Mathematics - 2015-08-17 (page 2 of 3)

user91500

09:23

@N3buchadnezzar math.stackexchange.com/questions/707256/…

N3buchadnezzar

Thanks

St3114

hi can someone say which one is homeomorphism ( no need a proof) $f_1:\Bbb R^*\to \Bbb R,f_1(x)=1$

$f_2:\Bbb R\to \Bbb R,f_2(x)=\frac x{2}$

Balarka Sen

$\Bbb R^*$ is $\Bbb R-\{0\}$?

Well, do you know what a homeomorphism is?

St3114

09:47

Yes. And yes f_2 should be homo. f_1 not but my book tells the exact opposite

Tobias Kildetoft

@St3114 Do you mean homomorphism or homeomorphism?

St3114

I made a typo. I didnt know these are different things :) it is homomorphism in algebra

Tobias Kildetoft

@St3114 Then indeed the first one is not and the second one is

St3114

Thank you

Balarka Sen

10:06

hi @Tobias

Tobias Kildetoft

@BalarkaSen Hi

Balarka Sen

I've started learning some singular cohomology and commutative algebra.

Tobias Kildetoft

nice

Balarka Sen

how're you doing?

Tobias Kildetoft

Good. Trying to write up a first draft of my grant proposal

Balarka Sen

10:09

ah. you found your research topic, then?

Tobias Kildetoft

@BalarkaSen Well, at least in general terms

Lie Algebra

Let $\mathfrak{h}$ be a subalgebra of $\mathfrak{g}$

I want to show that $\mathfrak{h}$ is an ideal of $N_\mathfrak{g}(\mathfrak{h})$.

Proof:

We want to show $[N_\mathfrak{g}(\mathfrak{h}),\mathfrak{h}]\subseteq \mathfrak{h}$

And by definition $\forall n\in N_\mathfrak{g}(\mathfrak{h}), \forall h\in \mathfrak{h}, [n,h]\in \mathfrak{h}$

So it sort of follows immediately, is that right?

Balarka Sen

@Tobias What's the geometric motivation behind Nakayama's lemma? If there are any, that is.

Lie Algebra

11:09

I don't see where I have used the fact that $\mathfrak{h}$ is a subalgebra

I mean the way I have written my 'proof' makes me think I could prove any Vector subspace is an ideal of it's normalizer

Tobias Kildetoft

11:34

@LieAlgebra If it is not a subalgebra, it will not be contained in its normalizer

@BalarkaSen I am not familiar with one

Balarka Sen

oh well, thanks anyway!

Lie Algebra

@TobiasKildetoft Oh, I wrote a question on main(and I answer to a similar point to yours) math.stackexchange.com/questions/1400167/…

@TobiasKildetoft Is $i\subset g$ a part of the definition of $i$ being an ideal of $g$? It's not in the book I am looking at

I mean it makes sense from standard algebra

Tobias Kildetoft

@LieAlgebra Are you sure it is not part of the definition?

Lie Algebra

11:50

Oh gosh, it is a requirement I messed up

Or it probably is

It says that for $i$ to be an ideal of $g$, when $i$ is a subspace of $g$

and I was trying to see that $i$ was an ideal of $N$ when $i$ was a subspace of $g$

Tobias Kildetoft

Right, we do not define what it would mean to be an ideal for something else (though it would sorta be a submodule of something instead)

Balarka Sen

@Tobias The fact that if $M$ is a flat, finitely generated $A$-module implies $M_\wp$ is free over $A_\wp$ for all prime ideals $\wp \subset A$ (and vice versa) seems suspiciously like an analogue for trivializations of vector bundles to me.

Tobias Kildetoft

@BalarkaSen I am not familiar with trivialization of vector bundles

user147690

12:06

Hey @BalarkaSen, Hey SemiTraditional

Balarka Sen

A vector bundle is a map $p : E \to X$ between topological spaces such that fiber over each point $x\in X$, i.e., $p^{-1}(x)$, has the structure of a real vector space s.t. for every point $x\in X$ there is an nbhd $U$ such that there is a homeomorphism $f : U\times \Bbb R^n \to p^{-1}(U)$ for which diagram consisting of this, the projection map $p$ restricted to $U$ and the map $U \times \Bbb R^n \to U$ - which is projection to the first coordinate - commutes.

