@tb Wait, I'm confused. What's $d'_0$ and where did $h_0$ come from?

@AsafKaragila The question is: You got $d_0$ by making a choice, right? What happens if you make another choice?

So, make another choice $d_0'$ instead of $d_0$.

@tb I don't know... I fail this course?

Similarly, you got $d_1$ by making a choice to obtain a commutative diagram. Make another choice $d_1'$ consistent with $d_0'$.

Hmm, and he didn't accept my answer 8-).

@JonasTeuwen bizarre behaviour. He preferred to copy your answer and accept that :)

@AsafKaragila the question then is how are $d_0, d_1$ and $d_0', d_1'$ related?

@AsafKaragila With me so far?

@tb I'm a bit lost, but I think I get you. If I show that the choice of lifting is irrelevant to applying $Hom(-,M)$ then it will magically imply that the map between $Ef$ and $Ep$ is an ismorphism.

So, you argued that you can find $d_0$ and $d_1$ such that $p_0 = f_0 d_0$ and $d_0p_1 = f_1 d_1$, right?

@tb Yes...

@AsafKaragila It will not quite be irrelevant to applying $\operatorname{Hom}{({-},M)}$ (the homotopies will still be there, but it will be irrelevant after passing to $Ef$ and $Ep$).

@tb Indeed I did.

So suppose you've another pair of maps $d_0'$, $d_1'$ satisfying the two commutativity relations.

Hmmm... I see what you're saying.

I can use distributivity and have that $0=f_0(d_0-d_0')$.

Exactly.

So $d_0-d_0'$ goes into the kernel of $f_0$ namely the image of $f_1$.

Which would then imply that $f_1(d_1-d_1')=0$ as well, which goes to a kernel of an injective map and therefore equals zero?

@AsafKaragila Yes, and $f_1$ is a map onto the image of $f_1$. So you can apply freeness of $P_0$ to find a lift $h : P_0 \to F_1$ such that $f_1 h = d_0 - d_0'$.

Ahhhh. Methinks.

Let me draw the first diagram.... again.

Now observe that $f_1 (d_1 - d_1') = (d_0 - d_0') p_1 = f_1 h p_1$.

Since $f_1$ is injective, this tells you that $d_1 - d_1' = h p_1$.

(here we're using that we're dealing with abelian groups, otherwise we'd have to iterate the process of lifting, etc).

Wait, I have reached the point where $d_0-d_0'$ is mapped into the kernel of $f_0$.

Now I can use the fact that $f$ is injective, apply $p$ from the right and $f^{-1}$ from the left and attain $d_1-d_1'$.

Yes, this sounds right. I prefer to do what I said, though.

Right, it makes more sense!

Now we can think about the map $d^\ast: Ep \to Ef$.

Note that $Ep = P_0 / \operatorname{im}{p}$ and $Ef = F_0 / \operatorname{im}{f}$.

What about verifying that $d_0^\ast$ and $(d_0')^\ast$ doesn't muddle the final result?

Well, apparently Asaf is getting more proofs done in this 8 hours than he did for the last 8 hours! Good, keep it going! ;-)

@KannappanSampath So... if I stay awake for three more days I'll finish my thesis in a week?

@AsafKaragila Looks like it! : )

Given $e,e' \in P_0$ such that $e - e' = p(x)$ (so they give the same class in $Ep$), we have that $d_0(e-e') = d_0p(x) = fd_1(x)$, so it makes sense to define $d^\ast[e] = [d_0 e]$, right?

@AsafKaragila we'll be right there, a few more minutes :)

This is tiresome. At least there's a huge prize at the end... set theory!

@tb Wait. How do you define $Ep$?

$Ep = P_0 / \operatorname{im}{p}$.

Oh, this is not what I defined as $Ep$... now wonder I was confused!

Sorry, I haven't applied $\operatorname{Hom}{({-},M)}$ (I forgot about that)

So what do we do?

@AsafKaragila Wait, we apply $\operatorname{Hom}{(-,M)}$ to $p: P_1 \to P_0$ to get a complex $p^\ast: \operatorname{Hom}(P_0, M) \to \operatorname{Hom}(P_1,M)$ and $Ep = \operatorname{Hom}{(P_1,M)}/\operatorname{im}{p^\ast}$, right?

Could someone look at trial 3 (and perhaps t2) here?

I get $-2$ but it should on positive $z$ -axis, something odd happening.

@tb Yeah.

And the maps $d_0: P_0 \to F_0$ and $d_1: P_1 \to F_1$ give us maps $d_0^\ast: \operatorname{Hom}{(F_0,M)} \to \operatorname{Hom}{(P_0,M)}$, and similarly for $d_1$ and the $'$ versions.

Yes.

