Mathematics - 2022-07-25

Mad Spaces

02:52

boo scary faces

BAYMAX

@robjohn Sorry for my late reply..I was away from computer..

Yes, so we will get two points $(x_{1},y_{1}), (x_{2},y_{2})$

but now as $a,b$ varies we wcan have two points, zero points

is it possible to keep track of the points in the plane as $a,b$ varies

Not sure of that makes sense?

robjohn

@BAYMAX Did you try my suggestion? Take the difference of the equations?

BAYMAX

03:08

Ok, I took difference of two equations

and we get

$y = x(b-a+1) + (b-a)$

and quadratic in $x$

$x^2(b-a+1) + x(2b-a+1) + (1+ab-a) = 0$

So now I have $x$ in terms of $a,b$

and $y$ in terms of $a,b$

but i dont see how to get $y$ in terms of $x$

robjohn

03:28

$y = x(b-a+1) + (b-a)$

But you are getting one or two points, not a curve.

BAYMAX

yes for each fixed $a,b$ we might one point or two points

now if we take another set of $a,b$, we will get another set of (x,y)

robjohn

yes

and the equation that you wrote and I repeated gives $y$ in terms of $x$

BAYMAX

how the collection of such points $(x^*,y^*)$ will look? as $(a,b)$ varies

Let $(x^*,y^*)$ denote the points that satisfy the two very first two equations

so is it that

$y^* = x^* (b-a+1) + (b-a)$

robjohn

@BAYMAX with two variables $a,b$, I would expect it to be a two dimensional shape, but I don't know what it would be without computing $x,y$ for different $a,b$

BAYMAX

hmm I see

Mad Spaces

03:45

generally. is Av = v^T*A, where v is vector A matrix?

leslie townes

try v = (1,0,0,...,0)^T

Mad Spaces

thx

Leslie dont you live in the uk, you are so early!

leslie townes

i live near los angeles but have a weird work/sleep schedule because a lot of our clients are elsewhere

Mad Spaces

ah okay

Mathematical Emergency

You know what the Grothendieck group construction typically looks like right?

It's what can go from $\Bbb{N}$ to a $\Bbb{Z}$ structure under comm. $+$

Well, I'm asking what happens when you form all non-empty subsets of $\Bbb{Z}_n$ or any finite abelian group

and then take the grouthendieck group of the elementwise monoid formed

$\{0\} \equiv 0$ is idenitty

So take $\Bbb{Z}_2$

We have $\mathcal{P}(\Bbb{Z}_2) = \{0,1, \{0,1\}\}$

I'm wondering if every thing doesn't just collapse.

leslie townes

03:58

you can have collapsing if the commutative semigroup you are doing the construction on is not cancellative. cancellation is i think exactly what you need to get the semigroup to embed in the group.

so {1} + {1,2} and {2} + {1,2} being the same thing is gonna collapse some stuff. assuming + is union or something like that.

Mathematical Emergency

Yes, collapsing is fine, I'm just worried about collapsing to $0$ lol

group

I'm thinking $\{0,1\}$ absorbs the whole group

so is a zero element and so everything collapses!

Darn it

But if you specify proper subsets as well, then the grothendieck group is the powerset itself or $\Bbb{Z}_2 \simeq$

But what about $n = 3$?

leslie townes

it might help to pin down what operations you are using, what the identity element is, etc.

the construction gets progressively worsely behaved in general outside of cancellative semigroups with identity

Mathematical Emergency

I think you can't get closure of elementwise unless you include the whole group

and the whole group elementwise absorbs every other set

to itself

Thus the Grothendieck group will alwayz collapse to trivial group

QED, I suck at exploring numbers

It's as if you can't recover ANY group structure from the subsets that isn't artificially transferred onto it (not naturally constructed)

Except for a monoid

Mad Spaces

Given a function $f(z_1,z_2) = (z_1-z_2,z_1+z_2)$ on $\mathbb{C^2} \rightarrow \mathbb{C^2}$ the base is set to be $v_1=(1,i), v_2=(2,0)$ i was asked to find the dual function, i found the dual basis $ v_1^* = (0, 1/i), v_2^*=(1/2,-1/2i)$
How can i continue now?

i googled how to find the dua function bad to no results

in my notes i have that the dual function should take a function from C^2 and send it to the composition on f , not sure hto

Mathematical Emergency

@MadSpaces define for us what the dual is

Do you mean

Mad Spaces

04:11

if $ f : V \right arrow W$ then $ f^* : W^* \rightarrow V^*$

Mathematical Emergency

$\circ f$ or $f\circ$ ?

Mad Spaces

Thats what i have in my notes

Mathematical Emergency

Correct your $\LaTeX$ please!

