Okay got it....

@Hippalectryon oh, something is wrong though .. (let me check)

Hey @iwriteonbananas

@Rememberme please stop writing $\{A_{\alpha}\}$

you need to be precise

okay ....

for all $\alpha$

@Tobias Is that fine

@Tobias, are you there?

I have a very funny proof of $\partial\;\rm{int}(A)\subset\partial\;A$.

@Rememberme what do you mean for all $\alpha$? There need not be some fixed element in all the intersections

@SohamChowdhury Yeah

it's this:

Hmm....

we know that $\partial A\cup\rm{ext}(A)\cup\rm{int}(A) = X$. ($X$ is the entire space, and $A$ is a subset)

this also implies that $\partial \rm{int}(A)\cup\rm{ext}(\rm{int}(A))\cup\rm{int}(\rm{int}(A)) = X$.

@Chris'ssistheartist Numerically it seems right though

now, these three parts are disjoint.

Is that even true? Take $X = \Bbb R$, $A = (0, 1]$. $\text{bd} A = \{0\}$ and $\text{bd}\, \text{int}(A) = \{0, 1\}$

Shouldn't the inclusion be reversed?

@Hippalectryon Yeah, but where all that negative quantity comes from? I check things now. CLARIFIED NOW

eh, I think I have to send him another erratum then.

First check if I am being silly.

I am not really paying attention.

simplifying, $\partial\rm{int}(A)\cup\partial A\cup\rm{ext}(A)\cup\rm{int}(A) = X$.

@Soham You can use the fact that $Bd(A)$ and $int(A)$ are disjoint

combining everything, $\partial\rm{int}(A)\cup\partial A=\partial A$. done.

@Rememberme I am using that.

Oh I didnt see that

I just saw the problem

of course I am being silly.

int of A is (0, 1), which has no boundary. bd of A is {1}.

that's the whole thing.

gotta grab some coffee

is that okay, B?

seems like it

I'd just do all this by hand.

okay, grab your cuppa.

@Soham want to hear about the covering space vs. galois theory thing?

I am free right now.

I would hear....

as in why one should study both AT and Galois theory at the same time?

But I dont know anything

yep

please

Me too then

sure.

you know what fields are, right? commutatives rings with identity and inverse.

is it your birthday today, B? you seem very happy.

:P no, it's not. i am being kind.

@BalarkaSen aye.

haha

@BalarkaSen always a good thing ;) continue.

ok. $\Bbb Q$, rationals, is an example of a field. consider $\Bbb Q(\sqrt{2})$. this is rationals with the solution to $x^2 - 2 = 0$ adjoined.

right.

okay I get this ...

i.e., field of elements of the form $a + b \sqrt{2}$ such that $a, b$ are both rationals.

addition and multiplications are obvious.

yes.

yup

it's easy to check it's a field (left as an exercise)

ok. let's move on.

@BalarkaSen ever the teacher.

$\Bbb Q(\sqrt{2})/\Bbb Q$ is a notation indicating that an algebraic number is adjoined to the base field $\Bbb Q$ to get a bigger field

huh?

that's a quotient?

how do you define adjoined @Balarka.. Sorry for the question if it is not for me

weird notation, yes, but not quotient.

so how is that different from the usual, uh, extension?

what's the difference between $\Bbb Q(\sqrt{2})$ and $\Bbb Q(\sqrt{2})/\Bbb Q$?

@Remember coming to that, wait a bit

@SohamChowdhury it indicates $\Bbb Q$ is the base field. just a notation.

it's not a new algebraic object or anything

so they're the same thing?

you can read that as "$\Bbb Q(\sqrt{2})$ is sitting over $\Bbb Q$"

no, not the same thing in the sense that $\Bbb Q(\sqrt{2})$ is a mathematical object, and $\Bbb Q(\sqrt{2})/\Bbb Q$ is a shortened statement

get it?

i.e. it just says that "$\Bbb Q(\sqrt{2})$ is an extension of $\Bbb Q$", right?""

yep

oh ...

okay, go on.

I get it

the person responsible for this notation deserves to be shot

@Hippalectryon can you prove it's $0$?

It confuses stuff .. doesnt it @Soham

B, go on.

unless you're writing a huge message.

ok, good. you can do this with general fields (re @Remember) : if $K$ is a field, an algebraic is defined to be a "solution" of a polynomial $a_0x^n + a_1 x^{n-1} + \cdots a_n$ in the ring $K[x]$

@Chris'ssistheartist Most likely not, even if I had enough time. I'm not that good :P

and then adjoin these solutions to $K$ to get a bigger field $L$

@BalarkaSen okay, I get it.

