@BalarkaSen I am little bit confused about finding inverses for the extension fields. Suppose we have F is some field and K is extension obtained by $F[x] / (p(x))$ suppose we have a an element of this field how do you determine its inverse ?

But how is this different from the union of two sets ?

@Adeek An element there looks like $f \pmod{p}$. An inverse is a $g$ such that $f \cdot g = 1 \pmod{p}$. This is all just like modular arithmetic.

@LeGrandDODOM Rudin's Real And Complex Analysis is pretty good. If you want to go deeper into measure theory, and you can read German (mathematics, not necessarily prose), I can strongly recommend Maß- und Integrationstheorie by Jürgen Elstrodt.

in a lot of ways, for example the set you obtain by pairing has always $2$ elements @Mahmoud

consider the set $\{\mathbb{N},\mathbb{R}\}$ compared to the set $\mathbb{R}\cup\mathbb{N}=\mathbb{R}$

@BalarkaSen so the inverse g if we pulled back to F[x] will be obtained as follows f(x)g(x) - 1 = p(x)s(x) so we have f(x)g(x) + p(x)s(x) = 1.

but here we have two unknowns no ?

Yes. But you dealt with this in arithmetic mod p. Euclid's algorithm.

hm ?

@Alessandro Crystal clear :)

what do you mean you dealt with this arithmetic mod p ?

Given a number n, how do you find it's inverse mod p? p is some prime.

we use euclid algorithm

you know that gcd(g,p)=1, by the extended euclidean algorithm there exist t and s such that gt+ps=1, mod p you ignore the second term and t is the inverse of g

oh I see so we use euclid algorithm too for this I guess but for polynomials

Yes.

oh ok

I now want to make a "Prime of the Ancient Mariner" joke, but I'm too lazy to figure out how.

Maybe Fermat ?

Probably for the best.

what does $A-b$ mean in set theory? $A\backslash\{b\}$?

it means $A\setminus b$, I think it's old notation and people tend to use the other one more, but I'm not sure

The whole $\frac{\mathbf{A} + i \mathbf{B}}{2}$ you do in a Lie algebra seems to be a way of saying $\mathbf{A}$ represents a real rotation, and $i \mathbf{B}$ a complex, i.e. hyperbolic, rotation, and they are distinct in the Lorentz group, weird

in measure theory $A-b$ reads as $\{a-b, a\in A\}$ (provided it makes sense)

the context is:

$$n>\text{sup}S-h$$

h is clearly an element

ah...

$\sup S$ is likely a number, depending on context

lol sup S too -.-

$\sup S$ is an element as well

yes!

@BalarkaSen is my inspiration

Hey guys, quick question. What's an example of a smooth homogeneous function which is not a polynomial?

Ah, I should say that I require homogeneity for negative scalars too

Homogenous means $f(ax)=a^kf(x)$ for some $k$?

Yes, and for all real numbers a.

What's the domain?

if i know the inflection point of a curve with the set equation of $y = x^4 + ax^3 + bx^2 + cx + 29$ how can i go about finding $a, b and c$

Also reals?

(I don't know much about this)

Yeah, f: R^n -> R.

For my application I do want a^k, not |a|^k.

(Though |a|^k is probably more standard)

@DavidZhang math.stackexchange.com/questions/1693412/…

Take any smooth function $g$ from the unit sphere to the (positive?) reals, and extend it to $\Bbb R^n$

In general there are lots and lots of smooth functions on RP^n.

Oh, right, you need to pay attention to opposite points on the sphere

Ya

Ah, of course that works. Should have been obvious. But what's an easy-to-write-down non-polynomial smooth function on RP^n?

if nobody minds helping me

@SylentNyte I would probably start with computing the second derivative

since that's related to inflection points

@DavidZhang Take a chart in RP^n, call it U. U --> R^n be the chart homeomorphism. Project to a coordinate to get U --> R. Extend to RP^n by making it zero outside a closed neighborhood of U by partition of unity.

Yea I've done that

I'm just not sure where to go from there

It's hard to write out as a function from R^n - 0 though. I don't see an easy function.

How many inflection points are there?

I think there's at most two?

Why does Ted have to write the worst exercises in his books :(

these are torture

If you don't like them, fine. But don't insult him.

