Mathematics - 2018-01-24 (page 5 of 6)

@EricSilva: So the key idea was to throw all the $dz^i$ into the differential ideal, so that $\bar\partial$ turns into $d$ mod the ideal. But then there's a question of what dimension the integral manifolds will have.

Semiclassical

namely, you should assume that all the springs act at the same point O

Abcd

9:23 PM

Yeah, that's fine.

Eric Silva

@TedShifrin hmm

Semiclassical

without that, things get complicated without being terribly interesting

Ted Shifrin

@CookieToast: How's the Chapter 1 stuff coming/going?

Semiclassical

ok.

So, the compression of spring B is pretty obvious

Abcd

Maybe that's why "particle" (instead of block) is given in the question.

Semiclassical

9:24 PM

yeah

Abcd

@Semiclassical yes, then?

Ted Shifrin

@EricSilva: Nirenberg's theorem requires that $\mathscr I \cap \overline{\mathscr I} = \emptyset$, which holds here, I believe.

Semiclassical

if you move the particle towards B by a distance $\Delta x$, the the compression is $\Delta x$. Nothing strange there.

Abcd

yes

and the extension of the two other springs is $\Delta x\cos 30^\circ$?

Cookie Toast

@Ted I've finished every homework problem except the challenge problems. Its frustrating, but I've had a lot of homework this week and I havent had more than an hour or two this week to work on it.
That being said, I'm pretty much going to use your book instead of my classes for the semester wherever possible.
My professor "proved" the vector equation for $\cos\theta$ by rewriting the law of cosines formula today. Our class' textbook is much the same

Semiclassical

9:26 PM

That's too quick.

Eric Silva

@TedShifrin mmmmk

Abcd

Being sarcastic? Am I right?

Ted Shifrin

@Cookie: You'll find that my course is a more seriously mathematically thorough course. I've taught plenty of engineering-style multi calculus courses, too :)

Semiclassical

You may be, but I'm not sure off the top of my head, and I'd prefer to do this more carefully

What're the new and old positions of the particle?

For simplicity, let's take the coordinate system to have B at the top i.e. on the positive y axis

Abcd

$(0,0) \to (0, x)$

Cookie Toast

9:28 PM

@Ted I am a bit farther in the videos though. I managed to get through lecture four before school started!

Arrow

Given a locally Euclidean (locally homeomorphic to some Euclidean space) topological subspace $X\subset\mathbb R^n$ and $p\in X$, let $\mathrm{T}_pX$ denote the *tangent set* of $X$ at $p$, namely the set of derivatives of curves in $X$ based at $p$ and differentiable at zero.

Are the following conditions equivalent?

1. $\mathrm{T}_pX$ is a linear subspace of $\mathbb R^n$ of dimension $\dim_pX$.
2. We have the following equality, where the limit is taken in a translated neighborhood in $X$ of $p$. $$\lim_{h\to 0} \frac{\pi_{(\mathrm T_pX)^\perp}(h)}{\|h\|}=0$$ (Here $\pi_{(\mathrm T_pX)^…

Semiclassical

actually, I'd say $(0,0)\to (0,x)$ based on the fact that $B$ is at the top rather than on the right

you can do it the other way, of course

Kasmir Khaan

@TedShifrin if A,B subsets of R, and A is open, is the intersection of A and the closure of B contained in the closure of A int B ?

Abcd

edited already...

Semiclassical

kk

Now, what are the positions of A,B,C?

Abcd

9:30 PM

that's what my question is...

$\Delta x \sin 30^\circ$?

Ted Shifrin

Glad you're still having fun, @Cookie :)

I dunno, @Kasmir. Is it?

Do you have a way of thinking of the closure other than the definition?

Kasmir Khaan

hmm

thats the thing Ted, I dont have any geometric thing in my head

Abcd

going to sleep now, good night. Please verify my answer if possible @semi. Bye...

Kasmir Khaan

so no intuition thats why I asked you :D you allways got a good pic :D

Semiclassical

@Abcd think about it in the morning

Cookie Toast

9:35 PM

It's causing some serious ethical dilemmas! Do I go to bed like a responsible person, or do I stay up and continue playing with the math? @TedShifrin Luckily I have coffee haha

Abcd

yeah, sure.

