Mathematics - 2023-02-13 (page 1 of 2)

Thorgott

00:01

look at $\pi_1$

monoidaltransform

Ah right. Sorry, I don't understand lens spaces, so is the $\mathbb{Z}_p$ action on $\mathbb{S}^3$ an isometry on $\mathbb{S}^3$?

Thorgott

yes, the action is by isometries with respect to the standard metric

monoidaltransform

Thanks!

ペガサス Seiya

01:07

@TedShifrin Where's the PDF of said lecture? What's it called?

Also, I might have injured my collar bone so I'm at a hospital right now

Thorgott

@Thorgott we also need to argue the function $\frac{f(x)}{x_n}$ is non-vanishing, but this is clear if $x_n\neq0$ by assumption, and if $x_n=0$, we know the consatnt term $\partial_nf(0)\neq0$, so the term in the sum has smaller absolute value for $x$ sufficiently near $0$. I guess this local statement suffices as we can argue in adapted coordinates around every point.

Ted Shifrin

@ペガサスSeiya That sounds bad. Did you trip? It’s on my UGA webpage, which I think is linked in my profile.

Thorgott

01:24

@Ted so, in the smooth case, closed hypersurfaces still induce line bundles. so we could set up a theory of divisors and get an association divisors -> line bundles. I've never seen anybody do this in the smooth setting. I guess it's not fruitful?
particularly, there doesn't seem any meaningful way to go back from line bundles -> divisors. in the complex/algebraic setting, the point as far as I understand is the existence of meromorphic/rational sections with restricted zero loci, but in the smooth case a generic section is transversal to the zero sections and so only has finite zero locus...

Ted Shifrin

Finite? No.

GratefulDisciple

@ペガサスSeiya It's here: Tidbits of Geometry through the Ages, linked from @TedShifrin's UGA homepage.

Ted Shifrin

A section of a line bundle is still a twisted scalar function. But real line bundles aren’t so interesting.

Thanks, Grateful.

Thorgott

oops, of course not finite

got my dimensions mixed up

so perhaps the analogous theory would work after all, but I guess it's as you say and just not necessary as we already understand smooth line bundles well enough

I don't really know what divisors are good for anyway. The only applications I've seen were to construct embeddings into P^n from ample line bundles. The analogous question in the smooth case has much better approaches.

Semiclassical

dumb grammar question

when using X(s) to denote something which may or may not be plural (e.g., "map" vs "maps"), is it "X(s) was" or "X(s) were"

there's probably a classy way to ask that question but heck it

Thorgott

01:40

can you give an example sentence?

Semiclassical

found a reference: english.stackexchange.com/questions/93940/…

Ted Shifrin

Linear systems are a crucial tool in understanding varieties.

Semiclassical

from there: "Selecting an appropriate study topic(s)" vs "Selecting appropriate study topic(s)"

that's not quite what i was asking but it's more general

Ted Shifrin

@Semiclassical “The function”?

Semiclassical

not X(s) as a function, but like singular vs plural

no math in sight

Tanner Swett

01:43

My general rule of thumb is that we use the plural unless the quantity is known to be exactly one.

Ted Shifrin

@Thor But on a connected topological manifold, there are either a single or two real line bundles.

Semiclassical

that's probably the way to cut the Gordian knot here

Ted Shifrin

@Semiclassical oh, I have a book(sj.

Tanner Swett

@Semiclassical ...ah, but if I understand you right, you're asking if a phrase such as "the folder(s)" (meaning "the folder or folders") should be treated as singular or plural. Is that right?

Semiclassical

i'll give the specific example i was running into

"In particular, note any changes in brightness or color and whether the images changed when the polarizer(s) was rotated."

Ted Shifrin

01:45

I would avoid the number issue:) Give me the folder or folders.

Semiclassical

simplest way to get around this is probably just to do "when you rotated any of your polarizers"

Ted Shifrin

You don’t know how many polarizers?

Semiclassical

they start out using just one, then eventually look through two at once

Ted Shifrin

Avoid passive, anyhow.

Tanner Swett

In formal writing, I'd probably change it to "when the polarizer or polarizers were rotated." In informal writing, I'd just treat it as plural: "when the polarizer(s) were rotated."

Ted Shifrin

01:46

Rotated your polarizer(s).

Semiclassical

that works

"image" feels like the wrong word here since they're not looking through lenses, but that's the most economical solution i could think of

Tanner Swett

"Appearance," maybe?

Semiclassical

maybe

Tanner Swett

A photograph, a projection, a reflection, and a sight through a lens are all images. If you're just looking at something, then what you see is its appearance.

Anyway, I came here to ask a little question of my own.

Semiclassical

a polarizer is sorta active in the same way as a lens, so i guess i'm fine with image for now

Tanner Swett

01:51

Is there a standard term that describes the relation between two theories S and T defined by the statement "the arithmetic statements that S proves are exactly the same as the arithmetic statements that T proves"? My best guess for the terminology was "S and T are arithmetically equivalent," but that doesn't seem to be standard terminology. (Or maybe it is; I didn't look very hard.)

