Mathematics - 2018-09-20 (page 2 of 4)

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6:04 AM

@Pig yeah, it's why it's usually best not to talk about it i guess

infinite is ftw

breaks distances so bad

sooo happy

1

Q: Let $A=C[x] $ prove there is no norm on A in which a C* algebra

Let $A=C[x] $ prove there is no norm on A in which it makes a C* algebra. i think this is true because the spec(a) is infinity for any $a\in A$ ? but im not sure how to prove it. I did try verifying the axioms for a norm but im not sure what a* in this case

c-star-algebras

been trying to figure this out forever

6:25 AM

and crushed this one too

0

A: Cycle edges cut of a 3 regular graph

The first thing to notice is that in the sub-graph remaining that it is a chain of even cycles the next thing to notice is that $|Y'|\equiv0 \mod 2$ Notice in the picture no matter what side you are on of the split of Y' in order to make a cycle you must end on the same side as you started so the...

RAWR need to sleep

last proof is almost incomprehensible actually surprised its correct

Tobias Kildetoft

7:04 AM

So all of "open cover", "cover letter" and "open letter" are things. But is there such a thing as an open cover letter?

3

7:20 AM

@TobiasKildetoft are you here?

Tobias Kildetoft

yeah

could you help me check a proof I wrote?

Tobias Kildetoft

possibly

Theorem: splitting field is normal

Let $F$ be a field and $f$ in $F[X]$ and $K_1$ be the splitting field of $F$. We shall prove that $K_1/F$ is normal.

To that end, let $b_1$ in $K_1$, and its minimal polynomial be $g$, and we shall prove that $g$ splits in $K_1$.

Let $K_2$ be a splitting field of $g$, so $K_2 = F(b_1, b_2, \cdots, b_n)$ where $g(b_j) = 0$.

1. Let K2j be a copy of K2. We have maps F(b1) -> K1 and F(b1) -> K2j, where the second map sends b1 to bj. Take K1 tensor K2j base F(b1), quotient by a maximal ideal, call it Ωj.

2. We have maps K2j -> Ωj, i.e. K2 -> Ωj. Take the tensor product of the Ωj as j goes from 1 to n, base K2, and then quotient by a maximal ideal, and call it Ω.

Tobias Kildetoft

this is getting to be unreadable when you don't use LaTeX

7:23 AM

3. We now have maps φj : K1 -> Ω and ψj : K2 -> Ω satisfying φj(b1) = ψj(bj).

1. Let $K_{2j}$ be a copy of $K_2$. We have maps $F(b_1) \to K_1$ and $F(b_1) \to K_{2j}$, where the second map sends $b_1$ to $b_j$. Take $K_1 \otimes_{F(b_1)} K_{2j}$, quotient by a maximal ideal, call it $\Omega_j$.

2. We have maps $K_{2j} \to \Omega_j$, i.e. $K_2 \to \Omega_j$. Take $\Omega_1 \otimes_{K_2} \Omega_2 \otimes_{K_2} \Omega_3 \otimes_{K_2} \cdots \otimes_{K_2} \Omega_n$, quotient by a maximal ideal, call it $\Omega$.

Tobias Kildetoft

So your map $F$ is not a homomorphism?

(but just $F$-linear)

$F(b_1)$ means $F$ adjoint $b_1$

Tobias Kildetoft

woops, right. I meant the map from $F(b_1)$

I have two maps from $F(b_1)$

both are field homomorphisms

Tobias Kildetoft

Ahh, right, irreducible polynomial so transitive Galois group

7:28 AM

I didn't assume any separability

Tobias Kildetoft

then why can you send $b_1$ to $b_j$?

$F(b_1) = F[X]/(\min(b_1))$

$\min(b_1) = \min(b_j)$

to give a map $F[X]/(\min(b_1)) \to L$ is to pick an element $y \in L$ satisfying $\min(b_1)(y) = 0$

Tobias Kildetoft

ahh, right. So this actually does prove that the Galois group is transitive, right?

well you need to be able to extend the map to prove that the Galois group is transitive

(which you can, but which I didn't prove)

Tobias Kildetoft

Ahh, right, this was just for the smaller field, not the splitting field.

