Hi people! Hi Ted!

I have some (stupid?) questions about polynomials.

If a polynomial is not solvable, are its zeros then irrational numbers?

Well I suppose not, since I cant insert the solutions into the polynomial and then it holds

In particular, yes. Unsolvable polynomials have roots which are algebraic but not expressible in terms of radicals

what are radicals?

HI @Rudi

What do you understand by "solvable"?

Well Galois sense

I want you to tell me what that is.

when its solutions do not have solvable symmetry

This could be a language issue.

OK. That's equivalent, by a deep theorem in Galois theory, to saying it's roots are not expressible as a bunch of addition, multiplication, division, and n-th roots of rational numbers.

That's what I mean by radical

OK, right, that is what I understood

so in that case how can solutions be expressed?

Series?

You can do that, yes. But there's no "finite" expression for it, whatever sense of the word you might want to take that as. Even though they are after all algebraic numbers.

that means?

Unsure what the question is here.

sorry I just check what "algebraic numbers" mean on WP. But thats circular. Solutions of polynomials

Uh, that's the definition

@Rudi: I don't understand your original question. Yes, the roots are irrational. But messy irrational. :)

Well I just started for the first time to really think about non-solvable polynomials, and thereby I started to wonder what kind of irrational numbers that might be.

@TedShifrin Or rather, there is at least one root which is a messy irrational.

It could be something like $(x^2 - 2)(x^5 - x + 1)$, since Rudi didn't specify irreducibility or anything

Hi @Ted @Balarka

What would be such a messy irrational number?

Hi @Alessandro

LOL, well, if you're going to play that game, you can get a rational root, too. :)

Hi, demonic @Alessandro

@Rudi Roots of $x^5 - x + 1$ is such a messy irrational.

True, @TedShifrin

I mean there are transcendental numbers that have names, like $\pi$ and so on,

@TedE, @ÍgjøgnumMeg: I got 6 pings last night during your conversation. @ÍgjøgnumMeg, you have to type TedE, because just Ted pings me (and him both, I presume).

Woops! Apologies

So I wonder if there are any famous / phsically important "messy" irrational polynomial solutions

Hi @ÍgjøgnumMeg!!

Servus @Rudi :)

The only (not messy) example that I know offhand is $\phi$, the golden ratio.

I am sure R. B. King would know where unsolvable algebraics pop up in chemistry

Agh, he was a colleague of mine @Balarka.

But that can be also the solution of a solvable polynomial right?

@TedShifrin Yeah, you told me when I mentioned him before. Odd coincidence.

Well, if you define a transcendental number, could you find a polynomial that has it as a root?

Of course not. But thats not the point

(Yeah, I don't know this subject so that was partially curiosity on my part)

Well, $x - \pi$ is a polynomial whose root is $\pi$. Transcendental numbers are numbers which are not roots of polynomials with coefficients in $\Bbb Q$ in general

The one King with graph theory?

I don't know, @Rudi. He has a book on quintics and solvability.

I once invited him to a conference, but he didn't come

Riemannian manifolds are confusing @Ted

He works in chemistry as far as I remember

@Balarka: I actually didn't know that. He fancied himself a topologist.

Huh

Yeah its him "topology in chemisty" he did very good and interesting work

Yes, he did inorganic chemistry.

Why so, @Alessandro?

I have no intuition for what they look like

Actually his book on quintics was my first foray into mathematics

It's beautifully written

at least in my opinion, I am also inorganic chemist, but many colleagues don't like maths so its a bit of a despearte field

He never mentioned that book to me, @Balarka. Odd.

I knew him and his wife from duplicate bridge, too.

Oh wow

@Alessandro: Pretty much every manifold you've ever seen has a Riemannian structure. Maybe every manifold.

The book's name is "Beyond the Quartic Equation" if you're curious

OK I 'll look it up!

Doesn't every manifold admit a Riemannian metric?

In my point of view he was the first on who really understood the Wade-rules for boranes (Wade however got the Nobel price for it)

His point is you always "see" a manifold equipped with some Riemannian structure, not as a pure flabby object

@Rudi Damn.

Whatever the Wade rule is

@Alessandro: Yes, abstractly. But my point is that any time you write down a submanifold of $\Bbb R^n$ it has a natural Riemannian structure that you're used to.

You mean by pulling back the standard metric on $\Bbb R^n$?

Yup.

You measure lengths and angles using the (restriction of the) Euclidean metric.

