Mathematics - 2020-07-01 (page 2 of 2)

6:00 PM

When $f$ is integrable, then we say that there exits a partition for a given epsilon such that
$$
U(f,P) - L(f,P) \lt \epsilon \\
\sum_{i=1}^{n} (M_i - m_i) (x_i - x_{i-1}) \lt \epsilon $$
Then, what is actually small in that summation? Is $(M_i - m_i)$ small (that is the difference between sup and inf of $f$ in an interval) or the length of interval $(x_i - x_{I-1})$ is itself small?

Drathora

Would C^1 imply that B contains a ball or am I mistaken?

Ted Shifrin

@Michael: I'm not sure I see why.

You don't know, @Knight. Think of it as sums of areas of rectangles.

Michael Albanese

I don't think it's all that obvious. One way to see it is to realise $\mathbb{Z}_2\ast\mathbb{Z}_2$ has $\mathbb{Z}$ as an index 2 subgroup, and know enough about 3-manifolds to know that the only closed orientable 3 manifold with fundamental group $\mathbb{Z}$ is $S^1\times S^2$.

Not satisfying, I know.

Ted Shifrin

That's actually a really weird condition, @Drathora. I was missing the link between $|s-t|$ and $1/p$.

Knight

@TedShifrin well in that sense both are very small

Ted Shifrin

6:03 PM

No, not necessarily. Think about our example from last night. There $M_i-m_i$ was about $20$.

Knight

@TedShifrin Thats exactly whats troubling me.

Ted Shifrin

It shouldn't trouble you.

So you make the base of the rectangle sufficiently small.

Knight

We tried to minimize the effect of a jump by taking a very small interval

Michael Albanese

You can also realise $\mathbb{RP}^3\#\mathbb{RP}^3$ as the real blowup of $\mathbb{RP}^3$. Then there is a bundle $\mathbb{RP}^1 \to \mathbb{RP}^3\#\mathbb{RP}^3 \to \mathbb{RP}^2$ by projecting to a hyperplane. Pulling back by $S^2 \to \mathbb{RP}^2$ gives the trivial $\mathbb{RP}^1$-bundle on $S^2$, namely $S^1\times S^2$ and the induced map $S^1\times S^2 \to \mathbb{RP}^3\#\mathbb{RP}^3$ is the desired double covering.

Knight

But what about the $inf$ of $f$ in that small interval?

Ted Shifrin

6:05 PM

We don't care. $M_i-m_i\le M_i$ whenever $f\ge 0$.

If you are doing a general function, what is the original assumption about $f$ when you define integrable?

So here's an exercise you should do. Prove that any increasing function (probably NOT continuous) is integrable on $[a,b]$.

Knight

It is bounded

Ted Shifrin

Right. So, if $|f|\le M$ to start with, you know what about any $M_i-m_i$?

Drathora

Yeah, the condition is a little strange Ted. I'm sure that there's some argument I'm missing where for any \epsilon you move around in a $1/p \le \epsilon$ ball around $t$ down the axis-aligned directions. Hopping from point to point in $B$ until you reach your desired $s$. And then because all of your moves are approximated by the partial derivatives your $f(s)$ and $f(t)$ have a different greater than or equal to what's required.

And such a path of "hops" exists because $B$ is a positive measure set with that property on the Jacobian

Making that argument work on the other hand, is a little harder than waving my hands like that ^

Alessandro Codenotti

hi everyone

Drathora

hey!

Alessandro Codenotti

6:18 PM

@BalarkaSen Or use $\pi_1$ of wedge is free product of $\pi_1$'s, seems easier here

Balarka Sen

@MichaelAlbanese @TedShifrin So we want examples of prime manifolds with a non-prime cover?

Michael Albanese

Yes.

Balarka Sen

@Alessandro It's more elementary to argue with the universal cover, without invoking SvKT

Alessandro Codenotti

I don't know, I never learned covering spaces properly, so I avoid them if I can

Ted Shifrin

Hi @Alessandro

Alessandro Codenotti

6:21 PM

Hi Ted

Michael Albanese

There are apparently examples among non-orientable three-manifolds, but I'd prefer an orientable example.

Balarka Sen

There are no orientable 3-manifolds of this sort, I presume, because essentially you want to have embedded RP^2's which separate

Let me think

Michael Albanese

Yeah, no orientable 3-manifold examples.

Alessandro Codenotti

Doe anyone want to think about some general topology regarding Baire spaces? I believe there is a missing hypothesis in a theorem in a book I'm reading

Ted Shifrin

You're better at this stuff than all of us, Alessandro.

Balarka Sen

6:31 PM

@MichaelAlbanese Yeah I cannot come up with an example. I will have to spend some time thinking about it.