Thorgott

@TedShifrin no, it's more complicated than that

they're parametrized by $H^1(M,\mathbb{Z}/2\mathbb{Z})$

Ted Shifrin

Yes, and …?

Oh, never mind.

leslie townes

improv night

Ted Shifrin

Happy supra bowl.

leslie townes

whee!

Semiclassical

02:08

@TannerSwett Google spits out the following paper: iphras.ru/uplfile/logic/log19/LI19_Strollo.pdf. a relevant sentence is this: "Again a natural expectation might be that ACA− and P A(S)− should be two arithmetically equivalent theories, sharing the same arithmetical content."

so that suggests it's a reasonable choice of language.

another relevant paper: projecteuclid.org/journals/modern-logic/volume-2/issue-3/…

the footnotes for that seem particularly important

Tanner Swett

@semiclassical Good to know, thanks!

Semiclassical

do look at the footnotes, they make me less sure about whether "arithmetically equivalent" is the right phrase

Silly Goose

hey all! I've got i think a simple question :P. I'm trying to prove injectivity of the function $f(x) = \frac{1}{x-a} * \frac{1}{x-b$ where $a,b \in \mathbb{R}$. I made the usual supposition that $f(x_1) = f(x_2)$ to see if it implies that $x_1 = x_2$. However, I've come to a halt in the algebra.

Tanner Swett

@Semiclassical Which footnotes do you mean? All 23 footnotes of the second paper you linked to?

Silly Goose

I have $x_1^2 - (a+b)x_1 = x_2^2 -(a+b)x_2$. I was wondering if I can treat $x_1^2$ and $x_1$ as being linearly independent. I think no because they are simply particular elements of the reals here, not variables $x$ and $x^2$, which are linearly independent

But if that's the case I am the nstuck on how to simplify further

Semiclassical

02:22

i guess that wasn't too specific. footnotes 4/12/16 and their associated text @TannerSwett

Tanner Swett

@SillyGoose It's been a couple years since I did that kind of math, but I feel like the next step is probably to complete the square on both sides.

Silly Goose

oh mayn

Semiclassical

Another way to proceed is to move all the terms to the same side of the equation and factor

Silly Goose

maybe im off track XD. the ultimate goal is to produce a bijective map between $(a,b)$ and $\mathbb{R}$. graphically this function looks like itll do the trick

leslie townes

silly: you're right that you can't treat x_1 and x_2 as variables there. side note, why do you expect that function to be injective? (there is a problem that is fairly close to the surface when a = b, for example)

Semiclassical

02:24

this sounds like an XY problem

leslie townes

yeah, i would work on other ways of doing this.

Semiclassical

also, it's not clear what you mean by a bijection between $(a,b)$ and $\mathbb{R}$

$(a,b)$ is just some point

leslie townes

one contribution i will toss into the pile: maybe first try to convince yourself that if you can solve it for one pair of distinct a, b, you can solve it for any pair of distinct a, b.

that will reduce the number of parameters you need to fiddle with.

Silly Goose

to show that the set ${c \in \mathbb{R} : a < c < b, a,b \in \mathbb{R}}$ has the same cardinality as $\mathbb{R}$ (not looking for the answer to clarify!) @Semiclassical

Semiclassical

oh, so $(a,b)$ as the interval

Silly Goose

02:26

yesyes

Semiclassical

there's a standard bijection i'd use, but i don't know how to hint it

Silly Goose

hm so is the simplestt (at the level of me which is an undergrad at the beginnings of an intro analysis course) method to define an explicit bijection?

Semiclassical

well, you can do it just using precalculus functions

Tanner Swett

@SillyGoose Well, you've managed to come up with a function with vertical asymptotes at x = a and x = b, but I think now you need to alter it so that it has a root between a and b.

Semiclassical

you want a simple invertible function that takes numbers on some finite interval and whose range is all reals

Tanner Swett

02:29

I think there are lots of ways to do it, and I see that Silly Goose is trying to do it using a rational function, and as far as I know, that's a totally reasonable approach and it'll work just fine.

Semiclassical

in which case your example makes a lot more sense

Silly Goose

hm well id like a function that looks like x^3 but has symptotes

Semiclassical

(tbh I thought you were wanting to do the bijection between $\mathbb{R}^2$ and $\mathbb{R}$, where the method would be way different)

you might just do the boring thing and plot the graph (e.g. a=-1, b=1) to see if it works first

Tanner Swett

@SillyGoose The rational function 1/(x - a)(x - b) won't work, but I think you're on the right track. You just need to take this function and change it so that it does work.

In order for the function to be a bijection, it needs to have a root, and that function obviously doesn't have a root. But you know how to alter it to add a root to it, right?

Semiclassical

the example i usually go with is trig functions. but the two approaches are not really different

Silly Goose

02:33

ah gosh

i see

Tanner Swett

Like I said, there are lots of ways to do it.

Silly Goose

hm how does having a root imply bijection?