@BalarkaSen its just that you need n+1 spherical harmonics to describe a regular polyhedron with n-vertices.

I see

except for the tetrahedron

@BalarkaSen really?

No :P

You got me

OK

I am relieved because that was not comprehensible

Well that makes sense, but my problem is with Riemannian manifolds that don't look like $\Bbb R^n$, I guess I'm just not used to think about them

Oh, interesting, I hadn't heard of Kiepert's algorithm. Pretty complicated, I gather.

Yeah

@Alessandro: What do you mean "don't look like $\Bbb R^n$"?

The key point is somehow that even if a quintic cannot be solved in terms of radical it can be solved in terms of elliptic modular functions

Must a Riemannian manifold locally look like $\Bbb R^n$ with the Euclidean metric? Must it embed into some $\Bbb R^n$?

If you have a Lie group, there's a canonical invariant metric on it, and then you induce Riemannian structures on homogeneous spaces $G/H$ that are very natural from an algebraic viewpoint.

@AlessandroCodenotti The former is false, the latter is a deep theorem of Nash

No, it won't look like the Euclidean metric unless it has zero curvature everywhere.

Ah I see

What is your course doing with Riemannian manifolds (besides a definition)?

I'm not sure, we only saw the definition and not much more so far

Seems like you're asking the right questions given you've only seen the definition

Just the musical isomorphisms and the generalization of the gradient of a smooth function to a Riemannian manifold

I was going to suggest you read a little about parallel translation in my surfaces notes. I thought you already had some of that as an undergrad. For most of the fancy stuff you can get intuition from a good course on surfaces.

Oh, just formalism, then.

That's just using the inner product to give $V\cong V^*$.

My next question to you would be "If a Riemannian manifold is not locally $\Bbb R^n$ with the Euclidean metric, what is the appropriate local model for it?"

(And induced stuff on exterior algebra.)

@Balarka: Most likely itself.

Yeah, in general, I guess. I was going to point out normal coordinates, and the Hadamard theorem in constant curvature case.

I thought you were heading towards constant curvature.

@Balarka: From a review of that book — This short book of 149 pages does read more like a journal article than a textbook. There are few examples and no exercises or computations. The author's goal is to make the material available to non-specialists and mathematically oriented readers, but is unsatisfying in that it has too much introductory material for someone who has seen this material before and not enough for someone who has not.

Yeah haha I got stuck trying to understand ideals and quotient rings, and switched to Dummit-Foote

Somehow I remember liking his bit on symmetries of the icosahedron and how it relates to quintics

He talks a bit about invariant theory there

I never read Klein's stuff on the icosahedron and the quintic. I really should.

Oh, we also proved that every manifold is metrizable using the fact that they admit a Riemannian metric, but I feel like that's a very inefficient proof

@Balarka: This is cool. Finally it is worth highlighting the geometry that connects the quintic and the icosahedron. Using a radical transformation, a quintic can always be put in the form $y^5 +5\αlpha y^2 +5\βeta y+γ = 0$. The vector of ordered roots of such a quintic lies on the quadric surface $\sum y_i = \sum y_i^2 = 0$ in $\Bbb P^4$ and the reduced Galois group A5 acts on the two families of lines in this doubly-ruled surface by permuting coordinates.

The $A_5$ actions on these families, parameterized by $\Bbb P^1$, are equivalent to the action of the group of rotations of an icosahedron on its circumsphere and the quintic thus defines a point in the quotients — the icosahedral invariants of a quintic.

@Alessandro: It follows more "directly" from something like Urysohn metrization. But meh.

Yeah I would have used Urysohn's theorem

Blah.

I mis-edited and time is gone.

These are not interesting questions, @Alessandro.

If you're going to work with manifolds, the metric space structure that comes from the Riemannian metric and geodesics is more interesting and important.

I see

@TedShifrin So this action of $A_5$ on $\Bbb P^1$ is the action of $A_5$ as group of symmetries on the icosahedron, imagined as 12 regularly spaced points on the Riemann sphere, right?

The quotient is the compactified modular curve $X(5)$, I believe.

What happens if you allow algebraic numbers instead of only rationals as coefficients for the polynomial? Are the solutions still algebraic numbers?

Yes, algebraic numbers are algebraically closed.

