Mathematics - 2016-04-23 (page 2 of 3)

Balarka Sen

17:00

My idea is to let $f_n$ be $1$ on $(r - 1/n, r + 1/n)$ whenever $r$ is rational and $0$ for irrationals sufficiently away from $r$ and in the middle a segment joining $0$ and $1$ in the graph. This seems very doable.

Mambo

@TedShifrin Yes. We have a Lie bracket in the latter set as $[X,Y] f = X(Yf) - Y(Xf) $. How do we relate our concrete Lie Bracket to this one ?

Balarka Sen

We just need $f_n$ to be $C^0$. In fact one shouldn't have much trouble applying the same idea for $C^\infty$ using bump function techniques either, if $C^0$ can be done.

Ted Shifrin

Well, then you're certainly not believing the result I'm telling you to prove, @Balarka.

Valery Saharov

Hi Robjohn. Care to check my solution? math.stackexchange.com/questions/1745355/…

Ted Shifrin

@Mambo: Differentiate $e^{tA}e^{tB}-e^{tB}e^{tA}$ at $t=0$.

Balarka Sen

17:02

@TedShifrin I admittedly don't see the correlation between the problem you asked and the result you told me.

Ted Shifrin

I am saying the limit function must be continuous on some nonempty subset.

Balarka Sen

Well, the characteristic function of rationals is continuous on the rationals.

Ted Shifrin

spanks @Balarka very hard

Mike Miller

What? No, that's totally and obviously false...

Valery Saharov

Tell me at least how to find a basis of a space that $P_i:=\frac{1}{2\pi i}\oint\limits_{\mathcal{C}(c_i, \varepsilon)} (A-z I)^{-1}\,dz$ projects onto if the dimension is known?

Mambo

17:04

@TedShifrin But how is that an operation on vector fields?

Balarka Sen

Op. I am confusing the popcorn function with this. I apologies.

Ted Shifrin

@Mambo: $e^{tA}$ is the left-invariant vector field corresponding to $A$ in the Lie algebra.

You raise a more basic question, @Balarka. Can there be a function on $\Bbb R$ that is continuous precisely on the irrationals? precisely on the rationals?

Balarka Sen

And the popcorn function is not continuous on the rationals either. On the irrationals. Eesh.

Ted Shifrin

(That's actually on my topology final I'm giving in a week or so.)

Mambo

@TedShifrin Can you please elaborate on that ?

Balarka Sen

17:06

@TedShifrin Popcorn function is an answer to the first.

Mike Miller

sigh

Balarka Sen

@TedShifrin Ah, now I see why I said such a thing.

Ted Shifrin

I slightly lied, @Mambo. Given an $n\times n$ matrix $A$, we get a left-invariant vector field on $GL(n)$ by taking $X(M) = \dfrac d{dt}\big|_{t=0} Me^{tA} = MA$.

Balarka Sen

I thought by continuous on a nonempty set you meant continuous when restricted to the set with the subspace topology.

Mike Miller

what

Ted Shifrin

17:10

No, I meant points of continuity, @Balarka, which is what everyone means.

Balarka Sen

Of course this is true for the char function on the rationals - it is identity restricted to the rationals hence cont.

@TedShifrin Right, I figured.

Mike Miller

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3

Semiclassical

ummmmm okay

Mambo

Is any left invariant vector field on $GL(n)$ given this way? @TedShifrin

Ted Shifrin

Yes, @Mambo.

Left- (or right-) invariant vector fields are in one-to-one correspondence with elements of the Lie algebra.

Balarka Sen

17:14

@MikeMiller (1) I didn't see the removed message. Did I say anything wrong? (2) "what?" I know, that's silly, but it didn't strike me at first that that's what Ted meant by continuous on a set.

Ted Shifrin

Mike got first and second confused whilst accusing you of the same.

Mike Miller

yeah

when i'm unnecessarily mean i try to do my best to also be right

sometimes it doesn't work so well

ADG

hello people, how to put n red and n blue balls into m bins with each bin having exactly same number of both coloured balls.

Balarka Sen

OK, then probably I am not as silly today as I am usually. One terminology misunderstanding - that's alright. Need to be more careful.

ADG

no one has any idea?

Ted Shifrin

17:18

What do you mean by "how to put"?

Mambo

@TedShifrin So, we can define the Lie algebra associated to any Lie group $G$ as the set of all left invariant vector fields on $G$.

Ted Shifrin

Precisely, @Mambo.