Tanner Swett

Other way around. Being a bijection implies having a root.

Semiclassical

yeah. i like the trig approach because then i can interpret it as stereographic projection

but trig functions become rational functions under a Weierstrass transformation, so again. not actually different

Silly Goose

@Semiclassical are you suggesting x/(x-2)(x-3) as a possibility

and then alternatively like some arccos parameterized properly would be the trig way?

leslie townes

02:42

there's no "the" trig way, there are lots of ways. this problem is maybe challenging because of how underdetermined it is

Silly Goose

oh i should say one other solution can come from defining a particular trig function arccos

oh oops i tagged the wrong person in my reply above i meant @TannerSwett

are you saying it needs to have a root if it is a bijection because 0 is an element of the codomain?

leslie townes

i also maybe wouldn't get too bogged down at the outset in specific formulas. you're looking for a bijection between a "finite" open interval (a,b) and R. i suggested that the problem does not change difficulty if you can find a bijection between any particular "finite" open interval (e.g. (0,1), or (-1,1), or whatever you like) and R. it might be worth thinking about that.

or, it would be enough if you could find a bijection between one "finite" open interval (a,b), and any "half-infinite open interval" (c, +infty) or (-infty, c), as long as you can also find a bijection between some "half-infinite open interval" and R. it might be worth thinking about that too.

you can come up with all kinds of formulas that maybe don't do too good a job of explaining a story behind how you might build one of these things.

as one example. it's not too hard to show that 1/x defines a bijection between (0,1) and (1, infty). it's also not too hard to show that x - 1/x defines a bijection between (0, infty) and R. and you can fiddle with those two subproblems to come up with a solution to your problem.

or use inverse trig functions or whatever else.

Thorgott

02:58

or exp

leslie townes

i think it makes sense to look for functions with formulas, but also helpful to think in the background, nothing about 'bijection' requires continuity. so you could break this into any other kinds of subproblems, e.g. by establishing that any one example of any interval of any kind, including kinds like [a,b), [a,b], (a,b], was in bijection with R

peek-a-boo

a history question: does anyone know who (and when) first proposed the solution of the Basel problem ($\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$) using the Fourier series of $f(x)=x$?

I’m able to find lots of history regarding Euler’s proof and his contributions, but less so on the other methods

Silly Goose

hm so like using some piecewise of ln(blah), some line connecting, exp(blah)

whose domain is a<x<b

well i guess i am seeing that we'd like asymptotic behavior at x = a and x =b (while still being a bijection), but it seems you have a different emphasis in mind on waht the problem is trying to say

Semiclassical

03:41

@peek-a-boo glancing around, it appears the Fourier-series proof indeed doesn't have as much documented history for it as the others

leslie townes

yeah, that's interesting. as a guess, it would probably go at least as far as whoever "proved" what we'd now think of as parseval's theorem (or equivalent). proved in quotes because this was likely at a time when maybe they didn't do details so well.

peek-a-boo

alrighty, looks like I’m free to make up history as I see fit when I describe the story to my students

Semiclassical

@leslietownes possibly. I don't know when Parseval's theorem was proven relative to the rest

tho as you say, "proof" is probably the wrong way to look at it

when was it formulated, etc etc

Cotton Headed Ninnymuggins

04:03

Does anyone have any hints for how to start proving the equality for all $k\in \mathbb{N}$?

Semiclassical

what you've said makes no sense

first, there's no sense in which you can establish this equality for various $k$. what would it even mean for $k=1$ here, for instance?

Cotton Headed Ninnymuggins

@Semiclassical why?

I messed up

Semiclassical

for all natural $n$, perhaps?

Cotton Headed Ninnymuggins

I meant $n\in \mathbb{N}$ yes

Semiclassical

okay. then let's see if it works for the first few cases. for $n=0$ there's no problem, since both sides collapse to just being 1

but what do you get if $n=1$?

hmm, n=1 is also fine. one moment

yeah, n=2 should break it

Cotton Headed Ninnymuggins

04:10

They should be equal unless my instructor is making us prove something that isn't true

Semiclassical

well, if $n=2$ then the relevant binomial coefficients are 2C0=1, 2C1=2, 2C2=2

... am i going senile. hmm

Cotton Headed Ninnymuggins

Semiclassical

yeah

Cotton Headed Ninnymuggins

The RHS is just $2^n\cdot 2^n$ I think since it's the cardinality of the power set times $2^n$

The LHS is what I'm struggling with

Semiclassical

yeah. i think that's what threw me: it certainly doesn't look like it should be true

Cotton Headed Ninnymuggins

04:15

That's for sure

Semiclassical

easiest way is to recognize the LHS as the binomial expansion

$(x+y)^n = \sum_{k=0}^n \binom{n}{k}x^k y^{n-k}$

what choice of $x,y$ gives you the LHS?

Koro

What is the meaning of Gal(f) ?

Semiclassical

Galois something of f?

Koro

How is it defined? Is it defined only for functions f with rational coefficients?

