Mathematics - 2016-02-14 (page 2 of 3)

I think the construction is done similarly in Hatcher, but I'm having trouble "translating" between the two (relevant section is 1.3, subsection "The Classification of Covering Spaces")

Balarka Sen

9:59 AM

I am struggling to understand your notation. First, how is your professor defining $E$? What are the elements?

Second, in your definition of $O$ above I don't get what (1) $H \cdot [\sigma * \tau]$ is (2) I can't parse "$\tau : I, 0 \to V, \sigma(1)$ a path"

Danu

Given a subgroup of $\pi_1(X,x_0)$, call it $H$, we defined $E$ as a set as follows $$ E:=H\backslash [(I,0)\to(X,x_0)]$$

@BalarkaSen Yeah sorry, I added parentheses in the last thing because it was not clear: $\tau:(I,0)\to (V,\sigma(1))$ is a path (that stays inside $V$)

$H$ acts on paths as $[h]\cdot [\sigma]=[h*\sigma]$, i.e. just precomposition with the loop

Balarka Sen

Ah, now that's starting to make sense.

And the projection $p : E \to X$ is defined as?

Evaluate paths at it's endpoint?

Danu

I think so (this was also not very clear to me)

Balarka Sen

Right OK I get it. $\tilde{X} = [(I, 0) \to (X, x_0)]$ can be topologized: consider the basis of open sets $O_{[\sigma, U]} = \{[\sigma * \tau]: \tau : [0, 1] \to U, \tau(0) = \sigma(1)\}$ where $[\sigma]$ is an element of $\tilde{X}$. So the basis open sets consists of homotopy classes of little paths growing out from $\sigma$.

The topology on $E$ is just given by the quotient topology, as $H$ acts on $\tilde{X}$ naturally as you mentioned above.

The projection map $p : \tilde{X} \to X$ is given by $p([\sigma]) = \sigma(1)$ (well-defined as homotopic paths must have same endpoints). So this naturally gives you a projection map $p(H \cdot [\sigma]) = \sigma(1)$ from $E \to X$.

Danu

10:16 AM

@BalarkaSen Understood

Balarka Sen

Now, let's try to understand continuity.

Danu

@BalarkaSen Also fine; now, why is it continuous?

Balarka Sen

Let us just work with $p : \tilde{X} \to X$ because paths are easier to understand than $H$-cosets of paths. We want to prove $p$ is continuous.

Danu

Hatcher discusses it in the last paragraph of 64/first paragraph of 65

Balarka Sen

Well, then, if you know why $p$ is continuous, you automatically prove that $E \to X$ is continuous.

Danu

10:23 AM

I don't quite understand his argument.

Balarka Sen

Let me have a look.

Danu

Not completely sure why $p^{-1}(V)\cap U_{[\gamma]}=V_{[\gamma']}$ (in his notation)

I see inclusion of $V_{[\gamma']}$ but not the other way around

Balarka Sen

Note that Hatcher is already choosing a basis for the base space $X$: it's the collection of path connected open sets $U$ of $X$ so that the inclusion induced map $\pi_1(U) \to \pi_1(X)$ is trivial. This defines a basis $\mathcal{B}$ for a topology on $X$, and then $\{O_{[\sigma], U} : U \in \mathcal{B}\}$ defines a basis for a topology on $\tilde{X}$.

Danu

That's the same basis I defined, by the way.

[though I did not yet formulate semilocal simply connectedness, that comes slightly later in the discussion as presented by my prof.]

Balarka Sen

Ah, alright.

@Danu $p^{-1}(V) \cap U_{[\gamma]}$ consists of things of the form $[\gamma * \tau]$ such that $\tau$ is a path in $U$ joining $\gamma(1)$ to some point in $V$, right?

So $\tau : [0, 1] \to U$, $\tau(0) = \gamma(1)$ and $\tau(1) = v \in V$.

But this means I can homotope $\tau$ to be entirely inside $V$, right?

Danu

10:37 AM

@BalarkaSen Can we?

Balarka Sen

Well, $U, V$ are both in $\mathcal{B}$.

