Mathematics

since I don't have the background to know if my question is trivial, I'll stick to MSE over MO. too much material in my motivation section to fit it into the homotopy room without hogging, although maybe I'll link there.

@AlexanderGruber what? I can think of two things you might be talking about, but neither is one I would guess you would reference without further context...

Balarka Sen

@AlessandroCodenotti Nah, I am forgetting everything I know and don't know. But I also don't care a lot since the rest of the things went good.

Mike Miller

@arctictern i'm watching for it

skill patrol

@BalarkaSen most of them are pretty nice guys though.

Balarka Sen

6:30 AM

I agree. but it's scary nonetheless.

Alexander Gruber

@arctictern From about a month ago. Looks like good content, it'd be a shame if people couldn't see it

arctic tern

the one I deleted on the question with the bounty?

Alexander Gruber

Right.

arctic tern

I was going slowly insane alternating between thinking it was a nice answer and meaningless gibberish

and guilty the OP might dislike it and feel swindled out of their bounty

link it and I'll undelete

Alexander Gruber

@arctictern here you go

arctic tern

6:33 AM

unbaleeted

Alexander Gruber

i'm all about those species man

We don't see enough of that on MSE

arctic tern

yeah, there's like one guy who regularly answers those questions

marco reidel or something approximately shaped like that

also, I never understood the discussion of the 15-puzzle in my group theory class, or after. it wasn't until I learned about groupoids that it suddenly made sense.

Alexander Gruber

I learned about them from Kerber's book Applied Finite Group Actions

alright alright, got my latest bounties up

Here they are if anybody's interested 1 2 3

3

I'm outy for the night y'all.

skill patrol

Cya pal, take care :-)

Tim The Enchanter

6:57 AM

Happy birthday Arctic.

arctic tern

7:09 AM

@MikeMiller @BalarkaSen @TedShifrin Okay, asked.

Balarka Sen

You can probably write S^7 as S^4 semidirect S^3 in different ways than the obvious Hopf bundle. Eg, Milnor's fibration in his exotic 7-sphere.

arctic tern

that'd be weird. I don't think it affects the question though since I'm only caring about the composition factors, not how they fit together necessarily.

Balarka Sen

Ah, alright.

arctic tern

wonder if there's a (numerology) tag...

nope

Balarka Sen

lol

I upvoted, btw.

Alessandro Codenotti

7:24 AM

You could create it. But that doesn't mean you should :P

Balarka Sen

Huh. I just realized that my Hopf bundle comment, if works, can be made into an answer for this.

Danu

12:30 PM

Does anyone here have access to e-books from the AMS? I'm trying to get a look at a chapter from "A celebration of the mathematical legacy of Raoul Bott"

Semiclassical

1:33 PM

@Danu Our university only seems to have it in print at our math library.

Akiva Weinberger

2:00 PM

@BalarkaSen How do you say "thank you" in Bangla?

(I'm thinking it would be nice to thank some local merchants in their native language)

nbro

any help for this question: math.stackexchange.com/questions/2200017/…?

Fawad

Q)Express 29 as a linear combination of4147 and 10672.

Answer is weird,can someone help

SteamyRoot

I'm going to guess that $29$ is the gcd of those numbers.

Fawad

@SteamyRoot yes

How should I come up with 551 and 9×58 ?

SteamyRoot

2:17 PM

Extended Euclidean algorithm

In arithmetic and computer programming, the extended Euclidean algorithm is an extension to the Euclidean algorithm, which computes, besides the greatest common divisor of integers a and b, the coefficients of Bézout's identity, that is integers x and y such that a x + b y = gcd ( a , b ) . {\displaystyle ax+by=\gcd(a,b).} This is a certifying algorithm, because the gcd is the only number that can simultaneously satisfy this equation and divide the inputs. It...

Fawad

@SteamyRoot thanks!

GPhys

Hello @AkivaWeinberger

Secret

3:02 PM

[A series question from the maths club I have been to]
The following is the original question:

Find the sum without using generating functions
$$\sum_{i+j+k=17}ijk$$

(NB I already looked at the answer, thus the following is a new version of it)

Find the sum without using generating functions nor the official answer
$$\sum_{i+j+k=17}ijk$$

Current thought in progress: trying to inscribe a sphere of radius 7 into a lattice of points

Almost forgot: i,j,k are integers

And of course, extended questions:

find the following sums:
$\sum_{\sum_{i=1}^n a_i=r}\prod_{j=1}^na_j$

$$\sum_{\sum_{\sum_{\dots _{\sum_{i=1}^3 a_i}}}}k$$

NB, I actually don't know if a sum with nested indices make sense

Maple SE - Area 51 Proposal

3:36 PM

Hello guyz I am not going to say anything obvious

arctic tern

lol

skull petrol

lol

Secret

4:02 PM

[to be done]Find $f$ such that $f(transcendental) \neq 0$ and $f(algebraic)=0$

Preliminary thought $\sum_{p \in \Bbb{C}[T] - \{0\}}[p(x)]^2$

However this is too complicated

thus more thought is needed

sorry typo: switch $\neq$ and = around in the criteria of $f$

SteamyRoot

What do you mean, find such $f$ ? Just define $f$ like that?

