Mathematics - 2015-08-18 (page 1 of 2)

Joel Reyes Noche

00:06

I remember there used to be a command that showed the word "MathJax" in black and green (similar to how the command \LaTeX shows $\LaTeX$). What was that command?

Brennan.Tobias

01:40

Could you recommend a galois theory book that includes a lot of computations?

PVAL

@Brennan.Tobias At what level?

Brennan.Tobias

I want to refresh enough to answer this question math.stackexchange.com/questions/1400925/splitting-field-of-x43

pjs36

I love those Galois questions when they have intuitive answers. While I've forgotten close to 100% of the Galois theory I once knew, I have a suspicion that the fourth roots of $-3$ should be as symmetric as a square, so I'd expect the dihedral group of order 8.

But I'm probably doing something wrong, who knows. I just remember it worked like that once :P

Brennan.Tobias

01:57

wow! solving that in your head is impressve!

pjs36

Nah, I literally just remember Kaj answering a similar question and thinking "Huh, those $n$th roots of [whatever] are as symmetric as a regular $n$-gon!" And, for all I know, I'm dead wrong :)

PVAL

I'm sure even DF does computations like that. The roots of that are going to be $3^{1/4}$ times odd powers of a primitive 8 th root of unity $\zeta$. which can be arranged to be $\Bbb Q[\zeta ^2,3^{1/4}\zeta]$. and the action on those two elements determines the automorphism... something something transitive all those actions deterimine an automorphism. something something $D_8$

Pedro Tamaroff

@Brennan.Tobias D&F is certainly a book that does computations.

PVAL

Maybe theres a $\Phi_4=x^2+1$ somewhere.

Brennan.Tobias

thanks!

PVAL

02:11

Unfortunately most books on Galois theory aren't very computational. Can you believe they usually only compute 2 elements of $Gal(\overline \Bbb Q, \Bbb Q)$???

Mike Miller

@PVAL How grotesque!

Mike Miller

02:23

@PVAL Did you see this question?

Mary Star

02:35

In a proof that I am reading I came accross the following sentence:

"The rational functions $X + UY$ and $X — UY$ in $t$ have poles only at $t = \pm 1$."

What does it mean hat the functions **have poles** at these points?

Lie Algebra

If I have $\def\m{\mathfrak}$ $\m{h}$ is a Lie subalgebra of the Lie algebra $\m{g}$ and $\m{i}$ is an ideal of $\m{g}$

And want to show that $\m{i}$ is an ideal of $\m{h}+\m{i}$

all this takes is showing that $\m{h}+\m{i}$ is a subalgebra, and then:

$[\m{i},\m{h}+\m{i}]\subseteq [\m{i},\m{g}]\subseteq \m{i}$

Right?

anon

right (although you're also using the obvious fact $\frak i\subseteq h+i$)

@MaryStar it means you can't plug t into the expression because then you would be dividing by 0.

google "pole of a function" to see the idea generalized to holomorphic functions in complex analysis

Lie Algebra

@anon Yes, thanks very much!

Mary Star

03:00

@anon Ok... Thank you!! :-)

Brennan.Tobias

How do I show that we will have a normal extension?

splitting fields are always normal?

Lie Algebra

If I have $\def\m{\mathfrak}$ $\m{h}$ is a Lie subalgebra of the Lie algebra $\m{g}$ and $\m{i}$ is an ideal of $\m{g}$

And want to show that $\m{h}\cap\m{i}$ is an ideal of $\m{h}$

all this takes is showing that if $i_1,i_2$ are ideals of $\mathfrak{g}$ then $i_1\cap i_2$ is an ideal of $\mathfrak{g}$

And then showing $i$ and $h$ are ideals of $h$(which they seem to be) and I am done

Right?

PVAL

03:25

@MikeMiller The first time I've had to use [ Initials] to abbreviate a source in an answer.

Brennan.Tobias

I think I can answer this galois theory question, if I explain it would anyone tell me if it's right?

Mike Miller

@PVAL: Let $V$ be the 4-mfld with intersection form $E_8 \oplus E_8$. How do you show that $V \setminus \{pt\}$ is smoothable? This should somehow follow from Freedman's work.

PVAL

with much pain and suffering

Kevin Driscoll

@MikeMiller Could you explain to me in a few sentences what the central premise of Galois theory is?

Mike Miller

Nah.

Someone else can do a better job.

PVAL

03:37

I can quote theorem at you @MikeMiller . Does that work?

Any open 4-manifold admits a smooth structure, so just apply that theorem to $V$.

Mike Miller

@PVAL: Is this one of the main theorems of his paper? I don't have access right now.

Is it easy to see?

Lie Algebra

@KevinDriscoll To determine when polynomials can be written in terms of radicals? But I haven't studied any, so that's all I know

Brennan.Tobias

maths.gla.ac.uk/~ajb/dvi-ps/Galois.pdf

look at the diagram on the first page "The Galois Correspondence"

PVAL

@MikeMiller I dont have FQ on me so I cant check right now, but iirc soon after the annulus conjecture they prove something like "any 4-manifold has a smooth-structure after removing any discrete set of points". I don't remember how one gets this from the annulus conjecture and I do not think it is immediately obvious.

Mike Miller

Sorry, "is it easy to see" is obviously a stupid question. I mean to ask if there's a good reason for this without thinking about Casson handles and stuff.

