobians is $E_2^{p,q}=({\rm R}^pG\circ{\rm R}^q F)(A) \Longrightarrow {\rm R}^{p+q} (G\circ F)(A)$

Thorgott

yeah, basically the same thing..

Balarka Sen

17:42

but i think the real relationship lies in the nonabelian setting of calculus of functors

Lukas Heger

oh yeah the Goodwillie stuff

Balarka Sen

its a fantastic idea

has a lot of geometric applications too

Lukas Heger

someone should write a book titled "A Course in Calculus" on calculus of functors

Serre basically pulled off the same joke with "A Course in Arithmetic"

leslie townes

is serre still alive?

haha, same thought

Lukas Heger

we need a whole sequence of math books with troll titles. We also already have Weil's classic "Basic Number Theory"

so for arithmetic, I suggest Serre, for algebra, I suggest Lang, for calculus Goodwillie and for basic number theory Weil

3

of course everyone reading a book called "Basic Number Theory" knows a what a Haar measure is

Thorgott

17:51

its more fun to assume the readers know something they dont than to assume they dont know something they do

user4539917

18:05

but the latter allows one to be comically pedantic

Koro

If A and B have the same cardinality, then free R-modules on A and B are isomorphic.

But I'm not sure how to prove that the free R-modules R^n and R^m are isomorphic iff m=n. Here R is commutative.

Thorgott

do you know this if $R$ is a field?

Koro

no R is a commutative ring.

Thorgott

read again

Koro

no, R is not a field in this exercise.

Thorgott

18:21

that's not what I'm asking

Koro

oh sorry.

Thorgott

I'm asking if you can answer the question when $R$ is a field

Koro

No I don't know this even if R is a field.

Ah wait.

No I do.

monoidaltransform

0

Q: Principal Bundle Definition Question Action restriction

Let $G$ be a topological group and $\pi: P\rightarrow M$ a fiber bundle with fiber $G$ (working in continuous category). We say that $\pi$ is a principal G-bundle if (1) $G$ acts freely on the right of $P$. (2) For each $p\in M$, there exists a neighborhood $U$ of $p$ and a fiber preserving home...

differential-geometry vector-bundles fiber-bundles

Koro

In that case, we have a vector space!!

Thorgott

18:22

exactly

Koro

Is the following correct?- $e_i$'s, i from 1 to n generate R^n. Suppose f is an isomorphism between them (from R^n to R^m) . We want to show n=m. f(e_i)'s are all distinct and linearly independent.

Thorgott

so I suggest you think about trying to reduce this case to the one where it is a field, perhaps by constructing some field from $R$

Koro

Suppose that n>m. $E_i$'s i from 1 to m form a basis of R^m. Not sure how to get a contradiction from here.

Jakobian

@LukasHeger this is gold, lol

Koro

R is not an integral domain so I am not sure how to get a field using R.

If it were then I could inject it inside its field of fractions.

Thorgott

18:33

all of this is true

Koro

18:44

0

Q: $R^m\simeq R^n$ iff $m=n$, where $R$ is a commutative ring and $R^m, R^n$ represent free $R-$modules.

If $m=n$, then it is true that $R^m\simeq R^n$. Now for the other direction, let $f:R^m\to R^n$ be an isomorphism. The standard basis $e_i$'s, $i=1$ to $m$ generates $R^m$ and therefore $f(e_i)$'s also generate $R^n$. But I'm not sure how to show from here that $m=n$. If $R$ is a field, then $R^m...

abstract-algebra modules

Thorgott

this will most certainly be flagged as a duplicate

without looking it up, I'm beyond certain this has been answered multiple times already

Jakobian

Uh. Could anyone help me with the proof that linearly ordered spaces are hereditarily normal?

I'm starting with some linearly ordered space $X$ and taking subspace $A\subseteq X$ then convex components of $A = \bigcup_\alpha C_\alpha$ I see that $C_\alpha$ are normal

leslie townes

12

Q: $R^n \cong R^m$ iff $n=m$

How can i show that two $R$-modules of finite rank are isomorphic if and only if they have the same rank, i.e., $R^n \cong R^m$ iff $n=m$.

abstract-algebra commutative-algebra modules

Jakobian

So... do I take $A$ to be open then each $C_\alpha$ will be open?

and it's known that for hereditarily normal it suffices.

okay I just figured it out

but I've never seen the proof that it's enough for all open subspaces to be normal so I'm going to do that first

Thorgott

yeah, hereditarily normal is equivalent to all open subspaces being normal

Jakobian

18:53

I was stuck at this proof, all the ones available online are pretty complicated to me. But this one I'm going for isn't that bad

Thorgott

I think it's just some trick, let me think

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