Mathematics - 2015-11-12 (page 3 of 4)

user174558

13:00

I know how to tell Lies.

Soham Chowdhury

no, no, we're all mathematicians here (leastways, that's what we tell ourselves)
reducing stuff to previously known theorems (e.g. the fire engine joke) is what we do.

Huy

I don't know what naive means

Soham Chowdhury

How Sophustic of you.

Huy

I know some Lie theory

Soham Chowdhury

No special meaning, @Huy.

Just meaning basic stuff. No Killing forms/semisimplicity and all that advanced stuff.

Huy

13:01

I need to work on my algebra though, don't remember much from exact sequences and I need them now

Soham Chowdhury

he only works with SU(n), SO(n), O(n), Sp(N) and the like.

Huy

SL

because almost everything is linear

Soham Chowdhury

matrix groups only. which is why it only requires you to know single-variable calc and linear algebra.

yeah, that too.

Balarka Sen

I have answered your question, @Soham.

Huy

well to prove some stuff you'll also need some topology and algebra

Soham Chowdhury

13:03

yeah, there is a bit of topology towards the very end, but he does it himself.

user174558

Is @SohamChowdhury going to be a mathematician?

user174558

I now know that @BalarkaSen does not know what Diwali is, lol.

Soham Chowdhury

he tells himself that, but in all probability a ceiling fan is going to fall on him and crush him before that happens.

Balarka Sen

A Lie group is just a group object in the category of smooth manifolds.

There, now you know everything about Lie theory.

Go study higher topos theory from Lurie.

user174558

I sense that Soham is quite sophisticated in the Eng lang.

Soham Chowdhury

13:05

A monad is just a monoid in the category of endofunctors.

There, now you know Haskell.

user174558

And Haskell is just a lang. There you know lang.

Balarka Sen

Yeah, now go study homotopy type theory from the HoTT book.

I love doing this.

Soham Chowdhury

Balarka is either very happy today or very angry.
a) birthday? b) broke up

Balarka Sen

Neither.

Soham Chowdhury

I can't logic dammit

user174558

13:06

I don't think Balarka has a gf.

Mats Granvik

@BalarkaSen Is it known what the lower bound for the spacings between Riemann zeta zeros is?

Balarka Sen

I am confirming that.

Soham Chowdhury

tbh I thought he was a girl when I saw his name for the first time.

Balarka Sen

That said, (b) is vacuously false.

Soham Chowdhury

why did you just rewrite my proof in the room, @B?

user174558

13:08

Maybe Soham or Balarka is Ramanujan reborn.

Soham Chowdhury

and you're the banana I ate two days back, reborn

seems legit, yes.

user174558

I expect another flag, lol

Soham Chowdhury

@Jasper, you might find this interesting:

iwriteonbananas

@SohamChowdhury Can you give me the banana peel?

Balarka Sen

@MatsGranvik Hm. I don't know, but I think so. Why don't you have a look at Iwaniec-Kowalski? I think they have a chapter on survey of gaps between zeta zeros.

lol @iwriteonbananas

lololol

@SohamChowdhury Because I can't make head or tails of your mucky proof.

You're confusing epsilons and deltas. I did too, until I knew the topological definition. It's just the open sets preimage is open definition reformulated.

Soham Chowdhury

13:09

nah. I know what's supposed to go where.

the sad thing is that I couldn't go back and edit.

Huy

how can eps and delt be confusing

Soham Chowdhury

@BalarkaSen they're isomorphic. I can't either, which is why it took me so long to realise this. :P

Huy

it's always the same way around

Soham Chowdhury

I used epsilon instead of delta and vice versa because I suck at typing LaTeX and thinking at the same time.

Balarka Sen

@Soham Here's a follow-up : prove that if a function preserves converging sequences, then it's continuous. Namely, the converse.

user174558

13:11

Just use alpha and beta, lol

Huy

@BalarkaSen hey, that's the Kowalski teaching at my uni!

Soham Chowdhury

and then I couldn't change it.

@BalarkaSen oh, sure.

user174558

I am going to take a nap, byes.

Balarka Sen

@Huy cool!

