@EdwardEvans "The torsion Summands are finite dimensional F-vector spaces of dimension deg a_i over F (assuming the a_i are irreducible over F)." actually can I get further explanation of this? The elements of F[x]/(a(x)) are p(x) mod a(x). where is F appearing in this?

the coefficients of the polynomial are in $F$

are u saying that (1,x,x^2,\dots) is then a finite basis? I m having trouble relating the two.

oh okay so because F[x] mods out a(x), the higher degree terms gets rewritten in the lower degree of a(x).

@Hawk: That doesn't look like a finite list.

Is it correct terminology to say "3 is the modular residue of 13 mod 5" or what would replace "modular residue"?

computer-science residue lol

@Hawk the second thing you said is correct

@MikeMiller thanks for the reddit recommendation, got some good answers

@EdwardEvans what are your thoughts on L-functions?

that's

a broad question

is it true that most of the theory of L-functions is conjectural?

It's kind of fragmented; there's a notion of L-function for many different arithmetic tools/concepts/gadgets

I imagine a lot of the theory depends on the Riemann hypothesis or some variant thereof

good point

@EdwardEvans is this considered to be an L-function

$ J(s,\chi)=\sum_{n=1}^\infty e^{-\frac{\chi(n)}{n^s}} $

I don't think so; an L-function should at least be a Dirichlet series, I think

oh okay

An L-function is somehow the analytic continuation of an L-series coming from some arithmetic object

yeah an L-function is the analytic continuation of a Dirichlet series right?

e.g. a cusp form $f(q) := \sum_{n = 1}^\infty a_n(f)q^n$ gives you an L-series $\sum_{n=1}^\infty a_n(f)n^{-s}$

(just take the coefficients and bung them into a Dirichlet series)

then you show that this has an analytic continuation to some left-half plane

and that object is what you would call an L-function

(The Hecke L-Function of a cusp form)

there are also Artin L-functions which come from the theory of Galois representations, but I don't know enough about that yet to say anything

so $J(s,\chi)$ might come from some arithmetic object?

err I don't know, maybe

I'm assuming $\chi$ is a Dirichlet character

yeah

actually there's no reason to think that it's anything like an L-function

I don't know, it's quite a broad thing to say "$J(s, \chi)$ might come from an arithmetic object"

I think the analytic continuation of $J$ is partly expressible as a sum of $L(s,\chi)$ where $L$ is an $L$ function

is there any notion of adding together $L$ functions that you know of?

err I'm not sure, I don't really know enough about L-functions to speak meaningfully about them

I just know a couple of constructions

should be able to talk a bit about p-adic L-functions after this semester though

nice, I know a lot less about htem

nobody knows shit about them

haha

@EdwardEvans Yeah you'll get a lot of junk sometimes but at least there are still grad students / postdocs who can answer some questions like that

All the answers were fairly good tbh. Even an arithmetic geometer said "you can probably get away with not doing algebraic geometry for now but sooner or later you'll end up needing it"

@BalarkaSen @EdwardEvans youtube.com/watch?v=YvyrzBLNYGI

getting those demisemiquavers synchronised must be ass

also cool af

I like the 7/16 rhythm

lol $\pi/e$ time sig

Adam Neely (another plug, I love him) has a lot of videos on weird time signatures

you've heard of the original right

Richter basically deleted a 16th note at the end of every bar

lol wtf

to me this increases the harshness of the winter coldness

I've heard the original but I don't know the piece like that

lol ok

if you like the harshness of the winter then you need to listen [insert name of literally any low-fi norwegian black metal band]

basically the 1st beat of the next bar to comes sooner than you expect

@EdwardEvans but I also like classical music :P

Do you know TwoSet Violin?

yeah

now when I hear the original I feel like it's slower :P

because of the "extra" 16th note

lol weird

what about TSV

Whether or not you know the channel lol

just cool videos

oh lol

Oh this representation theory was by far the most shittiest exam I have ever written

I felt like a toddler scribbling symbols on a paper

isn't that how mathematics is done

@Sayan how come so bad?

If $p : \widetilde{X} \to X$ is a covering map of CW complexes. Then, does its restriction to 1-skeleton map $\widetilde{X}^1$ to the 1-skeleton of the base $X^1$?

