Mathematics - 2015-05-31 (page 3 of 4)

barto

16:00

@Lembik Sage seems a topic broad enough to have its own site. Also, some questions may be off-topic on SO due to a too prominent mathematical component.

Soham Chowdhury

I have never seen this many people online.

user147690

16:26

@SohamChowdhury I have ;). There was a glitch once that stopped people from getting kicked and we had over 84 people get kicked off at once when it was fixed

user147690

@SohamChowdhury Why are you not asleep young man!

pjs36

To follow up on what barto said, the Sage Proposal needs upvoted questions, the threshhold being 10 upvotes, and it's considered "good" - so if you care, find those questions near the threshold, and game the system a bit!

user147690

@pjs36 There's 4 more over the threshold

user147690

I would do the full gaming effort with some of my other accounts, but I would probably get in trouble ;P

pjs36

Way to go, @AlexClark ! e-highfives

Thiago Balbo

16:35

Anyone would have some thoughts on the sample size N on this question? math.stackexchange.com/questions/1306626/…

Soham Chowdhury

16:49

@AlexClark It's hardly 10:30, man. Why are you not in bed?!

It's almost 3!

user147690

@SohamChowdhury Because I have those two assignments due today

Soham Chowdhury

Are they still due?

user147690

They are due at midday(its monday now)

Soham Chowdhury

Get a nap.

user147690

I can't

user147690

16:51

I haven't started the second one yet

Soham Chowdhury

Damn.

I feel bad for you :(

user147690

I am almost done with algebra

Balarka Sen

Start doing the second one, @AlexClark

user147690

Nah so close to finishing

Soham Chowdhury

@BalarkaSen Sage advice :P

user147690

16:52

Got all of my $\Bbb Z[x]$ done, got my $\Bbb Z[i]$ as a PID, and all my maximals of $\Bbb R[x]$

user147690

Just need my continuous function ring crap

Balarka Sen

oh, btw, @AlexClark, I think I will eat my words.

Soham Chowdhury

@AlexClark: "I got 99 problems but $\Bbb Z[x]$ ain't one"

user147690

@BalarkaSen What do you mean?

Balarka Sen

continuous function ring look like the most interesting ring of the world now :P

user147690

16:53

Oh hahahha really?

user147690

Maybe you'll write me a proof that $M_c =\{f\in R|f(c)=0\}$ is not finitely generated lol

Balarka Sen

yeah, although I've literally never worked with them. I picked up some survey on etale stuff, and they're saying that it's actually interesting.

Soham Chowdhury

@AlexClark You're going to think that up now?

God. College is hard, it seems.

user147690

@SohamChowdhury Unfortunately I am...

Balarka Sen

hello @AlexWertheim

user147690

16:55

@BalarkaSen Hello @BalarkaSen

Soham Chowdhury

empathetic pat on the back

Balarka Sen

ok, two Alexes are getting confusing quickly.

Alex Wertheim

Hello @Balarka. How are you?

user147690

@BalarkaSen Hahahahhaha

Soham Chowdhury

Why don't one of you change usernames in the, uh, public interest?

Balarka Sen

16:56

good idea.

user147690

I better get back to the grind(-ing of my brain into mush)

4

Soham Chowdhury

yeah.

good luck man.

user147690

Bye for now

Balarka Sen

you still have to prepare for the presentation, @AlexClark

:P

Alex Wertheim

Lol. There can only be one

user147690

16:56

@BalarkaSen Oh damnit I forgot about that :\

user147690

More work puhhh

Balarka Sen

sorry for being a jerk and reminding you :P

user147690

:P

user147690

I may not do it actually, I don't think I will be able to

Balarka Sen

@AlexWertheim I'm good. I think I have progressed with my list of problems to think about.

What about you?

Alex Wertheim

16:58

That's very good news, @Balarka. :)

I'm pretty good as well. I read a very beautiful proof yesterday.

