Mathematics - 2021-01-21 (page 2 of 3)

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Alessandro Codenotti

6:00 PM

$(x^2-2)(x^2-3)(x^2-6)$ has a root modulo every prime

Your example has infinitely many primes for which it does and doesn't have roots btw

@Krijn right about this

my other question- is it true?

Alessandro Codenotti

@Krijn which example?

@AlessandroCodenotti x^2 + 1 has roots depening on p mod 4, but both cases (1 and 3) are infinite

Alessandro Codenotti

ahh I see now

6:02 PM

@AlessandroCodenotti I like this one, that's exactly an answer to my question :-)

Alessandro Codenotti

I wrote finitely instead of infinitely above

To get elementary nonequivalent ultraproducts you want polynomials with a root modulo INfinitely many primes and no root modulo infinitely many other primes. Then you can choose which of the two sets of prime to put in $U$ and get roots in $F$ or not accordingly

6:14 PM

Hey @AlessandroCodenotti do you think you could define the natural numbers in there?

Alessandro Codenotti

I have no idea to be honest, but I think in general it's not easy to define the naturals in a field

It can be done in $\Bbb Q$ by a miracle, but already in $\Bbb R$ it shouldn't be possible

Koenigsmann very strong

you guys ruined algebra

wot

Looks like you guys were discussing something like one of my favorite algebra facts. "Almost every" polynomial that is irreducible /$\Bbb Q$ is reducible mod every $p$. The easiest examples are $x^4+1$ and $x^4-10x^2+1$, if I remember correctly.

6:16 PM

Morning chat

Looks like I fell asleep with this tab open

Morning @Fargle

Alessandro Codenotti

Do you have a simple example of a polynomial that has no roots modulo $p$ except for at most finitely many $p$ @Ted by any chance?

Too busy inaugurating?

@Alessandro no roots, but not necessarily irreducible?

Alessandro Codenotti

yeah I'm happy with no roots

i'm far from an algebraic number theorist, but this sounds suspicious.

Alessandro Codenotti

6:21 PM

@EdwardEvans I summon thee

This sounds like a decent question to put on main.

Or to search for?

Alessandro Codenotti

I suppose there's either an easy example or a simple proof that there are none

I should think for a moment about whether Tchebotarev density (which gives the "almost every" in my previous statement) is remotely relephant.

@TedShifrin What?!

nods @Krijn

6:25 PM

@TedShifrin this made my brain crash. how is this... what?

Yeah, BigSocks, it's a good exercise for you to do those examples :)

but Tchebotarev... that sounds like a russian that shows up at the end of the book

and "almost every"... that sounds like analysis!

There is density in number theory, too.

Actually yeah it is starting to make sense

I mean if you think about large degrees

But the fact that it's for every p is still weird

hey Ted, do you know about growth rates of balls in Riemannian manifolds?

6:29 PM

Well, but I think it's true for every degree individually (4 and larger).

That's not my thing, @Thor, but "know about" is vague.

I wanted to know if there are examples of polynomial growth of non-integer exponent

So not polynomial?

A wise men once said nothing ......

I think it doesn't exist @TedShifrin @AlessandroCodenotti

write P = XQ + c, then $P(cn) = c(nQ(cn) +1)$

That would be my guess, @Astyx.

6:31 PM

So if you take n the product of the primes

Whoa. Slow down, @Astyx. Too fast for me.

lmao

yeah, I guess calling it polynomial is unusual terminology, though that's what we use

when pupil beats the master

I mean the balls of radius $R$ around some point (doesn't matter which) grow depending on $R$ like $R^{\alpha}$ does (up to a multiplicative and additive error above and below), where $\alpha$ is non-integer

6:33 PM

So suppose P only has roots in $\Bbb F_p$ for $p\in E$, where $|E|\lt +\infty$

OK, for now I'm listening to the algebra discussion.

Ah damn you're right

You're using $P$ too many times.

OK.

Thanks.

