Mathematics - 2015-11-14 (page 2 of 4)

user174558

13:00

@skillpatrol Thanks Professor.

M.S.E

@Jasper people with anxieties like social anxiety, ocd, etc are normal people. they are not mental patients

@Jasper and no i don't need to look up a dictionary :)

skill patrol

@M.S.E It depends on the severity

Daniel Fischer

@Jasper Varying. Sometimes better, sometimes worse.

user174558

@M.S.E Same goes for 'religion'.

M.S.E

@Jasper the definition that Balarka Sen gave was way out of what religion means.

user174558

13:01

Like Professor @skillpatrol says, it always comes down to definition.

skill patrol

:-)

iwriteonbananas

@DanielFischer For $M\subset B(H)$ a von Neumann algebra, $M'$ denotes its commutant. A vector $\Omega\in H$ is called cyclic for $M$ if $M\Omega\subset H$ is dense. It is called separating for $M$ if $T(\Omega)=0$ only for the zero operator $T=0$.

M.S.E

@Jasper a porn addict, people with sleeping disorders, anxieties like social anxiety, or OCD in your case are normal people :) not mental patients.

user174558

@M.S.E Well, you should know that many people have very different ideas of what that word means.

Balarka Sen

@M.S.E I don't see a difference :P Anyway, I think we all agree that athiesm is irrational and illogical, as well as theism.

skill patrol

13:03

@M.S.E One can even have a defining moment in life...

iwriteonbananas

So I want to show if $\Omega\in H$ is separating for $M'\subset B(H)$, then it is cyclic for $M$.

user174558

@BalarkaSen Also depends on definition of irrational and illogical.

M.S.E

@BalarkaSen @Jasper okay ^_^ now let's forget about that religion topic

skill patrol

Define: define.

M.S.E

@skillpatrol one can have a defining moment in life? I'm sorry, where does this come into the story?

user174558

13:04

@M.S.E Have you heard of that term?

Balarka Sen

@Jasper Ah. Well, rational and logical have a universal mathematico-logical definition.

M.S.E

@Jasper which? Sorry, there is too much going around. what are u referring to ?

user174558

@M.S.E Defining moment.

skill patrol

^

Balarka Sen

@M.S.E Yes, exactly what theists say when their ideas are questioned, lol.

user174558

13:05

@BalarkaSen Well, anyone who has studied logic will know about different logics, lol.

Balarka Sen

But anyway, I need to concentrate on my work.

See ya

M.S.E

@BalarkaSen I'm an atheist. we went onto talking nonsense about defining words.

iwriteonbananas

Let $P$ be the projection onto $\overline{M\Omega}$. So if we could show that $\operatorname{Id}-P\in M'$, then we could conclude that $\operatorname{Id}-P=0$ from the fact that $(\operatorname{Id}-P)\Omega =0$, @DanielFischer.

skill patrol

@BalarkaSen later pal

Balarka Sen

@M.S.E I'm simply saying that atheism is illogical, in the sense that you are assuming something as an axiom which cannot be deduced logically.

user174558

13:07

I am going to take a nap, good night Professor @skillpatrol.

Balarka Sen

Of course, you can mend this, by letting your ambient logical system to be "Axiom 1: God exists/does not exist".

M.S.E

@Jasper defining moment. What comes to my mind is that an unusual event that happens in life, that have a defining moment of their life. A game changer?

skill patrol

@Jasper later pal

Balarka Sen

I'm fine with that, but it's extremely silly.

skill patrol

@M.S.E if you view life as a "game" yes

M.S.E

13:10

@skillpatrol okay so with that definition, what has that got to do with anything we were talking about?

skill patrol

nvm

just don't be too quick to judge

enough said

Daniel Fischer

@iwriteonbananas Aha. The commutant was the algebra of operators that commute with all $T\in M$, wasn't it? Writing $Q = I - P$, we have $QM = 0$, so we want to see $MQ = 0$, right?

M.S.E

@skillpatrol Hmmmmmm

Daniel Fischer

@BalarkaSen Define atheism.

M.S.E

@DanielFischer starred nice one :D

Balarka Sen

13:13

@DanielFischer The logical system obtained from the assumption "God exists"?

Anyway, defining that is @M.S.E's job.

Because I'm not an atheist.

iwriteonbananas

@DanielFischer Yes, that's the correct definition of the commutant. Why is $QM=0$?

