Mathematics - 2017-10-09 (page 3 of 5)

Balarka Sen

14:05

@AlessandroCodenotti Actually, RP^3 embeds in R^5. RP^3 is, as I said, the unit tangent bundle of S^2. But the tangent bundle $TS^2$ embeds in $\Bbb R^6$; this is because $S^2$ sits inside $\Bbb R^3$ as the unit sphere and each tangent plane $T_p S^2$ sits inside $T_p S^2 \oplus N_p S^2 = \Bbb R^3$.

So the embedding is $TS^2 \hookrightarrow \Bbb R^3 \times \Bbb R^3$, $(x, v) \mapsto (\mathbf{x}, \mathbf{v})$ (sorry for bad notation) where $\mathbf{x}$ is a unit vector in $\Bbb R^3$.

But then, the unit tangent bundle maps under this embedding in a way so that $\|\mathbf{v}\| = 1$ in the second factor of $\Bbb R^3$.

So, in particular, $\|(\mathbf{x}, \mathbf{v})\|^2 = \|\mathbf{x}\|^2 + \|\mathbf{v}\|^2 = 1$. So the embedding restricts to an embedding $T_1S^2 \hookrightarrow S^5$, where $S^5$ is the unit sphere in $\Bbb R^6$.

Mike Miller

@BalarkaSen Characteristic classes help obstruct but don't give a complete answer. The only ones I know off the top of my head is that $\Bbb{RP}^{2^n}$ needs twice the dimension and everything else can get away with a little less.

Balarka Sen

Ah

Mike Miller

i don't know if there is a complete answer. doubt it

Balarka Sen

If memory serves I think Milnor-Stasheff counted the dimension of the trivial subbundle in $T\Bbb{RP}^N$

Or something of that sort

Mike Miller

You embed in a Euclidean space and use the formulas to see that your highest SW classes vanish if you embed in R^n+k

k < n

Balarka Sen

14:17

Oh, I see, it's a stabilization problem. How much does $k$ need to be so that $T\Bbb{RP}^n \oplus \underline{\Bbb R}^k$ is trivial?

Mike Miller

right, and you use the multiplicativity of SW ans the fact that SW(normal) stops at degree k

Kasmir Khaan

14:35

@LeakyNun Hello !

Balarka Sen

@MikeM Thanks, I did the computation. $w(T\Bbb{RP}^n)$ should be $(1 + w_1)^{n+1}$ because of it being stably $O(1)^{n+1}$, which, if $n = 2^k$, is $1 + w_1 + w_1^{2^k}$, and coefficients of inverse of that does vanish for all degree after $n - 1$.

Sha Vuklia

Guys let $R$ be a commutative ring, and let $I=\langle X+Y,X^2+1\rangle$. My book claims we have the following isomorpshim:
$$
\big(\mathbb R[X][Y]/\langle Y-(-X)\rangle\big)\big /\langle\overline X^2+\overline 1\rangle\cong\mathbb R[X]\big/\langle X^2+1\rangle.
$$
So I know that $\mathbb R[X][Y]/\langle Y-(-X)\rangle\cong \mathbb R[X]$. But now it seems that $\mathbb R[X]/\langle \overline X^2+\overline 1\rangle\cong \mathbb R[X]/\langle X^2+1\rangle$. Could someone explain that to me? Btw, $\overline X^2+\overline 1=X^2+1+\langle X+Y\rangle$.

this is the context btw

Balarka Sen

The isomorphism is given by sending $Y$ to $-X$. Under that isomorphism the ideal $(\bar{X}^2 + 1)$ of $\Bbb R[X, Y]/(X + Y)$ is sent to the ideal $(X^2 + 1)$ of $\Bbb R[X]$.

Sha Vuklia

14:52

@Balarka sorry, but what do you mean by sending $Y$ to $-X$? I'm assuming we're considering an isomorphism between $\big(\mathbb R[X][Y]/\langle Y-(-X)\rangle\big)\big /\langle\overline X^2+\overline 1\rangle$ and $\mathbb R[X]\big/\langle X^2+1\rangle$, right? If so, aren't the elements we send of the form $\overline a$? (or something like that)

or have you already "simplified", and considering a different isomorphism?