1

Q: When are the units of R[x] exactly the units of R?

I (Anton) have edited this question to be the question Pete and Zeb discuss in the first few comments. What conditions on a ring $R$ imply that the units of $R[x]$ are exactly the units of $R$?

Soham Chowdhury

in other words, i've switched over to using a pencil

screw ink skipping.

Alex Wertheim

The proof I'm familiar with is Lam's. The one Anton gives is very elegant though. I wonder if it is his... if not, I would very much like a reference.

Balarka Sen

@AlexWertheim yeah, that thing I was talking about yesterday turned out to be indeed bordism homology. I have proved that it's a homotopy invariant and that it satisfies the long exact sequence using my definition, so I'm happy up until now. I'll try to prove the excision axiom, which looks like a mess.

@AlexWertheim Isn't that a theorem in Atiyah-MacDonald?

theorem/problem/whatever

Alex Wertheim

That's great, @Balarka. Sounds like good progress.

It's a problem, but Atiyah's hint is to prove it the way Lam does.

Balarka Sen

ok, let me see it.

Alex Wertheim

17:03

Which is a mess when generalizing to a ring in multiple variables. Anton's generalizes very easily, with almost zero modifications.

Hello, @MikeMiller.

Balarka Sen

morning, @Mike

Semiclassical

hi chat

Mike Miller

morning

Balarka Sen

wonders why Mike is always so grumpy

user147690

Is Mike always grumpy?

Balarka Sen

17:05

you'll find Ted agreeing with me, at least.

Clarinetist

0

Q: Proving that for a random vector $\mathbf{Y}$, $\text{Cov}(\mathbf{Y})$ is nonnegative definite.

I have already seen What is the proof that covariance matrices are always semi-definite?. Note that I am self-learning this topic. Suppose $\mathbf{Y}$ is a random vector with covariance matrix $\text{Cov}(\mathbf{Y})$. I would like to show that $\text{Cov}(\mathbf{Y})$ is nonnegative definite...

matrices statistics

user147690

@BalarkaSen He's only often grumpy, but he is also often helpful :P(not to say they are exclusive)

Balarka Sen

"being helpful" doesn't imply "being not grumpy"

I agree that he's helpful, though.

user147690

Indeed, I didn't mean to suggest any dichotomy

Semiclassical

@bolbteppa: hope we can continue our discussion about KP/Grassmannian stuff at some point. the learning curve when it comes to stuff like Babelon's book is steep for me

user147690

17:08

Just tired person phrasing

Soham Chowdhury

@Clarinetist You're going real fast :)

@AlexC There is always a relevant xkcd

Clarinetist

@SohamChowdhury Fast, but dang... this stuff is tough to learn on my own. It's gonna require a lot more time than I expected. But I'm so looking forward to it

Soham Chowdhury

Did you see my link? This one?

Karim Mansour

hi guys

user147690

@SohamChowdhury hhahahaha(I always love the scrollover comment)

Balarka Sen

17:10

linear algebra sure is cool.

Soham Chowdhury

@AlexC Another

Clarinetist

@SohamChowdhury I'll check it out later

@SohamChowdhury A former coworker showed me this last night, btw

Ismail Lumanovski - Clarinet solo / live in Moda Sahnesi - Istanbul

►

user147690

@SohamChowdhury Less funny but good still

Karim Mansour

hm

user147690

@KarimMansour Hey

Soham Chowdhury

17:11

One of the best ever, @AlexC. Take it from a connoisseur.

user147690

@SohamChowdhury Too long for me to read until after I sleep haha

Karim Mansour

isn't the $l_1$ norm of the following $x_n = \frac{2}{3^{n - 1}} = 9$

Soham Chowdhury

@AlexClark don't forget.

user147690

@KarimMansour $l_1$ norm i.e. the taxi cab norm?