Semiclassical

@TedShifrin My preferred way to do these problems is always to start from the fact that $\vec{F}_{net}=0$ at equillibrium

Arrow

@MikeMiller may I ask for your intuition?

Ted Shifrin

@Semiclassic: Sure, me too. That's tacit in how he's thinking.

Semiclassical

eh. not if you include $\ell$

Ted Shifrin

9:36 PM

@Kasmir: Have you guys not discussed limit points?

Mike Miller

Go for it.

Semiclassical

The equillibrium length is always a statement about a hypothetical scenario involving only that spring; it has nothing immediately to do with the actual equillibrium which involves all three

Arrow

Do you think the conditions in the above question are equivalent?

Ted Shifrin

Funny guy, @Cookie :)

Semiclassical

What matters it that $\vec{F}_{net}(\vec{x}+\Delta \vec{x})=\vec{F}_{net}+\Delta \vec{F}_{net}=\Delta \vec{F}_{net}$

Mike Miller

9:39 PM

Oh I have no idea.

Kasmir Khaan

@TedShifrin the way I Think about it is that p is a limit Point if for any nbhd of p ,it contains Point of our set

Arrow

@MikeMiller not even a hunch? :(

Mike Miller

That last thing in particular is too wild for me to have an intuition for. But I have essentially never thought about tangent spaces to nasty bois.

Kasmir Khaan

Point different from p i should add

Mike Miller

Maybe the Whitney umbrella once in my life, or a Zariski tangent space...

Arrow

9:40 PM

Why is that last thing wild? It's just asserting tangency of the tangent set

Ted Shifrin

@Kasmir: How do you see what the closure of $B$ is in terms of limit points?

Mike Miller

I rarely understand what formulas "mean" at a glance

Ted Shifrin

@Arrow: But what does $(T_p X)^\perp$ mean if $T_pX$ isn't a linear subspace?

Mike Miller

I guess he takes the linear closure

Arrow

@TedShifrin sorry, I should have asked for the mere existence of some linear subspace satisfying this condition. I got ahead of myself and wrote it's necessarily the same creature.

Mike Miller

9:42 PM

"There is a vector subspace V so that..." ?

Arrow

Yes

Kasmir Khaan

@TedShifrin the closure of B = B union B' ( the limit Points )

kinda of an open set and then we closed it

Ted Shifrin

OK, @Kasmir. Good. Use that to think about your question.

Mike Miller

Yeah, I dunno. Seems interesting but too hard for me to say anything useful immediately

Kasmir Khaan

@TedShifrin thanks Ted :D

Semiclassical

9:43 PM

And then that's just $-k(\Delta \vec{x})_A-k(\Delta \vec{x})_B-k(\Delta \vec{x})_C$

Kasmir Khaan

ill try to solve em now and come bacl later :)

Ted Shifrin

OK.

Arrow

@MikeMiller my problem is that I can't see any way of moving between them at all. In fact I can't even show that orthogonal projection is a locally about $p$ a homeomorphism, which I feel it must be.

@TedShifrin rewrote the question following your remark.

Given a locally Euclidean (locally homeomorphic to some Euclidean space) topological subspace $X\subset\mathbb R^n$ and $p\in X$, let $\mathrm{T}_pX$ denote the *tangent set* of $X$ at $p$, namely the set of derivatives of curves in $X$ based at $p$ and differentiable at zero.

Are the following conditions equivalent?

1. $\mathrm{T}_pX$ is a linear subspace of $\mathbb R^n$ of dimension $\dim_pX$.
2. There's a linear subspace $V\subset\mathbb R^n$ for which we have the following equality, where the limit is taken in a translated neig…

Mike Miller

This is also my problem. :)

Arrow

Aww :(
Hello @BalarkaSen

Ted Shifrin

9:47 PM

In my usual way of thinking, the inverse/implicit function theorem would tell you that you can locally represent $X$ as a graph over the tangent space and then Taylor's theorem gives your limit.

Balarka Sen

I can't sleep.