Write $P = XQ + c$ where $c$ is the constant term

OK.

6:35 PM

And let $n = \prod _{p\in E}p$

OK.

then $P(cn) = cn Q(cn)+c = c(nQ(cn)+1)$

Hello guys.

OK.

Now $nQ(cn)+1$ is not a multiple of any of the $p\in E$

6:37 PM

hello @abenthy

hello Ted :)

Granted.

What topological picture is Ted's profile pic ?

it's 2021 and this guy's still copying Euclid

Shaddup, @Thor.

I made a small mistake, I want n to be of the form $k \times \prod_{p\in E}$

6:38 PM

I have a question: If $A$, $B$, $C$ are topological spaces and $A \times B \to C$ is a finite covering map, then if we know the Betti numbers of $A$ and $B$, can we deduce the Betti numbers of $C$? Or does that depend heavily on the covering map?

Euclid , father of geometry , has my respect

not just the product of the primes in E, but any multiple

If P is not constant, then Q is not zero, thus there exists an n of this form such that $Q(cn)\ne 0$

Thus $nQ(cn)+1$ is a multiple of a prime $p\notin E$

It follows that the reduction modulo $p$ of P(cn) is zero, thus there's a root

Gotta go eat, brb

@abenthy You get nothing from this alone

Ah, I see the gist. Good. @Alessandro: You concur?

I can cook up arbitrarily bad situations

6:41 PM

@MikeMiller Okay, I feared so. This is because finite coverings can do many different things with the homology, right?

I was going to comment, @abenthy, that having $A\times B$ rather than just a topological space seems irrelevant.

Yes there is basically nothing you can say about the homology of a covering space without advanced machinery

I was hoping the Künneth theorem can help in this product situation.

And the advanced machinery is a fancy way to say "Meh we don't know much"

ROFL @MikeM

Sayan Chattopadhyay

6:43 PM

Hello chat

OK, back to @Thor. So I'm being stupid. What do I do about just a sphere or the hyperbolic plane? I get $\sin^2 R$ and $\sinh^2 R$ (up to constants).

the first one is constant, the second one is exponential (up to quasi-equivalence)

So you're not counting multiplicities when $R$ is large in the positive curvature case.

@MikeMiller: Thank you for your answer. Just to make sure, do you think you could even cook up bad situations for, say, Riemannian manifolds of finite volume and non-positive curvature? I'm not asking for concrete examples, just trying to get a feeling for the "advanced machinery" :)

yes

6:48 PM

There are theorems (these are things I haven't studied) by people like Ballman and Croke and others relating the volume to Ricci curvature, I believe.

the advanced machinery is group cohomology and spectral sequences

Ah okay, I see

you have to understand how $\pi_1$ downstairs acts on the homology of the covering space

If you have a concrete handle on that you might be able to extract some information

If the action is trivial you're in great shape but that's basically never true

Would that maybe apply in the situation of quotients $\Gamma \backslash X$ for some arithmetic lattices $\Gamma$ in a Lie group acting on a symmetric space $X$?

@MikeMiller "if you have genus $\geq 1$ you might be able to extract some information"

6:51 PM

@abenthy I don't know anything about that but I'm going to go with "Not a chance"

Yeah, I guess so too. Thanks

But maybe I'm wrong!

I'm only interested in growth from a coarse geometric picture. I'm looking at increasing functions $f,g[0,\infty)\rightarrow[0,\infty)$ and say $g$ quasi-dominates $f$ if there are constants $a,b$ such that $f(r)\le ag(ar+b)+b$ for all $r\ge0$ and call them quasi-equivalent if they quasi-dominate each other. So this allows a multiplicative and an additiv error, both on the value and on the argument.
This is so coarse that it doesn't differentiate between $r\mapsto a^r$ and $r\mapsto b^r$, but it still manages to tell apart $r\mapsto r^a$ and $r\mapsto r^b$ for $a\neq b$.

Yeah, I can't really help, but google those names I gave you.