The more torture they are, the more you'll have learned once you manage to find them ^^

He didn't insult him, just his problems

There's a difference

@BalarkaSen Perhaps I should clarify that I'm writing some expository notes, and my "application" for such a function is to fill in the following sentence: "Not all homogeneous smooth functions are polynomials. For example..."

(And I don't think it's really an insult)

So I'm looking for an example I can write down with the minimal number of characters :P

Oh wait, I can just take something like sin(x_1) on RP^n and extend as you described. Never mind!

Yeah, I was just about to suggest something sine-y

Though, note that this is only non-polynomial for dimension bigger than one

don't be whiny; be sine-y!

because for dimension one, x_1 can only be two possible things

@meow-mix Shiny.

@AkivaWeinberger theres only one, sorry for the late reply $(3,2)$

I was curious about those books by Ted you all talk about, but there seem to be none in my uni's library

the first derivative must equal 0, and the same for the second when $x = 3$ so I was trying something from there but i didnt get far

Ted might be able to send you a free copy. He allowed me to have one as long as i promised not to distribute @Alessandro

@SylentNyte So the second derivative is zero only when $x=3$?

The second derivative is quadratic, right?

@AkivaWeinberger yes

thats all the information ive been given

Try factoring the quadratic. If its only root is 3, what should the factorization look like?

ooohh

Why am I so bad ...

(x-3)(x^3+x^2+x-7)

if im not wrong

the determinant is a mapping $\mathrm{det}: GL(n,\mathbb{R}) \to \mathbb{R}$, right?

itll look something like that

ill try that and see where i go

@SylentNyte Well, I was getting at the fact that the second derivative (the quadratic) should look like something like $C(x-3)^2$ for some constant $C$

I don't know how that translates to the original function, though

@meow In fact a map to R - {0}.

why is that?

@AkivaWeinberger actually i dont think thatll help

@meow-mix it's a function $M_n(\Bbb R)\to \Bbb R$ which restricts to a group homomorphism ${\rm GL}(n,\Bbb R)\to\Bbb R^\times$

What does determinant of a matrix being zero mean?

$AA^{-1} = I$, take $\det$ on each side

err

@SylentNyte No? The second derivative is $12x^2+6ax+2b$, right? The above means that it should equal $12(x-3)^2=12x^2-72x+108$. This tells us what $a$ and $b$ are.

it maps $R^n$ to a lower dimension or something

I don't think $c$ changes where the inflection points are.

Wait

We also know it's the point $(3,2)$, right?

@meow-mix That's good intuition. Convince yourself that means it's not invertible.

So we can choose $c$ so that the value at $3$ is $2$

wait

oh wait

why does $12(x-3)^2 = 12x^2 - 72x+108$

i understand the $12x^2 - 72x + 108$ bit, but not the first

the determinant is $0$ means that the column vectors are linearly dependent, right?

I think it does, anyway

@meow-mix Mhm.

yes @meow

hmm

aha, ok this all makes sense

@SylentNyte If it's quadratic, begins with $12x^2+\dotsb$, and has $3$ as its only root...

...then it has to be $12(x-3)^2$, right?

but it isnt a root

It's a root of the second derivative, no?

what's your definition of determinant by the way?

And that's what we're looking at

oh wait

so youre saying that (x-3) is a root of the second derivative?

Yeah

ah okay yeah now i get ya

never saw it like that before tbh

And the second derivative is a quadratic, right? You worked that out

So we need another linear factor

And I think it should be $(x-3)$ also or else we get another root

okay

Remember that quadratics look like $A(x-B)(x-C)$ when you factor them

(The $A$ is just the coefficient of the first term)

yea okay

im trying something else

but its a very dodgy method

test: $\sin x$

$1+\cos x$

yea no it led me nowhere

$1+\tan^2 x$

why do you think the second linear factor is $(x-3)$

phew, finally got automatic chatjax on all my browsers

$(x-3)$

@Brody some script or how?

github.com/vyznev/chatjax

@Null ^

@SylentNyte thanks very much :)

@Null The "render ChatJax" command was the only one that worked before, so I had to click it every time new math was posted. But I got the "start ChatJax" to work now :)

@Brody that must have been PITA lol^^

From the bookmarks in the link to the right "$\LaTeX$ in chat..."