PVAL

03:44

It's one of the technical things Quinn proved using the disk theorem of Freedman. I think Stong published two papers (in the Annals) of corrections to some of these technical things, so they can't be that easy!

Well one of them is in topology I guess.

I don't know where the other is.

Mike Miller

OK. I will take it for granted.

Brennan.Tobias

If anyone has a moment could you check the correctness of this for me? math.stackexchange.com/a/1401004/262437

PVAL

@Brennan.Tobias something something transitivity

user262695

04:11

Would anyone know when every colouring of a graph is frozen?

anon

04:21

what's a frozen coloring?

Ted Shifrin

04:32

Re the discussion of Galois theory, I'm actually quite fond of Ian Stewart's book Galois Theory. It has lots of concrete examples, IIRC.

Hi, mr @anon.

anon

hello

Ted Shifrin

I hope you've had a good summer.

Kevin Driscoll

@TedShifrin Maybe you'll know better. Could you give me the sort of 5 sentence explanation of what Galois theory is and why it's studied?

I just can't grok what's on the Wikipedia page

Ted Shifrin

Symmetry groups of the roots of polynomials.

Abel and Galois developed it to understand which polynomials could, for example, be solved by radicals and which could not.

Indeed, Galois's paper was the origins of group theory as we have come to know it.

Fargle

I really like the implications for ruler-and-compass constructions, myself.

Also hi @Ted.

Ted Shifrin

04:37

There's no Galois theory needed there, @Fargle, just dimension of vector space.

Fargle

True.

Ted Shifrin

Unless you want to understand which numbers of degree $2^n$ are actually constructible. Then there's some Galois theory.

Fargle

But it falls out in the preliminaries.

Kevin Driscoll

So, what could be symmetric about the roots of a polynomial? I'm not comprehending what the natural operations on a root that transform it to another root would be.

Ted Shifrin

I don't consider dimension of a vector space to be Galois theory at all!

But hi, formerly inoperable @Fargle :P

anon

04:38

@KevinDriscoll Better: symmetry groups of number systems.

Kevin Driscoll

It's maybe more obvious for something like roots of unity, so there's just some discrete rotations or something. But for arbitrary polynomials I'm at a loss.

Fargle

@KevinDriscoll Note that if $a + bi$ is a root of a polynomial in $\Bbb R$, so too is $a - bi$. So conjugation would be a symmetry of the roots in that case.

Ted Shifrin

@anon is making the point that you want to think of mappings of field extensions that preserve the base field, but I was trying to keep this as low-brow as possible.

@Kevin: These symmetries are rarely just inherited from symmetries of the complex plane.

anon

If you have a field $F$ then a symmetry is a function $\phi:F\to F$ such that relabelling all of the elements of $F$ that appear in true equations yields true equations. It is both necessary and sufficient that $\phi(ab)=\phi(a)\phi(b)$ and $\phi(a+b)=\phi(a)+\phi(b)$, IOW ring automorphisms (which will be field automorphisms).

Ted Shifrin

They're symmetries in the sense that they map roots of the polynomial to other roots of the polynomial.

<--- gracefully retires from the discussion

@Fargle: Do you have any research to report to me yet? :)

anon

04:41

@TedShifrin not every permutation of a zero set is induced from a Galois symmetry though; it is unavoidable that one must speak to the arithmetic operations of the (polynomial's splitting) field

Fargle

Not yet. I'm still refreshing and learning some stuff.

Soham Chowdhury

What kind of research is @Fargle doing?

Fargle

@SohamChowdhury Numerical methods on particular classes of ODE systems.

Ted Shifrin

Yes, @anon, I know. But sometimes to give an idea to someone you are better off not being 100% rigorous. Teaching isn't always about giving all the truth.

hi @Soham

Soham Chowdhury

So you write nice code and make pretty pictures? :P

Hello, @Ted.

Kevin Driscoll

04:42

Okay, so was I correct that the roots of unity are the simplest example?

Lie Algebra

So I want to show that for $\mathfrak{h}$ a Lie subalgebra of $\mathfrak{g}$ and $\mathfrak{i}$ an ideal of $\mathfrak{g}$ that we have:

$(\mathfrak{h}+\mathfrak{i})/\mathfrak{i} \cong \mathfrak{h}/(\mathfrak{h} \cap \mathfrak{i})$

Any ideas? These are both subalgebras I can see

Fargle

Well, not quite yet. I still have to learn numerical analysis, but that's partly going to come with the research.

anon

@KevinDriscoll quadratic number fields are probably simpler

@LieAlgebra both subalgebras? of what?

Soham Chowdhury

Yeah, somehow I think that the high point of numerical analysis is making programs that crunch real-world data. So, essentially, a lot of programming.

But what do I know.

Lie Algebra

@anon of $\mathfrak{g}$

anon

04:44

@LieAlgebra neither are subalgebras of $\frak g$. neither are even subsets of $\frak g$, any more than $\Bbb Z/n\Bbb Z$ is a subset of $\Bbb Z$ for instance.

both are quotients of subalgebras of $\frak g$ though

Lie Algebra

Hmmm The image of the quotient map(where quotienting by an ideal) gives you a subalgebra I thought

I guess they are subalgebras of h+i and h respectively then

anon

no, they are quotients

it's important you understand the difference between subthing and quotient thing

Lie Algebra

Okay haha

anon

are you familiar with groups for instance? do you agree $\Bbb Z/n\Bbb Z$ is a quotient of $\Bbb Z$ but not a subgroup?