@Hippalectryon for all even power we get nice closed forms.

elements of $L$ are a vector space over $K$ with basis being solutions of that polynomial that is (@Remember you should be able to understand this, as you know vec. spaces)

okay its just solutions with set of field

@BalarkaSen hey, so do I. :P

ok, even better, then

Yes I get it

go on.

let's do some examples. $\Bbb Q(i)/\Bbb Q$ is an example, where $\Bbb Q(i)$ is the set of all rational complex numbers : $a + bi$ with $a, b$ rational

or $\Bbb R[i]/\Bbb R$ directly. note that $\Bbb R[i]$ is the same as $\Bbb C$

there are higher things, $\Bbb Q(\sqrt[3]{2})/\Bbb Q$, etc.

yes isomorphic to $\Bbb{C}$ if I am not wrong

yep

okay.

important note : elements of $\Bbb Q(\sqrt[3]{2})$ are of the form $a + b\sqrt[3]{2} + c\sqrt[3]{4}$.

@TobiasKildetoft I noticed for a couple of times having depreciative remarks to me, but I might have a different attitude the next time you try that again. Remember this well, and learn some more stuff the next time you wanna talk to me.

obviously (I guess)

ok. so I'll just get to galois groups without further ado

why that c suddenly @Balarka

eeek

@Chris'ssistheartist Lux Aeterna should be playing right now. ;)

@Rememberme basis is $\{1, \sqrt[3]{2}, (\sqrt[3]{2})^2\}$

Oh okay...

eh, go on

Ya now you can go on

if you wanted to make up a field of elts $a + b\sqrt[3]{2}$, then that'd be impossible (why? hint : square this element)

not closed under the field ops.

right, right, right

hmm. continue.

yes it is not nice!!

Axler teaches a bunch of fieldy stuff before starting vec. spaces.

Sadly Hoffmann didn't but Peterson did

ok, so consider the extension $\Bbb Q(\sqrt{2})/\Bbb Q$. galois group is the group of field automorphisms (i.e., auts of the commutative rings lying below which preserves inverses and identity) of $\Bbb Q(\sqrt{2}$ fixing $\Bbb Q$ pointwise.

let me elaborate on this :

Family of closed forms that Mathematica is not able to recognize $$\frac{\sqrt{\pi }}{2 e},\frac{3 \sqrt{\pi }}{4 e},\frac{11 \sqrt{\pi }}{8 e},\frac{53 \sqrt{\pi }}{16 e},\frac{345 \sqrt{\pi }}{32 e},\frac{2947 \sqrt{\pi }}{64 e},\frac{31411 \sqrt{\pi }}{128 e},\frac{400437 \sqrt{\pi }}{256 e},\frac{5927921 \sqrt{\pi }}{512 e},\frac{99816515 \sqrt{\pi }}{1024 e}$$

elements of $\mathsf{Gal}(\Bbb Q(\sqrt{2})/\Bbb Q)$ are self-(field)isomorphisms $\sigma : \Bbb Q(\sqrt{2}) \to \Bbb Q(\sqrt{2})$ such that $\sigma(x) = x$ for every rational $x$.

Automorphism: Bunch of all isomorphisms .. I guess @Soham

@Rememberme no

no, they are self-isomorphisms

aut = iso of something to itself

Okay get it now

@BalarkaSen hmm, I think this might have interesting uses

let's look at an example of an element of $\mathsf{Gal}(\Bbb Q(\sqrt{2})/\Bbb Q)$

wait, lemme try

Oh .. Okay

ok. there is a simple element.

find it out.

$\sigma:a+b\sqrt{2}\mapsto a-b\sqrt{2}$

exactly right

this is called a conjugation (for obvious reasons)

yeah.

let me do it also @Soham you are fast

it's sorta like $z\mapsto1/z$

$\sigma(a) = a$ for all rationals $a$, obviously, as the term containing $\sqrt{2}$ is zero.

go on.

@Remember there is only one other element, unfortunately :)

$\text{id} : a + b \sqrt{2} \to a + b \sqrt{2}$

hahahaha

Noooooo.. :p

ok, so we'll prove that these are all the elements :

$\sigma$ be an element of the galois groups. then $\sigma(0) = 0$, as fields aut preserve identity.

AMAZING, the numbers in numerator are generated by $$\frac{e^{2x}}{\sqrt{1-2x}}$$

this means $\sigma((\sqrt{2})^2 - 2) = \sigma(0) = 0$

@BalarkaSen oooh, that was cool.

hmmm

Nice

$\sigma((\sqrt{2})^2 - 2) = \sigma((\sqrt{2})^2) + \sigma(-2) = \sigma(\sqrt{2})^2 - \sigma(2) = \sigma(\sqrt{2})^2 - 2 = 0$

wait.

is $x$ rational?

okay.

@Chris'ssistheartist who are you talking to? and is that a generating function?

ok, so this means $\sigma(\sqrt{2})$ is a solution to $x^2 - 2 = 0$ as well

there are only two such solutions, so $\sigma(\sqrt{2})$ is either $\sqrt{2}$ of $-\sqrt{2}$

yes I guess wasnt that obvious @Balarka

these give, individually, the identity map and the conjugation map.

@SohamChowdhury with Hippalectryon

Yes so only two maps exists

@Chris'ssistheartist oh, cool.

