Mathematics - 2015-04-28 (page 1 of 3)

12:00 AM

I had a long discussion with Pedro a year ago about this ... it's like twisting coefficients, like comparing a Möbius strip to a cylinder. That's precisely what goes on with tensoring vector bundles.

Incurrence

Have you dealt with Tensor products of groups? There seems to be very little documentation on these

Apparently they[tensor products of groups] were 'created' in 1989

Ted Shifrin

You can also talk about differential forms on a manifold, or differential forms with values in some vector space or vector bundle. The latter two are obtained by taking sections of a tensor product bundle.

I only know them as tensor products of modules (i.e., for abelian groups).

Incurrence

Oh okay

Tensor product for non-abelian groups was introducted in 1989 apparently

Ted Shifrin

no clue

Disciple of Barney

I believe he is suppose to talk about tensor products of arbitrary finite groups

Ted Shifrin

12:02 AM

Yeah, I have absolutely no idea.

Incurrence

Indeed

Ted Shifrin

My motivation is from much more geometric things :P

Incurrence

Well I can talk solely about abelian groups I suppose

But they must be finite

Ted Shifrin

Ask your professor who gave you that topic for resources and why they're interesting.

Disciple of Barney

Shoot for the stars @Incurrence

Incurrence

12:03 AM

@TedShifrin He won't give me any, because this is meant to help teach us to research things on our own xD

and I picked, apparently the hardest topic

Ted Shifrin

shrug

I guess I have no idea even what the tensor product of free groups is ...

Incurrence

So can I ask a stupid question: What the heck is a tensor product of abelian groups?

Disciple of Barney

If I know anything about Ronald Brown, they were probably created to do some higher dimensional van Kampen theorem. Why don't you read around a lot, it sounds like you are suppose to do a bit of research @Incurrence

Incurrence

@DiscipleofBarney I am, but since I am just starting the research I feel everything is opaque and I can't get any room to wriggle around

Ted Shifrin

Do you know about modules, @Incurrence?

Incurrence

12:06 AM

@TedShifrin No, nothing at all yet

Perhaps that is the starting point?

Alex Wertheim

Modules are (I would say anyway) a more natural setting for tensor products.

Ted Shifrin

Well, start with vector spaces, then. Bilinear maps on $V\times W$ induce well-defined linear maps on the tensor product $V\otimes W$. That's the main idea.

Incurrence

If I learnt tensor products for modules from D&F, will I find this much easier?

Ted Shifrin

Hell if I Know, @Incurrence. :)

hi @AlexW :D

Alex Wertheim

Hey @Ted! How's it going? :)

Ted Shifrin

12:07 AM

Well, my last class was anticlimactic, @AlexW ... but we survived.

Incurrence

I am going to go do some D&F. I'll talk later

Ted Shifrin

bubye, @Incurrence ... bonne chance.

Alex Wertheim

Lol, what happened @Ted?

Incurrence

@DiscipleofBarney Also you are right I am fairly sure, it was to do with van Kampen theorem from memory

@TedShifrin Wait why was it anticlimatic before I leave haha

Ted Shifrin

Just reviewed what they needed to know for the final, @AlexW ... tried to put winding numbers and Gauss's law in perspective (since they come, analogously, from cohomology of $\Bbb R^2-\{0\}$ and $\Bbb R^3-\{0\}$.

Reviewed a bit of linear algebra and a bit of Stokes's Theorem. Reminded them they needed to know how to prove continuous + compact domain $\implies$ uniformly continuous $\implies$ continuous functions on a rectangle are integrable, etc.

Disciple of Barney

12:09 AM

At least glancing at some of the definition of tensor products for these groups it looks like you would benefit from going though some examples and seeing how semidirect product works, and work with tensor products in easier situation (like vector spaces) @Incurrence

Alex Wertheim

Very beautiful stuff @Ted. Anticlimactic? :)

Ted Shifrin

Seems ridiculous to try to understand some hugely abstract group construction without first being an expert on semidirect products and group extensions, @Disciple @Incurrence.