Balarka Sen

xkcd's insomnia comics were always a classic

Soham Chowdhury

17:13

@Clarinetist The trumpet in the video is modified with an extra piston for quarter-tones. Arabic influences ftw.

Clarinetist

@SohamChowdhury Whoa, didn't know you could do that on trumpet

Karim Mansour

The $l_1 = \Sigma |x_n|$

@AlexClark

I don't know its nick name though

user147690

@KarimMansour Oh I am too tired to help actually sorry haha

Soham Chowdhury

Listen to it as soon as you can, it's beautiful.

@AlexClark Work or sleep. Get off chat. /momvoice

user147690

@SohamChowdhury Working :P

Remember me

17:14

Hi@SohamChowdhury

Paul Plummer

@AlexClark You are up late

Karim Mansour

okay @AlexClark you should sleep !

user147690

@PaulPlummer 2 assignments due in 9 hours

Clarinetist

@KarimMansour $$\sum\limits_{i=1}^{n}\dfrac{2}{3^{i-1}} = 2\left(1 + \dfrac{1}{3} + \cdots\right) = 2\left(\dfrac{1}{1-1/3}\right) = \dfrac{2}{2/3} = 3$$

Remember me

Real work@AlexClark

Balarka Sen

17:15

I dunno if you saw it, @Mike, but I pinged you about something while you were away. Let me re-ask : are we familiar with $\widehat{\mathsf{Gal}(\overline{\Bbb Q}/\Bbb Q)}$ well enough? (this is the profinite completion of the absolute galois group, to clarify the ugly notation)

user147690

Algebra is super close to completion and complex analysis is unstarted @Paul

Karim Mansour

good @Clarinetist

Paul Plummer

@MikeMiller Have you shown any outermorphisms descend to trivial morphisms in the abelianization?

(for the class of groups you are looking at

Mike Miller

@BalarkaSen I've never heard anybody talk about it, so either we know it well enough in the sense that we don't care to know more, or it's completely and totally unassailable. I don't know which.

@PaulPlummer: I haven't tried. Last night I spent most of my time trying to find any actual examples of $\phi$ that would work for my purposes.

(i.e., that I can even use for the sake of the real question I have)

Karim Mansour

@Clarinetist any good music suggestions ?

Balarka Sen

17:21

@MikeMiller i see, ok. it seems to be the correct thing that could get the whole solenoid business work, thus my question.

Mike Miller

ok

Clarinetist

@KarimMansour I'm listening to some of Zedd right now. His stuff is a mixed bag; EDM in its traditional sense is too repetitive for me, but Zedd writes some gems.

Zedd - Daisy (Audio + Lyrics)

►

Paul Plummer

Yah, thought about it a bit, and I have a hard time believing that they always descend to a trivial morphisms. Later I will attempt to construct some counter example, and hopefully it would extend to a large class of $\phi$. Is the presentation of these things normally not given in the "HNN" extention form? Do you have a simple example, you don't know the answer for? @MikeMiller

Mike Miller

@PaulPlummer: Usually, of course, they won't descend to trivial morphisms. I'm only looking for one case where it does. The presentation is always given in HNN form.

Karim Mansour

nice

Mike Miller

17:26

There is a condition on $\phi$ for it to be admissible; e.g. if I could find a $\phi$ such that all autos of the extension descend to trivial things on the abelianization, but it didn't satisfy my condition, it wouldn't suffice. The condition is very difficult to describe and again I don't actually have a single example of one (though I know they exist and there are quite a lot)

Paul Plummer

Do you think any descend?

Clarinetist

@KarimMansour Another one

Zedd - Spectrum (Lyric Video) ft. Matthew Koma

►

Karim Mansour

whitman.edu/Documents/Academics/Mathematics/klipfel.pdf

very nice paper that introduces hilbert space

Mike Miller

I don't know. I kind of think there probably are examples.