@BalarkaSen right, go on.

let's call identity id and conjugation $\sigma$

sure

then $\sigma(\sigma(a + b\sqrt{2})) = \sigma(a - b\sqrt{2}) = a + b \sqrt{2}$

so $\sigma \circ \sigma = \text{id}$

idempotent?

I'm OUT. An important meeting.

thus, we have an isomorphism with $\Bbb Z/2\Bbb Z$

wait wait wait

then all nonid elements are order 2

@BalarkaSen I was going to say that. :P

$\mathsf{Gal}(\Bbb Q(\sqrt{2})/\Bbb Q) \cong \Bbb Z/2\Bbb Z$

@BalarkaSen so the obvious conjecture is: replace $2$ with $p$. works?

Let that be an exercise

Now, I presume, you see the relevance of that extension notation?

yes, I guess.

@SohamChowdhury $\mathsf{Gal}(\Bbb Q(\sqrt{p})/\Bbb Q) \cong \Bbb Z/2\Bbb Z$ too (left as an exercise)

similar logic works.

Okay....

@BalarkaSen identical logic, yes.

continue.

I'll leave it as an exercise to compute $\mathsf{Gal}(\Bbb R/\Bbb Q)$ (this is nontrivial!)

@SohamChowdhury right. field lattice translates to group lattice.

@BalarkaSen Sorry I didn't get that

it was meant for Soham, never mind.

okay

go on

@BalarkaSen well, this seems fun

anyway, now there are some special kinds of extensions called galois extension (if you do the exercise above, you'll see why $\Bbb R/\Bbb Q$ is not a galois extension. a nessesary condition for being galois is to have the bigger field as an algebraic extension of the base field. clearly that's not the case here, as $\Bbb R$ has transcendental elts)

I am not elaborating on what kind of extensions they are, as most you'll encounter will be galois.

is the answer $\Bbb R$?

The answer is the trivial group. :)

ah. no auts.

you'd have to prove that, though

it's slightly hard.

why the hell did I write that? anyway, go on.

hmm... galois extension need to think about them more

@Rememberme we don't even know what they are, as B would say :P

Yes we don't (refer DF)

ok. so $\mathsf{Gal}$ gives a bijective correspondence between intermediate extensions $K/E/F$ (i.e., extensions $E/F$ such that $E \subset K$ for some larger field $K$) and subgroups of $\mathsf{Gal}(K/F)$

@BalarkaSen like the correspondence thm?

for @Soham : the former forms a category, and so does the latter. $\Gal$ is in fact a functor between them, and in fact an equivalence of the categories

and then $\sf Gal$ is an aut-like functor

yes, like a correspondence theorem

ah, you got there first

:P

ok, so this was a rapid introduction to galois theory. i haven't told a lot of stuff, skipped a lot of stuff, handwaved a lot of stuff

so don't consider this to be an accurate survey

okay.

anyway, here's the topological side of the story :

Now I will get it :)

@Rememberme hey, you know more algebra than me.

No I don't

eh. linear algebra ;)

I have no idea about Category theory,

and many more

Yes linear algebra .. you can say

@Rememberme is not algebra

B, are you writing?

there is a map $f : \Bbb R \to S^1$, where $S^1$ is the circle embedded in $\Bbb C$ as the unit circle, given by $t \mapsto e^{2\pi i t}$

Okay get it

@BalarkaSen hmm

note that $f(n) = f(0)$ whenever $n$ is an integer (why?)

$1$.

integer number of rotations.

Yes 1.

right. so this map looks like this :

Hatcher!

I have seen this in munkres

It is a covering map?

i took this from Hatcher, but it's everywhere. in Munkres, Dummit-Foote.

yeah, this is like "looping" $\Bbb R$ up onto $S^1$.

yes, it's a covering map.

@BalarkaSen okay, what is that (exactly)?

Yes same here

okay, found out.

it's because for any point in the base circle, the preimage is an infinite set of points. for example, when you take the point $1$, the preimage is the set of all integers.

(this is not an accurate reason, but well, alright for you guys)

in particular, the top $\Bbb R$ "covers" $S^1$

ok

it's actually an infinite sheeted cover : because the preimage of a point in the base circle has infinite cardinality.

@BalarkaSen go on now.

if every point in the codomain has a neighbourhood evenly covered by the function then it is a covering map

the sheet has more structure than a set. it's in fact the group $\Bbb Z$ (not literally correct!).

so "you can say" that the circle $S^1$ has fundamental group $\Bbb Z$

@BalarkaSen Thats some machinery

@BalarkaSen I like the other argument better. The "proper" one :P

anyway, continue.

yes continue ...

this is also a proper argument. a more useful one in alg. geo.

@BalarkaSen yes, but I'm not familiar with this one, is all.

ok, so the circle has other covers too. $S^1 \to S^1$ given by $z \mapsto z^n$

right

you can imagine it similarly : cut the circle to make a string, loop the string around $n$ times, and paste the ends to get the circle "winding" $n$ times and then project to $S^1$.