Mathematics - 2015-11-29 (page 1 of 3)

Balarka Sen

12:00 AM

@PVAL That's nice. Can you refer me to a proof?

I didn't know about that fact.

Mike Miller

@Pedro: Reading, or more accurately, staring at a page.

PVAL

@BalarkaSen It's SL(2,R)/SL(2,Z) whoops. The proof is on page 84 of Milnor's K-theory book.

Mary Star

When an equation has a solution in K(x), does it then always have also a solution in K[x] ?

Balarka Sen

@PVAL Thanks. I am already terrified by the word "K-theory", let me see if I can understand the proof.

Mike Miller

There's no need to worry, since K-theory is not a word.

Balarka Sen

12:09 AM

kaytheory.

There it is.

Mike Miller

In any case, given the trefoil group, that fact should not surprise you.

Balarka Sen

I am unsure how the trefoil group hints at SL_2(R)/SL_2(Z). :/

Mike Miller

The trefoil group is PSL_2(Z). Now compute the fundamental group of that space.

Balarka Sen

The trefoil group is not PSL_2(Z)? It's the braid group on 3 strands.

Mike Miller

You're right, sorry. But it does admit a description in terms of PSL_2(Z).

Balarka Sen

12:17 AM

I mean, PSL_2(Z) is Z_2 * Z_3.

Ok, I agree.

Mike Miller

A Z-extension or something. I've said one too many wrong things to say anything with confidence.

Pedro Tamaroff

@MikeMiller Are you stuck?

I have to start reading more algebraic topology.

Mike Miller

Locally, no. Globally, yes.

Balarka Sen

No, you're right on that one, I think. $B_3 = \langle x, y | x^2 = y^3 \rangle$ admits a surjective morphism onto $\Bbb Z_2 * \Bbb Z_3$. The kernel is just $\langle x^2 y^{-3} \rangle \cong \Bbb Z$, right?

Mike Miller

Fair enough.

Balarka Sen

12:25 AM

@PedroTamaroff How much have you studied so far? Any fact you've encountered as interesting?

Pedro Tamaroff

@BalarkaSen I'm reading Spanier. He's pretty thorough and detailed, and I guess this makes the reading a bit slower. I have to catch up with the swifter (yet less detailed) pace the course I'm taking has. =)

I will scan Hatcher a bit, see what examples and results I can gather.

Balarka Sen

Ah, I see. I've never read Spanier. Is he functorial?

Pedro Tamaroff

Hahaha.

PerplexedGuest

I have a question - anybody know how to solve systems of $n$ linear equations with $n$ variables?

If there's some way I can do this in Mathematica or something I'd love to know. xD

Pedro Tamaroff

Yes, yes.

Balarka Sen

12:29 AM

Uh-oh.

Pedro Tamaroff

@PerplexedGuest There's Gauss elimination. That's standard and easy to do.

PerplexedGuest

Ooh, excellent. My only problem is that the equations constantly change as the size of $n$ increases and I'm not sure how to handle that. xD

Pedro Tamaroff

@BalarkaSen In the first chapter he focuses on H-spaces and H-(co)groups, and uses Eckmann Hilton, which is nice. The Eckmann Hilton argument is something that can be explained to a freshman in an introductory Algebra course. More people should be aware of it.

Balarka Sen

Ah, well. As long as he doesn't do the groupoid van Kampen theorem, I'm fine with it. Some basic general nonsense puts together the stuff in shape.

@PedroTamaroff Yeah, Eckmann-Hilton is nice.

Pedro Tamaroff

For example, the concatenation of paths in $\pi(X)=[S¹,X]$ for any space $X$ comes from the cogroup structure on $S¹$; since $S¹$ is a suspension. This means $[S¹,S¹]$ is abelian, and in fact the concatenation of paths is homotopic to the product of them, coming from the group structure of $S¹$.

Spanier uses that to show the degree map from $\pi(S¹)$ to $\Bbb Z$ is a group (iso)morphism.

Its a very nice argument, I think.

Balarka Sen

12:33 AM

I agree that people should be aware of it, but I don't know of a better motivation. To me, the motivation is that $\pi_1$ preserves products, so it takes group objects to group objects. So once the usual structure on $\pi_1(G)$ matches with the structure coming from the multiplication on $G$, we can conclude it's abelian as group objects in Grp are abgrps.

@PedroTamaroff Ah, interesting.

Mike Miller

Well, the iso part is the nontrivial part.

Balarka Sen

Wait a second, are we seriously not using the covering map R --> S^1 here?

One needs that one way or another to conclude isomorphism, I guess, as Mike pointed out.

Mike Miller

I'm not sure who should be aware of Eckmann Hilton but isn't.

