Mathematics - 2017-10-12 (page 4 of 9)

Alessandro Codenotti

07:03

What are you trying to do?

Daminark

Write an $n$-cycle as a product of 2 elements of order 2

It's... going poorly, to say the least

Tobias Kildetoft

@Daminark And I presume you want to make sure those elements do not involve any other numbers than the ones occurring in the cycle?

Daminark

Yup

Alessandro Codenotti

Elements of order $2$ are only products of disjoint transpositions, right?

Daminark

Yeah

Alessandro Codenotti

07:05

Are you sure that's always possible?

Daminark

Yeah, in office hours this came up and it's apparently supposed to be doable

Our prof said to try it on cycles up to 5 or 6

And find the pattern from there

But I'm completely stonewalled on doing it for a 5-cycle even

Actually hold it

$(12345) = (12)(23)(45)(35)$?

Nope nevermind

For God's sake...

Alessandro Codenotti

This is strange, one usually uses the dihedral groups to show that a product of a pair of order 2 elements can have arbitrary order

Daminark

So, the point of the problem is to show that any element in $S_n$ can be written as the product of two elements of order 2

My idea was, okay if we can do it for cycles by only using elements within, we can just take disjoint cycles and toy around with them

Well, this cycle idea is... much more easily said than done

Tobias Kildetoft

@AlessandroCodenotti That looks like a good strategy for this

Hmm, what is the easy way to see that Cayley-Hamilton implies that a vector space is the direct sum of the generalized eigenspaces for some matrix?

Alessandro Codenotti

07:28

@TobiasKildetoft oh, right $D_n$ is a subgroup of $S_n$, that works here. Won't spoil it for Dami though :P

Daminark

@Tobias So one result we did in linear algebra is that if $f$ and $g$ are coprime polynomials whose product is the minimal polynomial of $T$, then $\ker(f(T)) \oplus \ker(g(T)) = V$. You may be able to generalize this in a convenient way?

Tobias Kildetoft

@Daminark Hmm, possibly

Daminark

Also @Alessandro I don't know if that is enough, like we need to get all of $S_n$

Also are you doing this to get JNF?

Tobias Kildetoft

@Daminark yeah

Alessandro Codenotti

Sure, but you can get a lot this way and hope that's enough for the rest of $S_n$ :P

In particular you can get an $n$-cycle and then your idea of writing stuff as a product of disjount cycles might work

Daminark

07:33

Well I mean, $D_n$ (which our book writes $D_{2n}$ because why??) is much smaller so this hope feels sketch

Tobias Kildetoft

@Daminark you just need to do it in the right way

Alessandro Codenotti

@Daminark because it has $2n$ elements I guess

Tobias Kildetoft

@AlessandroCodenotti But we don't write $S_{n!}$ for the symmetric group on $n$ elements

Daminark

And being able to say $D_n \le S_n$, you know?

Alessandro Codenotti

I'm not saying that's a good choice, just guessing a possible reason, I use $D_n$

Tobias Kildetoft

07:36

Also, we call it $I_2(n)$, not $I_2(2n)$

Daminark

I say "why??" not wondering a reason so much as, why is DF so garbage?

Okay I got the 5 cycle for real this time

$(12345) = (12)(35)(25)(34)$

Alessandro Codenotti

$(01234)=(14)(23)(01)(24)$ if I got it right from the dihedral group

It should be the same

Daminark

You start from 0?? You should be ashamed of yourself

And hmm, is that the sort of thing that you can intuit out of the presentation of the dihedral group? Or would you need to play with the geometry of it to figure it out?

Alessandro Codenotti

We constructed $D_n$ as a subgroup of $M_2(\Bbb Z/n\Bbb Z)$

Daminark

Also hey @Leaky and @Tasty

Alessandro Codenotti

07:49

Geometry, those are two reflections whose composition is a rotation by $2\pi/5$

TastyRomeo

Morning

Leaky Nun

@Daminark hoi

Daminark

Yeah we did them geometrically instead and it's screwing me up hard

Leaky Nun

@AlessandroCodenotti I will construct D8 as the Galois group of x^4-2, anytime

TastyRomeo

Awful suggestion: construct the bouquet of rings with disks that has $D_8$ as its fundamental group and work on that space :P

Leaky Nun

07:51

@Daminark look, D_(2n) is the symmetry group of a regular n-gon

Alessandro Codenotti

@Dami let $f_{a,b}:\Bbb Z_n\to\Bbb Z_n$ be the map $x\mapsto ax+b$. The dihedral group has as elements the maps with $a=\pm 1$. Think about which are what geometrically

Leaky Nun

you can express every element with a reflection and the power of a rotation

@AlessandroCodenotti are you going to say $\Bbb Z_n \rtimes C_2$ next?

