'Do you have any expert intuition on what cohomology should mean as a whole? Also do you think of things algebraically or in terms of E-M spaces?

I come from the algebra side

I work with algebras, not groups, and sadly there are no E-M spaces for them

Oh really? So things like Brauer groups?

(I hope I didn't make an ass of myself there)

Hochschild (co)homology

Oh--the later, scary part of Weibel.

that's to algebras as groups cohomology to groups

haha

That's pretty cool man. I definitely want to learn all of that stuff in due time!

Algebra is just too cool

it's easy to neglect other studies

homological algebra attains its coolest when you are able to connect it with other things

so neglecting other sujects is not a good idea :)

I'm really excited next term because I am going to be doing Riemann surfaces and learning Dobeault Cohomology! I doubt there will be that much different than singular cohomology--or is it...

I'll be giving talk to undergrads in a couple of weeks about how Hilbert's 3rd problem, which deals with cutting tetrahedra and pasting the pieces back, is solved using Hochschild cohomology

That sounds pretty awesome!

yeah — quite amazing thing :)

Hell, I was impressed when I saw the Cohomological Hilbert's Theorem ninety actually make calculating H^1(Z/nZ,A) useful!

There was a question on MO very recently (yesterday?) about the difference between Dolbealy and de Rham

heh

in fact, the most useful aspect of the whole group cohomology thing is that it allows you do do proofs by induction

You can't say that and not explain...

in the enormous majority of situations, you really only case about H^0

but dealing with it alone is too hard: so one constructs the whole cohomology and does tricks like shifting degrees and so on

Oh, oh, oh ok. I'm with you

For me the place that homological algebra (not group cohomology) has been most useful

is in commutative algebra

not that the whole cohomology is ever not useful at once

like proving A is flat if and only if A is a flat R_p-module for all p

class field thiery can be set in a purely cohomological form, and then you need the whole thing at once

yup

Yeah! I'm taking that next spring. Have you ever heard of Larry Washington?

an extraordinarily beautiful example is the proof of Serre's theorem

characterising local regular rings as those of finite global dimension

Right! That is very cool

Homological dimension of rings is fantastic

it is a canonical example of using homology as a tool to do induction

Is there any particular book you reccomend?

about?

Either straight homological or group cohomology

hilton-stammbach's

Hahaha, that's the only book I've never looked at.

maclane's, because it explains the ideas

That great?

MacLane's notation makes me sad.

H-S is pretty dry but it is impeccably correct

I feel like, being an expert, you may dislike it

Weibel's to know where to go afterwards

MacLane's?

But for purely pedagogical reasons I have liked Rotman's book the best. I like how he covers everything, including the basics of sheaf cohomology

I love that book :)

Yeah, MacLane's notation

Rotman's very good too

the one book I have with me all the time is cartan-eilenberg

Never looked at that first. I guess historically it's a must.

people keep publishing papers where they discover things that are problems in that book...

That have previously been solved, or the problems are just that hard?

no, it is just that those people should have read C-E :)

by problems I mean exercises!

Haha, that must be embarrassing for them.

for "classical" homological algebra there has not been a lot of improvements in exposition since it was written, really

for the newer kind, there is Gelfand-Manin, for example

which is, well, a Russian book

but there is yet no really good text

That book is perpetually checked out of the library (damn professors). I've heard that it's extremely clear and concise. Is it worth getting?

it is two books, actually

Methods

I think

in theory, two editions of the same one

is that one I am talking about

but the symmetric difference of the contents is enormous :)

yup, that one

but it got republished as a volume in springer's encliclopedia of math

and that version is different to the original

Haha. You know what I find annoying? Being of the more algebraic persuasion I really like to get into the nitty gritty of all the arguments. My advisor for this hom. alg. course is a topologist and LOVES Weibel. But, Weibel is so handwavey at times. When I try to complain to my advisor he just goes "it's obvious! Why do you need to actually construct it!"

it i like a club

it is very annying to spell out details in written

so we handwave

people find it therefore hard to follow

but at somepoint they finally see it

and we lose all our air of mystery

Right. I just find it slightly unsettling to just appeal to a metatheorem every seven seconds

haha

Do you actually ever think in general abelian categories, or even if you are doing a problem in that genearlity do you just think in $R\text{-}\mathbf{Mod}$?

