@Secret hi

Oh hey @LeakyNun

I have a fun problem about Galois theory for you

well, don't leave us in suspense :P

Let $f$ be a separable degree $4$ polynomial over a field that is palindromic, show that $\operatorname{Gal}(f)$ is a subgroup of $D_8$

Neat. (I don't understand galois theory.)

@MatheiBoulomenos palindromic?

in this case, it means that it is of the form $ax^4+bx^3+cx^2+bx+a$

say what

so $x^4 p(1/x)=p(x)$.

it is closed under reciprocal

that's an important observation

@Semiclassical :c sniped

you run into that condition in orthogonal polynomials on the unit circle

though I never understood that well enough

@MatheiBoulomenos if there is one complex root, then it generates all other roots (V4 Galois group)

i guess the importance here is that if $p(x)=0$ then so does $p(1/x)$

hah, now i got sniped

@LeakyNun I don't think that's correct

provided the modulus ain’t 1

$x^4+x^3+x^2+x+1 = \Phi_5$ is palindromic and it's Galois group is $\mathbb{Z}_4$

should that be understood as "polynomial...that is palindromic"?

yes

@MatheiBoulomenos sniped

the relative clause was not really well-placed, I'll admit that

np

if I'm thinking of this momentarily as a polynomial with real coefficients

by the way @LeakyNun the problem works over an abstract field as well, so strictly speaking you can't reason about complex or real roots etc.

yeah

so the following doesn't cover the full scope

@MatheiBoulomenos complex just means degree 2 for now

Hi @BalarkaSen

Hi @Mathei

is there a solution without casework?

Just came back home after the ODE lecture

well, you don't have to do casework on the roots or something

a subgroup of S4 whose order is divisible by 3 must contain a 3-cycle

but reciprocal is an involution

@Semiclassical was right that the important observation is that if $p(x)=0$, then $p(1/x)=0$

the non-degenerate cases are: the roots are all real and come in reciprocal pairs; there's a reciprocal pair of real roots and the others are $x=\pm i$; there's 4-group of complex roots closed under reciprocals and conjugation

that's in the real case.

@Semiclassical i?

yeah, so that 1/i=-i

thinking about reciprocal pairs is the right approach

once you go to a generic field, though, I would imagine that the conjugation aspect goes away

but 1/exp(ix) = exp(-ix) in general

so yeah, reciprocals

it doesn’t have to be i

it does if there's to be exactly four roots, two of which are real

I disagree

wait, I agree

:)

no, I disagree

lol

if there's exactly two complex roots, then they must be closed under both conjugation and reciprocation

that can happen when the modulus is 1

x^2+3x+1 for example

I think that the whole conjugation/real/complex/modulus thing is distracting

@MatheiBoulomenos did you see my rant above about 3-cycles?

Blah, you’re right

Let the 3-cycle fix a. Then, r(1/a) != 1/a

as 1/a is a root, so it gets teleported

but then r(a) r(1/a) != 1, contradiction

oh, that's a clever approach

I think that works

unless a=1 lmao

but it’s separable

so 1 can’t be doubled

@MatheiBoulomenos what’s yours then

(cc @BalarkaSen for a third approach

i dont know what the question is

wait, no I think that doesn't work

how do you quote in chat?

and palindromic means $f(x) = x^4f(1/x)$?

yes

@MatheiBoulomenos warum nicht?

ach nein, funktioniert doch, ich bin dumm ^^

lol

I was unsure if there is a subgroup of $S_4$ that doesn't contain any 3-cycle, but is not contained in a dihedral group

if it’s divisible by 3, it must have an element of order 3 (Cauchy), which is a 3-cycle (cycle type must be 3)

But it follows from the Sylow theorems that this is not possible

so it cannot be divisible by 3

but then 24 taking away factors of 3 leave you with 8

D8 is the only group of order 8 that embeds into S4

@MatheiBoulomenos Cauchy :)

if the order is not divisible by 3, then it is a 2-group thus contained in the 2-Sylow subgroup which is D_8

oh, nice!

this is a cool question but i am too tired to think about algebra right now

that’s a shortcut with Sylow

hi guys

anyway what was your originell approach?

Well, the original approach was a bite more tedious than the argument with $3$-cycles

I have a quick question. Do you think khan academy is good for learning algebra, or should i find a book on it? and if so, what book?

