Mathematics - 2020-07-13

user76284

04:45

@geocalc33 A torus is the product of two circles $S^1 \times S^1$, or the quotient of $\mathbb{R}^2$ by $\mathbb{Z}^2$.

More generally $T^n \cong (S^1)^n \cong \mathbb{R}^n \mathbin{/} \mathbb{Z}^n$ for an $n$-dimensional torus.

Mary Star

08:16

Hello!! If a, b are sides of a triangle and it holds that a<b then does it hold for the respective heights that $h_a>h_b$ ?

Leaky Nun

@MaryStar area = base x height / 2

Mary Star

We have that $A= a \cdot h_a / 2$ and $A= b \cdot h_b / 2$. Since $a<b$ so that the results are both equal to A it mst be $h_a>h_b$ right? @LeakyNun

Leaky Nun

right

Mary Star

If the ratio of areas of two similar polygons is equal to 6, then the ratio of their perimeters is also equal to 6, right? (because of similarity) @LeakyNun

Ah no the ration of perimeters is then $\sqrt{6}$, or not? @LeakyNun

Leaky Nun

08:31

$\sqrt6$

Edward Evans

09:01

Let $M, R^\prime$ be $R$-modules and $m \in M$, $m \otimes 1 \in M \otimes_{R} R^\prime$. Is $\operatorname{Ann}_R(m) \otimes_R R^\prime \cong \operatorname{Ann}_{R^\prime}(m\otimes 1)$? (Just by $r \otimes r^\prime \mapsto rr^\prime$ ?)

Leaky Nun

@EdwardEvans can you not use $R'$

also what is $1$

-- Leaky Nun 2020

Edward Evans

hahaha

$1 \in R^\prime$

sorry

Leaky Nun

but $R'$ is not a multiplicative monoid

Edward Evans

$R^\prime$ is actually a ring and $\varphi : R \to R^\prime$ is a ring hom

forgot that

Leaky Nun

oh, I didn't know that

Edward Evans

09:05

yeah

Is it just extension of scalars? :P

Thorgott

10:38

what if R' has lots of torsion or something

Edward Evans

$R'$ aint got no torsion because it's flat (which I ALSO forgot to mention)

lol

imagine if I just stated the problem properly

Thorgott

ok then it sounds believable

but idk which SES to fit Ann in, actually

Edward Evans

yeah the problem is to show that $\varphi(\operatorname{Ann}_R(m))R^\prime = \operatorname{Ann}_{R^\prime}(m\otimes 1)$

$\varphi(\operatorname{Ann}_R(m))R^\prime = \operatorname{im}\lbrace \operatorname{Ann}_R(m) \otimes_R R^\prime \to R^\prime\rbrace$

errrrm and

right, the flatness of $R^\prime$ shows that $\operatorname{Ann}_R(m) \otimes R^\prime \to R^\prime$ is injective (coming from the inclusion $\operatorname{Ann}_R(m) \hookrightarrow R$)

so $\operatorname{Ann}_R (m) \otimes_R R^\prime \cong \varphi(\operatorname{Ann}_R(m))R^\prime$

so I'm just hoping the lhs is $\operatorname{Ann}_{R^\prime}(m\otimes 1)$, which sounds reasonable

maybe there's another way to do this

Thorgott

$\operatorname{Ann}_{R^{\prime}}(m\otimes 1)=\operatorname{ker}(R^{\prime}\rightarrow\operatorname{End}(\langle m\otimes 1\rangle))$

(I just wanted to write that, no clue if it's useful)

Edward Evans

hahaha

that's actually given as a hint

so probably is useful

but not sure how to use it

Thorgott

10:47

does tensoring with flat modules commute with kernels

it does, ofc

so uh

Edward Evans

tensoring is exact for flat modules right?

Thorgott

ye

you have $\operatorname{Ann}_R(m)=\operatorname{ker}(R\rightarrow\operatorname{End}_R(\langle m\rangle))$

now tensor it to get $\operatorname{Ann}_R(m)\otimes_RR^{\prime}=\operatorname{ker}(R\rightarrow\operatorname{End}_R(\langle m\rangle))\otimes_RR^{\prime}=\operatorname{ker}(R\otimes_RR^{\prime}\rightarrow\operatorname{End}_R(\langle m\rangle)\otimes_RR^{\prime})$

Edward Evans

oh nice

Thorgott

it remains to figure out that $\operatorname{End}_R(\langle m\rangle)\otimes_RR^{\prime}=\operatorname{End}_{R^{\prime}}(\langle m\otimes1\rangle)$ (hopefully)

Edward Evans

yeah this algebra course is mostly comprised of hope

Drathora

10:57

Thorgott do you think the coarea formula could be adapted to work for something like $\text{sgn(sin(}x))$ by glueing functions together with indicator functions?

