Mathematics - 2022-10-18

Ted Shifrin

02:58

@user346760 yes?

user

04:52

Is this working?

leslie townes

depends on what you mean by 'working'

user

It’s working! Yay! I missed you all. Been 13 years.

Jasper Loy and Asakiri are cool. Asakiri isn’t his name. He’s a set theorist. Think he knows JM. Can’t remember. Quaicjiu Yuan is cool, too.

And Bill Dubuque (water under the bridge)

I’m sure no one remembers me, though. I’m easily forgotten.

Anyway, see you later, folks

one potato two potato

Schwartz reflection principle in the unit circle can be proven using power series expansion..?

$f\in\mathcal{H}(\Bbb D)$ then $1/\overline{f(1/\bar{z})}\in\mathcal{H}(\Bbb C\setminus\overline{\Bbb D})$

Just saying naively ${\partial\over\partial\bar{z}} =0$

copper.hat

05:19

with a distinct user name how could one be so easily forgotten?

Ted Shifrin

05:34

@copper.hat All personality.

@onepotatotwopotato So how is this power series?!?!

one potato two potato

@TedShifrin I mean expressing the reflection function as a quotient of some power series and saying 'well there's no $\bar{z}$ term here, so holomorphic'

Ted Shifrin

05:48

I wouldn’t mention power series. Of course, you have to look carefully at boundary points.

one potato two potato

@TedShifrin Then how? Using the definition of holomorphic function?

The original Schwartz reflection principle in a symmetric domain w.r.t. real line can be proven easily using power series so I was expecting something similar here.

Lemon

07:24

If I have a manifold $M$ and an open set $U$ of $M$. It doesn't necessarily follow there is a coordinate chart $\phi$ defined on $U$ right?

leslie townes

right.

Not Tfue

27

Q: How do you prove Well-Ordering without Mathematical Induction? (and vice-versa)

Here is my attempt to prove the Well-Ordering Principle, i.e. that any non-empty subset of $\Bbb N$, the set of natural numbers, has a minimum element. Proof. Suppose there exists a non-empty subset $S$ of $\Bbb N$ such that $S$ has NO minimum element. Define $A = \left\{n\in \Bbb N : (\forall s...

I don't know why well ordering and induction not equivalent?

I think Induction implies well-ordering

But don't know why well-ordering doesn't imply induction

Lemon

@leslietownes i know my question is super open ended in the first place, but what information is required for there to be a coordinate chart on any arbitrary open set $U$?

Not Tfue

I don't get it why "Suppose $P(0)$ is true, and $P(n+1)$ is true whenever $P(n)$ is true. If $P(k)$ is not true for all integers, then let $S$ be the non-empty set of $k$ for which $P(k)$ is not true. By well-ordering $S$ has a least element, which cannot be $k = 0$. But then $P(k-1)$ is true, and so $P(k)$ is true, a contradiction." is wrong. Why is P(k-1) wrong?

leslie townes

lemon: i don't know. note that taking U = M this includes or is the question of when an n-manifold embeds in R^n, which is already not elementary even for surfaces.

if you don't want it to hold for all U but just want to know 'how big' U can be relative to M without being the whole thing, that also seems interesting, but i don't know.

math.stackexchange.com/questions/322027/… is semi related (maybe not what you are thinking about) but involves a smooth structure.

Lemon

07:40

Does anyone know why in Rieman normal coordinates one requires an isomoprhism $B$ determined by an orthononormal basis $(b_i)$ for $TpM$? That is Normal Coordinates defined by the coordinate chart $B^{-} \circ Exp^{-1}$ (assume Exp is a local diffeomorphism) where $B$ is defined by $B(x^1\dots x^n) = x^i b_i$ (all notations are borrowed from Jack Lee (p131). It seems to me orthonormal isn't that important.

@leslietownes I just realized my problem. $U$ (doesn't have to be all the way $M$) is at least the intersection of two open sets that ARE from coordinate charts. BUt none of those maps $\phi_i$ have to be defined on all of the set $U$.

Not Tfue

10:25

anyone?

is it illegal to take induction reverse?

robjohn

@NotTfue what do you mean by "induction reverse"? Do you have an example?

Not Tfue

Suppose P(0) is true, and P(n+1) is true whenever P(n) is true. If P(k) is not true for all integers, then let S be the non-empty set of k for which P(k) is not true. By well-ordering S has a least element, which cannot be k=0. But then P(k−1) is true, and so P(k) is true, a contradiction.