Hi @Arrow

Mike Miller

As usual eh

What happens when you lie in bed? You're physically tired, but the sweet release of... sleep never comes?

Ted Shifrin

metaphorically hits Balarka over the head to knock him asleep

Balarka Sen

@MikeMiller I like the parsing of that sentence

I am never physically tired, is the key point

;)

Ted Shifrin

Just spiritually?

Semiclassical

9:51 PM

tbh the only time I find it easy to fall asleep is when I've just woken up

Arrow

@TedShifrin I am not able to fill out your proof sketch. Could you help me some more? I don't see any $C^1$ (or locally differentiable) function on which to apply the implicit function theorem to locally represent $X$ as a graph. I also don't understand the bit about Taylor's theorem giving the limit.

Eric Silva

@TedShifrin idk if Balarka should sleep. He might have a metaphorical concussion.

Ted Shifrin

Well, if I get to assume I start with a $C^1$ manifold, then the inverse function theorem tells me that the manifold is locally a graph over its tangent space. Taking the tangent space to be the $x_1\dots x_k$-plane, then you write the manifold as $y=f(x_1,\dots,x_k)$ ($y\in\Bbb R^{n-k}$) and the first derivatives of $f$ all vanish at the origin.

@EricSilva: Granted.

Arrow

I am not assuming anything here except the ambient Euclidean space is a $C^1$ manifold. Moreover I want to avoid the $C^1$ assumption, but there's still the everywhere differentiable inverse function theorem. At any rate, I don't think I understand.

Ted Shifrin

So I'm thinking your second criterion doesn't imply the first. For example, you could take $V$ to be all of $\Bbb R^n$. What happens if you start with a cone and work at the cone point?

Balarka Sen

9:58 PM

I don't think the tangent set is a vector subspace

At the cone point

Arrow

In this case how is the second condition satisfied? There's no vector space which is tangent

Balarka Sen

It should just be... a cone again

@TedShifrin The multivariable content of the question is, suppose $f : \Bbb R^n \to \Bbb R^m$ is a function such that $Df_v(0)$ exists for all $v$. Moreover assume $\{(v, D_vf(0)) \in \Bbb R^n \times \Bbb R^m\}$ form a vector subspace of the same dimension as the graph of $f$ (which is $n$)

Is $f$ differentiable at the origin?

Ted Shifrin

Answer: NO.

Arrow

@BalarkaSen could you please explain why this is the heart of the matter?

Balarka Sen

What's an example?

@Arrow I have repeated this multiple times and am tired of doing it again

Ted Shifrin

10:02 PM

You've done examples. All directional derivatives 0 but function not differentiable.

Balarka Sen

@Ted $f$ is continuous, I mean.

Arrow

Understood. @TedShifrin so the conditions of my question are not equivalent?

Balarka Sen

Can you have all directional derivatives zero staying continuous?

I do not think so

Ted Shifrin

Sure I can, @Balarka.

Balarka Sen

Hm, maybe take $f(x, x^2) = x$ and $f = 0$ otherwise and bump it up along the parabola?

Idk

I'd like to see a formula

Ted Shifrin

10:07 PM

Let me ponder a moment.

Kasmir Khaan

@TedShifrin I don't understand the definition of a perfect subset

Ted Shifrin

Why not?

Kasmir Khaan

i mean if E is closed , dont that mean every point of E is a limit point of E no ?

Ted Shifrin

No...

You can give me a non-empty closed set with NO limit points.

Balarka Sen

@Ted OK, maybe you can bump the parabola up by decreasing the compact support closer and closer to the origin

That should work

Kasmir Khaan

10:11 PM

hmm

Balarka Sen

I fee like that should constitute a counterexample, @Arrow

Arrow

@BalarkaSen thank you. Unfortunately I don't understand what is meant by bumping up the parabola by decreasing the compact support.

Kasmir Khaan

@TedShifrin could it be a set with bounch of Points far from each other?

Ted Shifrin

Sure. What's the easiest example you know?

Arrow

If it has to do with the graph of $f(x, x^2) = x$ and $f=0$ otherwise, it seems to not be locally Euclidean about the origin. (Sorry if this is completely unrelated.)

