Mathematics - 2015-07-03 (page 4 of 4)

hi, @morphic. you're Eyes?

Remember me

Hello @Bal

@BalarkaSen I have a problem for you.

Chris's sis the artist

oooooooooo, what an amazing result I got!!!

Suppose that $\eta :\Bbb Z[[X]]\longrightarrow \Bbb Z[X_1,\ldots,X_n]/(f_1,\ldots,f_d)$ is a nonzero ring morphism. Then the image of $X$ under $\eta$ is nilpotent.

I think this was asked here in MSE. Don't look it up! =D

7:20 PM

@PedroTamaroff With probability one I wouldn't be able to do this.

No point in asking me algebra anymore.

@BalarkaSen well, think of what would happen otherwise: you get X going to something nonnilpotent, then what happens with all of the infinite power series in X?

I think I gave a clever proof there is no nonzero map from prod(Z) (inf many copies) to Z, this feels similar

@anon I nice proof is noting that the LHS has noncountably many units of the form $1+\sum \varepsilon_i X^i$.

Where $\varepsilon_i=0,1$.

I have no idea what's going on.

This maps to countably many units, and then two must repeat. This gives that the image $x$ of $X$ satisfies $x^n\nu=0$ for $\nu$ a unit.

oh, interesting, @PedroTamaroff.

7:26 PM

that's succinct, I like it

By the way, @anon, I made a post on the (now proven) Brauer-Thrall conjecture in my blog.

what's that?

me and mike talked about mapping class groups and braid groups on main

It says that if an artin Algebra has infinite respresentation type, then it admits indecomposable modules of arbitrarily finite long length.

We proved that in Mariano's class when $A$ is a finite dimensional algebra over an algebraically closed field.

@PedroTamaroff That is fairly easy in the wild case, right?

If something is easy in the wild case doesn't that defeat the purpose of calling it wild?

7:29 PM

Hehe.

@anon Did you have time to think about $\pi_1(S^3 - K) \cong B_3$? /reminder

@TobiasKildetoft We didn't really dwell into wild algebras, Tobias.

I think he did talk about how isoclasses of indecomposable modules in certain cases are paremetrized by projective spaces, though.

@PedroTamaroff That sounds a bit like the tame case

We did prove Gabriel's theorem, @TobiasKildetoft!

@BalarkaSen yeah, but with nothing to show for it. I tried to figure out what the generators are. One weird thing is that K is so symmetric but the generators are not (not every permutation of them gives an automorphism). googling led me to an interesting post about modular flow on lattices that I'll have to study more deeply later though.

7:30 PM

@PedroTamaroff Not sure what that is

@anon People are saying that $\Bbb C - \{x^2 = y^3\}$ has a nice description in terms of modular lattices.

It says that if $Q$ is a connected finite quiver then $Q$ has finite representation type if and only if the underlying graph of $Q$ is Dynkin $A_n,D_n,E_6,E_7,E_8$ and in such case there is a bijection between the indecomposable modules of $Q$ and the positive roots of the quadratic form associated to $Q$.

@PedroTamaroff Ahh, nice

In particular, it seems this thing deformation retracts onto $SL_2(\Bbb R)/SL_2(\Bbb Z)$ using some fairly elementary modular theory

what's a modular flow?

geodesic flow in a moduli space

7:33 PM

oh?

i'm hardly familiar with either of those words.

(says someone who never groked moduli spaces)

geodesic flow: things flow along geodesics (curves that minimize distance in curved space). moduli spaces are spaces of objects (equivalently, spaces whose points continuously parametrize some kind of family of geometric objects)

@TobiasKildetoft We also proved the theorem of Brenner and Butler, about tilting modules. And the lemma of Bongartz. Quite amazing things, really. Of course, I should pick a book eventually on these matters and learn them in more detail, in particular I should learn to compute things and apply the transformations and other techniques we saw in class properly.

I know the informal definition of moduli spaces. I don't know how to formalize it.

I have to go, be back in 20'.

for instance, every lattice in C is conformally equivalent to <1,tau> for some tau in the upper half plane, so you can use H to parametrize lattices.