That's where we are going to need the triviality of $i_*$ assumption.

Danu

I don't like that very much, since in my notes we have continuity before making this assumption.

Balarka Sen

I was merely referring to why Hatcher says $p^{-1}(V) \cap U_{[\gamma]} = V_{[\gamma']}$. I don't understand your professor's proof to be honest, it doesn't seem right.

JKnecht

How do you undo an edit you have made to a post?

Danu

Edit it again?

@BalarkaSen Sigh :\

I don't think he uses semilocal simple connectedness here

Balarka Sen

10:49 AM

You always ask me about horrendous details of the most ucky theorems in topology which I can never give satisfactory answers to, @Danu :D

Danu

Yeah, the details are usually the only tricky part :P

Balarka Sen

Details are friendly. Horrendous details aren't.

:)

JKnecht

@Danu If you edit again you need to change 6 charachters to make it a new edit...you cant completely undo it that way

Danu

@BalarkaSen So there is just one thing we need to check. The claim made in the first sentence on page 65... Does it still hold if $p$ is only surjective?

I think it does, and then we don't need s.l.s.c.

Note how he only says onto

Balarka Sen

I am not sure what claim you are referring to. non semilocally simply connected spaces don't have universal cover.

Danu

10:53 AM

The first sentence on page 65.

Balarka Sen

$\tilde{X}$ is simply connected?

Danu

@BalarkaSen Yes, but the point is that one needs this only for injectivity of $p$

@BalarkaSen He hasn't established that yet at that point in the text---plus I'm thinking about a modification with nontrivial char. subgroup anyways.

I think $p$ is still open and continuous without s.l.s.c.

Balarka Sen

Oh. Um. I am not sure.

I have to leave now, but you should ask about this to someone else. Mike, maybe.

Danu

Meh, I doubt that he'll be willing to invest the time to understand my setup ;)

Balarka Sen

Sorry for not being very helpful.

Danu

10:55 AM

(and I can't blame him)

Thanks anyways!

Balarka Sen

If you haven't figured it out after I return, I can have a look.

See ya later.

Forever Mozart

hi

eveybdys

@BalarkaSen

Danu

11:17 AM

@BalarkaSen I think I'll be able to do it, but I'll need to resort to writing things down. Urghhhh

Tobias Kildetoft

@Danu Ohh no, not the dreaded "writing things down"

Danu

@TobiasKildetoft I try to minimize paper waste, is how I like to think about it :)

(I work almost exclusively in TeX)

Huy

no black-/whiteboard? :P

Tobias Kildetoft

I only work in Tex when actually writing stuff up

But I do use most parts of each piece of paper for small calcultions

Danu

@TobiasKildetoft Most things I have to do are so simple that one doesn't really need to do anything else (I'm still a student).

Evinda

12:09 PM

Hi @TobiasKildetoft
I want to calculate the center of $Z(S_n)$. Could I ask you something about it?

Tobias Kildetoft

@Evinda You mean the center of the symmetric group?

Evinda

Yes @TobiasKildetoft

Tobias Kildetoft

@Evinda Well, it is trivial, which is fairly straightforawrd to show

Evinda

@TobiasKildetoft We let $g$ be a transposition $\in S_n$, right? So $\tau=( 1 m)$ for a fixed $m$.
Also let $\tau = ( 1 n), n \in \mathbb{N}$.

Then we want to check if $g= \tau g \tau \Leftrightarrow ( 1 m)= (1 n) (1 m) (1 n)$

(1 n) (1 m) (1 n)=(n m)

Right so far?

Albas

Hello@TobiasKildetoft

Evinda

12:52 PM

@Huy Happy Valentine's day :D

dash

1:27 PM

Hi. Is there a difference between prime and irreducible polynomial?

Balarka Sen

@Danu Hahaha.

@ForeverMozart Hiya.

@Evinda "Happy" and "Valentine's day" makes a bit of an oxymoron don't they?

Evinda

@BalarkaSen Don't we say it like that?

Danu

@BalarkaSen Only for some!

Balarka Sen

Depends on "we". I prefer unhappy valentine's day, because that's grammatically correct.