M.S.E

@skillpatrol did you read that? I deleted just in case he sees.

skill patrol

yes

4 mins ago, by skill patrol

enough said

M.S.E

@skillpatrol ok :)

skill patrol

:)

M.S.E

13:16

@BalarkaSen Yay! I finally found myself a job :D if only I could find a real job this easily.

Balarka Sen

@M.S.E Am I to take your silence as evidence that you agree that atheism is illogical? (DanielF pointed out that we don't know a good definition for atheism. But that's more evidence for the whole thing being illogical).

Daniel Fischer

@iwriteonbananas Oops, not that.

M.S.E

@BalarkaSen no I was into the Jasper talk after the religious talk ended.

@BalarkaSen Oh, did you ask me that? Atheism is not at all illogical.

@BalarkaSen Atheism is the lack of belief in religion (Something which can't be proved). If I don't believe your claim that $1=2$, that does not make me illogical.

Daniel Fischer

@BalarkaSen Well, if you say something like "atheism is an irrational belief system", it would be helpful to know what you mean by that to know whether you're right or blatantly wrong. Or something in between. That depends on what your definition of atheism is.

Balarka Sen

I've mention what my definition of atheism is.

M.S.E

13:20

@BalarkaSen you might not be having a correct definition. Here's the definition:

What is atheism?

Atheism is the lack of belief in any god or gods. That's it. While this should be obvious, many people seem to think it implies other things. It doesn't. The only generalization you can make about all atheists is that they don't believe in any gods.

The term atheism often invites misconceptions. Unlike a typical -ism, atheism does not refer to a particular positive ideology, and there is no school of thought behind it. It is used as a blanket term to refer merely to the absence of something others may happen to have. To be very clear, atheism is not a philosophy or a move…

Daniel Fischer

6 mins ago, by Balarka Sen

@DanielFischer The logical system obtained from the assumption "God exists"?

That ^^, @BalarkaSen?

Balarka Sen

should have been "God does not exist".

@M.S.E "Atheism is the lack of belief in any god or gods." What's the logical basis in that belief?

skill patrol

21 mins ago, by skill patrol

It always comes down to a matter of definition.

Balarka Sen

Or are you taking it is an axiom?

Daniel Fischer

Okay. Not my definition of atheism, though.

Ghost

13:21

is an Atheist

but are fascinated with trigonometry functions raised to multiple powers

Daniel Fischer

@BalarkaSen "lack of belief in existence $\neq$ belief in nonexistence"

M.S.E

@DanielFischer starred exactly

Balarka Sen

What's the logical basis of the former? :P

M.S.E

@BalarkaSen like I said previously. If I don't believe your claim that $1=2$, that does not make me illogical.

@BalarkaSen It's just a state of mind to ignore something that has no proof......doesnt mean atheism is proving that God doesn't exist!

Daniel Fischer

@BalarkaSen In the absence of convincing evidence, we choose to suspend judgement. Not "logical" as such, but a good rational strategy.

Balarka Sen

13:24

@M.S.E @DanielFischer Sorry, that's not atheism. That's agnosticism.

Which is of course perfectly logical.

Ghost

could the natural log be expressed as a power? (probably a dumb question)

M.S.E

@BalarkaSen Isn't Agnosticism a flavoured type of Atheism?

Balarka Sen

Nope.

If you think so, you're confused.

M.S.E

@BalarkaSen Ever heard of Agnostic Atheism?

@BalarkaSen en.wikipedia.org/wiki/Agnostic_atheism

Daniel Fischer

@BalarkaSen M.S.E has given his definition of atheism as "lack of belief". If you prefer to call that agnosticism, fine.

Balarka Sen

13:27

Well, @DanielFischer, M.S.E. never clarified his defn of atheism before!

Daniel Fischer

@BalarkaSen Hm. Agosticism is "ignoramus et ignorabimus". M.S.E's atheism is content with the "ignoramus". Looks like from M.S.E's viewpoint, agnosticism is a flavour of atheism.

Balarka Sen

Well, M.S.E.'s definition of atheism is agnosticism!

M.S.E

I think this conversation won't end here.

Daniel Fischer

My version of atheism is "I don't care about existence of gods".

M.S.E

@BalarkaSen Seriously? After all that I have said?

Ghost

13:30

is it possible to get an answer to my question?

Balarka Sen

Yes, @M.S.E. As I have said above, you are confused.

M.S.E

@BalarkaSen Okay we have a linguistic problem here.

Balarka Sen

@DanielFischer Sure, that's my version of agnosticism.