Nice question

Ted Shifrin

@Drathora If you don't assume the function is actually $C^1$ or at least differentiable, then nothing is approximated well by partial derivatives.

Michael Albanese

I was thinking of asking on MSE or maybe MO (I never know which one), but I just wanted to make sure I wasn't overlooking a simple example.

Balarka Sen

Also, your S^1 x S^2 -> RP^3 # RP^3 is the only example of a prime 3-manifold covering a non-prime 3-manifold, I believe

Drathora

Ah okay. That's frustrating

Balarka Sen

Yeah you should ask

Well, it's more likely that I am overlooking a simple example than you, so maybe you shouldn't take my word on that

Drathora

6:34 PM

The thing is, this holds for a function such as f(x,y) = (\lvert x \rvert + y, 2x + y) for example

Ted Shifrin

@Drathora: One of my challenge exercises in my book is to give a function all of whose directional derivatives at the origin are $0$ that is unbounded in any neighborhood of the origin!!!

Drathora

Interesting. I might give that a go in a moment, to see if it helps me think about this problem a little more clearly

Ted Shifrin

Well, probably not, but non-differentiable functions can be horrible, even if they have directional derivatives.

Alessandro Codenotti

Ah nevermind, I solved my issue, no missing hypothesis, I was missing the fact that if $U_n$ are open dense in $X$, then $\bigcap U_n$ is comeager, regardless of whether $X$ is Baire. The intersection might even be empty, but it is still comeager

Ted Shifrin

Hmm, I guess I don't know what comeager means.

I thought complement is nowhere dense.

Alessandro Codenotti

6:39 PM

comeager means that the complement is meager, and meager means that it is a countable union of nowhere dense sets

Drathora

I still think the result holds in the general case. But a restricted version is interesting in its own right I guess.

This is the last part of a proof I'm working on stating that such a function $f$ is "non-nullifying. I.e. the preimage of nullsets is null under the Lebesgue measure

Ted Shifrin

So how is the empty set comeager, @Alessandro?

Thorgott

preimage of nullsets being null amounts to saying it's Lebesgue-Lebesgue-measurable, hmm

Alessandro Codenotti

For example if you enumerate $\Bbb Q=\{q_n\mid n\in\Bbb N\}$ and set $U_n=\Bbb Q\setminus\{q_n\}$, the intersection is empty, but this is fine because the whole space is meager, being a countable union of singletons with empty interior

Ted Shifrin

Without differentiability (or preferably $C^1$) I'm not sure you can claim that, @Drathora, but I dunno.

@Alessandro: But what about a complete metric space?

Wait, @Drathora. That's not right. Did you change from injective to surjective derivative?

Thorgott

6:44 PM

well, if you have $C^1$, this ought to be true, no?

Alessandro Codenotti

The intersection is dense then, because complete metric spaces are Baire

Ted Shifrin

If I include $\Bbb R\subset\Bbb R^2$, the preimage of null certainly needn't be.

Knight

Ted is it the partition that really makes a function integrable? I have a reason for thinking that, in the summation $\sum_{i=1}^{n} (M_i-m_i}(x_i - x_{i-1})$ if we can make the length of interval small enough (that’s simple just take large amount of points) then summation will be very small and hence our function will integrable?

Drathora

Oh sorry, the overall property I'm showing is that if a function $f$ has the Jacobian property almost everywhere then it's non-nullifying

Ted Shifrin

I see, @Alessandro. I thought you were saying that in general the empty set is comeager. My apologies.

Drathora

6:45 PM

The Jacobian property being "linearly independent rows"

Alessandro Codenotti

The fact that $A\subseteq X$ is comeager iff $A$ contains the intersection of a family of dense open sets is true without assumptions on $X$, this is simply because $B\subseteq X$ is nowhere dense iff $X\setminus \overline{B}$ is dense

Ted Shifrin

Not true, @Knight. Take the function that is $1$ on rationals and $0$ on irrationals. Work it out.

Thorgott

hmm, then Ted's counterexample works, no?

no, nvm

you are asking for linearly independent rows, not columns

Ted Shifrin

@Drathora: Let's sort out the linear algebra. Now you're saying the different gradient vectors are linearly independent, rather than the different partial derivative vectors?

Knight

@TedShifrin for any partition $L(f,P)=0$ But I don’t know about upper sum

Ted Shifrin

6:48 PM

Linearly independent rows means the function is surjective, not injective, on the linear level.

Well, @Knight, think about it.