David Wheeler

One of the Oblique Strategy cards says: "See repetition as a form of change"

Mike Miller

17:27

@PaulPlummer: In some sense "Most" things should satisfy my condition

Chris's sis

hmmm, no answer so far

9

Q: Evaluating $\int_0^1 \frac{z \log ^2\left(\sqrt{z^2+1}-1\right)}{\sqrt{1-z^2}} \, dz$

What kind of real analysis tools would you employ for this integral? $$\int_0^1 \frac{z \log ^2\left(\sqrt{1+z^2}-1\right)}{\sqrt{1-z^2}} \, dz$$ EDIT: Here is a supplementary question, the cubic log version $$\int_0^1 \frac{z \log ^3\left(\sqrt{1+z^2}-1\right)}{\sqrt{1-z^2}} \, dz$$

calculus real-analysis integration definite-integrals

@Hippalectryon ^^^

Karim Mansour

very nice @Clarinetist

Hippalectryon

@Chris'ssis I've seen that one yesterday, already upvoted :-)

Chris's sis

@Hippalectryon My feeling is that it has a very nice closed form (hard to explain the feelings though).

Clarinetist

@KarimMansour Last one by Zedd

Zedd - Find You - [Lyric Video] ft. Matthew Koma & Miriam Bryant

►

Semiclassical

17:29

hi @chris'ssis. turns out the integral I showed you isn't actually the one I needed to compute :/

Clarinetist

@KarimMansour I've also been into some "progressive bluegrass" lately; interested?

Chris's sis

@Semiclassical Hi. Really? That one worked pretty nice (for some values). :-)

Karim Mansour

yeah

@Clarinetist

pjs36

Progressive, no Roscoe Holcomb?

Semiclassical

yeah. the one i gave you was equivalent to an integrand of the form $\frac{\log(1+e^{a x})}{1+e^{2x}}$. But it should've been $-i \frac{\log(1+e^{i a x})}{1+e^{2x}}$

Alex Wertheim

17:33

@MikeMiller: do you have any opinions on Tammo Tom Dieck's algebraic topology book?

Clarinetist

@KarimMansour

Forgotten

►

Balarka Sen

@AlexWertheim it's a good book, but too category-theoretic for my tastes.

not sure what Mike thinks, though

Semiclassical

which has poles on the positive real line, and so the integral instead should instead be the difference (up to a factor of two) of the integrals along $\mathbb{R}^+ \pm i0^+$

Karim Mansour

@Clarinetist not available

Alex Wertheim

Hmmm. Thanks @Balarka, good to know. Not sure where I stand yet on books that emphasize category-theoretic language, but I'll keep it in mind.

Clarinetist

17:35

@KarimMansour Dang, do you have access to Spotify?

Karim Mansour

no

Balarka Sen

If you like visualization, Hatcher is probably the best thing you want, @AlexWertheim

Semiclassical

so one has to play with the contours / integrand a little bit more to take the contour as along the real line

Clarinetist

@KarimMansour Here's a different track, does this work?

Front Row Boston | Punch Brothers – Familiarity (Live)

►

Karim Mansour

yeah

Balarka Sen

17:37

also, tom Dieck is really advanced. it's better to already have some exposure in basic alg top before picking that up.

Mike Miller

Seems like a fine book. I have some objections to a homotopy-first approach, but if that's the course one wants to teach, it's probably a good choice.

Semiclassical

@Chris'ssis: i can probably get the correct form without too much work, but i'm feeling lazy right now :P

Alex Wertheim

From what I've read of Hatcher (which is, admittedly, not so much), it's good, but it veers a bit on the side of conversational for me. I'm not ashamed to admit I tend to prefer the definition, theorem, proof style of things.

Chris's sis

@Semiclassical contour integration should work nicely.

Semiclassical

yeah

Mike Miller

17:38

Qual syllabus is based on the first three chapters of Hatcher.

@AlexWertheim: It'll be interesting to see what you decide to do smooth manifolds from. Lee is one standard choice and is far more conversational than Hatcher could even try to be.