Oh, and the map $\Bbb R^n \to p^{-1}(x)$ given by $r \mapsto f(x, r)$ is an isomorphism of vector spaces.

user147690

Hi @Rememberme, didn't like your avatars?

Balarka Sen

hi @AlexClark. what's up?

Remember me

No I don't have an idea why they suddenly started changing @AlexClark

Hey@Balarka How is it going?

user147690

@BalarkaSen Got my first Advanced algebra assignment, so just reading over the questions before sleeping. Gotta do boring Cardano's method stuff, and then some group theory proofs on simplicity and subgroups of symmetric group stuff

Balarka Sen

12:12

@Tobias As an example, consider the mobius strip bundle over the circle. $M$ be the moebius strip, $M \to S^1$ is the bundle projection (projection onto the center circle) with fibers having canonically the structure of $\Bbb R^1$. Clearly for every point $x\in S^1$, there is a neighborhood the preimage of which looks like $U \times \Bbb R^1$, and the diagram commutes.

@AlexClark that's nice. sounds like an easy assignment for you.

user147690

@BalarkaSen Definitely xD. He said it should be revision, but the pre-req algebra classes were out of sync, and so each semester has had to do a revision period that means we lost important knowledge(like sylow theorems)

Balarka Sen

but you can ignore my rants if you don't care about it, @TobiasKildetoft. it's probably just a superficial analogy.

user147690

Hey @SohamChowdhury how are you?

Balarka Sen

where were you for two days, @SohamChowdhury? not that asthma again?

user147690

I.e. we have missed you

Balarka Sen

12:19

@AlexClark uh-oh.

user147690

He is teaching us them though I mean, but he made this assignment easy pretty much as a result of making sure we are all good

Balarka Sen

what will you do in this advanced algebra course? ring/field/galois?

oh, ok

user147690

"Topics from groups, rings, fields, algebraic number theory, category theory & homological algebra, with applications to quantum algebras."

Balarka Sen

very nice!

user147690

Usually they do actually cover all of the things written in the course description, so it should be good

user147690

12:20

Whether I survive is the real question xD

Balarka Sen

You'll do fine. You're pretty strong at algebra already.

user147690

I hope I do :)

Soham Chowdhury

Hey.

Remember me

Hey@Soham Where were you?

Balarka Sen

I actually spent this morning doing exercises from D-F on homological algebra

Soham Chowdhury

12:21

I was sort of sick, yes. But it was my internet mysteriously vanishing that's to blame. :P

Tobias Kildetoft

@BalarkaSen Ohh, I do know what a vector bundle is. I am just not familiar with trivialization of them (and unfortunately I need to go now).

Balarka Sen

oh. well, it's just that neighborhood and that map $f$. see ya.

Soham Chowdhury

Where did you learn about vector and line bundles, Balarka?

Balarka Sen

@Soham Hatcher has fiber bundles. Vector bundles are just a special kind of fiber bundles.

Not that I have done any computations with them. I just know what they are and how the homotopy long exact sequence goes.

Soham Chowdhury

Ah.

Whatever that is.

Nice to see that people actually act like they care when I'm missing. :P

user147690

12:24

Act like xD?

Remember me

Is this fiber bundles which you are talking about @Balarka related to the fibers we use to construct a quotient group?

Balarka Sen

nah

fibers are a synonym for preimage of a point, as I had already told you before

user147690

You gave me NeO xD(which led me to mors principium est)

Remember me

Yes

@SohamChowdhury Act like?:p

Lie Algebra

@TobiasKildetoft Can I have a hint for proving that the normalizer of a subspace $S$ in $\mathfrak{g}$ is the largest subalgebra of $\mathfrak{g}$ containing $S$ as an ideal?

I want to prove the 'largest' part

I didn't mean to kill the chat.

Remember me

12:37

Algebra questions are pretty fun and easy..

user147690

@LieAlgebra If $S$ is an ideal of $\mathfrak{g}$, then $[S,\mathfrak{g}]\subseteq S$ and so $\mathfrak{g}\subset N(S)$. Pretty sure that's right, tell me if not haha

user147690

@Rememberme Not sure what that means @Rememberme haha, the difficulty is entirely up to you, so the easy part doesn't make sense

Remember me

Well @AlexC all the questions I have encountered till now don't tale up more that 2 min- 5min to solve. And they are fun as well

user147690

What problems are you doing?