Because what I wrote here: $\mathbb R[X]/\langle \overline X^2+\overline 1\rangle\cong \mathbb R[X]/\langle X^2+1\rangle$ doesn't seem to match to me,

because how can we have $\mathbb R[X]$ on both sides, while the ideals are in different rings

Balarka Sen

Fine, if you want to be super-rigorous, the isomorphism $\Bbb R[X][Y]/(X + Y) \to \Bbb R[X]$ is given by sending $\bar{X}$ to $X$ and $\bar{Y}$ to $-X$.

That's an unimportant piece of rigorousness

The point is under this isomorphism $\bar{X}$ is sent to $X$ and $\bar{1}$ is sent to $1$, hence the ideal $(\bar{X}^2 + \bar{1})$ is sent to $(X^2 + 1)$.

Sha Vuklia

oh okay, I've never thought of ideals being sent to other ideals, but I guess I get it then

also, thanks for the details anyways, because that helps

Balarka Sen

No problem. I remember being confused about various homeomorphisms of quotient rings when I learnt ring theory; I think I always feel comfortable to think about ring homomorphisms $A/I \to B$ as homomorphisms $A \to B$ with kernel $I$.

Maybe that's a useful thing to keep in mind, I dunno

Sha Vuklia

oh okay, I will think about that, thanks

Balarka Sen

It's basically the first isomorphism theorem of ring theory.

Sha Vuklia

15:10

hm yea, I have to admit I personally don't see that, because the problem I have is that for some $\overline a=\overline b$, we might not have $a=b$ - but I guess if one is more comfortable, they don't really mind that or something

Balarka Sen

ya the thing is $\bar{a} = \bar{b}$ means "$a = b \pmod{I}$"

Sha Vuklia

oh, and $I$ is the kernel anyways, so who cares

Balarka Sen

yup, exactly

HsMjstyMstdn

15:40

Hello everyone, is it true that if you look at any graph close enough, it can be said to resemble a straight line ? This isn't true of fractals, though, right ?

Sha Vuklia

@HsMjstyMstdn No, I think that only holds for differentiable functions. Think of $\vert x\vert$ for instance. That looks like a "V", and no matter how close you get to zero, the graph will always have a kink at $x=0$.

HsMjstyMstdn

@ShaVuklia Oh yeah, forgot that. Forgot about using differentiability as a test, too. Thanks.

Balarka Sen

Graph of a differentiable function at any point resembles it's tangent line, right. But the original function need NOT be differentiable for the tangent line to it's graph to exist.

Think about $f(x) = x^{1/3}$. Not differentiable at $x= 0$, yet it's graph has a well-defined tangent line at $x = 0$.

Sha Vuklia

why did I skip wiki :P

HsMjstyMstdn

Isn't there a term for this ? Smoothness, I think ? But my term is more on the "zooming in"

Or am I just forgetting my definition of a limit

Sha Vuklia

15:54

@Balarka wait, but what is the definition of tangent line? Because wiki seems to use the derivative in its definition

@HsMjstyMstdn A smooth function has derivatives of all orders, but you just need differentiability at order 1 (or this thing about the tangent line, as Balarka says)

Balarka Sen

@ShaVukila There are many many ways to define it, I don't care enough to define a sufficiently geometric notion. Wiki's definition works only when the function you're taking the graph of is differentiable.

Like I said, graph of $y = x^{1/3}$ has a well-defined tangent at $x = 0$. Draw it.

Sha Vuklia

you mean the line $x=0$? :P

Balarka Sen

Yes.

Sha Vuklia

right, ok well in that case, you could say then those functions with either finite or infinite derivatives

Balarka Sen

One way to define the tangent of a curve on the plane at some point $p$ is to locally parametrize it as $\gamma : (-\epsilon, \epsilon) \to \Bbb R^2$ with $\gamma(0) = p$ and looking at the line spanned by $\gamma'(0)$.

@ShaVuklia Well... you need to be careful about saying "infinite derivative", because it's possible for $f'(x)$ to tend to $+\infty$ as $x \to x_0$ yet the graph of $y = f(x)$ to not have a well-defined tangent line at $(x_0, y_0)$.

Think about $y = x^{2/3}$ :P

Sha Vuklia

16:04

huh, you take $(x_0,y_0)=(0,0)$?