Evinda

Ok... Which is the difference grammatically? :p @BalarkaSen

Balarka Sen

1:37 PM

@Danu Collection of persons for whom valentine's day is happy is a set of measure 0. So they don't count.

Danu

I prefer to measure by happiness.

Evinda

@Danu So are you celebrating the day?

Balarka Sen

@Danu You're isolated.

Danu

@Evinda No, not really.

Evinda

Hey @DanielFischer !!!
We have $Z(S_n)=\{ c \in S_n: cg=gc, \forall g \in S_n \}$.
We let $g$ be a transposition $\in S_n$, right? So $g=( 1 m)$ for a fixed $m$.
Also let $\tau = ( 1 n), n \in \mathbb{N}$.

Then we want to check if $g= \tau g \tau \Leftrightarrow ( 1 m)= (1 n) (1 m) (1 n)$

(1 n) (1 m) (1 n)=(n m)

$(1 n)=(n m) \Leftrightarrow 1=m $

So we deduce that $Z(S_n=\{ (1)\})$, right?

Valentin Tihomirov

1:48 PM

I see $|\vec v \vec v\rangle$ in a physical text. What does it mean?

I guess that they square a vector inside a ket. Why do you normally do that?

Daniel Fischer

2:10 PM

@Evinda That's too quick. $S_n$ also contains permutations that aren't transpositions, and besides there are transpositions that leave $1$ fixed (if $n > 2$, but it's quick to see that $S_2$ is abelian). It suffices to look at $\tau$ of the form $(1\,m)$ because $S_n$ is generated by transpositions of that form. If you take a cycle for $g$, say $g = (a_0\, a_1\, \dotsc\, a_{k-1})$, what is $(1\, m) g(1\, m)$?

@ValentinTihomirov What does "square a vector" mean? Usually, you don't have a multiplication on a vector space. Could it be just a typographical error and it should be $\mid \vec{v}\rangle$? Would that make sense?

Valentin Tihomirov

@DanielFischer I do not know. I see |a_0 a_0> at page 3 of short paper for simple mortals arxiv.org/pdf/1212.5214v2.pdf and trying to understand what does that check mean.

Daniel Fischer

2:28 PM

@ValentinTihomirov Well, umm. It's physics, it isn't supposed to make sense? More seriously, you're looking at a system consisting of two thingamajigs, and measure some properties, where the measurement can result in two outcomes. Perhaps $\mid a_0 a_0\rangle$ means the situation that measuring $A$ results in the outcome $a_0$ for both thingamajigs?

JeSuis

@DanielFischer Grüß dich! If we have an analytic function $f: \Bbb{C}\setminus A\to \Bbb{C}$ where $A$ is countable and bounded. Is $f$ constant ? ( if $A$ is closed it worksn but not sure for the general case).

Daniel Fischer

@JeSuis Is $A$ bounded, or $f$?

JeSuis

@DanielFischer sorry, $f$.

Daniel Fischer

Okay. Now, if $A$ is not closed, what does it mean that $f \colon \mathbb{C}\setminus A \to \mathbb{C}$ is analytic?

JeSuis

@DanielFischer $f$ can be written as a power series (locally), not sure I understand your question

@DanielFischer hm right

it doesn't make sense

Valentin Tihomirov

2:37 PM

@DanielFischer Ok, but how do I compute em mathematically? Are they just vector length squared?

Daniel Fischer

@JeSuis That's a bit strong. It's customary to say that $f$ is analytic on $M$ if there is an open $U \supset M$ and an analytic $F\colon U \to \mathbb{C}$ with $f = F\lvert_M$. That interpretation works here, but it puts you back into the situation of a closed exceptional set $A_0 \subseteq A$.

JeSuis

@DanielFischer Do you think we can prove the result with only complex analysis tools ? (Like Cauchy's formula, or something else)

Daniel Fischer

@ValentinTihomirov Vector length squared wouldn't make sense. I have no idea how to compute that, sorry.

Valentin Tihomirov

Yes, they have a vector |Ф> = (|00> + |11>)/√2 is projected upon basis |a_0>, |a_1| and somehow recover |Ф> = (|a_0 a_0> + |a_1 a_1>)/√2. I fail to see how second |Ф> equals the first one.