Remember me

Quotient groups and cosets from DF

For example this
$SL_n(F) \unlhd GL_n(F)$ @AlexC

Lie Algebra

12:41

@AlexClark Thanks, that's really clean. I am sad I missed it!

Remember me

These are the kind of questions

user147690

@Rememberme Wait until group actions xD, in my opinion that is where is starts taking more time

user147690

@LieAlgebra It's fine, I think we did it in class :P

Remember me

Hmm... I feel it will get trickier in rings and stuff. I don't see why group actions will take time @AlexC

user147690

I found them hard to understand for awhile for some reason, maybe it was just me

user147690

12:47

Harder than rings for me

Remember me

Rings seem very fascinating and challenging to me for some reason

Clement C.

Hi -- I have a naive question about cross-posting...

user147690

They are pretty nice, I definitely want to learn more about them

user147690

Ask away @ClementC.

Clement C.

I.e., is there any setting iin which it's not frowned upon? I have asked a question a while ago, witrhout answer; then I put a bounty on it, that expired this morning -- with no answer.

Since it's a probabilities/statistics question, I am now tempted to post it on Crossvalidated to see if it'd be more suited to people there.

4

Q: Finding a hidden "heavy" subset of random variables.

Let $X_1,\dots, X_n$ be independent non-negative random variables (with finite expectation and variance), and $0 < m < n$ be a fixed integer such that there exists a subset $S\subseteq [n]$ of size $\lvert S\rvert =m$ with $\sum_{k\in S} \mathbb{E}[X_k] \geq 100$; and $\sum_{k\notin S} \mathbb...

user147690

12:53

People are reasonable, I think they will understand(although people will complain regardless if they notice). What I mean is, it'll be frowned upon regardless, but I would just do it if it were me(and have done before on my old account)

Remember me

@AlexC How much about rings do you know about?

user147690

Actually I triple cross-posted one design question :P.

user147690

Noone noticed despite I believe 2 of them hitting the hot list

user147690

6

Q: Better way to display three pieces of data on one graph for a website

Note the oldest answer was referring to my graph prior to this last edit. It was a great answer, please don't discount it due to my edit. I want to display how much study I have done, and how many pages I have completed throughout the week, broken up by day, and I have done so as shown below...

data-visualisation

user147690

14

Q: Displaying three pieces of information on a graph

Note: 50 points of raw data are attached now. I want to display how much study I have done, and how many pages I have completed throughout the week, broken up by day, and I have done so as shown below: I have had people tell me that they can't understand the graphs, but I have no idea how e...

data-visualization

Dan Rust

12:55

Woah, seeing the name Alex Clark is a bit surreal.... My phd supervisor's name is Alex Clark.

user147690

@DanRust Ahahaha I have looked him up before, it made me a little sad xD(Because it will steal my thunder one day)

Dan Rust

haha

Clement C.

@AlexClark Thanks

Dan Rust

yeah that might make things difficult once you start publishing :P

Just steer clear of dynamical systems and algebraic topology I guess :P

user147690

Oh god haha

user147690

12:57

Algebraic topology looks interesting xD

Dan Rust

haha it is (my current area)

user147690

I think I will be going into Algebraic something, but still no idea. Alg and number theory look interesting, but what I have done with Lie algebras is also pretty cool

Balarka Sen

hi @DanRust!

Dan Rust

Hey @BalarkaSen!

long time :P

Balarka Sen

indeed! how have you been?

Dan Rust

13:07

Pretty good

been very busy with travelling

M.S.E

@Chris'ssistheartist hey, how do I evaluate $\int \frac{1}{\sqrt{e^{2x}+e^x+1}} dx$

Balarka Sen

what kind of math have you been thinking about then?

Dan Rust

spoke at the summer topology conference in Galways a few weeks ago

Galway*

Balarka Sen

oh, very cool.

Dan Rust

and then spoke at the young topologists meeting in Lausane even more recently :P

Recently, I've mostly been trying to find a way of converting certain types of problems in algebra into corresponding problems in topology/dynamics

Balarka Sen

13:09

i see. how has your research on dynamics been?

