Mathematics

12:00 AM

that it put it on pdf when it is over

Actually, the thing I had envisioned kind of breaks apart.. $(yx_i)x_i=x_{i+1}x_i$ where $y(x_i^2)=x_{i+2}$, so, it has to hold that $x_{i+1}x_i=x_{i+2}$

Leaky Nun

codecogs.com/latex/eqneditor.php @Jacksoja

Rithaniel

I've been using Overleaf for latex stuff.

Leaky Nun

oh and it isn't $C_2$, it should be $\Bbb Z$

Jacksoja

@LeakyNun Thanks !

Rithaniel

12:02 AM

Integers under addition, yeah, and $y=x_0=1$

Leaky Nun

nah it's just the trivial group, but I don't want to think about this question anymore until you give us enough context

Rithaniel

It was just something that came into my mind while thinking about groups.

Zee

^^ BS

Rithaniel

I did the same thing with topology when I was taking a course on it. While thinking about the topic, I come up with what I think might be a valid construct, but then find that I don't really understand it, or if it's even valid.

Talking about it with people helps me understand where my thoughts went off the deep end.

Rithaniel

12:19 AM

Though, I came up with that one by trying to describe a mechanic in a video game (Opus Magnum) via a group. In this game you can combine two lead to get a tin, combine two tin to get an iron, ect. You can also combine a quicksilver with any metal to get the next highest level.

Sergey Dylda

Hello chat
Is there any way to show that $End(V)$ (for $V$ - vector space) is a manifold, without choosing a basis on $V$?

Rithaniel

(Combining tin with lead doesn't give anything, though, so I guess that breaks closure, making it not a group)

Ted Shifrin

@Sergey: It's a vector space itself, isn't it?

Sergey Dylda

@TedShifrin
It is. How to define a topology on it, whitout a choice of a basis?

Ted Shifrin

Well, I don't know how to give a chart for a vector space without choosing a basis for it. Why can we not choose a basis?

Sergey Dylda

12:39 AM

@TedShifrin
I mean, a basis is a thing that always exists for vector spaces, but the choice of it, in general, is completely arbitrary. So I wondered what happens if we would not choose any basis at all.

Jacksoja

@TedShifrin Hi Ted

@TedShifrin for K and F fields, what does K/F mean ?

Ted Shifrin

An isomorphism $\Bbb R^n\to V$ is a choice of basis, so I don't see how to avoid it.

@Jacksoja: It means that $K$ is a field extension of $F$ (or $F$ is a subfield of $K$).

Jacksoja

it is not the same as the quotient right ?

Ted Shifrin

Not usually.

Jacksoja

as in G/H for groups

Ted Shifrin

12:43 AM

Without context one can never be sure.

Jacksoja

Galois theory

Ted Shifrin

Like $K/F$ is Galois? Then it's what I said.

$K$ is a Galois extension of $F$.

Jacksoja

@TedShifrin ok thank you

But if K is a field, the quotient by anything does not make sense right?

because the only ideals of any field are zero ideal and the whole thing

Kasmir Khaan

@TedShifrin Ted ! My man !

Ted Shifrin

right, @Jacksoja, unless you meant quotient as abelian groups ... but meh.

hi @Kasmir

Kasmir Khaan

12:57 AM

@TedShifrin Long time no see ! how are you ?

Ted Shifrin

Doing OK, and you?

Kasmir Khaan

not bad not bad

same as always ._.'

Ted Shifrin

It's way past your bedtime

Kasmir Khaan

@TedShifrin Haha yeah getting ready to bed =p

@TedShifrin Good night Sir !

Ted Shifrin

night!

IDUTDP

1:24 AM

Good evening.

Ted Shifrin

Good evening.

IDUTDP

Hi.

If i were to embark in a quest to find a non-recursive formula that would give me the number of ways i can associate a word, given a non-associative binary operation; do you think i would be able to succeed in 2 weeks or less?

Or is a harder problem than that.

What do you think?

Ted Shifrin

I have no knowledge thereupon.

IDUTDP

I think they are called the Carmichael Numbers . . .

Haven't looked up in case of spoilers . . .

I would like to do this if i can, but i don't like to spend too much time on just it . . .

Ted Shifrin

Carmichael numbers are related to pseudoprimes. Nothing to do with what you said, as far as I can tell.

IDUTDP

1:31 AM

Or maybe they were called the Catalan Numbers . . .

Ted Shifrin

Perhaps more plausible.

IDUTDP

It's the "Ca" at the beginning . . .

Ted Shifrin

shrug

IDUTDP

1:42 AM

Well; have a nice evening.