Thorgott

but then the claim is true, certainly

rank theorem implies there's locally a smooth change of coordinates making the function look like a projection and both diffeomorphisms and projections are Lebesgue-Lebesgue-measurable

(in the $C^1$ case, that is)

Drathora

Okay, so say I have a function $f$ on variables $x1,...,xm$, and $f(x1,...,xm) = (e1,...,en)$

Ted Shifrin

Yes, there was row/column confusion in the post.

That makes no sense, @Drathora.

Drathora

Where ei are real expressions

Knight

@TedShifrin $$U(f,P) = \sum_{i=1}^{n} M_i (x_i - x_{i-1}$$ by density of rationals we have $M_i =1$ But we can make $x_i -x_{i-1}$ as small as we please and hence upper sum will be zero

Ted Shifrin

6:50 PM

To me those are standard basis vectors. Horrid notation.

No, @Knight. Stop and write things out carefully. You're being sloppy.

Thorgott

I really should spend some time figuring out what the non-$C^1$ version of the rank theorem based on the non-$C^1$ version of the IFT looks like some day

Drathora

Oh, sorry. I use $\epsilon_i$ for these expressions normally, but I was being lazy

Ted Shifrin

And we have $m\ge n$. I always map $n$ to $m$, so this is doubly confusing to me.

Thorgott

Mapping $m$ to $n$ is fairly common

Ted Shifrin

I agree that if a linear map is surjective, preimage of null is null.

Thorgott

6:51 PM

You get an $n\times m$ matrix that way

Ted Shifrin

@Thorgott: Everywhere in my training and teaching we have $m\times n$ matrices.

I mean, occasionally I switch it up, but very rarely.

Thorgott

interesting, they're usually $n\times m$ for me

Ted Shifrin

Just shows you're a backwards algebraist.

Just like Herstein, mappings on the right.

Thorgott

oh god no

I've seen smooth maps $N\rightarrow M$ fairly often too

Ted Shifrin

LOL, not I.

Although that would give you an $m\times n$ matrix :P

Thorgott

6:54 PM

when I think of manifolds, I use $M$ first and then $N$, but when I think of natural numbers, I use $n$ first and then $m$

Ted Shifrin

You didn't respond to my $m\times n$ matrix, I see.

Balarka Sen

It depends on what manifold I am focusing on when I am writing. If I am going to write "Let M be a manifold, and consider a covering space of M: ..." I will write N -> M

Notation should be intuitive, not rigorous

Ted Shifrin

Not $E\to B$?

Thorgott

well, $N\rightarrow M$ would be the choice consistent with your matrix convention, then

Balarka Sen

What is B? M was my manifold

Ted Shifrin

6:56 PM

LOL

Fiber bundles, covering spaces, whatever.

Thorgott

Ted stole your manifold $M$ and replaced it with the manifold $B$

Ted Shifrin

B for base, of course.

Balarka Sen

But then the manifoldiness of B is not present

B seems like some random space

Drathora

Are we in agreement about what the linearly independent vectors are now? The ith vector is (d\epsilon_i/dx_1 , d\epsilon_i/dx_2,..., d\epsilon_i, dx_m)

Ted Shifrin

Thankfully, Guillemin-Pollack used $X$ and $Y$ for manifolds (which is good for local coordinates) and use $k$ and $\ell$ for dimensions typically. Much better.

Knight

6:57 PM

@TedShifrin Sir I tried but couldn’t find where I was wrong

Drathora

Or you can replace epsilon with f if you prefer to think of it that way I guess

Thorgott

$X,Y$ are plain topological spaces, not manifolds

Ted Shifrin

@Drathora: So the linear map goes across the rows. That's good. Linear functionals are row vectors.

Balarka Sen

Let $M$ be an $m$-dimensional manifold, $m \in M$ be a point and $(m_1, \cdots, m_m)$ be local coordinates around $m$

Ted Shifrin

smacks Thorgott

@Knight: Simplify the algebra. If all the $M_i=1$, what is the sum?

Drathora

6:58 PM

Yeah, I updated the post after you pointed out my mistake last night

Thorgott

jokes aside, $X,Y$ are schemes

Ted Shifrin

No scheming today.

Knight

@TedShifrin $b-a$ (a and b are points of interval on which f is bounded)

Ted Shifrin

Correct. It doesn't matter how small the $x_i-x_{i-1}$ are.

Knight

@TedShifrin It may be annoying to you but please explain what we exactly did when we proved the integrability of $g$ (it was $f$ all the way but at $c$)

Thorgott

7:00 PM

The result is probably true for totally differentiable functions that are not necessarily $C^1$

Knight

We made a new partition such that the interval containing $c$ became extremely small

Anonymous

Can someone point me to a rigorous proof of the fact that $\cup_{n \in \mathbb N} (-n, n)$ is a cover for $\mathbb R$? I guess the key point would be Archimedean property but I'd still be interested in the whole proof. Actually, can we also say $\cup_{n \in \mathbb N} (-n, n) = \mathbb R$?