@DanRust ah? if you want to elaborate, i'm listening

Dan Rust

So as an example, suppose I give you a primitive matrix over the integers and define the group $G$ to be the direct limit of that matrix. Can you find a 1-dimensional continuum whose first Cech cohomology is $G$?

Balarka Sen

now sure what you mean by direct limit of a matrix.

Dan Rust

a direct limit of a matrix is the colimit over the diagram $\mathbb{Z}^n\stackrel{M}{\to}\mathbb{Z}^n\stackrel{M}{\to}\cdots$

Balarka Sen

ah, i see

so you're taking the direct limit over all the Z^n's with maps being multiplication by that matrix. makes sense

Dan Rust

yeah

Balarka Sen

13:14

hmm. how do you propose to approach this question topologically/dynamically?

Dan Rust

so if $n=1$ and $M=2$, then a solution would be the dyadic solenoid

Balarka Sen

yeah, true. (in general, if $n = 1$ and $M = [p]$, the result is a $p$-adic solenoid)

Dan Rust

it turns out the general solution was also pretty simple in the case of $M$ primitive

you just need to use some tricks from symbolic dynamics

Balarka Sen

interesting

Dan Rust

so a collaborator and I tried to reprove some known results by using this conversion trick

but it ended up being harder than we initially thought (spent a couple of days without much luck)

Balarka Sen

13:16

that sounds pretty fun

Dan Rust

I'm still hopeful though :P

Balarka Sen

is it known which groups appear as cech cohomology groups of continua?

Dan Rust

I have no idea haha

that's a good question though

Balarka Sen

yeah, i have no idea either. the question should be reformulated by replacing a continua with something more pathological, like solenoidal spaces, though.

well, good luck with what you are doing, @DanRust. i hope you come up with something interesting!

Dan Rust

a solenoid is a continuum

a contiuum is just a compact connected metric space

Balarka Sen

13:24

by solenoidal spaces, i mean inverse limit of compact connected metric spaces.

a n-adic solenoid is just an example of a solenoidal space

Dan Rust

isn't the inverse limit of continua still a continuum?

Balarka Sen

sure

but not every continuum appears as inverse limit of continuums.

Dan Rust

what about itself under the identity map?

Balarka Sen

duh, of course. plug in "nontrivial maps" in the definition of solenoidal space.

Dan Rust

so a non-injective map?

Balarka Sen

13:27

i always confuse continua and write down continuums instead.

um, yeah, something which is not just an inclusion.

@DanRust do you know much about ergodic theory?

Dan Rust

not too much

I try to avoid it where i can :P

although that's becoming harder and harder recently

Balarka Sen

hehe. well, i am going to be in a talk related to ergodic number theory this wednesday, and i know nothing about it

Dan Rust

haha have fun with that one

right, I have to go

Balarka Sen

i doubt i will

Dan Rust

fun chatting though @BalarkaSen

Balarka Sen

13:35

ok, bubye

you too

user147690

13:47

Hey @BalarkaSen are you still around?

Balarka Sen

yeah, @AlexClark.

user147690

I am still unsure about something we talked about when I was working on central extensions.

If I am mapping from $\mathfrak{sl}(2,\Bbb C)\to \mathfrak{so}(3,\Bbb C)$, these have three basis elements, do I write a piecewise mapping from each basis element? Or do I write it as a map from $(a,b,c)\to (a,b,c)$ in the different basi?

Balarka Sen

i dunno what those curvy sl's mean

user147690

I.e. is it fine to write: $$\begin{bmatrix}a&0\\0&-a\end{bmatrix}\mapsto \begin{bmatrix}0&a&0\\-a&0&0\\0&0&0\end{bmatrix}$$

user147690

Oh just ignore the curliness and treat them as matrix groups

user147690

13:52

The curliness just adds the commutator bracket $[X,Y]=XY-YX$ I probably should have not added it

Balarka Sen

ok. you can write the map by where it sends a generic element of sl(2, C) to a generic element of so(3, C)

writing it out by what it does to the generators is fine too

but you should mention it.

user147690

So I'll write it as $\begin{bmatrix}a&b\\c&-a\end{bmatrix}\mapsto \begin{bmatrix}0&a&b\\-a&0&c\\-b&-c&0\end{bmatrix}$

I was just worried I was meant to write it as $(a,b,c)_{B_1}\mapsto (a,b,c)_{B_2}$ or something. I remember I was confused with short exact sequences with the discrete heisenberg 3 group

user147690

Okay going to eat a super late dinner, thanks

Balarka Sen

nah, that's fine.