Daniel Fischer

@JeSuis You mean if you have a bounded analytic function $f\colon \mathbb{C}\setminus A \to \mathbb{C}$ where $A$ is a closed countable subset of $\mathbb{C}$, prove that $f$ is constant using only complex analysis tools? Hmm, depends on what counts as "only complex analysis tools". The only ways I can think of now use also a bit of real analysis or set theory.

Valentin Tihomirov

2:48 PM

I understnad. I think |a_0 a_0> is what you call "a tensor product". It just says that I have a vector of vectors or "concatenated vector", I think. Then, obviously, |a_0 a_0> + |a_1 a_1> = |00> + |11>, given |a_0> = |0> and |a_1> = |1>.

JeSuis

@DanielFischer I would say Cauchy's formula, or everything as a theorem of complex analysis ( with a bit or real analysis or set theory it's not a problem), but a proof that use a big part of complex analysis will be cool :).

Valentin Tihomirov

It remains to understand why, given |b_0> = 1/2 |0> + √3/2|1> and |b_1> = √3/2 |0> - 1/2 |1>, we have |b_0 b_0> + |b_1 b_1> = |00> + |11>.

Evinda

@DanielFischer With permutations that aren't transpositions you mean that we could have more than two numbers, for example ( 1 2 3) ?
If it suffices to look at $\tau$ of the form $(1\,m)$ because $S_n$ is generated by transpositions of that form, why isn't g also of that form, given that it is also an element of $S_n$ ?

If we have $g = (a_0\, a_1\, \dotsc\, a_{k-1})$ can it be that $a_i=1$ for some i?

I'm an artist

A crazy day here (highly overloaded), teaching, researching, teaching, researching ...

Agawa001

reaching and stretching and stitching

I'm an artist

2:54 PM

Let me see the news ...

2

Q: Another way of doing integration

What's your option for calculating this integral? No full solution is necessary, it's optional as usual. Calculate $$\int_0^1 \frac{2 \zeta (3)\log ^3(1-x) \text{Li}_2(1-x) }{x}-\frac{2 \zeta (3) \log ^2(1-x) \text{Li}_3(1-x)}{x}+\frac{ \log (x) \log ^5(1-x)\text{Li}_2(1-x)}{x}+\frac{\log ^4(1-...

Albas

I guess Sunday is a day for research. Very less traffic on the chat today

I'm an artist

@Albas Sunday is always a great day for work. :D

Huy

@ValentinTihomirov: they mean $|0,0 \rangle$, $|a_0, a_0 \rangle$ etc. that's not a multiplication or something.

Daniel Fischer

@JeSuis Method 1, use Cauchy's integral formula. You have $$f(z) = I(R,z) := \frac{1}{2\pi i} \int_{C_R} \frac{f(\zeta)}{\zeta - z}\,d\zeta$$ in the interior of $C_R$ - which is a circle of radius $R$ with centre $z_0$ - for small enough $R$, so that $A$ doesn't intersect the closed disk determined by $C_R$. On the other hand, $I(R,z)$ makes sense for arbitrary $R > 0$ and $z$ with $\lvert z-z_0\rvert < R$, and for fixed $z$, $I(R,z)$ depends continuously on $R$.

That gives you $f(z) = I(R,z)$ for all $R > \lvert z-z_0\rvert$. Thus you get an extension of $f$ to all of $\mathbb{C}$. Liouville then says constant.

Huy

usually, $|x_1, x_2 \rangle$ means $| x_1 \rangle \otimes | x_2 \rangle$

Valentin Tihomirov

3:01 PM

@Huy I need to know the usual meaning of \otimes :)

Huy

@ValentinTihomirov: Tensor product.

JeSuis

@DanielFischer very nice!

Huy

@ValentinTihomirov: If I recall correctly, the Bell's basis is a basis over $\mathbb{C}^2 \otimes \mathbb{C}^2$, so you're working with a finitely-dimensional vector space. For the sake of simplicity, I think you can just use the Kronecker product.