Balarka Sen

Pedro mentioned freshmen in intro algebra courses should be aware of it.

Pedro Tamaroff

@BalarkaSen I didn't say they should be aware of EH.

I said they should be made aware of EH =)

I guess it might be mildly useless for them at first, though.

Balarka Sen

12:40 AM

Fair enough.

Pedro Tamaroff

@BalarkaSen Yes, of course Spanier uses that.

Balarka Sen

Phew. I was scared, because I haven't seen a proof otherwise (well, not literally, there are proofs using homology...)

Mike Miller

It is the one justification for teaching people first learning algebraic topology about groupoids.

Balarka Sen

Can you actually compute the fundamental groupoid of S^1 without serious machinery like Brown's van Kampen theorem?

Mike Miller

I'm not sure that's serious machinery, but that's what I'm talking about. Note that I'm not defending the introduction of such, just stating that it is the only reasonable justification.

Balarka Sen

12:46 AM

hm, ok. Admittedly I haven't studied that theorem, so I should suspend judgement about whether that or the covering space proof is easier :)

And I'm biased.

Mike Miller

I also don't prefer it to the covering space proof. But it exists.

Balarka Sen

re "But it exists": Such is fate. (Higher) category theory will take over the world after all. :(

Mary Star

Does someone of you know what a constant field of a field is?

Mike Miller

Didn't I give you homework?

Balarka Sen

$\pi_0 Diff(S^n) \to \Theta_{n+1}$?

Mike Miller

12:54 AM

I don't remember. Maybe. Who knows.

Balarka Sen

lol.

Mike Miller

But if you think so, then I'll still be spiteful if you hadn't done it. :)

Balarka Sen

1:06 AM

Well, I defined the map, did I not? You take a diffeo $f$ of $S^n$, and form $D^{n+1} \cup_f D^{n+1}$. If $f, g$ are two of them with a diffeotopy $f_t$ between them, then you can just look at $S^n \times [0, 1]$ and cap off by $D^{n+1}$'s at both ends using diffeomorphism $f$ on $S^n \times \{0\}$ and diffeo $g$ on $S^n \times \{1\}$.
The resulting thing is diffeomorphic to $D^{n+1} \cup_f D^{n+1}$ and $D^{n+1} \cup_g D^{n+1}$ both as you can isotope $f$ and $g$ inside the enlarged manifold and then extend to a diffeomorphism by isotopy extension.

It's clearly surjective by h-cobordism theorem. Choose a disk in the exotic sphere, surger it out. The thing remaining is diffeomorphic to $S^n \times [0, 1]$ with a disk glued to $S^n \times 1$ by identity, hence is $D^{n+1}$. That's the same as two disks glued along by a diffeomorphism.

For the injectivity part, I do not know.

Mike Miller

You hadn't explained the first part. Good argument. The word "surger" is not really relevant here. You just mean delete it.

For injectivity, google the pseudoisotopy theorem.

Balarka Sen

Yeah, I remember now. It's an argument borrowed from Hatcher chapter 0. :)

L33ter

hm

Balarka Sen

@MikeMiller OK, thanks. I will google it tomorrow morning.

But I like to say 'surger' :P It sounds so... pharmaceutically cool.

Ah, puns at 6 in the morning is bound to get sad. I'm off to bed! Thanks again.

L33ter

conventionally is $S_1$ unit circle ?

Pedro Tamaroff

1:20 AM

@L33ter You want $S¹$.

Mike Miller

@BalarkaSen: You can say surger if you're actually doing surgery.

L33ter

alright thank you @PedroTamaroff

PerplexedGuest

2:32 AM

I love math but sometimes it makes me wanna tear my hair out. xD

PerplexedGuest

3:02 AM

I've ended up at point in my life where I have to find $a_{n}$ such that for $n \geq 1$:
$$\sum_{i={n+1 \choose 2} +1}^{{n+2 \choose 2}} a_{i} = a_{{n+1 \choose 2}}$$

L33ter

3:25 AM

hey

I'm mostly just an idiot

4:38 AM

Hi @L33ter

Where can I learn a little category theory. I want to understand what an initial object of a category is

Looking at 'an introduction to category theory' by Harold Simmons in a second

Question 3: In category theory, if we have a bunch of objects, where there is an object that 'starts the morphisms', do we call that object the initial object? I.e in the category of rings, I am told the initial object is $\Bbb Z$, am I to take it we have a bunch of chains coming out of $\Bbb Z$?

idonutunderstand

In the theory of abstract algebra, and there other kinds of "structure-preserving" maps besides homomorphisms and isomorphisms?