OK :-)

If I recall right, what they do is find invariants of this $A_5$ action on $\Bbb P^1$ on the homogeneous coordinate ring $\Bbb C[z, w]$ of $\Bbb P^1$, which turns out to be $\Bbb C[X, Y]$ for some polynomials $X(z, w)$ and $Y(z, w)$ and if $Z(z, w)$ is the polynomial of degree $12$ whose roots are the vertices of the icosahedron, then it turns out that $X^3 + Y^2 + Z^5 = 0$

And the homogeneous coordinate ring of the quotient turns out to be $\Bbb C[X, Y, Z]/(X^3 + Y^2 + Z^5)$

That's the modular curve

Yeah, I'm disappointed in myself for not having paid attention to this decades ago.

I had the impression that this stuff was too niche; I remember presenting this to Mahan once

This is when he told me learn about covering spaces, because I intuitively knew what $\Bbb P^1 \to \Bbb P^1/A_5$ being a branched covering means but didn't know the definition :p

The history of a @Balarka :P

more like how I wanted to learn Galois theory but got tricked into doing topology instead

i need to learn galois theory seriously for next semester

we have a course in it

Is there any simple way to describe the structure of solvable groups?

I rem. @LeakyNun had once a go in explaining to me, but its all gone again

What I kind of like is "the next nicest after Abelian ..."

yeah they are groups you can "break up" into abelian groups

like in product sense?

Almost

sigh

It would have been too nice

it means there is a tower of subgroups $1 = G_0 \leq G_1 \leq G_2 \leq G_3 \cdots \leq G_n = G$ such that $G_i$ is normal in $G_{i+1}$ for all $i$, and each $G_i$ is abelian

so $G$ is "almost" like $G_0 \times G_1/G_0 \times \cdots \times G/G_{n-1}$

but except those products are "twisted products"

@BalarkaSen whats that?

just some vague term i made up

its like the difference between taking an annulus and a moebius strip if you wish

simple example?

i just gave you one :P

an annulus is product of a circle and an interval, the moebius strip is a twisted product of them

can we have a small finite group that is a "twisted product" of two other groups?

Yes. Consider $S_3$, and the normal subgroup $\Bbb Z_3$ given by the 3-cycles. The quotient $S_3/\Bbb Z_3$ is isomorphic to $\Bbb Z_2$

So you're tempted to say $S_3$ is a product of $\Bbb Z_2$ and $\Bbb Z_3$, but it's not because it's nonabelian. The resolution is it's a "twisted product"

For googling purposes for @Rudi, it's officially called a semidirect product.

This situation is not a full picture because the "twisting" in some sense is "controlled" here (it's what is known as a semidirect product of $\Bbb Z_2$ and $\Bbb Z_3$), but it can be more complicated and untractable

Ted sniped me

Well, of course

@TedShifrin Ah!

@BalarkaSen ok. I suppose its somehow connected with the relation between conjugation and commutation.

@BalarkaSen wait, the multiplication of the two groups is non-commutativ, right?

no that doesn't make sense

Z_2 and Z_3 are abelian groups. Not sure what you meant.

So what would be the direct product of $\Bbb Z_2$ and $\Bbb Z_3$?

Z_6

I see :-)

I have to work that out in detail and think about it

Sure $\Bbb Z_6$ factors out nicely.

That only works, though, when $2$ and $3$ are relatively prime. It wouldn't work for $\Bbb Z_2$ and $\Bbb Z_2$ or $\Bbb Z_2$ and $\Bbb Z_6$.

(Interrupting yet again)

Yes makes sense.

This is (one of the approximately $\aleph_{17}$ versions of) the chinese remainder theorem (still for googling purposes)

Alessandro exaggerates.

Chinese remainder theorem is one of the greatest theorems in mathematics.

nods

What is the essential difference between S3 and Z6 without mentioning "abelian".

does $S_3$ have an element of order $6$?

@BalarkaSen for some reason I always hated it, maybe thats why I didn't study maths

@TedShifrin yes

what's the element?

@Rudi_Birnbaum no

S3=S6

Huh?

=S720

Well the full permutation group of 3 elements is sometimes called S3

I think, right?

That's what we're talking about.

Then you said something about $S_6$, which is a group of order $6!=720$.

@TedShifrin No that was @LeakyNun

No, you wrote S3=S6.

exactly

And I repeat ... Huh?

Since we started with comparing S6 and Z6

No, we were talking about $S_3$, not $S_6$.

but then I switched to the notation S3

You can't invent your own notation here. The symmetric group on $n$ letters has $n!$ elements.

