okay maybe you have to worry a bit about transcendence degrees in the latter case

ok he just means that there are two square roots of -1 and you can't tell which one is which

lol

seems pretty canonical to me

let's say I have the polynomial X^2+1 and splitting field R(α,β)

can you construct an isomorphism from C to R(α,β)?

okay I see why you call it non-canonical

but the non-canonicalness is very small

lol

you always have two choices for a choice of isomorphism and you don't need Zorn to prove that an isomorphism exists

for $\Bbb Q$ you have uncountably many choices and without choice, you can't prove that two algebraic closures are isomorphic

I see

2. and 3. looks like it should just be some general result in group cohomology

really

not quite sure, but it looks like it

$H^1$ certainly has a lot do with semidirect products

hmm

I know that you can show that $H^1$ is isomorphic to some subgroup of the automorphism group of a semidirect product

hmm

@Mathei I think you should look at Hovey's book for model categories; it might be nice to know something about simplicially sets. There is a really great paper called "six model structures on dg-modules" by Bartels, May, and Riehl

Riehl's book on abstract homotopy theory is very good and probably up your alley

So say you have $1 \to A \to A \rtimes G \to G \to 1$ where $A$ is abelian. Then $H^1(G,A)$ is isomorphic to those automorphisms of $A \rtimes G$ that fix $A$ and reduce to the identity in $(A \rtimes G)/A \cong G$

@MatheinBoulomenos how tf is the galois group big enough to contain the units of its finite extensions, this makes |no sense at all|

I think that the model categorical perspective is not truly crucial for most basic algebraic topology, but May's second book on the subject gives an intro

that's a classic in result in group cohomology and you can find it proved in Rotman (Weibel has it as an exercise)

(Speaking from a position of ignorance I think it's probably not really crucial until you talk about the homotopy theory of operads or stuff like that, where comparing homotopy theories is particularly important)

@LeakyNun not sure how helpful that is, but it seems related

at least that's what I could think of

hmm

@MikeMiller thanks for the recommendation

I feel less comfortable making recommendations in front of actual experts even though that's silly

I don't really know what do next in terms of algebraic topology. I covered the stuff which I feel is standard, i.e. fundamental groups, covering spaces, (co)fibrations, Eilenberg-Maclane spaces, Postnikov systems, the usual (co)homology theories (singular, simplicial, cellular, axiomatic, I know some stuff about sheaf cohomology and de rham), Brown representability and some duality stuff for topological manifolds

Do you know what you care about? A reason to adventure further?

A goal is always helpful

Well, there is some recent stuff going by Hesselholt, Scholze and others which seems to be on the intersection of AT and arithmetic geometry

maybe understanding that

eventually, at some far point in the future

Also if I understand correctly, Lurie et al. have variants of commutative algebra, noncommutative algebra and algebraic geometry, those $A_{\infty}$ and $E_{\infty}$ stuff and spectral schemes

math.stackexchange.com/questions/2806547/… Can you please check whether my comment correct or not?

that seems really interesting

@MikeMiller so I guess, learning about simplicial sets might be some thing that's relevant and in reach?

Oh definitely. I don't know a reference to suggest

Lurie recommended May's book somewhere on MO

Their combinatorics is quite important

I have things to say about your further suggestions but I would need to think about what exactly to say instead of just rambling

@MatheinBoulomenos "Simplicial objects in..."?

yes

I've never read it but I like the table of contents

@MatheinBoulomenos do you think there is abstract proof 2 <-> 3 using the definition of a semidirect product being a split epimorphism?

btw I don't really see how what you said earlier is relevant

@LeakyNun what is a 1-cocycle in terms of category theory?

I frequently feel from what I know/read that one of the thrusts of modern homotopy theory is localization as a tool and object of study (I have never found a good general reference on the subject written from this POV, though). May & Ponto "... Localization and Completion" is a good book for learning the simplest kind (localization of spaces at a set of primes) from my memory, it's not written to my taste but I bet it is written to yours

oh, that sounds like commutative algebra, nice!