Trying to think my way through it but struggling

Thorgott

11:16

sure

that's a locally constant function modulo a nullset, so very well-behaved

Drathora

I guess it should read (x+y) inside to make it multivariable

My concern here is that we're no longer Lipschitz, and we can't just rewrite the function by changing it on a null-set to make it Lipschitz

Thorgott

@Edward the map should just be $\varphi\otimes r^{\prime}\mapsto\varphi\otimes r^{\prime}$ (where the first one is an element in the tensor product and the second one is a tensor product of maps and $r^{\prime}$ represents left-multipolication with $r^{\prime}$); this looks like the identity, so it surely is a bijection (pray)

@Drathora when you remove the null set where it's 0, the function is locally constant, so locally Lipschitz

I mean, I don't even need the coarea formula or anything remotely fancy to tell you that a locally constant function is Lebesgue-Lebesgue-measurable

Drathora

Okay instead how about just a function that jumps, like maybe (floor(x)+y,x) as an example

I guess that's still sort of trivial in ways, but you get the idea

Something with a bunch of jumps that isn't constant

If locally lipschitz is good enough then yeah we're good

Thorgott

wait, actually, what were we trying to show again?

Drathora

11:33

That the preimage of any null-set is null under a function $f$ with the property

Where the property is something like the determinant of the Jacobian multiplied by its transpose is greater than zero almost everywhere

Alongside some other requirement we need to decide on

Thorgott

but that's clearly false for something like sgn(sin(x+y))

Drathora

Ye agreed, I should have used a better example

I just wanted to use sgn as an easy way to make these big discontinuities

Thorgott

huh, I just realized I was full of shit when I said that your property is equivalent to asking for Lebesgue-Lebesgue-measurability, it's a stronger property

Drathora

5

Q: Class of Lebesgue-Lebesgue measurable functions?

A function $f:\mathbb{R}^n\to\mathbb{R}^m$ is Lebesgue-Lebesgue measurable iff inverse images of Lebesgue measurable sets are Lebesgue measurable. Seems to me that since projections* and arithmetic operations are Lebesgue-Lebesgue measurable surjective linear transformations have to be Lebesgue-...

That's what this answer claims also though right?

Thorgott

Yes, but it's wrong

that said, the approach in that answer is strong enough to still be helpful with figuring out your property

Drathora

11:39

As in the answer is wrong?

But yeah even if that statement in the answer is wrong, as far as I can see the proof still works fine

For me, although idk about the original question

Thorgott

Lebesgue-Lebesgue-measurable means that the preimage of any Lebesgue-measurable set is Lebesgue-measurable, but the function is already Borel-Borel-measurable (since it's continuous), meaning that the preimage of any Borel-measurable set is Borel-measurable. Now, any Lebesgue-measurable set is the union of a Borel-measurable set and a null set. The preimage of the Borel-measurable set is a Borel-measurable set, hence Lebesgue-measurable.
So what's missing for such a function to be Lebesgue-Lebesgue-measurable is that the preimage of a null set is Lebesgue-measurable. That is implied by, but…

just look at the constant $1$ function, it's the opposite of non-nullifying - the preimage of a singleton is the entire space -, but it's very much Lebesgue-Lebesgue-measurable - in fact, the preimage of any set is measurable, it's either the empty set or the entire space

not sure what was riding me when I initially claimed it's the same notion, sorry

Drathora

ah

Haha, I posted a bounty on my original question and immediately an obvious counter-example was pointed out, and now I'm really confused how I missed it

Thorgott

that said, the coarea formula approach from the answer still works as intended

the real issue with the sgn(sin(x+y)) example is that its jacobian is 0 a.e.