Oh sorry

my mistake

I got cofused

And I asked you if $S_3$ has an element of order $6$.

It's OK.

I am sorry!

You told us not to mention the word "abelian."

So I asked a different question.

yes

OK in S3 we have a generator element of order 6

Think about the symmetries of an equilateral triangle (that's the group $S_3$).

We're not in $S_6$.

corrected

No, no, you're not thinking of the right group.

oh?

I see elements of order 3 (rotations) and elements of order 2 (reflections). I don't see anything else (well, the identity).

in $S_3$

The CRT is great, I feel like I'm not using it enough though

@Balarka I remember some theorem of the form "every group in this large class embeds into one with two generators" or something along this lines, do you have any idea if I'm remembering something sensible or if I'm already too old to trust my memory?

@TedShifrin in 2D or 3D?

@Thorgott: If you ever use $R/IJ \cong R/I\times R/J$ when $I+J = R$, you're using it :P

in 3D it would be $D_{3h}$.

The dihedral group of order 6 is the same as $S_3$.

@AlessandroCodenotti I know every hyperbolic group which is not virtually cyclic contains F_2, but I don't know about embedding in F_2

Free groups, @Alessandro

Oh, you said embedding in some quotient of F_2

That shocked me when I first learned it, studying for my algebra qualifying exam.

I like the fact but maybe some larger class is intended by Alessandro

@TedShifrin Yeah and in chemistry we call that "S_6" thats part of my confsion...

In chemistry these groups are called $S_{2n}$...

Terrible notation. Regardless.

Schönflies

So if chemists ever try to study group theory from a mathematician or mathematics book, they're hopelessly confused.

For us $S_n$ is the symmetric group on $n$ letters, so all possible permutations of $n$ objects.

I know that, I had a collaboration with @mercio ...

Anyhow, $\Bbb Z_3\times\Bbb Z_2$ has an element of order 6, and $S_3$ does not.

@BalarkaSen hey you're back

Does a @Balarka ever leave? :D

Well we call that element "improper rotation"

Shockingly, I'm not using that fact all that often

Well, it's probably cause I'm not doing much ring theory

gernot-katzers-spice-pages.com/character_tables/S6.html

@BalarkaSen did you watch Rosen's 0-0# video?

Yeah lol

@BalarkaSen hmm maybe it was something like "every group with two generators and some conditions on words is hyperbolic"

@LeakyNun lol I saw that too

you play chess also?

Its a compostion of a rotation by 120° and an inflection.

I have no idea what you're talking about.

Though I guess I've used the structure theorem for f.g. abelian groups recently and that's based on the CRT

@LeakyNun I mostly blunder pieces on lichess but I guess you could call that playing

@TedShifrin en.wikipedia.org/wiki/Improper_rotation

@AlessandroCodenotti you wanna play?

Sure

time control?

I usually play 5+3 but whatever you prefer

snap

@AlessandroCodenotti lichess.org/61p5aEso

I know about rotatory reflections, but for a triangle I don't see how you get more than 6 elements for the symmetry group.

im watchin it yall

@TedShifrin right.

This is the group I described 15 minutes ago — 3 reflections, 2 rotations, and the identity.

OK, right.

@BalarkaSen I can see that

3 elements of order 2, 2 elements of order 3, and the identity. But $\Bbb Z_2\times\Bbb Z_3 = \Bbb Z_6$ has the identity, 2 elements of order 6, 2 elements of order 3, and 1 element of order 2.

@TedShifrin I think your picture is for 2D objects while "our" $S_6$ is for 3D "dihedrons".

@TedShifrin OK.

Its different groups, clearly.

@Alessandro wow thats stronk

But abelian versus nonabelian is the most basic observation.

Yeah sure. So how does this semi-direct product work?

So using generators an example would be

I suppose the direct product would be where $ab = ba$ instead of $aba^{-1}=b^{-1}$, right?

gg

noice

the kingside attack was crashing

@BalarkaSen wait I wasn't looking at the chat, what was that about?

@BalarkaSen we're having a rematch if you want to watch

when queen moved to attack the rook and pawn at the same time

@LeakyNun I'd love to but I need to read some Galois theory lol

what kind?