I shall, though I'm likely to not understand anything

The curious and non-trivial fact motivating my question is that if our Riemannian manifold is the universal cover of some compact Riemannian manifold and the growth rate of its balls is polynomial, then it's automatically polynomial of integer exponent.

6:58 PM

Another book to look at is Gray's Volume of Tubes. This is integral geometry stuff (going back to Hermann Weyl) which ties into things I did early in my life.

He's mainly interested in large scale, though, and the tube theory is small scale

Yeah, but the volume of balls comes into that, of course. But you're right. I have no feeling for the large issue.

I have always said I am not and have never been a Riemannian geometer. This sort of question makes that clear.

I'm only asking out of curiosity, so I'll probably just put this up on main later if I can't find anything

Hello

user image

Sorry not to be more help, @Thor.

Why have we turned into Physics.SE chat?

7:05 PM

A user physicsismylife recommended me to ask Physics doubts here if someone is not present in the physics.SE

why did they do that???

no problem at all

If you can answer then do it otherwise it’s fine.

Because you and I were somewhat helpful, @copper.

dang!

7:07 PM

I don't even understand the statement, @srijan. How am I moving with a force with respect to the ground. This makes no sense.

I don't get it, are the two person colliding?

@Alessandro never responded to @Astyx's construction. It looked good (in principle) to me.

It says they collide, @Astyx. But I don't understand the whole premise. Momentum is not force.

I'm not sure whether they collide with the ground against each other

Ok

So as I wrote I am having confusion in understanding this as well. What I meant was by the state tan you commented on is that if a body is having a mass and acceleration. Then can I say it has a force w.r.t ground

@TedShifrin is it clr

@Astyx yes. They collide after they touch each other.

No. Ordinarily, we want to talk about momentum of colliding bodies.

7:12 PM

Ok. So I can’t write the statement like that @TedShifrin

I should say momentum of the Boyd w.r.t ground ?

What is the actual exact question you were given?

the question is very unclear. what does 'moving with force' mean?

If you could make a drawing of what's happening it could clear things up

Karl Kroningfeld

Is the motion wrt the ground? Is there a spherical cow assumption that the 'ground' is a line?

goes to pop corn

7:14 PM

Battery down. Sorry . I’m back

@KarlKroningfeld After flat earthers, line earthers

Ok.@Astyx I will send an image in a min.Pls wait

Karl Kroningfeld

@Astyx Being under lockdown, I approximate it with a point.

user image

I can be wrong with statements

Wow, that's a dirty white board

What's the word before "force" ?

7:21 PM

Mathematicians, you have here a question and answer that you may to check for correctness: ai.stackexchange.com/q/2729/2444

the user had originally used the term "entailment"

@Astyx Looks like "equals"

You're telling us we need to check something? Huh?

well, if you want

My wording wasn't the best

i must ask for a raise from mse...

So my take is the two persons are running towards each other with different accelerations, and you want to find the terminal velocity at collision

We need more info, for instance the original distance between the two persons and their initial velocity

7:25 PM

elastic, inelastic, friction, blah, blah, blah,...

the user used the symbol $\vdash$, but didn't he want to prove $f_1 \iff f_2$? It's been a while I had to do with logic

@copper.hat Pretty sure physics people are points in a vacuum

nothing in the question makes sense. moving with force is a nice poetic term but meaningless in terms of translating into symbols we can use.

do they have mass, where is the force coming from, etc, etc.

think about what it is you want to ask.

It would be great if we could have feedback from the person asking...

Karl Kroningfeld

One should allow the critics to review one's artwork without interference from the artist.

7:30 PM

lmao

Ok.

I am back

connectivity’s issues

Maybe what I wrote in sentences is not making sense

but what I wrote in values is .

