Mathematics - 2019-07-10

J. Doe

1:34 AM

Who took my long chain of heys, hellos, his and howdys!

Ajay Mishra

1:44 AM

@Tanvir my method turned into dead end, and at last that was way harder.

Tanvir

2:02 AM

@AjayMishra it is not as simple problem as you think. quite many people gave it a try but no answer so far. So of them have much better profile. As I need the solution, I am posting it as much as I can.

CroCo

2:13 AM

Given V(x,y) = (x^2 + y^2 -1)^2, is V(x,y) positive definite?

It seems not because V(0) > 0, am I right?

Ajay Mishra

2:24 AM

@Tanvir I just told you that was way harder than I thought. I ain't by any mean telling you that that problem is easy.

user76284

4:05 AM

One can find a natural continuous extension of the harmonic numbers:
\begin{align}
H_n
&= \sum_{k=1}^n \frac{1}{k} \\
&= \sum_{k=0}^{n-1} \frac{1}{k+1} \\
&= \sum_{k=0}^{n-1} \int_0^1 x^k \,\mathrm{d}x \\
&= \int_0^1 \sum_{k=0}^{n-1} x^k \,\mathrm{d}x \\
&= \int_0^1 \frac{1-x^n}{1-x} \,\mathrm{d}{x}
\end{align}
One can also find a natural continuous extension of the factorials, namely the gamma function. Is there a natural continuous extension of number-theoretic functions like the Möbius and prime-counting functions?

Secret

5:24 AM

Well the problem is the moebius function is too random to fit some reasonable curve to it

Square free integer terms in the expression of the prime counting function in terms of the meobius function do not necessary cancel out, thus you might need to have a continuous extension of the moebius function before you can have one for the prime counting function

Alessandro Codenotti

7:00 AM

@RyanUnger You might have a point

PrincePolka

7:45 AM

I'm trying to fit a curve to a regular pattern

ÍgjøgnumMeg

Mornin' all

Rithaniel

8:00 AM

Morning

Perturbative

8:21 AM

Does being a tutor help on a CV?

1010011010

8:50 AM

Howdy

I have a question about notation

Suppose I wish to constrain an index $k$ by the condition that the elements of two sets coincide

I thought of something but I think it's wrong:

$\{k^0\}:=\{k|\{\xi_k\}\cap\{\underline{\xi_k}\}\}$

And suppose I would want to have only the element that I excluded, then I thought

$k^+:=\{k|\{\xi_k\}\setminus\{\xi_{k^0}\}\}$

But again I'm not sure about this

I've never done set notation and now I need it for some idea in physics so my question is likely to be a bit dumb :p

@Perturbative It's helped me

MatheinBoulomenos

9:46 AM

@BalarkaSen Wherefore hath I been summoned?

ÍgjøgnumMeg

ha wherefore

MatheinBoulomenos

Hi @ÍgjøgnumMeg

ÍgjøgnumMeg

varför in Swedish, hvorfor in Norwegian, but hví in Faroese

which is where why comes from!

Hey @Mathein

MatheinBoulomenos

Ah I see, Alessandro answered the question that Balarka summoned me for

Alessandro Codenotti

10:02 AM

Mwhahaha

I don't know if that's the question Balarka wanted to ask you though

ÍgjøgnumMeg

10:15 AM

that's a long distance snipe

intercontinental ballistic snipe

3

user193319

12:21 PM

Does it make sense to speak of a splitting field of a polynomial if the polynomial has repeated roots?

ÍgjøgnumMeg

Sure, for a splitting field $E/F$ of a polynomial $p(X) \in F[X]$ you have $[E:F] \leq (\deg p)! $

and the inequality happens because you might have repeated roots

user193319

Ah, so the linear factors into which the polynomial splits don't have to be unique?

user193319

12:38 PM

Also, what does it mean to say "Let $\omega$ be the primitive sixth root of unity over $\Bbb{Q}$?"? More generally, given a field $F$, what does it mean to say that $\alpha$ is a $n$-th root of unit over $F$?