@Null PITA stands for?

@Brody pain in the ass

Ah, yeah. But you get used to it

But the looping render is much better oc

i now have to do a whole other excercise for linear algebra, lets see reading

I realized something wonderful $$\text{A set is determined uniquely by it's elements.}$$

Suppose $$f(x)=\dfrac{ax+b}{cx+d}.$$ For which $a,b,c,d$ does $\forall x, (f\circ f)(x)=x$?

Another one: a cake is baked with its indigrients??

(not to say it's something bad to remember)

as cake is awesome

and so are sets!

I assume the function's domain is all real numbers. So I was able to derive a system of equations: $a^2=d^2; \;\; c(a+d)=0; \;\; b(a+d)=0$. The book agrees with this, but I got different results. (in form, at least)

that's not completely obvious actually, you need an axiom (extensionality) guaranteeing that

@Brody how does the proof look like? sketchy i mean

I parametrized the solutions, so I said $u,v\in\mathbb{R},[(a,b,c,d)=(u,0,0,u)]\,\lor\, [(a,b,c,d)=(u,v,v,-u)]$. But the book has $$\dfrac{ax+b}{cx-a}\ne \dfrac{a}{c}\,\text{ or } \, a^2+bc\ne 0.$$ Question is... Is my answer equivalent?

a typo there (its)*

@Null No "proof" involved really. Just deriving the appropriate $a,b,c,d$ from the resulting equalities.

@Brody it reminds me a lot at Euler's $e$.

but maybe im dumb haha.

@Null I won't call you dumb, lol. I'm curious why you're reminded of $e$ though

@Brody well, if $f(x)=e^x$, then the deriviative is also $f'(x)=e^x$. Here you dont work with deriviative, but it's similar line of thinking imo.

@Null Hmm, I can sorta see the thought process, more like $f(f(x))=f(x)$ instead of $f(f(x))=x$ though

Okay! I figured I'm close but wrong, since $(u,0,0,u)$ for freely varying $u\in\mathbb{R}$ includes the 4-tuple with all zeroes, but that makes $f(x)$ not defined. Similar issue for $(u,v,v,-u)$. The book solution is like mine but removes these "problem points"

@AkivaWeinberger have ya gotten anywhere/

@Brody That sounds good to me.

@BalarkaSen That exercise was a bit tricky. I guess the issue was forgetting to check for extraneous solutions!

@BalarkaSen I want to clarify something. We say that $\alpha$ isn't algebraic over field F if there doesn't exist both an extension K of F and polynomial $p(x) \in F[x]$ for which $p(\alpha) = 0$ when evaluated in K right ?

@SylentNyte I think I already described where to go from here

@Adeek It doesn't make sense, per se, to say "[blah] is not algebraic over F"

Use that equation to find $a$ and $b$

I mean, what even is your $\alpha$?

Where does it live?

oh I see

and then use the fact that $(3,2)$ is a point on the graph to argue that $f(3)=2$ and use that to find $c$

@Brody I haven't checked your calculations though, but I believe you. Are you aware of what those $f$'s do, geometrically?

I think Spivak is a bit careless with the notation here. He asks $$\text{Prove that }f=f^2\text{ if and only if }f=C_A\text{ for some set A.}$$ But he himself stresses the distinction between a function $g$ and its value $g(x)$. Isn't $f(x)=f(x)^2$ more correct?

So, I see. So, $\alpha \in K$ where K is field extension over F for which there is a polynomial $p(x) \in F[x]$ for which $p(\alpha) = 0$ when evaluated in K.

I don't know the context but it could mean $f\circ f$

@BalarkaSen I just know they're rational functions. One "line" divided by another.

What's $C_A$?

the category for which the objects are morphism into A :D

haha

yes

@BalarkaSen If $A\subseteq \mathbb{R}$, then $C_A(x)$ is $1$ for $x\in A$ and $0$ for $x\notin A$.

@Brody Doesn't $f = f^2$ mean that $\forall x, f(x) = f(x)^2$ ?

so, we have $\alpha \in K$ transcendental when we have there doesn't exist a polynomial which represent it.

@Brody Then he means $f(x)^2$, yeah.

it would be weird proving something is transcendental

seems hard

@Balarka: It's past your bedtime again :)

good evening @Ted

@Astyx I don't know. He's written to distinguish a function $f$ which a set from the function value $f(x)$, but here he's not even doing that as far as I see.

