Mathematics - 2024-04-20

12:12 AM

a bit of polynomial ring stuff i've forgotten. suppose i show that $f(x)=0$ modulo some polynomial $q(x)$ where $f(x), q(x)$ each have integer coefficients. If I now let $x$ be some integer $k$, does it follow that $f(k)=0$ modulo $q(k)$?

it feels like i'm doing something improper but i don't know what

Jakobian

@Semiclassical $f(x) = q(x)h(x)$ so put in $x = k$

leslie townes

semi does "f = 0 modulo some polynomial q" mean that there is a polynomial p with f = pq

right

yeah so you're good

hmm

12:14 AM

if f is a multiple of q in integer polynomial land, f(k) is a multiple of q(k) in integer land

Semiclassical

i guess that assuming $f(x), q(x)$ (and thus $p(x)$) are integer polynomials is the key point

context, fwiw: math.stackexchange.com/questions/4902162/…

leslie townes

the approach looks sound to me (i haven't checked the algebra)

Jakobian

@Semiclassical that $f(x)$ and $q(x)$ are integer polynomials is needed so that $f(k)$ and $q(k)$ are integers

Semiclassical

12:29 AM

right

(i mean, if they're not you could still end up with $f(k)$ and $q(k)$ coincidentally being integers for some choice of $k$. but that's hardly generic)

monoidaltransform

Assume $X$ is a compact metric space and $Y$ is a metric space. Assume $f:X\rightarrow Y$ is continuous. Define the map $g:X\rightarrow \mathbb{R}$ by $g(x)=d(f(x),Y\backslash f(B_{r}(x))$. Is this map continuous?

Jakobian

@monoidaltransform how do you know $g(x)$ is a real number?

monoidaltransform

@Jakobian it's bounded below?

Jakobian

You have to assume $f(B_r(x))\neq Y$ for all $x$

monoidaltransform

Yes sure

is it not always continuous?

Jakobian

12:49 AM

I'm not sure

monoidaltransform

what if $f(B_r(x))$ is open

Jakobian

1:08 AM

If you let $Y = [0, 1]\times \{0\}\cup \{0\}\times (0, 1) \cup 3S^1$ and $X = [0, 1]$ and $1 < r < \sqrt{2}$ and $f(x) = (1-x, 0)$ then $g$ will have a sudden jump

@monoidaltransform

monoidaltransform

I see. When is this function continuous then

Jakobian

sorry the example is slightly wrong it should be something like

$X = Y = [0, 1]\times \{0\}\cup \{0\}\times [0, 1]\cup 3S^1$ and $f(x) = x$

the point is that it suddenly begins to cover the inner part and so $g$ has a jump

monoidaltransform

Is this function ever continuous?

Jakobian

Yes of course

monoidaltransform

what natural conditions will make it so

X=Y$ here are disconnected

so maybe if connected?

Jakobian

1:17 AM

I don't know I'm going to sleep

monoidaltransform

Ok

Obliv

2:45 AM

anyone know if it makes sense to do the curls of the grad operator first in $\nabla \times \nabla \times \vec{A}$

is this thingy associative? my physics class is very unmathematical

Silly Goose

6:38 AM

I am looking at group homology and the following boundary map

is there a more compact way to define this boundary map?

leslie townes

looks pretty compact to me

here is a discussion of motivations on the cohomology side

11

Q: An intuitive explanation for group cohomology via cochains?

I'm fairly new to topology, and so far I've understood cohomology via cochains. First we build an object called a cochain ($C^n$), then define a differential map that takes you from $C^n$ to $C^{n+1}$. Then all the types of cohomology groups I've encountered can be easily defined as: \begin{equ...

Alessandro Codenotti

6:53 AM

@Jakobian vaguely

I've read about those before, but I never worked with them

Silly Goose

7:09 AM

okay thank you

is $R$ here supposed to be $\mathbb{R}$? (asking since $R$ is used in these notes to stand for an arbitrary ring).

leslie townes

i don't see any reason for R to be the reals in that image excerpt

Silly Goose

hm well it describes them as linear functionals on $\mathfrak{g}^n$ but i can also accept that $R$ can be some other more general object

Soumik Mukherjee

I was going to say the same, as they used the word functional

leslie townes

"functionals," to me, seems as consistent with R just being some base ring as it does with R being the reals

Silly Goose

i see, so you are thinking maybe we are seeing $\mathfrak{g}$ as being a module or smth with ring $R$

leslie townes

7:15 AM

and as far as i can see, there is nothing in any of those formulas that would not make sense if R is just a ring