Alex Wertheim

I'm just doing some grazing right now, though, so it's not terribly pressing. I'll keep in mind that tom Dieck's book is a bit more sophisticated.

@MikeMiller: Reading Lee's topological manifolds right now is driving me crazy. He's so chatty.

Balarka Sen

haha

I tend to prefer chatty books, on the other hand.

Mike Miller

I'm not sure if that's correct. It does things in a different order than Hatcher, but it's still an introductory book.

Semiclassical

in retrospect, i should've noticed that those poles weren't there (physically they correspond to what are known as Matsubara frequencies, which arise in finite temperature calculationsw)

Alex Wertheim

It's a good book, don't get me wrong. I'm enjoying the material, and he writes well, but sometimes I wish that things would be explained in half the space.

Balarka Sen

17:41

For basic fundamental groups-covering spaces stuff, I can recommend you the second part of Munkres's topology book, @AlexWertheim

but admittedly the exercises in there are rather easy

Soham Chowdhury

@Clarinetist That is a Magritte painting!

Mike Miller

@AlexWertheim: I have the same objection. TM should be about half the length it is.

And I rather like his conversational style.

Balarka Sen

looks like I am the only one here who prefers an intuitions-first style

Alex Wertheim

Well, I think I probably won't be hopping books from Lee just yet @Balarka. But I'll keep your recommendation in mind, thank you. :)

Mike Miller

Lee's book works perfectly well as an alternative to Munkres.

Chris's sis

17:44

@Semiclassical Interesting.

Back in 90-120 min (visiting some relatives).

Semiclassical

kk

Balarka Sen

which reminds me that I should get started on multivariable calc.

Silent

@robjohn, how to show that "every rational number can be expressed in the form $a/b$ where $a$ and $b$ are integers, at least one of which is odd", without using prime factorization theorem?

Alex Wertheim

@MikeMiller: agreed. I haven't much enjoyed Munkres from what I've read of it. But I tend to be particular about what books I like (as so many are, I suppose).

At any rate, thanks for your input. It might be better to just use Hatcher when the time comes. But that won't be for a bit, so I'll know better then. (I was also leafing a bit through Bredon.)

Mike Miller

I got a lot out of Bredon after already learning a bunch from Lee and Hatcher.

Alex Wertheim

17:50

Interesting. Do you mean Smooth Lee, or Top Lee?

Mike Miller

Smooth.

Alex Wertheim

Hrm. That's definitely some time away then. Good to know though, thanks.

Mike Miller

I don't really remember how often he takes a derivative or whatever but surely he uses things that are quite closely tied to smooth manifolds.

Alex Wertheim

Often enough that it's important to know, I'm sure.

Mike Miller

Hypothetically it suits as an introduction, in practice I wouldn't be able to use it as one.

robjohn

17:53

@Silent it is the definition of rational numbers that we can write them as $p/q$ for $p,q\in\mathbb{Z}$. The other part follows because it both are even we can divide both by $2$.

Silent

@robjohn, but how does that guarantee that ultimately we will get an odd number?

Alex Wertheim

I was reading it for a little while just to see how far I would get before I hit something too advanced. I think coming back to it after Lee and another algebraic topology text (maybe Hatcher) sounds about right, based on where I ended up.

Mike Miller

What have you been up to lately?

bulbasaur

yo people

Silent

yo

Balarka Sen

17:57

@MikeMiller How much alg top/smooth stuff do you think I need to study, say, some serious homotopy theory (stable stuff, maybe) if I want to?

robjohn

@Silent For an even $n\gt 0$, we have $n/2\lt n$, and there are only finitely many even integers less than $n$. The chain must be finite.

Alex Wertheim

@MikeMiller: Just been pushing on through Lee and A&M mostly, lately. Onto the start of chapter 3 in Lee, and just got to the exercises in chapter 2 of A&M. Just cracked open Lang and Serre a little bit too.