@Brody No, he's right. He means the functions are equal, which means their values are equal at each point of the domain.

@Adeek Precisely. Nah, there are some things which are definitionally transcendental. Eg the element $x$ living in $k(x)/k$.

Hi again, @Alessandro

Sup, @Ted

or maybe $k(x)$, @Balarka :P

Rehi, DogAteMy :)

Edited, @Ted.

I had forgotten you were in Brooklyn Heights near my old friends.

*Hello, Dr. Shifrin

LOL

Yup

@TedShifrin Hmm. For $f:A\to B$ I imagine a set of ordered pairs, so $f^2=f\times f$. I dunno, the notation is confusing

And I'm by the library again (going home)

They've been posting about the park on FB, of course.

Yup.

It's on my block

@Brody It should be "clear from context" :)

@BalarkaSen I see

(Now you all know almost exactly where I live)

@Brody: Spivak does a good job of getting you used to functions the way mathematicians work with them. Standard calculus books are very sloppy in their notation, writing $y=f(x)$, $dy/dx$, etc.

But he's a bit sloppy here, no?!

How is he sloppy?

@Ted Could we quickly talk about the question I had yesterday (or the day before, I can't remember) about the subsets of $U$ to which $\overline D$ has a deformation retract ?

Sure, @Astyx. My suggestions were only slightly misleading, I guess. Didn't Balarka help you out?

I didn't see a follow-up.

Haha no issue

Oh, you were insisting on the half-plane picture. I decided you were probably right that that gives better intuition.

My suggestion was to work on the upper half plane, which compactifies to disk.

Regardless, vas-y, @Astyx.

@TedShifrin "Prove $f=f^2$ iff $f=C_A$ for some set $A$." But it should be $f(x)=f(x)^2$. The function value equals its square, but $f$ equal to $f^2$ looks misleading when earlier he stresses $f$ denotes a function, i.e. a collection of ordered pairs.

But he's defined products of functions, @Brody, $fg$, so this is $ff = f^2$. What's your problem?

And yeah, I know this is very nitpicky. :p

@Ted My question was more about he formalisation of the existence of a deformation retract from the disk to a closed proper connected subset of $U$, as I could not find a "simple" way of writing it

No, you're just wrong :D @Brody

There is no simple way to write it, @Astyx. There needs to be a piecewise definition, and you probably want to rotate the picture and put it in some particularly nice position first.

Ok so this answers my question then :)

Just do it on H^2, and then go back to D^2. There are very nice homeomorphisms from D^2 minus boundary pt to H^2.

Thanks

@TedShifrin Ohhh, I see it now. You're absolutely right. sorry Spivak D:

But even the halfplane case is hard to write down "neatly," Balarka. You definitely need it by cases.

Sorry if I was a bit blunt to you, @Brody :)

I don't mind your being nitpicky when you're right, though :)

@TedShifrin Np. I know you're not nasty :)

Wanna bet? Just ask Balarka :D

hmm, who needs a smack today? :D

Well, in the halfplane you can retract the whole thing to the x-axis and then retract to an interval, can you not?

Certainly not I.

I did it quite painlessly with $\{e^{it}, t\in [0, \pi]\}$ hoping I could go from this particular subset to all of them, but it didn't work out that way

eh, is "the function value equals its square" not equivalent to $f(x)=1$ for all x?

Oh, I guess you're smarter than I am, @Balarka. But I was thinking of the disk case slightly differently.

Random question: With regards to sphere eversion, is there a polyhedral version? Starting with a polyhedron with triangular faces homeomorphic to the sphere, and moving the vertices continuously such that no dihedral angle becomes zero and such that it ends inside-out.

No, @Astyx, so let's make sure you have the picture right. Say the interval is a quarter circle $\pi/4\le\theta\le 3\pi/4$. How're you going to map to it?

@AkivaWeinberger Jeez. Interesting question though.

Or, alternatively, such that no dihedral angle leaves $(\pi-\epsilon,\pi+\epsilon)$ for any $\epsilon$ (stronger)

We should check whether Smale's original theorem holds in the PL category. I would bet yes. @MikeM might know.

