Mathematics - 2019-12-23 (page 1 of 2)

Mike Miller

00:00

@TedShifrin No

Julius

Hi. Let $G \subset H$ be $\sigma$-algebras and suppose that $E[X \mid H]=E[X \mid G]$. It is not implied that $E[|X| \mid H] = E[|X| \mid G]$. Is there perhaps some weak assumption that, when combined with the first equality, would give the second? Doesn't seem so..

Leaky Nun

00:20

oh man 3b1b just uploaded 2 new videos?

it's 3 days too early

Ted Shifrin

Howdy, mr @Leaky.

Leaky Nun

hi

nbro

01:13

This Wikipedia article en.wikipedia.org/wiki/Notation_in_probability_and_statistics says

> The probability is sometimes written $\mathbb {P}$ to distinguish it from other functions and measure P so as to avoid having to define "P is a probability"

But isn't $\mathbb {P}$ itself a measure (more specifically, a probability measure)? I don't understand this statement in the Wikipedia article.

Mike Miller

read a book not wikipedia

21

nbro

Wikipedia is easy to find. Books are hard to choose and often you spend a lot of time to find what you want

Mike Miller

the notation there is irritating, let's say the probability measure is mu on your probability space X

then P is indeed a measure, but not on X; it's a measure on the real numbers

P(Z in A) is the measure mu(B) where B is the set {x in X such that Z(x) is in A}

yes instead of investing time in finding a book you have instead offloaded that time onto everyone you ask notation questions to lol

nbro

@MikeMiller So, your probability space is $X=(\Omega, \mathcal{F}, \mu)$?

Mike Miller

i named it X

and my random variable is named Z

and A is a subset of the real numbers

B is a subset of X whose definition depends on the random variable Z and the set A

i didnt bother naming the sigma-algebra

nbro

01:20

But my first doubt is: why do they talk about $\mathbb{P}$ and $P$?

Mike Miller

bold P is what I call P above

P is what I call mu above

i dont like that they are both named P so i named one something else

nbro

@MikeMiller I don't understand what it means for x to be X, if X is a probability space, so not a set, but a triple

Maybe X is the event space?

Mike Miller

lol

nbro

If Z is a random variable, then x in X must be an event, given that you use the notation Z(x), so X is the event space.

I think I got confused. I thought random variables were functions from the event space, but they are apparently functions from the sample space

Alessandro Codenotti

random variables are measurable functions $\Omega\to\Bbb R$

nbro

01:31

I never studied measure theory before, but measure theory is required to really understand probability theory, IMHO

Yesterday, someone here defined a random variable from a probability space to the real numbers, but then someone else defined it only from the sample space to the real numbers

I think that sometimes people use the expression "probability space" when the expression "sample space" is required

And I got confused that a random variable was a function from the event space

Any more doubts, sir? :P

Can you guys tell me what people mean by "probability distribution" in the following article: medium.com/tensorflow/…? Are they talking about pdfs, cdfs or probability measures? They use the notation P(x), which is widely used in statistics and machine learning, but a guy told me that this notation does not make sense! I think they mean probability measure and P(x) is a shorthand for P(X=x)

I think they mean probability measure because they use the concept of KL divergence under the hood to approximate the "posterior" (which should then be a probability measure), given that variational inference is based on KL divergence (and alike), which should is used to calculate the dissimilarity between probability measures, whatever that really means!

anakhro

03:38

@MikeMiller I am a little confused about some terminology: a "characteristic class" is what exactly? I thought that a characteristic class of a vector bundle $E$ was an element of the image of the Chern-Weil homomorphism associated to $E$. But then it seems that there is a functorial description where a characteristic class is a natural transformation between the isomorphism class of fixed rank vector bundles (seen as a functor on manifolds) and $H^*$.

These seem like distinctly different things and I don't know if they are referring to different things or there is a way to reconcile the notions.

Oh I think I see it.

I THINK I GOT IT.

So nevermind! :P

They satisfy naturality.

The polynomials, that is.