Drathora

ye

Just really struggling to figure out what nice-to-verify but broad as possible condition I can put on $f$

I'm a bit confused by part of that answer above actually

Since I'm pretty sure the coarea formula requires Lipschitz

But $C^1$ only implies Lipschitz on a closed interval right/

Edward Evans

12:00

@Thorgott I guess it looks like the identity but if I want to check the kernel of that map out I want to know when $\varphi \otimes r^\prime$ is the zero map, i.e. when $(\varphi \otimes r^\prime)(m\otimes 1) = \varphi(m) \otimes r^\prime = 0$

hrmrmrmrmrm

ah

$r^\prime \cdot 1 = 0$

lol

Thorgott

$C^1$ implies Lipschitz on compacta, but that should suffice

split up and countable additivity

maths researcher

12:16

Suppose $F:N\rightarrow \mathbb{R}$ is smooth. Then for $p\in N$ and $X_p\in T_pN$, $dF_p(X_p)=X_pF$ .

My attempt:

Let $X_p\in T_pN$. Let $c:(-\epsilon,\epsilon)\rightarrow N$ be a curve such that $c(0)=p$ and $c'(0)=X_p$. Hence, $X_pF=\frac{d}{dt}|_0(F\circ c)=dF\circ c_0(\frac{d}{dt}|_{t=0})$ = $(dF_p\circ dc_0)(\frac{d}{dt}|_{t=0})=dF_p(dc_0(\frac{d}{dt}|_{t=0}))=dF_p(c'(0))=dF_p(X_p)$.

I feel there's something missing... not sure what though

Drathora

My concern is something like $1/x$ on (0,1] for example

Thorgott

I think all you need is that the limsup of the difference quotient is bounded everywhere

In this particular case, it's easy: just split up $(0,1]$ into $[1/2,1],[1/3,1,2],[1/4,1/3],...$ and do it on each, I think that should work

Drathora

Ah, I see

Thorgott

@mathsresearcher I don't follow. What's $c_0(\frac{d}{dt}\vert_{t=0})$?

maths researcher

I meant $d(F\circ c_0)(\frac{d}{dt}|_{t=0})$, @Thorgott

Thorgott

12:28

@Drathora the general fact is that whenever you have finite limsup of difference quotient everywhere, you can decompose your set into a countable union of measurable sets, on each of which the restriction of your function is Lipschitzian

this is a theorem in Federer

Drathora

I see

Thorgott

@mathsresearcher $d(F\circ c)_0$?

Drathora

So following that, it seems like we at least can say that if we have a countable number of points where the function is non-Lipschitz, then we can deal with that

maths researcher

oops again, yeah @Thorgott

Drathora

Which I'd be satisfied to be honest, I don't feel the need to stretch as far as "almost-everywhere"

*satisfied with

Later

13:37

@robjohn Hello. I need some technical help about the JavaScript code used for rendering LaTeX codes in chat rooms. Can you help me? (If you think talking about that may be off-topic for this room, we can talk in another room)

Drathora

14:06

@Thorgott that theorem about decomposition into countable unions of measurable sets, is there an easy way for me to look it up without having immediate access to a copy of Federer?

Oh wait nevermind, I think I understand, we're just decomposing into a countable union and then interchanging a sum with an integral

So whenever we can do that countable decomposition, we're good

Thorgott

I don't know any other reference

maybe in other texts about geometric measure theory of specifically rectifiability

Drathora

That's okay, I think I'm almost done with this actually now

Vio Ariton

@TobiasKildetoft ah you are right - it was right under my nose. I don't know why I always overcomplicate things. Thanks!

Drathora

Final part is the actual verification that a function and a set can undergo this decomposition

robjohn

@Later what's up?

I haven't looked at the code for a long time, so I may need to review what was done.

Later

14:57

@robjohn Thanks for your response. I can use your JavaScript code "Start ChatJax" on some IPad browsers such as Safari and Chrome easily without any problem. But, some browsers cannot run that code in chat rooms while they work well on your webpage. Can that JavaScript code be modified to solve the problem?

Edward Evans

15:48

@Thorgott Idk if I'm making this too complicated; if I have $R \to M, r \mapsto r\cdot m$ and tensor this with $R^\prime$ I get $R \otimes_R R^\prime = R^\prime \to M \otimes_R R^\prime, r^\prime \mapsto ???$,

cuz $\operatorname{Ann}_R(m) := \ker \lbrace R \to M, r \mapsto r\cdot m\rbrace$

Drathora

@Thorgott got access to Federer now. Don't suppose you happen to know where that theorem was in this giant book? I'm looking through it now but there's a lot of material D:

Thorgott

16:05

@Drathora 3.1.8.