Semiclassical

i mean

the fact that i told you "something" rather than an actual thing should suggest that i don't know enough to be helpful :P

Koro

04:27

I have a set of exercises wherein there are exercises of the following type: Give an example of f such that Gal(f)= $Z_2$

@Semiclassical np :)

Semiclassical

that said, Wikipedia does include a definition for Galois group of a polynomial

it's as clear as mud to me tho

that said, the page does include a construction via cyclotomic polynomials to get Gal(f) = Z/pZ for prime p

the case for Z/2 seems particularly simple

field theory is where i stopped understanding abstract algebra tho

Cotton Headed Ninnymuggins

@Semiclassical I don't know what $x,y$ would give me the RHS since the binomial expansion doesn't have an $n$ power, but an $n-k$ power. However, since I know the RHS is $2^n\cdot 2^n=4^n$ then the LHS is equal to that because $x=3$ and $y=1$ right?

Koro

I have a definition but it is not yet clear to me: Gal (f)= {$\sigma \in S_n: (a_1,a_2,…,a_k) $ and $ (a_{\sigma 1},…,a_{\sigma n})$ are Q conjugates $\}$.

Semiclassical

@CottonHeadedNinnymuggins right

Koro

where a_i’s are distinct roots of f.

Semiclassical

04:33

so $(3+1)^n = \sum_{k=0}^n \binom{n}{k} 3^k$

but what's $(3+1)^n$?

Cotton Headed Ninnymuggins

What $x,y$ would work for the RHS using the binomial theorem?

@Semiclassical $4^n$

Semiclassical

right. which is definitely $2^n \cdot 2^n$

Koro

@Semiclassical ohh

Semiclassical

but you can also view it as $(2+2)^n$, which would give $x=y=2$

what's the powers in the binomial expansion in that case?

Koro

Galois developed all this at the age of just 18?

Everiste Galois

Semiclassical

04:35

@Koro yeah, it's kinda mindblowing

Koro

Now I want to learn to speak French.

Cotton Headed Ninnymuggins

Ohhh, so $x=y=2$ is what works in the binomial theorem. I see thank you so much.

Semiclassical

@CottonHeadedNinnymuggins in short, you're using the binomial expansion on $4^n$ in two different ways

Koro

:D

Semiclassical

it's a funny problem because it's way more obvious once you generalize it

Silly Goose

04:37

heh ive finally come up with a solution i think that i have convinced myself well enough of

Cotton Headed Ninnymuggins

yes, $(3+1)^n=(2+2)^n$, makes perfect sense.

Semiclassical

namely: $m=(m-a)+a$, so $m^n=\sum_{k=0}^n \binom{n}{k} (m-a)^k a^{n-k}$ which is true for any $a$

if you want an example which looks absurd, then, pick an even integer $m$ and compare the $a=m/2$ and $a=1$ examples

Cotton Headed Ninnymuggins

04:49

That's pretty snazzy

Semiclassical

coming up with strange-looking examples is easy when you know where you're starting from :P

Cotton Headed Ninnymuggins

05:17

indeed

Tanner Swett

05:52

@SillyGoose I don't think x/(x - 2)(x - 3) would help you very much, but I think (x - 2.5)/(x - 2)(x - 3) would be good.

And yes, a bijection whose codomain includes 0 must have exactly one root.

Koro

08:21

0

A: How to find discriminant of a cubic equation?

You are going in the wrong direction. I don't know what is the vertex point of a cubic. However, it is true that the sum of all roots of that cubic (real or not) is $-\frac ba$. Dealing with $-\frac b{3a}$ is useful because if you replace $x$ by $x-\frac b{3a}$ in your cubic, then you get a new ...

Can anyone please explain why the discriminant is as it is claimed in the answer? $((r_1-r_2)...(r_3-r_1))^2$

one potato two potato

09:03

Jost's Riemannian geometry textbook is quite good

ペガサス Seiya

09:28

@TedShifrin Yeah, I got hit by a car that was reversing. Not the driver's fault. I just like to run

Thorgott

12:00

@Koro well, what's your definition

Koro

nvm, I did a lengthy calculation to show the equality.

The Tralfamadorian

14:05

can anyone suggest me a book about the classification of clifford algebras? I want to understand the - much simpler - case of complex clifford algebras, but most books I found start by classifying real clifford algebras and afterwards explain complex clifford algebras via complexification

GratefulDisciple

@ペガサスSeiya I have a question about Boku no Hīrō Akademia in the Anime chatroom. Sorry to hear about your run-in with that car. I hope the X-ray of your collar bone turns out okay.

@TedShifrin Do you have a textbook recommendation for Statistics that emphasize proofs and conceptual understanding like your Multivariate textbook? For Probability I'm following your recommendation in your Math 4600 course syllabus.

Sine of the Time

14:25

@Thorgott I think I've understood the geometrical meaning, but how to formally write a retraction from $\Bbb{P} ^n \setminus p$ to $\Bbb{P}^{n-1}? Do you have any ideas?

