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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
@Semiclassical $f(x) = q(x)h(x)$ so put in $x = k$
semi does "f = 0 modulo some polynomial q" mean that there is a polynomial p with f = pq
yeah so you're good
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
i guess that assuming $f(x), q(x)$ (and thus $p(x)$) are integer polynomials is the key point
the approach looks sound to me (i haven't checked the algebra)
@Semiclassical that $f(x)$ and $q(x)$ are integer polynomials is needed so that $f(k)$ and $q(k)$ are integers
12:29 AM
(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)
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?
@monoidaltransform how do you know $g(x)$ is a real number?
@Jakobian it's bounded below?
You have to assume $f(B_r(x))\neq Y$ for all $x$
Yes sure
is it not always continuous?
12:49 AM
I'm not sure
what if $f(B_r(x))$ is open
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
I see. When is this function continuous then
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
Is this function ever continuous?
Yes of course
what natural conditions will make it so
X=Y$ here are disconnected
so maybe if connected?
1:17 AM
I don't know I'm going to sleep
1 hour later…
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
4 hours later…
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?
looks pretty compact to me
here is a discussion of motivations on the cohomology side
Q: An intuitive explanation for group cohomology via cochains?

pyroscepterI'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...

6:53 AM
@Jakobian vaguely
I've read about those before, but I never worked with them
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).
i don't see any reason for R to be the reals in that image excerpt
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
I was going to say the same, as they used the word functional
"functionals," to me, seems as consistent with R just being some base ring as it does with R being the reals
i see, so you are thinking maybe we are seeing $\mathfrak{g}$ as being a module or smth with ring $R$
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
Isn't functional only used when the codomain is a field?
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
Hm right
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.
8:55 AM
You could post it as a reference request question on MO, if it is written down somewhere people there will know
I've already wrote it on the site. It would be nice to know if a reference exists though
2 hours later…
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?
only @leslietownes and @XanderHenderson make sense...
11:48 AM
@LuckyChouhan I did what now?
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
12:35 PM
Hey guys, anyone on a tip on how to show that the sets are equal?
2 hours later…
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.
1 hour later…
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
1 hour later…
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
@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
Uh I've never heard about a different definition used in topology
The least $\sigma$-measurable cardinal is measurable by the way
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
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
5:04 PM
@Obliv zero operator
@YourLordJoyBoy Typically in long division we start by dividing the leading terms so in this case $4x^3$ by $x$
A: Is $(\nabla \times \nabla)$ an operator?

zupermanBy 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...