Soham Chowdhury

@Jasper "the Force is strong with you"

Huy

13:12

you're only allowed to say that when you've seen the Star Wars movies, @Soham

Soham Chowdhury

I have, once and all in one day

Huy

I doubt that

Balarka Sen

@SohamChowdhury I am going to give you harder continuity exercises later.

Soham Chowdhury

well, I think I saw three. not a fan or anything. there was some kind of Star Wars festival going on on TV.

Huy

hopefully.

Balarka Sen

13:15

watch Unbreakable.

Soham Chowdhury

Need I remind you?

Jul 15 at 13:48, by Balarka Sen

rock climbing is useless. everything excluding math is useless.

@BalarkaSen okay.

Balarka Sen

sure, but I have only stated that they are useless. what stops me from doing useless things?

Huy

your brain, hopefully

Soham Chowdhury

Oscar Wilde's "All art is quite useless.", etc.

Balarka Sen

@Huy How's your health?

Huy

13:17

been working again since Monday

we're doing some knot computation now in the course about manifolds

reidemeister moves, alexander polynomials etc

not quite sure yet how it links with everything else

Balarka Sen

knots are good.

Huy

what for though

Soham Chowdhury

> not quite sure yet how it links with everything else

Balarka Sen

@Huy If you care about classifying 3-manifolds, knot complements in S^3 are 3-manifolds.

Soham Chowdhury

> links

Huy

13:19

pun intended

Balarka Sen

It's a large class of 3-manifolds.

So you should also care about classifying knot complements as a special case of the whole programme.

Huy

ok

there's no classification of knots yet, right?

Balarka Sen

Nope, there is no classification upto homeomorphism of the complement yet.

@Huy Interesting fact: If $f : D^2 \to M$ is a PL map from the disc into a 3-manifold such that $f(\partial D)$ has a neighborhood homeomorphic to a open annulus, then there is an embedding $g : D^2 \to M$ such that $g(\partial D) = f(\partial D)$. If you get to know the proof of this one, let me know!

It's called Dehn's lemma, if I recall correctly.

Huy

PL map?

Balarka Sen

Piecewise Linear.

Paradox 101

13:33

Here why does $c$ approach $b-$ instead of $b$?

Huy

because you're only locally integrable on $[a,b)$

Balarka Sen

no, u r locally integrable on [a, b).

Huy

true dat

Paradox 101

@Huy a function is locally integrable only if the function is not bounded?

JeSuis

14:41

somone for "field theory" ? :)

Balarka Sen

hi @SamuelYusim

Mike Miller

@BalarkaSen I'm sorry, in what way is this true? There's an algorithm for telling if two knots are equivalent and it's existed since the 60s.

Balarka Sen

15:03

@MikeMiller Oh? I wasn't aware of that, I thought the problem of determining which knots are unknots is open too.

Apparently I am wrong. I stand corrected.

I'm certain I have heard somewhere that this is open in some sense, I probably must have misinterpreted it.

Mike Miller

People like better algorithms.

Balarka Sen

Of course, you have told me about the peripheral structure and that it's an absolute invariant for knots. Not sure why I made that statement.

Probably I was thinking of efficient algorithms.

robjohn

@skillpatrol what about it? It appears that the things shown without Pythagoras are just reproving the Pythagorean Theorem in particular instances.

Mike Miller

@Huy: We concluded that there is a classification of knots, in the sense that there's an algorithm to determine if two of them are the same.

Balarka Sen

@Huy Maybe you shouldn't ask me geometric topology questions, lest I tell you false facts, as above.

@MikeMiller Hmm, but the peripheral structure is not quite an algorithm, is it? Can you determine the subgroups of fundamental group of a knot which represents the meridian and the longitude, efficiently?

Sounds like a not-very-obvious question about presentation of groups to me.

Mike Miller

15:30

I didn't say that's the algorithm.

Boni

15:51

Does it make sense to form an exact sequence of additive and multiplicative groups?

For example, 1 --> M --> G --> A --> 0.

anon

can you explain what you mean more?

do you just mean you have an exact sequence triv->M->G->A->triv, where M is multiplicative and A is additive?

Boni

Basically, yes. I'm not sure if I can explain it. It was part of a construction of a Lie algebra in a seminar.