I was thinking why is this wrong P(k−1) ?

The comment section got me confused

"@AndrewH.Hunter You are correct, and the proof above is wrong. Induction and well-ordering is not equivalent."

robjohn

10:44

This is a proof by contradiction, not induction in reverse. The argument is more complicated than it needs to be this way, but it works.

Peter Balabanov

hello. maybe someone happens to know what Pr means here? math.stackexchange.com/questions/503245/…

seeing this for the 1st time

Not Tfue

I know it is contradiction but didn't understand why it is wrong but seems justified.

Peter Balabanov

11:03

@Peter judging by the context it must be conditional probability, but it's strange way to denote that. maybe the author meant projection on sigma-algebra generated by the latter which may lead to the same statement

Thorgott

11:33

I would assume Pr means probability, yes

the vertical bar is standard notation for conditioning

@Lemon that condition simplifies the components of the metric at $p$

user

11:50

Not complaining (sorta complaining…$ but it’s hard to read thin words in a sentence without a clear magnifying glass or Bette glasses

better glasses*

@copper.hat I don’t know if I know what you meant.

@robjohn can you advise me how to fix up my tensor question? I b struglin

Peter

13:13

Who wants to join in the search of a twin prime pair ?

What is the smallest $s$ such that $$10^{5000}+9^{5000}+s\pm1$$ is a twin-prime pair ?

copper.hat

14:11

What is the fascination with twin primes?

user346760

14:49

@TedShifrin One might be walking with a bunch of little canines, one of them happens to be a female and, well, there's that... really gets one to appreciate the collective efforts aimed at the common goal of synergizing all instances of such cases, doesn't it?

Jai Sri Krishna

16:26

Thank you @robjohn

Shinrin-Yoku

@TedShifrin can you give an example of problem in lin alg

where

one can naturally think

"oh if i had a alternating multilinear function on matrices i would solve this problem"

I.e. Is there a synthetic motivation of the det

Thorgott

16:53

not gonna happen

Shinrin-Yoku

@Thorgott

why?

Thorgott

cause it's a silly idea

it's not how mathematics develops

if you knew what you needed to a solve a problem before solving it, there'd be almost nothing left to do

Shinrin-Yoku

so the only way to motivate the det is using volume?

Im wondering if there is a synthetic motivation not a geometric one

@Thorgott

Ted Shifrin

Here's a stupid one, but it is what you deserve. Test (without solving a system of linear equations or using row operations) whether $n$ vectors in $\Bbb R^n$ are linearly dependent/independent.

Asking if $v_1\wedge\dots\wedge v_n = 0$ or $\ne 0$.

Shinrin-Yoku

how does a multilinear alternating function arise from that?

Ted Shifrin

17:04

It's wedge product or ... determinant.

As I said at some point earlier, the high school motivation for determinants is Cramer's rule. Maybe you should settle for that.

Shinrin-Yoku

cramers rule only motivates for 2x2 3x3

i like the oriented volume interpretation

Ted Shifrin

No, it works in all dimensions. How could you not realize that?

Shinrin-Yoku

It works, but how on earth would you motivate it?

Ted Shifrin

The same way you do in the $n=2$ case.

You want the value of some particular $x_i$ when you solve $Ax=b$.

Shinrin-Yoku

right and then you define the nxn determinant as what?

Ted Shifrin

17:08

The unique alternating multilinear function ...

I think you need to get past this obsession. I'm done with it.

Shinrin-Yoku

i agree @TedShifrin

i like the volume approach in your lectures

@TedShifrin

Ted Shifrin

If it didn't show up naturally with differential forms and integration theory, I doubt we'd be as interested as we are.

Thorgott

there's plenty of motivation, I am simply objecting to the notion that it should be possible to materialize the right notion out of thin air

in mathematics, concepts are usually best understood in their applications and, in the same way, their applications usually precede their definition in the first place

Ted Shifrin

Yes, abstraction usually follows after some concrete implementation.

Shinrin-Yoku

only problem is that the natural motivation i get from that is that the det is the unique function preserved under column ops, etc. we can then derive that it is multilinear and alt. but then why most texts start with multilinear and alt. instead of preserved under column ops and all that. @TedShifrin

^from the vol interpreattion

Ted Shifrin

17:11

Because you don't need matrices to talk about multilinearity.