JC574

got a basic Galois question :)

Balarka Sen

(I am not sure if I want to do it -- just asking to make sure)

Mike Miller

I think we spend more time in this chatroom talking about books than doing math. I think the answer to most of these questions is "Just do it and stop if you run into serious trouble". In any case, if you really want to learn more interesting homotopy theory, Hatcher's book still has plenty of interesting things.

Alex Wertheim

17:58

Btw: officially in Weyburn studio for the year. Shame I'll be getting there just as you're leaving. ;)

Mike Miller

And it's worth knowing cohomology beforehand, because there are serious connections between cohomology and homotopy (they're dual!)

Balarka Sen

@MikeMiller hah, fair enough.

Mike Miller

The extra topics in chs 3, 4 have lots of stuff to do.

Semiclassical

@MikeMiller dual to homotopy or to homology?

Balarka Sen

@MikeMiller that's interesting.

JC574

18:00

the structure of intermediate extensions $\mathbb{Q}(\omega,2^{1/3})$. It's the splitting field of $x^3-2$ over $\mathbb{Q}$

Silent

@robjohn, thanks

JC574

I can see we have $\mathbb{Q}(\omega),\mathbb{Q}(2^{1/3})$ etc

Balarka Sen

I am planning to start on cohomology after learning a bit of mult analysis, though.

Mike Miller

@AlexWertheim: Gotcha, gotcha. There's plenty to do there.

JC574

I've seen we have $\mathbb{Q}(\sqrt{\Delta})$

Mike Miller

18:01

Weyburn is nice, but the price was right for me to leave... I'm moving out in a couple weeks.

JC574

but isn't $\Delta = -108 $ or something?

Alex Wertheim

Agreed, @MikeMiller. But I think it's the right place for me this year. I think I'd only feel good looking for greener pastures once I'm older and (hypothetically) wiser. =P

JC574

so $\mathbb{Q}(6i \sqrt{3}) $ ?

Mike Miller

I don't entirely regret being here, anyway. Being close was very nice.

Alex Wertheim

Yeah, there's a lot of good stuff in there. I'll certainly have my hands full for the next while.

JC574

18:03

am I being silly here?

Alex Wertheim

On that note, time for me to get serious and get into the woods. I won't be coming around chat for the next while. See you on the other side, friends.

Mike Miller

See you.

JC574

I'm not sure how $\mathbb{Q}(\sqrt{\Delta})$ is meant to be an intermediate extension here

Mary Star

Hello!! Is someone of you familiar with fluid theory?

Hippalectryon

@MaryStar Fluid mechanics or fluid theory ?

Mary Star

18:09

I meant fluid mechanics... @Hippalectryon Are you familiar with that field?

JC574

@BalarkaSen you're a boss at Galois Theory right? help a bro out?

Hippalectryon

@MaryStar A little

Mary Star

Do you know what a stress tensor is? @Hippalectryon

JC574

or @MikeMiller? can you help me?

sorry this should be easy for me

Mike Miller

I don't know the context and I'm working right now, sorry

JC574

18:11

no problem :)

Mike Miller

You should take your question and consolidate it into one message so that it's easier for people to see and respond to

JC574

yeah fair point

Hippalectryon

@MaryStar Not at all, I've never used those.

Mary Star

Ok...no problem... @Hippalectryon

Hello @robjohn !! Are you familiar with the stress tensor?

robjohn

@MaryStar Don't think I've seen it before.

Mary Star

18:17

Ok... no problem... @robjohn

Soham Chowdhury

@MaryStar try asking in the physics chatrooms.

Mary Star

I asked... But I got no answer yet... @SohamChowdhury

Hippalectryon

@MaryStar The active users of The H Bar usually log in a bit later

Mary Star

Ok.. I see... Thanks for the information!! @Hippalectryon

Andrew Thompson

18:30

Anyone comfortable with jet spaces here?

user147690

I guess Algebra is sufficiently done minus the presentation

Mike Miller

@AndrewThompson: That would be Ted Shifrin you want to ask.