@ValentinTihomirov: is this something you're reading to understand the EPR-paradoxon or something different in QIT?

Daniel Fischer

@JeSuis Method 2, a countable closed set contains isolated points. By Riemann, you can remove the singularities there. That gives you a countable closed set $A_1$ (which is the set of limit points of $A$) and a bounded analytic function $f_1 \colon \mathbb{C}\setminus A_1 \to \mathbb{C}$. If $A_1$ happens to be empty, you're done (Liouville), but in general it isn't. But you can repeat, $A_1$ has isolated points … Repeat.

Let $$A_{\omega} = \bigcap_{n \in \omega} A_n\quad\text{and}\quad f_\omega = \bigcup_{n\in \omega} f_n.$$ Repeat further, for a successor ordinal, if you already have $A_{\alpha}$ and $f_{\alpha}$, let $A_{\alpha+1}$ be the set of limit points of $A_{\alpha}$ and get $f_{\alpha+1}$ from $f_{\alpha}$ via the removable singularity theorem.

For a limit ordinal, let $$A_{\lambda} = \bigcap_{\alpha < \lambda} A_{\alpha}\quad \text{and} \quad f_{\lambda} = \bigcup_{\alpha \in \lambda} f_{\alpha}.$$ Then $A_{\omega_1} = \varnothing$, and $f_{\omega_1}$ is hence a bounded entire function.

I'm an artist

I'm working on a proof that I expect to be on 25 pages or so. It's related to many results, and explaining each one it takes a lot. It will be my longest proof so far I think.

I was optimistic saying that, it might even take up to 30 pages.

JeSuis

3:13 PM

@DanielFischer I need to learn to understand this one, I am unfamiliar with "ordinal", but thanks a lot, juste need to learn :)

Brennan.Tobias

Hi

I set a bounty on my prime gaps question

I'm an artist

It's a wonderful thing to be optimistic. It keeps you healthy and it keeps you resilient.

Daniel Kahneman

Daniel Kahneman (Hebrew: דניאל כהנמן‎, born March 5, 1934) is an Israeli-American psychologist notable for his work on the psychology of judgment and decision-making, as well as behavioral economics, for which he was awarded the 2002 Nobel Memorial Prize in Economic Sciences (shared with Vernon L. Smith). His empirical findings challenge the assumption of human rationality prevailing in modern economic theory. With Amos Tversky and others, Kahneman established a cognitive basis for common human errors that arise from heuristics and biases (Kahneman & Tversky, 1973; Kahneman, Slovic & Tversky, 1982...

(by this guy)

Vrouvrou

@DanielFischer hello, to prove that if A is connected, then $\oversline{A}$ (the closure) is also connected, can i say: $A$ is connected then for two open sets A_1, A_2 we have : $A=A_1\cup A_2, A_1\cap A_2=\emptyset \Rightarrow A_2=\emptyset, or A_1=\emptyset$, suppose that $A_2=\emptyset$ we have that $\overline{A}=\overline{A}_1\cup \overline{A}_2$ and $\overline{A}_1\cap \overline{A}_2=\emptyset$ then $\overline{A}$ is connected ?

Daniel Fischer

@Evinda For example. Or $(1\,2\,3)(5\,8\,7\,9)$ as a different example. An arbitrary $g\in S_n$ is a product of transpositions of that form, but it is not a priori clear that some product of such transpositions can't be in the centre although none of the transpositions themselves lies in the centre. There are groups with nontrivial centre that have generating sets where none of the generators lies in the centre.

I'm an artist

Question for all: how long (in pages) is the longest proof you have ever written?

Daniel Fischer

3:16 PM

Yes, it can be that $a_i = 1$ for some $i$. And that $a_j = m$ for some $j$. You make a case distinction, a) none of the $a_i$ is $1$ or $m$, b) exactly one of the $a_i$ is $1$ or $m$, c) there are $i$ and $j$ with $a_i = 1$ and $a_j = m$.

You will find that in the first case, $g$ commutes with $(1\,m)$, in the second it doesn't commute with $(1\,m)$, and in the third, it commutes with $(1\,m)$ if and only if $g = (1\,m)$. But if $n \geqslant 3$, we can always choose an $m$ such that $g$ doesn't commute with $(1\,m)$.