And the volume interpretation is about vectors, not matrices.

I care a lot more about differential forms and exterior algebra than I do about determinants of matrices :D

Except for the Vandermonde determinant, maybe :D

Thorgott

yeah, the exterior algebra is the conceptual approach

Shinrin-Yoku

@TedShifrin ah cuase it works for matrices in general fields as well

Ted Shifrin

general commutative rings, in fact. Someone made a big point of this last week when you were harping on this.

Thorgott

it's a pretty genius idea in all honesty, but I don't think it's outlandish to attempt and understand a linear map not just by its action on vectors, but also by its actions on parallelotopes spanned by those vectors

Ted Shifrin

I actually knew a lot of the history when I wrote the linear algebra book 25 years ago, but I'd have to reread the historical stuff at this point.

Thorgott

17:20

I don't know much history, but I do know that these ideas go all the way back to Grassmann 178 years ago

he was more or less postulating anticommutativity of the exterior algebra as a consequence of orienting parallelograms

Ted Shifrin

Determinants were first developed (by Cardano!) to solve systems of equations. Leibniz wrote in a letter to L'Hôpital in 1683 that in the 3x3 case the determinant would determine whether $Ax=0$ has a nontrivial solution. Cramer did Cramer's rule in 1750 in an appendix of a treatise on algebraic curves. Lagrange noticed the link with volume later in the 18th century.

Jacobi elevated the theory in 1841.

Surprise, surprise. He "discovered" the Jacobian determinant.

Thorgott

that's the point where the volume interpretation really shines

it gives an intuitive explanation of the change of variables formula, I always loved that

Eminem

Given 3 normals to 3 planes in a 3d world n1,n2,n3 where n = {a,b,c}, how can I find the origin of this coordinate system?

Normally the origin, o = {0,0,0} but this is true only for the "default" coordinate system

Ted Shifrin

The origin is arbitrary. How can I know from normals to planes?

Eminem

Im trying to use a projection function, which requires a point, normal to the projected plane and origin. If the normal to the plane changes (i.e the coordinate system changes), wouldnt the origin paramer change as well?

Ted Shifrin

17:28

Go back to lines in the plane. Make your question explicit.

Projecting a point onto a line does not depend on coordinates. But it depends on the geometry (orthogonality).

If you change coordinates and compute both ways you’ll get the same answer.

Eminem

Im not sure what you mean

Ted Shifrin

The projection of a point onto a line (plane) doesn’t depend on the coordinate system.

Eminem

Hmm that is kinda weird since the function does get an origin parameter :)

Ted Shifrin

Because you're using equations for the line (plane), so you need vectors and an origin. What I'm saying is that if you change the origin (and get different equations) or change the orthogonal coordinate system, you will get different numbers everywhere but the same geometric solution of your problem.

Just draw some pictures (2D is easier).

Eminem

Again, since the function gets both normal and origin as parameters (and uses them), this seems weird that changing one or the other wont change the final result.

monoidaltransform

17:47

0

Q: Riemann Extension Theorem proof question zero set contained in smaller ball

$B_{\epsilon}(0)$ denotes polydisk in $\mathbb{C}^n$, where $\epsilon= (\epsilon_1,...,\epsilon_n)$. Statement of theorem: Assume $f:B_{\epsilon}(0)\subseteq \mathbb{C}^n\rightarrow \mathbb{C}$ is holomorphic and $g:B_{\epsilon}(0)\backslash Z(f)\rightarrow \mathbb{C}$ is holomorphic and bounded....

complex-analysis complex-geometry several-complex-variables

Ted Shifrin

The formula for the answer looks different. The geometric point is the same.

Try doing a simple example where you move the origin and rotate the coordinate system by some angle.

Ted Shifrin

18:00

@Thor Do you think it's OK for me to "forget" to respond?

monoidaltransform

18:12

Can any one help with my question?

user10478

18:24

If there is a real world condition that may or may not be present, i.e., a medical condition ($C =$ condition, $N =$ no condition), and an indicator signaling a prediction about that condition ($+ =$ condition predicted, $- =$ no condition predicted), then there are four unconditional probabilities, $\Pr(C)$, $\Pr(N)$, $\Pr(+)$, and $\Pr(-)$, and eight conditional probabilities, $\Pr(C | +)$, $\Pr(N | +)$, $\Pr(+ | N)$, $\Pr(- | N)$, $\Pr(N | -)$, $\Pr(C | -)$, $\Pr(- | C)$, and $\Pr(+ | C)$.