Incurrence

@DiscipleofBarney Okay, I'll do that, thanks. I have very little experience with the semi-direct product(despite proving every one of it's equivalent conditions)

Ted Shifrin

Anticlimactic for a last class ... just bits and pieces of reminders, @AlexW :)

Incurrence

I don't know what group extensions are at all :\

Disciple of Barney

12:11 AM

@Incurrence Did you do that exercise?

Incurrence

It may be something to do with short exact sequences though from memory

Ted Shifrin

Semidirect products are a special case. Direct products are a specialer case.

Incurrence

@DiscipleofBarney Indeed

Disciple of Barney

The one with Heisenburg and showing some exact sequence does not split

Incurrence

@DiscipleofBarney It took me a week, but I got it in the end

Alex Wertheim

12:11 AM

Fair enough, @Ted. Did your students do anything special?

Incurrence

Argh so little that I know

Ted Shifrin

BTW, @Incurrence, did you ever finish that thing about nilpotent?

Nah, @AlexW ...

not even polite applause ... that's fine :D

Incurrence

@TedShifrin Not yet :\. But I am learning about torison in groups so I am getting closer

Dave

@AntonioVargas hows that

Ted Shifrin

I sort of feel like saying the same thing I've said to Balarka a thousand times, Incurrence. You're better off learning foundations solidly instead of tilting at fancy windmills.

4

Incurrence

12:13 AM

@TedShifrin But I am not self-studying! This is graded stuff!

pjs36

Here here, Dr. Shifrin!

Incurrence

@TedShifrin That group nilpotency question was an assignment question that I was meant to solve 2 weeks ago!

Disciple of Barney

@Incurrence Don't you have to do two presentations? what is the second one?

Ted Shifrin

The tensor product of groups is way off the deep end, @Incurrence. I'm more in favor of your actually understanding the homework exercises and learning all there is to learn.

Antonio Vargas

@Dave Just fine, except don't write #/1

Incurrence

12:14 AM

@DiscipleofBarney We have to make two, but only have to present 1

Disciple of Barney

Okay

Dave

@AntonioVargas oh yeah ! good point. but one issue is i tested this on the graph

Incurrence

@TedShifrin :\ my presentation is on the tensor product of finite groups, for my assignment - due in 8 days

Antonio Vargas

@Dave and do you really need that many decimal places?

Dave

if the number is +1.9 etc its okay but negative and the line is off

Alex Wertheim

12:14 AM

Aw. I would've gotten you a cake @Ted! Do you like cake?

Ted Shifrin

Can you pick something more accessible and doable, @Incurrence?

Incurrence

@TedShifrin Yes many things

Ted Shifrin

LOL, @AlexW. You might have poisoned it.

Incurrence

Free groups

Simple groups

Central extensions

Lattice of subgroups

Feit-Thompson Theorem

Are my other options^

Ted Shifrin

Really working out semidirect products and how they show up in Sylow applications would be good, @Incurrence

pjs36

12:15 AM

Simple groups and Lattice of subgroups are super-tangible

Antonio Vargas

@Dave well you solved the equation correctly, so I don't know what to tell you. I just double checked.

Ted Shifrin

Central extensions would be good ... but you should still do semidirect products first.

Dave

okay and no i don't need that many decimals but they didn't say to round it off so i'll keep it as accurate as i can :P

Disciple of Barney

@Incurrence I would probably go for central extensions or free groups or what you are doing now.

Ted Shifrin

Much of what I know about free groups I deduce from algebraic topology. But some can be done just algebraically.

Incurrence

12:16 AM

But I thought if I could spend heaps of time getting the hardest one down, it would impress the lecturer(who I would love to do research with in a year and a half), and it would motivate me learning a heap of prerequisite knowledge

Alex Wertheim

I would never do such a thing, @Ted! Then again, I haven't bombed one of your tests... ;)

Ted Shifrin

Trying to impress lecturers is not the point, @Incurrence. Impressing him with good learning, command of examples, and a good presentation is what you should do.

Incurrence

@TedShifrin Fair enough

pjs36

That's the gamble: What if you can't get the hardest one down in 8 days? There are a few up there you could really knock out of the park, which is more impressive than doing a "meh" job on the hardest topic

Incurrence

@pjs36 True - good point

I guess I just assumed that it wouldn't be undoable given it is on the list

Ted Shifrin

12:17 AM

I'm much more impressed by solid mastery of stuff, rather than bullshit misunderstanding trying to impress me.