I'm an artist

@Huy @TobiasKildetoft

@JeSuis

Valentin Tihomirov

@Huy Just EPR paradox. I would like to know if non-matematicians can understand it.

Huy

@ValentinTihomirov: what do you know about mathematics? in principle, what you do is compute a few traces.

JeSuis

@I'manartist it depends we prove everything or if we use theorems..

God

are philosophical questions permitted here?

I'm an artist

3:19 PM

@JeSuis No matter how you did it, I'm only curious what is the longest proof you got, how many pages. I reached a point where have to use and prove many auxiliary results which I cannot skip.

Vrouvrou

@DanielFischer?

Huy

@ValentinTihomirov: check out the original paper on it, it's not too technical, but I don't know if you can actually understand what's happening without doing the maths

@ValentinTihomirov: Teleporting an Unknown Quantum State via Dual Classical and Einstein-Podolsky-Rosen Channels

Valentin Tihomirov

@Huy Just some linear algebra. But not as far as Tensor Product. I am already failing at change of basis.

I'm an artist

@Huy 4-5 pages?

Huy

@I'manartist: I haven't proven anything lengthy, I'm a terrible mathematician

@ValentinTihomirov: just use the Kronecker product since everything's finite-dimensional.

@ValentinTihomirov: also, don't believe anything I say right away. I learnt these things 4 years ago and don't remember them very well.

@Evinda: happy valentine's day to you too. don't do too much maths!

I'm an artist

3:22 PM

There is a problem where I suppose I might use almost 50 pages, but this is just a rough approximation, it might be more or less, of course, a new problem related to multiple series.

Daniel Fischer

@Vrouvrou No, that doesn't work. You don't know that $\overline{A}_1 \cap \overline{A}_2 = \varnothing$. Start with a partition of $\overline{A}$ into two disjoint (relatively) open sets. Look at what that partition does with $A$.

I'm an artist

At any rate, I won't try it soon since I'm pretty involved in other kind of stuff. However, writing proofs of hundreds of pages sounds appealing, interesting.

Vrouvrou

@DanielFischer if $A_2=\emptyset$ then $\overline{A_2}\cap \overline{A_1}=\emptyset$

Evinda

@Huy I am not celebrating today... Have you bought a gift for your girlfriend?

Huy

@Evinda: no, we have our anniversary on Thursday, so I invited her for a fancy dinner then.

Evinda

3:26 PM

@DanielFischer Could you explain further to me why we pick t of the form (1 m) but not g? I still haven't understood it...

@Huy nice

Daniel Fischer

@Vrouvrou Ah, right. But nevertheless, you can't start from $A$, you must start from $\overline{A}$, since a priori not every partition of $\overline{A}$ must come from a partition of $A$.

I'm an artist

@Huy think positively, I don't think you're terrible (in a negative sense) at all.

Back to writing some more on the proof.

BBL

Daniel Fischer

@Evinda We want to check whether $g$ commutes with everything. For that, it suffices that $g$ commutes with a set of generators of the group, since you can move every product of generators to the other side of $g$ factor by factor then. But something can commute with a product without commuting with each factor, so we have to view $g$ as an atomic entity.

Evinda

@DanielFischer Don't we have to check if $\tau$ commutes with everything?

So you mean that we can write each element of $S_n$ as (1 m) for some $m \in \mathbb{N}$? Or have I understood it wrong?

Daniel Fischer

@Evinda No, we check whether $g\in Z(S_n)$.

Balarka Sen

3:39 PM

@I'manartist 100 pages.

Daniel Fischer

And every $g\in S_n$ can be written as a product of such transpositions, $$g = \prod_{i = 1}^r (1\, m_i)$$ for some $r$ and some sequence $(m_i)_{1\leqslant i \leqslant r}$.

Balarka Sen

Just write one word on each page.

I'm an artist

@BalarkaSen Really? How many were blank pages?