ADG

number of ways of putting inside the bins @Ted

Ted Shifrin

Are the bins distinguishable or not distinguishable?

Mike Miller

@Ted I answered Alex M.'s question, though I don't know why he didn't do it himself.

Mambo

17:20

@TedShifrin Then for a general Lie group $G$, what is the counterpart of exponential ?

ADG

@TedShifrin both are distinguishable

actually i'm working it out for every case (total 4)

Ted Shifrin

Still called exponential, @Mambo. Comes from solving a differential equation.

Mambo

Is it called one-parameter subgroups? @Ted

Ted Shifrin

This is a standard counting problem, @ADG. I don't see why the two colors of balls are particularly relevant. Can you do it with just red balls?

Yes, @Mambo.

Balarka Sen

Yes, now I see why my construction of why $\chi_{\Bbb Q}$ cannot be limit of the $f_n$'s I proposed. Given any irrational $x$, there is a rational inside $[x, x+1/n]$. A $1/n$-nbhd of that rational will contain $x$, hence be $1$ on $x$. My $f_n$'s are all identically $1$, I think.

ADG

17:23

@TedShifrin I think you are taking this question casually. for n red balls into m boxed both distinguishable i know it's $m^n$

but it is difficult to include that each box must have same number of both coloured balls

also order inside box isnot important, that's also a problem

Ted Shifrin

@ADG: I'm not really interested in working on the question. So, yes, I am taking it casually. So since I don't really want to think about it, I withdraw any comments.

ADG

@TedShifrin OK. And I am sorry for my rudeness. Thanks for helping.

Ted Shifrin

No problem.

Balarka Sen

I am surprised I didn't note that my construction fails trivially.

What was I even expecting.

Mambo

@TedShifrin Thank you

Balarka Sen

17:37

@TedShifrin Interesting. I am starting to believe your claims now. Cool.

Mike Miller

I think this guy wants a brief and straightforward exposition of homology, probably so that he can get to work on Alexander polynomials. Not sure if there are any good references but just saying "read chs 1-2 of Hatcher".

Which is not very brief.

Balarka Sen

Why does he need to know chapter 1 to learn homology?

Ch 2 in Hatcher is quite independent of ch 1 in my opinion. For motivations and intuitions, probably needed. But not to get the job done, in my opinion.

Mike Miller

Probably would be wise to know about fundamental groups if you're doing knot theory.

Balarka Sen

Fair enough, I thought OP already knows about those since he only mentioned homology.

Mike Miller

I don't think that's a fair assumption.

Ilan Aizelman WS

18:08

How are you @TedShifrin? :)

Balarka Sen

18:34

Nice, I can integrate things just fine.

Except that I haven't really read spherical coordinates too much because it looks tedious to me.

Akiva Weinberger

Hola

Balarka Sen

Hi @Akiva

What's your favorite way of proving $\int_{0}^\infty e^{x^2} dx =\sqrt{\pi}/2$?

Akiva Weinberger

Uh, I haven't heard of very many

I think I might only know about the polar coordinates one.

Balarka Sen

Ah.

Akiva Weinberger

Maybe I've seen another one, but I've since forgotten it

Balarka Sen

18:48

Nevermind then. What's up?

Akiva Weinberger

So, my not-Hatcher book quickly became too much for me

and I need to do some Hatcher exercises to have any hope of progressing there

Balarka Sen

Then slow down a few notches. Don't give up.

Akiva Weinberger

Chapter 3 section 1, by the way

Balarka Sen

Did you do chapter 2.2. problems?

Akiva Weinberger

Yeah

Balarka Sen

18:50

How did you prove pasting two disks along S^1 by a degree 2 and degree 3 maps resp gives S^2?

Akiva Weinberger

Wait, no

Not the later ones

Anubhav.K

@BalarkaSen how r u?

Balarka Sen

alright

Akiva Weinberger

I think that one I looked up, actually.

Sorry to disappoint.

Balarka Sen

I don't think that has a right answer anywhere on the internet.

Akiva Weinberger

18:51

No?

Balarka Sen

Well, right and elementary answer.

Anubhav.K

How is your algebraic geometry study going on?

Akiva Weinberger

Huh.

Balarka Sen

Of course you can just Whitehead the shit out. That's probably on the internet.

Akiva Weinberger

I don't think I know about Whitehead

(except the book he wrote with Russel, which is very unrelated)

43 secs ago, by Anubhav.K

How is your algebraic geometry study going on?