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
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
Thank you!
5:13 PM
@Obliv just zero?
@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.
@YourLordJoyBoy What do you need to multiply to $2y$ to get $4y^4$?
@SoumikMukherjee Lemme think.....$2y^4?$
@Obliv as long as the n-th derivative is continuous
@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
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.
$2y\cdot 2y^4 = 4y^5$
all of which is to say that I don't really know how to meaningfully answer the question
@Obliv Yes, it was actually $2y⋅2y^3$
My bad
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
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
@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
@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.
this works when dividing by any polynomial of degree $1$
@Jakobian Excellent, excellent! :D
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)
6:35 PM
@Jakobian Yes, I can see that.
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
@Jakobian Yes, it makes sense.
Yes as in yes we didn't have derivatives yet?
@Jakobian Simply stating this makes sense.
so you did have derivatives?
6:37 PM
Wait, derivatives? O.o
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 $p'(a) = r'(a)$
@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.
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
@Jakobian Alright, alright, I think I get what you're talkin about.
$(h(y)(y-a)^n)' = h'(y)(y-a)^n+nh(y)(y-a)^{n-1} = (h'(y)(y-a)^n+nh(y))(y-a)^{n-1}$
6:41 PM
@Jakobian YES!
the term $h_0(y) = h'(y)(y-a)^n+nh(y)$ you can just blackbox
@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 Yes
7:44 PM
@XanderHenderson nothing sir, but I admire you :')
@LuckyChouhan Are you able to vote in the election?
8:17 PM
the closing crew seems to be on vacation.
frantically going through your history to serially downvote/close your answers to PSQs
not on my watch, copper
8:51 PM
@LuckyChouhan Hrm... Then I must be doing something wrong. Gotta get better at sh*tposting...
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.
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
@psie try libgen
@psie what do u define as free
open access maybe
@psie try libgen
9:11 PM
can you trust that site?
with a high degree of accuracy, yes
@Obliv what do you define as free, lol
@psie why are you learning topology anyway
@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
idk I'm not that much into analysis
@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
You don't need general topology for that
Urysohn function in a metric space is easier to construct
9:18 PM
I have never seen the term Uryshon function
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 $A$
@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
oh wow, I was just about to ask
@Obliv What are the others?
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
9:25 PM
@SoumikMukherjee Because the name is Urysohn, not Uryshon.
@Obliv Oh I have them in hardcopies, except the tears one
I don't have a copy of any topology book
That's a surprise
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
@Jakobian You read all the things from pdfs? Isn't that stressful for your eyes?
$\partial_x A$
@SoumikMukherjee not really
9:33 PM
@AlessandroCodenotti Things are going for a 4-way tie, the final round will be crazy
Indeed, tomorrow it's going to be great
I hope Fabi makes it @SoumikMukherjee
I hope that too
He will surely win today
Does your eyes strain often while reading pdfs?
that's why I don't read from pdfs much
@SoumikMukherjee you could read from iPad (Adobe Acrobat Reader) and put on dark mode
or dark mode on Adobe Acrobat Reader on computer
9:37 PM
@psie My phone is full on dark mode, even the wallpaper is pitch black:P
@SoumikMukherjee maybe its that
your wallpaper is black so it contrasts your screen
@SoumikMukherjee The engine give him a smallish advantage, but that position looks so scary for Pragg
@Jakobian But I maintain the brightness accordingly, and it feels better in dark mode
oh you mean phone wallpaper
so you read white letters with dark background?
@AlessandroCodenotti Also Pragg has 25mins to make 16 moves in that horrible position
@Jakobian Yes
9:45 PM
@SoumikMukherjee yeah it looks extremely likely for Fabi to win
@SoumikMukherjee I heard studies that its bad for focusing
its better to have white background with black letters
@Jakobian 😱
Nepo is offering draw
Fabi is up an exchange
But it was a very good knight traded for a very bad rook
Yeah, not sure why he traded
@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).
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 but a metric space is normal, so no need to worry
@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
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 $x\mapsto d(x, A)$
@Jakobian is that kinda standard? I guess I can explicitly say that's what I'm referring to
@Obliv yes
I got something on my right eyelid
I'm wondering if I hit my eye, or maybe a bug bitten me
does it itch
no it almost doesn't feel
first time this happens
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
I don't know. I don't really want soap near my eyes
maybe it's a stye, quite common I think, can last for a couple of month without doing any harm
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 $\textbf{A}$
maybe it is stye
how did I get that though... hm
quite common, probably touched something and then rubbed ur eye
just keep it clean and apply warm compress occasionally if u want
10:23 PM
yeah I was rubbing eyes recently
@psie a month? That's annoying
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 $\textbf{A}$
yeah :) you could go to a doctor and then they could cut up so all the gland goes out, like a pimple
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
I don't eat a lot of vegetable soup, but I would if I could
do u guys prefer $\textbf{A}$ or $\vec{A}$ to denote a vector/vector field
I'm starting to dislike arrow notation
10:56 PM
between those choices, i'd go with bold
if not just those choices, what would u go for
i think a lot of that arrow stuff came about primarily because it's easy to do in handwritten notes
wait are u trolling, u don't even render chatjax lol
i don't really like giving vector stuff special notational status, but i like consistency in choice of notation across types
oh true, if handwritten, bold is probably not an option
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
11:16 PM
@Obliv I would use neither
@Thorgott yeah
@leslietownes an uppercase Latin letter would probably be my choice
@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
@Jakobian we don't always agree, so when we do, i feel we must both be even more right than we usually are
hi all, would you agree that $\text{SE}(3) / (\{0\} \times \text{SO}(2))$ is diffeomorphic to $\mathbf{R}^3 \times S^2$?
11:26 PM
what's SE( ) ?
orientation preserving rigid motions of R^3?
$\text{SE}(3) = \mathbf{R}^3 \rtimes \text{SO}(3)$
does that candidate for the quotient work out dimension wise?
why not? basically i'm taking $\{0\} \in \mathbf{R}^3$ and $\text{SO}(2) \subset \text{SO}(3)$
oh, i didn't mean to suggest i was skeptical, i was just asking the first question i would ask before thinking about it
ofc the embedding $\text{SO}(2) \subset \text{SO}(3)$ is not canonical
11:29 PM
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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