Incurrence

I am willing to do 8hrs a day for the 8 days

Ted Shifrin

Well, stop chatting and get to work,@Incurrence. Good luck :)

Disciple of Barney

Central extentions of free groups :D @Incurrence

Incurrence

Fine I'll check out central extensions and see if I like it!

Ted Shifrin

Not much center to free groups, @Disciple :P

Incurrence

12:19 AM

Cya later(proabably 14 hrs from now for Ted) and 2 hours from now for Disc

Disciple of Barney

But you can extend them

pjs36

I'll just say that if I'm close to being good at anything (and that's a stretch), it's finite group theory. Most of that stuff you have is over my head, and I doubt I could throw together a presentation in 8 days with my background :)

Ted Shifrin

A lot of it's over my head, too, @pjs36, but I am far from an algebraist.

Incurrence

@pjs36 Haha it doesn't have to be amazing of a presentation

Bye!

Disciple of Barney

Bye

Ted Shifrin

12:23 AM

Learned anything good of late, @AlexW?

pjs36

Yeah, I learned my math from a bunch of algebraists by name, @TedShifrin, but in actuality they were finite group-theorists (disciples of Marty Isaacs, if you will). So my "algebra" knowledge is sadly more narrow than it should be.

Alex Wertheim

Depends on your definition of good, @Ted. :) I think it's very beautiful. Just continuing with manifold topology and commutative algebra.

Ted Shifrin

Yeah, finite group theorists form quite a clique, but there's a lot more algebra to know that gets used by a lot more mathematicians. :)

I won't fight "good" on that, @AlexW :)

Alex Wertheim

It's pretty elementary stuff, but I don't mind the slow walk.

Ted Shifrin

Commutative algebra has a lot of sophisticated stuff in it, @AlexW ... as does the background material on differentiable manifolds.

Alex Wertheim

12:26 AM

Oh, no doubt @Ted. The stuff I'm doing right now though is definitely beginner stuff.

Ted Shifrin

I'm not offended by "beginner stuff."

Alex Wertheim

Me neither. I'm very much in favor of the Grothendieck metaphor, with the water gradually opening the walnut.

Ted Shifrin

Hmm, haven't heard that one.

No water in CA, so therefore no learning.

Alex Wertheim

I can't stand pushing through math without understanding it well. Not that I claim to understand anything particularly well.

LOL @Ted. An unpleasant thought.

Ted Shifrin

Sometimes you have to take some things for granted, see how powerful they are, and then come back and understand them more deeply.

Spiraling is actually an effective technique for learning mathematics, as opposed to strict linearity.

Alex Wertheim

12:28 AM

No doubt. Right now I'm reading a bit about the spectrum of a ring, and I definitely don't grasp the depth of it yet.

Ted Shifrin

Yup, no learning and also no showering ... :D

Ah, that's the set up for the topological space structure underlying algebraic varieties.

Alex Wertheim

Indeed! That was the very problem I just worked on.

Ted Shifrin

I guess Jasper has really disappeared from here. I hope he's ok.

Alex Wertheim

That is, showing that Spec A is a topological space. Very nice.

He was just in the math mods' office the other day, @Ted. Did something happen?

Ted Shifrin

You'll get to the Nullstellensatz eventually, @AlexW.

He kept saying he was going to stop coming to chat. I dunno.

Alex Wertheim

12:30 AM

I've seen it once or twice before, but never really fully grasped it's strength. It's supposed to be (in some sense) a generalization of the FTA, isn't it?

Ted Shifrin

Yes, @AlexW

Alex Wertheim

Oh dear. He seemed to spend a lot of time here. Hopefully he's alright.

Ted Shifrin

Ultimately, tells you about what's irreducible among varieties ...

OK, I'll leave you guys to do some work. See ya later.

Alex Wertheim

Night, @Ted. Always a pleasure. :) Congratulations on your last day!