Lie Algebra

For sure

anon

04:47

yes, $\Bbb Z/n\Bbb Z$ does not sit inside (i.e. map into) $\Bbb Z$, but rather $\Bbb Z$ maps onto $\Bbb Z/n\Bbb Z$.

Brennan.Tobias

Could anyone give me a little help with showing the galois group of $\mathbb Q(\sqrt{2 + \sqrt{2}})$

is 2Zx2Z ?

anon

the categorical imperative tells us to think in terms of maps between objects, and from that perspective subthings and quotient things are "dual" (one has maps into, one has maps out of)

Brennan.Tobias

I've written out the proof that the extension is galois already

Lie Algebra

@anon Well how do I approach my problem? Via isomorphism theorem?

anon

@LieAlgebra good idea.

Lie Algebra

04:50

That was what confused me, since my book says "Ker is an ideal of g_1, Im is a subalgebra of g_2"

and an ideal is a subalgebra, so they should both be subalgebras(if first iso is how I am meant to go about it)

anon

subalgebras of what?

Lie Algebra

$\mathfrak{g}_1$ and $\mathfrak{g}_2$ respectively I guess( which I think in this case are both just $\mathfrak{g}$)

anon

why do you think in this case they're both just $\frak g$?

Lie Algebra

Well it's that or $\mathfrak{g}_1=\mathfrak{h}$ and $\mathfrak{g}_2=\mathfrak{h+i}$

anon

let's see if you can use iso-1 and then we'll see what the subthings and quotient things are

Lie Algebra

04:54

Okay I'll give it a shot

Kevin Driscoll

Alright, so let me see if I have this right. The point is to consider some more general operations, that is more general (or maybe just different) from ordinary $+$ and $*$ such that a certain group of numbers, for example the rationals plus the squareroot of a square-free number (plus presumably some other numbers that follow from the operation and your choice of root?), are closed under those operations.

And the claim is that there's a connection between such operations/fields and certain symmetry groups

anon

@KevinDriscoll that's more or less what I mean by "number system," yes - a field

@KevinDriscoll saying "there's a connection between fields and symmetry groups" is too indirect - we are defining symmetry groups of fields

it's like saying "there's a connection between regular n-gons and certain symmetry groups" - we are directly defining symmetry groups of regular n-gons (to get cyclic or dihedral groups)

Kevin Driscoll

Okay, so they're not necessarily related to the finite groups we're already familiar with that have nice geometric interpretations, say as the labelings of points on polyhedra?

anon

there's no obvious geometric meaning to Galois symmetries, no

Kevin Driscoll

They;re something new and more complicated

Okay, this at least sounds plausible. Although, totally intractable in a systematic way!

anon

04:59

it's more tractable than you think!

in order to specify such a symmetry, you only need to determine where it sends certain generators (that's where roots of polynomials come into play)

Kevin Driscoll

Ah okay, so there's some way you can make such things into a vector space and then you don't have to do as much work as it seems like

This GREATLY increases the space of groups in my conceptual library. I mean I thought basically all groups were specified by the finite simple groups and compositions of those and then some infinite Lie groups

that's all I know about

anon

@KevinDriscoll well, by Jordan-Holder, "compositions" of simple groups yield all finite groups, yes. if you allow infinite groups, there are many more than just lie groups though. and there are many different ways lie groups may "arise in nature / the wild," too. a finite group may be the fundamental group of some space, or the Galois group of some field extension, or the symmetry group of some figure in space, etc.

usually Galois groups are finite, or at least profinite, so they're limited too (in the class of infinite groups)

every finite group arises as a Galois group of some field extension (which is easy to prove when you grasp the basics), but it's an open question if every finite group is specifically the Galois group of some finite extension of Q (i.e. a number field). it's known as the inverse galois problem.

Kevin Driscoll

And do you know what, if any, finite extension of the rationals correspond to the sporadic simple groups?

or, I should say, is even ONE known?

anon

some are known, yes. I don't know the details.

Kevin Driscoll

okay, very interesting.

anon

05:12

@Brennan.Tobias are you sure the galois group is klein four?

Brennan.Tobias

I think I proved it in my answer

based on counting degree, it's the most shaky part of the answer

maybe it is wrong

you introduce a square root (of 2), then another (of 2+sqrt(2)). So you have two order two automorphisms

anon

replacing sqrt(2) with -sqrt(2) is not enough to specify an automorphism of Q(sqrt(2+sqrt(2))) though

so you don't know if it has order two in the top Galois group

Brennan.Tobias

oh yes you're right, that's a problem

anon

even if it restricts to an order two automorphism of Q(sqrt(2))/Q

the V4 extensions of Q correspond to biquadratic extensions Q(sqrt(a),sqrt(b)). that might be helpful.

indeed, suppose $\Bbb Q(\sqrt{2+\sqrt{2}})$ has Galois group $V_4$, so in particular has multiple intermediate extensions.

any intermediate extension would be quadratic hence of the form $\Bbb Q(\sqrt{r})$