Александър Гьорев

14:12

anyone interested in euclidean geometry ?

Boni

15:49

When a vector space is said to be finite, does that mean that the vector space has finitely many elements? I just want to confirm that, in case it is taken to mean finite dimensional as a convention for some reason.

Soham Chowdhury

2 hours ago, by Balarka Sen

hehe. well, i am going to be in a talk related to ergodic number theory this wednesday, and i know nothing about it

you have a nice life

Samuel Yusim

16:05

@Boni: It depends, unfortunately. I've seen it in both contexts: it could mean a finite-dimensional vector space over a finite field, or it could just as easily mean any finite-dimensional vector space.

you'll have to just figure it out based on other stuff the author has been saying

Boni

16:25

@SamuelYusim Thank you. I guess I can make it obvious by saying it is a "finite and finite dimensional vector space".

Or... just as you said, finite dimensional over a finite field. That is much more explicit.

Tiago Coelho

hey guys

can someone help me ?

with a easy problem?

cos 55ºx-8=x

x-cos 55ºx=8

why?

why is x- and not -x?

Fargle

17:34

@TiagoCoelho: it seems someone made a typo. It should indeed be cos 55ºx - x = 8.

Just to clarify: is it cos (55ºx) or (cos 55º)x?

Tiago Coelho

(cos55º)x

Fargle

Okay. Just wasn't sure.

Mike Miller

Morning.

Fargle

Hi @MikeM.

Tiago Coelho

so no solution ?

Fargle

17:40

It should have a solution.

Balarka Sen

18:31

@SohamChowdhury you have no idea. what kind of math have you got done today, btw?

Balarka Sen

18:41

hi again, @TobiasKildetoft

Tobias Kildetoft

@BalarkaSen Hi

Balarka Sen

i cooked up a proof by myself of the thing about localization of projective modules being free over localization of the base ring i mentioned above.

Tobias Kildetoft

nice

Balarka Sen

feel a bit more confident about commutative algebra now, haha

turned out I needed Nakayama's lemma in the proof.

Brennan.Tobias

19:23

Hi all

Would anyone like to check my answer on a number theory identity?

Lord_Farin

19:47

Such silence.

Remarkable.

Pedro Tamaroff

20:05

@Lord_Farin screams

@BalarkaSen If you used Nakayama's lemma, perhaps you proved f.g. projective modules are free over local rings.

Lord_Farin

@PedroTamaroff My heart jumps in fright, for I neglected to turn off the chat's sound.

:)

Balarka Sen

@PedroTamaroff Yeah.

Hello, Pedro, btw.

Pedro Tamaroff

@BalarkaSen The claim is true without assuming finite generation, however. The proof is a bit more complicated.

Balarka Sen

Oh?

Pedro Tamaroff

@Lord_Farin Then I have achieved my purpose.

@BalarkaSen Yes, google "Kaplansky's theorem."

Lord_Farin

20:06

@PedroTamaroff One would assume the moderators have more important stuff to attend to :).

Or perhaps I need to start thinking about my notoriety on MSE.

Balarka Sen

Thanks, @Pedro!

Pedro Tamaroff

@Lord_Farin At the moment there are no pressing matters. =)

Lord_Farin

@Pedro But seriously, how's life here these days?

Balarka Sen

Just asking : Do you know anything about the analogy between this fact about f.g. projective modules and vector bundles?

Pedro Tamaroff

@Lord_Farin Oh, I'll take a dare and say it's going well.

Lord_Farin

20:08

I have far too little time to spend here (or one might say that my priorities have shifted).

Balarka Sen

As I have mentioned above, this thing just looks like local trivializations for vector bundles.

Pedro Tamaroff

@BalarkaSen ncatlab.org/nlab/show/Serre-Swan+theorem

@Lord_Farin Yes, lately I haven't been around the site.

Balarka Sen

hmm, I remember Mike mentioning that.

Pedro Tamaroff

I mean, I do my mod duties here and there, but I'm not that active as a common user.

Balarka Sen

thanks again!

Miguelgondu

20:10

@Pedro

's a bot

Pedro Tamaroff

I have also been writing in a blog, @Lord_Farin. So that takes some time.