M.S.E

@Ghost whats your question? is it related to atheism? if not, the answer is no :P

^^^ just kidding

Ghost

no it is maths, surprisingly

Balarka Sen

13:32

Nov 5 at 16:49, by Balarka Sen

Maybe we need to change the name of this chat into "Hodge Theory and Topological Quantum Field Theories"

Daniel Fischer

@BalarkaSen Point is, people have different concepts and definitions of things like religion, atheism, agnosticism etc. If you believe everybody means the same thing when using such a word, you're wrong most of the time. Pretending there is one correct definition of such words is erroneous.

Balarka Sen

I lost all my credibility for saying that.

M.S.E

@Ghost oh yeah mate. Look, we are having a Theology class over here right now. Your question would be better suitable as a question on math stack exchange.

@BalarkaSen haha :D starred ;)

Balarka Sen

@DanielFischer Yeah, I don't think there's any point discussing this until we have a common, well-defined definition.

skill patrol

An agnostic is one who believes it impossible to know anything about God or about the creation of the universe and refrains from commitment to any religious doctrine. An atheist is one who denies the existence of a deity or of divine beings.

3

M.S.E

13:34

@skillpatrol thank you SIR. let Balarka figure out now the difference.

Daniel Fischer

@BalarkaSen Start with something easier. Establish a common universally agreed on definition of $\mathbb{N}$.

Ghost

@M.S.E nevermind then

M.S.E

@BalarkaSen A serious question: Can we like change the name of the chat room for a few hours? That would be really cool.

Balarka Sen

@M.S.E Compare skill's definition of agnosticism with Daniel's definition.

"In the absence of convincing evidence, we choose to suspend judgement."

"[...] and refrains from commitment to any religious doctrine".

skill patrol

^

Ghost

13:36

Stupid me for asking a maths question in a maths chat room - what an idiot I am for assuming that it'd be on topic

Balarka Sen

Looks similar, although one is being called atheism and the other agnosticism, lol.

skill patrol

God is unknowable.

M.S.E

@Ghost sorry its off topic

@BalarkaSen also heard of Agnostic Atheism?

@BalarkaSen See here en.wikipedia.org/wiki/Agnostic_atheism

skill patrol

Agnosticism

Agnosticism is the view that the truth values of certain claims – especially metaphysical and religious claims such as whether or not God, the divine or the supernatural exist – are unknown and perhaps unknowable. According to the philosopher William L. Rowe: "In the popular sense of the term, an agnostic is someone who neither believes nor disbelieves in the existence of God, while a theist believes that God exists, an atheist disbelieves in God". Agnosticism is a doctrine or set of tenets rather than a religion as such. Thomas Henry Huxley, an English biologist, coined the word "agnostic" in...

Ghost

wow...

Balarka Sen

13:37

grr, I don't want to talk about this anymore. theism is a religious piece of self-styled rubbish.

M.S.E

@BalarkaSen If you did not know, there exist something called Agnostic Theism.

Balarka Sen

atheism is also a nonreligious piece of self-styled rubbish.

M.S.E

@BalarkaSen This is why I said that Agnostic is a flavour.

Balarka Sen

everything is a piece of self-styled rubbish.

go away :P

M.S.E

@BalarkaSen See here en.wikipedia.org/wiki/Agnostic_theism

@DanielFischer does this room get so fun/active like this and confusing this often? I'm not very active here.

Ghost

13:39

So, to ask my question again (as it is on topic for a maths chat room) - can a natural log be expressed as a power?

Chris's sis the artist

Looks like I'm the only one here believing in God (100% sure of God's existence). :-)

Daniel Fischer

@Ghost What do you mean with "express the natural logarithm as a power"? Something along the lines of $\ln x = x^a$? That isn't possible.

Ghost

@DanielFischer thank you, that's what I was asking

M.S.E

@Chris'ssistheartist You might be surprised, the small conversation we started, went all over the place.

Balarka Sen

I am concluding that discussion about religion or non-religion is an utterly bogus piece of self-styled rubbish, and a complete and an irreversible waste of time.

Ghost

13:42

Actually, @DanielFischer - I do appreciate the answer to my maths question, which is on topic here, despite what @M.S.E stated to me

Chris's sis the artist

@BalarkaSen As an atheist, yeah, it is for you an irreversible waste of time, not for all. For a believer is almost all.

Balarka Sen

I'm no atheist, @Chris'ssistheartist

Daniel Fischer

@Ghost "for both general discussion & math questions alike" <- Yes, maths is also on topic here.