Mike Miller

04:30

@anakhro A characteristic class is an assignment of vector bundles to cohomology classes that satisfies naturality for pullback yeah

user400188

04:47

2

Q: For vectors $(x_1,x_2,\dots,x_n), (y_1,y_2,y_n)$, is "$\forall k$ s.t. $x_k>y_k$, $x_k-y_k>C$ vacuously true when $x_k\leq y_k \ \forall k$?

Where C is some positive constant? Basically, I am confused regarding vacuously true statements and trying to use this constructed example to clarify. The restate the example in slightly more detail, suppose we have two vectors denoted $(x_1,x_2,\dots,x_n), (y_1,y_2,y_n)$. The statement in con...

Akiva Weinberger

05:41

@BalarkaSen If $w=[\ell_1,\ell_2][\ell_3,\ell_4]\dotsb$ then $S(w)=[\ell_2,\ell_3][\ell_4,\ell_5]\dotsb\ne w$

but $w+S(w)$ is a root of $S(x)=x$ I guess

Leaky Nun

05:52

@AkivaWeinberger what is that?

Akiva Weinberger

yesterday, by Akiva Weinberger

> Meet the Hawaiian mapping torus - the mapping torus of the shift map. First singular homology is infinite cyclic generated by inner loop. Can you find the non-trivial elements of H_2? If you look at the image, it might feel H_2 is trivial since there is no "enclosed space!"

Thinking about this question

(We know the answer but we were thinking over some details)

Secret

Is this thing look like some kind of puffed up spiral?

Akiva Weinberger

It's homotopy equivalent to the complement of a spiral in 3-space, the outer arm of which goes to infinity and the inner bit spirals ever closer and closer to a limiting circle

(The limiting circle is not part of the spiral and so is contained in the complement)

Secret

I see

Leaky Nun

06:17

twitch.tv/maskenissen/clip/ObedientBenevolentBasenjiNinjaGrumpy

@AkivaWeinberger

Rithaniel

Carlsen showing off

Leaky Nun

@Rithaniel play?

Rithaniel

How long will you be on? I'm doing some sprite work at the moment

Leaky Nun

@Rithaniel when are you available?

Rithaniel

Maybe an hour from now? Maybe thirty minutes

Leaky Nun

06:27

ok

kauray

Axlers book defines the additive inverse of a complex number z as -z which satisfies z +(-z) =0. From that definition the book goes on to define subtraction in terms of addition as w-z = w+(-z). My question is what is the connection between the additive inverse statement and the subtraction definition?

Ted Shifrin

What do you mean? The definition is the connection.

Think about numbers. $5+(-3) = 2$ is the same as $5-3=2$, which is the same as $5=2+3$. (Now you can add the additive inverse of $3$ and get back to the original.)

kauray

Yes that's what I was asking, that the definition is the connection. I did understand in terms of numbers but not in algebraic terms

Leaky Nun

06:55

@Rithaniel

Rithaniel

Yeah, just about there. If it takes you any time to set things up, might be a good idea to do that

Leaky Nun

I'm just pinging you for the puzzle lol

Rithaniel

Ah, gotcha

Hmmm

So, we want a white move that doesn't lock black into mate? Because I believe I see a few

Leaky Nun

that's the point lol

there is exactly one move that is not mate

Rithaniel

I wanna say g6 to d6 isn't mate

Oh wait, the other bishop

Wait yeah, g6 to c6 and then b7 takes h7

It's interesting to find a board state where it's actually more difficult to lose than to win :P

Leaky Nun

07:08

nice

@Rithaniel so, are you ready?

Rithaniel

Yep

Leaky Nun

@Rithaniel what time control?

Rithaniel

I can do 10 minutes. Nice and relaxed, but still a time limit

Leaky Nun

increment?

Rithaniel

I'll leave that up to you

Leaky Nun

07:11

@Rithaniel lichess.org/soPEHkAt

Leaky Nun

07:34

@Rithaniel I can't send you a game as black unless you register

due to some bug

Rithaniel

I gotcha. I can still offer a rematch after I look through what went wrong that game

Leaky Nun

looks like I can't go back to the page where I can accept your rematch

creating an account is free

Rithaniel

Ah, okay, well, when I'm done looking at things I'll try clicking rematch. If you don't receive any kind of notification on that, then I'll create an account, just don't push me into it

Leaky Nun

sure

Rithaniel

It seems that the computer thinks moving the queen out is generally a bad idea. That seems counter intuitive to me. It's one of my best pieces. Why wouldn't I want to develop it?

