@GFauxPas Well, it was of the form "___ was ___ the whole time"

Those are the best twists

math.stackexchange.com/questions/2047334/… no one likes my question :(

Though according to The Knot Book, I can exchange twist for writhe.

@Akiva, South Park by any chance? There was a major twist in the latest ep of the current season

"What's your favorite branch of mathematics?" "Knot theory" "Me neither"

@KajHansen No

Bummer

But it is of the form "cardinal-direction place-type" @KajHansen

@AkivaWeinberger I got it.

im very confused..

its not working, i dont know why

oh

@Fargle I'm not on the most recent episode, though, I'm referring to something that happened in an earlier episode

@KajHansen you look a little like Matt Damon from far :D

If we consider the homomorphism from Z_12 to Z_12 (of which there are many) and we consider the case where |phi(Z_12) | = 2 (since this is a subgroup) why must phi(1) where 1 is in Z_12 be mapped to an element of order two in Z_20?

@Null, a number of people have said that in here

Hey @Kaj

got it

Z_12 to Z_20 * for the homomorphism

can there be an inverse matrix for a non square matrix?

there can be a left-inverse or right-inverse @SylentNyte

but "inverse" without qualification means inverse=left inverse=right inverse

the identity matrix whne you multiply it on the left by the left inverse would be a different size than the identity matrix you get when you multiply it on the right by the right inverse

ah okay, thank you

np

it's related to the fact that a function is bijective iff it is surjective and injective

if a function is surjective but not injective it has a right inverse

Hey @Cbjork

if a function is injective but not surjective it has a left inverse

@Kaj Can you find a sequence of polynomials $\{f_n\}$ with rational coefficients such that for for all $n$ $\max\{f_n(x):x\in[0,1]\}\leq 1$, $\min\{f_n(x):x\in[0,1]\}\geq 1$, $f_n(\frac{\pi}{4})\in(1-\frac1n,1]$; $f_n(1-\frac{\pi}{4}),f_n(\frac{1}{2}+\frac{\pi}{8})\in[0,\frac{1}{n})$.

bijective iff invertible, I meant

I think you have \geq and \leq swapped @Cbjork

LOL kaj I do

max is $\leq 1$, min is $\geq 0$

hey @TedShifrin

So basically the graph of the function is in $[0,1]\times[0,1]$

kaj have you done representation theory ?

So function evals in 0, 1 are bounded by 0, 1 @Cbjork ?

can someone explain here dim(V) is dim(W) right ?

@kaj yea

not dim(V) this is a mistake or is there something I am not understanding ?

@KajHansen what do you think ?

Anyone know?

I just started on it @Adeek. I learned what decomposible, reducible, faithful, and regular representations are today

oh

$g$ is a really bizarre function. For example, we can find arbitrarily long intervals for which it is constant.

(So I've read)

@KajHansen found good set of notes about representation theory

math.tecnico.ulisboa.pt/~ggranja/daniel

I will read it when I get home.

brb going to home

Is the factor ring Z/10Z the same as Z_9 since Z/10Z = {0, ..., 9}?

@fluffy_muffin No. $\Bbb Z_n$ is defined to be $\Bbb Z/n\Bbb Z$

@fluffy_muffin You can also represent Z/10Z as {1,2,3,4,5,6,7,8,9,10} and have 10 be your additive identity. But in Z_9 (or Z/9Z), 10 and 1 are the same thing.

@Maks Yes, griedient of the function is the direction of maximum increase of any function

Mind is neccessary to understand something?

@TheGreatDuck then I'm sorry. Thanks, if you need something in the future tell me.

@KajHansen is a linear function $f:\mathbb{R}^2\to \mathbb{R}^2$ a $2\times 2$ matrix?

It can be represented as such @Null

"The best thing you have going for you is your willingness to humiliate yourself." haha such a great movie quote

Thanks for the share @Adeek. I like using multiple sources when self-teaching

@KajHansen can the same be said about a matrix? or only if it meets some requirements?

Yes, every matrix correspond with a linear function @Null

Transformations on a vector space are determined by their action on a set of basis vectors.

The standard basis in R^2 is $(1, 0)$ and $(0, 1)$

The first column of a matrix is the image of $(1, 0)$ and likewise for the second column and second basis vector

@user243301 fair enough. I just wanted to clarify that for you. Thanks for the effort though. :)

@KajHansen i have to determine wether a linear function with 2 given points exists. would you calc that out with a matrix?

@KajHansen SAVE ME D:

^and that, pls

T_T

@usukidoll, a damsel in distress :O

I have like some confusion over abstract algebra problems . I felt like I did it already but idk can you check?