M.S.E

@Chris'ssistheartist A practically useful question for me: The god you believe in One day when I die, if he really exists like you say for sure, will he reject me from heaven for not believing in him?

^^^ This is all I care for.

Chris's sis the artist

Every time I see $$1+2+3+\cdots =-\frac{1}{12},$$ to add math into discussion,I remember God's existence.

Balarka Sen

13:44

shit, there we go again

Ghost

@DanielFischer yes, I know that, @M.S.E needs to be educated on that, especially when said to me:

7 mins ago, by M.S.E

@Ghost sorry its off topic

M.S.E

@Ghost I was joking. @DanielFischer I was just kidding, I thought it was obvious.

Daniel Fischer

I know.

Ghost

@M.S.E not really obvious, no

M.S.E

@Ghost It was a joke. Wasn't it obvious? O.o

@Ghost Wow :) okay. sorry

@Chris'ssistheartist ;)

@Chris'ssistheartist an answer for my practical question? :(

Ghost

13:46

Can't read minds, nor read 'jokes' in a chat room

and by the way, I am an Atheist

M.S.E

@Chris'ssistheartist cause I dont want to end up in hell.

Daniel Fischer

@Ghost Always work on the assumption that people aren't serious. That's safer.

Chris's sis the artist

I almost mixed up the result with another one.

M.S.E

@DanielFischer starred! Now that's an awesome life advice for the day :D

Ghost

@DanielFischer not really a safe measure

Chris's sis the artist

13:47

@M.S.E I also have tons of questions that you might never answer. Just to know that. I need research for such questions.

M.S.E

@Chris'ssistheartist Questions from whom? Me? Ask away :D

Daniel Fischer

@Ghost Not safe, but safer. If people talk nonsense, it's better to assume they're joking than otherwise.

Ghost

if you say so

Chris's sis the artist

@M.S.E ;)

M.S.E

@Chris'ssistheartist And the first question is :D

Ghost

13:49

at least I got an answer to my question - so saved me wasting my time on the board

Chris's sis the artist

@M.S.E $$\sum _{k=1}^{\infty } \sum _{n=1}^{\infty } \frac{\Gamma (k)^2 \Gamma (n) }{\Gamma (2 k+n)}((\psi ^{(0)}(n)-\psi ^{(0)}(2 k+n)) (\psi ^{(0)}(k)-\psi ^{(0)}(2 k+n))-\psi ^{(1)}(2 k+n))$$

M.S.E

@Chris'ssistheartist oh nooooooo :O I thought its gonna be on God and stuff :P

Ghost

I wonder if there are papers about modelling with the natural log of trig functions raised to a power

Chris's sis the artist

@M.S.E :D

M.S.E

@Chris'ssistheartist I see what you did there. Hmmmmm

Chris's sis the artist

13:54

@M.S.E Some problems require some specific research. Without (personal) research I'm powerless (maybe it's not the case for others).

M.S.E

@Chris'ssistheartist are you doing the research on that? :) Talking to God in anyway?

Chris's sis the artist

@M.S.E The wisdom comes from God too.

M.S.E

@Chris'ssistheartist Btw this brings up an interesting question, do you also believe that there was an actual God called Namagiri that spoke to Ramanujan ? Or do you think like me (atheists) that it was just all in his head, and a result of his genius?

Ghost

this is a waste of time

Chris's sis the artist

@M.S.E Indeed, Ramanujan believed in God, in his way, it's a known fact that he was pretty religious since very young. Euler was very religious too. It's hard to tell what happened there.

M.S.E

14:00

@Chris'ssistheartist yes I know they were religious, and he claimed that. So I was asking you if you believe that claim.....for which you responded that its hard to tell :) fair enough

@Chris'ssistheartist or was the answer :indeed?

Chris's sis the artist

@M.S.E No, it wasn't.

@M.S.E I also believe in some sense what Ramanujan believed, but, well, no more talk about myself. ;)

M.S.E

@Chris'ssistheartist hehe :) modesty ;)

Chris's sis the artist

@M.S.E (wisdom comes from God even when calculating integrals - to say it in short)

M.S.E

@Chris'ssistheartist Interesting, I see :) I wouldn't know though

Chris's sis the artist

@M.S.E I consider my calculation ability (no matter how it is) a gift from God. I created so far thousands of problems and solutions without a math background.

I'm full of research ideas every single day. Question: where our thoughts come from? It's not about creativity, it's about extreme creativity, plenty of ideas.