MatheiBoulomenos

Hi @LeakyNun

Alessandro Codenotti

Why?

Leaky Nun

@MatheiBoulomenos guten Tag

@AlessandroCodenotti it looks like a semiproduct to me lol

@TastyRomeo @MatheiBoulomenos help me I’m in a dilemma

I’m torn between $D_4$ and $D_8$

TastyRomeo

07:56

For...?

MatheiBoulomenos

I use $D_8$

But I can see the point of $D_4$

Leaky Nun

solution: write $D_{4,8}$

MatheiBoulomenos

:D

TastyRomeo

I use $D_n$ for the symmetries of the $n$-gon

Leaky Nun

I actually think my suggestion is good

Alessandro Codenotti

07:57

@TastyRomeo Another sensible person! $D_n$ has $2n$ elements

Daminark

@LeakyNun The issue is that my visual processing capabilities are nonexistent

Leaky Nun

@Daminark then frigging draw it out

MatheiBoulomenos

just write $\operatorname{Gal}(\mathbb{Q}(\sqrt[4]{2},i)/\Bbb Q)$

Daminark

Like aside from understanding definitions, any time I've ever tried a problem that involved rigid anything, I've utterly failed to make progress on it

Leaky Nun

8 mins ago, by Leaky Nun

@AlessandroCodenotti I will construct D8 as the Galois group of x^4-2, anytime

TastyRomeo

07:58

@LeakyNun Pick like, $n = 5$

MatheiBoulomenos

Ah I see

Leaky Nun

@MatheiBoulomenos :p

TastyRomeo

draw a pentagon on a sheet of paper and number the vertices

and cut out another pentagon and also number its vertices

Leaky Nun

@TastyRomeo wrong ping?

TastyRomeo

and then play with what every group element does

Oh, yeah, was meant for @Daminark

(also, on the pentagon you cut out, number the vertices on both front and back side so you can reflect)

Leaky Nun

08:00

Exercise: find the minimal $k$ such that $D_{n,2n}$ embeds in $S_k$, in terms of $n$.

Alessandro Codenotti

$k=n$?

Leaky Nun

my conjecture is that k=n but you never know

@AlessandroCodenotti how do you prove it?

Alessandro Codenotti

$D_n$ has elements of order $n$ so you have $k\ge n$

Leaky Nun

@AlessandroCodenotti 2+3=6

MatheiBoulomenos

$S_5$ has elements of order $6$

Leaky Nun

08:02

@MatheiBoulomenos sniped :D

MatheiBoulomenos

well, you wrote $2+3=6$

Alessandro Codenotti

Ah, right, that's true

Leaky Nun

@MatheiBoulomenos that’s the cycle type

MatheiBoulomenos

okay

Alessandro Codenotti

Well you don't need more than $n$, that's for sure

Leaky Nun

08:03

sure

wait

how do you embed $D_{5,10}$ in $S_5$?

MatheiBoulomenos

Well, $D_{10}$ acts faithfully on a $5$-element set

If you use the geometric definition

Alessandro Codenotti

$D_n$ can be constructed as a subgroup of $M_2(\Bbb Z_n)$ and all the elements of the former are also in $S_n$

Leaky Nun

ok, go on

Tobias Kildetoft

@AlessandroCodenotti I actually think Daminarks exercise will show how to embed it in smaller $S_n$

So if we take an element of order $k$ in $S_n$ and write it as a product of two elements of order $2$, $a$ and $b$ then $\langle a,b\rangle$ is $D_k$

Leaky Nun

for example?

Alessandro Codenotti

08:08

@TobiasKildetoft fair

Leaky Nun

@MatheiBoulomenos I got too carried away with <(1234),(13)>

ok so it is <(12345),(25)(34)>

<(123456),(16)(25)(34)>

MatheiBoulomenos

$D_6$ embedds into $S_5$ I think

Leaky Nun

ich denke auch aber ich kann nicht es bezeugen

MatheiBoulomenos

Well obviously $S_3 \times S_2$ embedds into $S_5$

I think that $S_3 \times S_2 \cong D_6$

Leaky Nun

:o wat

Tobias Kildetoft

08:16

@MatheiBoulomenos Yeah, I think that is the same embedding I gave above with $k=6$ in $S_5$ (up to some choices)

Leaky Nun

wait that’s right lol

@TobiasKildetoft I hear choices

so in general S2 x Sn = D(2n,4n)?