Are we going to repeated the same process as before?

Then we'll have $h^\ast\colon Hom(F_1,M)\to Hom(P_0,M)$?

Sorry noise, stupid mistake...fixing.

A small modification of what I said above gives you that you get a well-defined map $d^\ast: Ef \to Ep$.

@AsafKaragila yes, exactly.

@hhh These images are not terribly easy to read. I add my voice to the suggestion of learning $\LaTeX$.

So we want to compare what $d_0$ and $d_0'$ do on the level of $Ef$.

That's what that map $h^\ast$ is good for.

So take $\varphi \in \operatorname{Hom}{(F_1,M)}$.

Right.

Then $d_{1}^\ast\varphi = \varphi d_1 = \varphi (d_1' + hp_1) = (d_1')^\ast \varphi + p_1^\ast h^\ast \varphi$.

So $d_1^\ast \varphi$ and $(d_1')^\ast\varphi$ differ by something in the image of $p_1^\ast$, so they give rise to the same class in $Ep$.

And therefore the function naturally defined from $Ef$ to $Ep$ is injective!

No, therefore the function from $Ef$ to $Ep$ does not depend on the choice of $d_0$ and $d_1$.

Since $d_0'$ and $d_1'$ give rise to the same map, by what I just said.

Hi @Matt

Hi Kannappan!

Ah.

The same argument applied to $c_0$ and $c_1$ shows that the same holds for those maps.

So now the isomorphism is going to be unique too?

@robjohn Is this Gradient correct? I may be doing the substitution in the wrong spot (look gradient for $x$ has only scalars...looks worrying, I think I did substitution too early)...

And similarly for $c_0 d_0$ and $c_1d_1$.

@tb You mean $c_i$ and $c_i'$?

So now I obtain two maps from $Ep$ to $Ef$ and back, which commute and are very well defined?

@robjohn ERR around second and third line... in the substitutino $\bar{x}=(0,0,-2)$ (nevermind the wrong point but is the idea correct?)

Executor needed

@AsafKaragila No, I mean the composition $P_i \xrightarrow{d_i} F_i \xrightarrow{c_i} P_i$

which gives rise to a map $Ep \to Ep$.

which must be the identity since the identity $P_i \to P_i$ also satisfies the commutativity requirement.

All this shows that the maps $Ef \to Ep$ and $Ep \to Ef$ do not depend on $c_i$, $d_i$ and are mutually inverse.

@robjohn for this kind of writing with mistakes, original writing is much more revealing than the LaTex...I will do LaTex on the final document or when things are more clear... until then I need help to find the error spots because I cannot find them...

I see.

@tb Still NARA. At first I thought this one involved Hawaiian earrings but it doesn't fit. You seem to have a lot of the thing starting with f in your life. Maybe you need to be more pacifist. Have you considered trying these? I'm taking these and it works like a charm.

Thank you @Matt

@robjohn basically the problem is to calculate the $\nabla$ in some point $\bar{x}_0$, I am not sure whether I should do the substitution when doing $$\partial_{x} f(\bar{x},t)_{\bar{x}=(0,0,1)},$$ or only $$\partial_{x} f(\bar{x},t)_{x=1},$$ ideas?

@MattN It does very much have to do with Hawaiian earrings...

(but the timeline is probably completely different from what you expect)

$Oh^2$.

@AsafKaragila The lemma we just discussed is the fact that given two resolutions $P_\bullet$ and $F_\bullet$ you get a map $Ef \to Ep$ which doesn't depend on anything but $A$, the two resolutions and $M$. This gives you a unique abelian group associated to $A$ and $M$ and now you have to think about what happens if you change $M$ to $M'$, but this is pretty much the same as what we just discussed. I think you can take it from here.

@robjohn moved this detail to new q here with LaTex. Hope easy to read.

@tb No no no...

I don't need to think about what happens when we change $M$ to $M'$...

$M$ is fixed in the entire story... :-)

All the better for you!

Tell me about it!

Now let's see if I can write this whole thing we just did here.

Yeah. I think it shouldn't be too hard...

Painful, maybe, but not too hard :)

Well I have been awake and actively engaged in diagram chasing for the past 28 hours.

I'm impressed.

With very little food and very little to drink as well... :-P

My productivity starts going down after 5 hours or so until it reaches 0 after about 8 hours.

I can pull these things when I insist.

You're a machine : )

You're talking to a person that had like two good night sleeps in the past decade.

Also I am hardly ever well fed and well rested.

Someone stole one hour of precious sleep from me today.

Enough reason to be sleepy and therefore very grumpy.

Thanks.

@tb Thanks chat.stackexchange.com/transcript/message/3953468#3953468

T_T

Already so many removed's

I have left a message for Iyengar asking him to stop spamming the site with stuff life "DOWN TO EARTH" explanation of something!

