Mathematics - 2020-02-28

user76284

05:33

What's the right notation to convey $A \vdash B$ but under intuitionistic predicate logic, not classical logic?

Leaky Nun

$A \vdash_{\text{intuitionistic}} B$

Mathphile

08:03

just a test please ignore

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6

Rudi_Birnbaum

11:04

Hi all!

I have a question. I am dealing with "divergenceless" vector fields $v$: $\nabla\cdot v = 0$

The question is how to call that property in English (an German).

We had started with "divergence-free" and "divergence free" then went to "divergence-less" and "divergenceless"

and at some stage I have used "fullfilling the continuity equation" (its time-independen cases).

And in the haste of writing I ended up with "continuous" which is now obviously plain wrong (right?).

Happy for suggestions and comments!

Thorgott

I would go with "divergence free" ("divergenzfrei") and I think in the case of fluids they're called "incompressible" ("inkompressibel"), but I'm no expert

Rithaniel

13:17

Is it possible to have a field that has both elements of finite order and elements of infinite order?

I feel like I might know that the answer is "no," but I'm unsure

Alessandro Codenotti

Hint: characteristic

Rithaniel

Yeah, I seem to recall that all elements of a field have to have the same order, and if that order is finite, then it has to be prime, but again, a proof doesn't immediately leap to mind

Wait, use the order of the identity. Got it.

Alessandro Codenotti

Right. If you want a somewhat more abstract argument for every ring $R$ there is a unique ring hom $\Bbb Z\to R$ ("$\Bbb Z$ is the initial object of $\mathsf{Ring}$") and it has kernel $n\Bbb Z$ for some $n$. That $n$ is the characteristic of $R$, from which you quickly get what you want by properties of homomorphisms

Thorgott

13:38

this is true for order as elements of the additive group, the multiplicative group can have both elements of finite and of infinite order

Rithaniel

Like $i$ versus $2$

Alessandro Codenotti

Sure, the multiplicative group of $\Bbb C$ contains elements of any prime order

Thorgott

of any order was correct

Rithaniel

All of which, if they are of finite order, sit on a single circle

Thorgott

for each natural $n$, $\mathbb{C}^{\times}$ contains exactly one subgroup of order $n$ and it is cyclic

Alessandro Codenotti

13:43

Every finite subgroup of the multiplicative group of a field is cyclic

Rithaniel

Fields are almost disappointing in that sense. You have all these interesting things in more abstract constructs. Once you get to fields, everything becomes roughly uniform

Alessandro Codenotti

Algebraically closed ones are even more boring, they're completely classified by characteristic and cardinality if they are uncountable

Thorgott

I mean, this is true for integral domains too

Rithaniel

Yeah, integral domains are still fun for other things, though. There are a lot of weird things with Krull dimension and rings of formal power series

Thorgott

You can still say some really interesting (and arguably surprising) things about algebraically closed fields, like Artin-Schreier

What's true though is that fields are a pretty badly behaved category

Alessandro Codenotti

13:58

I'm a little confused. Let $\Omega\subseteq \Bbb R^n$ be as nice as you wish (open, bounded, smooth boundary, whatever). Of the inclusions $C^\infty_c(\Omega)\subseteq W^{1,p}_0(\Omega)\subseteq W^{1,p}(\Omega)\subseteq L^p(\Omega)$ which are dense? I'm interested especially in the case $p=2$

Thorgott

Isn't $C_c^{\infty}(\Omega)\subseteq L^p(\Omega)$ dense already?

Alessandro Codenotti

Hmm maybe you're right (and that would make sense, from a result I'm reading it does seem like they should all be dense)

Thorgott

I think $C_c^{\infty}$ is dense in $C_c$ by mollifying and $C_c$ is dense in $L^p$ by some tedious approximation arguments, but take this with multiple grains of salt

Alessandro Codenotti

I agree with the first part and I think I read about the second one a long time ago

Thorgott

14:14

I can sketch an argument for the second part in a couple minutes if needed. Not sure how much niceness is needed on $\Omega$ for both parts to go through, but I feel like openness should work.

geocalc33

14:33

Thorgott what's up

What's stronger than a diffeomorphism?

