@Faust not bad faust, good luck :D

ill be doing rep theory and analysis

also logic in march

but what is this with advisor? how does it work

Well in the summer i am doing a reasearch project that essential just amounts to me trying to learn a course in C*-algerbras and non-commutative geometry

but i get paid which is kind of cool

that is neat

My advisor is writing a textbook and he wants me to go through all the exercises and learn the material form his book and give feedback i can also ask him questions when i get stuck etc

we do also have summer classes but they are mostly courses that are extra stuff

Anyone think they can answer a quick question about Implicit Function Theorem?

give me your advisor name :D

hes trying to write a textbook for an undergraduate class in the topic area

Heath Emerson at UVIC

tell him there is a hungry student who also wants to be in on this

lol

inphsise the hungry part:D

i actually am trying to find a second person to do it with me cause its really hard material

u got me

also like those topics :D

or at least love to learn more of em

the thing is theres like 4 undergraduate 3rd year classes that are required for it at our university and even then there still some graduate level material on functional analysis i will have to learn on my own before the research project starts

so its going to be a ton of work

you got half a year before that

thats plenty of time :D

it starts may 1st so not really

you got my email , ill be happy to work some of that with ya

oh

we finish june

do you know any topology?

middle of june sometimes

taking course this semester :D

we finish around april 18th

ok

Hi @Kasmir, long time no see

@MatheinBoulomenos mathein :DDDDD

Happy to see you :D

Yes it has been ! been busy with family and stuff

How's it going?

How are you ?@MatheinBoulomenos

yeah we should defiantly do some functional analysis i got a textbook need to meet with my advisor and figure out what i need to learn then ill let you know

Pretty good, thanks

so far so good ! gonna start study again on monday so need to be in shape!

@KasmirKhaan round is shape ^^

@Faust all righty =p you can mention that kasmir is handsome also

lol

Ah I see. what are you taking?

@MatheinBoulomenos I have some questions about algebra mathein , hope you are not busy :D

Feel free to ask

faust gotta help a teacher

to make a book, he will be reading it and give him notes what seems good / bad , hard / easy

thanks Mathein, i need a min to phrase it properly

I heard about that, yes. It's on non-commutative geometry, right?

and C* algerbras

though i think they are very related from my understanding of the book

I thin Mathein would be best fit to do that faust , come to hink about it :D

anyway mathein, can you help me understand the quotient group properly ?

I know when we do G/N

we making a parting of G

into cosets, each of size N

those cosets ( seing them as elements) form a group if N is normal

so we can talk about Na Nb =Nab

that I understand, but the idea of modding out when we do examples

Can one continue the Euler product for the Riemann zeta function to the complex plane by simply multiplying the Euler product with something?

sometimes its not easy to identify the cosets

If you're having trouble with examples, it would help me to see these examples

i think its usually easier to use the FIT Kasmir

what is FIT ?

find a homomorphism that makes N

R/Z mathein for example

as its Kernel

so in that example you want to send every integer to zero

can you think of a homomorphism that does that?

hmm dont I need to understand the quotient group before iso theorems?

R>0 ---> R/Z

we can send a --> [a]

oups

anyway, the idea of modding out by a normal subgroup

Okay, this is really informal, but the way I sometimes think about quotient groups (or rings) etc. is that we deliberately "forget" about something or that we force something to be the same. For example, when we look at Z/2Z we force all even integers to be the same as 0 (from this it follows that all odd integers need to be the same, too), so Z/2Z has two elements.

do we use the fact that a R b, if ab^-1 is in N ?

and the notation of that

Z/2Z

is it because 2Z is acting like a unit ?

in that factor group

In that case the map you want is $\phi : \Bbb R \to \Bbb C^{x} $ with multiplication in the second group

2Z is the part which we forget about or which we "collapse to one point"

