Mathematics - 2017-03-28 (page 3 of 7)

hahah, I don't know @GFauxPas I actually did Algebra for two weeks straight, neglecting all my other stuff, but now I'm in a "multivariate calculus" mood, so everything else is boring

SteamyRoot

Algebra is by far the least boring thing in all of maths

Sha Vuklia

5:10 PM

and that's good then! @Astyx

SteamyRoot

(totally unbiased opinion cough cough)

GFauxPas

I'm taking an Algebra course now

but so far all we did were groups and group actions

Astyx

What isn't algebra anyway ?

GFauxPas

is that like the most important algebraic structure? or just the easiest?

SteamyRoot

Well, groups are nice enough to seriously work with

Astyx

5:11 PM

Monoids ..?

Magmae ?

SteamyRoot

But there's not too many restrictions on them, which allow for a certain "richness"

Sha Vuklia

No but seriously, I just asked a fellow student why $m \overline a=\overline{ma}$... and I'm done XD Algebra makes me feel stupid, while multivariate calculus makes me fly XD

GFauxPas

whats that notation mean sha?

SteamyRoot

modulo

Sha Vuklia

$\mathbb Z/n\mathbb Z$

additive group of integers modulo $n$

GFauxPas

5:13 PM

oh, so why $m[[a]] = [[ma]]$?

Sha Vuklia

you can write $\overline a+\dots+\overline a$

Astyx

Once you get the hold of it, algebra is really beautiful

SteamyRoot

now that is a notation I've never seen

Astyx

(or most of it)

Sha Vuklia

and the group operation is defined as follows:

$\overline a+\overline b=\overline{a+b}$

GFauxPas

5:14 PM

that's the notation for the equivalence class of $a$

Sha Vuklia

so you just put everything under the line

GFauxPas

but yeah it would be just a definition, wouldnt it?

Sha Vuklia

and get $\overline{a+\dots+a}=\overline{ma}$

SteamyRoot

I've never seen double brackets for equivalence classes

I've seen people use $[a]$, though

Astyx

Not really, you have to prove the equivalence classes modulo $n$ are compatible with the operations in $\Bbb Z$

SteamyRoot

5:15 PM

but in the case of modulo operations (or working on quotient groups), having a bar over it is quite common

Astyx

Same for me @Steamy

Sha Vuklia

@Astyx that's directed at me?

Astyx

No, at @GFauxPas

GFauxPas

well closer to what I've seen is $[\![a]\!]$

Astyx

Well you can read it too, I have nothing against that

GFauxPas

5:16 PM

but i was too lazy to do the \!

Astyx

@GFaux I've seen that notation for the integer interval between $0$ and $a$

GFauxPas

well i like how $[\![ \cdot ]\!]$ looks :)

Balarka Sen

@AlessandroCodenotti You can retract the solid torus minus it's core circle to the boundary of the solid torus

SteamyRoot

@GFauxPas Actually, isn't the group you start with isomorphic to $\mathbb{Q}/\mathbb{Z}$ ?

GFauxPas

I guess it would be, because $e^{2\pi i n} = 1$

my mother warned me about this :(

she said i'd take someone's $i$ out if I werent careful

SteamyRoot

5:24 PM

Hmmm...

GFauxPas

so find $\phi: \mathbb Q / \mathbb Z \twoheadrightarrow \mathbb Z/ 4\mathbb Z$

or whatever

SteamyRoot

Why the square?

GFauxPas

I was thinking of the group operation itself, sorry

Astyx

What does that double arrow mean ?

GFauxPas

surjectivity

Astyx

5:26 PM

That's a cool notation

GFauxPas

and $\hookrightarrow$ is injectivity

SteamyRoot

oh, I've seen that one differently

GFauxPas

yeah I think you can also split the end instead of curling it

SteamyRoot

$$\rightarrowtail$$

is what I've often seen for injectivity

GFauxPas

yup i've seen that one too

Balarka Sen

5:28 PM

@GFauxPas My mamma said, to get things right / you'd better not mess with someone's $i$

SteamyRoot

The one you write I tend to use for inclusion (where it's obvious/canonical how to do the inclusion)

GFauxPas

but $\hookrightarrow$ is what my professor uses so i guess ill use that. its easier to write

well I guess that makes sense if you think of inclusion as "the" injection

SteamyRoot

Anyway, I guess there is no such $\varphi$

GFauxPas

proofwiki.org/wiki/Definition:Injection has both

how can you prove there isnt one?