Does it mean there exists a (smallest?) field extension $E$ of $F$ such that $E$ has an element whose $n$-th power is $1$?

Alessandro Codenotti

$\omega^n=1$ means that $\omega$ is an n-th root of unity, $\omega^m\neq 1$ for $m<n$ means that it is primitive

user193319

Ah, okay. And what I said about "roots of unity over a field" is correct?

user193319

1:16 PM

Are simple extensions always galois?

Perhaps simple extensions via the $n$-primitive root.

Alessandro Codenotti

Yes, cyclotomic extensions of $\Bbb Q$ are Galois

user193319

I think $x^n - 1$ would be a separable polynomial which splits in the simple extension, right?

Alessandro Codenotti

Because they are the splitting field of the $n$-th cyclotomic polynomial

Curio

How should I consider the domain of the cotangent? != pi* k/2 or != pi*k?

Alessandro Codenotti

@user193319 this is false in general. What's the standard example of a non Galois extension?

user193319

1:20 PM

Ah, okay.

$\Bbb{Q}(\pi)$?

Alessandro Codenotti

An algebraic non Galois extension

(also note that every finite degree extension of $\Bbb Q$ is simple by the primitive element theorem)

user193319

I'm not sure. I was going to say $\Bbb{Q}[\sqrt{2}]$, but I think that's finite.

...a finite extension.

Alessandro Codenotti

That's a Galois extension

user193319

The example you have in mind uses rationals, right?

Of the form $\Bbb{Q}(\cdot)$?

Alessandro Codenotti

Yes

Every such extension is separable for free, so normality has to fail

user193319

1:26 PM

Hmm...I'm not sure. I'll have to think about that.

user193319

1:55 PM

If $\omega$ is the primitive 6th root of unity over $\Bbb{Q}$, how does one argue that $1,\omega,....,\omega^5$ are linearly independent in $\Bbb{Q}[\omega]$?

user131753

@MatheinBoulomenos: I must say that your blog posts in Category Theory are truly awesome.

user131753

2:09 PM

However I was wondering, is there any characterization of adjoint functors based on "categorical homotopies" as defined here?

Leaky Nun

2:29 PM

@user193319 by noticing that $\omega^3 + 1 = 0$

well actually, $\omega^2 - \omega + 1 = 0$

user193319

2:52 PM

@LeakyNun Doesn't that show they are linearly dependent?

ÍgjøgnumMeg

In the general scenario, if you have a linear dependence of the elements $1, \alpha, \dots, \alpha^{n-1}$, say $a_{n-1}\alpha^{n-1} + \cdots + a_1\alpha + a_0 = 0$ with the $a_i$ in your base field, then this induces the congruence $a_{n-1}X^{n-1} + \cdots + a_1X + a_0 \equiv 0 \bmod p(X)$ (this is assuming $p(X)$ is the minimal polynomial of $\alpha$ with degree $n$). But this means that $p(X)$ divides this polynomial, and so you have a degree $n$ polynomial dividing a degree $n-1$ polynomial

user193319

@ÍgjøgnumMeg So, is $x^6 - 1$ the minimal polynomial of $\omega$?

ÍgjøgnumMeg

No, because that guy is reducible

Alessandro Codenotti

Cyclotomic polynomials is the keyword to look up here

ÍgjøgnumMeg

^

Thomas Shelby

3:26 PM

Hey guys, can someone give me an example of an irreducible representation of a group on a complex Hilbert space(not one-dimensional)? Otherwise, any references to find one example?

user193319

@LeakyNun How were you able to deduce $\omega^2 - \omega + 1 = 0$?

Leaky Nun

that's the 6th cyclotomic polynomial

ÍgjøgnumMeg

@user193319 it's an irreducible factor of $X^6 - 1$

user193319

@ÍgjøgnumMeg $x^2 + x + 1$ is too, right?

MatheinBoulomenos

@user170039 thank you :)

user193319

3:31 PM

So, why doesn't $\omega^2 + \omega + 1=0$?

