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17:00
@EricGregor Good for you. : )
Next crisis to solve: didn't do proper shopping. Now stranded without proper food to cook because shops are closed.
@MattN you're probably right. I took the fact that Asaf never returned emails a bit too personally.
@robjohn You shouldn't take that too personal. He's quite a peculiar guy with things not completely in order :-).
Asaf is a prickly fellow
@robjohn I can understand that. OTOH, you're so old that you should know not to. And yes, I'm always right. : ) Unless I do maths.
@MattN Everything is math hence you are never right.
17:03
Aha. : )
@robjohn Sorry I had to leave suddenly
Isn't it true that: 1+floor=ceiling?
@RajeshD no problem. I just pinged you and saw your gravatar fall.
@tb yes
Ignored, sigh
@someone Please tell me if I am right: 1+floor=ceiling?
@KannappanSampath what if you start with an integer?
17:10
@KannappanSampath no; what tb said
OK. I did not think about integers at all! Thank you.
Need. To. Take. Rest. And. Come. Back.
@KannappanSampath remember the last part :-)
@tb can i bounce my proof off of you?
17:29
@EricGregor sure, sorry, you've got bad timing today :)
A submersion $\pi:M\to N$ is a smooth map of constant rank $k$, so we can apply the Rank Theorem for manifolds. At each point $p\in M$ there exist smooth coordinates $(x^1,\dots,x^m)$ centered at $p$ and $(v^1,\dots,v^n)$ centered at $\pi(p)$ such that $\pi$ has coordinate representation $$\pi(x^1,\dots,x^n)=(x^1,\dots,x^k,0,\dots,0).$$ Let $X=X^i \partial/\partial y^i$ be a vector field on $N$, and let $Y=X^i \partial/\partial x^i$ in $M$ be the lift under these smooth coordinates. This is a locally a smooth lifting of $X$ onto $Y$.
@tb no worries
This looks good up to the last sentence: you think that it should be a smooth vector field? What prevents you from verifying it?
I think it should be smooth because the $Y_\alpha$ are smooth and so are the partition of unity functions. Then I guess to see it's a lift we just have to show that the map under $\pi$ maps back to the vector field $X$
Can there be a function $f : \mathbb{R}^2 \to \mathbb{R}$ which is continuous everywhere except at one point $p$ where it jumps (as in a jump discontinuity) by different amounts along different (directions) unit tangent vectors at $p$ ?
$d\pi(Z)=d\pi(\sum_\alpha \phi_\alpha Y_\alpha)=\sum_\alpha \phi_\alpha d\pi(Y_\alpha)=\sum_\alpha \phi_\alpha X=X$?
does this work, @tb?
like the absolute value function, @Rajesh?
17:37
@EricGregor yes, this works. Smoothness is also okay.
great, thank you @tb.
@Eric : don't know what you are talking about...is it $|xy|$ ?
the modulus
$\sqrt{x^2+y^2}$
@RajeshD try $\dfrac{xy^2}{x^2 + y^4}$
a bit different because the amount of jump could be different in different directions !
17:38
@tb does my suggestion not work?
The absolute value is pretty continuous at zero.
lol, ok right
@tb : doesn't it blow up at $(0,0)$ ?
You wanted it to be discontinuous at one point, right? Define it to be zero at $(0,0)$ or whatever else you like.
but i need it to jump there
@tb I don't want it to blow up
@tb : consider a curve passing along through $p$, if we take $f$ along the curve, then this one dimensional function should have a jump discontinuity
17:45
@RajeshD my example does satisfy that.
wait a bit
@tb : it blows up along the $y$ axis when approaching zero
there is $y^3$ in numerator where as $y^4$ in denominator
@tb I'm not sure. I searched for care bears and then I thought I'd post this since it seemed non-sensical given the (non-)context.
@RajeshD I don't follow you. The function is zero on the $y$ axis.