@Edward yeah, that should do the trick

it was silly of me to bring in endomorphism modules

Drathora

thanks, will take a look now

Thorgott

good luck trying to untangle his notations and definitions

Alessandro Codenotti

I open the chat hoping that there is something nice being discussed and I see Federer instead

Edward Evans

@Alessandro do you know the album Catch Thirty-Three ?

also, yo

Alessandro Codenotti

The Meshuggah one?

Edward Evans

16:10

yeah lol

The 47 minute continuous suite

Thorgott

why wouldn't you love Federer

Drathora

haha

My only query is, what's the "ap" before the lim sup

Thorgott

approximate

Drathora

ah

robjohn

16:43

@Later I am not sure what the problem is, so it is hard to suggest a fix. If it works on the LaTeX on the installation page, I don't know why it wouldn't work elsewhere. Some browsers might be excluding certain JavaScript commands, but if it works on the installation page, it seems that they are not blocking the JavaScript.

Drathora

Nice I think I've got a reasonably nice statement proven now

Suppose $f : \mathbb{R}^m \rightarrow \mathbb{R}^n$ is s.t. det($Jf(x)Jf(x)^T$) > 0 for almost all $x$ in $\mathbb{R}^m$ and that $ \text{lim}_{x \rightarrow a} \text{sup} \frac{\lvert f(x)-f(a) \rvert }{ \lvert x-a \rvert } < \infty $ for all $a$ in $\mathbb{R}^m$.

Then the f-preimage of any null-set is null.

Now I guess the final thing I'm wondering about is does the first precondition tell us something about the second holding

Thorgott

in the $C^1$-case, the second is always satisfied

Drathora

Oh, and I guess the statement doesn't actually need to be for the whole of R^m where that's mentioned

We can restrict to a subset

Oh, and the lim sup condition only needs to hold for almost all $a$ in the domain of f

Later

16:58

@robjohn The browsers I have tested allows the JavaScript code to run in chat rooms, but the MathJax codes are not rendered. I guess if there were some hyperlink in a chat room, like the one in your installation page, then the MathJax codes would be rendered in the room. I added a hyperlink in the description of a test room, but the problem is when you tap on the link, a new page will be opened.

I meant a new window.

Sophie

19:17

for a square matrix is there a way of encoding the operation $\sum_i\sum_j\frac{|a_{ij}-a_{ji}|}{\sum_h a_{ih}}$ in standard operations?

maths student

(a) If $a, b$ are positive quantities such that $(a<b)$ and if $a_{1}=\frac{a+b}{2}, b_{1}=\sqrt{a_{1} b}, a_{2}=\frac{a_{1}+b_{1}}{2}, b_{2}=\sqrt{a_{2} b_{1}}, \ldots, a_{n}=\frac{a_{n-1}+b_{n-1}}{2}$
$b_{n}=\sqrt{a_{n} b_{n-1}}, \ldots$ then show that $\lim _{n \rightarrow \infty} b_{n}=\frac{\sqrt{b^{2}-a^{2}}}{\cos ^{-1} \frac{a}{b}}$

Sophie

Can you prove that $a_n$ and $b_n$ converge to the same number?

maths student

19:33

@Sophie no.

Sophie

I think the proof follows from there. You have to use the am-gm inequality repeatedly

maths student

but why the factor of cos inverse in final limit

@Sophie can you explain few steps

Sophie

I haven't proven it, so I don't know the details. I just gave you an outline

robjohn

20:38

@Later There shouldn't be a new page opened when the bookmark starts with javascript:

perhaps there is some security setting that is not letting javascript alter the page.

Ted Shifrin

Heya, @robjohn. So glad your script still works faithfully!

Have you ever seen the result in maths student's post ten lines up? I know how to prove convergence easily, but I don't see what the limit actually is!!!

Thorgott

huh, looks like a weird variation of the standard AGM thing

Ted Shifrin

Yes, I've seen the double sequence before (although the indexing is slightly off here). I've assigned it for homework. But I've never pondered the limit.

Thorgott

I think in the regularly indexed version, the limit is only expressible via elliptic integrals or something of the sort

interesting that one index shift gives something actually computable

Ted Shifrin

Yeah, the index change does totally change the limit.