Thorgott

If we can get a version of the rank theorem for totally differentiable not necessarily $C^1$ functions, it surely pans out

Ted Shifrin

We showed that given any $\epsilon$, there is a partition $P$ with $U(g,P)-L(g,P)<\epsilon$.

Knight

Yes. But the way we proved it was very non-intuitive to me.

Thorgott

7:02 PM

@S.D. The Archimedean property already gives you everything you want. And yes, the equality is precisely what it means for the sets $(-n,n)$ for $n\in\mathbb{N}$ to be a cover of $\mathbb{R}$

Ted Shifrin

@S.D. What more do you want to say than for any real number $x$, the statement $|x|<n$ holds for some (in fact infinitely many) $n$.

@Knight: It's totally intuitive. You use the fact that $f$ is integrable to get a partition $P'$ that works for $f$ (with $\epsilon/2$) and then we modify $P'$ to obtain a $P$ that works for $g$. This is good mathematics.

Since you don't know what $f$ is to start with, you can't do a calculational proof like you would for $f(x)=x$ or $f(x)=x^2$.

Knight

@TedShifrin Would you mind if I give you what I did after your explanation and where I got stuck?

Ted Shifrin

OK

Knight

Just a minute I’m writing it

Bhavay

@TedShifrin Hi. May i ask something too(as u told me u were a multi -tasker) ?

Ted Shifrin

7:09 PM

If it's quick.

Bhavay

Okay . The question states A point moves in x-y plane such that sum of distance from two mutually perpendicular lines is 3. The area enclosed by locus of point is ? In the solution they took two perpendicular lines to be x and y Axes . How can i prove it for any perpendicular lines?

Ted Shifrin

Why would you want to? Rotate the picture. Rotation doesn't change distance or area.

Anonymous

@Thorgott I think for a cover we just have $\subseteq$ rather than $=$

Ted Shifrin

But here you're doing the entire space.

So you can't have $\subsetneq$.

Bhavay

@TedShifrin right. Never occurred to me thanks.

Anonymous

7:14 PM

@TedShifrin Umm, what does "doing the entire space" mean?

Thorgott

depends on your definition of cover, but it doesn't make a difference in this case as Ted notes

he means we're covering the entire space ($\mathbb{R}$)

Anonymous

I don't see why $\cup_{n \in \mathbb N} (-n, n)$ can't be a larger set than $\mathbb R$ (without further justification)

Thorgott

then you need to revisit some material

how do you define $(-n,n)$

Ted Shifrin

Hint: Think about what "subset" means.

Anonymous

@Thorgott It's the open set in $\mathbb R$ with LUB $n$ and GLB $-n$?

Anonymous

7:17 PM

Oh, and it's a subset of $\mathbb R$ of course

Ted Shifrin

Aha. Open set in $\Bbb R$ :)

Thorgott

that's not a good definition

Knight

Should I present my work, it's big

All right, we have $ g:[a,b] \to \mathbb R$
$$
g(x)= \begin{cases}
f(x) & x \neq c \\
\alpha & x=c
\end{cases}
$$
(given that $\alpha \gt M_k$)
We're given that $f$ is intgerbale, that means there exist a partition $P=\{x_0, x_1 , \cdots x_n\}$ for a given small $\epsilon/2 \gt 0$ such that
$$
U(f,P) - L(f,P) \lt \epsilon /2\\
\sum_{i=1}^{k-1} (M_i - m_i) (x_i - x_{I-1}) + (M_k - m_k) (x_k - x_{k-1}) + \sum_{I=k+1}^{n}(M_i -m_i)(x_i - x_{I-1}) \lt \epsilon/2
$$
$[x_{k-1}, x_k]$ is the interval in which $c$ lies.

Ted Shifrin

It shouldn't be too long, @Knight.

Knight

Let's see what's difference between upper and lower sum is for $g$ on partition $P$
$$
U(g,P) - L(g,P) = \sum_{I=1}^{k-1} (M_i - m_i) (x_i - x_{I-1}) + (\alpha - m_k) (x_k - x_{k-1}) + \sum_{I=k+1}^{n} (M_i - m_i) (x_i - x_{I-1})$$

We can see that it's the term $(\alpha -m_k)(x_k - x_{k-1})$ thats ruining our job.

Thorgott

7:18 PM

$(-n,0)\cup(0,n)$ is also an open subset of $\mathbb{R}$ with sup $n$ and inf $-n$

but it is not the same thing as $(-n,n)$, of course

Anonymous

@TedShifrin Ah, so you mean the union of subsets of a set can't be larger than the set itself? (Because union of subsets is a subset)

Ted Shifrin

@Knight: Your (given that $\alpha>M_k$) is totally out of place.