I usualluy deal only with modules

and subcategories of modules, and graded versions, and such variations

I am fond of studying concrete algebras

One last question, and then I'll let you go.

Do you think of a chain complex of modules as a literal chain complex, or as a graded module?

well, my modules are very often themselves graded

:)

Haha, fair enough.

I like to keep the homological degrees separate

but then when you have a differential graded algebra

It's been very nice talking to you man! Thanks for answering all my questions!

(go on_

the concepts of module and complex end up being one

so you stop worrying, really :)

Haha, I will. Thanks mom :P

Good night!

heh

good night :)

@MarianoSuárezAlvarez

Ah you are here!!

@MarianoSuárezAlvarez Can I ask you something?

Suppose you have a ring $R$ such that

hello

$A \subsetneqq R \subsetneqq \operatorname{Frac}(R)$

Let $g$ be the inclusion map from $A$ into $R$

@MarianoSuárezAlvarez Hi

Let $S$ be the multiplicative set of all units in $A$

@MarianoSuárezAlvarez I an trying to see why any element in $R$ is of the form $g(a)g(s)^{-1}$ for some $a\in A$, $s \in S$

that is not true

oh crap

for example, take $A=\mathbb Z$, $R=\mathbb Q[\sqrt2]$

(so that $R=Frac(R)$)

yes

@MarianoSuárezAlvarez But I have the condition that $R \subsetneqq \operatorname{Frac}(R)$

Well, take $R=\mathbb Q[\pi]$ :)

ok

wait but still $R = Frac R$?

no

$1/\pi$ is not in $\mathbb Q[\pi]$

(because $\pi$ is not algebraic)

ah ok

whereas we had $1/\sqrt{2} = \sqrt{2}/2$

ah ok

so

That's why $Q[\sqrt2]=Q(\sqrt2)$

yea

@MarianoSuárezAlvarez :math.stackexchange.com/questions/137876/…

well, that's not why, only part of the reason :)

I am trying to show that $R \cong S^{-1}A$

So that's why what I asked you cropped up

@MarianoSuárezAlvarez Now I don't even know if that's true the isomorphism

/me looks

if $A\subseteq R\subseteq F(A)$ and you want $R$ to be $S^{-1}A$, then you need to take $S$ to be the set of elements of $A$ which are invertible in $R$, no?

if you localize A at its set of units, you just get A

ah crap

wait but why is it that if I do what I picked for $S$, I get that $S^{-1}A \cong A$?

if you localize at a set of invertible elements, you do not change the ring

pffff

localization constructs a ring where the elements at which you localize are invertible: if they already are, then nothing happens

yes because then $a/s$

would be just $as^{-1}$

yes

which is just multiplying two elements in the ring

face palm.............

ahhhhh

every fraction is equivalent to exactly one of the form $a/1$

@MarianoSuárezAlvarez :P :P :P

@MarianoSuárezAlvarez so

we take $S$ to be the set of all $a \in A$ that are invertible in $R$

then $S^{-1}A\subseteq R$

and they have the same set of units

you probably want now to show they are actually the same ring

yes

but then I want to apply this corollary in AM

$g(s)$ is a unit for every $s \in S$

$g$ is the inclusion map from $A$ into $R$

So

The map $g$ is injective so if $g(a)$ is zero in particular there is $1\in S$ such that $1\cdot a = 0$

@MarianoSuárezAlvarez I still can't see why every element in $R$

is of the form $g(a)g(s)^{-1}$

ok: you have an inclusion $g:A\to R$ and every element of $S$ is mapped to an invertible element of $R$; the general theory gives you an injection $S^{-1}A\to R$

yes

so you I need the surjective part

since everything is happening inside $Frac(A)$, w can simply write $S^{-1}A\subseteq R$.

pick $r\in R$.