Lie Algebra

Okay so look at arbitrary elements $h_1+i_1 \in \mathfrak{h+i}$ and $h_2+i_2 \in \mathfrak{h+i}$ and look at them modulo $\mathfrak{i}$ and we get:

$$h_1+i_1 {\pmod i} \equiv h_1$$
$$h_2+i_1 {\pmod i} \equiv h_2$$

And these are the same when $h_1-h_2\in \mathfrak{i}$. These are also in $\mathfrak{h}$ so then $h_1-h_2\in \mathfrak{h\cap i}$

How does that help? I don't see how that solves my problem?

anon

05:21

@LieAlgebra this is good work so far

you've conclude that elements of $\frak h+i$ mod $\frak i$ look like elements of $\frak h$ mod $\frak h\cap i$!

indeed, the point of the first isomorphism theorem is that isomorphisms $B/A\cong C$ correspond to short exact sequences $A\to B\to C$ (i.e. onto maps $B\to C$ with kernel $A\subseteq B$). so in this case, pick which quotient you want to call $A/B$ and which you want to just call $C$

just looking at the top terms $\frak h$ and $\frak h+i$, which way is there an obvious map? well, $\frak h\to h+i$, the inclusion. so consider $\frak h\to (h+i)/i$. can you prove it's onto? what is it's kernel? what does iso-1 tell us then?

Lie Algebra

Ahh I see, thanks I'll see how I go with that, that's really helpful

Brennan.Tobias

hmm im not quite seeing this

anon

BTW if I suddenly stop responding to comments at some point, it may be because my electricity will go out

Brennan.Tobias

I feel like there's a gap in my knowledge about working with automorphisms of the roots.

anon

@Brennan.Tobias seeing one of the things I said, or seeing how one of the things I said helps?

Brennan.Tobias

05:31

I suppose just the tools I need to form this argument in a rigorous way

anon

what argument? I haven't given one yet. just made some observations.

Brennan.Tobias

I mean that I want to come up with one

anon

kk

Brennan.Tobias

ah I see what I'm missing: How to determine the automorphisms in full

we can say that there's an automorphism that maps sqrt(2+sqrt(2)) to -sqrt(2+sqrt(2)) by transitivity, but what does it do on everything else? I don't know how to answer that

the lattice for Q(sqrt(2),sqrt(3)) that you mentioned is different than the lattice i have in mind for Q(sqrt(2+sqrt(2)) (it looks the same as the cyclic groups one)

Lie Algebra

@anon Thanks very much, I got it!

Brennan.Tobias

05:48

this strongly suggests that the group is really C_4, but I don't see how to show that

user1524837

How do I alter this graph's equation so that it passes through the points it show? : desmos.com/calculator/18daigxbor

Lie Algebra

y=10?

Oh nvm there are more points

user1524837

heh

Lie Algebra

Via interpolation?

user1524837

man i really gotta find out what the word means, second time I've been asked it

Lie Algebra

06:03

Interpolation

In the mathematical field of numerical analysis, interpolation is a method of constructing new data points within the range of a discrete set of known data points. In engineering and science, one often has a number of data points, obtained by sampling or experimentation, which represent the values of a function for a limited number of values of the independent variable. It is often required to interpolate (i.e. estimate) the value of that function for an intermediate value of the independent variable. This may be achieved by curve fitting or regression analysis. A different problem which is closely...

You literally just build a curve to fit the data

See the polynomial section

user1524837

oh, yep via interpolation

sort of, i mean i know i have to round something to the lowest number

dunno if rounding is part of interpolating

or is allowed*

is it?

Brennan.Tobias

@anon, got it - thank you very much for the help

user1524837

@LieAlgebra Ah actually, because of rounding, then no not via interpolation

user1524837

06:34

a

Balarka Sen

07:03

@KevinDriscoll yeah, last time I heard, a lot of sporadic simple groups are known to appear as Galois groups over Q

in fact, I think there's only one that's causing the trouble. I forgot.

Tobias Kildetoft

@BalarkaSen Hmm, that is the sort of thing I would have expected to be resolved by now (I mean, it is a finite set of groups we want to find)

Balarka Sen

but they're huge!

Tobias Kildetoft

@BalarkaSen Sure, but so are out computers nowadays. Though I have no idea how good we have gotten at this sort of thing. I guess the problem is that picking random polynomials will almost always just give us the corresponding symmetric groups.

Balarka Sen

yeah.

Tobias Kildetoft

Also not sure how precise anyone has been able to make that sort of statement (i.e. that we will tend to get symmetric groups)

Balarka Sen

07:07

I think there was some question about what happens when we pick random polynomial.

Let me find it.

Ok, this is it. Mike asked that one.

Tobias Kildetoft

@BalarkaSen Ahh, nice

Balarka Sen

@Fargle psst : look at Terry Tao's blogpost on a geometric method of proving that an angle cannot be trisected by ruler and compass. it secretly depicts the analogy with covering spaces and galois theory

the (field-theoretic -- I am not prepared to call it Galois theory either, like Ted) proof exploits the fact that compass and rulers are inherently "quadratic". it's a charming proof - reading it feels like you should have done it by yourself.

at least, it did so for me.

Soham Chowdhury

07:34

47

Q: Fractal behavior along the boundary of convergence?