Balarka Sen

Ya

Sha Vuklia

how is its tangent line not $x=0$?

if you're using your parametrised definition, then I'll just take your word for it

as in, I won't check it I guess:d

Balarka Sen

@ShaVukila Hmm. I am not super confident but imagine some parameterization $\gamma$ of the curve near $0$ (so $\gamma(0) = \vec{0}$), and consider $\gamma(h)/h$ and $\gamma(-h)/(-h)$ as $h \to 0$. These should converge to the derivative $\gamma'(0)$. I think limit of the two expressions converge to oppositely oriented vectors on $x = 0$.

So $\gamma'(0)$ should indeed not exist.

Hi @mercio.

mercio

hi

I haven't been here in a while

And now I want some KFC o..o

Balarka Sen

Indeed.

lol

Oh man I didn't even notice it's on the starboard. That's like my worst joke of the week

mercio

16:16

eats the axioms of KFC

Sha Vuklia

@Balarka hm, that could be true, but the graph of $x^{2/3}$ still looks like a straight line if you zoom in close enough, so it would rather be a counter example for your tangent line proposal, instead of for my derivative-at-infinity proposal. but in any case, I guess it doesn't really matter:P I just had to give my final response.

Balarka Sen

@ShaVuklia Well, no, if you zoom in close enough it looks like half an axis, like $[0, \epsilon)$, not $(-\epsilon, \epsilon)$.

That's the thing; negative half of the y-axis is far far far away from the graph. That's why it's not a tangent line.

I never thought about this though, it's pretty cute.

Sha Vuklia

well, depends on what HsMj... meant then. He said line, but he might have as well said half line, in which case $\infty$ differentiability works :P

Balarka Sen

OK, that is fair

Sha Vuklia

16:45

@Balarka can I ask you sth real quick? it's kind of dead here, so otherwise I'll just post my question on the main site

Balarka Sen

OK

Sha Vuklia

So say we have a commutative ring $R$, and $I$ and $J$ ideals, such that $I+J=R$. In the proof of the Chinese rem. thm, they use the following implication: $a\in I\cap J\implies a\in\ker(\phi_1)$ (and $a\in\ker(\phi_2)$). But I don't really see why this holds, and it isn't even intuitive to me:l

oh wait

I have to add some notation

So $\phi_1\colon R\to R/I$

and $\phi_2\colon R\to R/J$

Balarka Sen

$\phi_1$ and $\phi_2$ have kernel $I$ and $J$ respectively.

Sha Vuklia

oh really:/

why did I miss that

(thanks tho:P)

Balarka Sen

kk

pilko

16:50

hi, are there problems with SE.imgur at the moment? I can't embed pictures in my posts (and never had issues before).

Sha Vuklia

I had that problem too an hour ago, but then I just tried 10 min later and it worked

pilko

@ShaVuklia ok, thx, I'll try again later then.

Diego

17:16

hi

Lucas Henrique

hey guys.

Diego

17:37

I was wondering, is there any simple formula to know the number of distinct integer solutions to x_1+x_2+..+x_n=100?

Albas

One approach is using the stars and bars method

Diego

Does it tell the number of distinct ones? I mean, I consider 1+2+97 the same as 97+1+2

I really don't know the stars and bars method. I've always made combinatorial problems by "intuition" and knowing the classical formulas

Albas

By stars and bars I mean think of 100 as written as 1+1+1+...+1, 100 times. Now you want to partition these into n different sections. You need to count how you would do that.

Diego

I've just read the wiki and , as far as I see, it counts the tuples, and not the groups.

For example, suppose n=5.

And suppose I want non-negative integer solutions.

Leaky Nun

Any polynomial of degree 5 has no solutions.

3

Diego

17:47

Then this is equivalent to put 100 balls in 5 numbered boxes.

whose formula is \binom{5+100-1}{100}

but again, this counts, for example, 1+1+1+1+96 as different from 96+1+1+1+1, and I don't wan't that.

want*

Albas

@LeakyNun in general no, there are 5 degree polynomials which have solutions

@Diego then I guess you have to use inclusion exclusion principle

Diego

@Albas How's that¡

?

@LeakyNun x^5-1 has 1 as solution.

Leaky Nun

18:04

I’m just being edgy

A correct statement would be “all rational polynomials of degree 5 having exactly real solutions are not solvable by radicals”

(for their Galois group $S^5$ is not solvable)

Faust

anyone know how to show that the area of triangle dbc is the same as triangle bec ?

Evinda

Hello @LeakyNun !!!