I wonder if that a good thing to do in retrospect!

I'm blushing. That's more personal info than I can take.

Bye, I'll be away. I have procrastinated enough already!

Byee!

Bye all of you! Take care.

And you!

@KannappanSampath Have fun!

I have to go take a nap. BBL

Hah, and now I'm downvoted on the bounty thing...?

@JonasTeuwen really?

@robjohn Yes!

@JonasTeuwen Gimme link, I'll compensate.

@robjohn Here.

@robjohn Or is this rigth of calculating the nabla?

Relating to this.

I need coffee. Either that or a coffin.

@MattN start with coffee

@JonasTeuwen wow, that is messed up. You got the bounty, chessmath got the acceptance. Then there seems to be a bit of smack-talk going on.

@hhh hang on

<--- I think this is non-sense, hence the new paper

<--- I get scalars for the gradient...

@hhh what part of "hang on" don't you understand?

@JonasTeuwen Do you think the iPad is worth buying?

@hhh: Can you translate the original problem? Is it asking you to solve for $x,y,z$ such that the integral is $\log 2$?

Oh my. Brian hasn't visited the site for 2 days. I hope he's alright. Last time he was here he complained about a cold.

FREEEEEDOOOOOMMMMMM!!!!!!!!!

Congratulations.

I'll be saying this in about 6 months from now, hopefully.

@Skullpatrol If you can easily afford it: Yes.

@robjohn Yes... I don't understand why he is in a defensive mode.

I was not meaning to attack him.

Asaf just left the room!

Gone to the bar?

To get so drunk his hair turns green?

But he never leaves!

Does this mean it's the end of the world?

Haha, @Asaf finally gets his freedom and then the world ends.

: )

Yay, @Srivatsan. Long time no see.

Hey, Matt

How are you?

Long time indeed. How are you?

Thanks, I'm alright.

I am good too.

@tb Sleep well. I sent you something to read for when you get up.

@Srivatsan Are you still in India?

Yes, still in India.

We don't have a tag additive-combinatorics.

Ah, surprising

When did you notice this?

Just now.

@JonasTeuwen some people seem to need to feel argumentative. For example, Didier seems a bit deprecating in this thread.

Do you think we should create one?

@MattN what is additive combinatorics?

: D

@robjohn It's the study of properties subsets of abelian groups.

@robjohn "...and the work is cut in half." :)

At least that's what I've gathered so far.

@Srivatsan I didn't see you there :-)

hi

:)

That's a nice response to DP's first comment

@Srivatsan If the other answer is so great, go ahead and post it. I don't want to copy that comment.

@robjohn other answer?

@Srivatsan Thanks :-) I had actually written out complete answer, but then saw that it was a homework, so I cut the answer back to a hint. So I never saw ADF's comment.

@robjohn Sorry, I don't understand what you are talking about. I meant that I liked your response to his comment.

@Srivatsan ADF's comment.

Ah, I see.

@Srivatsan I know, I was talking about why I don't want to make my answer a copy of ADF's comment as Didier seems to think I should.

@Srivatsan: how are things there? It is raining here.

Things are going ok.

Since you mentioned weather: it's sunny here, just like any other day of the year in Chennai.

: )

@Srivatsan I take it you are visiting family. Tell me if I am getting too personal :-)

Does LA get a lot of precipitation? Definitely not like Seattle, right?

@MattN I still have to write my thesis.

@robjohn viiiiiiisiting, yes. :)

Hi Srivatsan, robjohn.

hi Asaf

@Srivatsan No, this is treasured weather here. We need all the rain we can get.

Hi @Srivatsan

woooo party time

@AsafKaragila Good morning :-)

@anon I'm not late for it, for a change. Yay.

@anon Ooh, I'll break out a new pencil! Paaartay!

@KannappanSampath you've returned. :-)

@Srivatsan LA is desert/mediterranean. Seattle is almost rain forest

@robjohn I had not heard this idiom (?) before...

@KannappanSampath hi Kannappan.

@robjohn Yes, I have! I am too distracted to keep SE away!

@KannappanSampath BTW, to answer your question, hopefully soon. I am not sure about the details yet, I will keep you posted as and when I do :)

@Srivatsan :-)

@Srivatsan breaking out something? It means opening a new or rarely used item.

that's not an idiom, just a metaphor

@robjohn But why break out a pencil?

Perhaps I am missing some other cultural reference. :) I don't know what a pencil has to do with a party...

@Srivatsan I was showing how exciting things are for a mathematician ;-) I was making a joke

Did someone read the comments I left for Iyengar? Should I delete or something? (I mean to ask: Is it offensive?)