I'm mostly just an idiot

In abstract algebra, not really I believe. In analysis definitely. I mean really the structure preserving always comes from being a homomorphism from my perspective.

We give it bijectivity and it is an isomorphism, we just have a homomorphism to the same object and its an endomorphism, and if its a homomorphism to itself that is bijective it is an automorphism.

idonutunderstand

What's an endomorphism?

nvm

Mike Miller

an initial object is an object $x$ such that for every $y$ there is only one morphism $x \to y$. instead of being told that for Ring it's $\Bbb Z$, prove it

I'm mostly just an idiot

@MikeMiller So there is a unique homomorphism to all rings, with domain $\Bbb Z$

I'll give it a shot, thanks

Mike Miller

4:50 AM

note that for this to be true rings must be defined to have a multiplicative unit

@idonutunderstand "structure preserving" pretty much just means homomorphism

I think there have been well-received questions on MSE about this notion

idonutunderstand

@MikeMiller @I'mmostlyjustanidiot aight

Mike Miller

eg here math.stackexchange.com/questions/1158323/…

Secret

https://www.facebook.com/BusinessInsider.Video/videos/719673064796775/

What is the underlying mathematical principle that allow match stick problems to be solved with so few moves. Is it because some kind of symmetry was exploited, if yes what's the umbrella term for the symmetry it exploits?

For example, we know the solution to this puzzle is as follows. and we know it is the solution because it fulfill the requirement stated in the question.

But what underlying princples allow the solution to exist?

I'm mostly just an idiot

Well I mean, we have flipped that donkey over $x=y$

Question 4: Is it to the authors discretion whether category $\rm{Rng}$ has unit or not?

Secret

Given an arbitrary matchstick problem of the form ("Initial configuration", move n matches so that "condition for final configuration") is there a more algebraic approach in solving it?

Mike Miller

5:06 AM

Authors should always clarify what they mean by the category of rings, but if no clarification is given, assume they have unit.

I'm mostly just an idiot

@Secret That depends on many things, it will become geometric and combinatoric if you have 'condition for final configuration' involving angles. If it just involves symmetries, perhaps you can think about the dihedral group

@MikeMiller Thanks

Secret

I see

Mike Miller

Most rings in nature are unital anyway.

L33ter

I don't understand something here

is the continuous map k here defined as qh ?

Mike Miller

No. There is a unique continuous map $S^1 \to S^1$ such that $kq=qh$. $k$ is defined to be that continuous map.

L33ter

5:15 AM

how do we know such a map exist?

Mike Miller

Great care. See if you can figure out why it's true on the level of sets first, then think about why it's continuous. (It's what he says: because $q$ is a quotient map.)

L33ter

5:30 AM

hm

idonutunderstand

5:44 AM

WTF is lambda calculus

some kind of formal system?

whose (elements?) can be evaluated

Mike Miller

Probably. What's it matter?

idonutunderstand

It was mentioned in a category theory lecture I saw on youtube.

Mike Miller

Fair 'nuff.

idonutunderstand

As an alternative to category theory

Or something like that

Or an alternate perspective on something fundamental

I don't know

@MikeMiller I assume you thoroughly understand it?

Mike Miller

Nope!

I know absolutely nothing about it but the name. :D

@Huy: Everything get settled?

idonutunderstand

5:54 AM

@MikeMiller do you study theory of computation?

Mike Miller

I don't know what that means, so probably not.

idonutunderstand

srsly????

turing machines?

automata?

Mike Miller

Sure, I've heard those words.

idonutunderstand

I gather these subjects aren't important in gauge theory and low-dimensional topology :D

Mike Miller

That seems like an accurate assessment, yes.

Huy

6:47 AM

@MikeMiller: Just woke up. Been reading a bit in Rosenberg (I think that's the standard textbook for it), will continue later today.

I have yet to find some software to manage/organize PDFs cross-platform.

Papers is amazing on OSX/iOS but the Windows version is really bad.

Mike Miller

Morning.

Huy

Evening.

Mike Miller

Once you start thinking about the connection Laplacian in much detail you're beyond me, so I hope the book helps.

Huy

Don't worry, I'll come back with geometric topology questions in a month or two.

Mike Miller

:D

iwriteonbananas

9:44 AM

Morning.

Balarka Sen

10:01 AM

hi @iwriteonbananas

iwriteonbananas

@BalarkaSen Long time. What's up?

Balarka Sen

Long time indeed. I am getting some schoolwork out of the way

iwriteonbananas

Same here.

Balarka Sen

Went to the math-physics conference. Learnt some good stuff.

iwriteonbananas

Nice. Anything particularly cool?