Playing with Mathematica, the sequence(s) converge incredibly rapidly in both cases.

Oh, I see what the index shift does. If we let $c_n = a_n/b_n$, then we get $c_n = \frac1{\sqrt2}\sqrt{1+c_{n-1}}$

I still don't see how to find the limit, though.

FruDe

21:06

Hello, this is my first time on mathematics chat

Ted Shifrin

Hello, @FruDe.

Edward Evans

Evening, @FruDe

robjohn

@TedShifrin It does look like a phase shifted AGM

Ted Shifrin

Yeah, it is, @robjohn, and I can use scale-invariance to reduce to $1,\lambda>1$, but I can't see how to get that crazy limit.

robjohn

i will look a bit

Ted Shifrin

21:09

I've never ever seen this. Not the end of the world shocking, I realize.

So, what brings you by, @FruDe?

Sophie

21:42

setting $b=1$ wlog and $a=\cos(\theta)$ yields $b_n=\cos(\theta/2)\cos(\theta/4)...\cos(\theta2^{-n})$

@mathsstudent @TedShifrin this also means that I was wrong before oops

Ted Shifrin

LOL, wrong about what?

I'll have to check your claim in a bit.

Sophie

$a_n$ and $b_n$ converging to the same number

Ted Shifrin

No, that's correct.

Sophie

really?

Ted Shifrin

Yes. I can prove it several ways.

Sophie

21:45

well the bit about using am-gm repeatedly was wrong at least, so the proof I had in my head didn't hold

Sophie

22:10

phew here's the proof by induction that $b_n=\prod_{h=1}^n\cos(\theta 2^{-h})$ and $a_n=b_n\cos(\theta 2^{-n})$: $a_{n+1}=(a_n+b_n)/2=b_n(1+\cos(\theta 2^{-n})/2=b_n\cos(\theta 2^{-n-1})^2$ and $b_{n+1}=\sqrt{a_{n+1}b_n}=b_n\cos(\theta 2^{-n-1})$ and $a_{n+1}=b_{n+1}\cos(\theta 2^{-n-1})$

FruDe

Hello, thanks for the welcome @Ted

I like using math stackexchange but I've never been on the chat, just wanted to see what goes on here :)

robjohn

22:27

@TedShifrin I think I have the limit...

Sophie

I'm only missing the proof that $\prod_{h=1}^\infty \cos(\theta 2^{-h})=\sin(\theta)/\theta$

robjohn

@Sophie That is $\cos(x/2^n)=\frac{2^{n-1}\sin(x/2^{n-1})}{2^n\sin(x/2^n)}$

and it telescopes

robjohn

22:52

I am just massaging my limit to see if it looks like the limit above

Sophie

@robjohn did you do it the same way I did?

robjohn

23:11

@Sophie I don't know. I haven't looked at your approach yet.

The recursion starts with
$$
\begin{align}
a_{n+1}&=\frac{a_n+b_n}2\tag{1a}\\
b_{n+1}&=(a_{n+1}b_n)^{1/2}\tag{1b}
\end{align}
$$
Combine $\text{(1a)}$ and $\text{(1b)}$:
$$
\frac{b_{n+1}^2}{b_n}=a_{n+1}=\frac{\frac{b_n^2}{b_{n-1}}+b_n}2\tag2
$$
Divide $(2)$ by $b_n$:
$$
\left(\frac{b_{n+1}}{b_n}\right)^2=\frac{\frac{b_n}{b_{n-1}}+1}2\tag3
$$
Set $c_n=\frac{b_n}{b_{n-1}}$:
$$
c_{n+1}^2=\frac{c_n+1}2\tag4
$$
Note the similarity between $(4)$ and the identity
$$
\cos^2\left(x/2^{n+1}\right)=\frac{\cos\left(x/2^n\right)+1}2\tag5

$(7)$ and $(8)$ give an explicit formula for $b_n$

Sophie

hmm they're largely equivalent

robjohn

@Sophie I am sure they are based on the same ideas

Ted Shifrin

Very clever, both of you. I was not persistent enough.

Interesting the difference the index shift makes.

robjohn

$b_n=\frac{\sqrt{b^2-a^2}}{2^n\sin\left(\arccos\left(\frac ab\right)2^{-n}\right)}$

Krijn

hi all

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$Mathematics$ Mathematics