Ted Shifrin

Thanks, I think :)

Alex Wertheim

12:32 AM

Hehe, damn, I should think a little more before I write these things. I meant it the nice way, I promise :D

ᴇʏᴇs

Hello @TedShifrin and @AlexWertheim

Alex Wertheim

Hello @ᴇʏᴇs. How goes it?

ᴇʏᴇs

@AlexWertheim Ok, you?

Alex Wertheim

Not bad. I have a fever right now, which is unpleasant, but otherwise things are fine lol.

ᴇʏᴇs

Sorry about the fever

Alex Wertheim

12:39 AM

Haha, that's ok, thanks for your sympathy. :)

Incurrence

@Disc A left R-module with set $M$ has as its first property that it must have an binary operation $+$ on $M$ in which $M$ is an abelian group. Is that literally addition, or it can be any operation in which it is an abelian group?

Hi @ᴇʏᴇs

Disciple of Barney

12:56 AM

The latter @Incurrence

Incurrence

@DiscipleofBarney Thank you

Disciple of Barney

What did you mean by "literally addition"

@Incurrence

Incurrence

@DiscipleofBarney I mean is it the operation addition

Can I treat it exactly as I would treat addition

ᴇʏᴇs

Hi @Incurrence and @DiscipleofBarney

Incurrence

@ᴇʏᴇs Hello

Disciple of Barney

12:59 AM

I am still not sure what you mean by that @Incurrence Like you thought $M$ could only be some subset of the complex number?

Hey @ᴇʏᴇs

Incurrence

Hmmm I haven't thought too much about it: I was just reading down the page in order, and that was something I wondered

It says a binary operation +

So I looked at wiki and wiki just says +

So I wondered if this could be treated like $(+,\mathbb{R})$

E.g. all the properites there hold

but we have $(+,M)$ so we are only dealing with the elements on $M$

Because that $+$ sign could very well represent multiplication for example

Disciple of Barney

Yah, it just has to be an operation that makes it into an abelian group. If you have not noticed already, its fairly customary to use $+$ to denote an abelian group operation (of course it doesnt have to be)

Incurrence

Oh I have never seen that before

Disciple of Barney

I have seen $+$ used for an operation that wasn't even a group operation though...

Incurrence

Usually they state it is an abelian operation and use $\bigoplus$

Disciple of Barney

1:07 AM

Oh, what book is this?

Incurrence

My lectures

In books I have barely ever dealt with this level of stuff haha

On a different note, can you give me a hint on showing that $\mathbb Q$ isn't finitely generated?

@DiscipleofBarney Oh and the book I am currently working in is D&F

Disciple of Barney

Of course $\oplus$ sort of comes from that idea, except they want to distiguish it. They may decide to switch from using that notation because direct sums customarily have that notation

Hint: Say you have a finite subset of $\Bbb Q$.... :P

Incurrence

Fair enough

@DiscipleofBarney Is this something to do with every finitely generated subgroup is cyclic?

Disciple of Barney

Think about what the denominators of of the elements in the group generated by this finite set have to be @Incurrence

Incurrence

Ok

Disciple of Barney

1:11 AM

@Incurrence You could also do it the way you mention

Incurrence

1:24 AM

Will I need to learn modules to understand central extensions? It seems to be a prerequisite in D&F, but wiki doesn't mention them at all

robjohn

Sorry to be gone so long today. A friend of ours was in the path of a fire today and needed some help evacuating. We needed to move her dogs and some valuables out for a while. The fire seems to be out, but we are still on alert for flare ups.

@TedShifrin Amen.

Incurrence

@robjohn Did the house get damaged in the end?

@robjohn :'(. These things are graded, so it isn't my fault, I love to build the foundations

robjohn

@Incurrence No, the fire seems to have passed her by, but as I said, we are still keeping an eye out.

Mew

Who is "we"?

robjohn

I hear fire engines, but they are to close to be for her place.

@Mew A friend and I

Mew

1:29 AM

oh

where did the dogs go to?

robjohn

@Mew They stayed in her car

ᴇʏᴇs

Poor things must've been so frightened

Mew

do u live near the bush or something

robjohn

@Mew she does. I don't.

Mew

what nation?