Maybe a better example ist $\Bbb R^\times /\{\pm 1\}$, here we just forget about the sign

hmm

its getting better let me try to make more sense of what you said :D

I like the example of $k\mapsto \exp(i 2\pi k/n)$

why did you tell him the map :\

$\Bbb R^\times /\{\pm 1\}$

i like this one

trying to make sense of it

i think i know i finnally know what it is

isint that isomorphic to R^+

need to remove zero as well

and no its definatly not isomorphis

we use R^+ to mean R\{0}

Be careful with "/" and "\"

the elements of the factor group are of the form

thats wierd

[n] = {-n ,n}

but eithier way u cant have zero but if you land on a negative number you could "factor it out" and mod it out

@MatheinBoulomenos fixed now :D

so you can have only positive non-zero numbers

thats the thing

those words mod out and factor out , dont make sense to me

like at all

so your landing in R remove zero and all negative with multiplication

yes $\Bbb R^\times /\{\pm 1\}$ is isomorphic to the group of positive reals under multiplication

Hi @Ted

Hi @Mathein, Kasmir, Faust.

wow theres no non intergers?

@TedShifrin Ted :D

@Faust fixed

oh thanlk god =)

@MatheinBoulomenos so what i said first was right?

hmm the thing is i did it using a map

@KasmirKhaan you said that you use R^+ to denote R\{0}. If you mean that, then no, that's not correct

perhaps what you thought was write but what you wrote wasn't

oups

wrong notation

R_>0

this is R^+

Yes, $\Bbb R^\times /\{\pm 1\}$ is isomorphic to that

:D

but the way i did it was not so obvious

i mean i had to find a map

a ---> [a]

hmm

Hi @TedShifrin

hi Balarka

Morning @TedShifrin

can you say in words, the intuition behind " modding out"

this idea did not stuck in my head

I understand it for modular arithmatic

there are at least two ways to see this: you can note that the restriction of the quotient map $\Bbb R^\times \to \Bbb R^\times /\{\pm 1\}$ on $\Bbb R_{>0}$ is bijective, or you can use the map $\Bbb R^\times \to \Bbb R_{>0}, x\mapsto x^2$

Z/5Z, we get any element we can reduce it to be less than 5

@MatheinBoulomenos that was a nice example never seen it before thats nice

i rpobally can't but someone might be able to if you post it

note that the way you implement mod 5 stuff is to say that $n-m\in 5\mathbb{Z}$ means that n,m are in the same equivalence class

Yes I know semi, but I lack the intuition to use that idea in general

The question is if i have 5 red balls, 4 green balls, 6 blue balls and we get 5 balls at random. What is the probability that AT LEAST one ball is blue.

like if the question was : what is R^X / {1,-1} Isomorphic to ?

I would be stuck

so what you do for general $G/F$ is to say that $a,b$ are in the same equivalence class when $ab^{-1}\in F$

@Shago can you figure out the probality that you get 0 blue balls?

@Kasmir an example which is dual to the other I gave. Can you figure out what $\Bbb R^\times / \Bbb R_{>0}$ is isomorphic two?

okay ill think about it now :D one second

So you want real numbers $a,b$ to be in the same equivalence class mod {1,-1} when $a/b\in \{1,-1\}$

it should be something like 9C5/15C5 @Shago

I have no idea how to get the probability of getting 0 blue balls

ok well you need 5 balls

none blue

there are how many balls that are not blue?

9 i guess

ok, now how many Ways can you choose 5 from those 9 balls?

126

@Faust unfortunate parsing there

lol

now how many ways can you choose 5 balls from 15?

3003

@MatheinBoulomenos should be isomorphic to the group +- 1

so 126/3003 is the chance of getting no blue ones

subtract that number from 1 and that gives you the chance that in your selection at least one of them is?

@KasmirKhaan correct. Note that the multiplication law in the quotient group basically amounts to the rules for signs that you learned way back in school "positive times negative is negative", "negative times negative is positive" etc.