@MatheinBoulomenos I don't really mean a categorical argument, I just mean an argument relying on the fact that a semidirect product is nothing more than an SES with a section

The technical book on model categories that my homotopy friends hate but find important is Hirschorn's "Localization of model categories" - I would guess that one important feature of model categories is that they are a way of presenting "homotopy theories" that can be localized

I don't know anything about that though

@MatheinBoulomenos instead of blindly unfolding definitions

I feel like there's something to semidirect products

@LeakyNun oh yeah that should work I think

heh, what should work

(infty,1)-categories a la Lurie can also be localized (I think the catchphrase there is Dwyer-Kan localization) and I remember this being important in a little of Nikolaus-Scholze I read

crossed homomorphisms (another name for 1-cocycles) are homomorphisms in a semidirect product

is there a homological argument?

is there a generalization in the dimension of the cohomology?

(the best test for my first question is the second question, I think)

is it an LES argument?

but my personal feeling is that infty-categories in general is young and technical and hard to get into, but will probably be important and simplified in the future

I did end up just rambling

@LeakyNun encyclopediaofmath.org/index.php/Crossed_homomorphism

@MatheinBoulomenos is it just a coincidence, or is there a higher connection between cohomology groups and semidirect products?

not sure what you mean by higher connection

it looks like a coincidence to me

nobody would care about group cohomology if it didn't sometimes "coincide" with stuff we actually care about

but I mean, the definition of the H1 has nothing to do with semidirect products to begin with

how do they just coincide

there must be something behind it

The definition of H^2 has nothing to do with Brauer groups

H^* is just Ext

or extensions of groups

how does Ext have to do with anything

btw it's W(K/F) not Gal(K/F) that is the domain

okay yeah, it still felt similar to what you want

and all these mir suggest that there is a connection between them

Is L/K finite?

@MatheinBoulomenos yes

I mean K/F

then what's the difference between W(K/F) and Gal(K/F)?

@MikeMiller rambling or not, I did find this helpful and interesting, so thanks!

@MatheinBoulomenos 1 -> K^* -> W(K/F) -> Gal(K/F) -> 1

W(K/F) := W(F)/[W(K),W(K)]

1 -> W(K)/[W(K),W(K)] -> W(F)/[W(K),W(K)] -> Gal(K/F) -> 1

hmm

that seems strange

I haven't seen those sequences

sunsite.ubc.ca/DigitalMathArchive/Langlands/pdf/…

sehe Seite 2

@LeakyNun there's a proof sketch how you show that 2. and 3. are equivalent on that page

but it's too specific

it's hard to believe that the correspondence holds when you don't impose the commutativity condition like on page 2

Gal(F/K) isn't mentioned at all in 3

math is hard

okay kids

:O

I need to understand cohomology as well

the Cech cohomology

no, math is easy, we're just all really dumb

@MatheinBoulomenos W(K/F) acts on T-hat through Gal(K/F)

hi @Daminark

have you eaten

actually I need someone to teach me Cech cohomology

I'll pay them even

just go to Czech

@LeakyNun I'm at the place where I'm getting some food, and I've finally had some water to drink so there's that

nice

DAMI

How's everything going with you? Also hey @Fr

@ForeverMozart, @GFauxPas, and @MatheinBoulomenos!

@LeakyNun but that action isn't mentioned in 2. and 3. In 2 and 3 Gal(K/F) is just any group. Are you sure you don't want the composed map $W_{K/F} \to \widehat{T} \rtimes \operatorname{Ga}(K/F) \to \operatorname{Gal}(K/F)$ to be the standard map you have?

just sent a paper off with some really 80's-retro stylized graphics

aesthetic is what I covet

wwwf.imperial.ac.uk/~buzzard/maths/research/notes/…

part 3 @MatheinBoulomenos

@LeakyNun is that your paper

no

@LeakyNun in part 3, I read "We're interested in continuous homomorphisms $W_F \to ^L T$ such that [insert diagram] commutes"

and then he goes on to state the equivalence

yes and he always works with "Weil representations" which do satisfy the commutativity condition

oh you need the diagram to commute

yes, that's what I've been saying the whole time

you didn't mention that

@ForeverMozart amazing

@MatheinBoulomenos sorry

Is Z-hat a subgroup of itself?

or a quotient of itself?

in some trivial ways, yes

@Daminark yes. the journal will probably reject. but not if they like neon ;)

@MatheinBoulomenos such as?