Disciple of Barney

1:40 AM

@Incurrence Not... maybe Dummit and Foote have applications in mind that use modules. Read the areas that talk about central extensions, or just extensions in general to get an idea of what you actually need to know

Or they have some application to modules in mind

Incurrence

group extensions get a mention on page 824, no mention of central extensions, they have pages on extensions of a map of ideals, modules or scalars

Oh they have finitely generated field extensions of groups

Disciple of Barney

How about semidirect product and exact sequences?

AlpArslan

Right, I'm sorry for what's possibly a really rather silly question, but,

Incurrence

They have some semidirect product and they have exact sequences, but they do exact sequences for projective, injective and flat modules...

I need a different textbook for this obviously

AlpArslan

Say I have a few generators, all with finite order, and I want to write a program to enumerate all the elements in the group.

Does anybody have any ideas?

Incurrence

1:47 AM

With what? The scripting?

Disciple of Barney

Not all groups that have only finite order elements are finite @AlpArslan

AlpArslan

Sure, I know.

pjs36

Aluffi's Algebra Chapter 0 have have some of that stuff, @Incurrence. It's free online, you could at least check

AlpArslan

But, essentially, I know that the final group is finite.

And I essentially need to multiply and then cross-check with the elements already listed.

Are there any efficient ways to do this systematically, basically?

Incurrence

@pjs36 Oh this looks nice, thanks!

Disciple of Barney

1:49 AM

@AlpArslan I doubt it, but I don't know

pjs36

@Incurrence You're quite welcome, it's quite the book! But I know it gets pretty category-theoretical, and I see this flat module business in regards to ... all that! :)

(I'm probably showing my ignorance now, I have a feeling I botched the language, so I'll shut up)

Incurrence

Haha, I am learning what a module is now so I wouldn't know if you are

Pages 228-242 in this book look like what I want to learn

Awesome

I will go now and learn, thanks!

pjs36

No problem, I hope you don't get sucked down the rabbit hole too far!

Incurrence

Haha hopefully I can start reinforcing the walls of the rabbit hole if I do

Mike Miller

morning

Incurrence

1:56 AM

Morning @Mike<

ᴇʏᴇs

afternoon

Incurrence

@MikeMiller What time is it there?

Mike Miller

6:56

Disciple of Barney

@Incurrence Also that problem I told you about, with the Heisenburg group, that was a central extension

And it doesn't split so it is not a semi direct product

Will Hunting

3:31 AM

@TedShifrin I am still alive. I am usually in another room these days, since I don't want to talk too much non-math in this room. Thanks for your concern.

Antonio Vargas

3:49 AM

Huh. Something happened to the pin your location map.

@BalarkaSen

Incurrence

4:50 AM

@AntonioVargas I couldn't find any way to revert it, perhaps the person who made the map has admin privileges or something?

Mew

7:09 AM

hello

any smart people on?

Mike Miller

No.

Mew

You're joking right?

You're smart aren't you?

Mike Miller

No.

I know the people in the users list right now, and I can confirm we're all quite stupid.

@Huy can back me up

Huy

I agree with @MikeMiller.

Mew

7:25 AM

That's a shame

Remember me

If you are here for finding smart people then you should first see how smart you are then you can think about others @Mew

Mew

I need assistance from a smart person

Remember me

Try it yourself

Mew

Already tried

that is why i need assistance

Anthony

@MikeMiller !

Mike Miller

7:28 AM

Hey @Anthony.

Unihedron

Hello, Mathematics.SE!

Anthony

I haven't seen your name in like, forever!

Mike Miller

Bad timing, I guess, @Anthony.

Anthony

I suppose. There's like weird waves of users in this room.

Mew

Hello Unihedron

Remember me

7:28 AM

If you want to ask something then just ask someone hopefully will answer your question but dont keep shouting and asking that are here any smart people because we are just too big idiots for you @mew

Mew

Remember me, i'll just go ahead and ask then

Remember me

hi @MikeMiller

ask

Mike Miller

I waste way too much time in here, @Anthony. You just don't do that when I do. :P

Mew

If I'm generating a random number, say using an algorithm Xn = (aX(n-1) + B) mod C

or any other algorithm, for instance Xn = sin (X(n-1))

How can I check whether the numbers generated follow a uniform distribution?