Balarka Sen

10:07 AM

Well, not in the realm of physics.

iwriteonbananas

Okay

Balarka Sen

@iwriteonbananas $M$ be a $n$-manifold. A Morse function $f : M \to \Bbb R$ is a function such that for every critical point $a$ of $f$, the Hessian $Hf$ is nondegenerate.

Fact : if $M$ admits a Morse function which has 2 critical points, $M$ is homotopy equivalent to $S^n$.

iwriteonbananas

What do you mean by nondegenerate here?

Balarka Sen

Just that the matrix $[Hf]$ is nonsignular at $a$.

iwriteonbananas

ok

@BalarkaSen Wow, that's a cool fact.

Balarka Sen

10:12 AM

:D

I'll be right back.

iwriteonbananas

sure

Balarka Sen

10:43 AM

back.

@iwriteonbananas What are you learning right now?

Amr

heeey

do all proofs of construction of measure on infinite products of measure spaces require some topological restrictions on the underlying measure spaces ??

this seems wired to me

iwriteonbananas

@BalarkaSen Boring analysis stuff. I'll be done in 1-2 hours and then I'm going to carefully review Hatcher's proof (and our discussion) of the ring structure on $\Bbb RP^n$.

Balarka Sen

Cool.

iwriteonbananas

I think we're going to do that in tomorrows lecture. We're finally not doing abstract stuff anymore, but a bunch of examples.

Lucas

Is anyone familiar with this notation concerning field extensions: "Let F -> K be a field extension"

Balarka Sen

10:50 AM

@iwriteonbananas Thank god.

iwriteonbananas

lol yes.

Balarka Sen

Funny: in the math-physics lectures, a person was introducing the projective spaces and said that there are no notion of parallels here because any two hyperplanes intersect in this world. I said in an undertone "you mean cup square is nontrivial", and (oddly) got approving looks from the mathematical people around my chair, which is the opposite of what I expected.

Maybe they were too bored by the physics lectures.

iwriteonbananas

@BalarkaSen haha, bold move.

physics lectures suck. i've been to a couple.

Balarka Sen

well, in my defense, I was bored!

iwriteonbananas

understandably.

Balarka Sen

11:00 AM

there was a good on on K-theory, but I didn't understand much because most of the time they were talking about hamiltonians rather than K-theory

there was some definite appearances of Bott periodicity in a different form.

but that's it.

iwriteonbananas

K-theory sounds interesting. by the way, i'm probably going to do my undergrad thesis on either bordisms or cohomology operations.

Balarka Sen

thumbs up from me. I don't know anything about cohomology operations other than that they are in bijection with $[K(G, n), K(G, m)]$.

iwriteonbananas

@BalarkaSen that's about all i know about 'em. but it's gonna be a while until i start working on it anyway. prof just suggested those two topics(and l^2-invariants)

Balarka Sen

oh no l^2

iwriteonbananas

but l^2-invariants dont seem as fun as the other two

Balarka Sen

11:04 AM

damn l^2.

(i am joking)

lol

I have to study the Thom-Pontryagin construction at some point of time.

iwriteonbananas

nah, i think they're somewhat cool. but it takes a lot of preliminary analysis work to get a grasp of them.

Balarka Sen

yeah, I had the impression that they are hard.

iwriteonbananas

@BalarkaSen oh, me too. but not high on my list of priorities.

Balarka Sen

what's on your list?

iwriteonbananas

@BalarkaSen the fact that Lück's book is really the only resource for studying them doesn't help of course.

@BalarkaSen i want to master everything done in my alg top, diffgeo classes, and functional analysis classes. and i want to pass all the other boring subjects i need to do (numerical analysis etc).

oh, and get a flavour of l^2 invariants.

Balarka Sen

11:10 AM

i have an idea. why don't you do your undegrad thesis on cobordism hypothesis and TQFTs? :P

iwriteonbananas

once that's all done i will have cleared some time to study lots of fun stuff.

Balarka Sen

@iwriteonbananas right

iwriteonbananas

@BalarkaSen lol, yeah right.

Balarka Sen

i'm gonna delete that, bad pun on good people.

iwriteonbananas

loll

Balarka Sen

11:13 AM

Lurie's works are practical, down-to-earth application of abstract nonsense.

iwriteonbananas

i like my pun but i'll delete it too

Balarka Sen

ok, so I want to study Milnor's paper on exotic spheres at some point of time.

Mary Star

Hello @anon @DanielFischer !! Could you take a look at my question math.stackexchange.com/questions/1548963/… and tell me why we take this y_p ?

Balarka Sen

but lacking in differential topology right now. gotta study char classes later on too. so I guess it'll take some time.

iwriteonbananas

@BalarkaSen really, it's applied mathematics.