Leaky Nun

07:38

because you would need to constantly defend it

Rithaniel

Yeah, but that's true about any piece I care about

Leaky Nun

so in general one should develop the minor pieces first

well not many pieces can defend the queen

Rithaniel

Fair

Leaky Nun

you can defend a knight with anything

oh unless it's attacked by a pawn

it's sort of reversed

Rithaniel

I kept up with you better that game than in the previous two

I never had the advantage, but at least I was hovering around 2 or 3 in your favor, instead of 5

Leaky Nun

07:45

oh and I found the rematch button

Rithaniel

07:58

That one was much worse. I lack the chess instincts I had back in high school

Like, I shouldn't have lost the queen that early, for sure, but my brain just didn't see the knight, for some reason

Leaky Nun

hmm

Rithaniel

Getting my ass kicked twice is enough for me, for now. If you're around in about 3 hours, though, I might be in a better mindset for another game

Leaky Nun

ok

Shootforthemoon

08:31

@TedShifrin Thank you very much! That's exactly what I was looking for

Quasar

09:15

-1

Q: Solve the initial value problem $(D^2+2aD+b^2)y=\sin \omega t$, $y(0)=y'(0)=0$.

Exercise 4.4, problem 24, page 144 of the book on differential equations by KKOP here. Solve the initial value problem $(D^2+2aD+b^2)y=\sin \omega t$, $y(0)=y'(0)=0$, where $a,b,\omega$ are real constants, $a<b$. Consider separately the cases $\omega \ne \sqrt{b^2-a^2}$ and $\omega = \sqrt{b^...

ordinary-differential-equations

While trying to find a particular solution to the inhomgenous differential equation, I get some ugly terms, so I would like to ask for someone's help in solving it.

ATR

10:07

1

Q: Can two ergodic measures on the same space give rise to isomorphic dynamical systems?

Consider $(X,\mathcal B, T)$, where $T$ is measurable, $X$ is a $d$-dimensional unit cube. If we pick two measures that are ergodic with respect to $T$, $\mu$ and $\nu$, do the systems $(X,\mathcal B, T, \mu)$ and $(X,\mathcal B, T, \nu)$ need to be isomorphic? Seems like I have the following cou...

dynamical-systems spectral-theory

If there are two orbits of different period of the same measurable transformations that support two different ergodic measures, will the dynamical systems defined by those measures, be isomorphic?

Astyx

10:42

Hi chat

Mike Miller

11:12

@BalarkaSen The mapping torus's homology comes from the non-geometric part of H_1(H) so idk why one expects to see said homology

I anticipate it looks something like this

Enumerate the circles as x_i, y_i

Let l_i be the loop [x_i, y_i]

Then prod_{i >= 1} l_i defines a loop on the Hawaiian earring

But so does prod_{i >= 2}, and they are not homotopic, as [x_1, y_1] is not null

But [x_1, y_1] is a commutator hence null-homologous, I can fill it out with a torus

This altogether should define a map $\Sigma_{2,2} \to H$ which sends the first boundary to prod_i l_i and the second boundary to prod_{i > 1} l_i, which is exactly what you need to glue up the boundaries as a map to the mapping torus

Sigma_{1,2} sorry. I am gluing in a torus to a cylinder

So we get a genus 2 surface representing the non-trivial homology class

Shootforthemoon

11:38

Hi, if a strictly monotonic function is also continuous, is it invertible?

Mike Miller

Needn't be surjectivd

Shootforthemoon

no?