7:34 PM

@PedroTamaroff Not sure what any of those results are

I assume by tilting you mean full tilting, rather than partial tilting?

@BalarkaSen well, don't worry about that, neither do I

@TobiasKildetoft Yes, yes. But Bongartz says that any partial tilting can be completed to a full tilting, for example.

For example, when you tell me that moduli spaces of all elliptic curves is H/PSL_2(Z), that's reasonable to me as elliptic curves are classified by their j-invariant, which is a modular form.

Mariano said a mathematician proved this can be done precisely in two ways.

Really basic predicate logic question: I'm not familiar with discrete math, but I am familiar with general proof writing. How do I do a "formal proof" of $(p \to q) \vee (p \to r)$ assuming $p \wedge q$, $\sim p \wedge r$, and $s \vee t$?

7:35 PM

@PedroTamaroff With what definition of the terms? I am used to situations where there are no full tilting, but plenty of partial

My impression is that I need to show $p \to q$ or $p \to r$.

@TobiasKildetoft Tilting means that ${\rm pdim}(T)\leqslant 1$, ${\rm Ext}^1(T,T)=0$ and there is a SEC $0\to A\to T'\to T''\to 0$ where $T',T''$ are sums of summands of $T$.

To show $p \to q$, I know that this holds as long as it isn't the case that $p$ is true and $q$ is false.

@anon ok. i dunno what i can do with "moduli spaces are spaces that classify things" though. not a reasonable definition to me :(

@PedroTamaroff and what is partial tilting in this context?

7:37 PM

Take away the last condition.

And similarly for $p \to r$, where it is NOT that $p$ is true and $r$ is false.

anyhow, modular flows sounds interesting

@PedroTamaroff Ahh, ok. This does not yield the definition of tilting I am used to, which does not need the category of modules to have any projectives at all

What is your definition?

@Clarinetist p^q by itself implies p->q which implies (p->q)v(p->r)

7:38 PM

@anon for the moment, I am trying to find out a way to define fibers in the context of galois theory of fields.

@PedroTamaroff In a highest weight category, a tilting object is one with both a standard and a costandard filtration

Hehe.

OK.

I have to go!

@anon That's it?

It would be nice if you could tell me about that in some minutes, @Tobias. =D

@PedroTamaroff depends on how many. I will need to go to bed at some point

7:39 PM

At most 20.

then I should still be here at least for a short while

@Clarinetist do you disagree or find my reasoning hard to follow?

@anon Let me see if my understanding's right. $p \wedge q$ true means that both $p$ and $q$ are true, hence $p \to q$ is true and $(p \to q) \vee (p \to r)$ is true?

mmhmm

K thanks :)

@anon Another question. Given $(a \wedge b) \longleftrightarrow a$, how do I show $a \to b$?

My thoughts: I think $(a \wedge b) \leftrightarrow a$ implies $a \to (a \wedge b)$

(I think $\leftrightarrow$ is like an "iff," but I'm not sure)

7:46 PM

@anon wait, what d'you mean by figuring out the generators?

the isomorphism is pretty apparent.

@BalarkaSen oh?

@Clarinetist you can do that one with a couple truth tables if you want (yes it means iff btw)

@anon This person I'm tutoring says his professor wants a "formal proof"

(and AFAIK, the assignment is already due, so I'm not doing his HW for him)

the usual way to compute knot groups is to hammer the knot flat onto the plane, adjoining generators according to the crossings, adjoining relators according to the homotopies between the loops winding around the crossings.

@BalarkaSen I am not familiar with this procedure

oh.

let me find where i have talked with iwriteonbananas about this

it's called wiritinger presentation, btdubs

7:50 PM

I did suspect there were generators corresponding to the "connected components" of the planar diagrams, of which there are three

nah, the isomorphism is easy to show. what i want to know is why must it be the braid group.