Thorgott

yes, translate the geometric idea into a formula

that's how geometry works

ペガサス Seiya

14:54

@GratefulDisciple Yeah the X-ray turned out fine. No fracture thankfully

Mats Granvik

15:08

It appears that we don't need analytic continuation to locate the first non-trivial Riemann zeta zero.

$s= 3/2 + 14i \\ M=10^{10000}$

$$\Re\lim_{n \rightarrow 100}
\left(
\left[
1-
\left(
\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\sum_{m=1}^{m=M} 1/m^{\left(\tfrac{k}{n}+s\right)}}
\Bigg/
\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\sum_{m=1}^{m=M} 1/m^{\left(\tfrac{k}{n}+s+\tfrac{1}{n}\right)}}
\right)
\right]^{-1}
+\frac1n + s
\right) = $$
`0.49999999999999999999999999999999999999999999999999999999999999999999\
821263876336076842 +
14.134725141734693790457251983562470270784257115699243175685567460149\

Kumar Shuvam

ABCD is square. Points E (4, 3) and F(2, 5) lie on AB and CD, respectively, such that EF divides the square in two equal parts. If the coordinates of A are (7, 3) then the coordinates of other vertices can be.. I'm having trouble with this question

Any hints would be helpful 🙂

Sine of the Time

@Thorgott I think I've the right idea to the retraction, the problem is the inclusion $i : \Bbb{P}^{n-1} \to \Bbb{P} ^n \setminus p$. I don't know how to define it in order to be well defined

Koro

15:53

It may not look like it but Z[x] is isomorphic to Q+ with multiplication.

@SineoftheTime does the following work? $i: [x_1,…,x_n]\mapsto [x_1,x_,…,x_n,0]$

Sine of the Time

@Koro this is the idea, the problem is that the codomain is $\Bbb{P}^n \setminus p$ so I don't think it's well defined.

Thorgott

use a different coordinate. recall the geometric picture. $\mathbb{P}^{n-1}$ is supposed to correspond to which lines?

Ted Shifrin

@Thor We are dopes. Just write $f(x,t) = f(x,0) + \int_0^1 \frac d{du} f(x,ut)du = t\int_0^1 \frac{\partial f}{\partial y}(x,tu)du$ (where we have coords $(x,y)$). End of story.

Koro

For well definedness, suppose that we have $[x_1,x_2,…,x_{n+1}]= [x_1’,…,x_{n+1}’]$, that is there exists a c such that $x_i’= cx_i$. So I think that it should work.

(I may be wrong though.)

With that, the images of these equivalent points will also be equal. I’m using the following definition of $P^n= (R^{n+1}-0)/R$*

Ted Shifrin

16:08

@GratefulDisciple That is a standard probability text, nothing special. I don’t endorse it particularly. And I doubt there exists a comparable stat book, but I certainly do not know.

parz

‘Lo.

Koro

@ペガサスSeiya Sorry to hear that. I hope you get well soon.

One day a car drove over one of my feet, thankfully nothing happened to my foot. I pulled my foot back as soon as the first tire got over the foot.

Ted Shifrin

Foots, eh?

seiya needs to pay more attention. I’m not sure how he can run loose — let alone drive — with his bad eyesight.

Koro

@TedShifrin fixed

Thorgott

@TedShifrin ahhh, so you're doing like a parametrized Taylor expansion in just the relevant variable

that's much better indeed :)

@Koro you are, but only because their $p$ is $[1\colon0\colon\dotsc\colon0]$

Koro

16:18

Ohh okay.

Obliv

16:29

for $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$ the equation of an ellipsoid it need not be equal to 1

it can be any constant $\ne 0$ right?

Akiva Weinberger

PNDas

If you have Evans book then could you please look at the theorem 2 of section 6.5? They used strong maximum principle but didn't assume connectedness of the domain.

Akiva Weinberger

Did you know that if $\operatorname{bitsum}(X)$ refers to the number of 1s in the binary expansion of $X$, then $\operatorname{bitsum}(X)+\operatorname{bitsum}(Y)-\operatorname{bitsum}(X+Y)$ equals the number of "carry" operations you would do to find the sum of $X$ and $Y$ in binary?

Obliv

@Koro that's wild, I guess the way that it rolls over the foot doesn't cause as much damage as say the same force falling directly on the foot.

Akiva Weinberger

A reasonable question would be: "if we color the point $(X,Y)$ black or white depending on the parity of this value, what does the resulting picture look like"?

PNDas

16:32

@Obliv certainly the constant can't be negative and if the constant is zero then it's the origin.

Akiva Weinberger

($(0,0)$ is on the top-left)

Obliv

not as much potential for degloving

Ted Shifrin

@Thorgott I think when we started, I'd forgotten that $f(x,0)=0$ for all $x$. Silly.

Akiva Weinberger

Works out that there are two equivalent recurrence relations for the picture you get

@AkivaWeinberger and the picture looks like this

Obliv

@PNDas But it can be like as large as you want, it just determines the size of the ellipsoid ?