In this order, a nice cyclic pattern emerges in the conditional probabilities, whereby one can solve for each subsequent conditional probability by alternating between Kolmagorov's Axiom of Normativity and Bayes' Theorem, with wrap-around at the end.

It is easy to see that there are two degrees of freedom among the unconditional probabilities, with the remaining two being taken care of by Normativity. It is also easy to see that the applications of Bayes' Theorem pull at least one of the unconditional probabilities into the conditional probability cycle.

My question is, how many degrees of freedom exist in such a framework overall? How many probabilities must be given to solve for all of them, and does it matter which ones are given?

Ted Shifrin

@monoidaltransform No. I find it confusing (confused) from the very first sentence.

monoidaltransform

I guess my question reduces to the following: If $f:B_{\epsilon_1}(0)\subseteq \mathbb{C}\rightarrow \mathbb{C}$ is holomorphic and $f$ has no zeros at $\partial{B}_{\frac{\epsilon_1}{2}}(0)$ then does it follow that $Z(f)\subseteq B_{\frac{\epsilon_1}{2}}(0)$?

@TedShifrin

Ted Shifrin

Well, of course that is wrong.

But the setting here is that we're looking at a hypersurface in $\Bbb C^n$ and we're fixing other parameters.

So the proof starts by assuming that $f(0)=0$ but that $V(f)$ does not contain the $z_1$-axis? So by the identity principle, it cannot contain an open neighborhood of $0$ in the $z_1$-line.

monoidaltransform

18:41

In the third paragraph how is he sure that $Z(f_w)$ is finite?

Ted Shifrin

Your "rendition" of this proof in your question is just very sloppy.

monoidaltransform

Yeah, that's why i'm asking about it here. So I can understand it better

Ted Shifrin

Whoever that is didn't define $f_w$ in that paragraph.

monoidaltransform

Huybrechts, $f_w(z_1)=f(z_1,w)$ for $w\in \mathbb{C}^{n-1}$.

and $w=(z_2,z_3,...,z_n)$.

Ted Shifrin

He's just using the Cauchy integral formula to do this. Why do you care if $f_w$ has any zeroes inside?

monoidaltransform

18:45

Because how else would he extend it to a holomorphic function on $B_{\epsilon_1/2}$ if not by the Riemann extension of one variable?

Alacrity

math.stackexchange.com/questions/4555623/… Probability question

Ted Shifrin

Riemann extension doesn't talk about zeroes; it cares only about boundedness.

monoidaltransform

It doesn't talk about zeroes, but it does talk about finite sets. In particular, if $f:B_{r}(0)- \{a_1,..,a_n\}\rightarrow \mathbb{C}$ is holomorphic and bounded then it extends to holomorphic $B_r(0)\rightarrow \mathbb{C}$

Here $B_r(0)\subseteq \mathbb{C}$, ofcourse.

Ted Shifrin

But he said $f(z)\ne 0$ if $|z_1|=\epsilon_1/2$ and $|z_j|<\epsilon_j/2$ for $j>1$.

Oh, so if we let $|z_1|<\epsilon_1/2$, there will be zeroes, but there cannot be infinitely many by the identity principle.

The zero-set of $f_0$ contains no neighborhood of $0$ in the $z_1$-line.

monoidaltransform

I still don't see why we can't have zeros of $f_{w}$ in $\frac{\epsilon_1}{2}<|z_1|<\epsilon_1$.

$f_{w}$ is, after all, a map $B_{\epsilon_1}(0)\rightarrow \mathbb{C}$, no?

Lemon

18:54

@Thorgott well that's one of the implications. BUt I m asking why does it start with an orthonormal basis in the first place? We could've just began with any basis $(b_i)$ and defined an isomoprhism $B(x^1\dots x^n) = x^ib_i$ and still define coordinates on the norml neighborhood and call this "normal coordinate". If $(b_i)$ is not orthonomral does that make B not isomoprhism?

Ted Shifrin

He may be omitting some stuff with the preparation theorem or something. He's assuming that $f=0$ intersects the $z_1$-line only at the origin. So it should just be a continuity argument to know that for small $w$, $f_w$ has no zeroes if we restrict to a small compact set.