Anonymous

That makes sense

Thorgott

so your definition isn't actually a definition

Knight

@TedShifrin I did it to make the problem easier. What should I do after that?

Ted Shifrin

7:19 PM

What's the biggest that $\alpha-m_k$ could POSSIBLY be, @Knight?

You don't need to nail down estimates best possible. Remember we said at the beginning that $f$ (and hence $g$) had to be bounded?

So what is the biggest possible value of $\alpha-m_k$ if $|g|\le M$?

Knight

@TedShifrin $M-m_k$

Ted Shifrin

Do better. I don't want $m_k$ in there.

Knight

if $|g| \leq M$ then we have $|\alpha| \leq M$ and hence the biggest $\alpha$ can be is $M$.

Ted Shifrin

And $\alpha-m_k$?

What's the biggest $-m_k$ can be?

Knight

$M$

Ted Shifrin

7:24 PM

So the answer to my original question is ... ?

Knight

so, the biggest $\alpha -m_k$ is 0

Ted Shifrin

Huh?

Knight

sorry

it would be $\alpha +M$

Ted Shifrin

Reread everything we've said.

(And when we're done with this step, you realize you don't care how $\alpha$ compares to $M_k$ and $m_k$.)

Knight

$$-M \leq \alpha \leq M \\
-M \leq m_k \leq M \\
0\leq \alpha - m_k \leq 0$$

Ted Shifrin

7:29 PM

You need to be a lot more self-critical than this.

Knight

I don't know what I'm doing wrong. It feels very bad when you cannot answer such a simple question

Ted Shifrin

If $x\le 3$ and $y\le 2$, then $x-y\le 1$? Let's try it: $x=3$ and $y=-5$, $x-y=8$.

user675453

You should take a break

Thorgott

multiplying by $-1$ flips inequality signs!!

Ted Shifrin

Notice that I specifically asked you for the biggest $-m_k$ could be, so that you would not fall in this trap.

Knight

7:32 PM

so, the answer is 2M

I'm feeling too dumb now

I ... just ...

Ted Shifrin

Cool, $2M$ is right.

So now how do you finish your proof?

Knight

$$(\alpha - m_k) \leq 2M \\
(\alpha - m_k) (x_k - x_{k-1}) \leq 2M (x_k - x_{k-1})$$

Ted Shifrin

Remember you want to add some new points to your partition. So what you have isn't quite what you want. Here $x_{k-1}$ and $x_k$ are already fixed in the original partition $P'$.

Knight

I... am.. just cannot do it Ted

I feel like...

Ted Shifrin

OK, go do something non-math.

Knight

7:37 PM

Thanks for all your efforts that you did today

Ted Shifrin

If you are taking a course, you should talk to your professor. If you're reading a book, maybe you need a book like Spivak that shows you more details and examples.

Knight

that "biggest possible value of $\alpha - m_k$" made me to feel so bad...

Ted Shifrin

Lunchtime for Ted. Bye for now, all.

Knight

I'm leaving bye

Michael Albanese

@BalarkaSen: I asked about it here.

Balarka Sen

7:41 PM

Upvoted, Michael. Thanks for letting me know

Thorgott

7:56 PM

urgh, I'm currently TeXing slides for a seminar talk and I just spent 3 slides proving auxiliary propositions using linear algebra only to prove a rather uninteresting technical lemma, because I'm not allowed to go for the 3-line proof using field theory instead

Drathora

:'(

Thorgott, I've just scrolled up and read your comments about my question, sorry, I had to deal with something

"The result is probably true for totally differentiable functions that are not necessarily $C^1$"

Thorgott

yeah, I would guess so

Drathora

We immediately lose access to $ \lvert x \rvert $ then right?

wait that wasn't phrased well

$f(x,y) = (\lvert x \rvert + y , x - 2y)$ doesn't have continuous partial derivatives around zero

Thorgott

it doesn't have a partial x-derivative at all at zero

Drathora

8:14 PM

oh sorry, I'm being foolish

Yeah so our set $B$ here would just be the space with 0 excluded

Thorgott

I'm not really thinking about your $B$, because I don't get it. What I know for sure is that $C^1$+Jacobian has linearly independent rows everywhere implies Lebesgue-Lebesgue-measurability. What I'm saying is that the same probably holds if we replace $C^1$ with just total differentiability.