I rhymed

Mad Spaces

i dont know why that box came to live

Mathematical Emergency

What is the dual of a vector space then?

Is it $W \to k$?

Mad Spaces

04:12

set of all linear functions from that to the field.

Mathematical Emergency

where $k = \Bbb{C}$ ic now

Okay so you're given a basis of $\Bbb{C}^2$?

Mad Spaces

yes as stated, i have figured out the "dual basis" but i dont have an algorithim to calculate the dual function!

Mathematical Emergency

You know how you reversed V, W there?

So clearly it's $\circ f$

or precomposition

Okay, so $f : V \to W$

You're given $V = \langle v_1, v_2 \rangle$

= $\Bbb{C}^2$, right?

Mad Spaces

in this case V = W = C^2

Mathematical Emergency

I'll assume it does, looks like it does

Okay, cool

Okay, so you have

So given $g \in W^* = \Bbb{C^2} \to \Bbb{C}$

We have that $g\circ f(z_1, z_2) = g(z_1 - z_2, z_1 + z_2)$ by definition of the induced map

So $\psi(g) = g\circ f$

But $g$ is linear since we're working in the category of modules over $\Bbb{C}$.

So $\psi(g)(z_1, z_2) = g(z_1, z_1) + g(-z_2, z_2) = z_1g(1,1) + z_2g(-1, 1)$.

$\psi$ is by definition equal to the dual of $f$

@MadSpaces does that help at all?

Mad Spaces

04:24

i got to go to the bathroom , a dozen of coffes since last night did not help at all

God have mercy

Mathematical Emergency

Letta rip

Mad Spaces

alright yes i see this is similiar to what i had in mind

since i found the dual basis, can i continue simplifying by saying $z_1g(1,0) + z_1g(0,1) + z_2...)$ and insert the dual basis vectors

To eventually find a closed form

I see we can not insert the dual basis vectors into g

Gwyn

09:16

Is this math.stackexchange.com/questions/4498529/… a bad written question or am I just being unlucky with answers? :(

jay

quite a long question @Gwyn

maybe its possible to narrow it down more precisely

Gwyn

09:43

@jay Thank you for the feedback, jay) unfortunately I am very new to the topic I'm asking and so, at least for now, I don't know how to shrink the question. Maybe I will edit it in the future, hoping to improve it.

polite proofs

12:51

@TedShifrin That is indeed much more clear and concise.

I completely forgot about using the sign of $f$ at $a$ to generate an interval to handle all of those problems.

Which is a nice feature of $f$ being continuous

Koro

14:06

Suppose that $\mathscr A$ is the set of all inductive subsets of R. Define $S= \cap_{A\in \mathscr A}A$.

Please treat \mathcr as italic curly A. I don’t know how to type that.

My question is: why is 1 the smallest element of S?

Clearly, the set$\{x: x\ge 2\}$ is inductive too and so is the set $\{x:x\ge 1\}$ and their intersection is the former set, which should be contained in S.

So 1 can’t be the smallest element of S.

This S is the set of all positive integers. I don’t understand how :(

Thorgott

14:22

define inductive set

Koro

I understood it now. Inductive set must contain 1 so {x:$x\ge 2$} is not inductive. :)

Thorgott

yup, nice

Koro

15:09

Considering only field properties of R, and order properties (i.e, 1) x>y implies x+z>y+z; 2) x>y,z>0 implies zx> zy), can it be concluded that 1>0?

Thorgott

yes, try showing that any non-zero square is positive using 2)

robjohn

15:35

@Koro is that universal? Looking at some definitions, I don't see that.

Ted Shifrin

16:34

@robjohn That's the definition I've always provided (shrug).

robjohn

@TedShifrin Okay. I was just looking around and the definitions required a partially ordered set (not even integers) and a successor function. I haven't really done anything with inductive sets, so I won't intrude.

Ted Shifrin

Well, intrude away. I've only encountered this at the level of undergraduate teaching (intro to higher math sorts of courses).

You were fortunate to miss my linking yesterday to three (I think) of the "typical" posts these days. Sad.

opio

17:50

Ughh I really cant see how the first line of Equation (11.1.48) from here www2.stat.duke.edu/~sayan/ambrosio.pdf is correct.

opio

19:04

how do people define $H^1(I,X)$ where $I$ is a closed and bounded interval of $\mathbb{R}$ and $X$ is a metric space

Ted Shifrin

I have no idea what you’re talking about. Not cohomology. Sobolev space?