Mike Miller

11:58

Cmon you can write down an example of a non-surjective strictly monotonic function

For instance, I'm sure you know a strictly monotonic function on R whose image is only the positive numbers

If you mean "the map is invertible onto its image" meaning that there's a function with domain Im(f) which is inverse to f, that's true

Jacksoja

12:37

@LeakyNun Hello

Leaky Nun

hi

Astyx

It doesn't have to be continuous for that to be true as well

Thorgott

More surprisingly, even without supposing continuity, the inverse function will be continuous (on a usually disconnected domain, however)

Shootforthemoon

12:58

@MikeMiller Yes, I meant so

solisoc

Given that there exists an enumeration of the set of finite sets of positive integers, am I correct in thinking that simply replacing each of the finite sets with its complement produces an enumeration of the set of cofinite sets of positive integers?

Leaky Nun

@solisoc yes

solisoc

Cheers

Also, consider the set of all sequences of one or more As e.g. A, AA, AAA, and so forth. Am I correct in my understanding that if we only allow finite sequences of As, this set is (trivially) enumerable, but if we allow infinite sequences of As, it is not?

Leaky Nun

@solisoc depends on what you mean by "infinite"

clearly AAA.... is just one thing

solisoc

My thinking is that the positive integers are already all sent to the finite sequences of As

There is nothing left to send to any infinite sequences of As

Leaky Nun

13:11

oh well

send 1 to the infinite sequence

2 to A, 3 to AA, 4 to AAA, etc

solisoc

gasp

Thorgott

by that logic, $\mathbb{N}_0$ wouldn't be enumerable either

solisoc

So what is the ambiguity in "infinite"? (Also... how do I get MathJax to display?)

Thorgott

some infinities are bigger than others

as for mathjax, check tinyurl.com/cfqcvpc

Rithaniel

Look in the top-right of the screen and you'll see some links to instructions on how to get Mathj-- dangit, ninja'd

anakhro

13:20

@TedShifrin is there an alternative way of motivating Chern-Weil theory without using Chern-Gauss-Bonnet? I know that's the historical way that Chern came up with it, but I was wondering if you knew of any intuition one could provide for why we'd even want to consider the Chern-Weil homomorphism? It seems like pure chance that it ends up being beautiful.

solisoc

@Thorgott @LeakyNun so is there another infinite sequence of As that is not AAA...?

Leaky Nun

never mind

Thorgott

Depends on how you define infinite sequence

Leaky Nun

I just thought that was the issue

because otherwise you would just have one infinite sequence

and if you add one thing to a countable set then you still get a countable set

solisoc

Sure, but you've piqued my interest now. Can you give an example of how "infinite" could be defined to give more than one infinite sequence of As?

Leaky Nun

13:27

the set of real numbers is not countable

so now I have a sequence of As indexed by the real numbers

maybe you wanna call that an infinite sequence of As also

but then it won't be the same as the countable sequence "AAA..."

solisoc

Ah cool, good point

I think in the context I'm considering that would be begging the question

Rithaniel

One big set theory result is that, no matter how many elements a set contains, even if it is infinite, it's power set always contains far more elements.

solisoc

The broader context, that is; obviously in answer to the question I posed above you're quite correct

Rithaniel

So, from that, you immediately can define infinitely many "sizes of infinity"

Shootforthemoon

I'm struggling to find the answer to this question: math.stackexchange.com/questions/3485394/…

Anyone can help?

Thorgott

13:37

Are you asking about how to derive the Inverse Function Theorem from the Implicit Function Theorem?

Shootforthemoon

yes, starting from that particular statement of the theorem

possibly without the Jacobian

solisoc

Thanks for your help @LeakyNun @Thorgott @Rithaniel

Leaky Nun

np

Rithaniel

Any time (Hopefully my bringing up set theory didn't cause any confusion)

Ultradark

math.stackexchange.com/questions/3483846/… Thoughts?

Thorgott

13:55

@Shootforthemoon the usual idea is to apply the Implicit Function Thm. to the function $(x,y)\mapsto f(y)-x$

working that out would be a good exercise, probably

as for the Jacobian, that is integral to the Implicit Function Thm., so I don't see how you wanna make do without it

Shootforthemoon

I'm gonna study it, saw it is defined when talking about a system of multivariable equations, so I thought it was not necessary for the case, but maybe it is implicitly present even there

@Thorgott Thanks, I'll follow your suggestion

anakhro

@LeakyNun I am not sure how good of an idea it is to call $(a_i)_{i\in\mathbb R}$ a "sequence".