Anthony

@MikeMiller I suppose so. Anyway, I think I'm goin to bed. Before I do though, are you any good with modular arithmetic/number theory/the weird stuff they shove into discrete math courses?

Mew

7:30 AM

is there a theoretical way to check?

Mike Miller

No, @Huy, but you can ask anyway.

Anthony

I have this weird question on one of my CS hw's, and I'm not sure what they're looking for. It's asking which quadratic residues in $\mathbb Z_{77}$ have only two roots.

Mike Miller

I think it's asking you to find which elements are the square of precisely two numbers.

Anthony

Oh, no, I get the problem statement. I just don't know if they want me to grind numbers, or if there's some nice way.

He gave us the result that the elements that are coprime with 77 have four.

So then there's only $16$ elements left to check, but that still seems like a silly problem. Nonetheless, I think I found them, and they're 11,22,44, 14,42,49,56, and 70.

Mike Miller

Note that those are precisely the nonzero quadratic residues that aren't coprime with 77...

Anthony

7:36 AM

Well sure- but like... Is there no better way then to just grind numbers?

I noticed that if we take the quadratic residues mod 7 and 11, multiply them by 22 and 14 respectively, I think we get those numbers... I think... But I doubt that's anything.

Mike Miller

I don't think "well sure" is the right answer to that. After having grinded the numbers, you see that there's actually a nice pattern here. Perhaps it's a shame that you had to grind the numbers in the first place, but you might be able to prove that these are precisely the ones without any grinding (now that you see the pattern), or maybe genearlize to $\Bbb Z_{pq}$.

Huy

Why did you highlight me? O.o

Mike Miller

I dunnoj.

Huy

j

Anthony

@MikeMiller When you say "these" do you mean the quadratic residues that aren't coprime with 77?

Mike Miller

7:39 AM

Yes.

Anthony

Again, though, that means that I'll have to calculate the residues? And I don't know a good way of doing that besides marching through the numbers, squaring, and reducing.

I don't know, maybe that's what we're suppose to do. Just seems silly.

Mike Miller

@Anthony: No.

Anthony

Oh. Why not?

Mike Miller

suppose $gcd(a,77) = 7$; argue identically for 11. Suppose $b^2 = a$. Take this equation mod 7. You see $b=0$. Take this equation mod 11. Because $a$ is not zero mod 11, $b^2 = a$ has two roots mod 11. Chinese remainder theorem shows that this has precisely two roots.

Now to see what the residues are. Suppose $b^2 = a$. No condition on $gcd(a,77)$. You see that $a$ has to be a quadratic residue mod both 7 and 11. Conversely if $a$ is a quadratic residue mod both 7 and 11, you can check by hand - this does not involve checking each number, you can give an argument that works in $\Bbb Z_{p \cdot q}$ - that $a$ is a quadratic residue mod 77.

Anthony

Why does $a$ have to be a quadratic residue mod both 7 and 11?

Mike Miller

7:48 AM

by reducing the equation $b^2 = a$ mod 7 or 11

Anthony

Is this something with the CRT?

Mike Miller

No, it's literally the thing I said.

Anthony

oh

crap

sorry. My heads been in the gutter today.

Mike Miller

If $a$ is a quadratic residue mod 77, there is some $b$ such that $b^2 = a$ mod 77. So clearly this equation also holds mod 7 or 11.

You should get some sleep. That's where I'm headed.

Anthony

Yepyepyep. Flossing my teeeeeth. Thanks for the help.

Mike Miller

7:50 AM

Sure.

Mew

8:10 AM

Is there a way of theoretically evaluating the randomness of a random number generator algorithm such as thte Linear congruential generator or is the only method by generating numbers and testing fi they are random enough?

using say a chi squared distribution

mathmath12

8:29 AM

I won't pretend I understand this, but I searched around a little this might apply:

"Algorithmic "Martin-Loef" Randomness (AR). AC and AP also allow a formal and rigorous definition of randomness of individual strings that does not depend on physical or philosophical intuitions about nondeterminism or likelihood. Roughly, a string is algorithmically ``Martin-Loef'' random if it is incompressible in the sense that it's algorithmic complexity is equal to its length. "

hutter1.de/ait.htm

I guess that would fall in this domain of Algorithmic Information Theory.