Mike Miller

morecontext would probably be valuable

anon

whether a group is understood additively or multiplicatively has nothing to do with its structure, it's notation that helps us keep track of where it came from. the multiplicative group of nth roots and the additive groups of integers mod n are the same group (up to isomorphism). of course, defining or writing out homomorphisms often depends on writing the groups in a sensible way to begin with.

Boni

I'll have a go. So, we started with a lattice L, with has the additive structure.

He wanted to construct a Lie algebra from it, which is the G in the SES.

So he created M to be Z/2.

And.. G was supposed to pop out.

anon

15:59

you said "construct a lie algebra G" before you said "G was supposed to pop out." your chronology seems off. anyway, how is the lie algebra supposed to be a group, how is it supposed to map into the lattice, how is Z/2Z (which is technically additive in that notation but w/e) supposed to map into the lie algebra?

maybe I'd be able to guess what's being talked about if I knew more lie theory. (presumably lie algebras are related to lattices via adjoint reps and roots or weights or whatever.)

Boni

Presumably, the commutator will give a Lie bracket on the group, which makes it into a Lie algebra.

The interactions of M and A is what is confusing for me.

anon

do you mean a Lie group, and differentiating the commutator to get elements of the lie algebra?

Mike Miller

@anon: Lattices are discrete so that wouldn't make sense.

anon

true

then I have no idea what is meant by commutator on a group giving a lie bracket. a commutator on an associative algebra gives a lie bracket...

Mike Miller

yeah, i don't really see where a lie algebra comes here.

Boni

16:04

Would it make sense if G somehow has a vector space structure?

Mike Miller

in the sequence you wrote G definitely does not have a vector space structure, but no, i would still be confused

Boni

Yea, I give up. This makes no sense me either.

Yep. I'll just wrote up the notes and try to see if the next seminar makes things clearer.

Thanks anyway.

iwriteonbananas

16:23

Let $$0\to M'\to P_{n-1}\to P_{n-2}\to\cdots \to P_0\to M\to 0$$ be an exact sequence of R-modules where the $P_i$ are projective. Then $\operatorname{Ext}_R^{n+1}(M,N)\cong \operatorname{Ext}_R^1(M',N)$.

@MikeMiller Do you know how I might prove that? I tried extracting all the SES's I could from the sequence and considering the correspoding LES's in Ext.

Mike Miller

Oh, probably not. I don't even remember how these things are defined.

iwriteonbananas

Ok, shame.

Mike Miller

When you do Ext do you start with a projective resolution, hom it, take homology?

iwriteonbananas

Precisely.

Mike Miller

Ok, so what I would probably do here is try to modify that into a projective resolution. Pick a projective resolution of M' and somehow "replace M'" in that sequence with the new resolution, I think.

Then you'll canonically have the (n+1)th term of your resolution of M be the 1st or 2nd or whatever term of your resolution of M'.

iwriteonbananas

16:28

You mean you want to chuck out $M'$ and extend the sequence to a projective resolution of $M$?

Mike Miller

Something like this.

Balarka Sen

(this is what I had suggested)

user174558

Is there any reason why Hatcher is so popular? I don't see any reason.

Balarka Sen

It's free.

OK, back to watching The Stalker.

iwriteonbananas

Ok, I'll play around with it a bit more...

Anthaas

16:34

Can anyone here answer a question about argumentation theory?

Mike Miller

"Ask; don't ask to ask"

Balarka Sen

@iwriteonbananas It's actually not really hard. Extend that projective resolution by augmenting with a projective resolution of M'. Ext^1(M', N) is precisely the homology of the n+1-th place of the hom-ed chain complex, which is just Ext^(n+1)(M, N).

I guess that's precisely what Mike was hinting at.

Anthaas

Given an extension of Dung's argumentation framework, and all definitions therein, let AF_1 = <A,Def> be a framework where A = { A, B, C, D, E, F, G, H } and Def = {A def G, D def C, D def E, C def F, F def C, E def F, F def E} where def is the binary defeats relation. The set {A, B, D, F, H} is given as the unique preferred and stable extension of AF_1. It is also given as a grounded extension, and I don't understand how/why. Why can't {A}, {D}, {B}, or {H} be grounded extensions?

iwriteonbananas

@BalarkaSen Hmm, yeah. Can you be a bit more precise about "augmenting with a projective resolution" ?