Drathora

Oh okay. $B$ is just the subset of $\mathbb{R}^m$ where the property holds

Thorgott

yeah, the argument I have assumes it holds everywhere

actually, I guess you can relax it to "everywhere except on a closed set of measure zero"

but I don't see myself removing the closedness assumption

Drathora

8:30 PM

hmm

Thorgott

but my argument is probably suboptimal, cause I'm thinking from a differentiable viewpoint

I should probably look for a more measure-theoretic argument

Drathora

I was honestly surprised that I couldn't find more literature on how the differential properties of a function affect its measure-theoretic properties

Took me quite a long time before I stumbled across the 1-dimensional case in a paper from the 60s on the chain rule

Did you see the actual question I posted by the way Thorgott, or just what I've asked in here?

Thorgott

I've seen the actual question, but I prefer thinking about whether the conditions imply Lebesgue-Lebesgue-measurability directly rather than your technical condition, which I don't have a good intuition for

user193319

Are there any examples of continuous homomorphisms from a topological group that isn't discrete to a discrete group?

Thorgott

ok, it's time to look into the ultimate source of wisdom

if there isn't something helpful regarding this in Federer, there may as well be nothing helpful at all

Drathora

8:37 PM

:')

user193319

Other than the trivial homomorphism, of course.

Thorgott

@user193319 how about $\mathrm{sign}\colon\mathbb{R}^{\times}\rightarrow\{\pm1\}$?

Anonymous

(This is likely a silly question, but I'm just starting with this topic.) How to prove that the intersection of two intervals like $(-\infty, b_x]$ and $(-\infty, b_y]$ (where $b_x, b_y \in \mathbb R$) is non-empty?

Anonymous

I do understand it intuitively

Thorgott

Why is it true intuitively?

Anonymous

8:46 PM

@Thorgott I'm just imagining two cuts on the real number line :P

Anonymous

There must be an overlapping region

Thorgott

Where do they overlap?

Anonymous

$(-\infty, \mathrm{min}(b_x, b_y)]$?

Thorgott

well, then you already have a proof :)

not only have you shown that the intersection is non-empty, you have actually calculate the intersection

Drathora

A nice way to consider the proof is to say w.l.o.g assume $b_x \le b_y$. Then $t \in (-\infty , b_x]$ iff $t \leq b_x$. Likewise $t \in (-\infty , b_y]$ iff $t \leq b_y$.

Anonymous

8:51 PM

I have a few confusions though. For instance, $b_x$ or $b_y$ couldn't possibly be $-\infty$ right?

Anonymous

I guess $\mathbb R$ doesn't contain $-\infty$

Anonymous

Otherwise $(-\infty, -\infty)$ would be empty (or make no sense)

Drathora

Yeah, our assumption is that $b_i \in \mathbb{R}$

Anonymous

@Drathora Right, that looks neat

Drathora

Yeah, and then you just finish off by showing that $t$ being in the first interval implies it's also in the second interval

And of course note that the first interval was non-empty to begin with

Anonymous

8:53 PM

@Drathora I see. So what if we consider the extended reals with $\pm \infty$ included? In that case is something like $(-\infty, -\infty)$ empty?

Drathora

Yes, as an open interval doesn't include its endpoints

That interval would be defined as follows:

Thorgott

Indeed, just as $(x,x)$ is empty for any real number $x$

Drathora

$I = \{ r in \mathbb{R} : -\infty \le r \le -\infty \}$

Thorgott

(this generalizes to intervals in any totally ordered space)

one more dollar

user193319

@Thorgott Thanks!

Anonymous

8:56 PM

@Thorgott Thanks, this makes sense

Drathora

Got there eventually haha

Anonymous

8

A: Why is the empty set considered an interval?

Given an ordered set $(A,\leq)$ an interval is a subset which is convex. Namely $I$ is an interval in $A$ if whenever $a,b\in I$ then for every $x$ such that $a<x<b$, we have that $x\in I$ as well. In the real numbers, because the order is complete, it follows that every bounded interval has end...

Anonymous

This answer apparently explains it well

Drathora

Is there a way to make chat appear in display mode?

Or are the rest of you just reading it as is like I am?

Or copying it into Tex or something?