I was just thinking about the roots in conjugate pairs $a,1/a,b,1/b$

tell us @MatheiBoulomenos

sorry for the noob question

@Novantix what algebra?

high-school algebra?

basic algebra, just 1 and 2

Khan is good for that

note that since $a \times \frac{1}{a} = 1$, and analogously for $b$, the Galois group must take a reciprocal pair to another reciprocal pair

okay great, thanks.

@MatheiBoulomenos hmm

The only permutations that do that are $\{ (), (1,3,2,4), (1,2)(3,4), (1,4,2,3), (1,3)(2,4), (1,4)(2,3), (1,2), (3,4) \}$

ich denke noch, dass mein approach besser ist :P

"approach" heißt "Ansatz" oder "Herangehensweise"

her-an-gehen-s-weise

warum, o deutsch

long words are fun :P

was bedeutet herangehen?

Naja, wenn man es wörterlich nimmt heißt "an etwas herangehen" sich einer sachen nähern, also "to go closer to something", aber es wird meistens im übetragenen Sinn verwendet, dann heißt es einfach "to approach something"

@MatheiBoulomenos danke

I just realized why I should do that. lol

$\Bbb Z/2 \times \Bbb Z/3$

basically $\Bbb Q(a,\sqrt2) \cong \Bbb Q(a+\sqrt2)$

exactly, we were talking about that yesterday

lol

do you have mehr Uebung ueber Galoistheorie?

Nice exercise: Construct a Galois extension over $\Bbb Q$ with Galois group $Q_8$

@BalarkaSen I thought about that

I wanted to ask that in MSE

but it turned out to be a duplicate

Pretty sure it's everywhere on the internet.

but I don't exactly rmb the answer so I'll think about it anew

"exercise" heißt eher "Übungsaufgabe" in diesem Kontext

uebung-s-auf-gabe

Can someone tell me how author got second equation from the first? I think it's algebraic manipulation and I tried it but couldn't reach equation 2 from 1.

du kannst auch einfach nur "Aufgabe" sagen

alright

last time I thought about it, I built to the $\Bbb V_4$ part just fine

I mean, $\Bbb Q(\sqrt2+\sqrt3)$ is trivial

Products and semidirect products are easy to build.

But Q_8 is neither

also

It is not a split extension of anything by anything

I'm not sure which direction I should be building

Is the top one $\Bbb Q$ or the bottom one?

I think it's the top one because it fixes only $\Bbb Q$

the top one is $\mathbb Q$

alright, so I built to the $\Bbb V_4$ part

now to extend it normally so that you can't take away other elements

Would $\Bbb Q(\sqrt{\sqrt2+\sqrt3})$ be it?

The correspondence is indeed given by $G \mapsto E^G$.

I'm not even sure what the smallest $n$ is such that $Q_8$ embeds into $S_n$

@Mathey 8

@MatheiBoulomenos it must be $8$

From Cayley, we get an embedding for $n=8$

@BalarkaSen why not 6?

so we need a degree 8 polynomial

This is an exercise in Dummit-Foote. I have forgotten how to do it.

It's probably a class equation argument

note that $-1$ is contained in every non-trivial subgroup of $Q_8$

Ah yeah that's it, Z/2 is the intersection of all the nondumb subgroups of Q_8

Or look at the 2-sylow in S6

<(12),(34),(56)>

Quote from book: "It is easy to check that the composite of two quotient maps is a quotient map map; this fact follows nicely from the equation $p^{-1}(q^{-1}(U)) = (q \circ p)^{-1}(U)$." Throughout the chapter, we have been assuming that $p$ is a mapping from $X$ to $Y$, so I figured that $q$ would be a mapping from $Y$ to, say, $Z$, but this doesn't seem the make sense; the domains and codomains don't match up.

What is the domain and codomain of $q$?

this means, if $Q_8$ acts on a set, then $-1$ is contained in every non-trivial stabilizer. but from orbit-stabilizer if $Q_8$ acts on a set of less than $8$ elements, every stabilizer must be nontrivial

thus the action cannot be faithful

@MikeMiller how do see that $Q_8$ is not a subgroup of that?