Kasmir Khaan

10:14 PM

hmm i would guess one point

Ted Shifrin

What about an infinite set?

Kasmir Khaan

well ><

that is the thing

Ted Shifrin

What infinite set did you know when you were 8 years old? :)

Kasmir Khaan

you told me that is can be viewd as open and closed ( that part comfused me being open and closed )

Natural numbers?

Balarka Sen

@Arrow The graph I have in mind is a modification of that. Multiply by a bump function $\rho$ near the parabola such that $\rho(x) \to 0$ as $x \to 0$.

Ted Shifrin

10:15 PM

No, a point is not an open set.

Yes, natural numbers. Or $\Bbb Z$. They are closed sets with no limit points.

Kasmir Khaan

hmm your first comment what does it mean

a point is not an open set

it should be closed to make my point on that statement

Ted Shifrin

The set consisting of a single point in $\Bbb R$ is a closed set but definitely NOT an open set.

Yes, it is a closed set, of course.

Kasmir Khaan

hmm yes i was trying to find a closed set that is not perfect

so you gave me N and Z

Ted Shifrin

Right.

As far from perfect as you can be.

Kasmir Khaan

I need to make the definitions straight in my head :D

so Ted, the idea here, is to make the terms precise right?

what one would mean by open and closed and limit point ect

Ted Shifrin

10:20 PM

Well, sure, that's the beginning.

Kasmir Khaan

so we get a common ground so to speak :D

Arrow

@BalarkaSen sorry for frustrating you, I'm just not able to follow your descriptions. Do you want the bump function to "flatten" the tilted parabola near the origin, thereby making the tangent set the $xy$-plane and yet leaving the resulting subset without a tangent plane at the origin in the sense of the second condition?

Kasmir Khaan

ehm nice

btw Ted , a notation here that is not very clear

in Rudin {E_alpha}

and sometimes Union_alpha E_alpha

from what I understood , its a set of sets

but the point here, the element of E_alpha comes from what set?

and the family of sets, do we consider the sets as elements or what inside those sets?

Ted Shifrin

@Balarka: Well, I can go back to your original statement. I can easily give you a non-differentiable function that is continuous and has directional derivatives in every direction at 0.

Each $E_\alpha$ is a subset of your master space $X$.

This is a confusing point in set theory. You have a collection of subsets of $X$.

Arrow

@TedShifrin I think $f(x,y)=\frac{x^3y}{x^4+y^2}$ works for this? (The graph is even locally Euclidean.)

Ted Shifrin

10:25 PM

That probably ends up being differentiable at $0$, though?

Arrow

No, because of $f(x,x^2)=\frac x2$ - this parabola has a tangent not within the $xy$-plane.

Ted Shifrin

Here's an easy one. $f(x,y) = |x|y/\sqrt{x^2+y^2}$. Continuous. Has all directional derivatives, but no tangent plane.

Oh, good, @Arrow.

Balarka Sen

Oh, I guess my reformulation of Arrow's question is flawed

Ted Shifrin

My issue with @Arrow's second condition is that I could just take $V$ to be the ambient vector space, so the projection on $V^\perp$ is null.

Arrow

Ah, so restrict $V$ to also have the right dimension.

Mike Miller

10:30 PM

I'm not really sure what it means for $V$ to have the right dimension

Arrow

$\dim V=\dim _pX$ since we're assuming $X$ is locally Euclidean.

Ted Shifrin

Me either.

Mike Miller

Oh, I didn't know we were asusming that of X

Ted Shifrin

So that would rule out taking the Zariski tangent space of larger dimension.

Arrow

Yes, I wrote it in the question I think.

Mike Miller

10:31 PM

My bad

Ted Shifrin

You did, @Arrow.

Arrow

Without this assumption we could take a paraboloid and an upside down paraboloid "tangent" at the origin: their union would satisfy the second condition but not the first I think.

Balarka Sen

The right question is probably this. Suppose $f : \Bbb R^n \to \Bbb R^m$ is a continuous function such that for any curve $\alpha$ going through the origin, $(0, (f \circ \alpha)'(0))$ exists and they form a vector space of dimension $n$ inside $\Bbb R^{m+n}$. Does that mean $f$ is differentiable at $0$?