SteamyRoot

I mean, there should be some $q \in \mathbb{Q}/\mathbb{Z}$ such that $\varphi(q) = 1$

GFauxPas

5:31 PM

0

SteamyRoot

But then you can't map $q/2$ anywhere

GFauxPas

why not?

we have to have $e \mapsto e$

that's non negotiable

no, in the domain it's $\times$

SteamyRoot

Woops, why did I edit that message

GFauxPas

well in the original question it's

SteamyRoot

Your original question is multiplication of powers of $e$ though

which translates to the addition of the rationals that appear in those powers, no?

GFauxPas

5:33 PM

$\{ e^{2\pi i q} | q \in \mathbb Q \} \twoheadrightarrow \{1,-1,i,-i\}$

SteamyRoot

so $e^{2\pi i q_1} \cdot e^{2\pi i q_2} = e^{2\pi i (q_1+q_2)}$

Sha Vuklia

Consider the additive group $\mathbb Z/n\mathbb Z$. Let $\overline a\in\mathbb Z/n\mathbb Z$, with $1\leq a\leq n$. It’s easy to show that:
$$
\operatorname{order}(a)=\frac{\operatorname{lcm(a,n)}}{a}.
$$
However, why is the following true:
$$
\frac{\operatorname{lcm}(a,n)}{a}=\frac{n}{\gcd(a,n)}.
$$

Astyx

Using prime decomposition this should be rather straight forward

Sha Vuklia

oh right

GFauxPas

yes steamy

Sha Vuklia

5:35 PM

I'm gonna try it

SteamyRoot

So the domain is just $\mathbb{Q}/\mathbb{Z}$ with $+$ as the operation

Astyx

Yup it is

However there should be more elementary ways of showing this @ShaVuklia

Sha Vuklia

@Astyx sure, but this will do:)

Hi @Ted!

SteamyRoot

Well, rather than a full prime decomposition, you could just write $a = a' \gcd(a,n)$ with $\gcd(a',n) = 1$

and similarly for $n$

Ted Shifrin

Hi @ShaVuklia @Astyx

Sha Vuklia

5:37 PM

@SteamyRoot oh ok, I'll try that too!

SteamyRoot

So anyway, @GFauxPas : once you agree that the operation is $+$, let $q$ be a preimage of $1$ and try to find the image of $q/2$ (there's only $4$ options, none of which will result in a morphism)

Ted Shifrin

I guess I've always argued the desired formula directly there.

Hi @Ted

Hi Balarka

Ooh

5:39 PM

Teddie bear!

GFauxPas

Very nice steamy :) ill share that with my professor

Danu

Teddy?

Ted Shifrin

LOL, hi, @Danu.

Robert Frost

Can anybody suggest a way to repose this question in a sensible way? math.stackexchange.com/questions/2204335/… I want to learn more about the circumstances in which an iterative process continuing becomes ever less likely, and the probability can be proven to converge to zero; similar to probabilistic method perhaps.

Danu

@TedShifrin I have cohomololololo question for you

I've got this (closed, oriented) manifold whose cohomology ring I know for $\Bbb R$ and $\Bbb Z_2$ coefficients. How much can I find out about the $\Bbb Z$ coefficient ring?

Ted Shifrin

5:41 PM

I'm pretty stooopid these days.

Only free and 2-torsion.

Balarka Sen

That's what I told him

Robert Frost

... for example if we select some n from the first 16 integers we can see the probability of it continuing past the mth step is less than if we select some n from teh first 32 integers.

GFauxPas

why cant we have $e^{2\pi qi/2} = 1$?

Ted Shifrin

Balarka knows more of this stuff than Teddy does these days.

Danu

@TedShifrin Nothing more? For instance, let's look at the following

Balarka Sen

5:42 PM

Nah

I know how to draw some pictures

Danu

I happen to know that $H^{2,3}(M;\Bbb Z_2)=\Bbb Z_2$ and $H^2(M;\Bbb Z)=0$. Now the L.E.S. associated to $0\to \Bbb Z \to \Bbb z \to \Bbb Z_2\to 0$ has a piece like this

SteamyRoot

@GFauxPas You mean, $\varphi(q/2) = 1$ in $\mathbb{Q}/\mathbb{Z} \to \mathbb{Z}_4$-notation?

Danu

$$0\to \Bbb Z_2=H^2(M;\Bbb Z_2) \to H^3(M;\Bbb Z)\overset{2\cdot}{\to} H^3(M;\Bbb Z)\to H^3(M;\Bbb Z_2)=\Bbb Z_2\to \dots$$

Robert Frost

...but as we know, there is always a point to which we can increase m, where the probablility diminishes to zero - at least if we draw n from a set less than 2^60.

Danu

I also know that the free part of $H^3(M;\Bbb Z)$ is zero

So I think I have that $H^3(M;\Bbb Z)$ is a finite Abelian group, which under the homomorphism $2\cdot$ has kernel $\Bbb Z_2$. That tells me that there is only one summand $\Bbb Z_\text{even}$, right?