Soumik Mukherjee

Isn't functional only used when the codomain is a field?

leslie townes

in some contexts, maybe, but what about this context is requiring that?

even if context requires R to be a field, i don't see why R would have to be the reals

Soumik Mukherjee

Hm right

Jakobian

8:09 AM

@AlessandroCodenotti basically I was wondering if in literature there exists a proof that a metrizable space is realcompact iff its of non-measurable cardinality

Gillman and Jerison list only one direction and I have a proof of the other using a result of Hirata

And does it make sense to post it on math.stackexchange

Well I think the answer to the latter should be yes.

Alessandro Codenotti

8:55 AM

You could post it as a reference request question on MO, if it is written down somewhere people there will know

Jakobian

I've already wrote it on the site. It would be nice to know if a reference exists though

Jakobian

9:14 AM

I've asked for a reference as well

Jam

11:20 AM

How to find how many ideals does the quotient ring $$Q[x]/<((x-2)^2)(x-1)>$$ has? I though of making it into a product from chinese rem theorem. So Q[x]/<x-1> is a field with 2 ideals but what about the other?

Lucky Chouhan

only @leslietownes and @XanderHenderson make sense...

Xander Henderson

11:48 AM

@LuckyChouhan I did what now?

Jakobian

12:16 PM

@Jam you can use \langle, \rangle and \mathbb

Using Chinese remainder theorem makes sense, you end up with $\mathbb{Q}[x]/(x-1) \oplus \mathbb{Q}[x]/((x-2)^2)$

I think the latter should have only three ideals, $(0), (1)$ and $(x)$

actually $(1+ax)$ for $a\in\mathbb{Q}$ might be distinct ideals as well

Oh okay I see now

Every ideal of $\mathbb{Q}[x]$ is principal since $\mathbb{Q}$ is a field so an ideal is of the form $I = (p(x))$. Then $I$ corresponds to an ideal of the quotient ring iff $p(x)|(x-2)^2(x-1)$.

So the ideals of $\mathbb{Q}[x]/((x-2)^2)$ correspond to either $x-2, (x-2)^2$ or $1$

so yeah three ideals

so the answer is 6

Mad

12:35 PM

Hey guys, anyone on a tip on how to show that the sets are equal?
https://math.stackexchange.com/questions/2335947/1-1-3-a-set-of-discontinuity-points-for-a-function-is-a-borel-set/2335986#2335986

YourLordJoyBoy

2:39 PM

@JakobianWhat Im trying to do is get that tag wiki edit badge.

More importantly however, I'm trying to solve dividing polynomial problems and I attempted this several times alone before asking here. (s^4-7s^3+2s^2-7s+8) / (s) Apparently I'm to Divide the polynomials. Express numbers using integers or simplified fractions. Check your answer by multiplication.

Ryder Rude

3:45 PM

homology considers maps from r-simplices to the manifold and homotopy considers maps from n-spheres to the manifold

is there a more general idea here or do these two maps just happen to be useful

YourLordJoyBoy

4:47 PM

Have tried this on my own, and have constantly found myself infuriated. Please someone help me with this problem? (4x^3+6x^2+3x+99) / (x+3)

$(4x^3+6x^2+3x+99) / (x+3)$

There we go.

the / is division btw

Jakobian

@AlessandroCodenotti so apparently there is a difference between what some topologists call a measurable cardinal and what set theorists call a measurable cardinal

and what topologists call a measurable cardinal is really called a $\sigma$-measurable cardinal

Alessandro Codenotti

Uh I've never heard about a different definition used in topology

The least $\sigma$-measurable cardinal is measurable by the way

Jakobian

Yeah. Gillman and Jerison use measurable but they mean $\sigma$-measurable

and all the theorems about realcompactness and measurability really refer to $\sigma$-measurability

so I've stepped on a mine there. Oof

A cardinal $\kappa$ is $\sigma$-measurable iff its greater than the least measurable cardinal

so the least measurable cardinal is really the only important one in those statements in topology

Obliv

I can't find this on wiki what is this called

$$\nabla\times\nabla = \hat{i}\left[\frac{\partial^2}{\partial z\partial y}-\frac{\partial^2}{\partial y\partial z}\right]-\hat{j}\left[\frac{\partial^2}{\partial z\partial x}-\frac{\partial^2}{\partial x\partial z} \right]+\hat{k}\left[\frac{\partial^2}{\partial y\partial x}-\frac{\partial^2}{\partial x\partial y} \right]$$

we have $\nabla^2 = \nabla \cdot \nabla$ the laplacian or hessian

Soumik Mukherjee

5:04 PM

@Obliv zero operator

Obliv

@YourLordJoyBoy Typically in long division we start by dividing the leading terms so in this case $4x^3$ by $x$

Soumik Mukherjee

3

A: Is $(\nabla \times \nabla)$ an operator?