I have a question... How can we show that $e^{-\frac{\ln{n} \ln{(\ln{(\ln{n})})}}{\ln{\ln{n}}}} \to 0$?

Leaky Nun

@Faust same base (BC) same height

@Evinda what the hell

Faust

@LeakyNun Counter example $x^5 =0$

Leaky Nun

@Faust good

Evinda

18:09

Yes, I have to show this :P @LeakyNun

Leaky Nun

good luck

Faust

@LeakyNun my goal is to show that DE and BC are parallel i believe using the fact that H is the same would defeat that purpose

Leaky Nun

@Evinda disregard the triple logarithm

Faust

can i make an arguement that angle BEC is a right angle?

Leaky Nun

@Faust mid pt thm (aka similar triangles)

@Faust no

Faust

18:11

Also i know how to show that DE is parrallel to it but i have to do it be showing that those 2 triangles have the same content

Evinda

@LeakyNun What do you suggest to do in order to show the limit?

mercio

@LeakyNun what about the polynomial $(x-1)(x-2)(x-3)(x-4)(x-5)$ ?

oh wait he already had a counter-example :x

Leaky Nun

@Faust parallel -> same height

@Evinda disregard the triple log

I stated the theorem wrongly also, lemme correct:

Theorem (Galois theory): any irreducible rational polynomial of degree 5 with exactly 2 non-real roots is not solvable by radical.

Faust

@mercio it is true that an arbitrary 5th degree or higher polynmial may have no solutions

Evinda

@LeakyNun We would get that $e^{-\frac{\ln{n}}{\ln{\ln{n}}}} \leq \frac{1}{e}$

Tobias Kildetoft

18:20

@Faust Depends on where you want the solution to be

Evinda

Right? @LeakyNun

Tobias Kildetoft

But it might not have a solution that can be obtained from the coefficients using the usual arithmetic operations together with taking $n$'th roots

Faust

@TobiasKildetoft is there always solutions in the complex numbers?

Tobias Kildetoft

@Faust Yes

Leaky Nun

@Faust yes. fundamental theorem of algebra

Faust

18:21

ah doh

Leaky Nun

the correct version reads “solvable with radicals” instead of “has solutions”

@Evinda mmhmm

Faust

Somethig wierd happens at 5th degree or higher on the reals though?

i mean lots of lower degree polynomials have no soltions on the reals but i thouht something wierd happened at or above 5

Leaky Nun

yes, Galois theory explains that

and there’s no simple way to explain it

Faust

oh i dont know what that is ^^

Leaky Nun

@Faust treat it as an analysis of solvability with radicals

radicals meaning x^(1/m) with m positive integer

Evinda

18:24

@LeakyNun I thought the following.

Since $\ln{n} \leq n$,we get that $\ln{\ln{n}} \leq \ln{n}$, so $\frac{\ln{n}}{\ln{\ln{n}}} \geq 1 \Rightarrow -\frac{\ln{n}}{\ln{\ln{n}}} \leq -1 \Rightarrow e^{-\frac{\ln{n}}{\ln{\ln{n}}}}\leq \frac{1}{e}$

Tobias Kildetoft

@Faust It is not really about the reals, it is about polynomials with rational coefficients and solutions of a certain form

Evinda

Am I wrong? @LeakyNun

Tobias Kildetoft

@Faust Do you know any group theory?

Leaky Nun

@Evinda sure

Faust

lots

@TobiasKildetoft

Leaky Nun

18:25

@Faust look up “solvable group”

Tobias Kildetoft

@Faust What happens is that $A_n$ is solvable for $n\leq 4$ and not for $n\geq 5$

Faust

intresting

Tobias Kildetoft

What Galois theory does is associate a group to the polynomial, and the polynomial is solvable by radicals iff the Galois group is solvable

Faust

ah ic.