i feel like there's a deeper connection

but since the planar diagram is rotationally symmetric, these generators have C_3 acting on them which extend to automorphisms of pi_1 if I'm not mistaken, but no such luck for the generators of the braid diagrams (I don't think you can cyclically permute them and extend to automorphisms of B3)

here

BTW in my original comment when I said generators I mean the loops in pi1 that correspond to the generators of the braid group

wiritnger presentation for the $\pi_1$ of trefoil knot complement is $\langle x_1, x_2, x_3 | x_1 = x_2 x_3 x_2^{-1}, x_2 = x_3 x_1^{-1} x_3, x_3 = x_1 x_2^{-1} x_1\rangle$

Karim Mansour

7:54 PM

@Semiclassical here?

so, oh, there you go about the symmetry

Karim Mansour

lol I like fowles and cassidy definitions sometimes

they sometimes define stuff as self-reference it is funny I noticed that in some physics definition

to actually get $\langle x, y | x^2 = y^3\rangle$, I guess you'd have to do van Kampen on the torus (the trefoil knot is a torus knot - you can embed it in a torus).

Hatcher has it. I've never tried doing it, as Wiritinger is much simpler

did you see the link above, @anon?

yeah

I'll read the discussion later

ok, sure.

computing knot groups turn out to be very easy for tame knots. however, what I want here is to have an explicit reason for why the knot group is the braid group on 3 strands.

i.e., there must be some direct connection between the trefoil and the braid group

8:00 PM

@BalarkaSen so do you know what the loops around K are that correspond to the three generating braids in B3?

not directly, no.

but you can find it out by explicitly coming up with an isomorphism of B_3 with the group i wrote down above

@BalarkaSen did you write down the isomorphism in the linked chat?

no, I didn't.

also, were you also going to say something about the symmetry?

well, yeah, the symmetry is apparent in $\langle x_1, x_2, x_3 | x_1 = x_2 x_3 x_2^{-1}, x_2 = x_3 x_1^{-1} x_3, x_3 = x_1 x_2^{-1} x_1\rangle$

8:04 PM

I see a group presentation and a suspicion that you can do kampen on the torus, but I don't see an isomorphism written down.

$x_1, x_2, x_3$ are precisely the loops that goes around the overpasses in the trefoil knot

@anon Last question here.

If line 10 is correct, the variable is declared.
The variable is declared or line 10 isn't correct.
Therefore, line 10 is correct or the variable is declared.

Let $p$ = “line 10 is correct” and $q$ = “the variable is declared.”
Is the claim to be proven here $(p \to q) \wedge (q \vee \sim p) \to p \vee q$?

@anon ok, so you want an isomorphism, right? remind me what's the standard presentation for $B_3$.

it'd just be tedious job, not hard.

why do I have to type the damn thing when it's on wikipedia? :P

@BalarkaSen Same as for $S_3$ without the two generators squaring to the identity

8:07 PM

there are lots of presentations for S_3. are you talking about the (12), (123) one?

@BalarkaSen if the loops in pi1(S^3-K) corresponding to the canonical generators of B3 are sufficiently crazy-looking, then I might start being suspicious there is an elegant and direct relationship between the torus and 3-braids.

i'll just look up wiki. sigh.

@BalarkaSen No, the one with the simple transpositions

i don't recall what the relators were

@TobiasKildetoft HAI.

8:09 PM

@PedroTamaroff Welcome back

oh, I see. that one.

well, there's the commutativity one if they're not right next to each other, and then there's the reidemeister move one

those are all the relations

looks like a complicated presentation

@anon In fact the so-called braid relations

Pulp Fiction Scene - Best Intentions

@TobiasKildetoft You were saying something about "...a highest weight category" and "...a tilting object."

►

8:12 PM

artin's presentation $\langle a, b : aba = bab\rangle$ is simpler

@BalarkaSen Right, that was the one I meant

@PedroTamaroff Yeah.

ok.

Have you seen the term highest weight category before?

let me whip up an isomorphism.

@TobiasKildetoft No, not at all.

8:13 PM

@PedroTamaroff Are you familiar with highest weight modules for semisimple Lie algebras?

No. =/
*Thinks this might be hopeless.*

@PedroTamaroff Hmm, then it might be a bit tricky. The definition itself is not terribly complicated, but the motivation will be missing

You would not happen to have heard of quasi-hereditary algebras?