@LeakyNun Nice game, I got worried I messed up when you captured my light square bishop with the knight

Plain vanilla

instead I messed up lol

Just doing all Dummit-Foote exercises

(I looked at SF afterwards)

@AlessandroCodenotti wow I haven't fallen for that in a look time lol

@LeakyNun I haven't checked it with the engine yet

@Ted ?

lol I missed f5 @leaky

@AlessandroCodenotti gg

pretty sure that endgame was winning if I knew how to play endgames :P

Maybe I should just not sacrifice bishops for dubious attacks though

maybe lol

oh wow you missed mate in 1 XD

@LeakyNun Anyway the engine should agree that taking the rook wasn't good there, because after Bxh1 Qxh1 castling is super dangerous for white since you can put a bishop on h3 and already have the queen on the long diagonal

@LeakyNun I often do that too :P

@AlessandroCodenotti oh I didn't know that

yeah I just checked with the engine, it says to just castle (which is what I did)

(the English is the one opening I know a little theory about)

@LeakyNun that's actually amazing because on move 63 I played Re7 hoping you wouldn't see that very mate, but I hadn't seen it a few moves earlier lol

rip

I have to go study some maths now

But I followed you on lichess, I'm often online if you want to play a game!

ok

me too

@BalarkaSen anything interesting?

Wondering what the ring of integers of $\Bbb Q(2^{1/3}, \zeta_3)$ looks like.

oh well

lol

i thought this is the stuff which sparks your interest

yeah

I think it's the obvious one

@BalarkaSen Is "ugly" among the possible answers?

I think I computed $\Bbb Q(2^{1/3})$ once

@Alessandro hah

ok let's look at the discriminant of $\Bbb Z[2^{1/3}]$

the different is generated by $(f'(a))$ where $f(x) = x^3-2$ and $a = 2^{1/3}$

so $(3 \cdot 2^{2/3})$

the norm of that is $2^2 3^3$

so $[\mathcal O_{\Bbb Q(2^{1/3})} : \Bbb Z[2^{1/3}]] \mid 6$

so we check $\frac12 \Bbb Z[2^{1/3}]$ and $\frac13 \Bbb Z[2^{1/3}]$

i.e. 2^3 + 3^3 = 8 + 27 = 35 element

using additive symmetry we can reduce 3^3 (i.e. {-1, 0, 1}^3) to 2^3 (i.e. {0, 1}) so 8 + 8 = 16 elements to check

$\{\frac12, \frac13\} \{0, 1, a, 1+a, a^2, 1+a^2, a+a^2, 1+a+a^2\}$

some of them imply other

clearly $\frac12 a^2$ and $\frac13 a^2$ are not integers (take norm)

so $\frac12 a$ and $\frac13 a$ and $\frac12$ and $\frac13$ are also not integers

I think taking trace eliminates all the $\frac12$ ones

so now it suffices to check $\frac13(1+a^2, a+a^2, 1+a, 1+a+a^2)$

$a+a^2$ is $\begin{pmatrix} 0 & 1 & 1 \\ 2 & 0 & 1 \\ 2 & 2 & 0 \end{pmatrix}$

determinant is 6, so eliminated

$1+a$ implies $a+a^2$ so $1+a$ is eliminated as well

$1+a^2$ is $\begin{pmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 0 & 2 & 1 \end{pmatrix}$

determinant is 5 so eliminated

$1+a+a^2$ is $\begin{pmatrix} 1 & 1 & 1 \\ 2 & 1 & 1 \\ 2 & 2 & 1 \end{pmatrix}$

determinant is 1+2+4-2-2-2 = 1 so eliminated

so $\mathcal O_{\Bbb Q(2^{1/3})} = \Bbb Z[2^{1/3}]$

now we add $x^2+x+1$ to it

comments wryly: This has to be the longest comment in history :P

@TedShifrin surely your calculations are as long if not longer :P

I have a question. If we are given an abelian topological group with countable neighborhood base chain $G_1\supset G_2\supset...$ of subgroups at $0$, how to prove that every Cauchy sequence must eventually lies in one of $g+G_n$?

so $f'(\omega) = 2\omega+1$ has norm $(2\omega+1)(2\omega^2+1) = 5+2(\omega+\omega^2) = 3$

In other words, eventually constant in every $G/G_n$.

This is used to construct the inverse limit of $...\to G/G_{n+1}\to G/G_{n}\to...\to G/G_1$.

you can prove something stronger

for a given Cauchy sequence, for every $n$ there is a $g$ such that the Cauchy sequence eventually lies in $g+G_n$