What about $\tilde{X}=X=S^2$ and $p$ the identity, but in the top copy of $S^2$ you throw in some extra 1-cells

What is the CW structure of your base?

Oh okay. Doesn't matter. You don't map the 1-cells upstairs to 1-cells downstairs.

@EdwardEvans Dude, he asked us to calculate irreps for random groups with our zoom video on and he kept on shouting that he wants to see our hands and faces on the video. Plus there were tricky questions and I am not that good at group theory.

Yeah, this is not true. Thanks.

@Sayan random groups or groups chosen at random?

hahaha

@feynhat I had the one 0-cell and one 2-cell in mind, but it's not very important

It's not immediately clear to me whether there always a CW-structure on the base s.t. the covering map is cellular though

If there were random groups, I would have to summon Balarka :p

lol

We have our exams in uni rather than at home

hasn't been a problem so far lol

Oh cases have gone down in germany ?

no I think they're going up, but the lecture halls are big enough that we can just spread everyone out

I haven't been following anything lately

Oh I see, that's good. I hate this online exam thing, it's like giving an exam while trying to balance a beach ball on your head

yeah it seems a bit silly

maybe good for like.. first year courses

Analysis 1, Linear Algebra 1 etc.

where there's like 300-400 participants lol

And the instructor keeps on slapping your face with rhetorical BS, and QAnon type conspiracies that you might be plotting to cheat on the exam

wtf

Yeah so some dude's video stopped for a moment, usual network errors, the instructor was like no you are plotting to cheat by fixing a photo and trying to fool me

lol he sounds paranoid af

Yeah he gets paranoid super fast.

We all are planning to now tell him that we would prefer a take home or open book kinda exam because it is getting very hard with network errors and him shouting to take exams

It sucks that people haven't been accommodating about this

@AlessandroCodenotti There certainly is if you're allowed to change the cell structure on the domain

If $X$ is a CW complex and $f: Y \to X$ is a covering map then $Y$ has a CW structure so that $f$ is cellular and in fact a covering map on each $Y^k \to X^k$

To construct the cell structure on $Y$ just write down all the lifts of the cell structure downstairs

does anyone know if we have a discrete metric space, the general proof to show that the set X in the matric space is open is to show that we have a ball around each point and the radius is less than 1, but what if we consider radius of say 1.5?

then your ball contains the whole space

contains the whole space?

wouldnt it just contain the points that are within 1.5

and what is the definition of the discrete metric?

d(x,y) = 0 when x equals to y

d(x,y) = 1 when x not equals to y

So are there any points of distance 1.5 or greater from each other?

can we consider the distances of 1?

I don't understand what that means

sorry let me rephrase that

i mean there are points of distance 1

so i can include them in the distance of 1.5 or greater

since 1.5 is technically larger than 1

"Distance 1" is not "Distance 1.5 or greater", because 1 is not greater than 1.5!

The ball of radius 1.5 includes all the points of distance less than 1.5 away

And that includes all the points of distance 1, because 1 < 1.5

if the balls contains the whole space, then that means that all the balls are open balls and the set is open again?

you need to refer to your definitions and carefully apply those definitions

you're trying to show that $S \subset X$ is open

what does it mean for a set to be open?

it means that the set contains none of its boundary points?

or is there another definition here that i am missing

but what is the main set? sorry if this all seems basic for you, i am still grasping the basic ideas

@JosephRock That's not a formal definition of open set, or at least not a useful one

S is a subset of X. X is a discrete metric space. You're trying to show that S is an open set

(Though maybe your course uses different definitions than I would)

"Open" to me means that for each point x in S, there is a ball of radius r around x which is still contained in S

For some radius r>0

The reason I was objecting to the 1.5-ball is because it's too big to be useful

We're trying to fit an open ball inside our set

> Consider covering spaces $p : \widetilde{X} \to X$ with $\widetilde{X}$ and $X$ being connected CW complexes, the cells of $\widetilde{X}$ projecting homeomorphically onto cells of $X$. Restricting $p$ to the 1-skeleton then gives a covering space $\widetilde{X}^1 \to X^1$ over the 1-skeleton of $X$. Show: (blah blah)

What does Hatcher mean by 'projecting homeomorphically onto'? Does the two-fold cover $S^1 \to S^1$ satisfy this condition?