Thank you very much : )

@KasmirKhaan every time you say isomorphic it makes me uncomfortable lol

@MatheinBoulomenos aha nice

why faust?

the way i saw it was, ab^-1 is in R^+

means a/b is positive

@Shago sometimes its alot easier to count the number of ways for something not to happen than it is to count it happening once , twice , three times , .... etc and suming them all together

so either a , b >0 , or a,b <0

Any time you see at least you should think im going to count none

it may not always work but its a good place to start

@Kasmir yes, that works

ok, thank you very much Faust : )

so a R b if they both have same sign

@KasmirKhaan i normally talk about $\Bbb R \to \Bbb R$ /(w.e) being homomorphic

@MatheinBoulomenos hmm is +- 1 the sign function ?

That doesn't make sense, Faust.

exactly

im trying to mod

its not working

sup chat

there we go

Like out of 100 things faust say , i only get 3 things and 2 of them I dont need _D

Homomorphic is not an adjective that is used.

you can use the sign function, that's group homomorphism from $\Bbb R^\times \to \{\pm 1\}$ which has $\Bbb R_{>0}$ as the kernel @Kasmir

You have a mapping. It's the homomorphism.

Hey @Eric

If it's a bijection, we call it an isomorphism. And we do have the adjective "isomorphic."

@KasmirKhaan lol

Hi @Eric

@MatheinBoulomenos thanks mathein :D

It's a symmetric relationship. Having a homomorphism from $G$ to $H$, you need to know the direction of the mapping. Saying two groups are "homomorphic" makes absolutely no sense.

hi Eric

@TedShifrin that doesn't mean it doesn't make me uncomfortable even if its completely correct

you could perhaps say "homomorphic mapping" but that'd just be unnecessary

I was grading a LA exam which asked if two groups where isomorphic. The student wrote "$G$ is isomorphic, but $H$ is not isomorphic" and he had long "proofs" of those facts

i mean the mapping i just dint want to type phi

the only reasonable way to define homomorphic would just make any groups homomorphic

we sometimes say that one group is a homomorphic image of another (which is just another way of saying it's a quotient)

if you can find a homomorphism then the original thing mod the kernel is isomorphic to where your map landed is how i prefer it

@TedShifrin

That's sloppily stated.

but thats the way we think about it

we dont go oh this is isomorphic to that

"Where your map landed" had better be the entire thing.

yes yes

@MatheinBoulomenos lol classic grading horror story

Sometimes isomorphisms don't arise from the fundamental isomorphism theorem, so get over it.

sorry its just my thought process

@MatheinBoulomenos ya this is fine phrasing i think

i usually say that it's the image under some homomorphism or smth

its not something i say out loud

@Faust: I can say $(\Bbb R,+)$ is isomorphic to $\Bbb R^\times$.

so when asked to show a group is iso to another

You can't say that?

we just need to find a map

but finding that map, should that be easy or what

not necessarily @KasmirKhaan haha

@TedShifrin you can say that, but it's wrong

haha because most of the examples on the book the map was easy to find

but when they do state what map we use

I typed the wrong thing, @Mathein. Good point.

i would not be able to come up with it

of course @KasmirKhaan "easy" is in the eye of the beholder :P

had it not been written

I meant $(\Bbb R,+)$ is isomorphic to $(\Bbb R^+,\cdot)$.

im trying to think of the bijection

@Antonios-AlexandrosRobotis by easy , i meant the most natural thing to do

anyway guys

@KasmirKhaan some isomorphism come really natural, some are really difficult to find

If i tell you what grade i got on this course

you wont belive me

you got 4?

@TedShifrin I'm not sure if I'm seeing an issue with one of Spivak's exercises. I was wondering if you were interested in giving me a hint or confirming my suspicion.

@Antonios-AlexandrosRobotis And "trivial" is in the eye of the reader, to whom it is left as an exercise

Get dab'd on

@TedShifrin ok i belive it

loolll @BalarkaSen

What's that, @CryinShame?

@BalarkaSen if i ever write a book ill include this in it somewhere

I got a B , questions were not on first part, iso thms and these type of group theory

were on sylow

group actions and symmetric group

and rings

there were two reall hard ring homomorphisms on my final

but i feel like I should not even have passed because much I dont know

we had a nice question, about chain of non proper ideals

@KasmirKhaan your understand isn't all that much diffrent from mine imo