Alessandro Codenotti

@Thorgott Nah I can believe it

I guess it's in Brezis if I really want to see the ugly details

geocalc33

I think that the problem with your question is that "diffeomorphism" is too weak a notion to be relevant to general relativity (GR). In GR a "spacetime" is a Lorentzian manifold, that is, a differentiable manifold equipped with some additional structure. You are pondering what transformations you can apply to a spacetime without destroying all this structure. Now, a diffeomorphism is not enough; it does preserve the differentiable structure, but it needs not preserve the Lorentzian one.

The "right" notion you are after is that of a conformal mapping of Lorentzian manifolds.

Lukas Heger

15:31

@Alessandro mollifying $L^p$ functions is not more tedious than mollifying $C_c$ functions actually. If you take the convolution of a locally integrable function and a $C_c^\infty$ function, then the result is $C^\infty$, so if you take an approximate identity you get that $C^\infty \cap L^p$ is dense in $L^p$ without more difficulty than showing that $C_c^\infty$ is dense in $C_c$.
If you actually want to $C_c^\infty$, you need to multiply by suitable cut-off functions, but that's not really that tedious: Write $\Omega = \bigcup U_i$ where the union is directed and all $U_i$ are open and b…

Leaky Nun

@LukasHeger did you see my message?

Lukas Heger

no

which one?

Leaky Nun

@LukasHeger show that Q has no Z-basis, using the most high-powered way possible

Lukas Heger

hmm

geocalc33

haha

most high powered way :)

Lukas Heger

15:41

upon taking Pontryagin duals wrt discrete topologies, this is equivalent to showing that $\Bbb A_{\Bbb Q}/\Bbb Q$ is not isomorphic to some product of $S^1$s

Leaky Nun

ah, I forgot about Pontryagin dual

geocalc33

Pontryagin duality?

Thorgott

I see, that's a good point.

geocalc33

Can you convolute the differential structure of

a real analytic man-fold

Leaky Nun

and we know that $\Bbb A_\Bbb Q = \Bbb Q \times \widehat{\Bbb Z} \times \Bbb R$? wait that isn't right

geocalc33

15:44

depends on how fat the man is!!

Leaky Nun

what's the decomposition? @LukasHeger

Lukas Heger

uhm, I don't think there's a decomposition like that for the additive adele group

Leaky Nun

ok

Lukas Heger

but I might be wrong

Leaky Nun

but we can embed $\widehat{Z}$ inside right

geocalc33

15:47

decomposition in what adele class?

Lukas Heger

@LeakyNun yes

Alessandro Codenotti

@LeakyNun Can you compute Ext^1(Q,Z) without knowing that Q isn't free Abelian?

(this is essentially the same thing as Lukas product of circles thing but on the other side of the duality to be fair)

Leaky Nun

0 -> Hom(Q,Z) -> Hom(Q,Q) -> Hom(Q,Q/Z) -> Ext(Q,Z) -> Ext(Q,Q) -> Ext(Q,Q/Z) -> 0

how does Ext work again

geocalc33

ah Hom(Q,Z)....

Alessandro Codenotti

@LeakyNun I never remember lol

geocalc33

15:52

hey alessandro

Alessandro Codenotti

Hi

Leaky Nun

looks like Ext(Q,Q) = 0

Alessandro Codenotti

You only need that Ext to be nonzero, you don't care what exactly it is here

geocalc33

Off topic w.r.t. the conversation atm but, can one convolute the conformal structure of a smooth and bounded lorentzian submanifold

Alessandro Codenotti

The point is showing that Q is not projective as a Z-module

Leaky Nun

15:54

so we want to show that Hom(Q,Q) -> Hom(Q,Q/Z) is not surjective

Hom(Q,Q/Z) = Hom(injlim (1/n)Z,Q/Z) = projlim Hom((1/n)Z,Q/Z) = projlim nQ/Z = ???

Mike Miller

16:29

@LeakyNun take any diagram of abelian groups, so that for any arrow f in this diagram, there is another arrow g with gf = 0, and furthermore every object has a morphism-out. then the colimit of the diagram is zero, because the map from each object of the diagram to the colimit is forced to be 0

there seems to be an issue with this argument lol

oh I see, none of the composites are actually 0 as I demanded

geocalc33

@MikeMiller do you have a minute?

geocalc33

16:48

well, if you feel like editing or answering feel free :) math.stackexchange.com/questions/3562412/…

ABC

18:43

Guys I need to prove this equality. z is a complex variable. o(..) is Peano notation.

someone can help me?