it's the quotient by the trivial subgroup

and the entire subgroup

oh, it is actually a subgroup of itself in some nontrivial ways

@Daminark proofwiki.org/wiki/Existence_of_Set_of_Countable_Ordinals can you think of ways to use more zfc axioms

maybe I can get all of them in one proof

So think of $\widehat{\Bbb Z}$ as $\prod_p \Bbb Z_p$

for any $n \in \Bbb N$, $p^n\Bbb Z_p$ is isomorphic to $\Bbb Z_p$ itself

so for any sequences of natural numbers, indexed by the prime numbers $(a_p)$, the subgroup $\prod_p p^{a_p} \Bbb Z_p$ is a subgroup that is isomorphic to $\widehat{\Bbb Z}$

@LeakyNun so actually, $\widehat{\Bbb Z}$ contains uncountably many diffrent subgroups isomorphic to itself

I see

@GFauxPas use large cardinals!

I think any open subgroup is isomorphic to $\widehat{\Bbb Z}$

(but those examples I gave are only open when $a_p = 0$ for all but finitely man $p$

@GFauxPas I thought we established that choice is not necessary

@LeakyNun let $X$ be a non-empty set. By the axiom of choice, there exists some $x \in X$

ok

i like elevator music and vaporwave

how do you prove that without choice?

you need to choose some element

> I make elevating music, you make elevator music

@LeakyNun are you a rapper?

nah

why is my name changed to Kenny Lau

I see why

@LeakyNun It's not hip-hop, it's pop

lol

@MatheinBoulomenos sure!

I think 8a's in the other room seems to really know his stuff

@LeakyNun choice isn't necessary but its easier

what's a large cardinal

im not even doing cardinal arithmetic, that's just what the set is called

the set of countable ordinals

would prefer it had a different name so I can use it without people thinking about ordinal arithmetic but that's what it's called :(

All four options are correct, right?

well if $\ell$ doesn't appear in the symbols where is it implied?

regarding (C) and (D)

i'm not answering your question, i'm asking a question back

oh, so c,d only says convergent, not convergent to $\ell$.

Thank you very much

sure

@GFauxPas, two spheres $(x-1)^2+(y-2)^2+(z-3)^2=1$ and $(x-3)^2+(y-1)^2+(z+1)^2=16$ do not have any tangents right?

Because when I draw their intersection with x-y plane, second circle 'contains' first one.

I can't tell by looking at the equation

/shrug

what would tangent mean?

One point of intersection?

Huh??

wha???

Lolll

@Zee my man!! (denzel washington voice)

@NicholasRoberts I think tangent plane

Anyway please check this: The number of spheres of radii $\sqrt2$ such that the area of each circle is $\pi$ is ...

I think $8$ is answer.

is this correct?

area of each circle? what does this mean in the context of the question

@NicholasRoberts no, this is new question!

I know

You have a sphere of radius radical(2) such that the area of each circle is pi. What does "each circle" mean?

oh, area of each circle of intersection with the three coordinate planes

Oh ok. Can you show your work to get 8?

The number of spheres of radii $\sqrt2$ such that the area of each circle of intersection with the three coordinate planes is $\pi$ is 8, right?

rewritten.

@NicholasRoberts oh, sorry!

wouldn't such a sphere be invariant under shifting and roation? then the answer is either 1 or $\infty$?

oh that's different

@NicholasRoberts i did not do it rigorously, drew a mental picture.

Please let me know is the answer correct?

the sphere must intersect all 3 coordinate planes simultaneously/>

?

yes

Hm, not sure how to go about it rigorously at the moment

@NicholasRoberts no, its fine, just 8 is correct or not? I have a test of multiple choice questions

so, rigorous proof not reqired

I dont know. Whats your intuitive reasoning?

@NicholasRoberts Its omething like sphere with center $(1,1,1)$ and radius $\sqrt 2$ so intersecting circles have same radius.

now i can take centers like (-1,1,1), so in total 8 speres

How do you know radius will be pi?

no, i don't know, i said 'something like'

i believe a sphere with center (1, 1, 1) and radius radical(2) will not intersect all 3 coordinate planes simultaneously

Is anyone here comfortable guiding me through some basic Galois theory?