Anonymous

@Drathora There's a script to render MathJax in chat

Anonymous

8:57 PM

math.ucla.edu/~robjohn/math/mathjax.html

Drathora

Oh I see it in the top right

thanks, that'll speed up the rate I can read in here haha

Alright, now back to thinking about the question

Thorgott

9:17 PM

@Drathora this is probably the right way to do this measure-theoretically: math.stackexchange.com/questions/1046947/…

your Jacobian is precisely made as to satisfy the condition given there at the end

so what's left is to investigate how we can minimalize the hypotheses necessary for the coarea formula to apply

Drathora

Hmm okay, I might need some minutes to digest this haha

This has already been helpful though. I've only ever seen these types of functions referred to as "non-nullifying" by a computer scientist and "measure-0-reflecting" by a category theorist

So another term to search for is always nice haha

Thorgott

lmao the category theorist

I've never heard "non-nullifying" before tbh

Drathora

dl.acm.org/doi/pdf/10.1145/2103621.2103721

^non-nullifying

Thorgott

but it's precisely what's missing for a Lebesgue-Borel-measurable function to be Lebesgue-Lebesgue-measurable, so it makes sense

Drathora

archive.numdam.org/article/CTGDC_1994__35_4_291_0.pdf

^ measure-zero-reflecting

Haha

Thorgott

9:36 PM

categorical measure theory is scary

Drathora

Yeah that paper is a little spooky

I'm thinking of trying to extend that paper actually. It only considers finite measures

I'd like to extend it to consider s-finite measures

(Yes that is an s, not a sigma)

Thorgott

I'm not gonna ask you what that means :P

Drathora

Haha, I honestly think they're a lot simpler than sigma-finite measures. They're literally just countable sums of finite measures

A much friendlier definition :D

Alright, I've went through the answer you linked

And it's looking very promising

Can you just verify for me what $Df(x) Df(x)^*$ is?

Oh

Thorgott

$Df(x)$ is the jacobian at $x$ and $Df(x)^{\ast}$ is the adjoint, i.e. the transpose in this case

Drathora

Yeah okay, adjoint was the notation I was forgetting.

You know, this condition actually might be nicer for my purposes than the linear independence condition

Ted Shifrin

9:49 PM

I'm not paying attention, but my conjecture is that "this condition" is totally equivalent.

Thorgott

it is, Ted

Ted Shifrin

I never saw the condition stated, but I guessed.

Drathora

I agree, but viewing it like this it seems a lot more "computable" to me, although maybe I'm wrong

So the only additional assumption here was that $f$ was $C^1$ from what I can see.

Thorgott

yeah, thought the $C^1$ case was already clear before

my hope is that the coarea formula approach generalizes to the case with weaker assumptions

though it does need some form of Lipschitz-ness conditions, I think, so only the existence of partials will not be handled by this either

Drathora

Mhmm, I'm just spending a moment now seeing where the assumption is actually used

Thorgott

9:56 PM

it's used to apply the coarea formula

Drathora

Is the continuity also used to justify that $f^{-1}(E)$ is Borel?

Thorgott

oh fuck

I just realized that I was assuming $f$ continuous throughout

but in the case of only partials this isn't even a given

hmm, but maybe only assuming existence of partials still guarantees continuity a.e.?

I've never thought about this

Drathora

I should clarify that having only partials isn't something I'm deadset on

Thorgott

are you willing to settle on totally differentiable?

Drathora

What I really want is to be able to perform analysis on a function like $f(x,y) = (\lvert x \rvert + y , y)$ for example

Thorgott

10:01 PM

that's $C^1$ a.e.

Drathora

Yeah

Thorgott

in which case, I'm pretty sure we're basically done already

yeah, we are

Drathora

Or something that maybe has a jump like $f(x,y) = (x,y)$ when $x \le 0.5$ and $= (2x,y)$ otherwise

For example

So basically continuous a.e. and $C^1$ a.e. would be nice, if we can't generalise further

Ideally though I'd like to keep this as general as possible, but this is definite progress haha

Thorgott

I think we can do a bit better

What's left is a 3-stage plan:
1. Look up the most general Lipschitz-type condition required to make the coarea formula work
2. figure out how much differentiability we need to ensure this condition by some MVT-type argument
3. see where we can an "a.e." in

sadly I don't have the time to think this through in detail rn

Drathora

Thanks a lot for your help. I'm going to take what you've written here and see what I can do with all this information.

I'll update you if I make any breakthroughs, I've suspected this result held for a year and a half now, so I'm too far in to go back now haha

Although most of that time wasn't spent thinking about it of course

Balarka Sen

10:17 PM

@TedShifrin Been thinking about a little complex geometry

Ted Shifrin

oh oh @Balarka

Thorgott

np, I'll probably think about this again tomorrow

Balarka Sen

If $\Sigma \subset M$ is a real hypersurface in an almost complex manifold $(M, J)$, consider distribution $\xi$ along $\Sigma$ consisting of the maximal complex subspaces, so $\xi = T\Sigma \cap J T\Sigma$. If it's given by the Pfaffian equation $\alpha = 0$, denote $\omega := d\alpha|_{\xi}$. Apparently $\Sigma$ is called pseudoconvex ($J$-convex?) if $\omega(X, JX) > 0$

Ted Shifrin

This is CR geometry.