Right, I was just thinking about $\alpha$ to be linear directions

That is not what his question is

Arrow

If I understand correctly, this might answer affirmatively?

Balarka Sen

I think it's Ted's question about if $f$ is continuous along every path through the origin then $f$ is continuous

Just ramped up in smoothness order

Julius

10:35 PM

Theorem 3.3.2 in maths.ed.ac.uk/sites/default/files/atoms/files/khwaja.pdf says "$|ph(1-z)|<\pi$ and $|ph(z+1)|<\pi$". Is it meant that only one of these conditions is enough? I'm having this doubt because taking $z < -1$ (which satisfies only the first condition) also happens to work. $ph$ is complex argument or phase

Balarka Sen

@Arrow Right, there you go. It's the same solution

Find a sequence of points along which $f$ is not differentiable, then interpolate them by a smooth curve

plus stuff near the origin

Ted Shifrin

Glad to know I've created some memories in your brain, Balarka :P

Arrow

@BalarkaSen I just don't understand how this proves the equivalence between my conditions, where seemingly no functions apart from the inclusions are involved.

Regarding what you propose - the problem with this interpolation is that I see no way of making the resulting curve actually within $X$.

Ted Shifrin

Use the ball in Euclidean space that your $X$ is locally homeomorphic to?

Not that I'm really following ... There's a ton of ruckus going on with repairmen in my house.

Arrow

Using a homeomorphism might not preserve the crucial differentiability of the interpolation curve at zero.

If indeed I could interpolate such a sequence to get a curve taking values in $X$ which is differentiable at zero, this seems to prove the first condition implies the second.

@TedShifrin I am desperate here :) Does your intuition tell you these condition are equivalent or not, by the way?

Ted Shifrin

10:41 PM

So how do you even define the tangent space if I only have a homeomorphism that isn't differentiable? How do I define differentiable curves?

Arrow

We're in a Euclidean space, so just use that.

It sits inside R^n

Oh, right.

This is annoying

I asked the question here by the way, if anyone cares.

Balarka Sen

10:45 PM

So you'd have to find a topological submanifold of R^n and a point p there such that no neighborhood of p is locally differentiably homeomorphic (with inverse also differentiable) to R^k. I think the bad points of Alexander's horned sphere is a good starting point :P

It has to be really bad I mean

Arrow

I don't think any bad point of that thing satisfies either of the conditions though

Balarka Sen

I don't see it well enough to be convinced about anything

Arrow

If you don't then certainly I don't :)

Balarka Sen

It could be potentially true that no curve passing through the bad point of Alexander's horned sphere is differentiable at the bad point as a curve in R^3

@Mike, @Ted, is this plausible?

Mike Miller

God I have no idea

I work with simple things so I don't need to think so hard, haha

Balarka Sen

10:49 PM

lol fair

If @Akiva doesn't know probably none of us knows

Ted Shifrin

I only think (thought) about singularities in the complex analytic/algebraic category.

LOL ... DogAteMy the guru.

Balarka Sen

the guru of infinitely botched things

that could be a good title

Arrow

Digressing, does anyone see how to prove the second condition implies $\pi_V$ is locally a homeomorphism about $p$? The locally Euclidean hypothesis will be needed otherwise a pair of "kissing" paraboloids will be a counterexample.

Ted Shifrin

I can't imagine there being anything differentiable about the horned sphere.

Mike Miller

Yeah get him

I think the reason you're having trouble finding people discussing this in the literature might be that it's 3hard 5them

Ted Shifrin

10:51 PM

$\pi_{V^\perp}$ gives you the representation of the object as a graph of (presumably) a function over $V$.

Rob Kirby would know the answer.

Or at least would have 40 years ago.

PVAL-inactive

11

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Let $M$ be a compact smooth manifold and $N$ be a compact smooth submanifold of $M$. The usual transversality theorem claims that for a generic diffeomorphism $f$ of $M$, the submanifolds $N$ and $f(N)$ are transverse. I am interesting in another point of view. The diffeomorphism $f$ is now fi...