I am new to this chapter. I maybe wrong at most points so pls just correct me there

i am not a physicist

question is simple

a person of mass 20kg is moving with 6m/s^2 towards right

I said it is force because that is what is force value F = ma

So what I am saying is another person is moving with 30*8

Ofc you can’t move with 210N

it is just my thought because of the values

person is moving with 6m/s_2 having force =120N

In PSE , someone told me it is w,r.t ground I should say

but you see what is better

OK, what's the question

ok. Then as the two persons are running towards each other

then

What should I say

how much force did person 1 exert on person 2

Is it 120N? If yes , then according to Newton’s third law , it exerted a 120N on itself also

That’s what I was asking

Did you get it

try to understand

Is that a problem from a textbook? From yourself?

Myself

7:42 PM

It doesn't make sense as such

Where does it not ?

try to understand, @Astyx

try to understand

Force is something that relates mass and acceleration

Ok

reaches for popcorn

7:43 PM

Then , @Astyx

You're way behind me, @JoeShmo.

I stole that joke from you, yes

I can make my questions better if you tell me where am I wrong @Astyx

it's an excellent joke. I'll mention you in the credits

Gee thanks.

7:44 PM

So if you say, for instance, that a train weighing 200kg goes from 20m/s to 0 in 2 seconds, with constant deceleration, you can compute the force applied on it

Ted, how do I open a private chat? I need you for a 20 seconds

Yes. Now I maybe wrong but that’s what I think.

@Astyx is it better if we can have a private chat

@srijannahar what problem are you trying to solve?

@copper.hat what did you not get from everything I wrote

What you need to do is make sense of what you mean by force when you ask that question, and realise that is not what force means in physics

7:47 PM

@JoeShmo There is no such thing as a private chat. But you have to create a room at the main chat page, I think. I've never done it.

@Astyx ok. I thought it did because by inserting values . It gives Force

force in this case is rate of change of momentum. unless you tell us what happens to the two objects afterwards one cannot even guess.

Ok.

@TedShifrin chat.stackexchange.com/rooms/118755/… so much for private

@copper.hat

After the distance between two person is 0

they collide

Why ? Because they were running so they exert force on each other

Then what is that value

that it

7:56 PM

Can you please try to make self contain statements in messages instead of taking 8 lines of messages to finish a sentence?

K

what does moving with an acceleration mean? i really do not understand what you are trying to ask? are there two people running and accelerating towards each other? is the collision elastic? what are their masses? when the collide you are presumably modeling the collision as an instantaneous sort of thing, so force is not the right thing to look at.

an instantaneous force is a meaningless sort of thing (unless impulsive, but that is beyond current discussion).

@copper.hat I have written Masses in the first line itself.

@copper.hat pls leave the question. I am not able to make you understand it more now. Its fine.

@copper.hat isn't instantaneous force equal to mass times instantaneous acceleration?

@robjohn yes, but if it is only applied on a set of measure zero it can have no effect...

unless we are into distributions or the like

what i am trying (clumsily) to say is that when there is a collision (in the context of this problem), there really isn't a useful notion of a force at the moment of collision.

8:09 PM

I agree

Velocity is continuous, so unless you model the collisions saying which forces are involved, you cannot find the force from just having the collision velocities

K

It could be that the two persons are going sufficiently fast to go through each other because their repelant forces are not strong enough to make a difference in the short timelapse of the collision

i think the op is trying to understand something but has not stepped back to ask a question that will help the understanding.

A black hole could go through the earth and leave it unchanged if it went fast enough

:-)

8:19 PM

Good night everyone

Have a great day for the ones who have morning

@TedShifrin did you do something? I just got a response

@srijannahar good luck.

did you bribe anybody with gourmet popcorn? confess

LOL ... no, but great!

It's an aye, btw :-)

(surprise, surprise)

8:22 PM

Great. Move forward.

8:36 PM

@Astyx does that mean a bullet could go through a person and leave them unchanged if it went fast enough?

yes, to some extent

In newtonian mechanics at least

in Constructive Feedback, 41 mins ago, by Shaun

The following question of mine was downvoted and I don't know why.

ir depends on how you model the interaction (sounds rather clinical).