It's certainly ordered, but there is no concept of "successor".

Leaky Nun

@anakhro I didn't want to introduce $\omega_1$

ÍgjøgnumMeg

@Balarka @Soham https://www.youtube.com/watch?v=EqAtk5D1R1Y

This is unbelievable

Shootforthemoon

@Thorgott Sry, in this case we should apply the thm in order to find the existence of y=g(x), right?

Thorgott

14:12

yes

Shootforthemoon

I fixed $F(x,y)=f(y)-x$, but don't see how I could verify that $F_y$ is different from zero for every $(x,y)=(f(y),y)$

Thorgott

the Inverse Function Theorem has assumptions, you should use those

Leaky Nun

@Shootforthemoon compute it

Shootforthemoon

@Thorgott Oh, true, so I need to put this into my hypothesis

Ultradark

thoughts? math.stackexchange.com/questions/3484084/…

Shootforthemoon

14:19

I was trying to verify the assumptions of the theorem before applying it XD

Thorgott

definitely

i mean, clearly not every map is locally invertible, so you do need assumptions

Shootforthemoon

I got that the implicit function is $y=g(f(y))$ for each $x=f(y)$ and that its derivative is $g'(x)=\frac{1}{f'(y)}$ @Thorgott

Is it correct?

Ultradark

14:35

Why is $x^{2n}-x^n-1=0$

Semiclassical

it's not in general?

I mean, for x=0 or x=1 it's clearly false

Ultradark

why are its roots always in terms of the golden ratio

I'm stymied

Leaky Nun

to get to the other side

Semiclassical

consider the substitution t=x^n.

Ultradark

$t=x^n$

Semiclassical

14:38

what equation do you get in terms of t?

Ultradark

oh

you get $t^2-t-1=0$

Semiclassical

right. and the roots of that are definitely related to the Golden ratio

Ultradark

Can you plot the integral of a function that doesn't have a closed form in elementary functions on wolfram alpha?

I'm trying to plot the integral from let's say 1 to $x$

Semiclassical

i sorta doubt it

wolfram alpha isn't quite that sophisticated

what function do you have, though

I've got mathematica, so I could conceivably do some computations

Ultradark

14:53

$\int_e^x \exp(1/\ln(t))~dt$

Semiclassical

mmkay

Ultradark

by the way I'm manually doing

that log-log plot

Leaky Nun

just use desmos

Ultradark

on desmos

Semiclassical

you may be better off doing that in wolfram alpha as the following differential equation:

Ultradark

14:54

how can I use desmos?

Semiclassical

let y(x) be that integral. then y(e) = 0 and y'(x) = e^(1/ln x)

adesh mishra

@Semiclassical I pinged you yesterday did you get notify?

Semiclassical

yes. but you didn't actually give a question.

adesh mishra

@Semiclassical I gave

Ultradark

wow that's so cool

adesh mishra

14:58

Should I write it down ?

Semiclassical

this one?

yesterday, by adesh mishra

If $\alpha, \beta, , \gamma and \delta $ be the eccentric angles of the four points of intersection of the ellipse and any circle, prove that $$ \alpha + \beta + \gamma +\delta$$ is an even multiple of $\pi$

adesh mishra

Yeah

Semiclassical

okay. where "eccentric angle" means "angle in the parametrization x=a cos t, y= b sin t" apparently

adesh mishra

@Semiclassical I waited for you yesterday for too long

@Semiclassical yes

Semiclassical

this seems unpleasant.

well, I say that

but it seems like it should boil down to the following fact: If the angle $\alpha$ is between 0 and pi/2, then the other angles are of the form $\pi-\alpha,\pi+\alpha,2\pi -\alpha$

adesh mishra

15:02

@Semiclassical what ? I didn't get you

Semiclassical

suppose one of your intersections is at (x,y) in the first quadrant

then the following should also be intersections: (-x,y), (-x,-y), and (x,-y)

which have the angles so specified

adesh mishra

Okay

Semiclassical

and therefore when you add them together you get $\pi+(\pi-\alpha)+(\pi +\alpha)+(2\pi -\alpha) = 4\pi$

adesh mishra

@Semiclassical How? How is the intersection of ellipse and circle is symmetrical?