Seconded

Balarka Sen

18:53

It says if $f :X \to Y$ is a based map between CW complexes which induces isomorphism on $\pi_n$ for all $n$, then $f$ is an htpy equivalence.

lol not that Whitehead.

I think this one's J. H. C. Whitehead.

Akiva Weinberger

Oh, hm.

How's algebraic geometry going for ya?

Balarka Sen

@Anubhav.K @Akiva Alright. Done with most of the material in Shafarevich chapter 1,2. Time to learn differential forms.

Anubhav.K

good

Akiva Weinberger

@Anubhav.K You know algebraic geometry?

I don't

Balarka Sen

@Akiva Gimme a map $f :X \to Y$ which is an isomorphism on $H_n$ for all $n$, but is not a homotopy equivalence.

$X, Y$ CW complex.

Anubhav.K

18:55

I don't, but I'll start after a few days...may be after 2 weeks...after my exam

But before starting AG, I want to do a few chapters of commutative algebra

Akiva Weinberger

Well, I know that $T^2$ and $S^1\vee S^1\vee S^2$ have the same homotopy, though I'm not sure how to use that

(I don't think I know commutative algebra, either. What is it?)

(Not as simple as just studying abelian groups, is it?)

Balarka Sen

@AkivaWeinberger Note that we have an explicitly map here. So just having two things with same homology groups is not helpful.

Anubhav.K

A bit more, where people do things over rings and modules

Balarka Sen

@AkivaWeinberger You study commutative rings.

Tobias Kildetoft

@BalarkaSen And their modules

Akiva Weinberger

18:58

Aha

Hello, Toby

Can I call you Toby?

Sorry

Tobias Kildetoft

@AkivaWeinberger Sure, though nobody usually calls me that, I don't mind

Balarka Sen

@TobiasKildetoft Right. Hi, by the way.

Tobias Kildetoft

@BalarkaSen Hi

Akiva Weinberger

Right, so, remember how $S^1\vee S^1\vee S^2$ is homotopy equivalent to a solid torus minus a trivial loop?

Map the torus to the boundary of that. Does that work?

Anubhav.K

What is your qes??

Akiva Weinberger

19:02

7 mins ago, by Balarka Sen

@Akiva Gimme a map $f :X \to Y$ which is an isomorphism on $H_n$ for all $n$, but is not a homotopy equivalence.

Anubhav.K

ohhhh....

Balarka Sen

@AkivaWeinberger Not sure if I get your map.

Mike Miller

@BalarkaSen Torus to boundary of solid torus minus loop in interior.

Akiva Weinberger

Yeah

Balarka Sen

Ah.

Akiva Weinberger

19:04

Sí, כן

Anubhav.K

but you need to get isomorphism in the homology group too

Mike Miller

In any case, that's clearly not an isomorphism on $H_1$. :)

Balarka Sen

Right, because look at image of a meridian.

Akiva Weinberger

Oh

Right, yes

Hmmm.

Unrelated: There's a movie being made called Deeper

I just read the script. It's pretty great

Balarka Sen

I rather look forward to the new Captain America movie.

Mike Miller

19:14

i just want good questions

Balarka Sen

Why does one need to blowup $y^2 = x^3$ ones but $y^2 = x^5$ twice to remove singularities? Given a curve, how do you know how many times we need to blow one singularity up to remove it? How do we know that number is finite at all?

Anubhav.K

Ok , recently I was thinking, how to prove that there exists no closed surface with $\pi_1= \mathbb Z$ , without using classification theorm?

Balarka Sen

How about that? That's something I have been wondering.

Mike Miller

@BalarkaSen Didn't you just learn the answer to these?

tylerl-uxai

Can someone please show me how 0 factorial is 1?

Balarka Sen

19:17

Someone told me "something something arithmetic genus", but I didn't get a satisfactory answer.

Mike Miller

That is the answer.

Balarka Sen

But it's not clear to me geometrically what is happening :( I'd like an overview on that.

tylerl-uxai

Sorry to interrupt what you were talking about

Mike Miller

@Anubhav.K What am I allowed to use?

Anubhav.K

You are allowed to use Homology Co-homology...may be properties of homotopy, but not the classification theorem

Balarka Sen

19:19

One can use things Mike used in his recent blogpost. But more or less one really proves the classification theorem that way.

I think

Mike Miller

@BalarkaSen Yeah, that's just a classification.