Since $R\subseteq F(A)$, there are $a$, $b\in A$ such that $r=a/b$.

hmm

yes

@MarianoSuárezAlvarez ??

/me thinks

@MarianoSuárezAlvarez If $b \in S$ we are done no?

yes

Because then we can write $1/b = g(s)^{-1}$

So the problem now is if $b \notin S$

hmmmmmmmmm

Did you guys just start a localization party?

@AlexYoucis Well yes

@AlexYoucis The problem is this

I have an inclusion map $g:A \to R$

All this happening inside the fraction field of $A$

I am trying to understand why every element in $R$ is of the form $g(a)g(s)^{-1}$

for some $a \in A$ and $s\in S$

$S$ is the multiplicative set of all elements of $A$ that are units in $R$

Is that always true/

yes that's the problem now; I am trying to know that

I don't know if that's always true (but it is 3 am). What did Mariano say?

he himself is thinking

@AlexYoucis math.stackexchange.com/questions/137876/…

that's where all this stems from

Let $A=\mathbb Z$ and $R=\mathbb Z[\tfrac23]$

@MarianoSuárezAlvarez You're saying this is not true?

What are the elements of $A$ which are invertible in $R$?

No, I am saying, if this is to be true, it should be true in this case

What is $R$?

@MarianoSuárezAlvarez that is the set of all elements $a + \frac{2}{3}b$ for integers $a,b$?

no

hmmm

it is the set of all polynomials with integer coefficients evaluated at $\tfrac23$

ah ok

the units

well

doing this example should show what to do in the general case :)

wait

the units are just $\pm 1$?

elements of $\Bbb{Z}$ that are units in $R$

you are claiming that my $R$ is the localization of $\mathbb Z$ at the set of elements of $\mathbb Z$ which are invertible in $R$

yes

well: find S in this case

I can't seem to even get say $1/2$ in S

write down a general element of my R, multiply it by an integer n and ask yourself: what can you say about n if the product is $1$?

well

a general element is just

$a_0 + a_1(2/3) + \ldots a_k(2/3)^k$

(use k not n :) )

sorry man

heh

ok

@MarianoSuárezAlvarez That s not possible

because

@MarianoSuárezAlvarez This is messed up

I would need that $n(a_03^k + 2a_13^{k-1} + \ldots 2^ka_k) = 3^k$

ok

and...

so we need $na_0$ to be odd

what does that tell you about $n$?

or if we look at it another way

you are looking at the tree: look at the forest!

I need

$na_k$ to be a multiple of $3$????

closer

you've found an integer number $w$ such that $n\cdot w=3^k$

what does that tell you about $n$?

ah

$n$ must be a power of $3$

pfffff

indeed

so the elements of $S$, in this case, are powers of $3$

yes

$\{1,3,3^2, \ldots \}$

Which powers of $3$?

up to $k$?

wait

don't make wild guesses

hmmm

oh wait

assuming that $a_0$ is non zero

we have that $S$ is negative powers of $3$?

hm?\

S is contained in A

which is Z

pfff

1 belongs to S

that's obvious

yes

because 1 is invertible in R

is 3 in S?

um

no

my question is equivalent to: do there exist $k\geq0$ and $a_0,\dots,a_k$ such that $3(a_03^k + 2a_13^{k-1} + \cdots+ 2^ka_k) = 3^k$ ?

can you take $k=0$?

and then

we would have $a_0 = 3^{k-1}$

what is k now?

ah

there we go

so, can we have $k=0$?

no

so k is at least 1

good

can we have $k=1$?

but then

we want to solve $3a_0 + 2a_1 = 3$

no, not that

look at the equation I wrote

@MarianoSuárezAlvarez We want to know if we can do $3(a_0 + 2a_1) = 3$

yes

sorry this should be correct

can we?

well $a_0 = 1, a_1 = 0$?

oj, no

sorry

huhuhuh??

the equation is $3(3a_0+2a_1)=3$.

one has to be careful! :D

ah yes

bahhh

yes

but then still we can do $a_0 = 1, a_1 = -1$

yes

so $3\in S$

is in the set S

Hey Mariano

so what is $S$?

powers of $3$?

are you asking or telling me?