Balarka Sen

None. :)

Evinda

Why? Don't we have $Z(S_n)=\{ c \in S_n: cg=gc \forall g \in S_n \}$?
So don't we check if $c \in Z(S_n)$, i.e. if cg=gc for all $g \in S_n$ ? @DanielFischer

I'm an artist

3:40 PM

@BalarkaSen Awesome, really. :-)

Balarka Sen

I know.

Daniel Fischer

@Evinda Then replace $g$ with $c$ in everything I wrote.

Balarka Sen

Speaking of long proofs, I think we should all remember that quantity < quality.

Evinda

Also why does it hold that each transposition starts with 1? @DanielFischer

Agawa001

@BalarkaSen serious ? 100 pages of one proof of one problem ?

Balarka Sen

3:44 PM

@Agawa001 Evidently you didn't read my next message.

I'm an artist

@Agawa001 It's not that incredible. I have a series for which I have to use 25-30 pages. I think it's normal.

Daniel Fischer

@Evinda It doesn't. But transpositions of that form generate $S_n$, so it suffices to look at those transpositions to see whether $c\in Z(S_n)$. $c\in Z(S_n) \iff \bigl(\forall m\bigr)\bigl( c = (1\,m)c(1\,m)\bigr)$.

Agawa001

oh lol

@I'manartist what kind of crazy series requires 30 pages proof ?

I'm an artist

@Agawa001 I'll show it to you at some point. It's about problems that split into many other problems and where you need to prove each result and that means space. If you do it very elegantly you might even write 35-40 pages.

Evinda

Could you explain further to me what you mean with: transpositions of that form generate $S_n$ ? @DanielFischer

Agawa001

3:46 PM

i mean, i should borrow 29 cerebrums to understand it in one day

I'm an artist

(elegantly to the reader - you explain alternative ways in different points of the main proof, say)

Daniel Fischer

10 mins ago, by Daniel Fischer

And every $g\in S_n$ can be written as a product of such transpositions, $$g = \prod_{i = 1}^r (1\, m_i)$$ for some $r$ and some sequence $(m_i)_{1\leqslant i \leqslant r}$.

Evinda

Yes, I have read it... But why does this hold? @DanielFischer
No matter what $m_i$ is , the transposition starts with 1, doesn't it?

Daniel Fischer

@Evinda But $(1\,m)(1\,k)(1\,m) = (k\,m)$, so you can write any transposition as a product of transpositions of the given form. If you already know that $S_n$ is generated by transpositions, you're done, otherwise you still have work to do.

Evinda

I see...How can we show that $S_n$ is generated by transpositions? @DanielFischer

Daniel Fischer

4:00 PM

$(a\,b)(b\,c) = (a\,b\,c)$, $(a\,b)(b\,c)(c\,d) = (a\,b\,c\,d)$ etc.

Thomas Andrews

From a practical view, you can think in terms of sorting a list. If you have a list in any order, you can "sort it" by swapping pairs of elements at a time. @Evinda

Evinda

So does the fact that the product of transpositions is equal to one permutaion imply that $S_n$ is generated by transpositions? @DanielFischer

Thomas Andrews

One way to do this: Given a permutation $\pi$. Then if $\pi(n)=a$ you have that $\pi\circ(n\,a)$ fixes $n$, so you can proceed by induction on this permutation of $n-1$ elements. @Evinda Then you get $\pi(n\,a_1)(n-1\,a_2)\dots(2\,a_{n-1})=e.$ Then $\pi = (2\,a_{n-1})(3\,a_{n-2})\dots(n\,a_1)$.

Evinda

@ThomasAndrews What do you mean with $\pi\circ(n\,a)$ fixes $n$?

Daniel Fischer

@Evinda The point is that you can write every permutation as a product of transpositions.

Thomas Andrews

4:12 PM

Do you know what I mean by $\pi\circ(n\, a)$? It is the composition of permutations. So this means the resulting permutation $\sigma = \pi\circ(n\,a)$ has the property that $\sigma(n)=n$.

Akiva Weinberger

Wait, isn't $\sigma(n)$ equal to $\pi(a)$?

Or am I reading the notation backwards?