The complex power series $$\sum_{n=1}^{\infty}\frac{z^{n^2}}{n^2}$$ has radius $1$ (Ratio Test) and is absolutely convergent along $|z|=1$. Recalling something that my calculus professor (Ray Mayer, emeritus of Reed College) showed me 15 years ago, I started looking at a "graph" of this function...

What the . . . ?

Balarka Sen

hi @Soham.

iwriteonbananas

07:57

hello everyone

Balarka Sen

hi @iwriteonbananas

iwriteonbananas

what's happening, what's new?

Balarka Sen

learning singular cohomology, finally

iwriteonbananas

great

how far have u gotten?

Balarka Sen

done universal coeff theorem, doing cup products and cohomology rings.

iwriteonbananas

08:00

i've had fever and simultaneously had to study for exams. tomorrow's my last one for a while

good, if u have a head start on me, then u can answer my dumb questions later

Balarka Sen

@iwriteonbananas ah, that sucks

@iwriteonbananas haha, sure.

Soham Chowdhury

08:22

A poetic proof of the Halting Problem thingy

7

skull patrol

Thanks for sharing @SohamChowdhury :D

anon

08:38

Composition of combinatorial species: $$(F\circ G)(A)=\bigsqcup_{\Phi\vdash A}F(\Phi)\times\prod_{B\in\Phi}G(B)$$ Trace for Schur functors: $$\chi_{\small\Bbb S_\lambda V}(g)=\sum_{\sigma\in S_n}\frac{\chi_\lambda(\sigma)}{n!} \prod_{i=1}^n\chi_V(g^i)^{c_i(\sigma)}.$$ These seem suspiciously similar to me.

Balarka Sen

I have no idea what the two are.

Tobias Kildetoft

@anon They are both a sum of products. That hardly makes them "suspiciously similar"

anon

@TobiasKildetoft that $\textrm{org}(F\circ G)=\textrm{org}(F)\circ\textrm{ord}(G)$ comes from Faa di Bruno's formula. compare said formula with the second formula I gave to make the similarity stronger / more apparent.

note the coefficients in front of the products in the schur trace formula only depend on the conjugacy class of permutation, which correspond to cycle types, which correspond to integer partitions of n, which are exactly what di Bruno's sums over as well

the composition formula for combinatorial species sums over set partitions instead of integer partitions, but still

my org/org/ord's should be ogf for ordinary generator function

Jordumus

09:11

Anyone brave enough to help me with this problem? define an isosceles trapezoid with roundes base

Krazii KiiD

09:39

hello

can someone help a friend with this?

0

Q: C# Rotate 2 controls around a center point at the defaul angle in WinForms

I was trying to rotate 2 windows form button as in the following image: During their rotation, the distance between them should be 0 and when you click a label, they should "rotate" 90 degree, as like: if (red is up and black is down) { red will be down and black will be up; } else { ...

c# winforms rotation controls point

i know it is based on trigonometry, but i don't know too much of it..

Mary Star

Hello!! What exactly is a sequence of polynomials?

Lord_Farin

@MaryStar A mapping $\Bbb N \to \Bbb Z[X]$, most usually.

That is, just a polynomial $p_n$ for each $n$.

Krazii KiiD

@MaryStar Hey, i guess yes.

Soham Chowdhury

@Lord_Farin I am sure that wouldn't help her, @Lord_Farin :)

Lord_Farin

@SohamChowdhury I figured, and therefore added a more understandable sentence after the first one :).

But it's good to be precise sometimes. This can clear confusion.

Mary Star

09:53

@Lord_Farin Could you give me an example?

Soham Chowdhury

@MaryStar $p_n(x) = 1 + x^n$, say?

Lord_Farin

@MaryStar $p_n = X^n$, so it starts $p_1 = X, p_2 = X^2, ...$.

Krazii KiiD

Can anyone help me to set the RAD offset for the parametric equation for a circle?

Soham Chowdhury

10:10

> Will you like this book? Here’s a simple test. What’s the rule that produces the sequence 1, 11, 21, 1211, 111221, 312211 . . . ?
> This is Mr. Conway’s “look-and-say” sequence, so called because each number (after the first) is what you get when you look at the previous number and say it aloud: “one one; two ones; one two, one one; one one, one two, two ones . . .”
> If that makes you laugh with surprise, as it did me, you’ll like Mr. Conway, and you’ll like “Genius at Play.” If not, you might want to quit here and go read something improving about the Greek debt crisis.

Greek debt crisis.

Lie Algebra

10:34

If anyone can give insight into my question, it would be greatly appreciated:

2

Q: Basis of Lie Algebra $\mathfrak{su}(3)$

With the matrices below, apparently $\{u_k = -\frac i2 \lambda_k| k=1,2,\cdots,8\}$ forms a basis of $\mathfrak{su}(3)$ How could that be true? $-\frac i2 \lambda_1$ shouldn't even be an element of $\mathfrak{su}(3)$, it isn't hermitian.

matrices lie-algebras

It seems that this really isn't a basis, or that it isn't a basis in the conventional sense

St3114

11:09

hello. can someone say why $\Bbb Z_{180}$x$\Bbb Z_{10}$x$\Bbb Z_{6}$ is isomorphic to $\Bbb Z_{36}$x$\Bbb Z_{30}$x$\Bbb Z_{10}$ not to $\Bbb Z_{45}$x$\Bbb Z_{60}$x$\Bbb Z_{4}$

Lord_Farin

@St3114 Because $\Bbb Z_n$ is isomorphic to the product $\Bbb_{p_i^{n_i}}$ where $n = \prod_i p_i^{n_i}$.

