Mathematics

usukidoll

12:00 AM

like integers mod n being mapped to integers mod m :/

$f(a_{n}+b_{n}) = a_{m}+b_{m}=f(a_{m})+f(b_{m})$ ?!

Obliv

no. the last terms should be $f(a_n) + f(b_n)$

usukidoll

ohhh
$f(a_{n}+b_{n}) = a_{m}+b_{m}=f(a_{n})+f(b_{n})$

Obliv

so when you're proving this, you should show that those 3 quantities are equivalent if you can

gotta go good luck

usukidoll

$f(ab) = f(a)(f(b))$
$f(a_{n}b_{n}) = a_{m}b_{m} = f(a_{n})f(b_{n})$

Darpan Ganatra

Challenge: Prove 1+1 =2

Balarka Sen

12:11 AM

Axioms can't be proved.

Darpan Ganatra

Why?

Balarka Sen

Because it doesn't make sense. Axioms are assumptions.

To prove something, you need axioms aka assumptions. 1 + 1 = 2 falls in one of the basic axioms of arithmetic.

You can't make a theory (eg arithmetic) out of axioms and then try to use that theory to prove the axioms themselves.

Darpan Ganatra

But you can if we're talking about real world problems right? Like if we have something physical we can prove that 1+1 = 2?

I know what you're saying is true, just trying to digest why

Balarka Sen

As soon as you say real world, we're leaving the realm of mathematics and entering the realm of philosophy and metaphysics. Better not get into that.

Darpan Ganatra

Haha fair

Balarka Sen

12:19 AM

In any case, 1 + 1 = 2 is not even an axiom. It's a definition of the symbol "2".

Kevin Driscoll

It works either way. One can define '2' this way, or one can define 2 as the next natural number after 1 and then prove that adding one to one gives the next natural number after 1 or something like that

Balarka Sen

I'm sure there are lots of theories out there which defines arithmetic. I have Peano arithmetic in mind

usukidoll

1:00 AM

writer's block D:

PVAL-inactive

1:10 AM

@MikeMiller Have you looked at Martelli's book at all?

usukidoll

I know there's no ring homomorphisms $f: Q \rightarrow Z_{m}$ but I can't put in words... just examples D:

PVAL-inactive

@usukidoll Then you don't know.

usukidoll

q is the set of rational numbers expressed as p/q

being mapped to a set of integers mod m. I have done this on scratch paper

it's like if we have mod 3 then 1/3 fails if f(1/3) = 0,1 or 2 for the homomorphism definitions

PVAL-inactive

What must $f(m)$ be?

What about f((1/m)m)?

usukidoll

thinking... I know I can't write a proof for a specific mod m in which case the other day it was mod 3 = [0,1,2]

$f((\frac{1}{m}m)$

that's like f(1) if the m's cancel

Balarka Sen

1:45 AM

@TedShifrin Re 2.1.11c: I am not quite sure how that parameterization ever gives me any ruling at all, unless perhaps the rulings $\beta(u)$ in $\alpha(u) + v\beta(u)$ are identified as curves (so eg you can vary the parameterizations $v$-wise)?

Which would make sense, but just doing a sanity-check.

Balarka Sen

2:10 AM

Hrm. I can't connect to the second fundamental form as well as the first.

I guess it's sort of the differential of the Gauss map.

Well, cross the sort of.

Adeek

2:25 AM

hey @BalarkaSen

0celo7

@BalarkaSen Usually there's a minus sign floating around

Balarka Sen

@Adeek Hi

@0celo7 Right, I noticed.

Adeek

@0celo7 @BalarkaSen do you guys know any intuition about the following theorem.

Let X be normed space, Y be closed subspace of X where $Y \neq X$, then for all $\theta >0$ there exists $x \in S_x$ such that $dist(x,Y) > (1 - \theta)$.

0celo7

We proved that in analysis.

Adeek

yeah

I understand this prove

I understand the proof.

0celo7

2:28 AM

Are you showing that normed spaces are not locally compact if infinite dimensional?