Leaky Nun

right, it’s better if you know group theory @Faust

Tobias Kildetoft

18:26

And the group will be a subgroup of $S_n$ where $n$ is the degree (and we can always get both $S_n$ and $A_n$)

Leaky Nun

so basically it is the group of permutations on the roots that keep their rational relations unchanged

eg the two roots of x^2-2=0 are a=sqrt(2) and b=-sqrt(2)

a rational relation is a+b=0

Faust

ic

Leaky Nun

and transposing b with a keep them unchanged

so its Galois group is S2

Evinda

Is it as follows?
$$\lim_{n \to +\infty} \frac{\ln{n}}{\ln{\ln{n}}}= \lim_{n \to +\infty} \frac{\frac{1}{n}}{\frac{1}{n \ln{n}}}=\lim_{n \to +\infty} \ln{n}=+\infty$$

So $\lim_{n \to +\infty} e^{- \frac{\ln{n}}{\ln{\ln{n}}}}=0$ @LeakyNun

Faust

hmm wonder why i havent learned about it

Leaky Nun

18:28

for (x^2-2)^2-3=0, you have four roots

experimeting tells you that its Galois group is V4

@Evinda jawohl

Evinda

Wie hilft uns das aber weiter? @LeakyNun

Leaky Nun

@Evinda well the triple log is too slow to do anything lol

Evinda

Yes, but how could we prove it more formally? @LeakyNun

Leaky Nun

lol

Faust

well im going to go back to slogging through NT ill ttyl leaky

Leaky Nun

18:31

@TobiasKildetoft could you give me problems on Galois theory?

Tobias Kildetoft

@LeakyNun Not off the top of my head, no

Been too long since I did it myself

Leaky Nun

thanks

it’s basically the automorphism group of the splitting field right @TobiasKildetoft

Tobias Kildetoft

@LeakyNun Yes

except when the base field is not the prime field, we further require that it fixes the base field pointwise

Leaky Nun

or else Gal(C/R) would be a mess?

Tobias Kildetoft

right, it would not care about the $R$

Leaky Nun

18:34

does it work for finite fields?

Tobias Kildetoft

Sure, it works for any field, one just needs to be careful about what sort of extensions one considers (in order for the fundamental theorem to be true)

Evinda

Do we justify it maybe as follows? @LeakyNun

$$\ln{\ln{\ln{n}}} \leq \frac{\ln{n} \ln{\ln{\ln{n}}}}{\ln{\ln{n}}} \leq \ln{n}$$

We know that $\lim_{n \to +\infty} \ln{\ln{\ln{n}}}=+\infty$ and $\lim_{n \to +\infty} \ln{n}=+\infty$ and so $\lim_{n \to +\infty} \frac{\ln{n} \ln{\ln{\ln{n}}}}{\ln{\ln{n}}}=+\infty$.

Leaky Nun

@Evinda sicher

GFauxPas

sum @Semiclassical

Leaky Nun

18:56

@BalarkaSen yes, indeed

GFauxPas

sup*

Leaky Nun

@BalarkaSen where "formal" means "pretend that it's correct" instead of "rigorous"

auto-antonym lol

@GFauxPas inf

@KasmirKhaan hi

@TobiasKildetoft are you here?

Evinda

Ok, danke @LeakyNun

Tobias Kildetoft

@LeakyNun Yeah

Leaky Nun

@Evinda kein Problem

GFauxPas

18:59

I get it

Leaky Nun

@TobiasKildetoft is it possible for $[\Bbb R:F] < \infty$ for some $F$?

Tobias Kildetoft

@LeakyNun I don't actually remember, even though I am sure I have seen the question before.

Leaky Nun

@TobiasKildetoft is there something called "algebraic basis"?

Tobias Kildetoft

@LeakyNun Possibly

Leaky Nun

would it help?

GFauxPas

19:01

can anyone give me a nudge in the right directio please:

Tobias Kildetoft

no idea

Leaky Nun

actually how is index defined?

Tobias Kildetoft

@LeakyNun dimension as a vector space over the smaller field

GFauxPas

given $\langle x_n \rangle$ a bounded real sequence such that every convergent subsequence coverges to $L$, prove $\langle x_n \rangle$ converges to $L$

Leaky Nun

@TobiasKildetoft as a vector space...

so I'm thinking if you take away b, you must also take away sqrt(b), and so on

so the index must be infinite... do I make any sense?

GFauxPas

19:03

its not saying every subsequence, just every convergent subsequence, so the standard theorem doesnt apply directly

Leaky Nun

@GFauxPas I'm about to prove this... you freaked me out

[btw I gave up, as I only know sequence]

GFauxPas

:o

Leaky Nun

@GFauxPas contrapositive with Bolzano

GFauxPas

no, because the question doesnt say there are any convergent subsequences

only that if there are, they all have the same limit

Leaky Nun

@GFauxPas there must be, by Bolzano-Weierstrass.