Semiclassical

I'm assuming that the ellipse is of the form x^2/a^2+y^2/b^2=1, but that seems safe since otherwise the parametrization x=a cos t, y=b sin t makes little sense

i mean, the picture you linked made that symmetry quite apparent

adesh mishra

15:06

Yes, we can of course use that form

Then?

Rithaniel

So, here's an interesting thought: A sequence of random variables $\{X_i\}_{i\in I}$ where $X_1$ is a given distribution, $X_{2n}\sim\min\{X_{2n-1},X_{2n-1}\}$ for $n\geq 1$ and $X_{2n+1}\sim\max\{X_{2n},X_{2n}\}$ for $n\geq 1$. Is there a criteria for when this sequence converges in distribution? Does it ever converge?

Semiclassical

I'm not sure what else you need. Your picture spells it out well enough.

If you've got an intersection whose eccentric angle is 10 degrees, what can you conclude about the other intersections?

just from looking at your diagram

adesh mishra

@Semiclassical I'm not able to develop an intuition how the second intersection is gonna occur at an eccentric angle of $\pi -\alpha$

Semiclassical

if there's an intersection at (x,y) in the first quadrant, where are the other intersections?

adesh mishra

It can happen at any point in the second quadrant

Semiclassical

15:12

yeah, no

look at your diagram

what kind of symmetry do you have for your ellipse

adesh mishra

@Semiclassical all four quarters are symmetrical, Am I right?

Semiclassical

Can you say more about that? That's a bit too vague

adesh mishra

all four quarters have same opening of angle (i.e. 90 degrees)

Semiclassical

okay. what can you say about points on those four quarters?

e.g., if (x,y) is a point in the first quadrant

then what point does that correspond to in the second quadrant, by symmetry

adesh mishra

Yes, there will be a point (-x,y)

Semiclassical

15:19

right. how about the other quadrants?

adesh mishra

Yes, there will be points (-x,-y) and finally (x,-y)

Semiclassical

right.

so, suppose there's an intersection between the circle and the ellipse at (x,y)

what happens at (-x,y) ?

adesh mishra

Okay

@Semiclassical I'm not sure what will happen

Semiclassical

If (x,y) in the first quadrant is on the circle and on the ellipse, what can we say about (-x,y) in the second quarant

Is (-x,y) also on the ellipse?

adesh mishra

Well it's sure that (-x,y) will lie on the circle but can't say that it will lie on ellipse too

Semiclassical

15:23

Why not?

If x^2/a^2+y^2/b^2 =1, then what happens when you replace (x,y) with (-x,y)

adesh mishra

Yes, I'm getting you

Semiclassical

okay. So what can we say about (-x,y) ?

adesh mishra

if (x,y) satisfies the equation $$\frac{x^2}{a^2}+\frac{y^2}{b^2} =1$$ then surely (-x,y) will satisfy it too

You're great (one again)

if (x,y) satisfies the equation $$\frac{x^2}{a^2}+\frac{y^2}{b^2} =1$$ then surely (-x,y) will satisfy it too

Semiclassical

right.

adesh mishra

You're great (one again)

Semiclassical

15:26

same for the circle, of course

adesh mishra

Yes

Semiclassical

so therefore (-x,y) will lie on both the circle and the ellipse, so it's also an intersection

adesh mishra

YEAH

Semiclassical

what about the other two quadrants?

we've done first and second quudrant so far

adesh mishra

@Semiclassica I have got you. It's solved. You have thrashed the problem like some....

Semiclassical

15:29

it's really just a matter of that diagram

adesh mishra

@Semiclassical The way Sally Bugs did to that man in The Irishman

Semiclassical

you can do this more algebraically, though. suppose your circle is of the form $x^2+y^2 = R^2$ and the ellipse is $x^2/a^2+y^2/b^2=1$

adesh mishra

okay

then?