Balarka Sen

What's $S_x$?

Adeek

But do you know why intuitively this is ?

@BalarkaSen X is normed space right so we can always define $S_x$ same way we define it for $\mathbb{R}^n$

$S_x = \{ x \in X : ||x|| = 1\}$

0celo7

eww

what weird notation is that

Balarka Sen

I mean, you didn't mention that anywhere.

Adeek

Yeah I should have mentioned it.

0celo7

2:31 AM

What does $x\in S_x$ mean

$x$ is not a part of its unit sphere

Adeek

it means it has norm one.

Balarka Sen

$S_x$ is terrible notation. Unit sphere does not depend on the $x$.

0celo7

I'm very confused by this notation...

Balarka Sen

Did you mean $S_X$?

Adeek

yeah $S_X$

0celo7

2:33 AM

It just means that you can find a point in the unit sphere that's "far away" from $Y$

It's only nontrivial/interesting in the infinite dimensional case

Balarka Sen

@Adeek It sounds straightforward. Understand it for the simplest normed space you can think of.

0celo7

but intuition for such spaces is...nonexistent

Adeek

I understand it for finite dimensional

but infinite dimensional normed space it gets complicated

Balarka Sen

Good luck with that.

Adeek

3:02 AM

Hey @0celo7 so let us say we have a normed space X and closed subspace of it let us say Y.

oh gosh

?

continue

we could then define a natural norm on $X/Y$ defined as follows $|| . || : X /Y \rightarrow \mathbb{R}$. $||\bar{x}|| = inf_{y \in \bar{x}} ||x - y|| = dist(x,Y)$

so what this norm does from intuitive point of view is that it calculates the smallest distance from x to Y.

right ?

Balarka Sen

That's it.

Elements of X/Y are equivalence classes x + Y. The origin is 0 + Y. You're defining "distance from origin" in the most natural way, as dist(x, Y). The only important bit to check is that this is independent of the choice of representative in your class.

But that's not hard.

Adeek

3:10 AM

yeah I see

that is good @BalarkaSen

It is kinda weird how the closed ball is not compact in infinite dimensional

Balarka Sen

Things are too huge

Adeek

yeah

0celo7

@Adeek Do you understand the intuition behind the proof?

Adeek

I am gonna do it as excerise I will prove it over this weekend

@0celo7 don't spoil it please

0celo7

Might be too much for an exercise, but ok

there's a trick

Balarka Sen

3:19 AM

Why am I awake at 9 in the morning?

0celo7

@BalarkaSen because

Adeek

I will try it first and then if I run into trouble I will get a glimpse look into prof proof

0celo7

$D^n/\partial D^n\to S^n$ is blowing my mind

Balarka Sen

meh

0celo7

I don't understand what "One can factor the projection of the disk to the sphere through this space." means

Adeek

3:21 AM

what is it again compactness in metric spaces?

what is the topological condition again ?

0celo7

finite subcovers

Balarka Sen

@0celo7 He's probably referring to the universal theorem of quotient spaces. Which page, which line?

Adeek

no not that one

0celo7

@BalarkaSen Bredon 41.

Adeek

yes this is universal theorem of quotient space

0celo7

3:22 AM

third of the way down the page

Adeek

factor through the space if you have a function that is constant on the equivalence class

0celo7

I don't understand what he means by it near the top of the page either

I don't know what he means by "factor" here is all

Adeek

@0celo7 this is the condition I was looking for sequence in X it has a convergent subsequence.

0celo7

@Adeek complete and totally bounded

every sequence has a convergent subsequence

Balarka Sen

@0celo7 He means the map $D^n \to S^n$ he constructed is a composition $D^n \to D^n/S^{n-1} \to S^n$.

Adeek

3:24 AM

yeah this one

how is that book btw @0celo7 I put it on my to read list

0celo7

what book?

@BalarkaSen ah ok

thanks

@Adeek In general if you want to show something is not compact in a metric space you'll try to disprove sequential compactness.