GFauxPas

19:07

oh

Leaky Nun

3

Q: Computing $\sqrt{2}+\sqrt[4]{2}$ over $\mathbb{F}_3$

I'm computing different minimal polynomials in $\mathbb{F}_3$. I found that I can look at elements of the form $\sqrt[4]{2}$ in $\mathbb{F}_3[\sqrt{2}]$ and in particular I found the four fourth roots of two in that field are $1+2\sqrt{2},1-2\sqrt{2},2+\sqrt{2},2-\sqrt{2}$ while the square roots...

How is this related to galois-theory?

Alessandro Codenotti

Galois theory is all about polynomials and field extensions

Leaky Nun

@AlessandroCodenotti you certainly don't need Galois theory to find minimal polynomial??

Alessandro Codenotti

No but it's very possible that this exercise comes from a Galois theory course

Leaky Nun

@AlessandroCodenotti could you give me any problem on Galois theory?

I don't care if it's like beginner level

Alessandro Codenotti

19:20

I don't know enough about Galois theory to be handing out exercises

@Tobias are you still there?

Leaky Nun

@AlessandroCodenotti he said he doesn't have any off the top of his head

Tobias Kildetoft

@AlessandroCodenotti Still here, but as said, no Galois theory exercises on hand

Alessandro Codenotti

Sure, I have another algebra question to annoy you with :P

Leaky Nun

@AlessandroCodenotti yay, algebra

Alessandro Codenotti

Ah, a cool thing to think about is whether $\text{Gal}(F/E)$ is normal in $\text{Aut}(F)$

And if that is not always the case when does this happen

(I know the answer but not how to work it out for the second part)

Leaky Nun

19:27

aren't they the same when $E$ is prime?

Alessandro Codenotti

They are

Leaky Nun

cool

good luck when $E=\Bbb R$

Tobias Kildetoft

@AlessandroCodenotti What algebra question did you have for me?

Leaky Nun

@TobiasKildetoft scroll up?

Tobias Kildetoft

The one about being normal?

Alessandro Codenotti

19:34

No, not that one

I'm turning on my laptop, just a moment

Ok so I have no intuitive idea of what conjugacy classes "look like" and especially why do we care about them

So I'm trying a few examples, in particular I want to calculate the conjugacy classes in $D_n$ (which is a dihedral group with $2n$ elements to me, so congruences of a regular $n$-gon)

Leaky Nun

@AlessandroCodenotti well, it's the orbits of the group action of the group itself to itself via conjugation, right

Alessandro Codenotti

I don't know about group actions so I guess so

Leaky Nun

you should study group actions first lol

although alright Sylow used conjugacy class instead of group actions

Alessandro Codenotti

Anyway let's start with $n$ odd because it's easier, all the reflections are in the same conjugacy class

Leaky Nun

basically partition under the equivalence relation "related by conjugation"

so $s$ is the reflection and $r$ is the rotation?

Diego

19:40

does group theory have application on combinatorics?

Leaky Nun

$rsrs=1$?

Tobias Kildetoft

@Diego Very much so

Leaky Nun

@Diego yes, especially via group actions

Alessandro Codenotti

@LeakyNun there are a lot of reflections and rotations so I'm not sure what the reflection and the rotation are

Leaky Nun

@AlessandroCodenotti just pick any, it doesn't matter

every $D_n$ is bigenerated

with $r^n=s^2=rsrs=1$ (amirite @TobiasKildetoft)

Tobias Kildetoft

19:41

@LeakyNun Sure, or generated by two reflections

Leaky Nun

@TobiasKildetoft oh, to hell with that

Diego

Does anyone have a recommendation on a book on groups that teaches applications to combinatorics? I already had a course on groups, but it was full abstract

Leaky Nun

well, $a^2=b^2=(ab)^n=1$?

it isn't very hell then

@Diego pick any group with group action

Tobias Kildetoft

@LeakyNun Right. That makes it a Coxeter group which is nice

Leaky Nun

@TobiasKildetoft oh, to hell with Coxeter groups

Tobias Kildetoft

19:43

(the group denoted $I_2(n)$)

Leaky Nun

@AlessandroCodenotti so, every element can be represented as $r^m$ or $r^ms$

GFauxPas

Leaky want to try another im having trouble with?