As in, take the indicator function of the algebraic numbers or so

Secret

find an $f$ in terms of polynomials $p$. The idea is to try to construct an indicator function for the algebraic numbers such that it is expressed in terms of polynomials

Alessandro Codenotti

I feel like I'm missing something obvious, any hint on how to prove that a finitely generated abelian group has a finite torsion subgroup?

SteamyRoot

Because such group is isomorphic to $\mathbb{Z}^n \oplus \mathbb{Z}_{p_1} \oplus \mathbb{Z}_{p_2} \oplus \dots \oplus \mathbb{Z}_{p_k} $ ?

with the $p_i$ prime powers

Alessandro Codenotti

That's the next thing to prove

SteamyRoot

4:11 PM

Oh

Well, I guess idea is the same: it's abelian, so you just have to consider the order of the generators

torsion subgroup will be the subgroup generated by all the generators with finite order?

Alessandro Codenotti

Ah, no, wait, the next thing is that such a group $G$ is isomorphic to $\Bbb Z^n\oplus T(G)$, but that's the same thing modulo specifying the structure of $T(G)$

@SteamyRoot oh, duh, it's all abelian of course, product of finite order elements has finite order

Ok thanks, I should be able to fill in the details

BAYMAX

Any good resource for studying Principal Ideal Domain ?

like video lectures?

Sophie

4:50 PM

Someone needs to teach Ukrainian farmers how tile circles

Akiva Weinberger

5:28 PM

I just noticed something annoying about homology: the simplex $abc$ is not homologous to $bca$. (They don't even have the same boundary; $\partial(abc)=ab-ac+bc$, but $\partial(bca)=bc-ba+ca$. The problem is that $ab\ne-ba$ and $ac\ne-ca$.)

It turns out not to matter; the above problem can't happen with cycles. That is, if we have two cycles that differ only by even permutations of vertices of some simplices, then they are homologous. This means that the homology we'd get by "declaring samely-oriented simplices to be equal" is the same as the usual homology. It's still annoying, though.

Mike Miller

I don't find it to be, particularly

s.harp

I don't get it, abc is a triangle and bca is another orientation on that triangle (right?), these spaces should be homeomorphic?

Akiva Weinberger

They're homeomorphic, but not homologous @s.harp. Homologous means that their difference is the boundary of some sum of 3-dimensional simplices.

(where the boundary operation is defined so that each term in the boundary has the same order as the original simplex,

s.harp

I see, you are taking them to lie in $\Bbb R^3$ right

Akiva Weinberger

so for example $\partial(ab)=a-b$, $\partial(abc)=ab-ac+bc$, and $\partial(abcd)=abc-abd+acd-bcd$

IIRC

and you can check that the boundary of a boundary is zero.

Mike Miller

5:40 PM

Of all of my complaints with the singularchain complex this is rarely one of them

Akiva Weinberger

@s.harp You can. In the more general case, these are singular simplices, or functions from $\{(x_0,\dots,x_n)\in\Bbb R^n:x_0+\dotsb+x_n=1\}$ to the topological space you're studying.

But you can think of this as formal sums of geometric simplices (with ordered vertices) in $\Bbb R^n$, I guess.

Mike Miller

I'm pretty sure s.harp knows what singular homology is but didn't understand your original phrasing

Akiva Weinberger

Oh, sorry

Right, yeah, didn't realize

s.harp

I wasn't thinking really about the context of singular complexes, thats why I was confused :)

s.harp

5:56 PM

Sometimes I see a "homology version" of the Cauchy formula, meaning if $\gamma,\gamma'$ are $C^1$ curves in some open $U\subset\Bbb C$ that have the same homology class, then $\int_{\gamma} f dz = \int_{\gamma'}f dz$.

I'm a bit confused by that, I wonder if there are any $C^1$ curves that are not null homotopic but have zero homology class

Mike Miller

yeah, this already happens for the twice punctured plane

s.harp

because its homotopy equivalent to the "$8$", ok that was simple :)

Mike Miller

the point is that CS really just wants to say that bordant curves give the same integral - you're just using Stokes

s.harp

because if $f\equiv R(f) dx + I(f) dy$ then Cauchy Riemann implies $df=0$, so indeed $\int_{\partial D} f = \int_D df = 0$

and bordant curves really is the same concept as homologous curves

SoumyoB

6:17 PM

I think I should join the stats stack exchange, though I really don't know much of stats

but I'll probably get more probabilists over there

An old man in the sea.

6:29 PM

Could someone help me with this simple question? math.stackexchange.com/questions/2201628/…

I think I used to know it, but somehow I forgot how to do it.

Thanks

arctic tern

there seem to be more MSE-like questions getting upvoted on MO, judging from their front page atm

SoumyoB

7:01 PM

mse?

arctic tern

the website you're on

SoumyoB

and what's MO?

Transcript for

$Mathematics$ Mathematics