Balarka Sen

Were you not supposed to read Hatcher, not Bredon?

0celo7

@BalarkaSen Yes I know

Adeek

Bredon @0celo7

0celo7

too hard

Balarka Sen

3:30 AM

I find Hatcher's writing more lucid than anything I have read. Tastes vary though.

I am glancing back and forth through his vector bundles notes when reading Milnor-Stasheff.

0celo7

@BalarkaSen What the heck is $S^\infty$ supposed to be

or $\Bbb RP^\infty$ or $\Bbb CP^\infty$

Balarka Sen

Think of S^n as a subspace of S^(n+1) (the equator). Take the infinite union, give it direct limit topology.

Mike Miller

What could it possibly be?

0celo7

@MikeMiller unicorns, hopefully

@BalarkaSen Is that the same as the cell complex topology described in Hatcher?

Balarka Sen

Yes.

0celo7

3:37 AM

Ok, I understand that, but it seems rather formal

Balarka Sen

I don't know what that is supposed to mean.

0celo7

Never mind then

Mike Miller

It's the unit sphere in an infinite dimension space.

0celo7

@MikeMiller Any infinite dimensional space?

0celo7

3:49 AM

@BalarkaSen Hatcher mentions that the topology on a product CW complex can be finer than the product topology

Balarka Sen

Yes. In fact product topology is the coarsest topology out there.

0celo7

Do product maps of products of CW complexes have the same nice properties of product maps in the product topology?

Balarka Sen

Which nice properties?

0celo7

Specifically:

$f:X_1\to Y_1, g:X_2\to Y_2$ continuous maps of CW complexes

Is $f\times g$ continuous?

Balarka Sen

I mean, since it's continuous in the coarser product topology, of course.

0celo7

3:54 AM

I understand that continuous in a finer topology implies continuous in coarser topologies

But I don't see why it should work the other way around

There are sets open in $Y_1\times Y_2$ as a product CW complex that are not open in the product topology

So it's not immediate that the preimage of such a set is open

Balarka Sen

Ok, I was thinking of the domain, but not the codomain. I think it's still a yes; these kind of things are in the back of Hatcher, in the appendix.

0celo7

But maybe it is in the product CW topology on $X_1\times X_2$

@BalarkaSen I'll check

Balarka Sen

In general I don't think one cares anyway upto homotopy. One can homotope $f$ and $g$ to cellular maps (i.e., one which sends cells to cells), in which case their product becomes cellular and continuous.

0celo7

OK well I was asking for a reason, because of something on page 9 (I think)

one sec while I check the back of Hatcher

Balarka Sen

@0celo7 In real life, the two topologies almost always coincide.

So w/e.

0celo7

4:01 AM

@BalarkaSen I was wondering if the map $f\times 1$ on page 9 is continuous.

Is the suspension of a continuous map continuous?

I might be overthinking this

Is $X\times I$ simply a product or a CW product?

Balarka Sen

It's a CW product.

Note your map is identity on the second component.

0celo7

Ah, but he says on the previous page that it's equivalent because $I$ has finitely many cells.

Balarka Sen

Sure.

That's why I said they coincide almost always in real life

0celo7

Sure.

@BalarkaSen What exactly does he mean by "the quotient map of $f\times 1$"?

Mike Miller

@0celo7 Effectively, yes. Hatcher's is the unit sphere in the xountavle dimensional vector space with the standard inner product.

0celo7

4:05 AM

Is it just the map that appears in the charateristic property of quotient spaes?

Balarka Sen

The one which comes from the universal property, yes.

You look at $X \times I \to Y \times I$, compose with the qt map $Y \times I \to SY$, and since this composition is constant on $X \times 0$ and $X \times 1$, gives you a map $SX \to SY$

@MikeMiller That's mildly interesting though. I'd have believed you if you said Hilbert spaces.

Mike Miller

Huh?

Effectively the same doesn't mean homeomorphic. It means you can do the same things with any of them. They all work.

Balarka Sen

Oh.

Sorry for the misunderstanding. That makes sense.