M.S.E

14:11

@Chris'ssistheartist thoughts come from neurons firing in the brain ? :)

@Chris'ssistheartist But you love mathematics, and you do it more than anyone with a mathematical background. Don't you?

Chris's sis the artist

@M.S.E I love mathematics very much, but I prefer not to compare to anyone (in the sense of being in a competition). I only prefer to be in a competition with myself.

My pleasure lies in discovering results, this gives the maximum amount of pleasure. :-)

M.S.E

@Chris'ssistheartist :) That is a good discipline to have.

Chris's sis the artist

@M.S.E Yeap, I think so. :-)

M.S.E

@Chris'ssistheartist Your lucky that you found your passion :)

@Chris'ssistheartist most of us don't even know what to do with our lives :P

Chris's sis the artist

@M.S.E Let me show you something, to give you an example.

M.S.E

14:16

@Chris'ssistheartist YES :D

Chris's sis the artist

@M.S.E Today I found a way to calculate the integral below without pen and paper (it's by Ramanujan) $$\int_0^1 \frac{t^{a-1}}{1-t}-\frac{ct^{b-1}}{1-t^c}\ dt$$ Now, be careful about it. Major part of people provide with the wrong answer to it. Try to do it alone without looking for an answer. You'll have much fun.

M.S.E

@Chris'ssistheartist without looking for an answer? :)

Chris's sis the artist

@M.S.E I mean not to try to look for an answer given in some papers, sites, or elsewhere.

M.S.E

@Chris'ssistheartist what is the result? I'll like to see the final result :) So it would be prove that $$\int_0^1 \frac{t^{a-1}}{1-t}-\frac{ct^{b-1}}{1-t^c}\ dt=?$$

@Chris'ssistheartist hehe oops I am doing the exact thing you told not to :P

Chris's sis the artist

@M.S.E let it be a surprise!!! :-)

M.S.E

14:21

@Chris'ssistheartist without a pen and paper? for real?

Chris's sis the artist

@M.S.E Absolutely.

M.S.E

@Chris'ssistheartist so it would be a few steps in your head? O.o

Chris's sis the artist

@M.S.E One line.

M.S.E

@Chris'ssistheartist for real ? :O

Chris's sis the artist

@M.S.E Absolutely. :-)

M.S.E

14:23

@Chris'ssistheartist okay show me show me :D I can't figure it out obviously.

Chris's sis the artist

@M.S.E It's one of the most deceiving integrals I ever met. Be very careful! :-)

M.S.E

@Chris'ssistheartist what happens when c=1 ? ;)

Chris's sis the artist

@M.S.E Well, if you know the integral, my story is in vain. :-)

M.S.E

@Chris'ssistheartist Didn't get you? :)

Chris's sis the artist

@M.S.E sure, it reduces to polygamma functions (digamma functions more precisely).

M.S.E

14:29

@Chris'ssistheartist I haven't learnt Digamma or polygamma functions :/

Chris's sis the artist

@M.S.E OK :-)

M.S.E

@Chris'ssistheartist But still how does the 1 line go :D

without a paper and a pen

Chris's sis the artist

@M.S.E You'see that idea in my book. Next year, if all is fine, it will be released.

M.S.E

@Chris'ssistheartist But you made me too curious :(

@Chris'ssistheartist our secret? :)

Chris's sis the artist

@M.S.E Oh, let me keep some stuff to my book. :D

M.S.E

14:34

@Chris'ssistheartist ok ok ^_^

Krijn

Is the whole book on these type of integrals that you always post?

Chris's sis the artist

@Krijn integrals, series and limits. I think the series I prepare to my book are the most amazing part.

Some simply blow up any mind.

Krijn

I'm not such an integral-y type of guy

However you seem very enthousiastic so I believe you when you say that they would blow my mind

Chris's sis the artist

Yes, I'm very enthusiastic, I live these things deeply, the most beautiful thing that has ever happened to me. :-)

This world (of calculating integrals, series and limits) is simply amazing, full of mind-blowing mathematical connections, you never get enough and want more and more.

Krijn

14:51

@Chris'ssistheartist Could you do an easy example for me to show me what's so amazing and mind-blowing?

M.S.E

@Krijn starred :D

Chris's sis the artist

@Krijn OK. Calculate $$\lim_{n\to\infty} \left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}\right)$$ that is a classical example.

Krijn

@M.S.E :D I'm really curious, because I've never really liked integrals and if someone is this enthusiastic about them I must have been missing something

Chris's sis the artist

@Krijn However, you need to work more to see the big picture (of beauty) I presently see.