Mike Miller

16:49

@iwriteonbananas: Literally replace $M'$ with its projecfive resolution $P_n'$. The map $P_0' \to P_n$ is obtained by composing with the maps to and from $M$.

Balarka Sen

^

iwriteonbananas

Ok, that makes sense.

Balarka Sen

And it's not hard to see it's a chain complex.

iwriteonbananas

True, true.

Well, it needs to be exact.

But I guess that's clear.

Balarka Sen

right, whoops. I guess you have already verified it though.

iwriteonbananas

16:54

Just to be sure I'm not being dumb: The map $P_n'\to P_{n-1}$ is the composition $P_n'\to P_n'/{\operatorname{Im} p_{n+1}'}=M'\hookrightarrow P_{n-1}$, right?

Damn it, I'm getting kicked out of the library. Brb

and thanks for the help.

Remember me

@Balarka I did not know you liked watching TV series. I thought you hated everything nonmath.

Balarka Sen

The Stalker is a philosophical movie by Andrei Tarkovsky.

I'm putting you back on ignore for not knowing this.

Remember me

Sorry

I wont say anything about it

@Balarka I thought you meant the tv series... Sorry for that

Mats Granvik

17:14

Is "infinitely often" more seldom than "always"?

Mike Miller

@Balarka: Don't be mean.

Balarka Sen

@MikeMiller That was a joke. But I am really ignoring him because he keeps googling every question I ask him and claims he came up with it all by himself.

user174558

Maybe I should ignore him too. He claims he is good at things he has barely learnt.

Balarka Sen

It has been 4 times since he did this.

Mike Miller

Well, that's not my business. I just don't think you should be so mean.

user174558

17:17

It's settled then.

Anthaas

What is a maximal set for set inclusion?

anon

are you asking what that phrase means?

user174558

Also ignoring user=Twink.

Balarka Sen

I am done being helpful and kind with Sayan.

Anthaas

Yes

anon

17:19

any collection of sets can be partially ordered by the relation of inclusion. one takes a maximal element of this partial order. google "partially ordered set" and "maximal element." in other words, a set in the collection which is not contained properly in any other set in the collection. (which is exactly what you should expect the phrase to mean).

Anthaas

And minimal set for set inclusion would the same, but where it is not a proper superset of any other set?

anon

yes

Anthaas

My follow up question is too long to post here...

Ill split it

Given an extension of Dung's argumentation framework, and all definitions therein, let AF_1 = <A,Def> be a framework where A = { A, B, C, D, E, F, G, H } and Def = {A def G, D def C, D def E, C def F, F def C, E def F, F def E} where def is the binary defeats relation. The set {A, B, D, F, H} is given as the unique preferred and stable extension of AF_1. It is also given as a grounded extension, and I don't understand how/why. Why can't {A}, {D}, {B}, or {H} be grounded extensions?

How can {A, B, D, F, H} be both a maximal admissible extension, and a minimal complete extension?

anon

can't say I understand any of that

Anthaas

I could explain the terms?

Jessy Cat

17:37

1

Q: Use definition of Lebesgue integral for nonnegative functions to show restriction of a Lebesgue integral to a subset of a measurable set

I am trying to show the following: Let $f$ be a measurable function on a set $E$, and let $A$ be a measurable subset of $E$. Then, using the definition of the Lebesgue Integral of nonnegative functions (given below), prove that $\int_{A}f=\int_{E}\chi_{A}f$. Definition of the Lebesgue In...

Anthaas

0

Q: Dung's Argumentation Framework

Given an extension of Dung's argumentation framework, and all definitions therein, let $$AF_1 = <A,Def>$$ be a framework where $$A = \{ A, B, C, D, E, F, G, H \}$$ and $$Def = \{A def G, D def C, D def E, C def F, F def C, E def F, F def E\}$$ where $def$ is the binary defeats relation. The set $...

logic set-theory artificial-intelligence

Anthony

17:56

Why is the irreducible polynomial of some element $\beta$ over some field of polynomials $F(\alpha)$ divisible by the irreducible polynomial of $\beta$ over $F$?

evinda

Hello!!!

Anthony

yo

user174558

18:20

@evinda Aha!

evinda

@Jasper :)

How are you? @Jasper

user174558

So so.

evinda

Any news? @Jasper

user174558

Nope.