MatheinBoulomenos

@user170039 not directly, I think. I'm planning on writing about adjunctions soon, including multiple perspectives. Natural transformations (which can be though of as categorical homotopies), play an important role, though, so there definitely a relation

user193319

@ÍgjøgnumMeg For my purposes, I guess it doesn't matter which one holds. In either case, we can conclude that $\Bbb{Q}[\omega]$ is spanned by $1$ and $\omega$, and clearly they are linearly independent.

MatheinBoulomenos

@user193319 if $\omega$ is primitive 6th root of unity, then $\omega^2+\omega+1\neq 0$

it's true for some 6th roots of unity (namely, the ones which are primitive 3th roots of unity), but not for the primitive ones

Consider the factorization: $(X^6-1)=(X-1)(X+1)(X^2+X+1)(X^2-X+1)$. The roots of $X-1$ are primitive 1st roots of unity, the roots of $X+1$ are primitive 2nd roots of unity, the roots of $X^2+X+1$ are primitive 3rd roots of unity and the roots of $X^2-X+1$ are primitive 6th roots of unity

just saying that $\alpha$ is a root of $X^6-1$ doesn't determine the minimal polynomial of $\alpha$, because $X^6-1$ is not irreducible, but saying that it's a primitive 6th of unity does

Alessandro Codenotti

3:59 PM

@Mathei is there a notion of "dual of an object" in a generic category?

MatheinBoulomenos

4:13 PM

@Alessandro well, going from the category to the dual category does nothing on objects, so I'd say no

in a monoidal category, it makes sense, however

Rigid category

In category theory, a branch of mathematics, a rigid category is a monoidal category where every object is rigid, that is, has a dual X* (the internal Hom [X, 1]) and a morphism 1 → X ⊗ X* satisfying natural conditions. The category is called right rigid or left rigid according to whether it has right duals or left duals. They were first defined (following Alexandre Grothendieck) by Neantro Saavedra-Rivano in his thesis on Tannakian categories. == Definition == There are at least two equivalent definitions of a rigidity. An object X of a monoidal category is called left rigid if there is an object...

Alessandro Codenotti

I see

user131753

5:09 PM

@MatheinBoulomenos I will be taking a look at your blog from time to time but just in case I miss, would you mind to let me know about your post on adjunctions as soon as you post it in your blog?

Poline Sandra

5:45 PM

hello if d and d' are not equivalent metrics and (E,d) is complete can i deduce that ( E,d') is not complete ???

@LeakyNun hello

Leaky Nun

I think you can let $E=\Bbb R$ and $d'(x,y) = d(x,y)^2$

@PolineSandra

Thorgott

Or $\mathbb{R}$ once with the Euclidean metric and once with the discrete metric

Leaky Nun

oh nice

Poline Sandra

but what I know is this : d and d; are equivalent then (E,d) complete if and only if (E,d') complete

I can't deduce from this

is there an equivalence ?

Thorgott

@LeakyNun the square of a metric usually isn't a metric (certainly not in the Euclidean case)

Leaky Nun

5:56 PM

oh right

but the square root is right

Poline Sandra

someone help me

$(R,|.|)$ is complete and $d'(x,y)=|\exp(x)-\exp(y)|$, |.| and d' are not equivalent

can I deduce that $(R,d')$ is not complete

Thorgott

yes, the square root works. I think that way we can actually construct infinitely many pairwise non-equivalent metrics that turn $\mathbb{R}$ into a complete metric space.

Poline Sandra

in fact $(R,d')$ is not complete but can I deduce it from the fact that |.| and d' are not equivalent

Thorgott

@PolineSandra not directly, as the counter-example above shows

Poline Sandra

ok thank you

Alessandro Codenotti

6:29 PM

Hi @Ted

Ted Shifrin

hi, demonic @Alessandro

Alessandro Codenotti

7:02 PM

Are you here @Balarka?

Ted Shifrin

I don't see a @Balarka

Alessandro Codenotti

Me neither

Ted Shifrin

There is a silent @ÉricoMeloSilva.

Daminark

7:14 PM

Hello

Alessandro Codenotti

Hi @Dami

Érico Melo Silva

7:36 PM

@TedShifrin rats, found out hiding in plain sight

Ted Shifrin

hi, nerdly Demonark.