Is it obvious that the map $(\phi,\theta)=((2+\cos(\theta))\cos(\theta),(2+\cos(\phi))\sin(\theta), \sin(\phi)) $ is an injection? i am about to embark on a tedious calculation of a proof of this, just wondering if i'm missing something obvious
Those who downvote posts without reasons, written out as a small comment, suck.
17:57
@EricGregor insert some blanks in your TeX
hopefully that did it?
that equation is slightly off
@tb assume both $x$ and $y$ tending to zero (no matter which direction we are approaching zero), the numerator goes to zero as a cubic polynomial, while the denominator goes to zero as a $4^{th}$ power, hence it goes to $\infty$ as we approach $(0,0)$ , now consider the function $\gamma(x) = f(x,mx)$ this function blows up to $\infty$, but i want it to have a jump discontinuity at $0$
ok, fixed
See you later!
18:01
actually this may be obvious, by noting that (\phi,\theta)=(\phi',\theta') implies that $\phi$ and $\phi'$ are in either both the upper half plane or the lower half plane (by considering $\sin(\phi)$), and likewise by considering $cos(\phi)$ there is equality only if they both lie in the same right plane or left plane.
an analogous argument works for $\theta$...i think
This looks like a screwed up parameterization of a torus.
it is a parametrization of the torus
err, it is screwed up
(some $\phi$'s and $\theta$'s are mixed up.)
i can't edit it now
i'll rewrite it
$((2+\cos \phi) \cos(\theta), (2+\cos \phi) \sin \theta, \sin\phi)$
@tb : according to me the function $f$ cannot be unbounded in every neighbourhood containing $p$ (the point of discontinuity)
18:04
so if this map is called $T$, then $T(\phi,\theta)=T(\phi',\theta')$ should imply that $\phi=\phi'$ and $\theta=\theta'$ by my argument. do you agree @tb?
i'm actually trying to show that this map is an embedding of $S^1\times S^1$ into $\mathbb{R}^3$, but given that it's an immersion, after showing injectivity i should be done
@EricGregor well, you do want to restrict your $\phi$'s and $\theta$'s to $[0,2\pi)$ or something like that.
yes, of course @tb, i was assuming that was understood
so my argument works
@tb : to put it plainly, the conditions i've put on $f$ does not allow to it to be unbounded in any compact set in $\mathbb{R}^2$.
@tb : i hope you get back to me later, I do not want to disturb your conversation with Eric.
@RajeshD the function I gave you has directional derivatives in all directions of the origin.
@RajeshD I don't think we're really conversing. he's just tolerating my obtuseness here and there. your question may be more interesting
18:14
@EricGregor yes, this works. You should really think about what the two parameters do.
(hold one fixed and look at what you get if you vary the other).
@tb I could be utterly wrong with my basics, but tell me isn't the function $\infty$ at $(0,0)$ then where is the point of derivative ?
@tb, yes that's what i was doing (i think)
Hold $(x,y) \neq (0,0)$ fixed. Look at $\displaystyle f(tx,ty) = \frac{t^3 xy^2}{t^2 x^2 + t^4 y^2} = \frac{txy^2}{x^2 + t^2y^4} \xrightarrow{t \to 0}{0}$.
@tb there is $t$ in numerator and $t^2$ in the denominator, hence the limit tends to $\infty$ as $t$ tends to $0$.
$\dfrac{t}{1+t^2}$ converges to zero, of course. Numerator converges to zero, denominator to $1$
18:30
Hmmm.......thanks for clearing me out (actually it is taking me time for this to sink in), but it tends to $0$ from all directions, but i want the limit to be different in different directions @tb
Then try what happens if you look at $f(t^2,t)$, for example.
@RajeshD like $\frac{x^2-y^2}{x^2+y^2}$?
There are also examples that limit to 0 on each straight line approach to $(0,0)$, but don't on some curved approaches.
Let me see if I can think of one :-)
@tb in this case the limit is $\frac{1}{4}$...I think i got what i am intending to ask..I want $\gamma(0+)$ and $\gamma(0-)$ to be different....am I right ?