Karim Mansour

19:28

hi @BalarkaSen

Balarka Sen

hello

Karim Mansour

I was solving this question that might interest you

show that $R^{\omega}$ in the box topology has infinitely many components

Balarka Sen

you mean infinitely many connected components?

Karim Mansour

yeah

Balarka Sen

sure.

Karim Mansour

19:31

just a sec @BalarkaSen making caffe and will come solve this problem

Balarka Sen

ok. i'm busy right now though, so if you have a complete solution, ping me with it and i'll check.

Karim Mansour

alright

bolbteppa

20:20

" The point about homology is that it measures defects/obstructions. Suppose I have a map of vector spaces, $f:V\rightarrow W$. What is the obstruction that prevents that being an isomorphism? There is the kernel of $f$, stopping it being injective and the cokernel of $f$ stopping it being surjective. Homology measures defects, and the Snake lemma tells you how those defects are related, it tells you that if you were looking at n-dim holes, you next need to take into account the n-1 dim holes."

https://www.physicsforums.com/threads/exact-sequences.164213/
Why aren't math books written like this :\

Balarka Sen

@bolbteppa Because it's not literally true.

I'd'nt call that "the point of homology". A homology is a concise invariant which comes up from piece-wise linear structure on your topological space.

Tobias Kildetoft

@bolbteppa Because the homology would not give you the kernel and cokernel

bolbteppa

?

Balarka Sen

But there is a major transition in this - computing homology of a chain complex. This is not an easy idea, and there is no easy way to explain it.

bolbteppa

This is homological algebra, which is basically linear obstruction theory

Tobias Kildetoft

20:24

@bolbteppa You could call it that. It would be a long way from what homological algebra really is though

Balarka Sen

I don't know what a linear obstruction theory is.

And homology measures defect, IMHO, is not a good way to put it.

Tobias Kildetoft

homology "measures" how far a complex is from being exact

Balarka Sen

Sure. But that interpretation gives me no idea whatsoever why it's an interesting invariant.

bolbteppa

In other words, it's measuring an obstruction to the solvability of a system of linear equations, just as cohomology measures the lack of solvability to a linear operator right?

Tobias Kildetoft

@bolbteppa Not really, no

you can probably make it do that if you define everything in the right way. But why would you want to?

homological algebra does a bunch of things that we can't just do in an easier way

Balarka Sen

20:29

@bolbteppa A much better way to put homology is the following : for smooth manifolds, you _could_ say that homology is the theory obtained from letting smooth submanifolds to be your "cycles" and cobordisms (<-- this has a very physical interpretation, all that stringy stuff, which I am sure you are familiar with) between the submanifolds to be your equivalence relations.
The $k$-th homology group is then "roughly" the group made up of cobordism classes of smooth $k$ dimensional submanifolds of your ambient manifold.

bolbteppa

"The easiest way for me to understand de Rham cohomology is by the way of explaining why certain differential equations do not have a solution. "

Balarka Sen

I find that much better than the naive "$k$-dimensional holes" explanation.

bolbteppa

quora.com/…

Balarka Sen

I don't know in what sense homology measures obstruction to the solvability of system of linear equations. Your interpretation of cohomology about when a differential equation has a solution is very close to being true, though.

iwriteonbananas

@BalarkaSen That's pretty cool.

Balarka Sen

20:32

@iwriteonbananas This is just bordism homology.

iwriteonbananas

@bolbteppa Only a physicist would say that.

Balarka Sen

We've discussed it before.

iwriteonbananas

Yeah, I remember

user174558

I like bananas.

user174558

Any recommendation for an elementary NT book.

Balarka Sen

20:35

Hardy-Wright.

Anthaas

0

Q: Dung's Argumentation Framework

Given an extension of Dung's argumentation framework, and all definitions therein, let $$AF_1 = \langle A,Def \rangle$$ be a framework where $$A = \{ A, B, C, D, E, F, G, H \}$$ and $$Def = \{A def G, D def C, D def E, C def F, F def C, E def F, F def E\}$$ where $def$ is the binary defeats relat...

logic artificial-intelligence

user174558

@iwriteonbananas Why did you delete?

user174558

@BalarkaSen Hardy is too old for me, lol.