MatheiBoulomenos

No, that can't be right

Leaky Nun

hmm alright

Daminark

Okay I think I might have gotten a general formula

Tobias Kildetoft

@LeakyNun Dihedral groups are only a direct product when they have odd degree

(I think)

Daminark

08:18

Okay so you look at adjacent transpositions

Let $(k-1\ k) = C_k$

Tobias Kildetoft

@Daminark That's just bad notation :) Those should be called $s_{k-1}$

Daminark

Lol, aight

I'll separate if $n$ is even, you want $(s_1\ldots s_{n-1})(s_2\ldots s_n)$

Their product will be some cycle

Secret

[Random]

$\prod_{n\in \aleph_0}S^n$

Daminark

So now I should just prove that this property is preserved under conjugation, which shouldn't be too bad

Tobias Kildetoft

@Daminark it is in fact pretty trivial, yes

Daminark

08:27

So if $x^2 = y^2 = 1$, we want to toy with $gxyg^{-1}$

Tobias Kildetoft

@Daminark You should be using one of the most basic properties of conjugation

MatheiBoulomenos

conjugating in $S_n$ is really just renaming things

Daminark

Yeah I know that but I'm trying to be a bit pedantic

TastyRomeo

08:48

@LeakyNun The smallest $k$ such that $D_n \leq S_k$, is the sum of the prime powers appearing in the decomposition of $n$

For $n \geq 3$, at least.

MatheiBoulomenos

@TastyRomeo how did you come up with that?

Daminark

Yeah I'll just roll with the renaming actually, I think there's a way to execute it properly

And with that this problem is finally done

TastyRomeo

By thinking how you would embed $D_n$ as a subgroup.

You need an element of order $n$ in $S_k$, and $n$ will be the lcm of the cycle lengths of this element.

So with a bit of work you can show that each of the cycles appearing in this element will be a prime power factor of $n$

Then you still need to find an element of order $2$ that acts on this element of order $n$ in the way you want, but that's not too difficult.

Tobias Kildetoft

09:03

@TastyRomeo It is not clear that the smallest $n$ such that you have an element of order $k$ uses only prime powers

you might need to group some of those prime powers together

Leaky Nun

@TastyRomeo can you embed $D_{12,24}$ into $S_7$?

Tobias Kildetoft

@Daminark Anyway, the two properties you needed for conjugation to make this precise is that it preserves order and that it is multiplicative

Secret

$\prod_{n\in\Bbb{N}}S_n$

mercio

(1234)(567) and (24)(67) @LeakyNun

TastyRomeo

@TobiasKildetoft If you have a cycle of length $ab$ with $\gcd(a,b) = 1$, you can replace that cycle by two cycles of length $a$ and $b$...?

Tobias Kildetoft

09:07

@TastyRomeo Right, but which uses fewer elements?

@TastyRomeo Hmm, actually, I guess you will always get fewer elements using the sum here

I was thinking of which element orders are possible, rather than the maximal one

Daminark

Oh yeah I forgot multiplicative

That checks out

TastyRomeo

Yeah, the sum will always be less than the product in this case

Leaky Nun

@TastyRomeo so you can’t embed $D_{16,32}$ in $S_{15}$?

TastyRomeo

@LeakyNun I can construct $D_{12} \leq S_7$ since $12 = 2^2 \cdot 3$ and $7 = 2^2 + 3$. I'm not going to waste my time doing it.

Tobias Kildetoft

@TastyRomeo Well, we already have a recipe for doing this

TastyRomeo

09:15

@LeakyNun Nope.

Well, yeah, it's not difficult, I just can't be bothered typing it out :P

mercio

you can't embed a 16-cycle in S15

Secret

09:56

(Testing wolfram alpha)

Leaky Nun

10:13

@mercio how does that work? I mean, how do you figure it out?

user84215

The first week of the Abstract Algebra Course will start at 9:30 GMT on Saturday, October 14, 2017 in this room.

Tobias Kildetoft

@MathematicsAminPhysics Who will be in charge of it?

mercio

figure what out ?

user84215

@TobiasKildetoft I have explained all details in this room.

Secret

Hmm... how to prove tan e is transcendental...

Alessandro Codenotti

10:27

When can a linear functional be written as scalar product with an element of the space?

Tobias Kildetoft

@AlessandroCodenotti well, always for finite dimensional spaces of course

Alessandro Codenotti

Is it always possible for continuous functionals? Not necessarily in finite dimension

Tobias Kildetoft

I think also always in a Hilbert space, but I forgot if there is some additional condition. Possibly bounded.