Andrew Thompson

Yes, I supposed so. Its a silly question really; merely a definition.

Mike Miller

I have a book I can check for the notions you're asking about later, but it would be in an hour.

Andrew Thompson

math.stackexchange.com/questions/1306715/… here is the question, at least.

If Ted drops by I'd appreciate it if someone could remind him.

JC574

18:33

no worries it was a really stupid mistake on my part, sorted it out :)

Gato

@Hippalectryon Salut

Hippalectryon

@Gato o/

Gato

@Hippalectryon ça va ?

Hippalectryon

@Gato Les premiers résultats sont le 2 (mardi) :-)

Pour l'instant ça va

Gato

@Hippalectryon Le stress monte ? :p

Hippalectryon

18:39

Oui !

Gato

@Hippalectryon On croise les doigts.

Hippalectryon

C'est ça :D

Andrew Thompson

My question might be rephrased as follows: given a manifold $M$ and submanifolds $N, H$ of the same codimension, what does it mean for $N$ and $H$ to be tangent of order $k$ at $p \in N \cap H$?

anon

19:02

@AndrewThompson Intuitively I would say the graphs $z=f(x,y)$ and $z=g(x,y)$ have contact of order $k$ at $(x_0,y_0)$ if the the Taylor series of $f$ and $g$ around that point agree up to terms of degree $k$, with the obvious extension to $>2$ variables. Presumably the same idea is in play here, but I'm no differential geometer.

(Note that this is a case of two 2-dimensional submanifolds of M=R^3. With general M, maybe chart the nbhd of the point of contact with an open subset of R^3 to employ this idea.)

Mike Miller

@anon: that's the correct intuition, absolutely; jet bundles are here to formalize the notion of taylor series when we're not just working in $\Bbb R^n$

I just don't know precisely what one wants to say about jet bundles to get the desired statement, as I know nothing about jet bundles other than the catchphrases

Andrew Thompson

Yes, I agree with everything being said. My suggestion in the question is almost definitely wrong, we need charts about $p$ satisfying the submanifold-property.

Hm, I might be able to work this out myself now, thanks for putting it into context for me!

pjs36

First it was symplectomorphisms, now it's jet bundles? You guys are having all the fun :(

Andrew Thompson

Well, not jet bundles yet, although I'll probably nag you guys about that in a couple of days :)

Mike Miller

in any case, @AndrewThompson, I checked the reference I promised to and didn't see anything you were looking for, sorry

Andrew Thompson

19:08

Thanks for checking, @MikeMiller. Our departments math.DG-guy is coming home tomorrow, so I'll nag him about it.

Mike Miller

haha, gotcha

Samuel Yusim

19:23

so topological groups are pretty cool

Mike Miller

mhm

PerplexedGuest

Hello all!

Samuel Yusim

it's nice to see these new ideas being applied to something I already know about, it's a good demonstration of their usefulness in my mind

is an infinite product of topological groups a topological group?

Mike Miller

yeah

Samuel Yusim

and how involved is proving that?

Mike Miller

19:31

not at all, you can definitely do it

that makes me realize i have no idea what the coproduct of topological groups is

Samuel Yusim

hey, me neither

but I've also never heard the term coproduct before

Mike Miller

haha

one can take the notion of product and find out that these hundred different notions of product we have: cartesian product of sets, product topology of spaces, products of groups, ... all come from a certain "universal property"

that being "maps in = product of maps in"

like maps to $X \times Y$ are just pairs of maps, one to $X$, one to $Y$

Samuel Yusim

yeah, I've sort of figured that out

abel

@MaryStar, what is your question on stress tensor?

Mike Miller

the coproduct is the other direction: maps out of $X \sqcup Y$ are just pairs of maps, one out of $X$, one out of $Y$

Samuel Yusim

19:35

so it's a disjoint union?