0

Q: If $G=⨝_{i\in I} H_i$ and $K\unlhd G$ is perfect, then $K=⨝_{i\in I}(H_i\cap K)$.

This is Exercise 5.11(b) of Roman's "Fundamentals of Group Theory: An Advanced Approach". According to Approach0, it is new to MSE. The Details: On page 72 . . . Definition: If $\mathcal{F}=\{H_i\mid i\in I\}$ is a nonempty family of normal subgroups of a group $G$, we write $$H_{(i)}:=\bigvee \...

group-theory normal-subgroups direct-sum derived-subgroup

my message "good night" was marked as spam/offensive. i must not be trying hard enough.

9:01 PM

@copper.hat wut

:-)

me approximating the identity with test functions

user image

4

@user2103480 sounds like a bumpy and convoluted road.

@robjohn with smooth sections

modern distribution theory

2

9:15 PM

@copper.hat looks like the troll got suspended

a suspension troll...

$\Sigma$troll

9:35 PM

🙃good one

9:46 PM

Is it ever possible to have algebraic subsets $X \subsetneq Y$ in $\mathbb A^n$ such that $I(X) = I(Y)$? I don't think it's possible but just want to double check?

$X$ is precisely the vanishing locus of $I(X)$

@Thorgott Right, but can you have a case like the one I described above?

the fact I just quoted gives you the answer, why?

@Thorgott So this says no such example exists?

yes

9:59 PM

I just thought, since we have the possibility of $I \subsetneq J $ with $V(I) = V(J)$, that it might be the case one could have $X \subsetneq Y$ such that $I(X)=I(Y)$.

If we have a variety $(X,\mathcal O_X)$, and $U\subset X$ is open, is $(U,\mathcal O_X\vert_U)$ then also an algebraic variety? I'm guessing no? When we talk about open sub varieties, I'm guessing we're adding conditions? (if $U$ is locally closed, things probably work out?)

No, @Sha, no hope, unless $X$ is disconnected and $U$ is one component.

Oh ok, thanks @Ted. My notes didn't really define what it means to be an open subvariety, so that's why I wondered. They define a quasi-projective variety to be one that is isomorphic to an open subvariety of a projective variety. So what you're saying is that such an open subvariety can only be given by a component?

Karl Kroningfeld

This kinda depends on who you ask. I lean more toward the definition of an abstract variety: ncatlab.org/nlab/show/algebraic+variety

Ye, I noticed that there are many definitions circulating. So how we define algebraic varieties is as follows: an affine variety is a $k$-space that is isomorphic to $(Y,\mathcal O_Y)$ where $Y$ is an algebraic set and $O_Y$ is its sheaf of regular functions. Now, an algebraic variety is locally an affine variety. We also have $(\mathbb P^n,\mathcal O_{\mathbb P^n})$, and any variety that's isomorphic to that one, is called projective.

Oh, I should have said $(Y,\mathcal O_Y)$ with $Y\subset\mathbb P^n$ open for the definition of a projective variety

10:14 PM

I have a function $f(t-hu)\in C^m$, which I'd like to (Taylor) expand around $t$. Is the error term then $\mathcal{O}((-hu)^{m+1})$? I'm verifying a solution to an exercise, and it suggests something different.

user image

Well, I would say $\mathscr O(|hu|^k)$, first of all. If you do the $k$th order Taylor polynomial at $a$, by definition almost, the error is $o(|x-a|^k)$.

That's almost impossible to read.

Hard to :)

@TedShifrin But why not $k+1$?

I have no idea what "little ordo" is supposed to mean. The O is little oh, as I suggested.

You're not distinguishing between big oh and little oh.

The little oh statement is what you know in general, actually. The remainder formula, if you have another derivative, for example, does give you an estimate in terms of $|x-a|^{k+1}$.

I'm more familiar with the big oh, and if you do up to order $k$ the error is $k+1$, no?