@Lemon Of course it's an isomorphism, but you for the applications of normal coordinates it's not convenient. You want the metric to be the identity at the center (for convenience) and to have some nice formulas for curvature, for example. If you pick a random basis, that's a mess.

I don't know, monoidal. My head isn't wanting to do this today.

monoidaltransform

No worries. I'll post a clearer question on main. Thanks in any case.

Ted Shifrin

You can look at other texts, too.

monoidaltransform

Yeah, I think i'll look at Griffith and Harris

Thorgott

@TedShifrin oh well, I really wonder what a path is

@Lemon no, it's still an isomorphism, but I don't get your point

the thing that is called "normal coordinates" are called as such, because that's how they're defined

I (and Ted) gave you a reason what makes the ones where the basis is orthonormal special

user4539917

19:27

Mathematician Yitang Zhang Claims to Have Proven Riemann Hypothesis Problem

7

He's only published 3 papers.

Ted Shifrin

@Thorgott Didn't it say $g$ was the path?

PseudoLooped

20:14

@TedShifrin He's still around, but life post-corona has had him extremely busy for quite a while as it disrupted school, work, etc. When life slows down in a couple of years or less he'll be back on to say hi

user4539917

Thanks for the update.

Ted Shifrin

Well, @Pseudo, if you are in touch with anon, please send him my kind regards.

PseudoLooped

I definitely will - I didn't realize how interconnected he was on this site until I found your profile/book on your profile and shared it with my math group and he then explained everything about MSE from years and years ago. Definitely a lot and interesting

Ted Shifrin

20:31

Oh, that's bizarre. Well, not that many years ago. I think I've been here around 9 years or so.

PseudoLooped

That's fair, to me anything before 2015 is years-and-years ago lol

Ted Shifrin

So you knew anon from school?

PseudoLooped

Yeah, we've met and became good/close friends when he mentored a Putnam/Honors Calculus study group in college, which he still does

Ted Shifrin

I'm sure he did a great job of that. He was quite an effective teacher in here before he spontaneously combusted :P

P.S. Which book of mine are you referring to?

PseudoLooped

The free PDF one on Differential Geometry

Ted Shifrin

20:40

Ah ... I should have guessed.

Fargle

21:26

When did I start showing up here? I want to say it was around 2013.

Maybe '14.

Ted Shifrin

21:46

I showed up around the same time.

According to my profile, around April 2013.

Goku

About a month ago for me

user

22:10

May I please have some help with my question about multilinear algebra?

It's here: math.stackexchange.com/questions/4554616/…

I'm trying to understand what tensor equations look like in abstract axiomatic constructions and I am using the Einstein field equations as an example to understand the concept of a tensor equation.

robjohn

@Fargle your profile says 2014-08-09

Fargle

I can only imagine that I was as beloved then as I am now.

robjohn

That was when you first used chat. That was a couple of months after you joined the math.SE.

Ted Shifrin

Write them in terms of vector bundle language, especially since you want to consider differential operators (which will map sections of one bundle to sections of another).

@robjohn is a spy — nothing gets past him.

I don’t think I remember you before UT, @Fargle.

Fargle

I started there in '13, so that tracks.

I am older than my behavior betrays!

Ted Shifrin

22:22

You behave?

robjohn

@TedShifrin I only give out secrets for the right price.

Ted Shifrin

Yeah, but you’re cheap.

Fargle

@TedShifrin Touché.

one potato two potato

22:43

cooold

Ted Shifrin

23:01

Way too hot here in sunny CA.

copper.hat

23:43

i am melting in albany

UnderMathUate

Can someone check my understanding of this:

The definition of associates in my textbook says: If F is a field, the units in F[x] are the non-zero constant polynomials. f(x) is an associate of g(x) if f(x) = cg(x) for some c $\neq$ 0, $c \in F[x]$.

Then, this questions asks to list all associates of $x^2 + x + 1$ in $\mathbb{Z}_5[x]$. So since the units of $\mathbb{Z}_5[x] are 1, 2, 3, and 4$, the associates of $x^2 + x + 1$ would be

$$x^2 + x + 1 = (1) \cdot x^2 + x + 1$$
$$2x^2 + 2x + 2 = (2) \cdot x^2 + x + 1$$

Akiva Weinberger

Finally read through the proof of Sard

Wow, is that a technical proof

I should review the implicit function theorem

Ted Shifrin

23:59

@UnderMathUate wrong …. $c$ must be a unit, i.e., a nonzero constant.

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