I need to use Taylor expansion, true?

expansion is in 0

Ted Shifrin

No, @ABC. You just need to use (the first few terms of) the geometric series $\dfrac 1{1-u} = 1+u+u^2+\dots = 1+u+o(u)$.

ABC

Perfect, thanks!!

Ted Shifrin

You're welcome.

Sophie

19:14

If an elliptic curve has rank 1 and torsion of order 3, then how does that mean that the rational points can be written as $p,p^2...$?

PortMadeleineCrumpet

20:09

Hi, might anyone be able to answer a super basic question about figuring out concavity for an MLE?

Ted Shifrin

21:00

@PortMadeleineCrumpet: It can't be super basic if I don't know what an MLE is !! :)

topologicalmagician

@TedShifrin Hello!

I was able to solve the problem but my wifi turned off! My apologies for not replying

math.stackexchange.com/questions/3563578/… May someone help?

The sequence isn't cauchy, but i'm not sure why that implies it has no convergent subsequences

that isn't true though as well

Its bounded, non cauchy and thus non-convergent

Ted Shifrin

22:04

@topologicalmagician Any bounded sequence in $\Bbb R$ has a convergent subsequence (probably more than one).

topologicalmagician

@TedShifrin yes, thats bolzano weirstrass

Ted Shifrin

Yes, that's one way to do it.

But you're not in $\Bbb R$.

The question you linked to already has a solution.

Indeed, you gave it.

What is your proof that the sequence doesn't converge?

topologicalmagician

@TedShifrin yes, but i'm not sure how to prove that the sequence I have provided has no convergent subsequences

$d(x_n,x_m)=1$ for $n\neq m$

not cauchy, thus not convergent

Simple

just wonder, how do i cite my or someone message in chat room

Amin Idelhaj

Hey everyone!

Ted Shifrin

22:07

Hi, Demonark.

@topologicalmagician: So now doesn't that same argument work for any subsequence?

Cite it where, @Simple? In here?

topologicalmagician

@TedShifrin what about the subsequence $(0,0,0.....0)$

?

Ted Shifrin

We're talking about the sequence of $e_i$'s, @topologicalmagician, so I have no idea what you just wrote.

topologicalmagician

OH @TedShifrin a sequence in $L^\infty$ is a sequence of vectors

Simple

yes here, want to know how to quote/ cite. i have seen people quote a message from a different chat rooms and messages that posted couple days before

Ted Shifrin

Right click on the left of the message, @Simple. And select Copy Link Address.

@topologicalmagician: I thought our sequence was the sequence of $e_i$ to start with. You know what each of those is.

Or else how did you make your argument?

Simple

22:13

oh, thanks ted

Ted Shifrin

22:26

By the way, @topologicalmagician, it's not good form to delete a question on which people have given input (comments, answers, etc.) just because you finally figure it out.

topologicalmagician

@TedShifrin I just undeleted

Ted Shifrin

LOL ... OK.

You can post your own answer once you're confident you have it.

topologicalmagician

Yeah, I misunderstood notation, but now the answer is apparent

Ted Shifrin

Yeah, from what you typed here ($d(x_n,x_m)$) I figured you were not understanding what things meant.

Ted Shifrin

22:42

LOL, good name, @Descartes.

Descartes Before the Horse

Thank you :) @TedShifrin

A question: does anyone know if Ramanujan ever studied functions $f(x)$ for which $\sum_{n=-\infty}^\infty f(n) = \int_{\mathbf{R}}f(x) dx$?

I swear I read a paper which mentioned that he did, but would rather not drudge through his notebooks hoping to stumble across where he does.

Ted Shifrin

Aside from obvious step functions, you mean? Do you want $f$ to be continuous?

Descartes Before the Horse

Yes.

Ted Shifrin

I do not know much of anything about what Ramanujan did or didn't do.

Descartes Before the Horse

Differentiable everywhere, preferably.

The classic function e^{-x^2} almost qualifies, though there is a tiny difference between the sum and integral.

Ted Shifrin

22:46

You could cook up smooth functions by using by using bump functions.

Descartes Before the Horse

I'm sorry--what are those?

Ted Shifrin

All you need to do is cook up a function on the interval $[n-1/2,n+1/2]$ that is smoothly $0$ at the endpoints and has integral equal to $f(n)$. Lots of ways to do that.