And courses get increasingly more demanding.

Lord_Farin

@PedroTamaroff Ah, interesting. I also see you've been catching some more advanced algebra?

Pedro Tamaroff

@Lord_Farin Yes, I would dare say that too.

Balarka Sen

huh. that's quite interesting.

Pedro Tamaroff

@BalarkaSen jstor.org/stable/1970252

That's the original article by Kaplansky.

I don't have access though.

@Lord_Farin I am trying to learn more geometry now.

Lord_Farin

20:13

@PedroTamaroff Hmm. These days I can only scratch the surface of these theorems.

Balarka Sen

I'm bookmarking it. I have an account in Jstor.

Lord_Farin

@PedroTamaroff Ah, aspiring towards algebraic geometry, are we?

Pedro Tamaroff

@Lord_Farin Hmm, not really.

Balarka Sen

So, what have you been thinking about, @PedroTamaroff?

Pedro Tamaroff

@BalarkaSen Things. Like a walk in the park.

Balarka Sen

20:15

You know what I mean. Mathwise.

Lord_Farin

@PedroTamaroff This implies you have a life outside of MSE. Lies.

Pedro Tamaroff

@Lord_Farin tsk tsk I really do. =)

@BalarkaSen Well, last two finals were Probability and Statistics, and Homological Algebra.

The P&S one was quite fun but is not what I'm heading towards.

skull patrol

Unless, you have someone to walk with? ;-)

Lord_Farin

@PedroTamaroff :) No surprise, even I managed to secure one...

Balarka Sen

ah. well, I spent the whole afternoon solving D-F's exercises on homological algebra.

Pedro Tamaroff

20:17

Homological Algebra was nice(r), and is probably where I'll end up in the thesis.

Balarka Sen

that is, some basic Ext and Tor computation.

Pedro Tamaroff

@BalarkaSen Really? What did you do?

Aha. For example?

Balarka Sen

Ext_Z(Z/mZ, Z/nZ), Ext_Z/p^2Z(Z/pZ, Z/pZ), etc.

Pedro Tamaroff

Ah. OK.

Balarka Sen

That Tor^1_Z(A, B) = Tor^1_Z(B, A) for abelian groups is a nice fact, I learnt from hatcher.

Pedro Tamaroff

20:19

Ah, that follows from the commutativity of the tensor product quite easily. =)

Balarka Sen

Right. Makes things easy for computational purposes.

Pedro Tamaroff

Although you want an isomorphism there.

Be wary about equalities and isomorphisms.

Balarka Sen

Yes. I am being lazy.

Pedro Tamaroff

Or at least take the time to see if the isomorphisms are natural or not!

Lord_Farin

@PedroTamaroff It's a natural one, I'd hope?

Balarka Sen

20:20

Of course they are not set-equal.

Pedro Tamaroff

@Lord_Farin Yes.

Balarka Sen

@PedroTamaroff hmm. good point.

Lord_Farin

Meh. Being out of my depth sucks :/.

Although I am working on brushing up my algebraic topology when I'm not at work.

Pedro Tamaroff

@BalarkaSen Can you compute ${\rm Ext}_\Bbb Z(\Bbb Z_n,M)$ for an arbitrary abelian group $M$?

Balarka Sen

I am doing commutative algebra and singular cohomology now, @Pedro.

@PedroTamaroff M/nM

been there, done that

Pedro Tamaroff

20:22

@Lord_Farin What does that mean?

Lord_Farin

@PedroTamaroff I haven't ever done anything close to homological algebra, so I don't really get any further than a remote reminder that I once knew what these Ext's and Tor's meant.

Pedro Tamaroff

@BalarkaSen Consider $A=k[x_1,\ldots,x_n,y_1,\ldots,y_n]$ and $B=k[x_1,\ldots,x_n]$ made into an $A$-module by making the $x_i$ act as usual and the $y_i$ by $0$. Compute the Exts and Tors, as much as you can.

Balarka Sen

you look at your exact sequence 0 --> Z --> Z --> Z/n --> 0. tensor with M and use the Ext long exact sequence. You'll get a multiplication by n map on the dualized third arrow, so you can forget about the previous ones. Higher Exts are zero, as all abelian groups have three termed projective resolutions.

Lord_Farin

That's what happens when you leave academia, I suppose.