7:32 AM

right

3. We now have maps $\varphi_j : K_1 \to \Omega$ and $\psi_j : K_2 \to \Omega$ satisfying $\varphi_j(b_1) = \psi_j(b_j)$. Actually, we only have one $\psi : K_2 \to \Omega$, and $\varphi_j(b_1) = \psi(b_j)$.

@TobiasKildetoft are you following?

Tobias Kildetoft

Somewhat. But why are you taking tensor products and quotients rather than just composita?

eh...

I just like it more

Tobias Kildetoft

Ok

Hi guys!

so in $K_2[X]$ we have $g = \prod (X - b_j)$

7:43 AM

Anyone experience with latex beamer?

4. and hence in $\Omega[X]$ we have $g = \prod(X-\psi(b_j)) = \prod(X-\varphi_j(b_1))$

I'd like to strike out terms in a formula succesively using pause

Tobias Kildetoft

@Rudi_Birnbaum It has been quite a while, but I did use it back when I presented my Master's thesis

any ideas on that?

5. Claim: for any extension $L/K$ and any two maps $\alpha_1, \alpha_2 : K_1 \to L$, $\alpha_1(K_1) = \alpha_2(K_1)$

Tobias Kildetoft

7:46 AM

@Rudi_Birnbaum I think you need to use the slightly more advanced feature for that, where you "timestamp" various parts of the code

That allows you to add parts that are only included at certain times (i.e. at certain iterations of the given slide)

@TobiasKildetoft should I prove 5?

Tobias Kildetoft

hmm

@TobiasKildetoft That rings a bell, thank you Tobias! (I think I did that once some aeons ago...)

Tobias Kildetoft

I tried to find my presentation from back then, where I recall doing this, but I can't find it

WLOG we show that $\alpha_1(K_1) \subseteq \alpha_2(K_1)$.

Note that $f = \prod (X-\lambda_i) \in K_1[X]$

So $f = \prod(X-\alpha_2(\lambda_i)) \in L[X]$

but $\alpha_1$ takes each $\lambda_i$ to a root of $f$, i.e. some $\alpha_2(\lambda_j)$

7:51 AM

@TobiasKildetoft do you remember any syntax detail of that is \pause[$n$] and some \timestamp{$n$} or something?

so the generators of $\alpha_1(K_1)$ is inside $\alpha_2(K_1)$

so $\alpha_1(K_1) \subseteq \alpha_2(K_1)$

@TobiasKildetoft Yes, if we let E be an open cover, since E is also a letter of the alphabet, E is an open cover letter.

@TobiasKildetoft ok?

Tobias Kildetoft

@LeakyNun Yeah, looks good

so now recall that we have maps $\varphi_j : K_1 \to \Omega$

oh snap, their images are all the same

Tobias Kildetoft

7:58 AM

@Rudi_Birnbaum Sorry, I don't recall any of the syntax. I do recall the user guide being quite readable though

and recall that $g = \prod (X - \varphi_j(b_1)) \in \Omega[X]$

the roots of $g$ in $\Omega$ are exactly $\varphi_j(b_1)$

@TobiasKildetoft Yes, the manual is several hundred pages long.

@TobiasKildetoft OK, thank you!!

all of which are contained in the common image of the maps

so $g$ splits in $K_1$, qed

@JasperLoy I guess I won't have to read it all ;-)

8:00 AM

beamer, pgf, and pgfplots all have very detailed manuals. For presentations, diagrams, and plots respectively.

powerdot, pstricks, and pst-plot will be the pstricks equivalent family to the pgf family.

Tobias Kildetoft

@Rudi_Birnbaum Looks like it is under "overlay specifications"

If $M$ is a smooth manifold, is the map $H_{dR,c}^n(M) \to H_{dR}^n(M)$ from the compactly-supported de rham cohomology to the usual de rham cohomology injective?

Tobias Kildetoft

and the syntax looks very simple

@TobiasKildetoft in fact I just found a compositum-proof that the Galois group is transitive

from the same text that I'm reading

8:22 AM

@LeakyNun Are you reading Lang's Algebra?

no

just a course note of the galois theory course in my college

Hi nice folks :-)

I have to proove that a subgroup H and its left costs partion group G

and the proof is this:

Looks like I was late to the Atiyah fiasco

suppose that the cosets Hg_1 and Hg_2 have one element in commen, which we can write either as h_1g_1 or as h_2g_2, for some h_1 and h_2 in H. Then:
Hg_1 = H h_1^-1h_2g_2 = Hg_2, since h_1^-1h_2 is element of H.

but what if h_1^-1h_2 is element of H? is it just that the expression H h_1^-1h_2g_2 can we rewritten as H h_whicheverelement g_2?