PNDas

16:37

@Obliv I mean if you put $d>0$ in place of 1, suppose $d=e^2$, then you can divide both sides by $e^2$ to get 1 on the right side. By size do you mean volume? Volume will become $\frac{4\pi}3e^3abc$ (if I'm not wrong) . So I guess if you keep on increasing $e$, volume will go to infinite.

May be Evans assumed connectedness but forgot to write it? Anyway I found another book where they assumed connectedness.

Mats Granvik

teorth.github.io/QED

ペガサス Seiya

@TedShifrin to be fair I wasn't expecting the car to reverse, it looked like it was empty to me. The moment I got close to it, the driver reversed quick, giving me no reaction time and hitting my left shoulder, causing me to fall backwards

Ted Shifrin

16:54

Who reverses at 30 kph all of a sudden?

ペガサス Seiya

@TedShifrin that driver who was kind enough to take me to the hospital as I couldn't even stand up after that

Ted Shifrin

shakes head

Sine of the Time

@ペガサスSeiya how are you? What've you done?🙄

@Koro take for example $\Bbb{P}^3\setminus [1,0,0,0]$, if you consider the inclusion $i: \Bbb{P}^{2}\to \Bbb{P}^3\setminus [1,0,0,0]$ it does not work since $i([1,0,0])=[1,0,0,0]$ that is not in the codomain

Koro

17:20

Hmm. Thanks.

Mats Granvik

I really struggle with exercise 5.1 in Tao's QED. I solved it earlier today in 12 lines, but the optimal solution is 2 lines.

Koro

@SineoftheTime Let’s take shifting inclusion: $[x_1, x_2,…,x_n]\mapsto [0, x_1,..,x_n]$.

Mats Granvik

Now down to 4 lines.

What is the connection between Formulas and Assumptions in logic?

Sine of the Time

@Koro am I allowed to used it to prove the retract deformation? because the definition given to me is:$i \circ r \simeq id$

Koro

Ohh, the definition isn’t $r: X\to A, A\subset X, r(a)=a$ for all a in A ?

Sine of the Time

17:35

@Koro yes this is the definition of retraction

basically I want to prove $\Bbb{P}^3\setminus [1,0,0,0]\simeq \Bbb{P}^2$

Ted Shifrin

Deformation retraction means you can homotop the identity map to the retraction.

Koro

Oh I thought that you wanted to show that P^2 is retraction of the other space when P^2 is seen as a subset of it.

Ted Shifrin

@Sine I assume you and Thor discussed the picture for this. You can draw the 2D picture easily. You want to project from the point to the plane (line).

@Koro First you want a retraction, yes. But then you want to show you can do it gradually. For example, every point $p$ in $X$ is a retract of $X$, but that's not very interesting.

Sine of the Time

@TedShifrin yes, but I'm trying to define $i: \Bbb{P}^{2}\to \Bbb{P}^3\setminus \{[1,0,0,0]\}$ but I can't

Ted Shifrin

Huh? If you have a subset, you have its inclusion map.

Sine of the Time

17:42

@TedShifrin but is it well defined?

Ted Shifrin

Take the $\Bbb P^2$ to be the plane $x_0=0$ for convenience.

leslie townes

@AkivaWeinberger is there a multicolored version of this for primes other than 2?

Ted Shifrin

What do you mean well-defined? A subset is a subset.

Koro

If we take [a,b,c,d]in P^3-p, then at least one of b,c,d is non zero. Say b is non zero. We can divide by b. So we can map this [a,b,c,d] to [a,c,d]. If this gives me a continuous map that is 1-1 and onto, then

Ted Shifrin

Do not write $(a,b,c,d)$ when you're working with homogeneous coordinates.

Koro

17:43

I want to show this map to be inverse of i that we tried earlier.

Ted Shifrin

That is entirely confusing. Write $[a,b,c,d]$ or some people write colons.

jrh

I've been working through "A Guide to Proof Writing", on page 497 of the proof in Example 3, the author assumes that b is in the intersect of X and Z but I am not sure how that's a safe assumption, couldn't Z be the empty set? That would make the proposition true but it would be a false assumption.

Ted Shifrin

You don't need to do this.

Define the map by sending $[x_1,x_2,x_3]$ to $[0,x_1,x_2,x_3]$. Then well-definedness is trivial.

jrh

The example is trying to prove "for all sets X, Y, and Z, if X is a subset of Y, then (X intersect Z) is a subset of (Y intersect Z)"

Koro

@TedShifrin: is the idea correct? Once we show these two maps to be inverses of each other, we have shown the homeomorphism and we are done :-).

leslie townes

17:45

jrh: Z could be empty, but in that case there is nothing to prove. the implication "if something is in X intersect Z, then ___" is true no matter what ___ is in that case.

Sine of the Time

@TedShifrin Ok, but this is still considered inclusion map? because to proof it's a deformation retract I've to proof $i\circ r\simeq Id$

Ted Shifrin

The inclusion and the retraction are never inverses.

Koro

@TedShifrin that’s our i.