Doesn’t make sense, either.

opio

yeh I think I misunderstood notation

this book doesnt state how things are defined...

link.springer.com/article/10.1007/s13373-017-0101-1 on page 126 they write $H^1([0,T],\mathcal{W}_2(\Omega))$

sorry $\mathbb{W}_2(\Omega)$

not $\mathcal{W}_2(\Omega)$

$\mathbb{W}_2(\Omega)$ is a metric space.

any idea what this means?

Ted Shifrin

19:20

They have to define their notation somewhere. The only intelligent guess would be Sobolev spaces, but then you need derivatives of functions with values in your metric space. If it's a Banach space, you might succeed. Go search the book. We certainly can't be mind-readers.

@Xander @robjohn @copper Is there some sort of standard comment that we're supposed to post to someone who solves a homework/exam question that showed no effort?

I guess it's clear that that answerer is answering all sorts of low-hanging fruit.

Xander Henderson

@TedShifrin I don't think that there is a standard comment, but anything which says "hey, maybe don't do that" and links to math.meta.stackexchange.com/q/33508 can't hurt.

Ted Shifrin

19:38

Does that explicitly discourage people from getting rep by answering PSQs?

Xander Henderson

@TedShifrin I don't think so?

But it explains that answering low quality questions is discouraged.

Koro

@robjohn I have always seen that definition so far.

does anyone know about the book 'what is mathematics'?

Xander Henderson

19:55

By Courant and someone else?

Koro

yes :)

Xander Henderson

Yes, I know that book.

Koro

this video youtube.com/watch?v=fdJOv6iirfU popped up in my suggestions

and I came to know about the book today.

Xander Henderson

Ah. It is a great book. I have a 1978 printing.

Koro

it has number theory also. I think I'll start number theory from here.

@XanderHenderson that's probably the last print?

Xander Henderson

20:00

No idea.

It is the first paperback printing by Oxford (according to the frontmatter).

Koro

The book seems to have complex analysis, geometry, topology too.

the book also has a section 'how to use this book'.

:)

Ted Shifrin

20:51

While you're collecting gems of books, don't forget Hilbert and Cohn-Vossen Geometry and the Imagination.

Xander Henderson

@TedShifrin That's a good one, too.

Thorgott

21:12

that one has a very nice illustration of how you glue up an octagon to get a genus 2 surface

too bad I'm awful at actually remembering that picture

Ted Shifrin

The octagon is easy. A little trickier is seeing the connected sum from the picture. But I drew that in my diff geo notes.

Maybe that's what you mean.

Not very tricky, now that I look back at it. :D

Thorgott

yeah, seeing connected sums at the level of fundamental polygons is easy and essentially gets you all the way to the base case by induction; that's how some proofs of classification even work, I think

but imagining the actual gluing process "by hand" to get a surface with $g$ holes is what I'm really bad at

Ted Shifrin

I'm not sure I understand your point.

When I figured this out when I was writing the notes, I first drew the boundary of the disk we remove as a little loop based at the common vertex of the polygons. But then I realized — duh — to have it go across the polygon.

Thorgott

I'm just saying my visual imagination in 3D is bad

Ted Shifrin

You don't actually see how to slice a 2-holed torus in its middle and get two 1-holed tori with disks removed?

Even you are not that much of an algebraist.

Thorgott

21:25

what I mean is this (illustration from Cohn-Vossen)

it makes sense when you look at it, but I'm bad at replicating that

the connected sum approach I do understand

Ted Shifrin

Wow. I don't think I've ever noticed that.

Thorgott

From a 16-gon to a genus 4 surface

►

Ted Shifrin

The pretzels are not something I've ever pondered.

I actually had to draw the hyperbolid 8-gon that tiles the hyperbolic plane. I tried to do it carefully, to scale. That took me a few hours.

Thorgott

oh lord

Ted Shifrin

And now I cannot draw any pictures like that, cuz I have no more Illustrator.

That animation is pretty cool, but I could have done without Swan Lake.

Xander Henderson

21:32

But Swan Lake is great!

Ted Shifrin

I'm not a huge Tchaikovsky fan, but whatever.

There's the bat-like octagon.

Xander Henderson

Yeah, early and mid-20th century Russian avant garde composers are a lot better than 19th century Russian romantics.

But Swan Lake is still pretty good.

Ted Shifrin

My dad doesn't even count as Russian, despite heritage.

Thorgott

@TedShifrin love it

polite proofs

22:44

Dr. Shifrin, when is one no longer a so-called beginner?

Ted Shifrin

22:57

@politeproofs It’s all relative. I’m still a beginner. And I’m past the end.

polite proofs

23:17

🤨

We can put the origin at high school mathematics then..

one potato two potato

23:32

@Thorgott The point of this video is seeing generators?

Thorgott

23:59

it just illustrates how the gluing process works

Transcript for

$Mathematics$ Mathematics