@BalarkaSen what are some of the requisites for that paper?

Balarka Sen

11:16 AM

@iwriteonbananas algebraic topology, characteristic classes, differential topology.

iwriteonbananas

@BalarkaSen those are also on my list of fun stuff to study later.

ok

Balarka Sen

And differential topology means Morse theory

iwriteonbananas

just googled that paper. it has like 7 pages.

Balarka Sen

Milnor is famous for being concise and his lucid exposition.

iwriteonbananas

right

have you used one of his books?

Balarka Sen

11:19 AM

no, but probably will study Morse theory from his book later on. for now, I need to get the implicit + inverse ft out of my way

I did the theory and some exercises of Guillemin-Pollack chapter 1 on the math-physics seminar under guidance from prof. had to take inverse FT for granted.

iwriteonbananas

how dare you.

Balarka Sen

you know the guilty pang when you prove something by quoting other theorems you don't know how to prove? that happened a lot.

@iwriteonbananas i know :( thus, i am getting back to mult. calc.

user174558

I can't find Ted's book on Genesis.

Balarka Sen

it's not available online

user174558

Because nobody uploaded a copy.

user174558

11:23 AM

I am using my super big desktop now.

iwriteonbananas

his book on genesis? whut?

Balarka Sen

@JasperLoy no sane person would do that.

user174558

@BalarkaSen Then there are many insane people online.

Balarka Sen

@iwriteonbananas I am getting confused by all the things I have to learn. Maybe I should remove everything except multicalc from my list.

user174558

@BalarkaSen Genesis has helped me so much, when I don't have access to a math library.

user174558

11:28 AM

@BalarkaSen You should go through the undergrad course yourself. For a guide, look at the Cambridge schedules.

Balarka Sen

what undegrad course?

user174558

@BalarkaSen The undergrad math course, of course.

Balarka Sen

nah.

user174558

Up to you. You are not Sayan, so you should know what you are doing.

iwriteonbananas

@BalarkaSen just study one thing and forget about the rest. when you're done, study another thing and forget about the rest.

i get overwhelmed when i look at all the stuff i need to know a few months from now.

user174558

11:29 AM

@iwriteonbananas Or is it forget the rest? LOL.

Balarka Sen

@JasperLoy I know what I am doing :) I am studying multivariable calculus.

iwriteonbananas

@JasperLoy what's this genesis book about?

Balarka Sen

I know what I need to learn. Namely, real analysis. I don't think I should go plan for 3 weeks and then do nothing at all.

user174558

@iwriteonbananas Genesis Library, not book.

user174558

@BalarkaSen I will ignore those people who ignore me from now.

iwriteonbananas

11:31 AM

@JasperLoy and what's the genesis library?

user174558

@iwriteonbananas Have you never heard of it?

Balarka Sen

As you wish. Maybe the people you think are ignoring you aren't ignoring you at all.

@iwriteonbananas good idea.

iwriteonbananas

@JasperLoy nope.

user174558

There are a few Ramanujans in this chat. Chris and Ethan.

user174558

@iwriteonbananas libgen.io

iwriteonbananas

11:33 AM

@JasperLoy i see

user174558

@iwriteonbananas 90 per cent of math books can be found there.

Balarka Sen

@iwriteonbananas go download Lurie's higher algebra from libgen and start reading it.

iwriteonbananas

good to know i suppose.

user174558

@BalarkaSen Is there such a book?

Balarka Sen

yeah.

user174558

11:34 AM

I never heard of it, LOL.

Balarka Sen

bible of higher topos theorists, or something like that

user174558

Not interested in topos.

iwriteonbananas

@BalarkaSen haha. of course.

im interested in tapas.

user174558

@iwriteonbananas I am interested in girls.

Balarka Sen

@JasperLoy But quantum field theory is all about $(\infty, n)$-topoi!

as Urs Schreieber would say

iwriteonbananas

11:36 AM

lol topoi

user174558

@BalarkaSen Also not interested in QFT.

iwriteonbananas

@JasperLoy go outside and talk to some.

user174558

@iwriteonbananas And give them a banana.

user174558

No need to be paranoid, LOL.

user174558

It is only a banana.

user174558

11:38 AM

Fruits are healthful.

user174558

Didn't your mother tell you to eat more bananas?

Ramchandra Apte

but then you'll go bananas

Balarka Sen

@iwriteonbananas Which complex manifolds appear as complex algebraic varities? What groups appear as fundamental group of complex algebraic varities? What's the connection between topology and algebra of a complex algebraic variety?

All open problems.

Hard, open problems.

iwriteonbananas

@BalarkaSen not that i know what a complex algebraic variety is, but it's cool that such seemingly simple questions are open problems.