Ted Shifrin

They are homotopy inverses, but not set inverses.

Koro

Ohh, but sine said that they want to show $P^3-p\simeq P^2$. I suppose that $\simeq$ was for homeomorphism, was it not?

Sine of the Time

17:47

@Koro homotopy, the're not homeomorphic I think

Ted Shifrin

Absolutely not homeomorphic. One is $2$ complex-dimensional, the other is $3$.

Koro

Ahh okay.

leslie townes

jrh: more generally to prove that a set A is a subset of set B, it's enough to prove for all a that if a is in A, then a is in B. A being empty doesn't 'break' this reasoning, A being empty is just the one case in which element-wise reasoning would not be needed.

jrh

That is fair, the empty set is a subset of any set. Another thing I don't completely understand about the author's method, it seems like the procedure assumes q in p -> q; though if p were false, wouldn't that give false positive proof results?

Novice

@onepotatotwopotato Not an important question, but what were your analysis and manifolds chops like before looking at Jost's RG book?

Sine of the Time

17:54

@TedShifrin do you have 5 minutes to check if my reasoning is correct?

Ted Shifrin

If it's correct, it should only take a few seconds, not 5 minutes :)

Sine of the Time

@TedShifrin yes, just give me the time to write it :(

Ted Shifrin

Are you starting with the copy of $\Bbb P^2$ I suggested?

leslie townes

jrh: if p is false, then p -> q is true. this is slightly different from "assuming" q. it's only noting that q is implied by something false.

Ted Shifrin

If Leslie always tells the truth, then the moon is green.

Sine of the Time

18:00

I want to prove $X \simeq Y$, where $X=\Bbb{P}^2$ and $Y=\Bbb{P}^3\setminus \{[1,0,0,0]\}$. Consider $i: X\to Y$ defined letting $i([x_0,x_1,x_2])=[0,x_0,x_1,x_2]$ and $r:Y \to X$ defined letting $r([x_0,x_1,x_2,x_3])=[x_0,x_1,x_2]$. Is it true that $i \circ r\simeq Id_Y$? If we consider $F:Y\times I \to Y$ defined by $F([x_0,x_1,x_2,x_3],t)=[tx_0,x_1,x_2,x_3]$ we should have the homotopy

Ted Shifrin

NO, your map is wrong.

Sine of the Time

@TedShifrin are you referring to $F$?

Ted Shifrin

You need to project from the point $[1,0,0,0]$ to the plane $x_0=0$.

I'm referring to $r$. You need to draw a picture.

Sine of the Time

but $[1,0,0,0]$ is not in the domain

jrh

Right, if p were false that'd be a trivial proof. So we're assuming X is a subset of Y, and then assuming b is in (X intersect Z), the only thing left to prove is that b is in (Y intersect Z). I guess that makes sense, when you let the trivial cases of subset and implication fall away. Thanks.

Ted Shifrin

18:02

That's the point. The picture of the mapping is that you're projecting from $p$. You send $q$ to the point where the line $pq$ intersects the image plane.

Sine of the Time

how can I formalize it?

Ted Shifrin

Write it down and the answer will be obvious when you get there. Intersect the line $[t(1,0,0,0) + (1-t)(z_0,z_1,z_2,z_3)]$ with the plane $x_0=0$.

Happy heartless, @robjohn.

robjohn

@TedShifrin Heartless?

Ted Shifrin

Well, the grimace negated the heart.

robjohn

I see...

Ted Shifrin

18:09

Back to the doctor this week?

BTW, @robjohn. Perhaps you could scold this person for removing almost every question he/she has posted after comments.

I found one remaining post and added a comment to that effect.

Sine of the Time

@TedShifrin give me 5 minutes

robjohn

@TedShifrin I get a test next week and then see the doctor again the next week.

just back from an appt this morning

@TedShifrin I will look into this.

Ted Shifrin

Darn. Still recovering. Well, I wish you nothing but the best.

These things take time.

robjohn

The meds I'm on make me drowsy, so I spend a lot of the day napping.

Ted Shifrin

Well, you're entitled.

Are you still running your course at UCLA or have you retired from that?

Sine of the Time

18:17

@TedShifrin I can't see how if we consider $r([x_0,x_1,x_2])$ we have the identity map

robjohn

@TedShifrin I don't run the course, I just wrote and now maintain the software for the logic courses taught in the Philosophy dept.

Ted Shifrin

I misspoke. I know in the past you'd spoken of needing to head down to campus for that responsibility.

@Sine You need to listen to what I've said twice already. The retract is the copy of $\Bbb P^2$ sitting as the plane $x_0=0$ in $\Bbb P^3$.

Akiva Weinberger

@leslietownes Not that I or my friend (who made that) have done

I bet @robjohn and/or @PM2Ring could do it

Ted Shifrin

We're looking at the plane $\{[0,x_1,x_2,x_3]\}\subset\Bbb P^3$.

Sine of the Time

@TedShifrin yes

Ted Shifrin

18:20

But that is not what you wrote.