@Robjohn : I am still trying to get my question right.
where $\gamma$ could be given as $\gamma(t) = f(t^2,t)$
@tb For this function the limit is different in different directions, but what i also need is it should jump along any tangent vector, i mean the left and right limits should be different, they should be different as we approach the point and as we receed away from the point along any given tangent vector
I guess there aren't any simple closed form expressions for what i am asking but i want to know whether is it conceptually possible?
@tb : I am sorry that my question initially sounded ambiguous, I think i got it right now.........but i am especially thankful for clearing my misconceptions, coming from a 1-D world, I am finding the 2-D world a strange place.
18:56
$f(x,y)=\frac{(x^2-y)^2}{(x^2+y^2)^4+(x^2-y)^2}$
on all straight lines to $(0,0)$ the limit is $1$.
on the curve $x^2=y$, the limit is $0$
@robjohn : I am finding it counter intuitive but let me figure out
Simpler is $f(x,y)=\frac{x^2-y}{(x^2+y^2)^2+x^2-y}$
@robjohn : Is (0,0) a critical point for both these examples ? if that is the case then i don't have to bother about it i guess
I haven't looked at the derivatives...
sorry i've edited
let me check
yes it is
the derivative of $y = x^2 $ is zero at $x = 0$ !
@robjohn : I don't mean the derivative of the function but the parameterization $y = x^2$ !
If that is not the case then then my guess is that the whole differential geometry would be in trouble!
19:10
Yes, the path approaches along the $x$ axis, but if you take the path $y=0$, the limit is $1$
@robjohn I hope you got my question : (keeping the property you mentioned aside) (that is by not letting the case where $\frac{dy}{dx} = 0$), Along any given tangent vector the limit should be different in the two case (i) as we proceed towards the point and (ii) as we receed away from the point along any given tangent vector.
@RajeshD okay, I am confused. what two cases?
you want a different limit coming in than going out?
@robjohn Given a point $p$ and a tangent vector at $p$ we approach this point along this vector and we receed away from this point along the vector, I want the limit to be different in two cases
19:26
I don't think that is possible. If your path is linear and in a given direction at a given point, then going one direction is the negative of going the other direction.
as we come and as we go away just as in case of jump dicontinuity for a 1-D function, the left and right limits exist and are different
I believe it follows because $(x-y)=-(y-x)$
yes only then we call it a jump discontinuity....I don't want the function to be continuous at the point
@rob
@RajeshD so you are coming in from the right and receding to the left?
.
the limit should be different in both cases
19:31
How about $\frac{x}{\sqrt{x^2+y^2}}$?
@robjohn along which line ?
BTW the left (coming in ) and right (going away) both limits should be different along any given tangent vectors at the point
On that function, the limit is $\cos(\theta)$
.
Do you mean $x = cos(\theta)$ and $y = sin(\theta)$ ?
it doesn't pass through origin then which point are you considering?
@robjohn : I believe there can't be any closed form expressions, just like we don't have a closed form expression for a step function which has a jump discontinuity at $x =0$.
No, if you approach along the line with angle $\theta$, the limit of the function is $\cos(\theta)$
yes but it is same in both directions (the left and right that is while coming in and going out) along any line....we need it to be different.
$\gamma(0+)$ should be different from $\gamma(0-)$...along any parametrization $\gamma$
19:51
No it isn't. coming into $(0,0)$ from the left, the limit is $-1$ and the limit leaving to the right is $1$
look at it: $\frac{x}{\sqrt{x^2+y^2}}$
sorry it should be $fo\gamma(0+)$ should be different from $fo\gamma(0-)$...where $\gamma : \mathbb{R} \to \mathbb{R}^2$
@Robjohn : my comment was not in reply to yours....let me look at it
@robjohn Yes Agreed ! Thanks for the example. the amount of jump here is $1-(-1) = 2$, but i want this amount of jump to be different as we approach from different directions (tangent vectors) in general.......could this be conceptually possible ? is my question
As I said the function is $\cos(\theta)$, so the jump can go from 2 (left and right) to 0 (up and down)
or anything in between
as we approach in different directions......let me see this
@Matt good afternoon, or whatever
@Kannappan same to you :-)
@robjohn Hey mean square : ) Good afternoon, it's 10 pm here. Just come home from having dinner.