Tobias Kildetoft

@PalSzabo Those are right cosets, not left cosets (thought proof doesn't change in any meaningful way)

8:36 AM

@TobiasKildetoft yep, sorry, typo

the hing is that H h_1^-1h_2 will still be H, yes?

every element is just shifted

Tobias Kildetoft

yes

alright

thanks

I'm trying to show that a set consisting of sequence is closed under a norm. I'm unsure as of how to start.

Alessandro Codenotti

9:03 AM

Tell us the exact problem! @Oskar

Tobias Kildetoft

9:25 AM

Speaking of Atiyah. Did anyone see the Abel lecture he gave at ICM? Almost embarassing to watch.

9:35 AM

We know that $\Bbb{R^2}$ and $\Bbb{R^3}$ are not homeomorphic. Sometimes arguments are given that you could remove a line and one will become disconnected whereas the other would still be connected. We say here that $\Bbb{R}$ could be embedded in some strange way in R^3 so that argument doesn't really hold. Is there an example of such a strange embedding of the real line in three space ?

10:25 AM

@AlessandroCodenotti: The problem goes as follows: "Let $E=\ell^1$. Consider $$X=\{x=(x_n)_{n\geq1}\in E:x_{2n}=0\ \forall n\geq1\}$$ and $$Y=\{y=(y_n)_{n\geq1}\in E:y_{2n}=\frac{1}{2^n}y_{2n-1}\ \forall n\geq1\}.$$ Check that $X$ and $Y$ are closed linear spaces."

I guess that I must the fact that a metric space is closed if and only if it contains all of its sequence points.

Alessandro Codenotti

Start with $X$, looks easier

I don't really see why a sequence should converge just because $x_{2n}=0$ for all $n\geq1$.

Tobias Kildetoft

@OskarTegby Why would you want to show convergence of the sequences?

Because a metric space is closed if and only if it contains all of its sequence points.

what is a sequence point of a metric space ?

(also metric spaces are always closed)

Tobias Kildetoft

10:31 AM

those are not rhe sequences that are relevant to that definition

Okay.

you have to show that if a sequence of elements of $X$ converges (for the l1 norm) to some element of $E$ then that element is in $X$

Hm... Okay.

If $\tilde{x}$ is a sequence of elements in $X$ which converges to some element $\hat{x}$ in $E$ in the $\ell^1$ norm, then we need to show that $\hat{x}$ in fact is in $X$. I wonder where to start.

Of course we can use the definition of convergence.

10:54 AM

The definition of convergence with the $\|\cdot\|_1$ norm is that we have that for all $\epsilon>0$ we have an $N\in\Bbb{N}$ such that $\|x_n-\hat{x}\|_1<\epsilon$ for all $n\geq N$. Here, I renamed the sequence from $\tilde{x}$ to $x_n$.

I don't really see how this helps us to show that $\hat{x}$ is in fact in $X$.

Tobias Kildetoft

@OskarTegby Isn't the projection to the sequence consisting of the even terms a continuous map?

ForeverInactive

@JasperLoy Great songs. Keep singing like that.

Projection from the convergent sequences to that? I'm sure it is, but I don't really see it.

11:38 AM

$\left|\begin{matrix}e^{-i2A}&e^{iC}&e^{iB}\\e^{iC}&e^{-i2B}&e^{iA}\\e^{iB}&e^{iA}‌&e^{-i2C}\end{matrix}\right|$

Find the value of determinant for triangle ABC.

I have expanded used euler's formula and got the right answer = -4

Is there any other trick method?