0celo7

@BalarkaSen I'm being silly. Don't you need it to be constant on all the fibers, not just those two?

Balarka Sen

@0celo7 Er, huh? If it was constant on each $X \times t$, that map wouldn't be very interesting. $SX$ is obtained from quotienting the top and bottom of $X \times I$.

0celo7

4:13 AM

Ok there's two quotient maps hanging around.

I'll work this out tomorrow, I'm going to sleep now.

Cheerio

@BalarkaSen Ok, got it.

Drawing some arrows helped.

Balarka Sen

good to hear

0celo7

Now I really am going to bed.

Adeek

5:24 AM

hey

@0celo7 here ?

Brody

5:38 AM

Greetings from the nether realm! (aka mobile)

Ted Shifrin

@Balarka: It is not parametrized as a ruled surface in the "standard" way.

Adeek

hi @TedShifrin I wanted to ask a question about shauder basis

I was wondering is there a way to determine if whether a space can or cannot admit a shauder basis ?

Brody

no TeX rendering, huh. Switching to desktop then

Mike Miller

No, there's not. If it's a Hilbert space it does. If not you'd better just find one by hand.

Adeek

I see

Mike Miller

6:10 AM

@PVAL I haven't yet but I probably will when I get a chance. The claim that he has a nice proof of the classification of homeomorphisms of surfaces is enticing.

Right now the new paper by Adam and Danny is higher on my priority list.

Brody

6:42 AM

We can take the limit $n\to\infty$ of the perimeter of a regular $n$-gon inscribed in a circle of radius $r$ and see that it is $2\pi r$.

What's the crucial difference when we have a circle of radius $r$ inscribed in a square with perimeter $8r$ and fold the corners of the square in? (to perhaps suggest $\pi=4$) Is the limiting process ill-conceived?

nvm Math.SE has pertinent question and pertinent answers

user228700

7:15 AM

Hi everyone :-)

user228700

I have a quick question regarding geometry. I'm trying to solve a question, in which it is given that a particle traverses $x$ units by making 120° with the positive x-axis (clockwise) to reach some other point. I represented this (very roughly, only to indicate the angle) using the following diagram:

user228700

user228700

So I took the slope of the line to be 60°, rather than -120°(since anticlockwise). Why is this incorrect..?

user228700

I guess what I'm asking is, isn't 60° and -120° basically the same?

Tobias Kildetoft

@Kaumudi It asks for 120 degrees clockwise. Your arrow indicates a direction opposite to this

so the angle is correct, but I would interpret it as going in the other direction

user228700

7:51 AM

@TobiasKildetoft Oh, yes, that's my mistake. But still, the slope is $tan(60°)$, right?

Tobias Kildetoft

yes

Brody

@Kaumudi If the path has/needs a particular orientation, then the instructor might want that the terminal side of the angle determine this.

Brody

8:08 AM

mm sticking my nose in resolved questions/discussions. boundaries must be learned

Lozansky

8:31 AM

Is $\overline{\bar{z}^{n}} = z^{n}$?

SteamyRoot

@Lozansky $\overline{z_1z_2} = \bar{z_1}\bar{z_2}$ and $\bar{\bar{z_1}} = z_1$ for all $z_1,z_2 \in \mathbb{C}$, hence...

Lozansky

Yes :)

user228700

@Brody @TobiasKildetoft :/ Well, in the solutions, they've definitely taken the slope to be tan(-120°).

Tobias Kildetoft

@Kaumudi but that is the same, isn't it?

Brody

@Kaumudi $\tan(-120\deg)=\tan(60\deg)$ so slope shouldn't be an issue. The question is, does the instructor desire the line to have a certain direction (path orientation)?

user228700

8:42 AM

@TobiasKildetoft Uhh, crap, sorry. Was in the middle of doing another problem when I typed that, so I didn't verify what I was typing. Yes, those two are the same, but...what the heck has my book done? :/

user228700

Hang on, let me check real quick (They haven't provided solutions :/)

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$Mathematics$ Mathematics