Krijn

@Chris'ssistheartist I imagined so

Is it equal to $\ln(2)$ or am I doing something completely wrong?

Chris's sis the artist

14:57

@Krijn Right. How you did it?

Krijn

Well, rewrite to $\lim_{n \to \infty} \sum_{i = 1}^n \frac{1}{n+i}$

Then you do the trick where you exchange the sum for an integral

Which I'm not sure why that works anymore, but I was just toying with it

Chris's sis the artist

@Krijn I see.

Krijn

And then you'd get that $\int \frac{1}{n+x} dx = \ln(n + x)$

Which you evalute at $n$ and $1$, so you get $\lim_{n \to \infty} \ln\left(\frac{2n}{n+1}\right)$

Which is $\ln(2)$

Chris's sis the artist

How about thi sone? $$\lim_{n\to\infty} \left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}\right)=\lim_{n\to\infty} ((H_{2n}-\log(2n))-(H_{n}-\log(n))+\log(2))=\log(2).$$

Krijn

$H_{2n}$?

Chris's sis the artist

15:00

@Krijn harmonic number

Soham Chowdhury

oh, seems like I missed an argument.

Krijn

I'm not sure that I understand what you are doing, how did you get the $\log$ in there?

M.S.E

@Krijn $$\sum_{k=1}^{n}\frac{1}{n+k}=-polygamma(0, 1+n)+polygamma(0, 1+2 n)$$

Chris's sis the artist

This one I call to be part of the beauty in a small example.

@Krijn your integral should be $$\int_0^1 \frac{1}{1+x} \ dx=\log(2).$$

Balarka Sen

hi @Soham

Soham Chowdhury

15:05

hai.

sup?

Krijn

@Chris'ssistheartist Oh I understand your approach now, thats quite nice!

Balarka Sen

@SohamChowdhury inf?

Chris's sis the artist

@Krijn :D

Krijn

(although I did need the wikipedia page for harmonic numbers)

I should probably save all these examples of what I'd call cute proofs

It would make a nice list

Balarka Sen

@Krijn coends are nice

Krijn

15:13

I'm not that far with category theory yet

Btw, we did CW structures yesterday, that was fun

Balarka Sen

cool.

Krijn

Amazing that they allow you to compute the homology groups of $\mathbb{C}\mathbb{P}^n$ so fast

Balarka Sen

yes. did you use cellular homology directly, or the long exact sequence?

i'll be right back

Krijn

Cellular homology directly

The prof wanted us to know why we would want to put all of this work in, so he showed us that example so that we have a goal in mind

As Ma

I have a question relate to metric space. Is anyone here who can help me

Krijn

15:18

Just ask; hope that someone here can help you

As Ma

d(x,y)=(x−y)2
isn't a metric on R. Its triangular inequality doesn't holds.
As I did
d(x,z)=((x−y)+(y−z))2=(x−y)2+(y−z)2+2(x−y)(y−z)
As, a2+b2<a2+b2+2ab ,so we can say that (x−y)2+(y−z)2<(x−y)2+(y−z)2+2(x−y)(y−z) this implies that d(x,z)>d(x,y)+d(y,z). Am I right ?

Krijn

What if $(x-y)(y-z)$ is negative though?

What you should do is just find a counterexample, that is enough to show that the triangular inequality doesnt hold

And what you have computed shows that your counterexample would work if $(x-y)(y-z)$ is positive, so just find $x,y,z$ such that that is true.

As Ma

I got your point

but issue is that our teacher said to solve this question generally

Balarka Sen

ok, back

@Krijn yes, $\Bbb{CP}^n$ is an ideal example for cellular homology.

Krijn

@AsMa Well, you could generally say that this holds if $x > y > z$

Balarka Sen

15:26

the deal is that if $X$ is a CW-complex, $H_n(X^{(n)}, X^{(n-1)})$ is the free abelian group generated by $n$-cells of $X$. so if you can arrange this in a chain complex, and take it's homology, that would be a more efficient homology theory. fortunately, it turns out we can arrange this in a chain complex and the resulting homology is the same old singular homology (this is not as surprising, by the classification of singular homology theories)

As Ma

ok, thanks @Krijn

Krijn

@BalarkaSen I see, I guess we'll work towards that the coming weeks

Balarka Sen

@Krijn Er? I mean, that's essentially what cellular homology is.

Not sure how you have computed homology of CP^n with cellular homology without constructing the cellular chain complex.