20:06
Thank you, BTW, you might want to know it is 1:35am here!
Most places were closed so we had to eat at a hipster bar.
@robjohn : yes for $y = mx$ the amount of jump is $\frac{2}{\sqrt(1+m^2)}$...thanks @Robjohn ....I feel good before going to bed you just made my day just before it is ending !
@Kannappan : its AM i guess !
@MattN I look forward to dinner. We have mexican food on Friday night. We get the same take out so that we know what to expect and don't have to think about what to get.
So, @MattN The time-lag between the places we live is not that much!! Good thing!
@robjohn Funny : )
20:07
@RajeshD Edited, thank you.
@MattN I guess life is funny :-)
@KannappanSampath Yes but weren't you going to rest?
@robjohn Hm. That certainly depends on the angle.
But at the moment it is. Since we were forced to eat at a bar we were also forced to have a few.
@MattN Hmm, I told you I am bored to death here. How long will you expect me to sleep without having to do anything for hours?
@KannappanSampath I could sleep all day. Last night I slept 8 hours and this afternoon another 2. : )
@MattN "forced"?
20:09
@robjohn "forced". : )
@MattN Not bad, even I can if it's that way. But not in this case. I have been on this bed with friends getting me food from outside, fruits, milk and all of that. I have literally had nothing to do.
(applause for robjohn and tb), i just had a wonderful and enlightening session of learning with them
Ok. Sounds as if you're bored.
I am tempted to add an n and change the n to an m in this title here.
We already have had a horror, why add to it?
Just being silly. : )
20:15
@robjohn had to tell me about that picture I posted. He has not yet heard me out on that.:)
@KannappanSampath Yo, know anything about asaf?
@KannappanSampath I don't know if I saw that picture...
@robjohn The one about Omnigraffle or whatever... !
@N3buchadnezzar Asaf has been silent.
@KannappanSampath ah
@N3buchadnezzar Probably playing egg search right now.
20:17
@robjohn : I thought of this for this question.....but OMG....if only i could be more disciplined to pay attention to comments on my own questions......the comment by Day Late Don here has everything you have just mentioned......I guess it could be made as an answer so that i can accept it.
@N3buchadnezzar Nothing at all. He would not tell us anything.
So is he still on the site or what?
Yes. Just not in this chat.
Yes, he writes answers for the site.
What would you suggest @robjohn ?
20:19
@MattN New pastime? Good he is still is contact with atleast one chat regular!
@RajeshD I have prompted DLD to answer.
I wrote elementary answers today, good enough I should say.
@KannappanSampath I wrote one hard answer overnight, and then just answering questions on chat since.
Oh, and responding to a couple of comments on answers :-)
@robjohn I see. So, did not cap today, then?
20:24
@KannappanSampath no where close yet, today.
Let me say : I was just too late to catch up with "Day Late Don". May be thats his speciality !
@KannappanSampath only 35% of the way to capping (70/200).
@robjohn Now a +1 ...
I shamelessly beg to star my last comment if that is not ridiculous.
2
What an unproductive day.
20:26
Noooooooooooooooooooo
my previous comment Sir
not the one in which i am begging ! Plzzzz
: D
The only thing I did today was to read one proof.
Oh no
@KannappanSampath Thanks, but I was talking about this :-)
@MattN That theorem in C'x analysis.
someone's trolling me bigtime
2
20:28
@KannappanSampath Yes.
@robjohn It got the first upvote from me. I read it carefully too.
Okay I am off to bed...Good night and sweet dreams !
Nighty-night!
@KannappanSampath Cool. There was one good comment about staying in the principal area for $\log$
@RajeshD sleep well
@robjohn Yes, in the one I upvoted recently. But I don't know C'x analysis at all. ;)
20:32
@KannappanSampath See; even I am getting mixed up :-)
@robjohn Never mind.