Balarka Sen

If I write $\Sigma$ locally as $\phi = 0$, I compute $\omega = -dd^{\Bbb C}\phi$ where $d^{\Bbb C}\phi := d(\phi \circ J)$. I am trying to see that if $\Sigma$ is geometrically convex, it is pseudoconvex. Should be a computation

@TedShifrin Ah ok I wasn't aware of the term

Ted Shifrin

10:25 PM

What does geometrically convex mean in a general manifold? Geodesically convex?

Balarka Sen

Ah I am sorry, I meant in the case $M = \Bbb C^n$.

Ted Shifrin

Aha. Yes. Then this is classical.

At some point you should check out the famous Chern-Moser paper on CR hypersurfaces (early 70s).

Balarka Sen

So I guess $\omega = -2i \sum_{i, j} d^2 \phi/d z_i d \overline{z}_j \; dz_i \wedge d\overline{z}_j$ if I am doing this correctly.

And the Hessian pops out

Ted Shifrin

You have to relate real and complex hessians, but yes.

And I hate it when $i$ is $\sqrt{-1}$ and an index in the same formula. :D

Balarka Sen

Ah, gotcha. $\Bbb{II}$ for $\phi^{-1}(0)$ is just the real Hessian, right?

Well, when $\|\nabla \phi\|$ is normalized.

Ted Shifrin

10:28 PM

Um, I guess. I would have to think and I have a headache.

Balarka Sen

@TedShifrin Noted. I am just beginning the basics, really

Yeah I'll just compute, no worries

Also apologies for the index! I should have noted that

Balarka Sen

10:40 PM

Yeah, alright. Generally, $\Bbb{II}(X, Y) = D_X \mathbf{n} \cdot Y$ and $\mathbf{n}(p) = \nabla \phi(p)/\|\nabla \phi(p)\|$, so $$D_X \mathbf{n}(p) = \frac{H^{\Bbb R}(\phi)(p) X}{\|\nabla \phi(p)\|} - \frac{1}{\|\nabla \phi(p)\|^2} \frac{\nabla \phi(p)}{\|\nabla \phi(p)\|} \cdot H^{\Bbb R}(\phi)(p) X$$

$H^{\Bbb R}(\phi)(p)$ denoting the real Hessian matrix at $p$; let's just abbreviate that as $H$

Uh. $D_X \mathbf{n}(p) = (\|\nabla \phi(p)\|^2 H X - (\nabla \phi(p) \cdot HX) \nabla \phi(p))/\|\nabla \phi(p)\|^3$. Hm

(Forgot the vector $\nabla \phi(p)$ in the first thing I wrote)

So $D_X \mathbf{n} \cdot Y = ((\nabla \phi \cdot \nabla \phi) Y - (\nabla \phi \cdot Y) \nabla \phi ) HX/\|\nabla \phi\|^3$

That's a triple cross product and evaluates to just $\|\nabla \phi\|^2 Y$, so I do get $H(X, Y)/\|\nabla \phi\|$ as second fundamental form

I mean, because $\nabla \phi$ and $Y$ are orthogonal

Don't exactly remember how to relate real and complex Hessians, but $4 d^2/dz_p d\overline{z}_q = (d/dx_p + id/dy_q)(d/dx_p - i d/dy_q)$ which has real part $d^2/dx_p^2 + d^2/dy_q^2$

Ah what am I doing, should be $(d/dx_p + i d/dy_p)(d/dx_q - i d/dy_q)$, so real part what I wrote and imaginary part $d^2/dx_p dy_q + d^2/dy_p dx_q$

i also switched the signs but alright i dont really understand anything beyond symbols at this point

user862

11:23 PM

Hello, I was wondering if someone could help me think of a non-solvable subgroup that is the product of some of its soluble subgroups - specifically soluble subgroups $H, K, L$ where $G = HK = HL, L \neq K$. I know that $A_5 = A_4C_5$, but are there any other subgroups $K$ such that $G = A_4K$ or $G = KC_5$? Clearly we would need $|K| \in \{5, 12\}$, so this can't be possible for $A_5$. Maybe $A_6$?

Balarka Sen

@user862 why do you need $|K| = 5, 12$? $A_5$ contains a copy of $D_{10}$ generated by a $5$-cycle and a $2+2$-cycle, so in particular contains your $C_5$

so $A_4 D_{10} = A_5$ as well

user862

Ah, I was assuming that such subgroups would need to have orders such that the product of the orders of the subgroups $= |A_5|$.

Balarka Sen

But of course, $|HK| = |H||K|/|H \cap K|$

you can take massive subgroups with lots of intersection

user862

Quite right, I hadn't considered a non-trivial intersection.

Thank you!

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