Krijn

He just explained CW complexes so far, and only told us that there is some theorem (which we will work towards the coming weeks) that connects the homotopy groups of $X$ to its CW structure

Balarka Sen

"homotopy groups" did you mean homology groups?

Krijn

15:35

Yes, sorry

Balarka Sen

right, ok.

@Krijn Do you see why $H_n(X^{(n)}, X^{(n-1)})$ is free abelian generated by $n$-cells of $X$?

Krijn

No not really

Although I'm not in optima forma today

Balarka Sen

let's see. there is a theorem connecting homology groups of $(X, A)$ with the homology groups of $X/A$. are you familiar with that?

[for nice pairs $(X, A)$]

Krijn

Relative homology groups?

Balarka Sen

Yes.

Krijn

15:38

Yes

Balarka Sen

What does it say?

Krijn

$H_n(X/A) \cong H_n(X,A)$

Balarka Sen

not true for all pairs $(X, A)$.

Krijn

Hmm, I don't know this from the top of my head, and I am not near my lecture notes now

Balarka Sen

e.g., take $X$ to be $[0, 1]$ and $A$ to be the subset $\{0\} \cup \{1/n : n \in \Bbb N\}$.

1st homology of $(X, A)$ and $X/A$ (which is a Hawaiian earring) do not agree

@Krijn what you need is $(X, A)$ to be nice. one such nicety condition is that there is a neighborhood $U$ of $A$ which strongly deformation retracts onto $A$.

Krijn

15:42

Hmm, so what is the general statement?

Balarka Sen

if $(X, A)$ has homotopy extension property, then $H_n(X/A) \cong H_n(X, A)$.

that neighborhood condition is a special case for which the pair indeed has the homotopy extension property, but is all you need for now.

Krijn

Okay, got that

Balarka Sen

if you're free, apply this to $(X^n, X^{n-1})$ and tell me what $H_n(X^n, X^{n-1})$ is. otherwise, think about this.

Krijn

I was working on elliptic curves and algebraic number theory today, but I'll look at it tomorrow

Balarka Sen

sure.

Mike Miller

15:47

It's good to see math in here.

4

Balarka Sen

@Krijn motivation in case you forget about this problem : cellular homology is a baby version of spectral sequence of filtered topological spaces, and spectral sequences are used in all your algebro-number theoretic and algebro-geometric context. :P

Krijn

We're proving Mordell-Weil Theorem at the moment, which looks very boring when you know nothing about elliptic curves, and look very amazing when you know a bit about elliptic curves

Balarka Sen

@MikeMiller i hate to be negative but the amount of math in this chatroom has dropped down to 0, though.

Krijn

@BalarkaSen Oh thats some good motivation! :D

Balarka Sen

I have forgotten what Mordell-Weil says. re-enlighten me?

Mike Miller

15:49

@Krijn: Oh, that's not an easy theorem. Must be a nice class.

Rational points on an elliptic curve are a finitely generated group.

Balarka Sen

elliptic curve over \bar Q?

Krijn

@MikeMiller Well, not the whole theorem, I guess we just prove the Mordell theorem (and some handwaving on what Weil did)

Mike Miller

@Balarka: No, $\Bbb Q$.

@Krijn: It's still not easy then, is it? I don't quite remember the details.

Balarka Sen

ok, so $E(\Bbb Q)$ is finitely generated, that's all you mean, right? ok.

Krijn

@MikeMiller No definitely not. It takes two lectures (so four times 45 minutes) to prove it

Mike Miller

15:52

Ok. I must be thinking of something else.

Krijn

The first lecture showed that $E(\mathbb Q)/2E(\mathbb Q)$ was finite

Balarka Sen

very interesting. rank of an elliptic curve makes sense now.

Krijn

Although apparently the proof is much nicer if you know Galois cohomology

Balarka Sen

can you give a brief sketch of the ideas in the proof?

Mike Miller

Oh, that's what I'm probably thinking of. A friend of mine sketched a proof with Selmer groups.

Krijn

15:56

A brief sketch of what I had so far: Define another ellpitic curve $E'$ and a function $q: E' \to \mathbb{Q}/(\mathbb{Q}^*)^2$ with some nice properties

Balarka Sen

bummer, I don't understand anything in the wikipedia page on selmer groups.

Mike Miller

Me neither!

Balarka Sen

@Krijn ok.

lol

Krijn

And then we know functions $\phi$ and $\psi$ between the elliptic curves

With $\phi \circ \psi \cong [2]$ I believe

Balarka Sen

what are those functions?