Descartes Before the Horse

Could you shoot me a reference

Ted Shifrin

They're used all through analysis and differential geometry. They're built out of $e^{-1/x^2}$ so that you can glue local smooth functions together to make global functions.

Descartes Before the Horse

Ah I see

thank you

Ted Shifrin

22:48

Without worrying about smoothness, just multiply by a function like an upside-down parabola.

FuzzyPixelz

If I have a linear scalar differential equation $y^{(n)}=a_0(t)y+\cdots +a_{n-1}(t)+b(t)$ and $(y_1, \dots,y_n)$ is a basis for set of homogenous solutions, and we denote by $W(t)$ the matrix$(y_j^{i-1}(t))$. Can you check this formula for the general solution? $$y(t)=c_1y_1(t)+\dots+c_ny_n(t)+\int_{t_0}^t[W(t)W^{-1}(s)]_{1,n}b(s)ds$$Where $c_1,\dots c_n$ are scalars and $[M]_{1,n}$ is the coefficient in line $1$ and column $n$ of $M$.

Ted Shifrin

But all I need to do is cook up a weird function on the unit interval whose value at the midpoint is its integral. And there are zillions of solutions to that.

FuzzyPixelz

(assuming all the usual conditions on the functions $a_i$ etc)

Ted Shifrin

What do you mean, @Fuzzy? Why don't you check?

You typed the ODE wrong.

FuzzyPixelz

Well I did, but when I tried solving an equation with it, I couldn't find the right solution

Ted Shifrin

22:51

The easiest way to do this is look at the system where the new variable is a vector $(y,y',y'', \dots, y^{(n-1)})$.

FuzzyPixelz

That's what I did, it's the first-order system

Ted Shifrin

Then write the matrix equation for the associated differential equation.

OK. I'm a little surprised that $W^{-1}$ is on the right instead of the left.

Maybe that's your problem.

topologicalmagician

23:05

Does the following statement make sense: Show that the completness of $\mathbb{R}$ is equivalent to the squeeze theorem?

Ted Shifrin

I suppose it depends on what "the squeeze theorem" is.

FuzzyPixelz

Ok so if we write $Y=(y,y',\dots,y^{(n-1)})$ and $Y'=A(t)Y+B(t)$ with well-ch0sen $A$ and $B$. Then $Y(t)=W(t)W^{-1}(t_0)Y_0+\int_{t_0}^t W(t)W^{-1}(s)B(s)ds$. In fact $$Y'(t)=A(t)W(t)W^{-1}(t_0)Y_0+A(t)\int_{t_0}^t W(t)W^{-1}(s)B(s)ds+W(t)W^{-1}(t)B(t)=A(t)Y(t)+B(t)$$ Using $W'(t)=A(t)W(t)$. Then we take the first component of $Y$. Is that correct?

topologicalmagician

@TedShifrin I presume the sandwiching of sequences between two convergwent sequences

Ted Shifrin

Now you're being inconsistent with the $(t)$'s, after all that fuss.

FuzzyPixelz

Well, It's my least concern now

Ted Shifrin

23:07

Ah, ok, @topologicalmagician. Although, how would you state the hypotheses if the limit of the controlling sequences weren't in the space?

I haven't thought about Wronskians in about 44 years, @Fuzzy, so it's hard for me to participate.

FuzzyPixelz

No worries, I'm not being very clear, I know.

Ted Shifrin

I still am puzzled how it ends up on the right.

But, yes, of course, you're going to read off the first component when you're done, because that's where $y$ is.

topologicalmagician

@TedShifrin thats what i'm not sure of. Does the question mean Show that $\mathbb{R}$ is complete if and only if the sandwich principle holds?

sandwich principle = squeeze theorem

Ted Shifrin

I didn't write the question, @topologicalmagician, so you should ask the person who did! But my complaint stands. If it makes sense to say that $\lim a_n = \lim c_n = L$ and we know that $a_n\le b_n\le c_n$, then saying that the squeeze principle holds tells us that $\lim b_n = L$, but if $L$ isn't in my domain, then the hypothesis makes no sense.

topologicalmagician

@TedShifrin yeah, i suppose ill ask the person who wrote it

Anyways, I got to go. Have a wonderful day, @TedShifrin

Ted Shifrin

23:19

Let me know tomorrow (or next week) what it means.

topologicalmagician

Alright, @TedShifrin. I'll send an email soon.

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