Have you intersected the line with the plane as I suggested? What is $r([z_0,z_1,z_2,z_3])$?

Sine of the Time

$[0,z_1,z_2]$?

Ted Shifrin

How do you intersect a line and a plane in $\Bbb P^3$ and end up not in $\Bbb P^3$?

Sine of the Time

$[0,z_1,z_2,z_3]$

leslie townes

akiva: do you know the relation between p-ary carry-counting and multinomial coefficients? i'm wondering if it explains the recursion/picture in terms of something more 'known'

ペガサス Seiya

@SineoftheTime I'm fine now. Just hurt myself thanks to my carelessness

Ted Shifrin

18:31

@Sine That's right.

ペガサス Seiya

"Authentic" Italian pizza

Obliv

Seiya do you live in japan?

I thought bad drivers were only localized to new jersey (where I live)

Sine of the Time

@TedShifrin so the point is when I consider the retraction $r:\Bbb{P}^3\setminus{[1,0,0,0]}\to \Bbb{P}^2$ I see $\Bbb{P}^2$ with four coordinates instead of 3

ペガサス Seiya

@Obliv Yeah I'm in Japan right now. And, once again, it wasn't the driver's fault, I was carelessly running when I shouldn't. That said, yes Japan can have bad drivers too

Obliv

would calling $\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 0$ an elliptic double cone be incorrect?

it's called an elliptic cone in my text, but is double cone reserved for another surface?

robjohn

18:37

@AkivaWeinberger That can be used in base-$p$ for Kummer's Theorem. I show this in this answer

Obliv

Not saying it wasn't your fault, but I typically look behind when I'm pulling out and I don't pull out that fast.

The Tralfamadorian

can anyone suggest me a book about the classification of clifford algebras? I want to understand the - much simpler - case of complex clifford algebras, but most books I found start by classifying real clifford algebras and afterwards explain complex clifford algebras via complexification

Obliv

Some people think looking at your rear view mirror is sufficient but there are blind spots on the diagonals

robjohn

@AkivaWeinberger this answer mentions the carry in base-$p$

ペガサス Seiya

@SineoftheTime I tagged you in Quid's room, I wanna show you something

Sine of the Time

18:40

@ペガサスSeiya didn't see the tag...

Ted Shifrin

@SineoftheTime I need to leave for a few hours, but this is correct. Reread everything I've said. You are embedding $\Bbb P^2$ in $\Bbb P^3$ and that's the copy you work with in the entire problem.

Sine of the Time

@TedShifrin ok thank you for your help

Koro

I have shown that the composition of two reflections is a rotation or a translation. How do I show that every rotation is a composition of two reflections?

I know that every isometry in R^n is a composition of orthogonal linear operator and a translation.

Akiva Weinberger

19:00

@robjohn Ah nice

Obliv

19:22

is the dot product of u and v the magnitude of the projection vector of u onto v?

and vice versa the magnitude of the projection vector v onto u

ペガサス Seiya

I remember proving the commutativity of the Dot product

That was fun

Obliv

Is the dot product the scalar projection of u onto v?

ペガサス Seiya

Have you learned projection in trigonometry?

@Obliv the dot product isn't a projection

@Obliv watch Ted's lecture video, it'll give you a great run down

Obliv

If Ted did a Ted talk would it be a Ted^2 talk

ペガサス Seiya

Holy shit that's genius

Why didn't I think of this pun???

Obliv

19:30

because it's awful and I should feel bad for making it

Sine of the Time

@Obliv but you're not feeling bad, instead you're laughing

Obliv

I did not laugh, but I did smirk.

if I had to find the equation of a plane that passes through three points P1,P2,P3 I make vectors out of them such that the vectors contain all three points, take the cross product and do the dot product of that cross product with

ペガサス Seiya

Ted's Ted Talk

Obliv

$n_1(x - x_0) + n_2(...) + n_3(...) = 0$

but what goes in the parentheses again

Sine of the Time

@Obliv you can also solve a linear system

Obliv

19:35

Hmm?

do those initial points have to be the points P1,P2,P3

Sine of the Time

@Obliv $\pi: ax+by+cz+d=0$ and you substitute the coordinates of the three points

Obliv

like one from each point

but what is $d$

Kumar Shuvam

A straight line passing through P (3, 1) meets the
coordinate axes at A and B. It is given that the
distance of this straight line from the origin O is
maximum. The area of triangle OAB is equal to... Can anybody help me with question..one way could be calculate the length of perpendicular from orgin and maximise it..but that would be quite lengthy...is there any other wa

Sine of the Time

@Obliv this is the general equation of a plane in $\Bbb{R}^3$

Obliv

$a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$

oh I thought u were asking me

man I'm tired lol gotta take a break 1 sec

Kumar Shuvam

20:26

Anyone😢

Sine of the Time

@KumarShuvam hi

Kumar Shuvam

Hi

Umm! Can u help me with the above question..?

Sine of the Time

@KumarShuvam are you referring to this @KumarShuvam

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$Mathematics$ Mathematics