Exactly what he's saying?

ah i see, yes my course does indeed use your definition as stated, i was just using the boundary definiton because it made sense more intuitively

but I can see how it fails because it becomes too big

@feynhat A cell is a map $\Phi: e^n \to X$

however since X would contain the whole space, couldn't we argue that X is actually larger than any set?

@JosephRock Yeah, but who cares? That's not what we're trying to prove :)

ok cheers mate, that helped #gobruins

Oh you're taking analysis at UCLA? Who with?

Hey, does anyone know the Frobenius series method for solving DEs?

i am taking it with Dr Madrid Padilla

i think he is relatively new..

@feynhat :55793473 if $\Phi_{\tilde e}: [0,1] \to \widetilde X$ is a characteristic map for a cell upstairs, then $p\Phi_{\tilde e}: [0,1] \to X$ is one of the characteristic maps downstairs

Yeah, looks like he joined after I graduated

Actually no he's been there since '18

I just never knew him

haha it's a small but big world i guess

@feynhat this is a category error, because $S^1$ doesn't have a pre-specified CW structure

you must get this a lot but do you have any advice on proving things you have no ideas about

if you take the cell structure with one 0-cell and one 1-cell downstairs then there is a cell structure upstairs with 2 0-cells and 2 1-cells, each of those cells projecting to the corresponding one downstairs

@JosephRock Write down very carefully and precisely what your assumptions are (not just the words like "open" but write down what that means. "Contains no boundary points" isn't even precise enough, because then you have to define what a boundary point is)

Then write down very carefully and precisely what you're trying to show

See if you observe links between those

aah. okay. So, my example doesn't make sense unless I specify the cell structure.

I'll look but then I have to go think about green's theorem for a while

An exercise.

1.3, 32.

He's just assuming it's cellular to be honest he doesn't need more than that $p(\widetilde X^1) = X^1$

Remember that if $p: X \to Y$ is a covering map then so is $p: p^{-1}(A) \to A$

when one speaks of an open set that is closed, does that mean it has to be open or closed first?

There is no time in math bro

It's open and it's also closed, both properties are simply true

You can write down whichever one you want

does being bros with Dr Miller requires high math skills

ok thanks i will try it out

I think this throws people off a lot, especially with implications. "If P then Q" I think often summons the idea that some time has passed between P and Q, but that statement simply says that "P being true also necessarily means that Q is true". If you want time in your logic you probably have to go read Hegel instead of doing real analysis

@JosephRock Nah you just have to make plenty of errors

Or do physics

@JosephRock Do you play Payday 2?

Is that game still around???

That was the SHIT when I was in undergrad

I do actually @feynhat lol it's one of my favorite games

Mike living in $t=0$

It's been so long

that must be a long time ago

lol

I probably last played when they somehow landed giancarlo esposito to play a character

There's no time in math and as you can see my sense of time passing is rather weak now

$z=1$

OK I need to go understand Green's theorem

See ya

see ya

and i am going to do my homework which seems to me are unsolvable

@MikeMiller I see you're lucky enough to have avoided the weird modal logics people

Tryhards

@MikeMiller whos your favorite?

I was probably usually Chains or Dallas, I was gone before they got back old Hoxton

I thought about getting back in at some point but I'd rather have 3 friends to do that with instead of cruising public servers

Pubbies can't do the stealth portions for shit

thats one of the problems about team games, including payday2

but you could join one of those servers with real decent players in them , then again it takes up way too much time to play lol

does anyone know how i can come up with sets that are open but not closed?

Most will be! Try to make your intuition of what open/closed means into pictures

OK no I'm gonan distract myself again

Peace

@JosephRock Think about the real numbers as a metric space

hmm, actually, what are spaces in which this doesn't happen

i.e. what are spaces in which sets are open iff closed

For a $T_1$ space that's obviously equivalent to being discrete

Not even I want to think about spaces that are not $T_1$, so the question is settled

true

There's also so called extremally disconnected spaces, which are those spaces in which every open set has open closure

(that's also a terrible name, since those spaces can be connected)

yuck

why does one care about those

They correspond to complete boolean algebras through Stone duality

Wikipedia also tells me that extremally disconnected spaces compact Hausdorff spaces are the projective objects in the category of compact Hausdorff spaces, which I guess some degenerates people would consider a good justification