Well, I'm going to raise that issue now.

They defined little-oh in that problem there.

If you have $f(x)=O(x^{k+1})$, then of course $f(x)=o(x^k)$. Is the converse true?

Consider, for example, $f(x)=|x|^{3/2}$. It is $o(x)$. Is it $O(x^2)$?

10:21 PM

Gimme a second.

@Ted do you know an example of a Riemannian manifold in which a ball of finite radius has infinite volume? can this even happen?

would have to be incomplete, of course

maybe something with unbounded curvature?

Hmm, that's interesting.

The obvious incomplete things I think of, I believe the radius likewise blows up.

@TedShifrin Working on understanding $f(x)=O(x^{k+1}) \implies f(x)=o(x^k)$.

If the ball is compact, then it can't happen, @Thor, right?

Just use the definition, @schn.

$|f(x)|\le C|x|^{k+1}$, so ...

If you mean open ball, @Thor, then compact closure.

@TedShifrin And $x$ is really small, right?

10:25 PM

Right. $o(x)$ means you have a limit as $x\to 0$.

@TedShifrin $|f(x)|\le C|x|^{k+1} \leq C|x|^{k}$ for small $x$, or?

OK, or look at $\dfrac{C|x|^{k+1}}{|x|^k}$ and take the limit.

right, cause small coordinate charts have finite volume and you can cover it by finitely many

The interesting question is the other implication.

so Hopf-Rinow tells us a counter-example would have to be incomplete

10:31 PM

But typically the distance will blow up. What kind of incompleteness are you envisioning?

@TedShifrin Right, so the error term as written in the image is correct if it is the little oh, with big oh it would be $O(|hu|^{k+1})$? The other implication would not hold, but that is only a guess :)

I'm telling you that you can't necessarily use big O. Look at the example I suggested. Yes, what was written there made it clear it was little o (even though it was illegible).

Why can't one necessarily use big Oh?

Come on, man. I've told you three times now to look at the example I wrote for you to look at.

I'm not able to envision an example. I'm not even sure whether I should expect such a thing to exist or not.

10:39 PM

I know two sorts of phenomena. One is a hole that you could fill in without a metric singularity (like poking a hole in something complete). The other is a singularity in the metric which causes geodesics to have infinite length even though they "look" bounded.

Like the hyperbolic metric on the disk.

@TedShifrin You used $f(x)=|x|^{3/2}$, but in the expansion $m$ is an integer. Wouldn't that make big Oh legitimate?

right, the first type clearly won't produce a counter-example and in the second case, we won't get something of finite radius, so it's unclear what a counter-example should look like

I'm talking about the first degree Taylor polynomial for that at $0$. What is it?

@Thor: This paper might have some interest.

aren't these lower bounds

Looks like it.

10:46 PM

@TedShifrin It becomes problematic with the absolute value, or? However, in the example I posted initially, $O(|hu|^{m+1})$ would also work as an error term, no?

You're not thinking.

Breathing.

The first degree Taylor polynomial of my function is $0$. The error is the function itself. Is it $O(x^2)$?

No. Since $O(x^2)$ is possibly greater than $0$?

What?

10:52 PM

It is $O(x^2)$!

But also $O(x^3)$, right?

No, you are screwing up. It is NOT $O(x^2)$.

Let's just talk about positive $x$. Is $x^{3/2}\le Cx^2$?

$\implies x^3 \leq Dx^4$.

Depends on $D$, or?

What was the point you wanted to make? :)

Is there any positive constant $C$ for which $x^{3/2}\le Cx^2$ for all small $x$?

No.

OK. So your way of writing the error is NOT correct unless you have further assumptions on the function.

10:58 PM

What further assumptions?

You need more differentiability.

Right, since the error term is derived from m+1 th derivative of $f$.

But the big Oh notation kind of hides that.

But that's more assumption than you need to get a Taylor polynomial. So the little-oh expression is the correct one without further assumptions.

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