I am getting bored to death here. Some one tell me what on earth can I do?
Not math, I am not that healthy either.
@MeanSquare: have you capped?
No, he has not.
just before going to bed...
Hi Henning!
20:34
@MattN Hi.
@KannappanSampath He's still missing 95 til cap... not sure we can cap him.
@HenningMakholm How are you doing?
@MattN Yes had the same feeling. : )
@MattN Fine, thanks. You?
@HenningMakholm Alright, thanks : )
Are there any regular MO users here?
This question concerns Gerhard Paseman signing his posts.
20:38
@KannappanSampath there was even an MO question on this :)
While the FAQ explicitly discourages that (?), don't mods tell him that or I mean, they could do something about it, no?
@tb Link please? Thanks for letting me know!
What do you mean, here? He left the site because people didn't like his adding "ask me about System Design" or whatever...
@KannappanSampath The question was closed and deleted.
Title "Tell me about system design!"
He seems to be friends with the guy who thinks he's funny. : )
And he sounds like a comp sci.
Come to think of which...
@tb I meant MO faq. Was he also on SE?
@KannappanSampath this should lead you to some links: meta.math.stackexchange.com/questions/585/…
@MattN oh, yeah, one of those guys who can compete with WJ
20:44
Was he also on SE?--Yes. Thanks for the link.
'Ello! Gigili
@tb and here is a post with that sig.
'Ello Tee-bee.
I have to answer a question tonight, or else I shoot.
@Gigili I'd very much prefer if you answered a question...
@Gigili Why?
20:49
Hiha. I'll try my best.
Oh. Ok. I think I know why.
Oh?
@Gigili Because it's a long weekend...
@robjohn That particular post is entertaining.
@tb I think I solved his Hangman, I think. Should I write a mail to confirm?
Cannot resist the temptation, you see. : )
Long and of course boring, tru dat.
20:52
@KannappanSampath well, don't miss to ask him about system design!
@tb Gerhard, what is system design?
Morning
Morning.
Morning.
I wonder why is this guy real crazy? He makes relevant signatures though--Like Still thinking or whatever....
@BenjaminLim Morning
20:53
@BenjaminLim Saturday morning there?
What exactly are these signatures?
I'm not sure why I'm asking.
@robjohn yeah 6.53am
That's still night for me.
@BenjaminLim It is still 1:53 PM here on Friday
@robjohn LA time
20:54
@BenjaminLim Indeed :-)
What you guys up to here?
@KannappanSampath
@MattN I believe the time difference is about 10 hours between us
Nothing serious at all. @Ben
@BenjaminLim Oh I forgot to react to your having all those field theory exams behind you! I hope they went well.
Just about signatures....
@tb Yeah it went well !!
20:56
@BenjaminLim 8 : )
Yes, so, you did both exams well! That's nice.
@tb They asked us to show why the sum of two algebraic numbers was algebraic
I thought of bashing the algebra out
then I realised all I needed to show was that if $\alpha$, $\beta$ are algebraic numbers then $[\Bbb{Q}(\alpha + \beta) : \Bbb{Q}] < \infty$ which was pretty easy
Woooo
Party beer chicks
20:57
I thought he left our site????
Oh. This one. I saw that earlier today. The OP seems to be a SWE or something. I hope he's not written any software to do lazer operations on people's eyes or similar.
@t.b. As per one of your comments, I flagged my last question to be de-wikified, and I succeeded.
@MattN Software engineer=SWE?
@KannappanSampath Yes.
20:58
@AméricoTavares That was surely going to happen, if you told some mod. :)
@AméricoTavares Very good! The CW status might discourage some answerers...
@KannappanSampath @MattN You guys are terrible taking a dig at engineers.....
@BenjaminLim Huh? What? I? No, I am usually kind and polite with them. ;-)
I'm not : )
@KannappanSampath @MattN It was just a joke :D :D :D
21:01
@t.b. Yes, I thought so. @Kannappan I was lucky.
@N3buchadnezzar How do you manage posting here an image?
@AméricoTavares do you see the upload button?