By equality of mixed partials, that is the fact that: $$ \begin{align} \frac{\partial^2g}{\partial y \partial z}&=\frac{\partial^2g}{\partial z \partial y}\\ \frac{\partial^2g}{\partial z \partial x}&=\frac{\partial^2g}{\partial x \partial z}\\ \frac{\partial^2g}{\partial x \partial y}&=\frac{\pa...

Obliv

do we always assume equality of mixed partials?

what is the zero operator expressed in a non fancy way then?

is it just the kernel? hope I'm not mixing up terminology too much

Soumik Mukherjee

Symmetry of second derivatives

In mathematics, the symmetry of second derivatives (also called the equality of mixed partials) refers to the possibility of interchanging the order of taking partial derivatives of a function f ( x 1 , x 2 , … , x n...

Just a remark: Note that in a broader sense, if you replace the derivatives with covariant derivatives on a Riemannian manifold (concider that everything is happening on a sphere, for example), the difference between $f_{xy}$ and $f_{yx}$ is the curvature (at least when differentiating vector fields). So, you can interpret the above identity as saying that $\mathbb{R}^n$ is "flat". — Peter Franek Dec 20, 2014 at 17:21

Obliv

Thank you!

Soumik Mukherjee

5:13 PM

@Obliv just zero?

YourLordJoyBoy

@Obliv Absolutely liked. But, what about this one? $4y^4+8y^3+2y^2-8$ divided by $2y+4$ ? And I did try this on my own first. Any time I'm seeking help on a problem, I try to tackle it alone at first.

Soumik Mukherjee

@YourLordJoyBoy What do you need to multiply to $2y$ to get $4y^4$?

YourLordJoyBoy

@SoumikMukherjee Lemme think.....$2y^4?$

Jakobian

@Obliv as long as the n-th derivative is continuous

Thorgott

@RyderRude the way these are constructed from maps is, however, significantly different

there are reasons for either construction, but there are also generalizations of either

there is a sense in which homotopy is dual to cohomology (rather than homology) that concerns homotopy classes of maps

Obliv

5:20 PM

@YourLordJoyBoy let's try it. $2y\cdot 2y^4 = $ multiply the constants and we get $4$ but multiplying the variable we get $y^5$ so ur off by 1.

$2y\cdot 2y^4 = 4y^5$

Thorgott

all of which is to say that I don't really know how to meaningfully answer the question

YourLordJoyBoy

@Obliv Yes, it was actually $2y⋅2y^3$

My bad

Obliv

correct, so now you multiply that out with the first term $2y^3(2y+4) = 4y^4+8y^3$ and bring that down

subtract, see what is left, rinse and repeat

@SoumikMukherjee the confusing thing is if $\nabla\times\nabla = 0$ then $0\times \vec{A} = -\nabla^2\vec{A}+\nabla(\nabla\cdot\vec{A})$

because i got that from $\nabla\times\nabla\times\vec{A} = -\nabla^2\vec{A}+\nabla(\nabla\cdot\vec{A})$

unless in this problem statement they forgot the parentheses

for $\nabla\times(\nabla\times\vec{A})$

yeah looking online they did

Derivative

6:23 PM

can somebody recommend me a reference for generalizations of Teichmuller space i.e. moduli space of {complex, Kahler, Calabi-Yau} structures? Especially if the moduli space has a smooth structure and finite dimension

Jakobian

@YourLordJoyBoy If you divide $p(y) = 4y^4+...$ by $q(y) = 2y+4$ you obtain $p(y) = h(y)\cdot q(y)+r(y)$ where $r(y)$ is the remainder, and its a polynomial of degree $< \text{deg}(q) = 1$, in other words a constant polynomial

This means that if you plug in $y = -2$ which is the only zero of $q(y)$, then $p(-2) = h(-2)\cdot q(-2)+r(-2) = r(-2)$

so that the constant $r(-2)$ can be determined by by evaluation of $p(-2)$

if you only need the remainder term then this is a fast way to obtain it

YourLordJoyBoy

@Jakobian This is great! Hoping this'll help on the final. And speaking of the final, I have a different kind of math problem this time.