So the single factor $5$ can go from $180$ to $6$ without problem, but you can't merge the factors $2$ from $10$ and $6$ to form $4$.

St3114

thanks. so it is not to $\Bbb Z_{27}$x$\Bbb Z_{25}$x$\Bbb Z_{16}$

actually it was cyclic. the first one isnt :)

Lord_Farin

@St3114 Exactly.

St3114

11:26

is this about merging to form new or just merging. ie $\Bbb Z_{18}$x$\Bbb Z_{12}$x$\Bbb Z_{50}$ and $\Bbb Z_{90}$x$\Bbb Z_{40}$x$\Bbb Z_{3}$ I guess neither of them is isomorphic to the first one

Lord_Farin

@St3114 For each prime, you can write the three multiplicities (for 7 and greater, three times $0$) of each. So for the first of your new groups, it's $(1,2,1)$ for $2$, $(2,1,0)$ for $3$, $(0,0,2)$ for $5$. For the original, it is $(2,1,1)$ for $2$, $(2,0,1)$ for $3$, $(1,1,0)$ for $5$.

Hence the $5$ factors prevent them from being isomorphic; the one has an element of order $25$, the other hasn't.

St3114

i see. for the second one there is no element of order 8 in $\Bbb Z_{180}$x$\Bbb Z_{10}$x$\Bbb Z_{6}$

Lord_Farin

@St3114 Exactly :).

St3114

thanks again. have a good day:)

Lord_Farin

11:52

@St3114 Thanks, you're most welcome :).

Balarka Sen

12:03

@PVAL, are you around?

Soham Chowdhury

Nikodym set

In mathematics, a Nikodym set is the seemingly paradoxical result of a construction in measure theory. A Nikodym set in the unit square S in the Euclidean plane E2 is a subset N of S such that the area (i.e. two-dimensional Lebesgue measure) of N is 1; for every point x of N, there is a straight line through x that meets N only at x. Analogous sets also exist in higher dimensions. The existence of such a set as N was first proved in 1927, by Polish mathematician Otto M. Nikodym. The existence of higher-dimensional Nikodym sets was first proved in 1986, by British mathematician Kenneth Falconer...

wat

Lie Algebra

Are you wat'ting the straight line part?

Or the combination of that with the area of $1$?

Moses

12:33

@DanielFischer Hi how's it going? I want to ask something short about the topic of Banach Space Valued Functions, it is stated that $S(I;X) := \text{normed linear space } X-\text{ valued step functions with the } \| \cdot \| \text{ norm}.$ And the completion $\overline{S(I;X)} = L^{\infty}(I,X)$. Are you familiar with this result?

Daniel Fischer

@Moses Isn't that the definition of $L^\infty(I,X)$?

Moses

@DanielFischer Not sure but from what I know I would define equivalence class $L^{\infty}(I,X) := \{ f \text{ measurable and } ||f(x)|| \leq M\text{ for some } M \in X\}$ and then endow it with ess supremum norm. What do you think?

Daniel Fischer

@Moses There are different ways to define things, some of them are equivalent. How are things defined in the text you're reading? From the appearance of things, my first guess was that the above is the definition of $L^\infty$ in that text.

Moses

@DanielFischer It's an introduction to operator theory set of notes. The section starts defining $S(I,X)$ and the completion $\overline{S(I,X)}$. It then states as a remark that $\overline{S(I,X)} = L^{\infty}(I,X)$.

@DanielFischer Is it common for authors to use $=$ to denote isomorphic rather than $\cong$? I'm not sure if this is what happened in this case.

Daniel Fischer

12:50

@Moses Doesn't help. Either that is the definition (informally), or they ought to say what they mean with $L^\infty(I,X)$. You have a reasonable guess what it may mean, but what does "measurable" mean? Bochner-measurable, Borel-measurable?

@Moses If you have a canonical isomorphism, it is rather common to write "=". Apart from that, people are often sloppy.

Moses

@DanielFischer I'm not sure. The interval $I=[a,b] \subset \mathbb{R}$ so I assumed the usual Lebesgue measure.

Daniel Fischer

@Moses The big problem is at $X$. I don't remember half of it, but if $X$ is not separable, then there are significant differences between various concepts of measurability.

Moses

@DanielFischer But so if I assume that the definition of $L^{\infty}(I,X)$ is as I stated previously, then is it clear that $\overline{S}(I,X) = L^{\infty}(I,X)$?

For any of the measures on $X$ that you mentioned.

Daniel Fischer

@Moses I'm not sure. If $X$ is separable, I think it's no problem. But if $X$ is not separable, it may depend on which sort of measurability you use.

Moses

@DanielFischer Kewl thanks.

Balarka Sen

13:09

morning @Mike. I have a question regarding cup products. Hatcher computes the cup product structure $H^1(M) \times H^1(M) \to H^2(M)$ on the genus $g$ surface $M$ by splitting it up into a $4g$-gon, picking appropriate dual basis of the corresponding basis on homology (which are just the meridianal and longitudal circles, represented by the edges of the polygon), and then trying to define the corresponding cocycle by defining it on the diagonals first.