Balarka Sen

a complex algebraic variety is just the zero set of a bunch of polynomials over $\Bbb C^n$, or over $\Bbb {CP}^n$ (homogeneous polynomials in that case)

user174558

11:42 AM

Well, there are many open problems, the question is whether they are considered interesting and important or not.

iwriteonbananas

from the looks of it, Lurie denotes the category of functors C-->D by Fun(C,D)

Balarka Sen

this one is interesting, certainly.

iwriteonbananas

@BalarkaSen ohh, yeah

Balarka Sen

wait, you really downloaded HA?

are you completely crazy or what??

:P

user174558

Is it a very hard book?

user174558

11:43 AM

@BalarkaSen What is a comathematician?

iwriteonbananas

LOL and he denotes something called the associate operad by Ass.

Balarka Sen

$\infty$-topoi are tough stuff

iwriteonbananas

@BalarkaSen i googled it. the first result is the book as pdf. :P

user174558

@iwriteonbananas An Ass is Fun.

Balarka Sen

@iwriteonbananas that's standard notation for a lot of things

@JasperLoy dual of a mathematician

iwriteonbananas

11:46 AM

@BalarkaSen fair enough, i was aware of only one meaning.

Balarka Sen

en.wikipedia.org/wiki/Associated_prime.

iwriteonbananas

mhm.

user174558

I have just installed Ubuntu Mate 15.10 and TeXLive 2015 on both computers.

user174558

@BalarkaSen OIC, LOL.

Balarka Sen

a cocomathematician is a mathematician if and only if the mathematician is finite dimensional.

user174558

11:51 AM

Jacobson's Lie Algebras actually contains a proof of Ado's theorem, very good.

Balarka Sen

you should start reading it without further Ado then.

I'm mostly just an idiot

@JasperLoy Who is Sayan?

user174558

@I'mmostlyjustanidiot Someone on this site. The rest is secret.

I'm mostly just an idiot

math.stackexchange.com/users/274536/sayantan-santra?

user174558

@I'mmostlyjustanidiot Nope, he changed his name.

Balarka Sen

11:53 AM

people star way too many stuff

iwriteonbananas

@BalarkaSen hahah

user174558

@BalarkaSen Where people means me, LOL.

I'm mostly just an idiot

Nov 26 at 0:38, by Jasper Loy

@morphic Remember Me

Balarka Sen

I think my puns are finally improving.

I'm mostly just an idiot

math.stackexchange.com/users/210387/remember-me

He is deleted I think

Balarka Sen

11:55 AM

Come on, please stop this starring.

2

user174558

Nobody is starring bad things.

Balarka Sen

I mean, it's distracting and annoying, starring every single message of mine.

2

Paradox 101

If we need to solve the Legendre equation about points $x=1$ and $x=-1$ should we assume that for the power series solution $x_o=1$ and $x_o=-1$? Can anyone please confirm this?

user174558

@I'mmostlyjustanidiot I did not know that, thanks for the info.

user174558

@Paradox101 Are you the user teadawg?

Paradox 101

12:04 PM

@JasperLoy teadawg?

Mats Granvik

@JasperLoy yiddishwit.com/List.html

evinda

Hi @DanielFischer !!!
Could I ask you something ? We have $||A||:= \inf \{ M: ||Ax||_2 \leq M ||x||_2 \forall x \in \mathbb{R}^n\}$ and we want to show that $||A||= \sup \left \{ \frac{||Ax||_2}{||x||_2}: x \in \mathbb{R}^n \setminus{ \{ 0 \}}\right \}$.

If we consider the definition $||A||:= \inf \{ M: ||Ax||_2 \leq M ||x||_2 \forall x \in \mathbb{R}^n\}$ we get that $\frac{||Ax||_2}{||x||_2} \leq ||A|| \Rightarrow \sup_{x \neq 0} \frac{||Ax||_2}{||x||_2} \leq ||A||$.

Then we let $M= \sup_{x \neq 0} \frac{||Ax||_2}{||x||_2}$ and then we get $||Ax||_2 \leq M ||x||_2 \forall x$.

I'm mostly just an idiot

1:05 PM

@evinda You can use $\|A\|$ rather than $||A||$ , it looks nicer: $$\|A\| \text{ vs } ||A||\text{.}$$

Daniel Fischer

@evinda Let $b(A) = \{ M : (\forall x)(\lVert Ax\rVert_2 \leqslant M\lVert x\rVert_2)\}$. Then $\lVert A\rVert := \inf b(A)$. Further, define $S := \sup \left\{ \frac{\lVert Ax\rVert_2}{\lVert x\rVert_2} : x \neq 0\right\}$. Then for all $M \in b(A)$ and all $x$ we have $\lVert Ax\rVert_2 \leqslant M\lVert x\rVert_2$, and for $x \neq 0$ we can divide to get $\dfrac{\lVert Ax\rVert_2}{\lVert x\rVert_2} \leqslant M$.