People thought about that a lot back in those days.

So I was able to prove the statement. It's interesting though. So characteristic (i.e. indicator) functions are really the only functions for which their function value and the square thereof are equal?

The question would be, how many vertices would we need?

or faces

@Null, no.

I am not sure how the dihedral angle thing is analogous to Smale's hypothesis though.

because clearly this is impossible for the tetrahedron, for example

Well, sure, @Brody, because $y^2=y$ has only ... as solutions.

@Ted Is "like a basketball" a good enough answer ? I can't see how to put it into words really

Can't you verbally describe the map depending on where in the disk you are, @Astyx?

$y=0$ or $y=1$. I had to use this in the proof. I'm just pleasantly surprised :p

Just wait 'til you get to the really interesting chapters, @Brody :P

@TedShifrin Anticipated. Will be huffing through Spivak and Shifrin at the same time.

@AkivaWeinberger Let me get this straight. If you want to inside-out, you have to go through the surface (self intersect). You're asking if this can be done without making nbhd of the vertices flat at some point?

@Ted I see it as such

(sorry for my poor drawing abilities)

Looks kinda like a basketball.

@BalarkaSen Yeah essentially

Wait hold on

Also, you're not restricting the moves of the faces? You can bubble it up and down so that it goes through itself? That seems to wreck the idea of doing it for polyhedra.

No, I'm confused

@Brody It does, doesn't it ? :p

"Flat" would be having the dihedral angle equal to pi, not zero…?

Right?

@Astyx: So, the question is: Can you do it with just straight lines in the picture? :)

@Akiva Oh.

Huh

Flat would mean two faces (and four vertices) are coplanar.

@AkivaWeinberger "$12(x−3)^2=12x^2−72x+108$" from where did you get $12x^2−72x+108$

From expanding $12(x-3)^2$? ^

Yeah, I misunderstood you. It should definitely be doable without making them 0.

so you assumed (x-3) is the other root?

@Balarka, DogAteMy: Based on what I'm reading in wikipedia (I never read Smale's original paper), I'm guessing it'll work PL. Homotopy groups vanishing tells you that there is a regular homotopy between the original embedding and its eversion.

I was just weirded out about how to do it without making it flat :P

@SylentNyte Don't we already know that $3$ is the only root? I thought that was part of the problem — $(3,2)$ is the only inflection point

Yeah, you need to think about what the appropriate analogue of immersion is for PL.

@TedShifrin A more interesting question is, what's the minimum number of vertices needed.

Yes, that's what Akiva is saying.

@TedShifrin I guess you could draw parallel lines going through the disk which would make a "beam" between the two boundaries of your arc, and connect the points of the disk not reached by this beam to the closest of the boundaries ?

@BalarkaSen What did you mean by this btw? With $f(x)=(ax+b)/(cx+d)$.

He doesn't want the self-intersection set to become "too bad" that two faces overlap.

Have you looked at any of the references in the wiki article, DogAteMy?

But the drawing wouldn't look as good :p

@Brody Look up "Mobius transformation"

@TedShifrin No

Will do later

Ok, will do

There you go, @Astyx. The stuff outside your beam just collapses to two points.

(Am walking home and it's dark out)

Hello I have a question, here en.wikipedia.org/wiki/Vieta_jumping#Standard_Vieta_jumping in the first step we assume that there are solutions for which k is not a perfect square, why can't why assume that there are solutions for which k is a perfect square and follow the same reasoning?

Right, however formalizing this seems like a nightmare

@user379685: It's a proof by contradiction. So if you're trying to prove the number is a perfect square, you won't arrive at a contradiction by assuming it is :)

@BalarkaSen Oh, isn't the thing that transforms the open unit disk into the upper half-plane one of those?

Not at all, @Astyx.

If $-a\le x\le a$, send $(x,y)$ to (x,\sqrt{1-x^2})$ and, if $a<x\le 1$, then send $(x,y)$ to ..., etc.

Actually, nevermind, it's probably trivially false if I try to do it through PL moves.

@Ted Really ? I'll look into it then .. Anyway thanks a lot (again) (and @BalarkaSen too !)

@Ted and this only works if we have less than half a circle such that no two distinct points on the arc have the same $x$ coordinate

oh hi @TedShifrin i completed the first two, now onto the daunting "masses" problem

True, @Astyx. You need a different picture in the other case, I guess.