He does so by drawing arcs and defines the values by the intersection number of the diagonals with those arcs. I'm a bit confused here : how is the intersection number relevant?

Mike Miller

rhink of $H^n(M)$ as $\text{Hom}(H_n(M);\Bbb Z)$ so you can restrict the number of simplices you have to think about, then think about the definition again.

If $\alpha_i$ is something that takes the curve $\gamma_i$ and spits out zero, the you want to apply the Defn of cup prosuct to $alpha_i \smile \alpha_j$

Balarka Sen

you mean for $n = 1$, right? otherwise you have an Ext term in there.

Mike Miller

Not for a genus g surfacd

Balarka Sen

oh, yeah, you're right

Mats Granvik

Has the OEIS become too big to handle? Quote: "The submissions stack has been at 400 or more for several months. Please voluntarily restrict your submissions and please help with editing. (We don't want to have to impose further limits.) "

Balarka Sen

13:15

Ext([something free], Z) vanishes

I'm not quite sure what you're trying to point me at, @MikeMiller (sorry, I realize you're on phone and it's hard to be explicit from there). If $\alpha_i$ eats a curve and spits out $0$, then $\alpha_i \smile \alpha_j$ is just $0$.

No idea how that says anything about intersection. I feel like I'm being silly.

Mike Miller

13:38

@BalarkaSen: Sorry I misspoke. $\alpha_i(\gamma_j) = \delta_{ij}$ is the desired cocycle.

With what I said before I gave you the zero cocycle and hence, like you said, there was nothing interesting to say. :P

Moses

13:54

@DanielFischer In the topic of contour integration, it is shown that the integral is independent of the parameterization of the contour. Why is the parameterization of a contour $\Gamma : [\alpha, \beta] \to X$ defined as some countour $\Gamma_{1}:= \Gamma \circ \gamma$ where $\gamma: [\alpha_{1}, \beta_{1}] \to [\alpha, \beta]$ is a countinuously differentable bijection with $\gamma(\alpha_{1}) = \alpha$ and $\gamma(\beta_{1}) = \beta$?

Daniel Fischer

@Moses Convenience, methinks. You can without any problem allow piecewise continuously differentiable changes of parameter. You can get more general, but then you actually have to do some work to define the integrals.

Moses

@DanielFischer What is the general idea? You are converting a given contour $\Gamma$ into a parametric function of $\gamma$, which gives you another contour $\Gamma_{1} := \Gamma \circ \gamma$, you have the same integral when integrating over either contour . Is this correct?

@DanielFischer What is the significance of this?

Daniel Fischer

@Moses I don't understand. $\Gamma$ is a parametrisation of a curve in the plane. The idea is that it doesn't matter how fast you traverse the curve, as long as you traverse it in the same direction. $\Gamma_1$ is a different parametrisation of the same curve, the composition with $\gamma$ changes how fast you traverse (parts of) the curve.

Moses

14:11

@DanielFischer Yes but I'm not following how it alters the speed at which you traverse?

Daniel Fischer

@Moses Say we have $\gamma(t) = 2t$, then you move twice as fast through the curve with $\Gamma_1$ than with $\Gamma$.

Moses

@DanielFischer I see what mean now

Chris's sis the artist

14:30

@robjohn I'm back. I wanna show you something very interesting I received from r9m (a problem in AMM).

robjohn

@Chris'ssistheartist okay... just back from the park

Chris's sis the artist

@robjohn I think the core of the solution lies in the simple fact that $$g(x) = e^x g(x^2)$$ that can also generate a relation between the coefficients of the power series.

robjohn

@Chris'ssistheartist do we know that those limits are even finite?

Chris's sis the artist

@robjohn The text of the problem is exactly as you can see it above.

robjohn

@Chris'ssistheartist yes, but if the limits are infinite, the constant could be anything.

Chris's sis the artist

14:34

@robjohn Yeap, it's not specified. I suppose they are finite.

@robjohn This guy proposed a lot of very clever problems with very nice solutions. I think I told you in the past about other problem of him.

It was a problem about finding the nth derivative of a function.

Remember me

Hello@Balarka

Chris's sis the artist

@robjohn Anyway, let me know if you see anything useful about using that relation.

robjohn

@Chris'ssistheartist I'm looking at it...

Chris's sis the artist

@robjohn OK, no hurry with that. I'm stuck there, but I have some ideas to check.

@r9m I'm in chat.

Soham Chowdhury

14:50

Cool, cool videos of mechanical linkages that are just stupidly smart

Balarka Sen

hi

@MikeMiller ah, ok. what are your $\gamma_j$'s again?

Mike Miller

Basis curves. Same as Hatcher's.

Did you prove or find a counterexample to the claim that maps (between CW complexes) are determined by their action on homotopy groups?

Balarka Sen

uh, lattitidual circles or the longitudal circles?

Mike Miller

Neither alone constitutes a full basis. Just try to understand it on the torus before you do it in general.

Balarka Sen

@MikeMiller As I had indicated in a message (not sure you read that), I suspect maps from $K(G, 1)$ spaces to simply connected spaces would do the trick (all zero in $\pi_n$, but certainly there are non-nullhomotopic such maps?). I haven't come up with an explicit example yet.