@MikeMiller when i think of open , it just appears to be many empty balls in R2 but then it it closed and then the ball just has boundarys, which doesnt help come up with an example of open and closed since they contradict

@MikeMiller haha we all need a break tho LOL

@AlessandroCodenotti whats T1 mate

Each pair of distinct points has neighborhoods which don't contain the other point

@JosephRock It's a so called separation axiom, you don't need to worry about those if you're studying metric spaces, they'll become relevant when you start dealing with topological spaces in general

unfortunately I have to as one of the book's problems that I am to do asks the precise qn of an open and not closed set

Nah don't worry about separation axioms, think about the real numbers. What's the first example of an open set that comes to your mind there?

hm (4,6) for example?

oh i see so (4,6]

that was cool

No wait, you were looking for a set which is open but not closed, right?

Is (4,6) open? Is (4,6) closed?

yes I am @AlessandroCodenotti

Why is not open? Why is not closed?

sorry I mean (4,6] is open and closed

since i can have a ball of radius r<1 in (4,6] for all points

Even for 6?

but at the same time I can't at 6 so it is closed too

ok so that example doesnt work

is there any way i can alter this interval such that it is not closed

No wait you're confusing not open and closed

(4,6] is not open, because no ball around 6 is entirely contained in (4,6]

Reminds me of this. (Must be a difficult watch for Thor because he actually knows what they're saying lol).

But that only tells you that (4,6] is not open, it doesn't necessarily mean that it is closed. What is the definition of a closed set?

@feynhat lol, still a classic

if for every point the ball contained has a radius r less than or equal to l

ok so it is not closed and not open

do u think i can alter (4,6] to make it open

@feynhat hahaha i saw that video before, it's pretty funny but too bad there weren't any examples in it

Sure you can. There is a single problematic point that prevents (4,6] from being open

so (4,6)?

Yep. Why is this interval open though?

because I can construct a ball of radius r < G for every point in (4,6)

Ok. Is this set also closed?

yes

Why?

@AlessandroCodenotti Hmm, I'm trying to figure this out. Say $X$ is a top space and we're looking at its lattice of clopen subsets. If $X$ is extremally disconnected, then clearly that lattice is complete as sup is just given by closure of union. But if that lattice is complete, is the space necessarily extremally disconnected?

What's the definition os a closed set?

because i can construct a ball of radius r<G where r= g too for every point in (4,6)

no wait it is not closed

since i can't do it at 4 and 6

well actually I can

4 and 6 are not members of (4,6). What's the definition of a closed set?

by closed set do u mean closed?

closed set is where I can construct a ball of radius r< G where r can be equals to g for every point in (4,6) but since 4 and 6 are not not members then

the set (4,6) is closed actually

@Thorgott It's been a long time since I read about Stone duality, I don't remember the details, sorry (and I'm already trying to understand uniformities at the moment so I can't work them out now)

so this is a set that is open and closed

sorry my bad it it not closed

since I cant construct on 4 and 6

but arent 4 and 6 not members so i dont have to care about them?

np

is there any way i can change (4,6) to be not closed?

it is not closed

it closed because i can construct a ball of radius r<g , r= g at any point in (4,6)

I don't know what g is, but that is not the definition of closed anyhow

I mean the set {x∈X:d(x,x0)≤r} where x is a member of (4,6) and r is the radius

that is satisfied right?

So you're saying that every point in $(4,6)$ is contained in a closed ball which is contained in $(4,6)$?

im trying to show that $\{(x,y) \in \mathbb{R}^2 : |x| + |y| - 1 = 0 \}$ isn't an algebraic variety, so in order to do so I showed that $\partial [0,1]^2$ isn't an algebraic variety, which is somewhat easier since if it was then it would be a subset of the zero set of some $f \in \mathbb{R}[x,y]$ and subbing in $f(0,y) = 0$, $f(1,y) = 0$, $f(x,0) = 0$ and $f(x,1) = 0$ and using that a single variable polynomial of finite degree has finitely many zeros gives that all coefficients of $f$ are zero

then noting that $\{(x,y) \in \mathbb{R}^2 : |x| + |y| - 1 = 0 \}$ is $\partial [0,1]^2$ if we apply a bunch of invertible linear transformations to it gives the result, since a polynomial $p$ composed with a linear transformation $\mathbb{R}^2 \rightarrow \mathbb{R}^2$ is still a polynomial

but i am wondering, is there a direct way to do it? starting from $\{(x,y) \in \mathbb{R}^2 : |x| + |y| - 1 = 0 \}$ and not arguing using linear transformation?