@MikeMiller that's not the 2-sylow in S6

@MatheiBoulomenos Q8 is not a subgroup of <(12),(34),(56)> because the latter is abelian

right

but as the 2-sylow of S4 is D8, the 2-sylow of S6 has to contain D8, so it cannot be abelian

a 2-sylow of S6 is <(12),(34),(1324),(56)>

because drawing 2-ary tree blah blah

can this be written as a wreath product?

it is $(\Bbb Z_2\wr \Bbb Z_2)\times\Bbb Z_2$

thanks

@anon I got in the shower right after sending and was cursed to know I said something wrong with no way to correct it

how do you see that Q_8 is not a subgroup of <(12),(34),(1324),(56)>?

calculate the index 2 subgroups :)

A lot of elements of <(12), (34), (1324), (56)> commutes right

the vast majority indeed

whereas center of Q8 is like {1, -1}

something like that should play out an argument

I just prefer to work in this much smaller group than S6 vOv

@Mathei About the palindrome problem, I think you can look at the tower of extensions Q(a, b, 1/a, 1/b)/Q(a+1/a, b+1/b)/Q. The middle extension/Q has Gal = Z/2 and the top/middle has Gal = Z/4 I believe

In which case you get an extension game again

kind of an attempt to simplify the tediousness in your calculation...

Is there a naming convention regarding the subscript variable indicating the size of a set? So for example M={m_1,...,m_n} , here n indicates the size of the set, but using letters here gets messy quickly, since I have a lot of sets of different size that I need to include in the formalization of my work

Sorry, might be weirdly put but I'm not sure how to explain it better

ah my brain is mush, it's actually a vector. Can't edit my message anymore though

If $n=n_kp^k+\cdots+n_1p+n_0$ in base-$p$, the $p$-sylow of $S_n$ will be $\bigoplus n_k(\Bbb Z_p^{\wr k})$, where $$nG:=\underbrace{G\oplus\cdots\oplus G}_n, \qquad \Bbb Z_p^{\wr k}=\underbrace{\Bbb Z_p\wr\cdots\wr\Bbb Z_p}_k$$

my internet conked out for the last so many minutes

Wow, that's really cool @anon

@BalarkaSen that should work, but it's not really that tedious to see which permutations preserve the reciprocal pairs

Uh, yeah, fair enough :)

I am just lazy

and you use trees to prove that? @anon

@MatheiBoulomenos yeah, IIRC you make $n_k$-many $k$-depth $p$-ary rooted trees; there will be a total of $n$ nodes at the bottom, and the direct product of their automorphism groups will be a permutation subgroup of $S_n$. cardinality tells us it's a $p$-sylow

Wow, I've never seen graph theory get used in algebra before

Morning

Good afternoon

After Morning?

There is no such thing =/

pre-noon

O.o

Whooo, I'm in a photo together with Cédric Villani :3

gratz @TastyRomeo

Which is more important for algebraic number theory: group cohomology or algebraic K-theory?

surely group cohomology, I believe

or rather, galois cohomology

thanks

@BalarkaSen how do you prove that $K(G,n)$ is unique up to homotopy?

Whitehead's theorem is the relevant fact

I was thinking that if $G$ is abelian, you can use Brown representability and Yoneda, but that doesn't work if $n=1$ and $G$ is non-abelian

@MatheiBoulomenos It does, you just gotta interpret $H^1(X; G)$ for nonabelian $G$ as a set, not a group

But it's actually easy to construct a map $ f: X \to K(G, 1)$ corresponding to a map $F : \pi_1(X) \to G$ such that $\pi_1(f) = F$

You do it skeleton-wise

Namely, construct $K(G, 1)$ as a boquet of circles corresponding to generators of $G$, add 2-cells corresponding to relators, and higher cells that kills higher homotopy

Now define $f : X^1 \to K(G, 1)$ by sending each of the circles in $X^1$ to paths on the graph $K(G, 1)^1$ as defined by the values of $F$ on the generators of $\pi_1(X)$

thanks

$f$ should extend to the 2-skeleton $X^2$. Because if $D^2$ is a 2-cell in $X$ glued to $X^1$ along some graph-path $\gamma$ in $X^1$, then $F([\gamma]) = F(0) = 0$, so $f(\gamma)$ should also be nullhomotopic in the image.

Define $f$ over $D^2$ by just completing the nullhomotopy; do a suspension.

In this fashion you should be able to construct $f : X^2 \to K(G, 1)$.

And so forth...

Fun fact: I was taking a topology course and an algebra course in the same semester and we checked maps for continuity more often in the algebra course than in the topology course

lol

hehe

only mortals care about continuity

so topologists are immortal?

No, we just construct maps that are obviously continuous :)

"obviously"

Indeed, obviously.

Mhm.

our professor constructed a bunch of maps from certain $\operatorname{Hom}$-sets in the compact-open topology that were "obviously" continuous

\|What exactly deifnes obviously continuous specifically?