Leaky Nun

so you only need to check 4 cases

Diego

@LeakyNun I've got Dummit and Foot and it doesn't have comb

Leaky Nun

@GFauxPas sure

Alessandro Codenotti

19:45

I think that for $n$ odd there's only two conjugacy classes, one for reflections and one for rotations

Tobias Kildetoft

@AlessandroCodenotti There are more

Leaky Nun

@AlessandroCodenotti rotations are definitely closed under conjugation with rotations and with reflections

[btw rotations is $r^m$ and reflections are $r^ms$]

Tobias Kildetoft

@AlessandroCodenotti Remember that not all rotations need to have the same order

Leaky Nun

[and don't ask me why I used "is" for the former and "are" for the latter]

Tobias Kildetoft

Also, don't forget the identity element

Alessandro Codenotti

19:46

hmmm, all reflections are in the same class

fair, the identity is in its own class

GFauxPas

Let $\langle x_n \rangle$ be a bounded sequence of reals s.t. $S = \{x_n:n \in \mathbb N\}$ is infinite

Leaky Nun

@AlessandroCodenotti I think you should work with a specific dihedral group first

GFauxPas

prove there exists a nested sequence of intervals $\langle[a_n,b_n]\rangle$ such that $[a_n,b_n] \cap S$ is infinite for all $n$ and $\langle b_n - a_n \rangle \to 0$.

Leaky Nun

btw $aRb \iff gag^{-1}=b \iff ga=bg$ for some $g$

GFauxPas

possibly useful theorem:

Leaky Nun

19:49

for $r^a$ and $r^b$, we hope to find $r^csr^a=r^br^cs$ [rotations are pretty much useless]

GFauxPas

if $\langle [a_n, b_n ]\rangle$ is a nested sequence of closed and bounded intervals, then there exists a point $z$ that belongs to all of the intervals. If $\lim_{n \to \infty} (b_n - a_n) = 0$, then such a $z$ is unique

thats a theorem that might be helpful

really not sure how to go abut this :(

Alessandro Codenotti

hint: this sequence has a converging subsequence

Leaky Nun

So $c-a=b+c$, so $a+b=0$? @TobiasKildetoft amirite

the $c-a=b+c$ is done under $\Bbb Z_n$ and is abuse of notation

GFauxPas

Aless, which one

Leaky Nun

@GFauxPas the one given by Bolzano [again!]

Alessandro Codenotti

19:52

@GFauxPas this one has a converging subsequences

Leaky Nun

for $r^a$ and $r^bs$, rip as the existence of $s$ is an invariant

under conjugation

GFauxPas

what's "this" one

Leaky Nun

@GFauxPas click the arrow

GFauxPas

$x_n$?

Leaky Nun

yes

GFauxPas

19:55

ah

Leaky Nun

for $r^as$ and $r^bs$, you want $r^cr^as=r^bsr^c$, whence $c+a=b-c$ (inside $\Bbb Z_n$), whence $c=(b-a)/2$

so all reflections are in the same class

GFauxPas

does it have two, one converging to the lim inf and one to the lim sup?

oh, but they might be the same

Leaky Nun

the identity is in its own class

and the other rotations are all paired up

@AlessandroCodenotti @TobiasKildetoft amirite

fun exercise

@GFauxPas you only need one

GFauxPas

okay, call it $\langle x_{p_n}\rangle$

Alessandro Codenotti

It seems that there are $4$ classes in total in $D_5$

Leaky Nun

19:57

@AlessandroCodenotti which matches my result

Alessandro Codenotti

@LeakyNun I'm pretty sure there are two conjugacy classes of reflections if $n$ is even

Leaky Nun

$D_n$ has $1+1+\dfrac{n-1}2$ classes where $n$ is odd [no, this is not the class equation]

@AlessandroCodenotti I'm only referring to $n$ odd

if $n$ is even then $(b-a)/2$ wouldn't make sense

GFauxPas

not sure how to get the endpoints of the intervals :(

Alessandro Codenotti

The two classes are reflections wrt an axis through vertices or sides when $n$ is even

Leaky Nun

for odd $n$, according to my result, the class equation is $1+\underbrace{2+2+\cdots+2}_{(n-1)/2}+n$

@AlessandroCodenotti you're completely right

@GFauxPas let the limit be $L$.

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$Mathematics$ Mathematics