Balarka Sen

@AntonioVargas yikes. @AlexanderGruber I think you got a map-fixing to do.

mathmath12

This seems relevant to that question: " We can then say that a random sequence is one such that the shortest algorithm which produces it is approximately (to be explained below) the same length as the sequence itself—no greater compression in the algorithm can be attained. This proposal, suggested by the work of Kolmogorov, Chaitin and Solomonov (KCS), characterises randomness as the algorithmic or informational complexity of a sequence."

plato.stanford.edu/entries/chance-randomness/#2.1.2

Chris's sis

8:54 AM

Greetings

Mew

9:27 AM

Hi sis

Can you help me

Incurrence

11:14 AM

@Balarka @Disc Not much math discussion tonight

Disciple of Barney

@Incurrence Yah not much

Tobias Kildetoft

@Incurrence You can always try to start it up.

Incurrence

I could indeed

Disciple of Barney

Did your reading up on group extensions, etc go well

Incurrence

@DiscipleofBarney Haven't done too much of it, but no not really

So we have a short exact sequence:

$$e\hookrightarrow A \hookrightarrow E \twoheadrightarrow G \twoheadrightarrow e$$

and $A\subset Z(E)$

And then there is something about cohomology

Which is about invariants or something for modules

Disciple of Barney

11:24 AM

I think cohomology (or maybe homology) by be used to classify these extensions, but I don't think it is necessary to play around or find groups that are extensions of some groups to get a feel for them

Incurrence

I was going to read the text from Aluffi's Algebra Chapter 0 tomorrow and see how I go

For now I am thinking about one of my other problems

Finding the number of conjugacy class of $GL_2(\Bbb F_p)$

Tobias Kildetoft

@Incurrence If you are just getting a feeling for such extensions, you should definitely wait with cohomology

Incurrence

@TobiasKildetoft Yeah that is what I thought haha.

Cohomology is to do with algebraic topology I would assume

Yep just verified it

I am doing Armstrong right now to prepare for that

(to prepare for my hatcher)

Tobias Kildetoft

@Incurrence No, group cohomology is not directly related to algebraic topology (you usually start with homology, rather than cohomology in algebraic topology, and not the group kind)

Incurrence

Ahh my mistake

Tobias Kildetoft

11:30 AM

homological algebra is a big field which is used in many branches of mathematics

but which is better understood once you have seen those special cases that were the motivation (and this is in particular algebraic topology)

Incurrence

Oh okay

So no cohomology for me for now

That's for the best

Balarka Sen

@Incurrence yeah, I'm too busy watching and reading Lurie's talks/notes on TQFTs :p

Disciple of Barney

@Incurrence I looked at the relevant section. After you play around and read a bit on these things (I looked at the relevant section in Ch0 and it didn't look like it had too much background needed) you should definetly try that problem I posed. Show that the 3 by 3 Heisenberg group, $H$, over the integers can be written as a a central extension, and show that this extension does not split. So you will have something like $1 \to Z(H) \to H \to L \to 1$.

Incurrence

@DiscipleofBarney Okay I will do this, is it enough to do as a presentation?

Tobias Kildetoft

@BalarkaSen I never got around to even seeing an actual definition of those (not that this stopped me from attending plenty of talks about them. I just never understood anything of them).

Disciple of Barney

11:33 AM

@BalarkaSen Ah watching it, are you liking the material? I watched the first one, it was pretty good with the intuition, given that I don't know what bordism are

Incurrence

Since there are more students than presentation topics, people have to multi present, so I will be doing a dual presentation

Balarka Sen

@Incurrence it is not necessary to understand singular cohomology to do group cohomology, although there is definitely a relation.

Disciple of Barney

@Incurrence Yah, I actually think it could make a good presentation.

Incurrence

@DiscipleofBarney awesome thanks, I will have a shower and go to bed, and see what I can do

I doubt the other guy will do that, so that's definitely a good start

Disciple of Barney

Yah, it is about central extentions

Balarka Sen

11:34 AM

@TobiasKildetoft well, Lurie talks about a definition just at the beginning. it's a pretty systematic invariant.