Balarka Sen

Niven-Montgomery-Zuckermann

bolbteppa

@Jasper there is a book by Flath recommended on MathOverflow which is the nicest NT book I've ever looked in

user174558

20:36

I know about Ireland's book and also Baker's new book.

Balarka Sen

Ireland-Rosen is not elementary NT.

user174558

Yeah, right.

bolbteppa

It's different to all the other books in that it actually explains some things :p

Balarka Sen

@bolbteppa Have you seen the message I pinged you above? I think it might be more upto your line.

user174558

@BalarkaSen I have decided to use Jacobson for algebra, finally.

bolbteppa

20:37

@BalarkaSen trying to find a good explanation on linking this stuff to systems of equations

Balarka Sen

good luck with that, @bolbteppa.

user174558

trillia.com only has a few texts. The project doesn't seem to be expanding.

bolbteppa

I'll just throw this math.stackexchange.com/a/117992/82615 out there and say wtf

(Amost unrelated, but very cool)

Balarka Sen

natural context for homology is the PL category, so I doubt you'll find anything related to system of differential equations, which is the smooth category.

natural home for homology - PL category
natural home for cohomology - smooth category

user174558

Natural home for me - MSE chat

Tobias Kildetoft

20:40

@BalarkaSen natural home for homology: category with enough projectives. Natural home for cohomology: category with enough injectives

Balarka Sen

lol!

iwriteonbananas

@TobiasKildetoft haha

Tobias Kildetoft

(the ones I look at the most only satisfy the latter, so I never end up doing homology)

user174558

Natural home for homotopy and cohomotopy?

Balarka Sen

I have never in my life understood injective modules.

Tobias Kildetoft

20:41

fortunately, one can compute Ext in either case

@BalarkaSen It is just the dual notion of projectives

user174558

I understand injections though. They are painful.

Balarka Sen

@Jasper Nobody knows. That's the hypothetical homotopical algebra.

Tobias Kildetoft

@Jasper depends on what you inject

Balarka Sen

@TobiasKildetoft Yes, but I cannot visualize injective modules.

user174558

@BalarkaSen Just think of needles.

Tobias Kildetoft

20:42

@BalarkaSen If you can visualize projectives but not injectives, you are either in an "ugly" case or a too nice case

Balarka Sen

To give an example, I think of projective modules as vector bundles.

user174558

To answer the question 'What is the purpose of this site?' on meta, the answer is: to have fun chatting in chat.

Balarka Sen

Is there an analogous visualization for injective modules?

Tobias Kildetoft

@BalarkaSen For example, for rational representations of an algebraic group, there are in fact no projective modules at all (in positive characteristic)

Chris's sis the artist

Calculate elementarily (and in one line) $$\int_0^{\infty} \frac{\sin^6(x)}{x^4} \ dx$$

user174558

20:44

Hi @Chris'ssistheartist!

Tobias Kildetoft

@BalarkaSen No idea. The only cases I know how to visualize modules at all are for reps of quivers

Chris's sis the artist

@Jasper HI!!!! JASPER!!! :-)

@Jasper :D How is it going?

bolbteppa

@BalarkaSen here's a better way to say what I'm trying to say about homology physicsforums.com/threads/motivation-behind-homology.193452/…

Balarka Sen

which answer are you referring me to?

bolbteppa

@BalarkaSen You asked about how homology links to obstructions to systems of linear equations right?

Balarka Sen

20:48

not quite, I just bade you good luck on finding on :P but sure.

Tobias Kildetoft

@bolbteppa That is pretty imprecise, which is presicely what mathematicians don't like

bolbteppa

I don't agree, the best mathematicians always describe these things in apparently imprecise ways which tell you what's really going on, Tao wrote a huge post about this worth reading

Balarka Sen

But I really don't know which answer you are referring me to

Tobias Kildetoft

@bolbteppa when I say imprecise I mean in a way likely to lead people to understand it in a way that is actually wrong

bolbteppa

@BalarkaSen this one

Balarka Sen

20:51

no, I mean in the PF question. there are lots of answers. are you referring me to #5?