Hello, someone have the idea of how we prove that $\lim_{x\to 0}\frac{\sin(x)}{x}=1$ using the definition of limit $\forall \varepsilon>0, \exists \delta >0, |x|<\delta\Rightarrow |\frac{\sin(x)}{x}|<\varepsilon$

Tobias Kildetoft

@Vrouvrou And using which definition of sin?

just don't use the derivative

Tobias Kildetoft

no, I mean you need to have some definition of sin to work with to get started

for some of them this is trivial, and for some you need to do some work

:46816167 why have you created so many sock puppet accounts

3

11:53 AM

@TobiasKildetoft I don't know

Tobias Kildetoft

@Vrouvrou then you have no way to proceed. You might as well have been asked to show that this held for some secret function $f$ that nobody will tell you what is.

for example which is the definition of sin that it makes this limit trivial ?

Tobias Kildetoft

At least as far as I recall it is quite easy to see from the series definition

ok, and is this property can give something $|\sin(x)|\leq1$

@user2646 ?? Look I dont think its allowed, thats why I am asking.

12:07 PM

calm down, pal

@user2646 Okay?

my reasons are personal

3

@TobiasKildetoft: Care to elaborate? :)

(Referring to your comment to my question.)

1:01 PM

@Abcd Not sure what you mean by "triangle ABC" here

Do you just intend that as "complex numbers A,B,C"?

Actually, if I expand out that determinant I get $-3+2e^{i(A+B+C)}+e^{-2i(A+B+C)}$

so evidently it's more constrained than just "three complex numbers A,B,C"

(are they the angles of a triangle?)

1:19 PM

@Semiclassical of course

@Semiclassical I mean yes they are the angles of a triangle. And if you susbtitute A+B+C = pi in your expansion you'll get -4.

I am asking for some trick method through operations etc.

Presumably there’s some clever way revolving around exp(i pi A)*exp(i pi B)*exp(i pi C)=-1

(Not that I see it immediately either. But there’s not a lot else to work with)

What I could see is either the eigenvalues being simple (with their product being the determinant) or that there’s a nice factorization of the matrix

okay, here's a good approach

First, to make things simpler to write out, let $\alpha =e^{i A},\beta=e^{i B}, \gamma=e^{i C}$

The fact that A,B,C are angles of a triangle then translates to $\alpha\beta\gamma= -1$

Therefore $\alpha^{-1}=-\beta \gamma$, and similarly for the other two reciprocals

bah, I changed it and make it wrong

$(\alpha,\beta,\gamma)=(e^{-iA},e^{-iB},e^{-iC})$

with that in mind, we can rewrite the determinant as $$\begin{vmatrix}\alpha^2 & \gamma^{-1} & \beta^{-1} \\ \gamma^{-1} & \beta^2 & \alpha^{-1} \\ \beta^{-1} & \alpha^{-1} & \gamma^2\end{vmatrix}= \begin{vmatrix}\alpha^2 & -\alpha \beta & -\alpha \gamma\\ -\alpha \beta& \beta^2 & -\beta \gamma \\ -\alpha\gamma & -\beta \gamma & \gamma^2\end{vmatrix}$$

Do you notice anything about the rows and columns of that latter matrix?

1:54 PM

Is there a common ratio for $\frac{1}{k(k+1)} $? Trying to find the geometric series so I can calculate the pmf

Take k=1,2,3 and check if they change by the same factor each time

(the more useful thing to do is to realize that 1/k(k+1) can be expanded in partial fractions)

If you're trying to do $\sum_{k=1}^\infty \frac{x^k}{k(k+1)}$, though, then the better way is to consider how the series behaves under differentiation

D; oy never thought of it in the partial fraction direction. Would that give me the common ratio? I need that to calculate for c for the probability mass function

Why is it that every rational is constructible number? anyone see the intuition behind this?

@usukidoll you're presupposing it's going to be a geometric series. don't

a geometric series involves coefficients of the form a^k

but there's no exponentials in sight here

So you shouldn't expect to be able to fit it into that box

@Semiclassical yo semi :D

2:08 PM

hi

long time no see :D how are you ?

still doing physics?

ya

@Semiclassical Are you also interested in integrals like Waiting?

Its getting late like after 4 am D;. Partial fraction method may be the way to go

Yay its jaspy

@JasperLoy not terribly

@usukidoll to be clear, what summation are you actually trying to perform?