Pedro Tamaroff

@Lord_Farin Right.

Balarka Sen

20:24

So Ext^1(Z/n, M) fits in the sequence 0 --> M --> M --> ? --> 0 where the map M --> M is multiplication by n.

Pedro Tamaroff

Yes.

Balarka Sen

Can't be anything other than M/nM

OK, seeing your other exercise.

Pedro Tamaroff

Well, you get that it is the cokernel of multiplication by $n$, and cokernels are unique-up-to-isomorphism.

Plus you get a natural isomorphism. =)

@BalarkaSen The exercise I gave you is not completely trivial. Think about it for some days if you need to.

Balarka Sen

right. a cool application of that, combined with projectivity of free modules, is that Ext_Z(G, Z) is the torsion part of G

@PedroTamaroff yeah, it looks horrendous. I'll ping you when I get it.

Pedro Tamaroff

@BalarkaSen Hmm, not always.

Balarka Sen

20:27

well, f.g. abgrps

sorry

robjohn

Finite Abelian Groups?

Lord_Farin

@robjohn Did you just call Pedro a ...?

robjohn

@Lord_Farin ;-) no, I was clarifying what Balarka was trying to say...

Balarka Sen

well, finitely generated.

robjohn

@BalarkaSen Ah, the "generated" was what I was missing :-)

Lord_Farin

20:30

@robjohn I'm sure it was entirely coincidental.

:)

Pedro Tamaroff

@BalarkaSen Are you sure of this?

Lord_Farin

@robjohn How's life these days?

Pedro Tamaroff

What is true is that ${\rm Tor}^A(-,Q/A)$ is naturally isomorphic to $\tau(-)$ for $A$ a domain.

robjohn

@Lord_Farin Doing okay. Work has a less than pleasant atmosphere right now.

Pedro Tamaroff

Where $Q={\rm Frac}(A)$.

Balarka Sen

20:32

@PedroTamaroff if G is a finitely generated abelian group, then G splits up into Z^r 'oplus it's torsion part.

Lord_Farin

@robjohn Why's that?

Balarka Sen

Take the Ext -- it splits up into Ext(Z^r, Z) and Ext of the torsion part

Ext(Z^r, Z) = 0

now Ext(Z/p^kZ, Z) is precisely Z/p^kZ.

Pedro Tamaroff

Yes, right.

Balarka Sen

so Ext of the torsion part of G is just the torsion part $\implies$ Ext(G, Z) is the torsion part of G

robjohn

@Lord_Farin Don't want to go into a lot of detail, but my boss is convinced that I am a doofus, when I think the problem is that his communication skills with employees needs improvement.

Balarka Sen

20:33

@PedroTamaroff phew.

@Pedro Constructing a short exact sequence of the whole resolution from a short exact sequence of the modules was fun. I did the part about extending to a chain map of resolutions from a map of modules, and finished the rest from hints on D-F.

It was weird to see at first how cheap cohomology and homology invariants come from considering resolution of modules.

Pedro Tamaroff

@BalarkaSen I never read any of the homological algebra in D.F., really. Those are very relevant properties, although quite elementary.

@BalarkaSen Example?

Balarka Sen

(especially because I wasn't familiar with any homology/cohomology theory other than the singular/simplicial ones)

@PedroTamaroff well, Ext and Tor

Pedro Tamaroff

@BalarkaSen Do D&F call those "invariants"?

Balarka Sen

@PedroTamaroff yeah, not really doing much homological algebra. needed some knowledge about Ext and Tors to do comm. alg. and alg. top., and neither Atiyah-MacDonald nor Hatcher talks much about them other than the definitions, and the exercises aren't good either.

Lord_Farin

@robjohn :/

Balarka Sen

20:41

@PedroTamaroff No, but I am. clearly if two modules are module-isomorphic, then so are it's Ext and Tor groups. they provide good invariants for modules, which in turn (as I got to know) translates into good invariants for groups by considering Tor and Ext over the group ring ZG (group homology/cohomology)

Lord_Farin

@robjohn I'm sad to realise that I can't easily come up with a way that this will end well :(.

Pedro Tamaroff

@BalarkaSen Yes, but that's quite trivial! Any functor sends isomorphic objects to isomorphic objects!

robjohn

@Lord_Farin Nor do I. I just try to do what I can, but it is not easy to cope.