Ted Shifrin

5:45 PM

But any p-torsion for $p\ne 2$ gives an iso there and you can't tell ...

GFauxPas

yes Steamyroot

$\phi(q/2) = \phi(q) = 1$?

and so on

Danu

Now I wanted to think about the cokernel of the $2\cdot $ map. Do I have the following? Since $0\to \operatorname{ker}\to A \overset{f}{\to} B \to \operatorname{coker}\to 0$, and in my case $f=2\cdot$ and the map going OUT of $H^3(M;\Bbb Z_2)$ is not necessarily zero, can I conclude that $\operatorname{coker} 2\cdot $ lies inside $\Bbb Z_2$?!

That seems pretty strong, no?

Balarka Sen

@Danu Oh, yeah, I guess that does mean it's a Z/2 module

SteamyRoot

Then $\phi$ isn't a morphism, since you want that $\phi(q/2) + \phi(q/2) = \phi(q/2 + q/2) = \phi(q) = 1$

Balarka Sen

I don't think you can say more tho

Ted Shifrin

5:47 PM

You just know that there must be $\Bbb Z\oplus \Bbb Z_2$ or other stuff hapoens in other terms?

s.harp

Here is a way to waste time: consider the free group generated by the alphabet of whatever language you are using right now and quotient out the following equivalence relation:
two words of the language are equivalent if they are homophones, for example see and sea.

now see if you can show that the language you are looking at is "homophonically connected" in the sense that the group you get after quotienting is the trivial group

Balarka Sen

@Ted The free part is zero according to him

'cuz H^3(M; R) = 0

GFauxPas

and $1 + 1 \ne 1 \bmod 4$

right?

thats why?

SteamyRoot

uhhh

GFauxPas

oh wait, thats the codomain

Ted Shifrin

5:48 PM

Why can't there be $\Bbb Z_3$?

Danu

I know that $H^*(M;\Bbb R)=\Bbb Z[\alpha]/\alpha^3$ where $\alpha$ is degree 4

SteamyRoot

If your language is homophonically connected, how the hell does anyone ever understand you?

Danu

@TedShifrin In $H^3$? I think there can be

SteamyRoot

@GFauxPas Yes, indeed

Danu

I just said there can only be one summand with $p$ even (and, as it was pointed out to me by @s.harp and @SteamyRoot, then it must be a power of 2)

Ted Shifrin

5:49 PM

Not even. Power of 2?

SteamyRoot

You can try out the other possible images of $\phi(q/2)$ too. You'll never get that $\phi(q/2) + \phi(q/2) = 1$

GFauxPas

and if I do $\phi(p/2) \in \{-1,i,-i\}$ I'll come up with similar problems?

only 4 possibilities, straightforward enough to do them all

thanks :)

SteamyRoot

I was working in $\mathbb{Z}_4$

s.harp

@SteamyRoot Maybe I expressed myself poorly, for example with "see" and "sea" you get the conclusion that $e\equiv a$, remember the group is generated by letters and not words

SteamyRoot

which is isomorphic to $\{1,-1,i,-i\}$

Danu

5:50 PM

@TedShifrin Those are even, too! :P

But whaddaya think about what I said regarding the cokernel

does that make sense?

SteamyRoot

But, you could translate the argument to that group, of course.

GFauxPas

so $\phi(q/2) \in \{1,2,3\}$

Alucard

@SteamyRoot Excuse me, but what does homophonically mean?

Ted Shifrin

$\Bbb Z_6\cong \Bbb Z_2\oplus\Bbb Z_3$.

SteamyRoot

@Alucard see @s.harp 's post

Danu

5:50 PM

@TedShifrin Yeah, as was pointed out to me already :D

GFauxPas

or rather $\{0,2,3\}$

?

@s.harp Oh, right.

I'm slow.

You're just fine :-)

@GFauxPas Yup. Now, try those as image of $q/2$, and see what happens

GFauxPas

5:51 PM

$2\phi(q/2) = 0 + 0 \ne 1$

Alucard

so homophonically connected means two words that are spoken out the same but spelled differently?

GFauxPas

2ϕ(q/2)=2+2≠1

2ϕ(q/2)=3+3≠1

yup, all absurdities

nice

s.harp

@Alucard That means two words are homophones, "homophonically connected" means (in math speak) that the free group generated by the letters of the language becomes trivial if you quotient out the homophones

SteamyRoot

@s.harp ncf.idallen.com/english.html

s.harp

The term is nothing standard, I just made it up

SteamyRoot

5:53 PM

I wonder how many relations you'd obtain from that poem :P

Danu

By the way @Ted, remember when you told me about the Grassmannian of oriented 2-planes? It's coming in handy now! :-)

s.harp

jesus christ

Danu