Jakobian

this works when dividing by any polynomial of degree $1$

YourLordJoyBoy

@Jakobian Excellent, excellent! :D

Jakobian

even if it doesn't have degree $1$, but you can write it as $q(y) = (x-a_1)...(x-a_n)$ where $a_i$ are distinct, then you can first obtain $r(a_n) = p(a_n)$ and then try to solve a system of linear equations (this would be the same method)

YourLordJoyBoy

6:35 PM

@Jakobian Yes, I can see that.

Jakobian

If $q(y)$ had a root occurring twice, then this won't be as easy, you'd have to take some derivative. I assume you didn't have derivatives yet

YourLordJoyBoy

@Jakobian Yes, it makes sense.

Jakobian

Yes as in yes we didn't have derivatives yet?

YourLordJoyBoy

@Jakobian Simply stating this makes sense.

Jakobian

so you did have derivatives?

YourLordJoyBoy

6:37 PM

Wait, derivatives? O.o

Jakobian

Well what I'm trying to get at is if $p(y) = h(y)(y-a)^n + r(y)$ where $n > 1$ then you can take derivative to get $p'(y) = h_1(y)(y-a)^{n-1}+r'(y)$

and then $p'(a) = r'(a)$

YourLordJoyBoy

@Jakobian Ah

Now about the problem I was talking about, its like this. Currently, my grade in algebra 1 is a 69 (actially 69.59%). Apparently the quizzes are 2% of the final grade. So I'm trying to see what adding a 100 on one of those plus a 70 on the final would give me. Since the final is 30% or 3% on the calculator.

Jakobian

so if you are dividing by a polynomial which has roots with multiplicity $> 1$ then you can still apply this method to find the remainder but you need to obtain new linear equations by taking derivatives

YourLordJoyBoy

@Jakobian Alright, alright, I think I get what you're talkin about.

Jakobian

$(h(y)(y-a)^n)' = h'(y)(y-a)^n+nh(y)(y-a)^{n-1} = (h'(y)(y-a)^n+nh(y))(y-a)^{n-1}$

YourLordJoyBoy

6:41 PM

@Jakobian YES!

Jakobian

the term $h_0(y) = h'(y)(y-a)^n+nh(y)$ you can just blackbox

YourLordJoyBoy

@Jakobian Seems right.

But yeah, and also the study guide I've been doing is a second 20 percent added to the grade. So I'm trying to figure out the total my grade will be with all of these additional grades added.

Jakobian

Exercise

YourLordJoyBoy

@Jakobian Yes

Lucky Chouhan

7:44 PM

@XanderHenderson nothing sir, but I admire you :')

Soumik Mukherjee

@LuckyChouhan Are you able to vote in the election?

copper.hat

8:17 PM

the closing crew seems to be on vacation.

leslie townes

frantically going through your history to serially downvote/close your answers to PSQs

not on my watch, copper

Xander Henderson

8:51 PM

@LuckyChouhan Hrm... Then I must be doing something wrong. Gotta get better at sh*tposting...

psie

9:02 PM

can anyone recommend free texts on topology? Looking for some supplementary notes to some books that I already have. As an aside, unfortunately the one book that I have, Introduction to Topology by Gamelin and Greene, is not available as an ebook, and there are certain features I enjoy about having access to a book as a pdf.

Obliv

What am I doing wrong, I am trying to prove $\vec{\nabla}\times(\vec{\nabla}\times\vec{A})=-\vec{\nabla}^2\vec{A}+\vec{\nabla}(\vec{\nabla}\cdot\vec{A})$ but I did the left hand side and got a vector.. Also the right hand side doesn't make any sense, a laplacian gives a scalar so adding $-\vec{\nabla}^2\vec{A}$ to a vector doesn't make sense

Soumik Mukherjee

@psie try libgen

Obliv

@psie what do u define as free

psie

open access maybe

Jakobian

@psie try libgen

psie

9:11 PM

can you trust that site?