Disciple of Barney

and get to show one that does not split

Incurrence

Okay talk tomorrow

Thanks

Disciple of Barney

Bye

Tobias Kildetoft

@BalarkaSen I guess I never really felt the need as it is not really my topic (I just happened to do my PhD at a center that specialized in them)

Balarka Sen

@DiscipleofBarney yeah, it sure sounds fun. i'll just see enough of it so that my skepticism about category theory flees.

@TobiasKildetoft I can't really say if I feel too interested about them, either. I am just reading the theory right now, I'd like to see the actual power of the invariant.

Tobias Kildetoft

11:37 AM

@BalarkaSen What are they an invariant of? 3-manifolds, or something more general?

Balarka Sen

They're diffeomorphisms invariants of smooth d-dimensional manifolds in general.

Tobias Kildetoft

I think the only one I hsve any idea of what is, is the one where you get a 3-manifold via surgery of a link and attach reps of quantum $sl_2$ to the components. to get something similar to the colored jones polynomial.

Balarka Sen

You kinda start by forming a category of smooth manifolds $\mathbf{Cob(d)}$ with objects being smooth (d-1) dimensional manifold, and morphisms being bordisms between them. A TQFT is just a symmetric monoidal functor between $\mathbf{Cob(d)}$ and $\mathbf{vect}(\Bbb C)$

@TobiasKildetoft that looks involved.

what i am talking about is Atiyah's definition of a TQFT. it's nice, since you can compute the invariants of a smooth manifold by cutting it out in small "easy" pieces.

Disciple of Barney

It looks like, at the same time it may be justifying higher cats to you @BalarkaSen

Balarka Sen

yeah

i'm just gonna get to extended TQFTs.

that's where the real category theory comes.

Disciple of Barney

11:43 AM

Why stop there? it looks like cool and fun stuff

Balarka Sen

i think i have got loads of other more pressing things to do than studying super-fancy invariants :p for example, i want to study de Rham cohomology first. it sure is cool and fun, though, so i'd like to know the basics. maybe the first two lectures, but not more than that.

i am also reading the notes. i guess i'd want to compute some invariants with it to see how powerful it is.

it's surprising that geometric topologists use such a systematic thing as an explicit invariant.

Disciple of Barney

When he computed the circle, I was like "that is neat" (of course that isn't really the meat of what it is about)

Balarka Sen

i guess it must be very powerful and very computable.

i'm interested in the folk theorem, actually, @DiscipleofBarney. that every commutative forbenius algebra is a 2-dimensional TQFT seems like a very nice fact.

yeah, that TQFT of S^1 has so much structure is nice to watch with a popcornfull of mouth.

2

Disciple of Barney

That was also neat, although I had never heard of a commutative forbenius algebra before I watched it so...

Haha

Well I am going to bed

Balarka Sen

g'night.

i am going to get back to the talk.

@TobiasKildetoft where did you do your PhD?

Tobias Kildetoft

11:55 AM

@BalarkaSen Aarhus (Centre for Quantum Geometry of Moduli Spaces)

(also called QGM)

Balarka Sen

ahh.

cool.

Tobias Kildetoft

Though I was just at that centre because my advisor is there, and he is only there because of some slight overlap of his work with what the centre does (and which is in a rather different direction than what I did)

Gigili

Hi

What is rank-1 approximation?

maja

1:39 PM

that's imberassing, but.. I completely forgot how to take this: $$\lim_{x\to\infty} \frac{1+2^n}{3^n}$$, can I manage this without splitting the limit in two?

Mew

2:31 PM

why don't you split into two?

Mike Miller

@BalarkaSen: It's a convenient framework. I don't know a single useful TQFT that is easy to define, and I don't know a single useful fully extended TQFT (but this could be just my ignorance). As I said before, I think the theory Lurie developed is beautiful - I would be surprised if your average geometric topologist could use it to get something practical to work with.

But the development was partially with this goal in mind.

I guess there's some serious effort to extend some well-known useful TQFTs that are almost, but not quite, TQFTs, but I don't know enough about this to say anything profound

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