Ted Shifrin

You too.

IDUTDP

:)

Indrasis Mitra

3:08 AM

Any comment on how to move forward with this problem would be really helpful. Really stuck here

0

Q: Help in building final solution after solving the separated Eigenvalue problems

I (with help from a MSE user) used the following substitution to seperate variables in a second order linear PDE $$\theta_w = e^{-\beta_hx}F'(x)e^{-\beta_cy}G'(y)$$ The following two ODEs (Eigenvalue problems) are a result of applying variable seperation to a system of three coupled PDEs \begi...

user131753

3:21 AM

@MikeMiller Would you mind to elaborate this comment a bit?

Zerix

If a function is continuous at a point then it is derivable at that point. If it's non continuous, is it okay to say it's non derivable

Fargle

4:10 AM

@Zerix You have the implication backwards: if a function is differentiable, then it is continuous. (There are functions which are continuous on the whole real line but differentiable nowhere!) Your question is the contrapositive of this statement, so yes, it would be okay to say that.

Zerix

4:37 AM

Okay. Thanks for the help sir!

user76284

5:11 AM

How should one refer to the "free group on $n$ generators" when one has redundant generators?

uhoh

5:41 AM

1

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user131753

6:05 AM

Since I have given some further though to my question I thought it would be better to write it down.

user131753

Let $F:(\mathbf{S},U)\to(\mathbf{T},V)$ be a concrete functor. For each $\mathbf{X}$-objects $X,Y$ consider the following full subcategory $\mathsf{PairFibre}_{\mathbf{S}}(X,Y)$ of $\mathbf{S}$ whose object class is the union of the $\mathbf{S}$-fibres of $X$ and $Y$. Similarly consider the full subcategory $\mathsf{PairFibre}_{\mathbf{T}}(X,Y)$ of $\mathbf{T}$ whose object class is the union of the $\mathbf{T}$-fibres of $X$ and $Y$.

user131753

Thus a concrete functor $F:(\mathbf{S},U)\to(\mathbf{T},V)$ is a concrete isomorphism iff for each $\mathbf{Set}$-objects $X,Y$, there exists an isomorphism $G_{(X,Y)}:\mathsf{PairFibre}_{\mathbf{S}}(X,Y)\to\mathsf{PairFibre}_{\mathbf{T}‌}(X,Y)$ such that the following diagram,

user131753

$$\require{AMScd}\begin{CD} \mathbf{S} @>{F}>> \mathbf{T}\\ @A{\text{incl}_{\mathbf{S}}}AA @AA{\text{incl}_{\mathbf{T}}}A \\ \mathsf{PairFibre}_{\mathbf{S}}(X,Y) @>{G_{(X,Y)}}>> \mathsf{PairFibre}_{\mathbf{T}}(X,Y)\end{CD}$$
commutes.

user131753

So, informally we can think of a concrete isomorphism as a "paired-local" version of an isomorphism in the sense that a concrete isomorphism is determined by focusing at the same pair of "structures" under two different guises. It is in this sense that so far we care only about the categorical properties, the $\mathsf{PairFibre}_{\mathbf{S}}(X,Y)$ and $\mathsf{PairFibre}_{\mathbf{T}}(X,Y)$ are "essentially same".

user131753

Hence, in particular when $X=Y$ we conclude that since $\mathsf{PairFibre}_{\mathbf{S}}(X,X)$ and $\mathsf{PairFibre}_{\mathbf{T}}(X,X)$ for categorical investigations an $\mathbf{S}$-structure can be completely substituted by a $\mathbf{T}$-structure.

user131753

6:08 AM

In other words for each categorical property of $\mathbf{S}$-objects and $\mathbf{S}$-morphisms there exists a logically equivalent formulation of the "essentially same" property by $\mathbf{T}$-objects and $\mathbf{T}$-morphisms and vice versa. Does this way of seeing things sound all right @KarlKronenfeld @MikeMiller?

Mary Star

8:27 AM

Hey!! Let $a\in \mathbb{C}$. I want to show that there is no irreducible polynomial $f\in \mathbb{Q}[x]$ such that $f(a)=f(a+1)=0$.

For that do we suppose that there is such a $f$ and so if $f=gh$ then g or h must be unit?

Or do we do something else? Could you give me a hint?

Leaky Nun

@MaryStar I have a hard time believing that statement.

aha I've convinced myself now

Mary Star

@LeakyNun Why? Does this not hold?