2:10 PM

I could take a pic of the problem. I have to solve for c to satisfy the pmf

I mean, if you're doing a moment generating function, I'd expect $\displaystyle \sum_{k=1}^\infty \frac{x^k}{k(k+1)}$

I was reading the conversation on Atiyah above. What happened to him?

though one would need an overall constant in order to have $E[1]=1$, which is perhaps what you mean

Why is every rational constructible?

ibb.co/inWxZz 2.12 I know I have to solve for c

2:14 PM

@SharathZotis First convince yourself that reciprocals of natural numbers are constructible.

Ok I've convinced my self of that?

does that give u all rationals, since we have all 4 operations?

Once you do that, it's easy since p/q = p*1/q

add,multiply, divide

I seee

Another approach, I suppose: Draw a line from (0,0) to the point (q,p)

that'll be the line y=(p/q)x

if you now intersect it with the line x=1, you get (x,y) = (1,p/q)

i see that's very clever

2:18 PM

@usukidoll Okay. So what's the condition for $c$, expressed in terms of a summation?

2:30 PM

I'm tired. I need to get back to this problem later. Its almost 5 am and I've been up for 22 hours.

Or less than 20 idk but it's at least 12

kk

3:11 PM

@Semiclassical its the determinant of symmetric matrix

hi @MatheinBoulomenos!

MatheinBoulomenos

hi @LeakyNun

@MatheinBoulomenos do you want to hear my proof that splitting field is normal lol

MatheinBoulomenos

okay

@Abcd that's true. But do you notice anything if you look, for instance, at the first column?

3:14 PM

so i'll actually prove that the three things are equivalent for a finite extension $K/F$:

@Semiclassical alpha beta gamma can be taken out as common factors

for each column in turn, yes

and you can moreover do the same for each row in turn

1. For any $L/F$ and $\sigma_1, \sigma_2 : \operatorname{Hom}_F(K,L)$, $\sigma_1(K) = \sigma_2(K)$.
2. Any irreducible $F$-polynomial that has a root in $K$ splits in $K$.
3. $K/F$ is the splitting field of some polynomial.

How can you use that for this determinant?

@Semiclassical wow yes it would make it very simple

3:16 PM

@MatheinBoulomenos 2 -> 3 is trivial, and we will prove 3 -> 1

well you already proved 3 -> 1

so I'll just prove 1 -> 2

It still leaves you a little work, mind: it gets you down to the determinant \begin{vmatrix} 1 & -1 & -1 \\ -1 & 1 &-1 \\ -1 & -1 & 1\end{vmatrix}

@Semiclassical not a big deal we will just be left with a determinant of 1s which is easy to evaluate.

Right.

@MatheinBoulomenos Let $g$ be an irreducible polynomial that has $b_1 \in K$ as a root.

Interestingly, it's not too bad to generalize that to an n-by-n determinant consisting of 1's on the diagonal and -1 everywhere else

That's a good exercise as well.

3:17 PM

Let $L/K$ be the splitting field of $g$.

Let $b_j$ be an arbitrary root of $g$ in $L$.

We have a map $\varphi : F(b_1) \to L$ sending $b_1$ to $b_1$

(Best way I know to do that is to work out the structure of eigenvalues/eigenvectors for the n-by-n matrix, and use the fact that the determinant is the product of the eigenvalues)

as well as a map $\psi: F(b_1) \to L$ sending $b_1$ to $b_j$

@MatheinBoulomenos ok it's beginning to sound like your proof, actually

Take $L \otimes_{F(b_1)} L$ and quotient by a maximal ideal to get $\Omega$ with $\varphi', \psi' : L \to \Omega$ satisfying $\varphi'(b_1) = \psi'(b_j)$

then we get $p,q : K \to \Omega$ satisfying $p(b_1) = \psi'(b_j)$, $\varphi'|_K = p$, $\psi'|_K = q$

So $\psi'(b_j) \in p(K)$

Anyone knows how to calculate Mod[2 ^ 100!, 101] in wolframalpha? I can get both Mod[2 ^ 100, 101] and Mod[100!, 101] to work, but WA is unable to recognize the first expression :/ Or is there any other website where the first expression can work?

(on a side note, chat.se should prompt a confirmation dialog in case someone accidentally clicks "delete" :|)

By assumption of $K$ we have $p(K) = q(K)$, so $\psi'(b_j) \in q(K)$

Tobias Kildetoft

@GaurangTandon Have you tried adding some parentheses to make sure it is parsed correctly?