Since that holds for all $M \in b(A)$, we get - for every fixed $x\neq 0$ - $\dfrac{\lVert Ax\rVert_2}{\lVert x\rVert_2} \leqslant \lVert A\rVert = \inf b(A)$. Now we can take the supremum of the left hand side as $x$ ranges over $\mathbb{R}^n\setminus \{0\}$ and get $S \leqslant \lVert A\rVert$.

On the other hand, for $x \neq 0$ we have $\frac{\lVert Ax\rVert_2}{\lVert x\rVert_2} \leqslant S$, and hence $\lVert Ax\rVert_2 \leqslant S\lVert x\rVert_2$. That last inequality also holds for $x = 0$, and thus $S \in b(A)$. But $S\in b(A)$ clearly implies $S \geqslant \inf b(A) = \lVert A\rVert$.

I'm mostly just an idiot

1:20 PM

A $K$-homomorphism $\sigma : L\to M$, that also is an isomorphism, is called a $K$-isomorphism. A $K$-isomorphism $\sigma:L\to L$ is called a $K$-automorphism.

Question 5: So a $K$-isomorphism $\sigma$ just requires that $\sigma|_K$ is a bijection? and secondly, in the case that $K$ is the prime subfield, $\Bbb Q$ or $\Bbb F_p$ depending on characteristic, $K$-isomorphisms and $K$-automorphisms are the same thing?

evinda

2:05 PM

@DanielFischer Great!!! I got it... I want to show that the following holds:

$\|A\|= \sup \left \{ \frac{\|Ax\|_2}{\|x\|_2}: x \in \mathbb{R}^n \setminus{\{0\}} \right\}\\ = \sup \{ \|Ax\|_2:\|x\|_2 \leq 1 \} \\ = \sup \{ \|Ax\|_2: \|x\|_2=1\}$

In order to prove that $\sup \{ \|Ax\|_2: \|x\|_2 \leq 1\}= \sup \{ \|Ax\|_2: \|x\|_2=1\}$ I have thought the following:

$\{ \|Ax\|_2: \|x\|_2=1\} \subseteq \{ \|Ax\|_2: \|x\|_2 \leq 1\} \\ \Rightarrow \sup \{ \|Ax\|_2: \|x\|_2=1\} \leq \sup \{ \|Ax\|_2: \|x\|_2 \leq 1\} $

Daniel Fischer

@evinda You write $x = t\cdot x_0$, where $\lVert x_0\rVert_2 = 1$, and $t = \lVert x\rVert_2$.

Balarka Sen

2:30 PM

@I'mmostlyjustanidiot Uh, a $K$-isomorphism by definition fixes elements of $K$ pointwise. So $\sigma|_K$ is the identity map.

evinda

We do this in order to prove that $\{ \|Ax\|_2: \|x\|_2 \leq 1\} \leq \sup \{ \|Ax\|_2: \|x\|_2=1\}$ ? @DanielFischer

Daniel Fischer

@evinda We do it to show that for every $x$ with $\lVert x\rVert_2 \leqslant 1$ we have $\lVert Ax\rVert_2 \leqslant \sup \{ \lVert Ax\rVert_2 : \lVert x\rVert_2 = 1\}$. Then, from this inequality for all considered $x$, we deduce the inequality between the suprema.

evinda

@DanielFischer Do we set $\lVert x_0\rVert_2 = 1$, and $t = \lVert x\rVert_2$ or do we get it from the equality $x = t\cdot x_0$ ?

Daniel Fischer

2:48 PM

Doesn't matter.

evinda

@DanielFischer So we just set it? Then we have $||Ax||_2=||Atx_0||_2$.

Can we say that the latter is $\leq ||A|| ||t|| ||x_0||$ ?

Krijn

Hi @BalarkaSen

I've got my project assigned: to define the cohomology groups of a topological space and to prove the universal coefficients theorem that relates homology and cohomology

It's a nice subject, I think

Daniel Fischer

@evinda What you need is that you can write every $x$ with $\lVert x\rVert_2 \leqslant 1$ as $t\cdot x_0$ with $\lVert x_0\rVert_2 = 1$ and $t \geqslant 0$. And you can see that that determines $t$ as $\lVert x\rVert_2$, and unless $x = 0$, that determines $x_0$ uniquely, as $\frac{1}{\lVert x\rVert_2}\cdot x$. And then you use what you know about scalar multiplication, linear maps, and norms.

evinda

So t is a real postive number, right? @DanielFischer
If x=0 then couldn't we also pick $x_0=0$ ?