Mike Miller

for spaces, it's the disjoint union; for sets it's the disjoint union; for abelian groups it's direct sum (but not direct product - these disagree wildly when we're doing it over infinitely many things)

Samuel Yusim

or, well, in the context of say, groups, that doesn't work so well

Mike Miller

for groups it's the free product which is annoying and complicated

Samuel Yusim

neat

Tobias Kildetoft

@MikeMiller Is there some general nonsense reason why topological groups should have a coproduct in general?

Mike Miller

19:36

@TobiasKildetoft not that I know of, I'm not much of a category theorist

I guess it could be a really crummy category

David Wheeler

for topological groups, the co-product is a little wonky, see here: math.stackexchange.com/questions/5095/…

Tobias Kildetoft

@MikeMiller It might be one of those cases where one needs to assume hausdorff or something like that

Mike Miller

my topological groups are always hausdorff

Tobias Kildetoft

@DavidWheeler Ahh, neat

Mike Miller

or rather, always have closed points, but these are equivalent for topological groups

Tobias Kildetoft

19:37

@MikeMiller Well, they do tend to become hausdorff very easily

Mike Miller

i guess if you like algebraic groups that's the wrong perspective

Tobias Kildetoft

(mine are not really topological anyway)

Karim Mansour

you know guys I have been reading Aposotol analysis

very nice

Mike Miller

yeah, you've got a different structure

Karim Mansour

I like it more than rudin

Tobias Kildetoft

19:38

@MikeMiller Right, those are "my" groups

@MikeMiller I suppose you want to throw a "compact" in there for niceness

Mike Miller

nah, I definitely don't, and I definitely don't just want Lie groups

I'd like infinite dimensional groups too

Karim Mansour

hmm

can you have a topological group that isn't compact

Mike Miller

$\Bbb R$

Karim Mansour

because in the definitions of a topological group I always seen them as being compact

Thiago Balbo

Hey guys, would anyone have some clue on finding the right sample size N on relative ab-test on this question? math.stackexchange.com/questions/1306626/…

Mike Miller

19:40

then your definition is absolutely impossibly incorrect

Karim Mansour

is it possible to define a invariant measure that is left and right invariant on non-compact topological group ?

Mike Miller

frequently you can only work with compact things, like you need to be able to integrate over the whole thing

Mary Star

I haven't really understood what it is... Could you explain it to me? @abel

Mike Miller

you can define a left-invariant one, and it won't have $\mu(G) = 1$, which is why you can't do much representation theory with it

you can't do averaging arguments

I guess you can still work with locally compact abelian groups just fine but I'm the wrong person to ask

Karim Mansour

I see

abel

19:44

what is your question. do you mean you don't understand the whole thing?

Mary Star

I am reading some notes in fluid mechanics and one part was about stress tensors but I haven't really understood that... Could you give me the general idea, the meaning of stress tensors? @abel

anon

Did half my brain fall out or isn't a translation-invariant function $\Bbb R^n\to\Bbb R$ a constant function? Or else what is the action of ${\rm Diff}(\Bbb R^n)$ on $C^\infty(\Bbb R^n)$ discussed here?

abel

if you look at point in the fluid and isolate a small parcel. what is the force acting on the surface that is orthogonal to a vector $n$. that is given by $\sigma n$ where $\sigma$ is the tensor matrix. the way the components of $\sigma$ are computed depends on fluid properties called the constitutive relations. the diagonal entries are called pressures and of the diagonal entries are called the viscous part of the tensor.

Mike Miller

yeah, oops

in which case yes, I agree it's obviously constant; it's specified by $f(0)$

abel

the whole point of the stress tensor is so that you can compute the surface forces that you will need to apply newtons law of motion.

r9m

19:58

@Chris'ssis do you know how to send solutions/stuff to AMM problems?

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