@MatheiBoulomenos Oh, there are point-set topological theorems which back the obviousness behind those maps

lol

Hom(X x Y, Z) \cong Hom(X, Hom(Y, Z)) etc etc

If you need a theorem, it's not obvious

and you need some conditions on $X,Y,Z$

Locally compact Hausdorff

we never stated the conditions

it was left as an exercise to figure out for which spaces the stuff works

Well thats not the same thing

That sounds like a pretty good exercise to me :)

I mean these kind of tedious point-set justifications do not belong to an algebraic topology course

I worked it out myself at some point of time

And that was waaaay after I learnt and actively used that result

I worked it out, emailed a friend who asked for the proof (that's why I worked it out in the first place lmao), and forgot about it.

but maybe I'm just too bad at topology lol

that's cuz you have an algebraic brain

that must be it

"It's only true for 'nice enough' $X$ (that is, $X=S^n$ for some $n$)"

well, I mean it's also true if all the spaces are one-point sets

It would be interesting to see a theorem that's only true for manifolds of high enough dimension, and it turns out that "high enough" means thousands

that would be hilarious

@Mathei It seems you only need compactly generated Hausdorff.

How do you visualize compactly generated?

i don't really think hom-spaces with compact open topology belong in a first course, and I would never want to teach a rigorous treatment, but if one is going to teach them they should be rigorous; this happens in some algebraic topology books (davis and kirk do it, as does May) where they have a discussion of compactly generated weakly hausdorff spaces and various results on Hom-spaces

I don't lol

@AkivaWeinberger something something concentration of measure

@MatheiBoulomenos I just think about locally compact tbh

e.g. arxiv.org/pdf/1310.1204.pdf

fair enough

it also includes all metric spaces though haha

so including any Hilbert space / manifold you care to think about

Diff(M) I think is metrizable and hence CGH

I think high dimensional in that context isn't on the order of thousands, though

it's just like, anything that doesn't seem stupid

that sounds like a physicist definition :P

@MikeMiller Well, M is compactly generated Hausdorff, so Hom(M, M) would be CGH, and Diff(M) with subspace topology would be CGH too, right? I think it's an open subset

everything w/ compact-open topo

anything that is not CGH is stupid?

so most schemes are stupid

Diff(M) does not have subspace topology

Oh. What topology do you give it?

@MatheiBoulomenos Schemes have much more than a topology :) The understanding of space in the two areas are very different.

@BalarkaSen One should also account for how close derivatives are.

@MatheiBoulomenos The Zariski topology is a dumb topology; only the structure sheaf is actually interesting because that is what parallels the structure of an algebraic variety

the Zariski topology is not dumb!

To us topologists, yeah, they are

They are the generic examples of bad spaces

Don't get me wrong, I love the Zariski topology, but just not from the topology POV

you're talking about it as an algebraic gadget, i think i mostly saw the day before. it's a useful one! but it doesn't present usefully when thinking about the shapes human beings are usually more likely to first visualize (which you might identify as being roughly like CW complexes)

it doesn't give me an interesting shape in any obvious sense vOv

so I am not inclined to start talking about mapping spaces or homotopies or whatever in that level of generality

I don't really care about visualizing shapes. I think that's also why I'm so bad at topology

we all have our various different tastes in math

Is it possible to tell an employee "You're fired" and be lying?

@MikeMiller Ah, I see

(to all degrees)

Yes, I am not sure why I completely deleted the information about derivatives, that's silly of me

I have seen this in Hirsch, something something Whitney topology

Thanks

It's a very easy and natural mistake to make

Hausdorff is too restrictive imo

the only Noetherian Hausdorff spaces are finite

it is too restrictive if you're inclined to care about Noetherian spaces lol

I'm pretty sure most CW complexes are not even an inverse limit of noetherian spaces

@Mathei Who cares?

CW complexes are not meant to be algebraically nice objects. They are meant to be topologically nice objects

affine schemes are an inverse limit of noetherian affine schemes

CW complexes are not meant to be affine schemes!

Not everything in the world is an affine scheme, man

affine schemes are useful and important and interesting objects, but they just don't exhaust the set of all useful and important and interesting objects out there

okay okay

profinite groups are CGH as well, so the category of CGH spaces includes some algebraically interesting objects, too

hi

hi @Diego

So, I had this problem of probability: In a bus there are 15 passenger and the bus makes 4 stops. Whats the probability that all 15 get off at the same stop?

In my first attempt, i said 'Well, each passenger, as a human being, chooses where to get off. So, for all to get off at stop 1, the probability would be (1/4)^15. Since there are 4 stops, the probability required would be (1/4)^14 which is approx. 0.0000003%