Tobias Kildetoft

For example, while lacking surjectivity is an obstruction to there being a solution, it just means there are some values for which there is no solution, while calling it an obstruction to a solution can also sound like it prevents a solution from existing

Balarka Sen

oh, I see, you're referring me to matt grime's answer.

bolbteppa

@BalarkaSen yeah sorry, posts 3 and 5 are good

Balarka Sen

I am sharing @TobiasKildetoft's opinion that answer #3 tries to explain what homology is in a completely incorrect way.

bolbteppa

@TobiasKildetoft Good point, I'm trying to get everything right, yeah I agree this stuff has the potential to mislead but it's aiming at a deeper understanding if done properly right?

Balarka Sen

20:54

Not really.

It's just a vague general-audience explanation of what homology means. It has no deeper meaning.

Homology is absolutely completely not about solutions of linear system of equations.

bolbteppa

@BalarkaSen if you are doing homological algebra it is the case, i.e. linear obstruction theory, which is what I'm linking to :\

Balarka Sen

(whereas saying cohomology is about solutions of differential equations has some truth in it)

Anthaas

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Q: Dung's Argumentation Framework

Given an extension of Dung's argumentation framework, and all definitions therein, let $$AF_1 = \langle A,Def \rangle$$ be a framework where $$A = \{ A, B, C, D, E, F, G, H \}$$ and $$Def = \{A def G, D def C, D def E, C def F, F def C, E def F, F def E\}$$ where $def$ is the binary defeats relat...

logic artificial-intelligence

Balarka Sen

No, homological algebra has no deep connection with linear system of equations :P

Tobias Kildetoft

@bolbteppa No, homological algebra is not about "linear obstruction theory" whatever that means

it is more closely about the failure of things being semisimple if you want an explanation in so few words

Balarka Sen

20:58

And answer #5 explains homology as dual of cohomology (which has explanation in terms of obstruction to solns of diff eqns).

There is no direct way to understand homology so simply. It's not a simple thing.

This had bugged me a lot when I studied homology, and the closest most satisfying explanation I got was the cobordism thing.

bolbteppa

Okay so now we see a link between cohomology, which studies obstructions of linear operators, and homological algebra, which studies obstructions to linear operators, where we're measuring the obstruction to a linear operator being an isomorphism (as mentioned here physicsforums.com/threads/exact-sequences.164213/#post-1293776 ) which is the whole reason for even defining the kernel and cokernel in an exact sequence, what is so awful about this?

The cobordism thing is just about finding holes though :\

Balarka Sen

@bolbteppa Where is the hole in $\Bbb{RP}^2$?!?!?

That hole thing works only in the restricted 2 dimensional setting of orientable manifolds.

I have no idea why people get so excited by that idea. In the 2 dimensional setting, $\pi_1$ measures "holes" too. So it does not really tell me what homology "is".

bolbteppa

@BalarkaSen I'm not sure yet sadly, I am really just trying to find out how what I'm saying makes sense of what you're saying, right now all I see is this weird fact about homological algebra and cohomology and now the Snake lemma is linked to that perspective which is cool, what is so objectionable to what I just said apart from the cobordism comment haha?

Balarka Sen

Yes, cohomology is directly related to finding obstruction to differential equations. There's no objection about that. But my point is that you can't find similar explanations for homology that comes naturally. Solns of linear system of eqns is very superficially related to short exact sequences (the rank-nullity). Why should I believe there is a proper connection with homology?

bolbteppa

Well as we've just seen for linear maps in homological algebra a similar obstruction interpretation comes about, I'm not sure if you still have a problem with that? I don't know why you call it superficial for example

Balarka Sen

21:11

I may be biased, but I think the cobordism idea has a lot more physics to it. You're looking at smooth submanifolds and cobordisms between them inside the ambient manifolds. This looks a lot like string theory to me when the submanifold is dimension $1$ (string topology, topological quantum field theories). But I might as well be wrong, as I don't know any physics :P

bolbteppa

Yeah the cobordism stuff comes up in a lot of physics, that is cool stuff I want to make sense of too haha

Balarka Sen

@bolbteppa I call it superficial because in homological algebra you work with R-modules. If you let R = field then you get vector space theory, which naturally leads you to solns of linear systems. That's where you are getting all those connections from. But since vector spaces have so less structure, they can't really be used to explain the subtleties of R-modules we deal with in homological algebra.