It's to the right of the box where you type
Why the sudden influx of ducks here?
I think these are chickens.
Did you grow up in a city?
@Benjamin Yes, now I've seen it.
@MattN oH yeah
@Americo Another suggestion: You can reply to people's messages by moving your cursor over a message you want to reply and clicking on a little arrow at the end there...
@AméricoTavares Like this one.
21:06
@AméricoTavares congratulations :-)
I am learning here from you, thanks!
@robjohn Thanks.
@AméricoTavares Slowly and steadily
@Americo We'd like to see you often here. :-)
@BenjaminLim Yes, that's my way.
@KannappanSampath OK. I will try come here every week.
This chat is like a black hole.
21:10
@AméricoTavares Thank you. Looking forward to seeing you.
Sucking in everyone : )
First I posted three nice chicks, here is a nice pussy =) Really cute!
@N3buchadnezzar so, after we educated Skull you're starting the spamming here? :P
21:15
Vacation! And cats stimmulate the mathematical part of the brain.
Virtually Abelian. The list of things I need to brush up on is growing faster than I can brush.
Even I have never heard of anything about "Virtual"ness in Group theory which means, I haven't done any! This makes me sad. :/
@KannappanSampath well, you probably have focused on finite group theory where everything is virtually trivial...
(old geometric group theorist's joke)
I see. Given I picked up group theory from Aschbacher after Dummit-- Foote, this is then true, I guess.
@KannappanSampath One of the most amazing theorems in group theory is Gromov's theorem on polynomial growth = virtually nilpotent
21:21
Heh. Power point slides.
@tb I see. So, this is related to Geometric group theory then?
Firstly, it is nice to see the use of the symbol for action by a Mathematician.
@JonasTeuwen I'm creepin' up on you : )
@KannappanSampath "pure" group theorists are also interested in virtual properties, as witnessed by Arturo's answer. I think they give rise to some interesting varieties.
But of course, growth estimates/properties are pure geometric group theory...
Hilbert spaces, Ricci flow on Riemannian manifold--Way over my head.
I didn't mean to imply that you should read that attentively. It was a major breakthrough a few years ago.
21:26
I have always felt that doing group theory, if you need to progress, you need to learn other branches of math as well. This happens with a few branches of Math only, I think.
@tb Oh, so, it was an expository lecture I was attending then? Good, that's what all talks I have attended had been to me. : D
@KannappanSampath I don't understand what you're trying to say here.
Arturo took 49 seconds to answer this one.
I meant to say that: If you wanted to do group theory, and make some progress, soon you realise that you need to know some representation theory-- to do that some linear algebra--then, some combinatorics--some finite geometries--then some designs and codes and then you get to know the existence of a class of simple groups and you thank Erdos.
@tb ^
@MattN We could have answered that as well, we will have proofs for those in the transcript for CA room, \me thinks!
Yes but I'm not in the mood : )
But, yes, Arturo is swift, we agree. :)
21:33
Far too lazy. (Me, not Arturo, obviously)
And I think I should go to bed.
@KannappanSampath Well on the one hand groups are ubiquitous... On the other hand, I don't think that group theory is any different to many other subjects with respect to interrelations to other topics.
@MattN Well, Good night, then. Sleep well.
@tb Hope the abuse you got today was not too harsh. Didn't expect mean square to join in. : )
I could take it... wait... whiiiine
@tb It's a feeling as I said. I do Math for all the fun involved in doing Algebra.
21:38
: )
An old story.
@MattN Right, thank you.
Welcome : )
@MattN Searching transcript works? What key word then?
@KannappanSampath It's not in the transcript.
Good night folks.
Hmph, then I would not get to know. Looks like a suspense with which going to bed is a difficult task.
@MattN Good night, Take care.
21:49
How do I get to the transcript for a particular day?
Holy, there seems to be no simple way. I have to go all the way there by clicking a day after another.
@KannappanSampath http://chat.stackexchange.com/transcript/36/2012/mm/dd/ (no leading zeroes)
@tb Thank you.

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