Balarka Sen

That's horse before carts, if I may say that. That it sends isomorphic objects to isomorphic objects makes it a functor, not the other way around!

I am probably just being childish, as I like to think about invariants of topological spaces.

Pedro Tamaroff

@BalarkaSen Yes, my point is that ${\hom}$ and $\otimes$ are (bi)functors, so their derived functors are well, functors.

Balarka Sen

20:48

you mean $\otimes$ instead of $\text{Tor}$. I totally agree. That it's a functor is very easy to prove, but it matches up with the way I think about these things. I like to think of them as invariants for modules (i.e., an easy way to tell whether two modules are isomorphic or not),like singular (co)homology are invariants for topological spaces.

Pedro Tamaroff

Does Hatcher talk about the relation between group cohomology $H^n(G,M)={\rm Ext}_{kG}^n(k,M)$ and classifying spaces?

Balarka Sen

Hatcher defines group cohomology as $H^n(K(G, 1))$. I learnt that it can be defined that way too.

Lord_Farin

@robjohn Well, all the best to you. You know, you, mixedmath and Maríano are the only mods still in office from when I joined. All of you have always acted well as far as I was involved.

Pedro Tamaroff

Right.

Lord_Farin

@robjohn Just felt like sharing that well-earnt compliment :).

Balarka Sen

20:50

The Ext construction for group cohomology is pretty geometric :)

robjohn

@Lord_Farin Thanks. Mariano is here once in a while, but I think he is more active on MO these days.

Balarka Sen

You immediately get a projective $\Bbb ZG$-resolution by considering how the fundamental group acts on the universal cover of $K(G, 1)$ (or more precisely, it's $n$-cells) by Deck transformation.

Lord_Farin

@robjohn I can relate. I hardly find time to contribute beyond the occasional rambling on meta. I feel like it has become harder to find interesting questions these days.

But perhaps that's just due to me being out of academia and my scope of comprehension slowly shrinking out of existence.

Balarka Sen

@PedroTamaroff Commutative algebra is pretty fun. I am reading the theory from Reid and doing exercises from Atiyah-MacDonald (at least, until Eisenbud arrives).

Lord_Farin

Alas, bed-time has come. See you around, @rob @Pedro.

Pedro Tamaroff

20:56

@BalarkaSen Yes, it is.

robjohn

@Lord_Farin Sleep well.

Balarka Sen

interpreting much of the theory as study of sheaves of polynomial functions on a variety is making a lot of things very geometric.

(even though I don't know how the complex analytic theory goes. that is, I know nothing about sheaves other than the definitions)

well, nice talking to you @Pedro. I have to head to sleep now.

g'night everyone

Pedro Tamaroff

Night. Sorry, I'm a bit busy now.

skull patrol

Later pal

Kevin Driscoll

@robjohn I see you cleared the room with some physicist-like notation, for example $\frac{1}{1 - D}$ for some differential operator $D$

Well played, sir

robjohn

21:08

@KevinDriscoll That was $\frac D{1-e^{-D}}$ which is just $D$ plugged into $\frac x{1-e^{-x}}=1+\frac x2+\frac{x^2}{12}-\frac{x^4}{720} +\frac{x^6}{30240}-\frac{x^8}{1209600}+\frac{x^{10}}{47900160}+...$

Kevin Driscoll

@robjohn AH, and presumably that just means $1 + \frac{1}{2} D + \frac{1}{12} D D + ...$

skull patrol

np

Kevin Driscoll

thank you

@robjohn Is there an easy way to know that $\frac{x}{1- e^{-x}}$ is entire?

(or maybe its not and then we've done some REALLY outrageous)

robjohn

@KevinDriscoll It is not entire... it blows up at $x=2\pi i$ for instance.

@KevinDriscoll It is not outrageous, these are asymptotic series, not convergent series.

Kevin Driscoll

@robjohn Actually now thinking about it, we do it for things which aren't entire too. My first example was $e^A$ for $A$ an operator, because then the series converges everywhere. But indeed if you're OK with asymptotic series then I am too.

robjohn

21:14

@KevinDriscoll They do converge as long as the fourier transform of the function is supported in the unit ball (polynomials for instance)

Brennan.Tobias

hello

skull patrol

Hi pal

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