That is why it is superficial.

bolbteppa

I don't see how that eliminates or makes irrelevant the fact we've just linked the idea of cohomology to homological algebra, all you've said is that systems of linear equations look more complicated when you allow the coefficients to be say functions from a ring, but the idea of obstructions to solutions hasn't gone away.

Whether I can talk about all of homology in such a cavalier way, I agree, is not obvious

Balarka Sen

OK, let me get this straight. You have a topological space $X$. How are you interpreting $H_n(X)$? Can you give me a completely rigorous analogy instead of the vague connections you have been talking about?

bolbteppa

@BalarkaSen are you are asking me to link this idea of obstructions to homology on arbitrary topological spaces?

Balarka Sen

21:20

Obstruction to solutions of linear system of eqns in where? In cohomology, the obstruction are to solve differential equations on your manifold.

@bolbteppa Homology of nice topological spaces, if you want. But yes, I want a rigorous link instead of vague conjectural connections.

Karim Mansour

ok @BalarkaSen I proved one implication

Given $R^{\omega}$ in the box topology prove that x and y are in the same component iff x - y is eventually zero

I proved <--

Balarka Sen

yep.

Karim Mansour

do you want to check my proof ?

bolbteppa

@BalarkaSen I can't do that, I'm not sure if it can be done, that's what I'm trying to find out - I am trying to find out the extent to which you can push this idea. It's most likely not true in general at all, I agree with your skepticism, but that's not the point. Right now there is this weird fact that cohomology is linked to obstructions for differential equations, and homological algebra is about obstructions to isomorphisms of linear operators, and you seem to have a problem with this...

evinda

Hello @Alessandro

Balarka Sen

21:25

@KarimMansour sure.

bolbteppa

I'm trying to find a way to see how you're not just trying to tell me the Jordan Normal Form is different in homological algebra than it is in linear algebra because we now use modules as an excuse tbh haha

On a more fundamental level, just why in the world does this happen

Balarka Sen

@bolbteppa The reason is that every short exact sequence of vector spaces split, whereas the same is not true for R-modules.

The story is way, way simpler in vector spaces than modules.

Karim Mansour

So suppose that $x,y \in R^{\omega}$ such that x - y is eventually zero, note x and y can be writtten as $x = (x_1,....,x_i,...)$, $y = (y_1,....,y_i,....)$. By hypothesis there exists $N \in \mathbb{N}$ such that $x_i - y_i = 0 \forall i > N$, so that means that $x_i = y_i \forall i > N$

now

consider the following map

$f : R^n \rightarrow R^{\omega}$

where $(q_1,q_2,...,q_n) \mapsto (q_1,q_2,...,q_n,x_{N + 1},x_{N + 2},....)$

such map is continous

and $x,y$ is in $f(R^n)$

since R^n is connected so is the image, but the image is then a component of x,y

right ?

Balarka Sen

What are those $x_{N+k}$'s?

Karim Mansour

well we know that after threeshold of $x_i$ we get $x_i = y_i$

so I am picking the values

for which $x_i = y_i$

Balarka Sen

21:31

ok, I see.

Karim Mansour

the other side seems difficult

Balarka Sen

What you did seems good to me.

Karim Mansour

ok perfect

I guess maybe the other side I can do it by contradiction

Balarka Sen

I'd try showing that $x$ and $y$ lie on the same path-component itself. Good idea about that $f$.

Karim Mansour

dunno yet

Anthaas

21:32

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Q: Dung's Argumentation Framework

Given an extension of Dung's argumentation framework, and all definitions therein, let $$AF_1 = \langle A,Def \rangle$$ be a framework where $$A = \{ A, B, C, D, E, F, G, H \}$$ and $$Def = \{A def G, D def C, D def E, C def F, F def C, E def F, F def E\}$$ where $def$ is the binary defeats relat...

logic artificial-intelligence

Balarka Sen

@bolbteppa As I have said, I have no problem with that interpretation of cohomology. "homological algebra is about obstructions to isomorphism of linear maps" is scandalous because it largely overlooks what homological algebra is really about.

And I have already explained why vector spaces aren't good enough to see homological algebra.

Karim Mansour

@BalarkaSen usually one uses maps with proves from algebra so it was quite natural haha

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