@MarianoSuárezAlvarez telling

then yes :)

but that's what I said like here:

Ok so now

you wrote that, but you did not know how to prove it

hahahahahahahaahah :D :D :D :D

I have to go now

my friend's mum is shouting at me to go

Will you be around this evening??

also, at that point we only knew that $S\subseteq\{1,3,3^2,\dots\}

but not the equality

or later today?

probably yes

Ok see you soon

:)

heading to my friend's dad's place

@MarianoSuárezAlvarez maps.google.com.au/…

ok

we are like 12 hours apart :)

Spam flags please

Hi! Is it true that a proper metric space (which means that any closed ball is compact) has to be separable?

(done)

Anybody? It is a yes-no question.

Hi @savi

Hi, @RajeshD

What about BV ?

@RajeshD BV?

en.wikipedia.org/wiki/… ?

functions of bounded variation I do not know they form a complete metric space but they are not separable

Wiki says it is a Banach space, so it is complete.

Nonetheless, completeness doesn't imply properness.

Do you know it is proper?

I do not know

I am afraid

Don't worry, thanks anyway.

Ha! So awesome: Now he puts exercises into his answers.

@savick01 Yes, of course, $X = \bigcup_{N=1}^\infty \bar{B}_N(x_0)$ and compact metric spaces are separable. Countable unions of countable sets are countable, too.

@RajeshD No infinite-dimensional normed space is proper.

Bah. Didier downvoted my answer, there goes my pure streak :)

@tb Of course, thanks!

and I know for the max -case $0\leq z \leq \sqrt{2}$ and $0\leq r \leq 2^{0.25}$ if $z,r\in\mathbb R$.

$$\int_0^{2\pi} \int_0^{2^{0.25}} \int_{r^2}^{\sqrt{2-r^2}} rz dz dr d\rho,$$ just this easy?

\mbox{} really is one of those things that gets Michael's goat...

been working on an answer. Didn't even notice I was dropped here.

Now, I've got to get ready to go to the Renaissance Faire shortly.

How is everyone here?

@JM I used Mathematica to plot a vector field to illustrate the solution to an ODE. The illustration came out looking nice; I don't know if the explanation helped at all.

@JM Which Michael? and why do people care what someone else does in LaTeX to get a particular output?

afk for a while. I will try to check in before we leave for the day.

@robjohn There's only one Michael who's overly obsessed with "the one true way to $\TeX$"...

@robjohn Neat. You're dressing up as?

(I think I already voted up that answer of yours with a vector field.)

@MattN Shades of Amitesh. I don't see that guy around as much.

Ahoy.

Does anybody here have access to the pdf at iopscience.iop.org/0025-5726/8/6/A01 ?

@robjohn, very nice answer!

thanks

@DylanMoreland Well. He's got at least one reason to not disappear completely: retagging questions that have been retagged incorrectly by me.

@teddy: Don't worry, I'll learn eventually. When I tagged this one I tried to stick to the tag description: Lp spaces are normed vector spaces so it seemed to fit the tag. What I didn't think of was that you want elementary things tagged accordingly. In this case this means tagging it measure-theory instead of functional-analysis.

what's the command to the disjoint union?

any one

I still don't understand integrals and why the rectangles are bound to the x axis

If I have an hyperbola that opens down and is in quadrant 1 why is the area not infinity?

if I have y = 5x-x^2 wolframalpha.com/input/?i=y%3D+5x-x%5E2

Then why is the area from 0 to 4 not infinite?

The only thing that I can think of is that the riemann sums are just rectangles of width dx and height f(x) and by definition that starts at y = 0

but I am not sure how that connects with everything else