Semiclassical

you can solve that for $x,y$ to get...

$$x= \pm a\sqrt{\frac{R^2-b^2}{a^2-b^2}}, \qquad y=\pm b \sqrt{\frac{a^2-R^2}{a^2-b^2}}$$

under the assumption $a>R>b>0$

as such, there's four solutions with four different sign combinations

which corresponds directly to the four quadrants

Balarka Sen

16:06

Consider the map $f : \Bbb R^2 \to \Bbb R^2$, $f(x, y) = (x^2 + y^2, 2xy)$, basically the realification of $f(z) = z^2$.

Thorgott

Wouldn't that be $x^2-y^2$?

Balarka Sen

Right

This has a rank $0$ isolated singularity at the origin. I want to perturb this somehow so that this has no rank $0$ singularities.

Maybe try $f_t(x, y) = (x^2 - y^2 + 2yt, 2xy)$

$Df_t = (2x, -2y + 2t|2y, 2x)$

Shootforthemoon

@Thorgott At the end I got that the implicit function is $y=g(f(y))$ for each $x=f(y)$ and that its derivative is $g'(x)=\frac{1}{f'(y)}$, should it work?

Balarka Sen

So $\det Df_t = 4x^2 + 4y^2 - 4ty$

The zero locus is a circle

Does it have rank 1 along this circle? Maybe not

Of course it has, right? Let's just do $t = 1$. $Df_t = (2x, -2y + 2|2y, 2x)$. This matrix is never the zero matrix

@Thorgott This looks good right?

Thorgott

@Shootforthemoon I think you should end up with $f(g(x))=x$, if you went with the naming convention I suggested

@BalarkaSen Yes

Shootforthemoon

16:20

@Thorgott ok, thanks very much!

adesh mishra

@Semiclassical My connection got broken that's why I couldn't reply to you. Thanks for helping me in a very elegant way.

Balarka Sen

So this means isolated complex singularities can be unravelled into circles worth of fold points

Let's try higher order maybe. $f(x, y) = (x^3 - 3xy^2, 3x^2y - y^3)$

Thorgott

The isolated complex singularities of? Surely not $z\mapsto z^2$

Balarka Sen

Yes, that

I perturbed the map a little bit such that the isolated rank $0$ singularity at the origin became a circle of fold points

Real perturbation, not holomorphic perturbation of course. Holomorphically isolated rank 0 singularities are the stable types

Thorgott

Are you using "singularity" to mean "critical point"?

Balarka Sen

16:32

Yes, wherever the Jacobian is not invertible

Oh, this is interesting. If $f_t(x, y) = (x^3 - 3xy^2 + ty, 3x^2y - y^3)$, then $Df_t = (3x^2 - 3y^2, -6xy + t|6xy, 3x^2 - 3y^2)$, and $\det Df_{\varepsilon} = 0$ is a leminscate if $\varepsilon > 0$

That is no bueno

I guess it is possible to perturb a little more so that the two halves of the leminscate come off to give two circle's worth of fold points

Thorgott

Next higher power you get something like a lemniscate but with three things and I have no clue what that's called

Balarka Sen

Oh you checked?

I also like what happens if I do $z^3 + ty^2$ instead of $z^3 + ty$

There's an appearance of a cusp

At least assuming you perturb a little more so that the cusp comes off of the fold line a little

Thorgott

Balarka Sen

Nice!

Thorgott

I wonder if we get one more teardrop for each power

Balarka Sen

16:46

We should

Thorgott

Ok, using higher powers of y with a factor of t gets weird

it's starting to look like a nuke

Balarka Sen

Lol

Hm, if I use $f_t(x, y) = (x^3 - 3xy^2 + ty, 3x^2y - y^3 + tx)$, then $\det Df_t = 0$ gives a circle.

That's what I want, a good manifold's worth of fold points

Thorgott

that's probably a better way then

this is $z^5+ty^4$

Balarka Sen

Nuts

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$Mathematics$ Mathematics