Leaky Nun

but the problem is, do you understand the statement

Mary Star

We have a rational polynomial that has two complex roots and we want to show that it cannot be written in the form f=gh where neither g nor h are units, or not? @LeakyNun

Leaky Nun

no that's all wrong

sure, that's the literal interpretation of the statement

but you don't see the meaning behind it

"f=gh implies g or h are units" is the definition of irreducible

but that's just it. a definition. it isn't the most useful

you can't keep resorting to definition

definition can only get you that far

you only use definition when proving basic properties

hint: you'll definitely need galois stuff

Mary Star

8:35 AM

How do we use here Galois? Could yyou expalin that further to me? @LeakyNun

Leaky Nun

hint: the statement is saying "if $a$ is algebraic then $a$ and $a+1$ cannot be conjugates"

hint: galois group acts transitively on the roots of an irreducible polynomial, i.e. conjugates

Mary Star

Ah I understand that so far... so we suppose that f is irreducible then from your hint we have that if a is a root then its conjugate must also be a root of f. Since it is given that f(a)=f(a+1)=0 a+1 must be a conjugate of a. To get a contradiction we have to show that a+1 is not a conjugate of a? @LeakyNun

Sir Cumference

Hey quick question, my book says that a subset $E$ is dense in $X$ if every point in $X$ is a limit point of $E$, or a point in $E$. So by definition every set is dense in itself right, due to it satisfying the latter?

Leaky Nun

@MaryStar let $K$ be the splitting field of $f$ over $\Bbb Q$. $\operatorname{Gal}(K/\Bbb Q)$ acts transitively on the roots. What does this tell you?

@SirCumference yes

Sir Cumference

So is this page using a different definition?

Dense-in-itself

In mathematics, a subset A {\displaystyle A} of a topological space is said to be dense-in-itself if A {\displaystyle A} contains no isolated points. Every dense-in-itself closed set is perfect. Conversely, every perfect set is dense-in-itself. A simple example of a set which is dense-in-itself but not closed (and hence not a perfect set) is the subset of irrational numbers (considered as a subset of the real numbers). This set is dense-in-itself because every neighborhood of an irrational number ...

Leaky Nun

8:47 AM

@SirCumference caveats: 1. usually $X$ is the ambient space, so you need the subspace topology to talk about a set being dense in itself; 2. a more concise definition would be $X \subseteq \overline{E}$ where $\overline E$ is the closure of $E$

Sir Cumference

@LeakyNun Hmm, I assumed dense in itself implies $E$ is dense when the universe $X=E$

Leaky Nun

I think that is a very unfortunate terminology

Mary Star

We have that a and a+1 are elements of K and $\operatorname{Gal}(K/\Bbb Q)$ maps a to a+1, or not? @LeakyNun

Leaky Nun

if $X$ is the universe then $X$ is always dense

@MaryStar that makes no sense, a group cannot map something to something; I think you really need to think about this whole galois stuff a bit more

Sir Cumference

@LeakyNun Which, the Wikipedia page?

Leaky Nun

8:50 AM

@SirCumference well from my google result "dense-in-itself" seems to be an actual mathematical term

Mary Star

@LeakyNun Oh I meant the elements of $\operatorname{Gal}(K/\Bbb Q)$, i.e. $r\in \operatorname{Gal}(K/\Bbb Q)$ then $r(a)=a+1$. Or am I thinking wrong?

Sir Cumference

@LeakyNun Ok, but for clarification it doesn't literally mean "dense in itself" by the above definition right?

Leaky Nun

@MaryStar that's also wrong.

@SirCumference right...

Sir Cumference

sigh this terminology sometimes

Leaky Nun

this whole topology thing has a lot of terms

Mary Star

9:15 AM

@LeakyNun Why is it wtong that for $r\in G$ we have $r(a)=a+1$ ? (where $G=\operatorname{Gal}(K/\Bbb Q)$ ?

ÍgjøgnumMeg

9:54 AM

@Mary if the assumption is that $a$ and $a + 1$ are conjugates then a more precise statement would be that there exists an $r \in G$ such that $r(a) = a + 1$ (since $G$ acts transitively on the roots of $f$)

Mary Star

Ahh ok!!

I have also an other quesion.. It holds that $Gal(\mathbb{Q}(\zeta )/\mathbb{Q})$ is isomorphic to $(\mathbb{Z}/n\mathbb{Z})^{\times}$. Does this mean that $|Gal(\mathbb{Q}(\zeta )/\mathbb{Q})|=|(\mathbb{Z}/n\mathbb{Z})^{\times}|$ ? @ÍgjøgnumMeg

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$Mathematics$ Mathematics