Daniel Fischer

3:07 PM

@evinda $t$ is non-negative. We can't pick $x_0 = 0$, we require $\lVert x_0\rVert_2 = 1$, since what we're given is $\sup \{ \lVert Ax\rVert_2 : \lVert x\rVert_2 = 1\}$. Although it doesn't matter much since $A0 = 0$. But you'd need to carry an extra case around.

evinda

3:23 PM

@DanielFischer A ok... So is it right as follows?

Let $x \neq 0$ with $||x||_2 \leq 1$. Then there is a $t \geq 0$ such that $x=t x_0$, where $||x_0||_2=1$ and $t=||x||_2$.
Then we have $||Ax||_2=||Atx_0||_2=t ||Ax_0||_2 \leq t \sup \{ ||Ax||_2: ||x||_2=1\}=||x||_2 \sup \{ ||Ax||_2: ||x||_2=1\} \leq \sup \{ ||Ax||_2: ||x||_2=1\}$.

Thus, for $x \neq 0$ we get $\sup \{ ||Ax||_2: ||x||_2 \leq 1\} \leq \sup \{ ||Ax||_2: ||x||_2=1\}$.

For $x=0$ we have $x=t x_0$ with $t=0$ and $x_0=\frac{x}{||x||_2}$.

Daniel Fischer

@evinda The first part is okay. But that also works for $x = 0$, there's no need to make the case distinction. However, when you make the distinction, just directly use $\lVert A0\rVert_2 = \lVert 0\rVert_2 = 0 \leqslant \sup \{\dotsc\}$. What you wrote for the $x = 0$ case breaks down at $\frac{x}{\lVert x\rVert_2}$. That doesn't make sense for $x = 0$. For $x = 0$, you can take an arbitrary $x_0$ with $\lVert x_0\rVert_2 = 1$ and write $0 = 0\cdot x_0$.

evinda

Ah I see... so we just use the first part with the only difference that we say at the beginning let $x$ with $||x||_2 \leq 1$. Right? @DanielFischer

Daniel Fischer

Yes.

Mike Miller

3:53 PM

Morning.

evinda

4:06 PM

@DanielFischer So now it remains to show that $\sup \{ ||Ax||_2: ||x||_2=1\}= \sup \left\{ \frac{||Ax||_2}{||x||_2}: x \in \mathbb{R}^n \setminus{\{0\}}\right\}$.

It holds that $\{ \|Ax\|_2: \|x\|_2=1\} \subseteq \left \{ \frac{\|Ax\|_2}{\|x\|_2}: x \in \mathbb{R}^n \setminus{\{0\}} \right\} $, right?

Thus $\sup \{ \|Ax\|_2: \|x\|_2=1\} \leq \sup \left \{ \frac{\|Ax\|_2}{\|x\|_2}: x \in \mathbb{R}^n \setminus{\{0\}} \right\}$.

For the other direction does it hold that $\frac{||Ax||_2}{||x||_2}=||A\frac{x}{||x||_2}||$ ?

Daniel Fischer

@evinda Well, $\frac{1}{\lVert x\rVert_2}$ is a positive real number, so tell me, using the properties of norms and linear maps, are the two equal?

evinda

@DanielFischer Yes, they are equal. But how can we use it? Or can't we?

Daniel Fischer

@evinda What's the norm of $\frac{1}{\lVert x\rVert_2}\cdot x$?

evinda

It is 1. But how can we use this? @DanielFischer

Daniel Fischer

4:22 PM

Look at what you want to show.

evinda

@DanielFischer Do we say the following?

$\frac{||Ax||_2}{||x||_2}=||A \frac{x}{||x||_2}||_2 \leq \sup \{ ||Ax||: ||x||_2=1\}$

This holds for all $x \in \mathbb{R}^n \setminus{ \{ 0 \} }$, thus: $\sup \{ \frac{||Ax||_2}{||x||_2}: x \in \mathbb{R}^n \setminus{\{0\}}\} \leq \sup \{ ||Ax||_2: ||x||_2=1\}$.

Daniel Fischer

Yes.

evinda

Nice.. so now we have shown that: $\|A\|= \sup \left \{ \frac{\|Ax\|_2}{\|x\|_2}: x \in \mathbb{R}^n \setminus{\{0\}} \right\}\\ = \sup \{ \|Ax\|_2:\|x\|_2 \leq 1 \} \\ = \sup \{ \|Ax\|_2: \|x\|_2=1\}$, right? @DanielFischer

Thank you very much!!!! @DanielFischer