@TobiasKildetoft Yes, I have.

@PedroTamaroff Well, then we can start there. A highest weight category is (it turns out, up to equivalence), the module category of a quasi-hereditary algebra

OK.

8:20 PM

Unfortunately, I do not recall the details of how one shows this, but in the category, we then get standard and costandard modules

one of each for each element in some poset

for $\lambda$ in that poset, the standard module is called $\Delta(\lambda)$ and the costandard is called $\nabla(\lambda)$

$\Delta(\lambda)$ has simple head $L(\lambda)$ which is the socle of $\nabla(\lambda)$ and all simple modules are of this form for a unique $\lambda$

@TobiasKildetoft What form does $\Delta$ have?

@PedroTamaroff Not sure what you mean by form

Err, I mean. When you say module I think module over a ring.

Are you defining a ring in terms of the poset?

@PedroTamaroff Right, these are modules over a quasi-hereditary algebra

OK. But where is the poset arising from? I thought I know what quasi-hereditary algebras where. I am reading they are defined in terms of chains of ideals that are QH. Perhaps the poset is obtained by looking at primitive idempotents with $e\leqslant f$ is $f$ has a refiniment that contains $e$ or something of the sort?

8:27 PM

I am stating some properties of the category

@PedroTamaroff I forgot where it comes from in general unfortunately

One way to get an example is to start with a poset and define an algebra from that (the incidence algebra)

Minimal elements would correspond to primitive idempotents.

@TobiasKildetoft Yes, I've met those in a combinatorics course.

@PedroTamaroff sounds about right

OK.

@PedroTamaroff I need to go now. Maybe we will have a chance to talk about this another time. Let me just mention that a source to read about this is a paper by Parshall and Scott called Derived Categories, Quasi-hereditary Algebras and Algebraic Groups

@TobiasKildetoft Cool. Thanks. Cheers.

8:33 PM

and that the motivating example is the category of modules for a semisimple Lie algebra or algebraic group

or at least a suitable type of subcategory of these

well, duh, @anon, the isomorphism is obvious.

$x_2 \mapsto a$, $x_3 \mapsto b$.

he says 20 minutes later

what are a and b?

oh, your a^2=b^3 thing

Hehe, anon being mean.

$a, b$ are generators of $\langle a, b | aba = bab\rangle$.

and that's B3?

huh

8:37 PM

yeah, that's artin's presentation of B_3

like I said, the Wiritinger presentation spits out B_3 in a second (not 20 minutes : that's just dumb me).

well, I think of braid groups using braids and braid relations

and was not familiar with the artin presentation

does it generalize to B_n in some way?

I found it here, where they have wrote down the generators in the braidish-fashion

I don't know.

oh derp, I am thinking of B4 not B3

B3 is generated by two braids

yes.

well, that was a waste of time and energy

8:41 PM

hi mr @Pedro

hello @Ted

Hello mr Ted.

hi @Balarka

Boy, @Pedro, embarrassed myself earlier with comments on a silly exterior algebra question. Jack Lee now thinks I'm totally stupid :P

ok, I'm not willing to think about this anymore. @anon let me know if you find something.

I have to smash my brains open on the fiber-in-galois-theory problem, as usual.

@TedShifrin Where?

Exterior algebras are nice. [something something, Koszul resolutions, something].

8:45 PM

Someone was trying to prove that $\Lambda^k T^*M$ was a smooth bundle by applying the rank theorem to the map $\text{Alt}: \otimes^k T^*M\to\otimes^k T^*M$. I suggested that was a silly way to do it but then embarrassed myself by saying that the map did, in fact, not have constant rank.

Oh. Peter doesn't know about tangent spaces.

Nonsense.

Well, actually, I was insinuating that the map $\otimes^k V\to\otimes^k V$ didn't have constant rank.

I was, it turns out, thinking of the bilinear map $V\times V\to \otimes^2 V$ rather than the linear map given. I should just stick to basic calculus :)

Hehe. Perhaps you were too worried about water running out in California.

No joke :(

Come to South America. We have plenty water. =D

8:50 PM

Plenty of it and heat and humidity in GA, too.