Evinda

@DanielFischer But if we suppose that $\tau=\prod_{i=1}^r (1 m_i)$ it doesn't hold that $\tau=\tau^{-1}$, does it?

Akiva Weinberger

No, don't think so

Thomas Andrews

Depends on the order in which you write permutations, I suppose, but I might have gotten crossed up. There are two ways to treat permutation composition, and I was thinking left-first, but then I shouldn't write $\pi(n)$ for the value of $\pi$ at $n$. @AkivaWeinberger

Akiva Weinberger

If the left-most one acts first, then $\rho:=(1\ 2)(1\ 3)=(1\ 2\ 3)\ne\rho^{-1}$

@Evinda

@ThomasAndrews It doesn't really matter, in any case. If $L_n:=(S_n,\circ_L)$ is the group with left-first composition and $R_n:=(S_n,\circ_R)$ is the group with right-first composition, then they're homomorphic. $f:L_n\to R_n,\rho\mapsto\rho^{-1}$ is a homomorphism. Right?

Thomas Andrews

4:25 PM

Well, it matters for comunication purposes, @AkivaWeinberger If we are talking about different $\circ$ operators, it is bound to be confusing, especially to a student just learning.

Akiva Weinberger

'Cause $(\pi\circ_L\rho)^{-1}=\pi^{-1}\circ_R\rho^{-1}$.

Right, yeah.

Daniel Fischer

4:39 PM

@Evinda For $\tau$, we would have $r = 1$.

Evinda

@DanielFischer So you mean that we pick $\tau=(1 m)$ and as g the product?

L33ter

hi @BalarkaSen

Balarka Sen

Hello.

Evinda

If so then we can do this, since we consider that $g$ is a product of permutations and so if we would want for example $\tau$ to be (1 2) (2 3), the term (2 3) could be a part of g, right? Or have I understood it wrong? @DanielFischer

L33ter

I am trying at the moment to improve my grammar for English.

I got 70 % in my English essay, so I gotta improve my writing.

I hate it you know because actually my high school in Egypt didn't really provide me with the good background, but I learned mostly everything by myself.

Balarka Sen

4:54 PM

Read some English l33terature.

L33ter

btw @BalarkaSen I ordered

Yeah I am doing that now as well as improving my grammar.

btw I ordered hatcher's book I like hard copy more than pdf books.

I am marking for vector calculus, so I am just gonna use that money on some math books.

Balarka Sen

Grammar cannot be improved by studying grammar textbooks.

L33ter

how can it be improved then ?

Balarka Sen

By reading literature, IMO. Language came before the grammar, not the other way around.

L33ter

Yeah, but then you would know actually how to write carefully without actually knowing where the logic came from and why you put things in a certain way.

Balarka Sen

5:01 PM

I can't parse that statement, but I can tell you I don't know much grammar. I don't write perfect English (especially in chat!), but I happen to think what I write classifies as good enough.

And the way I learnt English is via reading English literature.

Albas

My grammar is much more worse than yours@L33ter

But I guess reading plays help a lot

Plays like Agamemnon

dash

5:14 PM

Hi. Why the order of a finite field must be always prime? Why $Z_6^*=\{1,5\}$ cannot be a finite field?

Thomas Andrews

It isn't closed under addition, for one. @dash

Also, the order of finite fields need not be prime, but the only finite fields of the form $\mathbb Z_n$ are prime. The other finite fields are of order $p^k$ for some prime $p$ and some positive integer $k$. @dash

Evinda

At the first case we have m->1->1->m and m->1->1-> m, right?
Why do we deduce that g commutes with (1 m).
In the second case we have m->1->a_1->a_1 and 1->m->1, right?
How do we deduce that it doesn't commute with (1 m) ?

At the third case we have 1->m->1->m, m->1->a_1->a_1

if it commutes with (1 m) only if g=(1 m), then it never commutes with (1 m) for $n \geq 3$, right?

L33ter

5:31 PM

yeah I agree @Albas

Tobias Kildetoft

6:49 PM

@I'manartist I think no more than 2 pages probably, and to get to even that I need to include the proofs of the needed lemmas.

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$Mathematics$ Mathematics