I need some basic help

@NicholasRoberts it will

@Cows whats the question? I just did some galois theory in grad school

ok well I have a polynomial

and it sounds like 8 is correct @Silent

let me write it down

hang on

@anon Thank you very much!

@NicholasRoberts $$2 t^6 - 2 t^5 - t^4 + t^2 + 1$$

So I want to factor this

I know that the fundamental theorem of algebra says I can do this in complex

but I wanted to construct a field extension of some sort

something like Q(something, something)

You want to construst a splitting field?

construct**

well I think the splitting field is well what it factors into no?

yes

The goal would be to factor this into two cubics, or a quadratic and a cubic. Then it will be easier from there

Did you check for rational roots?

I am not quite sure how to do this

I know it has complex roots

I think it should have rational roots no?

Do you know the rational roots test?

Is this what eisenstein criterion does?

No, eisensteidn is a quick test to see if a polynomial is irreducible

I don't know the rational root test

You take all factors of the constant term and all factors of the leading term

let me google it really quickly

and then consider all possible quotients. In this case, possible rational roots will be plus/minus 1 and plus/minus 1/2

ok, i see

hang on one sec, let me run through the test too

I checked it has no rational roots.

ok

so you were checking against all possible p/q

ok, so does this leave it to just complex roots?

Yes, in this case there are only 4 possible rational roots. This is often a quick way to see if a cubic polynomial is irreducible

yeah?

No, it could still have real (irrational) roots.

ah yes of course

There may be a clever way to factor this, but i dont see it at the moment

Man , I can’t decide if I should read some Wikipedia or jack it and go to sleep

What exactly is the question regarding this polynomial?

yes, I have pondered this for a while

My man!

well

I want to be able to attack an evil integral from hell

$\int dt \sqrt (2 t^6 - 2 t^5 - t^4 + t^2 + 1)$

My strategy is to exploit the splitting field in some manner

or something like that

the whole integrant is under the square root?

U sub biatchhhh

lolz

integrand*

yes

@Zee i can't tell if u sub was a serious remark

lol

Damnnnnnnn

yeah

What would a splitting field do for you here? This really only useful in computing galois groups of polynomials

I have a plan

Not serious

But why do you even care about this integral

Could you maybe get the integrand in the form of $\sqrt{1-u^2}$

then its an easy trig sub i think

@Zee the integral holds some secrets I care about. This is probably the best way I can explain it

Actually probably not the best approach.... hmmm....

@Zee is a crackpot mathematician. Do not listen to him!

hehe

will this integral get you laid?

if you solve it

hahahahahahahahahahahahahaha

Well , I could exert effort and think about this but since am not getting paid nor am I provided with any intrestic motivation (since it’s a “secret”) am afraid I cannot help

@Zee It will bring me happiness

What if this function has no antiderivative?

I am pretty sure it can be reduced to a collection of elliptic integrals

I am not sure how to immediately do it

Damnnnnn! (gta voice)

Happiness is within yourself not in an integral

My idea was to first factor a this

Happiness is computing galois groups

GF(?)

So let me see

@NicholasRoberts so you are saying this polynomial is irreduceable right?

No, we do not know that. All we know is that it has no rational roots.

It could possibly be factored into a product of two cubics, or a product of a quadratic and quartic with coefficients in Q

I am hoping to capture multiplicity of 2 somewhere so that I have some stuff slipping out of the root

then If I have a cubic or even quartic, I'd know how to beat it up

Yeah, im not sure how to factor this beast

it is mildly challenging

@Jan can you help us sort this wrinkle?

vaporwave is so good youtube.com/watch?v=OeAItlnXGNo&t=1471s

does anyone else here like V A P O R W A V E ?

ok

@BalarkaSen

I summon you

By the power of vaporwave

What is the dimension of $U=\{(a,b,c,d):a+b=c+d\}$ where $U\subset\Bbb R^4$?

@LeakyNun

@Daminark he listens to vaporwave?

Hmm, so the immediate guess would be 3 because you can sorta specify $a$, $b$, and $c$ to be anything you want, then you're forced on what $d$ should be

Formally we should try to write a basis this way

I've been listening to synthwave/80s retro for a long time, but only recently discovered this :(

its so good though

So first attempt at a basis is $(1,0,0,1)$, $(1,0,1,0)$, $(0,1,1,0)$

@Zee did you get any inspiration, or ideas about how to explore this thing plaguing me?