But I do want to visit South America. Want to learn some Spanish first. By the time I get around to it, you'll probably be in North America.

what about water?

California has none @Balarka

wut

@TedShifrin As always, you're ever-welcome. =)

I'm gonna work on learning Spanish, partly thinking it might help a bit tutoring in San Diego.

8:52 PM

Googled. Didn't know you people were having a water-crisis.

I'm not there yet, @Balarka, but I shall be in three weeks.

I wonder how that happened. No rain?

@TedShifrin ah, true. yikes.

No rain, not enough snow in the mountains, and abuse by various people in industry. And, to a minor extent, residents still wasting water.

very uncool.

Pedro, where in South America are you from? just curious

8:55 PM

@Cristopher Argentina.

So, @Pedro, you now 100% a confirmed algebraist?

Oh, that's great. Me too :P

@TedShifrin Well, I am trying not to be consumed by the dark force, Ted.

But you're failing?

Just partially.

8:56 PM

I sure miss the days when people had informative profiles on here.

My profile has a link to my blog.

My blog says I am from Argentina.

I think that's in my profile, too.

I'm not talking about you, @Pedro. It's most everyone else.

@Christopher doesn't really have an informative profile.

Besides, for better or for worse, I already know who you are.

I've got an informative profile.

8:58 PM

@TedShifrin I just replaced the Wilson shoes we bought in Athens.

Soles wore out. =(

Minimally so, @Balarka. Well, @Pedro, a year isn't bad when you play so much!

Yes, that's what I thought too.

But now you have a new place to visit me :P

I'm gonna try to find some people to play tennis with ... And will be playing bridge seriously.

sanity check : there's a bijective correspondence between fibers of $p : \widetilde{X} \to X$ and morphisms $\widetilde{X} \to \widetilde{X}$ in $\mathsf{Cov}/X$, right?

I don't? Well I guess I don't, hehe

9:01 PM

I have no idea what your final notation is, @Balarka, but, yes, I believe that's part of the homotopy lifting theorem.

@Cristopher Do you study mathematics?

$\mathsf{Cov}/X$ is the category of covers over $X$, with objects being covers, and morphisms being continuous maps between covers such that the obvious triangle with $X$ below commutes.

ok, good.

@PedroTamaroff Not really. I'm studying to become an actuary but my studies involve a lot of mathematics, especially statistics. what about you?

@Balarka: They really use the notation / rather than $\text{Cov}(X)$?

I think the / is used to be read as "over X".

9:04 PM

yeah. I guess that's because it's a slice category (whatever that is)

Weird, though.

$X$ stays fixed or $X$ varies in this category?

$X$ is fixed

I see, so that makes my notation inappropriate.

@Cristopher Ah. I'm studying in FCEyN, mathematics.

9:05 PM

These days, there's hardly any math at all in actuarial science. But we won't get @Clarinetist started.

@PedroTamaroff Wow. I would really like to study mathematics at UBA, too :P Pure math or applied math?

Pure.

@TedShifrin I don't know how the htpy lifting comes in. The bijective correspondence I have in mind sends a point $x_1$ in $p^{-1}(x)$ to the morphism $\widetilde{X} \to \widetilde{X}$ that sends a point $\widetilde{x}$ to $\tilde{\gamma}(1)$, where $\gamma : [0, 1] \to X$ is the path corresponding to $\widetilde{x}$, such that the lift $\tilde{\gamma}$ is based at $x_1$.

Path lifting then. I didn't stop to think. But it's part of that argument, yes.

Awesome.

9:07 PM

oh, sure.

Besides, I'm losing my math brain entirely, @Balarka. Soon I'll be down to sophomore level.

What subjects are you studying now?

hm. guess that's not hard to analogize in the galois theoretic setting.

@TedShifrin nah.

@Cristopher Probability, real analysis.

BTW, @Pedro, did you get my ping days ago with the correct way to do that probability question you asked?

9:13 PM

@TedShifrin Yes. I worked it out.