Jakobian

with a high degree of accuracy, yes

@Obliv what do you define as free, lol

@psie why are you learning topology anyway

Obliv

@psie if u give me ur email i can send u some

@Jakobian I don't wanna say something that the all seeing xander might smite me for. Also can u take a look at what I wrote and tell me if that equality holds

as it is, or do I have to modify it

Jakobian

idk I'm not that much into analysis

psie

@Jakobian well, I always stumble upon some theorems that use some topology and then I need to study a little topology, like recently, I studied that $C_c(\mathbb R)$ is dense in $L^1(\mathbb R)$ and the proof uses a Urysohn function. I had never heard about this before and until I read a little more about Urysohn's lemma, which is topology

Jakobian

You don't need general topology for that

Urysohn function in a metric space is easier to construct

Soumik Mukherjee

9:18 PM

I have never seen the term Uryshon function

Jakobian

If $A, B$ are two non-empty disjoint closed sets then you set $f(x) = \frac{d(x, A)}{d(x, A)+d(x, B)}$

where $d(x, A) = \inf\{d(x, y) : y\in A\}$ is the distance from $x$ to $A$

Obliv

@psie the offer stands. I have 4 topology texts recommended to me I can send u, i think one was gamelin and greene though

so maybe 3

psie

oh wow, I was just about to ask

Soumik Mukherjee

@Obliv What are the others?

Obliv

actually it is in pdf form same with the others. colin adams Introduction to topology: pure and applied, topology without tears(not really a textbook i don't think)

and munkres

Soumik Mukherjee

9:25 PM

@SoumikMukherjee Because the name is Urysohn, not Uryshon.

@Obliv Oh I have them in hardcopies, except the tears one

Jakobian

I don't have a copy of any topology book

Soumik Mukherjee

That's a surprise

Obliv

I wish there was a better way to shorthand partials of a function. If one uses $A_x$ to mean $\frac{\partial A}{\partial x}$ then we can't use $A_x$ to denote the $\hat{x}$ component

Soumik Mukherjee

@Jakobian You read all the things from pdfs? Isn't that stressful for your eyes?

Jakobian

$\partial_x A$

@SoumikMukherjee not really

Soumik Mukherjee

9:33 PM

@AlessandroCodenotti Things are going for a 4-way tie, the final round will be crazy

Alessandro Codenotti

Indeed, tomorrow it's going to be great

I hope Fabi makes it @SoumikMukherjee

Soumik Mukherjee

I hope that too

He will surely win today

Jakobian

Does your eyes strain often while reading pdfs?

Soumik Mukherjee

yes

that's why I don't read from pdfs much

psie

@SoumikMukherjee you could read from iPad (Adobe Acrobat Reader) and put on dark mode

or dark mode on Adobe Acrobat Reader on computer

Soumik Mukherjee

9:37 PM

@psie My phone is full on dark mode, even the wallpaper is pitch black:P

Jakobian

@SoumikMukherjee maybe its that

your wallpaper is black so it contrasts your screen

Alessandro Codenotti

@SoumikMukherjee The engine give him a smallish advantage, but that position looks so scary for Pragg

Soumik Mukherjee

@Jakobian But I maintain the brightness accordingly, and it feels better in dark mode

Jakobian

oh you mean phone wallpaper

so you read white letters with dark background?

Soumik Mukherjee

@AlessandroCodenotti Also Pragg has 25mins to make 16 moves in that horrible position

@Jakobian Yes

Alessandro Codenotti

9:45 PM

@SoumikMukherjee yeah it looks extremely likely for Fabi to win

Jakobian

@SoumikMukherjee I heard studies that its bad for focusing

its better to have white background with black letters

Soumik Mukherjee

@Jakobian 😱

Nepo is offering draw

Alessandro Codenotti

Fabi is up an exchange

But it was a very good knight traded for a very bad rook

Soumik Mukherjee

Yeah, not sure why he traded

psie

@Jakobian well, we need to know that the space is normal to get such a function, right? And that is what Urysohn's lemma states (if and only if).

Soumik Mukherjee

9:51 PM

It was on a light^2 and it would take Pragg some time to reroute his knight for a trade

Now Fabi has a bad bishop

psie

@psie but a metric space is normal, so no need to worry

Alessandro Codenotti

@SoumikMukherjee Pragg's position definitely looks easier to play now

I guess Fabi will play to prepare f5? I don't see many other options now that the position is so closed

Jakobian

10:12 PM

@psie yes statement of Urysohn lemma is in fact equivalent to being normal

but that Urysohn lemma for arbitrary normal space is harder to prove

its like creating something from nothing

you shouldn't use hard results on blind faith, there is a simple more elementary proof accessible to you right now

you can use classic concepts from metric space theory such as the distance function $x\mapsto d(x, A)$