@meow: I'm not saying you have to do all of them :P

Somebody gave me back directions :(

Of course, this is the extension of $f(x)$ to complex numbers. I don't even want to try Spivak's exercise with $x\in\mathbb{C}$ and complex coefficients.

and I walked blocks in the wrong direction in the cold in the dark

Which exercise now, @Brody?

@TedShifrin

@meow: You're being way more pedantic than I would have expected, but that's cool :P

@TedShifrin None for now. I solved 3.8 earlier, and 3.9 was the whole deal about $f=f^2$.

@Brody: So what do complex numbers have to do with it?

if its a pain to read :P

@TedShifrin Yes, it does.

ted i still dont get it, let's say i don't know if k is or is not a perfect square, what's wrong with following the reasoning given in the wiki but assuming that there are solutions for which k is a perfect square

It's not quite the definition of det, @meow, but it's a theorem. :)

And given two arcs is there no (simple) homeomorphism from the disk to itself such that the image of the first is the second ?

@Balarka, DogAteMy, see the link Mike just posted.

Oh, that's clever, @Astyx. Sure there is.

@TedShifrin Ah sorry, 3.8 was the exercise I was talking about. If $f(x)=(ax+b)/(cx+d)$, for which numbers $a,b,c,d$ is $f(f(x))=x$ true? I imagine allowing everything to be complex makes the exercise significantly harder.

As per Möbius transformations being brought up.

That's what I meant when I said I isolated the case of the upper half of the circle to try to generalise to any arc

But keep in mind you need to replace the tangent bundles etc with PL microbundles, which are the natural kind of bundle in the combinatorial setting.

Here's a neat intuition for sphere eversion though. Immerse an RP^2 inside R^3 (say the Boy's surface). Take the normal bundle and look at the unit circle bundle (t = 1). That's an S^2 inside R^3 because that's what double covers RP^2. Now slowly make t smaller. At t = 0 it agrees with the zero section, RP^2. Now move it backwards till t = -1. That's the sphere, "everted". I think this can be made rigorous in some sense.

And what would the expression of such a homeomorphism ?

@MikeMiller Cool, I'll have a look.

@MikeM: So Akiva is asking how many faces the polyhedron needs to be sure you won't go through flat angles ...

Sorry, can you rephrase the question?

@TedShifrin Not flat, zero angles. That two faces won't collapse onto each other.

He doesn't want dihedral angles becoming either 0 or $\pi$, I think.

Immersions should rule out either.

Also what applications does the Sperner lemma have (other than proving Brouwer's fixed point theorem) ?

I can't answer that one at this stage of my life.

@BalarkaSen I am extremely skeptical that the immersion of $S^2$ you just produced is isotopic to the standard embedding.

I'm going, good night/day to all of you !

@TedShifrin so the affine span of $n$ affinely independent points in $\mathbb{R}^n$ is simply $\mathbb{R}^n$ itself?

what does affinely independent mean? not contained in some affine hyperplane?

Mike's question should be a hint, @meow.

@MikeMiller Given a polyhedron in R^3, can you homotope it through polyhedron (aka1-parameter family of PL-embeddings of a polyhedron) so that the end map is the same polyhedron, embedded so that orientation is reversed, and that you don't have to "push one face onto another" at any stage in the middle. (by which I guess you don't want to fold a paper through a line so that both the square faces fall into themselves).

This is not possible for the tetrahedron say, visually speaking. If you push a vertex to the base, everything falls flat onto the base.

well affinely dependent points in $\mathbb{R}^n$ are contained by an affine subspace of dimension less than $n$

G'bye Astyx

Be careful about that, @meow. Check the definition.

so i mean, i guess you could say that

This, I believe, is Akiva's question.

Two affinely independent points determine a line, no?

yes

Three?

do three affinely independent points determine a line? no

they determine a plane

yes

Right ... so continue inductively :P

oh wait

hmmm

@BalarkaSen When you're working with PL immersions, you'll be subdividing the triangulation you started with as you go forward in time. This is the natural thing to do. So as long as you're willing to further subdivide the tetrahedron, yes, you can do this. (Recall that a PL immersion is just a locally injective PL map.)