Anubhav.K

I mean you cannot use the fact that orientable manifold has eular characteristics 2-2g either

Ohh, Mike has a Blog...I didnt know that

send me the link

Mike Miller

OK, so we may as well assume the surface is orientable, by passing to the double cover. One version of Poincare duality implies that the cup product pairing on $H^1(\Sigma;\Bbb Q)$ is nondegenerate. This implies it's even-dimensional.

So your surface does not exist.

Anubhav.K

yeah

Akiva Weinberger

@tylerl-uxai So

Anubhav.K

19:23

and $H_1$ torsion free implies it has to orientable

tylerl-uxai

hey!

So 0 times anything is still zero.. not 1

Mike Miller

@Anubhav.K That's true by classification, but it's not obviously true to me without it.

Akiva Weinberger

We know that 3!=6 because that's how many ways we can rearrange three objects:

tylerl-uxai

ok

Akiva Weinberger

(1 2 3), (1 3 2), (2 1 3), (2 3 1), (3 1 2), (3 2 1)

Anubhav.K

19:24

No, that's an application of non-deg of cup product only

Akiva Weinberger

That's six

Mike Miller

Oh, no, I'm being silly.

Anubhav.K

that for an non-orientable closed manifold $H_{n-1}$ is not torsion free

Akiva Weinberger

Similaely, 2!=2 because we can rearrange two objects in two ways:

(1 2), (2 1)

Mike Miller

That's not a cup product thing. That's just that $H_2 = \Bbb Z$ plus universal coefficients.

Akiva Weinberger

19:25

And 1!=1 because we can arrange one object in one way:

(1)

Now, for zero objects: How many ways can you rearrange that? Just one:

( )

Another argument:

You can imagine every product as having an infinite along of *1s at the end.

3! = 3 * 2 * 1 * 1 * 1 * 1 * …

2! = 2 * 1 * 1 * 1 * 1 * 1 * …

Anubhav.K

@MikeMiller send me your blog's link please

Mike Miller

I don't know it.

Akiva Weinberger

1! = 1 * 1 * 1 * 1 * 1 * 1 * …

Mike Miller

I have to google for a while to find it every time I want to write.

Balarka Sen

assorted details following

that's the name

Mike Miller

19:28

Thanks.

tylerl-uxai

Okay back @AkivaWeinberger (I had to say hello to relatives)

Akiva Weinberger

@tylerl-uxai So, continuing the pattern:

Balarka Sen

No problem. It's a good blog, more people should know about it.

Akiva Weinberger

0! = 1 * 1 * 1 * 1 * 1 * 1 * …

Balarka Sen

In particular, you should know about it, @MikeMiller ;D

tylerl-uxai

19:29

that's really cool that you can think of multiples as... rearrangement

or factorials I mean

Mike Miller

I need to figure out how to set it up so it only shows the first few paragraphs of a post.

tylerl-uxai

So does that mean it's undefined since it goes to infinity?

There are no objects?

So there's an undefined number of arrangements

Akiva Weinberger

I gave two proofs; one for each definition

Factorials are usually defined in two ways: In terms of rearrangements, and in terms of multiplying stuff

@tylerl-uxai No, there's one arrangement.

tylerl-uxai

sorry, but 1*1*1 is probably just 1 eventually

Akiva Weinberger

( )

tylerl-uxai

19:30

unless you multiply it by 0 eventually

Akiva Weinberger

@tylerl-uxai It is. That's why 0! = 1

tylerl-uxai

not trying to sound arrogant by arguing, but I really don't think you can multiply 0 times something to get 1

why are you multiplying 1*1*1?

Akiva Weinberger

I'm not. Did I ever do that?

I was continuing the pattern.

4 mins ago, by Akiva Weinberger

3! = 3 * 2 * 1 * 1 * 1 * 1 * …

tylerl-uxai

so it stops at 1 and repeats

instead of counting back

Akiva Weinberger

4 mins ago, by Akiva Weinberger

2! = 2 * 1 * 1 * 1 * 1 * 1 * …

4 mins ago, by Akiva Weinberger

1! = 1 * 1 * 1 * 1 * 1 * 1 * …

3 mins ago, by Akiva Weinberger

0! = 1 * 1 * 1 * 1 * 1 * 1 * …

In the first, I multiply everything from 3 to 1 (three things), and then a bunch of ones

Balarka Sen

19:33

@Anubhav.K A more convoluted proof: Let $\pi_1 = \Bbb Z$. Then universal cover is noncompact. It has $\pi_1 = 0$, so by Hurewicz $\pi_2 = H_2$ which is also $0$ as it's a noncompact 2-fold. So $\pi_2$ of the base is also zero. Hurewicz + vanishing of higher homologies keeps getting continued, so $M$ is a $K(\Bbb Z, 1)$. That's $S^1$. No orientable 2-manifold can ever possibly be htpy eq to $S^1$, because of $H_2$.