i.e. going straight to all coefficients of some polynomial $p$ whose zero set contains $\{(x,y) \in \mathbb{R}^2 : |x| + |y| - 1 = 0 \}$ are zero

@Thorgott yes, for example at 4.01 i can construct a small ball around it which will still be contained in (4,6)

that is true, but has nothing to do with being closed

isnt that what the definition of being closed says @Thorgott

I don't know where you are getting your definitions from, but it certainly shouldn't be saying that

oh, that is closed because you are just taking {(4,6)} with the relative topology induced by the euclidean metric

here x needs to be a member of {(4,6)} which is just a singleton

although yes, being closed has nothing to do with finding an open ball around each point in the set

oh never mind, maybe I gave too much benefit of the doubt, why are you talking about 4.01

i am talking about 4.01, because finding a closed ball around each point in the set would satisfy my definiiton of being closed..

what is the set?

(4,6)

you mean $\{4,6\}$ ?

no i mean the interval (4,6)

oh sorry

you are saying that is closed in $\mathbb{R}$?

yes

with the euclidean topology?

why do you think so

because given the defn of the closed ball that i have stated earlier on

i see that this defn is satisfied

by closed you do mean is the complement of an open set, right?

(c)Foranyx0 ∈Xandr>0,thentheballB(x0,r)isanopen set. Theset{x∈X:d(x,x0)≤r}isaclosedset. (Thissetis sometimes called the closed ball of radius r centered at x0.)

this is the defn that I used for a closed set

by closed I mean a closed set in the above defn

you are saying the closed ball of radius r is closed, that isn't a definition

it is true but you aren't defining anything

erm what do you mean

in the above I have an interval (4,6)

and i want to see if it is a closed or open set

so since for every point in (4,6) i can find a closed ball

then i say (4,6) is a closed set

but the defn of closed being complement of an open set is another defn that works too I think

but your definition , for a metric space just defines open sets , it doesn't at all define closed sets

What you're quoting is not the definition of a closed set, it's an example.

if your definition is O is closed iff for all o \in O, {x \in X : d(x,o) <= r} \subset O for some r > 0, this is just another way of saying O is open

although you might want to say 'if O is open iff for all points in O there is an open ball centered at point contained in O' so you can define 'O is closed' by just replacing open ball in that definition with closed ball, that isn't how it works

indeed

@porridgemathematics when you said this is just another way of saying O is open , but isn't this a closed ball?

ok I see now that my defn of a closed set is wrong

my apologies thor

yeah, the definitions aren't the same, but they imply each other

since every closed ball contains an open ball and vice versa

if there is a closed ball that is a subset of some set, there is a open ball of smaller radius guaranteed to be in it too, if there is an open ball that is a subset of some set, there is a closed ball of smaller radius guaranteed in it too

^

the fundamental connection between open/closed is that they are complements of each other

you can have sets that are both open and closed, but in general you won't find these in connected spaces , besides the entire space and the empty set

(this is actually a characterization of connected space)

Interestingly, EVERY open subset of the rationals under the order topology is closed

in fact I think this is true for every space whose point-set is countable

This is false, $\Bbb Q\setminus\{0\}$

but the complement of (4,6) would be (-infinity, 4] union [6, infinity)

which is an closed set

@AlessandroCodenotti which is?

Alessandro oh shoot you're right

@KrullDimension there are even hausdorff connected countable spaces

Like what?

aren't the rationals with the order topology just the same as the rationals with the subspace topology?

@porridgemathematics yes but since you have d(x,o) <= r in the defn then this a closed ball as it contains the boundary so why do you say O is open?

oh i guess not

uh wait, actually i don't see why they aren't the same

are they?

I think so

so you just mean the rationals with its inherited topology?