Call the first one $\alpha_1$, second one $\alpha_2$, and third one $\alpha_3$

We only need to specify linear independence, because the dimension already has to be strictly less than 4

I approve of daminark music

So @cows

yes

. .

So idk much physics but

Am reading about field theories

@Daminark Wow! thank you very much.

And am wondering what is the general mathematical structure

Underlying all of them

So $x\alpha_1 + y\alpha_2 + z\alpha_3 = (x+y,z,y+z,x)$

Are they just PDEs?

@Zee bro . . . I am not sure about anything at this point anymore, there is a lot of mathematics involved beyond the standard surface level stuff that is taught

If that's $0$, then $z=0$. So $x\alpha_1 + y\alpha_2 = (x+y,0,y,x)$. So $y=0$ if that's $0$. Then $x\alpha_1 = 0 \implies x = 0$

So yeah you're done

How do we know that $x^3+7x^2+6x+5$ has exactly one negative real root?

Product of the roots is $-5$

So the only possibility we need to eliminate is that all the roots are negative

@Daminark so is vaporwave still a thing?

I don't really think so? I kinda only realized it was a meme after the meme died, but I dunno, the music just sounds sorta nice

@Daminark but product of complex roots can't be negative?

Well, you know there has to be a real root

And complex roots come in conjugate pairs

oh, yes

so why three real roots not possible?

What is the theorem that says that if $f:A\rightarrow R$ is a continuous function and $A$ is compact then it attains a maximum and a minimum?

in $A$

wierstrass?

@Piyush in the case where $A$ is a closed interval I've seen this described as extreme value theorem. In general it's just that continuous functions map compact sets to compact sets, not sure if there's much to be named

Well, if all 3 roots are real, either two are positive and one is negative or all 3 are negative. Then in principle you could look at the other terms. Sum of the roots is $-7$, sum of pairwise products is $6$

Thanks I wanted it to have a name so that i can invoke it without having to write the statement.

@Daminark thank you very much

No problem!

Also I dunno for sure if that's the right approach

This is purely instinct

@Daminark, i promis this will be the last question: how many solns are there for $x\equiv 1\pmod {163}$?

How many solutions in $\mathbb{Z}$?

no, sorry, in $Z_{163}?$

Just 1 solution

Namely 1

how?

What other solutions would there be?

@Daminark no other way to see that only one solution? really can't see why other number should be soln or not

i can see eg for 2,3

So, the reason I asked for clarification earlier is because I want to know what we're inputting for $x$. For example, it makes sense that $x$ be any integer, because we can ask, for example, is 164 congruent to 1 mod 163? It is! Is 166 congruent? No

no, 166 is congruent to 3

Now if we're asking this question about elements of $\mathbb{Z}/163$, then there's no solution. Formally, you can note that 163 is prime, so you're solving an equation $x-1$ over the field $\mathbb{F}_{163}$. Well, it's a degree 1 polynomial over a field, so it has one root

If you're just asking how many solutions in $\mathbb{Z}$ there are to a modular equation, the answer is infinitely many, $n(163) + 1$ for $n\in \mathbb{Z}$

thank you very much.

[Random]

An example where pictures are misleading:

There exists no subset of reals that is a continuum union of disjoint intervals

Proof by contradiction showed if it were, there would have to be uncountably many rationals in the reals, which is absurd

Pictures are misleading here because you can put () forever in a real line and spaced out such that they are disjoint, and it will not be obvious that it is countable

If we have a differentiable function in compact set then can we say that derivative is continuous. And hence imply intermediate value property?

I feel like differntiablity at each point doesn't mean it is continuous.

That's not necessarily true

Derivatives have the intermediate value property but aren't always continuous

Try to think of an example

So I have $h(0) = 1, h(1) = 2, h(3)=3$ a continuous and differentiable function. I could prove that 1l. There is a d such that $h(d)=d$, 2. c such that $h(c)=1/3$

Final Question a c such that $h'(c)=1/4$

Sorry typo: $h'(c)=1/3$

@LeakyNun Thanks