4:45 AM
@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.

3 hours later…
8:16 AM
Hello!! If a, b are sides of a triangle and it holds that a<b then does it hold for the respective heights that $h_a>h_b$ ?

@MaryStar area = base x height / 2

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

right

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

8:31 AM
$\sqrt6$

9:01 AM
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 $r \otimes r^\prime \mapsto rr^\prime$ ?)

@EdwardEvans can you not use $R'$
also what is $1$
-- Leaky Nun 2020

hahaha
$1 \in R^\prime$
sorry

but $R'$ is not a multiplicative monoid

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

oh, I didn't know that

9:05 AM
yeah
Is it just extension of scalars? :P

2 hours later…
10:38 AM
what if R' has lots of torsion or something

$R'$ aint got no torsion because it's flat (which I ALSO forgot to mention)
lol
imagine if I just stated the problem properly

ok then it sounds believable
but idk which SES to fit Ann in, actually

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

$\operatorname{Ann}_{R^{\prime}}(m\otimes 1)=\operatorname{ker}(R^{\prime}\rightarrow\operatorname{End}(\langle m\otimes 1\rangle))$
(I just wanted to write that, no clue if it's useful)

hahaha
that's actually given as a hint
so probably is useful
but not sure how to use it

10:47 AM
does tensoring with flat modules commute with kernels
it does, ofc
so uh

tensoring is exact for flat modules right?

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

oh nice

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

yeah this algebra course is mostly comprised of hope

10:57 AM
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

11:16 AM
sure
that's a locally constant function modulo a nullset, so very well-behaved

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

@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

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

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

11:33 AM
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

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

Ye agreed, I should have used a better example
I just wanted to use sgn as an easy way to make these big discontinuities

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

5

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?

Yes, but it's wrong
that said, the approach in that answer is strong enough to still be helpful with figuring out your property

11:39 AM
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

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

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

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.

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/

12:00 PM
@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

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

12:16 PM
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

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

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

Ah, I see

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

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

12:28 PM
@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

I see

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

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

oops again, yeah @Thorgott

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

1 hour later…
1:37 PM
@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)

2:06 PM
@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

I don't know any other reference
maybe in other texts about geometric measure theory of specifically rectifiability

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

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

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

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

2:57 PM
@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?

3:48 PM
@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 $\operatorname{Ann}_R(m) := \ker \lbrace R \to M, r \mapsto r\cdot m\rbrace$

@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:

4:05 PM
@Drathora 3.1.8.
@Edward yeah, that should do the trick
it was silly of me to bring in endomorphism modules

thanks, will take a look now

good luck trying to untangle his notations and definitions

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

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

The Meshuggah one?

4:10 PM
yeah lol
The 47 minute continuous suite

why wouldn't you love Federer

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

approximate

ah

4:43 PM
@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.

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

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

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

4:58 PM
@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.

2 hours later…
7:17 PM
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?

(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 $\lim _{n \rightarrow \infty} b_{n}=\frac{\sqrt{b^{2}-a^{2}}}{\cos ^{-1} \frac{a}{b}}$

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

7:33 PM
@Sophie no.

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

but why the factor of cos inverse in final limit
@Sophie can you explain few steps

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

1 hour later…
8:38 PM
@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.

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!!!

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

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.

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

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.

9:06 PM
Hello, this is my first time on mathematics chat

Hello, @FruDe.

Evening, @FruDe

@TedShifrin It does look like a phase shifted AGM

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.

i will look a bit

9:09 PM
I've never ever seen this. Not the end of the world shocking, I realize.
So, what brings you by, @FruDe?

9:42 PM
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

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

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

No, that's correct.

really?

Yes. I can prove it several ways.

9:45 PM
well the bit about using am-gm repeatedly was wrong at least, so the proof I had in my head didn't hold

10:10 PM
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 $a_{n+1}=b_{n+1}\cos(\theta 2^{-n-1})$

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 :)

10:27 PM
@TedShifrin I think I have the limit...

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

@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

10:52 PM
I am just massaging my limit to see if it looks like the limit above

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

11:11 PM
@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 $b_n$

hmm they're largely equivalent

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

Very clever, both of you. I was not persistent enough.
Interesting the difference the index shift makes.

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

hi all