Mike Miller

14:57

If you didn't ping me with it I didn't read it.

Balarka Sen

I have done this with the torus here. Can you check if it's alright?

Mike Miller

No, not for a few hours.

Balarka Sen

yeah, I didn't ping you. your avatar was just hanging in the user panel.

Mike Miller

It does that sometimes.

Remind me about your question around noon PST and I'll look at it.

Balarka Sen

alright.

r9m

15:05

@Chris'ssistheartist hi :)

Chris's sis the artist

@r9m Hey! Did you manage to make some progress on it?

r9m

@Chris'ssistheartist nope .. I don't have much clue how to proceed :|

Chris's sis the artist

@r9m OK

r9m

@Chris'ssistheartist it's a tough one .. seems like a hard nut to crack!

Chris's sis the artist

@r9m Did Tauraso find a solution to it? On that page he posted anything to it.

r9m

15:12

@Chris'ssistheartist I guess he doesn't post solutions to anything unless the last date of submission is over .. :)

Chris's sis the artist

@r9m Ah, I see.

@r9m I don't think it is that hard, AMM does not accept very hard problems. Maybe there is a small trick we simply miss.

r9m

@Chris'ssistheartist probably! but the problems often are more than just tricky

Chris's sis the artist

@r9m you say that until you see the authors' solutions. Of course, sometimes trying different paths things become pretty hard.:-)

r9m

@Chris'ssistheartist well authors' solution often doesn't answer our personal curiosities :-) (solves the problem but at the cost of our fun :P )

Chris's sis the artist

@r9m ^^ by the same author, I showed you this one some time ago.

r9m

15:19

@Chris'ssistheartist oow yeah!! I love that one! :D

Chris's sis the artist

@r9m hehe, kind of!

@r9m That author has very nice questions and very clever solutions, and problems like the one above I couldn't ever forget.

r9m

@Chris'ssistheartist I have seen G. Stoica's older problems too! They are really nice indeed and ingenious ones!

Chris's sis the artist

Yeap.

Mike Miller

@BalarkaSen: Yes, your thing for the torus works fine. It's the exact same computation that's used in general.

Balarka Sen

@MikeMiller Hatcher does something with intersection of curves for the general g genus surface case, though, which I don't understand.

Mike Miller

15:30

Again, it's definitional. Plug in $\alpha^i \smile \alpha^j$ into the definition when the curves they correspond to don't intersect and you'll see it's zero.

@BalarkaSen: The point, to clarify, is that $\alpha^i$ can be defined by saying that $\alpha^i(\gamma_j)$ is the intersection number of $\alpha_i$ and $\gamma_j$, where $\alpha_i$ is a certain curve.

This gives you $\delta_{ij}$ as desired.

Balarka Sen

I'm sorry, keep getting disconnected in here.

@MikeMiller Uh. What's the motivation behind defining something like that?

Mike Miller

So that you can intersect them.

Balarka Sen

For example, I wouldn't be able to come up with such a cocycle.

Mike Miller

There's a picture next to the computation.

That's precisely what his dashed lines are...

If this still gives you a queasy stomach, ignore it, and again, plug into the definition.

Balarka Sen

15:46

Yes, I don't understand the picture. He defines $\varphi_i$ to have values the intersection number of the edges with a particular curve $\alpha_i$. Why pick $\alpha_i$ to be that curve, and $\varphi_i$ to be the intersection number in the first place? This feels so confusing.

I guess I should just leave and stare at it for a while, and as you said, think about the definition.

Mike Miller

To be honest, I don't really understand what you mean by 'think about the definition'. This is a plug 'n chug exercise. I think if you do it for that genus 2 surface in the picture you should see why the intersections show up, and understand his comments...

Balarka Sen

I am not entirely sure what you mean by "plug and chug". Plug what into what?

Mike Miller

You write down a representative for every element in a basis of $H^1(M)$, and then you plug these representatives into the definition of the cup product, and then you evaluate it on the sum of all the simplices. Hatcher explicitly details how to write down a representative of every element of the basis of $H^1(M)$. So you just plug these representatives into the definition.

The intersections are the geometric interpretation of his definition of the representatives.

Semiclassical

16:06

hey chat

Balarka Sen

hmm. oh. so you mean that we can just write down some other representative regardless of any geometric interpretation by intersections?

Semiclassical

out of curiousity, what picture are we talking about?

Balarka Sen

Hatcher pg. 207

nigelvr

16:33

anyone here good with Lie groups? math.stackexchange.com/questions/1400783/…

Balarka Sen

morning @Fargle

Fargle

Good morning @Balarka

Balarka Sen

did you see the message I pinged you earlier?

Fargle

I don't think so.

Now I see it.

Mike Miller

16:56

@Balarka: Yes, you just write down a representative. It actually needs to be a cocycle. Intersecting with a subcomplex that meets with the complex in a nice way (like his dotted curves) will always be a cocycle. Don't worry about that.

Balarka Sen

OK, phew.

Semiclassical

it's not at all surprising, but all of that is quite reminiscent of complex contour integrals on branched surfaces

Balarka Sen

what is not surprising, @Semiclassical?

Semiclassical

the fact that it's similar

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