Obliv

@Jakobian is that kinda standard? I guess I can explicitly say that's what I'm referring to

psie

right

Jakobian

@Obliv yes

I got something on my right eyelid

I'm wondering if I hit my eye, or maybe a bug bitten me

Obliv

does it itch

Jakobian

no it almost doesn't feel

first time this happens

Obliv

10:19 PM

just keep it clean and dry and see what happens i guess

wash it with soap, maybe antibacterial soap if u got

Jakobian

I don't know. I don't really want soap near my eyes

psie

maybe it's a stye, quite common I think, can last for a couple of month without doing any harm

Obliv

keep ur eye closed obviously :P u dont have to put a lot, just a tiny bit

how about if i need to take $\partial_x A_x$

I wish I denoted $A_x$ to mean $\partial_x A_x$ but I already reserved it to mean $\hat{x}$ component of $\textbf{A}$

Jakobian

maybe it is stye

how did I get that though... hm

Obliv

quite common, probably touched something and then rubbed ur eye

just keep it clean and apply warm compress occasionally if u want

Jakobian

10:23 PM

yeah I was rubbing eyes recently

@psie a month? That's annoying

Obliv

nvm I think $A_i$ to mean $i$ component of $\textbf{A}$ is more useful cuz then I can do $\partial^kA_i$ to mean $k$-th partial of $i$-th component of $\textbf{A}$

psie

yeah :) you could go to a doctor and then they could cut up so all the gland goes out, like a pimple

Obliv

eww

that's worst case probably. it should heal on its own

sooner than a month I'd guess since @jakobian is young and eats lots of vegetable soup

Jakobian

I don't eat a lot of vegetable soup, but I would if I could

Obliv

lol

do u guys prefer $\textbf{A}$ or $\vec{A}$ to denote a vector/vector field

I'm starting to dislike arrow notation

leslie townes

10:56 PM

between those choices, i'd go with bold

Obliv

if not just those choices, what would u go for

leslie townes

i think a lot of that arrow stuff came about primarily because it's easy to do in handwritten notes

Obliv

wait are u trolling, u don't even render chatjax lol

leslie townes

i don't really like giving vector stuff special notational status, but i like consistency in choice of notation across types

Obliv

oh true, if handwritten, bold is probably not an option

leslie townes

10:57 PM

so e.g. if you wanted to use capital roman letters from the beginning from the alphabet, just do that

and don't use boldface or italics or arrows

or some people will adopt the convention that things like u, v, w, near the end of the alphabet that aren't x, y, z are vectors, and if you are consistent with that, you wouldn't need to give them additional distinguishing things beyond letter choice

sometimes in physics and other contexts you will see people making notational choices that force you into having to decorate, e.g. using a "decorated" r to denote a position vector, and regular lowercase r is the length of the position vector, where you might need both r's in the same expression

and IMVHO people should just not do that

Jakobian

11:16 PM

@Obliv I would use neither

Ryder Rude

@Thorgott yeah

Jakobian

@leslietownes an uppercase Latin letter would probably be my choice

Ryder Rude

@Thorgott i dont like that we randomly consider maps and make them into a group. is there a systematic idea which generalises both homology and homotopy

but yeah, the constructions of these groups r very different

leslie townes

@Jakobian we don't always agree, so when we do, i feel we must both be even more right than we usually are

gf.c

hi all, would you agree that $\text{SE}(3) / (\{0\} \times \text{SO}(2))$ is diffeomorphic to $\mathbf{R}^3 \times S^2$?

leslie townes

11:26 PM

what's SE( ) ?

orientation preserving rigid motions of R^3?

gf.c

$\text{SE}(3) = \mathbf{R}^3 \rtimes \text{SO}(3)$

yes

leslie townes

does that candidate for the quotient work out dimension wise?

gf.c

why not? basically i'm taking $\{0\} \in \mathbf{R}^3$ and $\text{SO}(2) \subset \text{SO}(3)$

leslie townes

oh, i didn't mean to suggest i was skeptical, i was just asking the first question i would ask before thinking about it

gf.c

ofc the embedding $\text{SO}(2) \subset \text{SO}(3)$ is not canonical

leslie townes

11:29 PM

:)

gf.c

no worries

i think it's a valid subgroup. what i'm ultimately considering is the bundle map $\text{SE}(3) \to \text{SE}(3) / (\{0\} \times \text{SO}(2))$

basically wondering if this makes $\mathbf{R}^3 \times S^2$ into a homogeneous space

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$Mathematics$ Mathematics