Ah, may be just me, but I found it a little obscure. =)

@KannappanSampath link?

@Srivatsan This one

And the previous 2 as well.

Reading this second edit... wtf.

@Ilya Can you elaborate on that. : )

@KannappanSampath Spamming? Really? I find it a bit harsh. (But I have not been around for a while, so I don't know if iyengar posted too many such questions in a short interval.)

@KannappanSampath Have you tried taking these?

Why does iyengar annoy you so much?

@Srivatsan He did not post too many of them certainly, but his questions are like "DOWN TO EARTH" approach to BSD and all of that!

Hi folks

@MattN He does not annoy me but certainly I am worried he is putting himself to unnecessary strain.

He has been told but he is stubborn, which kind of hurts me.

Why don't you just ignore him?

@KannappanSampath I left him a comment you might like.

@AsafKaragila Where, exactly? On the same thread?

@MattN But, I think he is enthusiastic and must focus his energies!

Yeah.

BTW, Kannappan, I don't find the question that wrong. I simply interpret the question as: "Tell me some easy-to-follow book that motivates cohomology well." As for "DOWN TO EARTH" (I don't know why this ought to be in caps ;)), I am just ignoring it...

Yay : ) Two upvotes! Asaf, why did you delete your answer?

@AsafKaragila Well Said. I will only hope that he heeds to the call!

@MattN I was thinking about the Pontryagin duality theorem :-D

@KannappanSampath Unlikely...

@AsafKaragila Hm...

Most indeedy.

I'm going to be away for an hour. See you folks in a bit.

@Srivatsan Well, I don't say that his questions are wrong. He says "even the basic concept of exact sequence is used widely in advanced mathematics". Is this not a reason enough to master this tool then? This is what I would say is implicit troll or some such thing!

@KannappanSampath Too many people have tried to reason with him. The only real change in his questions is better language, and less of the "Prof. Dr. Riemann Sir"'s. :)

Anyway, the point is I am not arguing with you, but just feel that it is ridiculous.

Okay. I'm awake about 31 hours by now.

I'm gonna take a nap.

@AsafKaragila Sleep well. : )

@KannappanSampath Yes. What I wanted to say was: If I take this question in isolation (not very justified, given the history behind the user), I feel it is an acceptable question. However, it is true that the user is not very willing to do the mathematics the way others do.

Well, let's forget and get to real things: How are you doing? (<--Change of topic)

Worse, he think that the mathematics community is out to get him, and deliberately obscure the simplest truths. :)

ok, change of topic.

I am doing ok. How are you?

@Srivatsan I am fine. : )

How are you guys faring with commutative algebra?

@Srivatsan Yes. We have almost finished chapter 1 of AM.

I rephrased it to sound nicer ;)

AM? Adkins? No, that is Adkins and Weintraub.

Then, we started studying the multiplicative systems and probably be led to Localizations.

AM=Atiyah-McDonald.

Oh yes. Should've guessed. I thought you were following something else?

@Srivatsan That was not bad either. :-)

@Srivatsan Yes, but Matt has an Exam to write this July which will test Matt from AM. So, we started it early.

@KannappanSampath Not bad, as in? Are you doing both? Or you changed course?

Awesome.

Does Martin Sleziak show up in chat these days?

@Srivatsan Almost never. : (

One of these days, I was solving some exercise in Artin. I came across a rather curious question. Can you look at it and tell me if it's a well-known problem?

Hi @Kannappan @Sri

Sure. I will.

@KannappanSampath The question is on modular Pythagorean identities: How many solutions does $x^2 + y^2 = z^2 \mod p$ have, where $p$ is an odd prime?

Hi @RajeshD

@Srivatsan Let me see. Where is this exercise in Artin?

This is not super-crucial or anything" It's just one approach I am pursuing for the problem.

Oh, so you are approaching a problem which requires to you answer this question. Is it?

I am not sure what its answer is! But, I can work it out.

Hi @MarianoSuárezAlvarez

@Srivatsan Here?

@Srivatsan: math.stackexchange.com/questions/115893/…

Off I go!

@robjohn Oh well, so some people are just hard to get along with :-).

Hi @Jonas

@RajeshD Hi.

@Jonas : please have a look at this and settle the issue

The question does not make any sense to me. Convergence in what norm?

In $L^2$ norm convergence of a Fourier series is easy. Pointwise for general $L^p$ functions is hard.

@Jonas the issue is about what the author means by a sentence.....the OP is not getting into such technical issue at that detail...the book he is referring to is written for non mathematicians

Have a look at page 54 on the book he linked, if possible

it is available there

I think he is talking about pointwise convergence

Okay, I have typed up something.

@Jonas An integral definition. I need some help. Will you please help me?