3:25 PM

@TobiasKildetoft I did, didn't work

Tobias Kildetoft

what did it say?

@GaurangTandon you can use up-arrow key to edit your message instead

@LeakyNun cool, thanks for the tip!

Try the following:
Use different phrasing or notations
Enter whole words instead of abbreviations
Avoid mixing mathematical and other notations
Check your spelling
Give your input in English
Other tips for using Wolfram|Alpha:
Wolfram|Alpha answers specific questions rather than explaining general topics
Enter "2 cups of sugar", not "nutrition information"
You can only get answers about objective facts
Try "highest mountain", not "most beautiful painting"
Only what is known is known to Wolfram|Alpha

just generic stuff

I also tried Mod[(2^factorial(100)), 101], doesn't work :(

Tobias Kildetoft

so WA being its usual useful self

@MatheinBoulomenos well by definition of $q$ we have $q(K) = \psi'(K)$, so $\psi'(b_j) \in \psi'(K)$, so $b_j \in K$ as required

done

3:28 PM

Mod[2^Factorial[100], 101] doesn't work either ...

@TobiasKildetoft So what happened to Atiyah? I did not follow the news.

@JasperLoy well he's going to present a proof of Riemann Hypothesis next week

Tobias Kildetoft

@JasperLoy I don't think anything specific happened. He has just gotten old, and his mind is not what it used to be

MatheinBoulomenos

@LeakyNun that's a nice proof. not really simpler than without tensor products, but I like it

Tobias Kildetoft

His ICM Abel lecture was mostly a confused old man going out of tangents and showing pictures of famous mathematicians

all the while referencing slides that were not currently being displayed

3:29 PM

@MatheinBoulomenos I have another proof that can deal with all roots at once

Oh OK, my mind is also not what it used to be...

@MatheinBoulomenos if you're interested

Tobias Kildetoft

Frankly, I think it was a silly mistake to let him give this lecture at ICM, given his record the previous years of increasingly outlandish and unsubstantiated claims. The committee should have been aware that no good talk would come of it (and their description of it afterwards is exceptionally disingenuous in its praise of his "wit").

MatheinBoulomenos

@LeakyNun sure

Well, all I know is that the Atiyah-Singer index theorem is a generalisation of the Riemann-Roch theorem and the Gauss-Bonnet theorem at the same time. =)

3:34 PM

Hi yall

hi

Leaky :D

Tobias Kildetoft

@JasperLoy Yeah, Atiyah has been one of the absolute greatest mathematical minds of the 20th century

@LeakyNun That's what I hope to do next decade. =)

@TobiasKildetoft That is so true Tobias

they should not have let him attend that talk

MatheinBoulomenos

3:36 PM

Hi @Kasmir

Nobody starred my message about open cover letter, sad...

@MatheinBoulomenos Mathein :D I did not want to say hi , not to bother you :D

@JasperLoy May I ask where you from Jasper?

@KasmirKhaan Yes, you may ask me anything, but my location is something I cannot share in this chat.

fair enough :D

I meant more like what country

not what appartment number :D

Tobias Kildetoft

@JasperLoy Is that because we know precisely how fast you are travelling?

MatheinBoulomenos

3:38 PM

@KasmirKhaan why should a greeting bother me?

@KasmirKhaan Yes, that is what I mean actually. I cannot share that.

@MatheinBoulomenos not in that case , i saw youi and leaky were discussing something =p

did not want to interrupt is all

Is any set a subset of the universe?

@JasperLoy fair enough Jasper ! :D

@Li357 yes

3:39 PM

So it's by definition and I don't have to prove it.

that would lead very quickly to the notion of zorn's lemma

Can we do math without Zorn's lemma?

@Li357 it isn't by definition

we can't @KasmirKhaan

hmm ._.

Most people use the axiom of choice without worrying about it in everyday mathematics.

In the typical presentation of ZFC, there are ten axioms, but some of them are redundant, meaning they can be deduced from some of the others.

It has been since galleleo and newton the same dilemma

of infinity

if you notice, most powerful theorems has relation to understand infinity

3:42 PM

I really like the book Introduction to Set Theory by Karel Hrbacek and Thomas Jech. I think it is the best elementary axiomatic set theory book in the world now.