Secret

Unbounded functional for at least one element of the space should cause diverging values, so I guess not allowed

Alessandro Codenotti

A functional is bounded iff continuous so that'd make sense

Secret

10:31

$\int g(x)()dx$ is an example of a functional in some Hilbert space

and we don't want that integral to diverge

Leaky Nun

@Secret tan e = sin e / cos e = (e^2ie - 1)/(e^2ie + 1)

{1,ie} is linearly independent so {e,e^ie} is algebraically independent

the rest follows

Secret

11:05

Now thinking... tan (-ie/2)...

mercio

11:16

@LeakyNun where does your proof go wrong if I look at tan(pi/4) instead of tan(e) ?

also I think you forgot an i

Tobias Kildetoft

@LeakyNun I joined the room you invited me to

Dattier

11:58

Hi, I'm looking for a challenge Sat to test a solver

user84215

Hi

Dattier

@Secret do you know a challenge sat, here satcompetition.org but I don't know where, because my english is very bad.

@LeakyNun can you help me please

Martin Sleziak

@AlessandroCodenotti Isn't this Riesz representation theorem

This is also mentioned in Wikipedia article on Hilbert spaces‌:

> The Riesz representation theorem affords a convenient description of the dual. ... An immediate consequence of the Riesz representation theorem is also that a Hilbert space H is reflexive, meaning that the natural map from H into its double dual space is an isomorphism.

I would also say that it is rather easy to show that the norm of the functional $\langle x,\cdot\rangle$ is equal to $\|x\|$. So you can't really expect such representation for unbounded functionals.

Alessandro Codenotti

12:13

I didn't know about that theorem but that's exactly what I'm looking for, thanks!

Martin Sleziak

BTW if you have time, feel free to stop by in functional analysis chat room sometimes.

Alessandro Codenotti

I'm definitely going to, I have a lot of functional analysis doubts to solve

Martin Sleziak

I will just remind that the above theorem is about Hilbert spaces, so some kind of completeness is probably needed.

I do not have a counterexample at hand, but my guess is that it should not be too difficult to find one. (Basically by taking some incomplete inner product space and an element from the completion.)

Alessandro Codenotti

Hm, yes, I think it can be done considering the finite linear combinations of a Schauder basis of, say, $L^2$

Martin Sleziak

I am not sure I follow the last comment. Anyway, there is a post on the main with some counterexamples, see the links I posted here.

I have to admit that Schauder basis (and some other types of bases in normed space) belong to topics I do not know much about, but would be glad to learn a bit about them. (If only there was enough time.....)

Alessandro Codenotti

12:29

To construct an incomplete inner product space consider the usual complete orthonormal family in $L^2((-\pi,\pi))$ and take the space of the finite linear combinations of the elements of that family

I think the space constructed this way works as a counterexample

Martin Sleziak

Well, this is the same as sequences having only finitely many terms considered as a subspace of $\ell_2$.

Alessandro Codenotti

Yes

Martin Sleziak

Yes, it is one of the standard examples of non-complete space.

And if I have a Hilbert space $X$ which has a subspace $S$ that is not closed (i.e., not complete), I can simply take some point $y\in X\setminus S$.

The functional $x\mapsto\langle x,y\rangle$ is an example of a functional on $S$ which does not have this representation.

Alessandro Codenotti

Right, that works too

Martin Sleziak

At least I think this might works as counterexample.

Probably I also need $\overline S=X$.

user84215

12:37

How does the (1,3)- curvature tensor behave under scaling the metric?

Eric Silva

@MathematicsAminPhysics it's invariant under scaling

user84215

@EricSilva Because the christoffel symbols and connections remain invariant. Right?

Eric Silva

yes

user84215

@EricSilva So that also must be true for the curvature tensor of the unit sphere. Right?

Eric Silva

sure it's true for the (3,1) Riemann curvature tensor of any riemannian manifold

user84215

12:51

@EricSilva The curvature tensor of the unit sphere is: $R_1(X,Y)Z=<Y,Z>X - <X,Z>Y$. Right?

Secret

in The h Bar, 9 mins ago, by Slereah

https://publications.mpi-cbg.de/Mayer_2010_4314.pdf

Using four complex parameters to build an elephant

Not sure what it will be useful for other than fulfilling a saying

Balarka Sen

@Brody Great. So if $a^k = 1$ for some odd $k$, and order of $a$ is even so that $a^\ell = 1$ for some even $\ell$, $k$ and $\ell$ are coprime, hence...

@LeakyNun lmao that is itself a consequence of the division algorithm aka Bezout.

anyone can quote wikipedia.

there is absolutely 0 credit in that

Eric Silva

@MathematicsAminPhysics something like that probably

user84215

@EricSilva But that is not invariant under scaling the metric.

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$Mathematics$ Mathematics