Akiva Weinberger

In the second, I multiply everything from 2 to 1 (two things) and then a bunch of ones. In the third, I multiply everything from 1 to 1 (one thing), and then a bunch of ones. So, for 0!, I multiply everything from 0 to 1 (zero things) and then a bunch of ones.

tylerl-uxai

thanks so much

Akiva Weinberger

(Where by "from 0 to 1" I mean everything less than 0 and more than 1; zero things satisfy that)

@tylerl-uxai Another argument:

tylerl-uxai

but like, is that also

could you say 1*1*1*1*1*zero=0?

since there's a zero object?

Akiva Weinberger

Yeah

tylerl-uxai

19:35

ok cool

thank god someone is agreeing with this

Akiva Weinberger

It's just that, in 0!, we never multiply anything by zero

Anubhav.K

@BalarkaSen this is also good

tylerl-uxai

Why do we always multiply by an infinite number of 1s in factorials?

Akiva Weinberger

In factorials, we never multiply by anything less than 1.

@tylerl-uxai I was just saying that you can consider every product to be like that.

It's always the same thing, with or without the ones.

Balarka Sen

It uses more tools than Mike's proof (namely, Hurewicz), so I thought against posting it. But then I changed my mind. This one can be modified appropriately to prove that no 3-manifold can have fundamental group $\Bbb Z$ either, IIRC.

tylerl-uxai

19:36

@AkivaWeinberger Does this make sense to you?
var start,
result;

for ( var n = factorial; n > 1; n = n - 1 ) {
if ( n === factorial ) {
// n * ( n - 1 ) is a start...
start = n * ( n - 1 );
result = start;
} else {
// after it loops once, substitute 'result' for 'n'
result = result * ( n - 1 );
}
}

Balarka Sen

$\pi_2 = 0$ is problematic for 3-folds though. Hmm.

Akiva Weinberger

So, when we define what it means to multiply zero things together (called the "empty product"), we define it to be one

Mike Miller

@BalarkaSen No 3-manifold can have fundamental group $\Bbb Z$, eh?

Anubhav.K

yes

Balarka Sen

compact, connected, orientable.

Anubhav.K

19:37

closed

Balarka Sen

without boundary

tylerl-uxai

I guess it's good that it doesn't keep multiplying by 1... Since that would cause it to loop until infinity

Akiva Weinberger

Here's what your program should do to also work with 0:

Mike Miller

@BalarkaSen No compact, connected, orientable 3-manifold without boundary can have fundamental group $\Bbb Z$, eh?

Balarka Sen

Oh sure

S^1 times S^2

Akiva Weinberger

19:38

Well, I don't know what programming language that is, so I'll explain in words

Balarka Sen

blah

oh come on.

Anubhav.K

yes

Akiva Weinberger

Let i be 0, and let factorial be 1

Anubhav.K

$Z^4$ be the case

Balarka Sen

@MikeMiller Ah, now I see what I need.

Sphere theorem. There's precisely two.

Akiva Weinberger

19:40

After each step, set i=i+1 (that is, increment it) and set factorial=factorial*i (multiply by the new i)

Balarka Sen

S1 \times S2 and it's nonorientable counterpart.

Akiva Weinberger

And, if i==n, stop.

Balarka Sen

Both bound embedded spheres representing nontrivial class in pi_2 but are prime.

Akiva Weinberger

So, essentially, at every stage, we have factorial equal to the factorial of i

Even at the very start

Mike Miller

That's correct. What's the extra condition you wanted in your argument above to deal with $\pi_2$?

Also that's not the sphere theorem

Akiva Weinberger

19:42

So it starts by knowing that 0!=1. Then it multiplies by 1 to get 1!=1, and it multiplies by 2 to get 2!=2, and it multiplies by 3 to get 3!=6, and… etc.

Until it gets to n

Balarka Sen

@MikeMiller They have to be irreducible.