After reading that book, you can go on to Set Theory by Thomas Jech, which is a monster of a book.

@JasperLoy oh nice ._. i really want to learn about sets

is that the simplest book you can recommend? the former?

@KasmirKhaan Not the simplest, but mathematics is not meant to be simple. But it does start right at the beginning, so you can read it.

Yes that is what i meant by simplest =p

for someone who has no previous knowledge about the topic :D

Okay thanks Jasper =p

Yes, you can read it. You just need to know some logical arguments like if, if and only if, and things like that. No formal logic is required.

good! =p

3:49 PM

@MatheinBoulomenos As before, let $g$ be an irreducible $F$-polynomial with a root $b_1$ in $K$, and let $L/K$ be the splitting field of $g$.
1. For each $b_j$, let $p_j : F(b_1) \to K : b_1 \mapsto b_1$ and $q_j : F(b_1) \to L : b_1 \mapsto b_j$, and let $\Omega_j = (K \otimes_{F(b_1)} L) / \mathfrak m$. This gives us $\varphi_j : K \to \Omega_j$ and $\psi_j : L \to \Omega_j$ such that $\varphi_j(b_1) = \psi_j(b_j)$.
2. Using the maps $\psi_j$, consider $\Omega = \Omega_1 \otimes_L \Omega_2 \otimes_L \cdots \otimes_L \Omega_n / \mathfrak m$. This gives us maps $\alpha_j : K \to \Omega_j \t…

hi folks :-) Are there any real n x n non-orthogonal matrices, with determinant 1?

plenty of them

are there names for those?

SL_n \ O_n

sorry, this is quite dumb, what is the \ stand for?

@LeakyNun thanks

3:59 PM

take away

oh ok

4:12 PM

> you do not have a choice to disobey us ----anonymous

4:24 PM

Wait

is Atiyah claiming a proof of RH?

yes

hmm

believe at your own discretion

Perhaps it's his seniority taking its toll finally

which is unfortunate

or maybe it's true!

lol

Tobias Kildetoft

@ÍgjøgnumMeg He has had several similar claims in the past years (though slightly less amazing), one about complex structures on $S^6$ and one about a simplified proof of Feit-Thompson

And his mind is definitely not what it used to be

4:28 PM

:(

Sad

Best not to talk about it too much

Riemannian hypothesis is obviously true, and the prime number conjecture is false, and who can said about Riesz criterion

what is the prime number conjecture?

Riemann hypothesis

In mathematics, the Riemann hypothesis is a conjecture that the Riemann zeta function has its zeros only at the negative even integers and complex numbers with real part 1/2. It was proposed by Bernhard Riemann (1859), after whom it is named. The name is also used for some closely related analogues, such as the Riemann hypothesis for curves over finite fields. The Riemann hypothesis implies results about the distribution of prime numbers. Along with suitable generalizations, some mathematicians consider it the most important unresolved problem in pure mathematics (Bombieri 2000). The Rieman...

O great I misquote

@Albas So, a homeomorphism taking R^2 to R^3 would take the x-axis to a properly embedded copy of R. We could one-point compactify to get a crazy circle in S^3.

4:36 PM

there isn't really a name on "distribution of the prime numbers"

prime number theorem?

Now we have algebraic topology at our disposal. There is a theorem called the Alexander duality theorem that relates the homology of a compact locally contractible subset of S^n to its complement

In particular for degree reasons you find that the reduced homology of the complement is 0 - the complement is connected

hi @loch

Hi @LeakyNun

@loch would you be interested in a proof that splitting field is normal?

4:43 PM

No

ok

Maybe when i have more time some other day:p

4:56 PM

Can someone explain this definition? "An orientation of a digraph is clique acyclic if every clique is transitive." What does it mean for a clique to be transitive? Transitive with respect to what?

existentialcomics.com/comic/153

me whenever someone asks me a question

@mercio but I am actually asking a semantic question.

I actually don't know what transitive means in the context of a clique of a graph

ah I see that digraphs are directed graphs

right

but i don't know what is an orientation of a digraph

I would think that a digraph is already oriented ?

5:08 PM

Ah true

and I know what a clique is in a context of graphs, and it could be different for digraphs

replace "digraph" with "graph

"

for cliques ?

hm?

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