Yeah

but I was wrong anyway

rightr

@KrullDimension ams.org/journals/proc/1953-004-03/S0002-9939-1953-0060806-9

even outside of that counterexample provided, any open ray (q,+infinity) isn't closed, since q is definitely a limit point ?

so none of the basic open sets are closed at all

does anyone know of a simpler way to solve this?

then noting that $\{(x,y) \in \mathbb{R}^2 : |x| + |y| - 1 = 0 \}$ is $\partial [0,1]^2$ if we apply a bunch of invertible linear transformations to it gives the result, since a polynomial $p$ composed with a linear transformation $\mathbb{R}^2 \rightarrow \mathbb{R}^2$ is still a polynomial

does anyone know if (-infinity,4] union [6, infinity) is a closed or open set?

it's the complement of (4,6), so it's closed

but why is (4,6) not closed?

or i mean why is (4,6) closed

(4,6) isn't closed

it's open, the complements of open sets are closed

but I mean i wanted a set that was open and closed at the same time

and aless said that (4,6) is both open and closed

but i can't quite figure out the details as the defns seem to be not right here

would be really helpful if someone could just clear this T.T

that is not closed

In R, the only sets that are both open and closed are the whole space and the empty set, no?\

yes

Dear All, is there any one familiar with SVM?

@porridgemathematics why would you say (4,6) is not closed, I mean we concluded that it is open but being open does not mean it it not closed

it isn't closed, because the complement isn't open

@JosephRock if you want examples of spaces that have non trivial subsets which are open and closed, you need to pick a disconnected space to start with, e.g. (0,1) U (2,3)

its complement is (-inf,4] U [6,inf) which isn't open so it isn't closed, or just notice that 4 and 6 are limit points of (4,6) and aren't in it

Is there a way to prove the generalized/ideal-sensitive/Atiyah version of the Cayley-Hamilton theorem, which states "a module homomorphism with its image in an ideal solves a monic polynomial with non-leading coefficients in that same ideal", WITHOUT explicitly invoking determinants?

anyone familiar on SVM? please help

Suppose $f$ is continuous complex-valued function, $g$ is analytic, and that $f \circ g$ is analytic. Can I conclude that $f$ is analytic? Do I need to impose other conditions on $f$?

Sheldon Axler gave a nice determinant-free proof of the version for complex vector spaces in "Down with Determinants", but all the proofs I've seen for the one that's used in proving Nakayama's lemma invoke the explicit determinant algorithm in showing that the characteristic polynomial has coefficients from the ideal

@user193319 take $f$ to be your favorite continuous complex-valued non-analytic function and $g$ to be constant

Damn...hmm...

I am trying to solve the following problem: Let $L$ be a line in the complex plane. Suppose that $f(z)$ is a continuous complex-valued function on a domain $D$ that is analytic on $D \setminus L$. Show that $f(z)$ is analytic on $D$.

I have the following theorem: Suppose that $f(z)$ is a continuous function on a domain $D$ that is analytic on $D \setminus \Bbb{R}$, that is, on part of $D$ not lying on the real axis. Then $f(z)$ is analytic on $D$

My thought was to reduce the problem to the case where $D$ is a disc centered at the origin, $L$ passed through the origin.

Then there exists $z_0 \in \Bbb{C}$ unit complex number such that $z_0 L = \Bbb{R}$. Define $g_{z_0}(z) = f(z_0z)$, and then apply the above theorem to $g_{z_0}$ to conclude that it is analytic, and then try to conclude that $f$ is analytic.

hmm....

Isn't any branch of the logarithm analytic except on one ray?

well, if you already got that, the rest is for free

since any line can be taken to $\mathbb{R}$ via a biholomorphism

@Thorgott Do you see anyway to reduce my problem to the theorem I cited?

(also, I don't see any reason for why the way you proved it for $D\setminus\mathbb{R}$ should not work to prove it for $D\setminus L$ analogously)

Oh, so because $z_0 \mapsto z_0z$ is a biholomorphism, we can conclude that $f$ is analytic from the fact that $g_{z_0}$ is analytic?

@user193319 do you see why this is true if $g$ is a biholomorphism

Yeah, because $(f \circ g) \circ g^{-1} = f$ and compositions of analytic functions are analytic functions...right?

yes

that's all you need

Sweet! Do I even need to consider the special case of $D$ being a disc centered at $0$ and $L$ passing through $0$?

no

Okay. I'll think about the more general case then. Thanks, bro!

np

(-infinity,4] union [6, infinity) is closed because (4,6) is open.. but (4,6) is not closed, is open because complements of open sets are closed, isnt this like some circular reasoning?