Akiva Weinberger

(Incidentally, != means "not equal" in many programming languages. So 0!=1 can either be read as "zero factorial equals one" or "zero does not equal one." And they're both true)

Mike Miller

Good. So what general conclusion did you just prove about irreducible 3-manifolds?

Akiva Weinberger

(Just a little curiosity)

Balarka Sen

I think we already discussed this before, but I forgot. I proved that there are no irreducible closed compact connected 3-manifold with $\pi_1 = \Bbb Z$.

Mike Miller

19:45

how did you prove this?

Danu

@MikeMiller Hmm... I don't think the prof. for gauge theory is that motivated to do very interesting/advanced stuff :\

Mike Miller

what's he gonna do

Danu

I think he's choosing audience over content..

I'm not even sure he'll do anything beyond mathematically formulating the standard model (of physics)

Mike Miller

oh

Danu

He seemed very hesitant to make any "promises" when I asked him about what more we can do

Mike Miller

19:47

Take the course & then read Salamon's notes on SW theory :)

Akiva Weinberger

So, I'm back (after discussing 0! with someone)

I still have that problem to do.

Balarka Sen

Look at universal cover. $\pi_1 = 0$. Since base 3-fold is irreducible, every embedded S^2 bounds a ball. If $\pi_2$ was nontrivial, some nontrivial element of it could be represented by an embedded sphere (this is sphere theorem, not?) which could not bound a ball by hypothesis, contradiction to assumption. So $\pi_2$ of base is $0$, hence so of the universal cover. $\pi_3 = 0$ by Hurewicz + noncompact Poincare duality. And so on. So base is a $K(\Bbb Z, 1)$.

Again $S^1$, nonsense.

Not sure if I am missing any condition.

Danu

@MikeMiller Hmm.. We'll see. I'm a bit disappointed :\

Mike Miller

That's perfectly fine. Can you now tell me what you have proved for an arbitrary irreducible 3-manifold with the exact same argument...?

Danu

The exercises also seem very simple, so I guess nothing much will be covered

Mike Miller

19:49

I'm sorry to hear that

Danu

I was very happy with myself after discovering that $\Bbb Z^2/(m\Bbb Z+n\Bbb Z)\cong \Bbb Z_{\gcd(m,n)}\oplus \Bbb Z$

tylerl-uxai

@AkivaWeinberger I was helping a family friend. For whatever reason, if you change the minimum i from 2 to 1 or 0, it will ruin the factorial

Danu

One of the first times I was inspired to do something creative by mathematics

tylerl-uxai

I can show you exactly the steps if you want, so you can see how computers do it

Akiva Weinberger

Check out what I was saying from here

Balarka Sen

19:52

@MikeMiller If an irreducible 3-fold (compact/connected/blah blah) manifold has infinite $\pi_1$, then it's a $K(G, 1)$, right?

Mike Miller

Yes, that's what I was trying to get you to say. Whence a million results on 3-manifold groups.

tylerl-uxai

omg this is tough haha

I hope it's okay that I did n = n - 1 (counting down instead of up)

Akiva Weinberger

I mean, they both work, for positive numbers

Right?

Pedro Tamaroff

@Mike @Balarka

Balarka Sen

@MikeMiller So the upshot is that whenever you have an infinite group you can easily check if it's a 3-fold group by group cohomology, yeah?

Akiva Weinberger

19:54

It's just a simple way to deal with the 0 case

Mike Miller

You have obstructions through group cohomology.

I would hardly say you can easily check it.

@PedroTamaroff How do you get blog posts on wordpress to only show the first few paragraphs before clicking?

Balarka Sen

Ah, you're correct.

@Pedro

Pedro Tamaroff

@MikeMiller Explain?

Balarka Sen

This. It shows every post in the full length in its front page. Mike's asking how to get it to show all the posts, but shorten its length to a few paragraphs on the front page.

Mike Miller

Guess yours doesn't do it either so nevermind.

Pedro Tamaroff

19:57

Yeah, mine doesn't. I guess you can throw in some bucks to get it? Wouldn't know.

By the way, me and my friends have opened a blog to collect problems.

What were you just discussing, about three-manifolds?

It is not all algebra, I promise.

Balarka Sen

Nice, @Pedro.

We were discussing infinite 3-manifold groups.

Pedro Tamaroff

@BalarkaSen Meaning?

Balarka Sen

"Which infinite groups are fundamental group of 3-manifolds?"

Pedro Tamaroff

@BalarkaSen This one is down your alley. Classify all connected three fold covers of a wedge of three projective spaces of dimension two.

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