Cool

My double integral was right. My measure wasn't. That's similar to a question I gave my class about breaking a stick twice and asking for the expected length of the longer piece of the shorter piece.

oh. given a galois extension $K/k$, I think the analogous notion of fiber is precisely the set of roots of the minimal polynomial of the primitive element of $K$ over $k$.

hmm

@Balarka: You might go back to the original thing I thought was right. I'm no longer sure it is.

it makes sense, as $\#$ of the roots of the minimal polynomial is precisely $[K : k]$, which fits with the fact that number of sheets of a cover is equal to the index of the top fundamental group in the bottom fundamental group

@TedShifrin hm?

9:19 PM

Told you. My math brain is dead.

I am sure it's a bijection. the first map I have already described, the other-direction map sends a morphism $f : \widetilde{X} \to \widetilde{X}$ to $f(pt)$ where $pt$ is the element in $\widetilde{X}$ corresponding to the constant path $[0, 1] \to X$.

hmm.

speaking of which, I should look at the construction of the algebraic closure once again. universal cover is constructed as the space of all homotopy classes of paths based at a chosen basepoint. is there any analogous thing done with algebraic closures?

then it should be possible to analogize the notion of paths and homotopies in galois theory of fields

9:37 PM

bah, impossible to translate all that algebra in the covering space language.

hi @PaulPlummer

Hello @BalarkaSen

how's your day?

Its going okay, that broken mouse sucks. Doing a problem in Hatcher, you can homotopy any two retracts (to a common subspace)

hm, I don't recall that one.

It is problem 13 chapter 0

9:49 PM

ohh.

i recall struggling a lot with it before coming up with the right thing

Yah I think I am getting close, but not there yet

let me know when you come to more interesting things like 16 or 18 :)

I will

(or rather don't. if you tell me about 18, i'll probably blabber your ears off about hopf fibrations)

i'm rather fond of the "eureka" moments i have had with those problems :P

Well I will tell you about 18 afterwords then :D Learning about hopf fibrations should be cool

9:59 PM

afterwards ... taking @Jasper's role here.

we'll see about that when I start talking :P

haha @Ted

It's past your bedtime again, @Balarka.

right, I think I am going to go to sleep.

Have you been looking at schemes? @BalarkaSen

I "know" what schemes are, but they are hard stuff.

I'd've to learn some of those at some point of time, but that time is not now.

10:01 PM

Then you have to know what sheaves are.

I guess I would have to.

Oh, well I was sort of looking around for different versions of van Kampen when you were talking about it, and it sort of looks like there is some van Kampen for schemes (I may have completely misunderstood what was going on)

no, I don't want to know if there is something like that.

That's for an algebraic sort of $\pi_1$, yes, @PaulP.

I am pretty sure what I am talking about has a complete rigorous description, but I want to figure it out myself.

10:03 PM

@TedShifrin Why did you say there's hardly any math in actuarial sciences? In fact in actuarial sciences we study analysis, algebra (concepts like matrices, subspaces), statistics (e.g. stochastic processes), numerical analysis, and even topology concepts

Ah so you want something different

well, in this country, @Cristopher, we used to have exams on numerical analysis and other math things, but ten years ago it changed. Now it's all statistics, economics, and business.

I just want to figure out if there's an analogy in the context of galois theory of fields

and I kind of want to figure out myself :)

@TedShifrin Oh I see. That's too bad. It's not like that here

Oh I understand the feeling

10:05 PM

Lots of things have gone downhill in the US, @Cristopher.

@Paul Yep. The point of doing mathematics is to rediscover (and sometimes really discover) things by yourself and enjoy doing so.

Oh well :( I guess I can consider myself fortunate to study in Argentina. Even though here some things have gone downwhill too

I don't know ... It seems Pedro has some decent math opportunities there.

Nevertheless, what you say brings hope that there might just be something like an analogy in the context of fields. Schemes are just the kind of things that represents top. spaces and rings/fields at the same time.

Thanks for letting me know!

No problem it sounds like it should be an interesting problem

10:09 PM

well, an interesting topic, as it's not a rigorous question or conjecture or anything.

Maybe you will rediscover schemes, and learn what they really are trying to figure out the problem :D

nah

Maybe. Good for him if that's the case :)