Akiva Weinberger

7:51 PM

So

Tutoring a rising sophomore in math starting Monday

Prepping for the SAT

/PSAT

(basically the same test)

for about a month

A bit worried because this is my first time doing this

I've done tutoring in math before but not specifically for the SAT

and also this is the first time I'm getting paid for it

So like I need to figure out how to fill 90 minutes twice a week for a month

("Rising sophomore" means she'll be a sophomore once the school year starts, for clarity)

anakhro

8:57 PM

My beloved Ted.

Ted Shifrin

9:08 PM

@AkivaWeinberger I recommend you get (or have her do so) one of the standard review books with lots of questions in it. Have her do a sample test. Then figure out what you need to work on with her. Remember that she probably won't be as interested in the depth of mathematics as you are :P

Ryan Unger

Practicing for the SAT before sophomore year? Wow

anakhro

@TedShifrin @RyanUnger what is your favourite derivative?

Ryan Unger

Ted: exterior

anakhro

e.g. Directional, covariant, Lie, etc.

Heh. I would have actually pinned Ted as a covariant kind of guy.

Ryan Unger

Ted is the differential forms guy

Have you not yet been accosted by him for not using forms enough?

anakhro

9:15 PM

Yeah, but I like to think he has a mysterious side to him.

Ryan Unger

Well you can't go wrong with the covariant derivative

Very versatile thing

anakhro

What is your favourite, Ryan?

Ryan Unger

I just said covariant

anakhro

YOU NEVER SAID IT.

Ryan Unger

It includes the partial derivatives

anakhro

9:18 PM

What other cool derivatives are there?

Ryan Unger

Darbeaux derivative

anakhro

Is Darbeaux the better looking one in the Darboux family?

Ryan Unger

Maybe I misspelled

anakhro

It's cute, don't worry.

Do you think derivatives like the Frechet derivative should be included in a question like this?

Ryan Unger

the Darbeaux derivative is $\omega_f:=f^*\omega_G$, where $f:M\to G$, $G$ is a Lie group and $\omega_G$ is the Maurer-cartan form

I guess? Whom is the question intended for

anakhro

9:22 PM

Yeah that is "Darboux".

Alessandro Codenotti

Cantor Bendixson derivative of course

Ryan Unger

lmao I meant to spell it correclty this time

it's Darbeaux from now on

anakhro

@AlessandroCodenotti is that Bendixson of Poincare-Bendixson fame?

Alessandro Codenotti

I don't know

anakhro

Wikipedia says it is.

Interesting, I didn't know he did anything in set theory.

Kalua

10:10 PM

Any $C^*$-algebraists here? I think I found an error in a paper: Given a $C^*$-subalgebra $A\subset B(H)$ for a Hilbert space $H$ and an operator $D\in A^1\subset B(H)$ for the unitization $A^1$ of $A$, with $0$ isolated in the spectrum of $D$, and $p\in B(H)$ the projection on $\ker(D)$. The paper claims $p\in D$ (by spectral theory). But functional calculus gives $p=1_{0}(D)\in A^1\setminus A$, where $1_{0}$ is the characteristic function of $0$.

I meant "paper claims $p\in A$"

anakhro

If it is an academic paper, I highly recommend posting it on mathoverflow.

Unless the error is not relevant to the paper.

Kalua

Okay, will do. It's this paper: personal.psu.edu/ndh2/math/Papers_files/… , page 351., the sentence starting with "Then D belongs..."

It's quite relevant, there was a lot of work involved constructing $D$ and $p$. I must be missing something, because this is such a glaring error

anakhro

There are many people on mathoverflow who will be able to tell you what is wrong if there is something you are missing or the paper is missing.

I personally don't know much at all about C*-algebras so unfortunately I can't help. :(

Alternatively, you can ask the people who wrote the paper if you know them.

But be nice about it and insist that you might be missing something.

Kalua

10:27 PM

I will ask my prof. / stackexchange before I would go to such lengths :D But thanks for the advice :)

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$Mathematics$ Mathematics