Krijn

15:58

Rational maps $E \to E'$ and $E' \to E$

Which we know explicitly

Balarka Sen

yes, but are the definitions too complicated?

Krijn

Oh no not at all

Balarka Sen

ok, then what are $\phi$ and $\psi$?

Krijn

Lets see if I can find them, hold on

Got them

So we write $E: Y^2 X(X + aX + b)$

$\varphi: E \to E': (x,y) \mapsto (x + a + b/x, y - by/x^2)$

Sorry, $E': Y^2 = X(X-2aX + (a^2 - 4b))$

And $\psi$ is just this same process (going from $E'$ to $E''$)

And then $E'' \cong E$

Balarka Sen

er. you have typos there, don't you? $E : Y^2 = X(X^2 + aX + b)$, you mean?

Krijn

16:04

And somehow $\psi \circ \varphi \cong [2]$

Correct

Now, $E(\mathbb{Q}) \to E'(\mathbb{Q}) \to \mathbb{Q}^*/(\mathbb{Q}^*)^2$ is an exact sequence

Balarka Sen

hmm, alright. the first map is $\varphi$ and the second is $q$?

Krijn

Where this second arrow is the $q$ map which sends $(u,v) \to [u]$

Balarka Sen

oh, alright.

Krijn

And $(0,0) \mapsto [a^2 - 4b]$

And $O \mapsto [1]$

You can show that it is a homomorphism of groups by calculation, but thats tedious

Balarka Sen

sure, continue. ($\Bbb Q^*/(\Bbb Q^*)^2$ looks suspiciously similar to the Zariski tangent space, although likely there's no connection)

Krijn

16:07

Now the image of $q$ is finite!

And so $E'(\mathbb{Q})/\varphi(E(\mathbb{Q}))$ is finite

Balarka Sen

@Krijn yep, makes sense.

Krijn

And by the same argument you can show that $E(\mathbb{Q})/2E(\mathbb{Q})$ is finite!

(You can do the same for $\psi$ and then you use the kernel-cokernel sequence

Balarka Sen

right

Krijn

Sometimes, mathematicians should rethink about how they name their machinery: en.wikipedia.org/wiki/Cox%E2%80%93Zucker_machine

Mike Miller

No, that was intentional.

Balarka Sen

16:13

:|

@MikeMiller really?

Mike Miller

They met as graduate students snd immediately decided to write s paper together.

Krijn

That's brilliant

Balarka Sen

hahaha

Mike Miller

@Balarka: don't forget I gave you homework.

Krijn

Oh, imagine the moment where they realised they could call their discovery a machine

Balarka Sen

16:18

@MikeMiller Right, thanks for reminding.

crl

16:30

If I have a rectangle, and dtop, dbottom, dleft, dright, respectively the distances to top, ... sides. If I want to check if a point is on top of the fist diagonal, I could use the euclidean distances (squared) dleft²+dtop² < dbottom²+dright² ok, but also it seems to work with the 'norm0' and 'norm1' distances dleft+dtop < dbottom+dright or max(dleft,dtop) < max(dbottom,dright) seems to give the right answer too, is it right they all work? I though Euclidean was the only right way for this

I can't find counter-examples at least

crl

16:53

trying it jsfiddle.net/crl/2ohg3umj

user174558

@crl Hi Cyril.

crl

hi

robjohn

@crl Comparing the euclidean distances to the corners will tell you which side of the line half-way between the corners the point is on, but not which side of the diagonal of the rectangle the point is on. For that, you'd need to use $$\left(\frac{\text{dleft}}{\text{width}}\right)^2 +\left(\frac{\text{dtop}}{\text{height}}\right)^2 \lt\left(\frac{\text{dright}}{\text{width}}\right)^2 +\left(\frac{\text{dbottom}}{\text{height}}\right)^2$$

user174558

@robjohn Your LaTeX is always so perfect.

robjohn

@Jasper I practice :-)

crl

16:56

ok, thanks, is this the only way just for curiosity?

user174558

@robjohn Over here, I practise instead of practice. =)

crl

takes a lot of practice to pratise

user174558

@crl You can only ask one question per day. =)

robjohn

@crl to get the diagonal, the other metrics won't work, if that's what you're asking

crl

I have actually access to the positions (x,y) of each corner, (and the ones of the point of course) maybe there's a simpler way with this, instead of using the distances like I thought

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$Mathematics$ Mathematics