Elaborate @Null

Sure

@KajHansen (a) i am currently working on

(und=and)

I think I did this before on the previous problem when I proved the sum version

Hm. That seems to be true by the mere definition of the product of ideals @usukidoll.

it is true but I feel like I'm double writing here

@KajHansen can i view f(x,y)=(w,z) as (x,y,w,z)?

Hang on @Null

Oh wait, not quite. $IJ = \{a_1b_1 + \cdots + a_nb_n \ | \ a_k \in I, b_k \in J \}$

I did that part already for sums

wait let me screenshot 3a

prntscr.com/dgavry

are they like the same thing or different?

because I did my closure under yada yada and it held

but when I saw 3b I'm like is that the same as to what I did earlier?

So it seems what we want to do is show that $a_1b_1 + \cdots + a_nb_n = xy$ for any given $a_k \in I$, $b_k \in J$ and some $x \in I$, $y \in J$

yeah but I felt like I've done this already on 3a?!!!!!!

Those are separate problems. In 3a, we only have $\{ \text{ big thing} \} \supset \{ab \}$

maybe proof by contradiction for 3b?

Let me think for a sec

the way it looks was like @____@ I'm confused

Ah, you can take advantage of the distributive property

?

So let's consider an element $a_1b_1 + \cdots + a_nb_n$, where $a_i \in I$ and $b_i \in J$

Since these are principle, any element in $I$ can be written as $rx$, for $r \in R$ and $x$ the generator of the ideal

like all of a distributed to b

?!

And likewise for $J$. It has a generator, call it $y$

OH!

like sy for s in s or maybe we can still leave it as ry for r in r

x is the generator for a and y is the generator for b

@KajHansen just ping me if you and bulbasaur worked his problem out hehe ;)

yep

Yeah @usukidoll, so $a_1b_1 + \cdots + a_nb_n$ can be rewritten as $s_1xr_1y + \cdots + s_nxr_1y$

Might be better to write it with a_i = x s_i

Now the $x$ can be factored out: $s_1xr_1y + \cdots + s_nxr_1y = (s_1r_1y + \cdots + s_nr_ny)x$

otherwise you presumably run into issues about commutativity

We're given commutativity @Semiclassical

woops

See where to go from here @usukidoll ? It's in the form we want at this point (why?)

I think x.x

why is $(s_1r_1y + \cdots + s_nr_ny) \in J$ ?

(y being J's generator)

x.x

@Null, let's look at the first one

because ys is in J... same with sy in J for r in R

Yep, and it's closed under addition

Remember that the matrix for the transformation can be gotten if we know the image of $(1, 0)$ and of $(0, 1)$ @Null

@KajHansen I would start with the definition of a linear function

It already gives us the image of $(0, 1)$

To find the image of $(1, 0)$, you can apply the definition of a linear function

Note: $f((1,0)) = f((1,1) - (0,1))$

@KajHansen in addition to show existence, I also have to say if this is the only one. Is this a lemma?

Linear transformations are uniquely determined by their actions on a set of basis elements @Null

$(1, 0)$ and $(0, 1)$ being the "standard" basis for R^2

@KajHansen (1,1),(1,0) is a basis too. so we are done for that part?

(not the canonical tho)

The matrix is harder to find without the standard basis though @Null

feels like (rs)(xy) = (sr)(yx) because of the ab format.. x.x

@KajHansen ah ok

If you just need to show existence, then sure

let's try to state it explicit for the spirit ;)

with we i mean me haha

?

It's sort of the same argument @usukidoll

Oh wait

different tomato

I thought you said closure under addition

I said closure under multiplication since we had IJ with the ab

the product of ab

not the sum

That's not bad. So $IJ = \{ab \ | \ a \in I, b \in J \}$. Is $rab \in IJ$? We have associativity: $rab = (ra)b$

that's what I've figured earlier abx = xab for x in R whoops I need x = r

@KajHansen: Hey man, wassup? I've been trying to contact you for some time now.

Hey there @FaraadArmwood. If you tried to call or text me, my phone is off basically 24/7. Sorry about that

rab=(ra)b in IJ for r in R?

The battery's dead, so it has to be plugged in to turn on at all

How's it going man? Still plugging away in ND?

Yeah @usukidoll, now argue that $(ra) \in I$ and $b \in J$

Yeah my phone is usually off too. Everything's okay and yeah. I still have a long way to go though. How's everything ur way? What ever happened to like Doug n everybody? I heard Sarah went to MIT.

$r_{1}a_{1}+...r_{n}a_{n} (b)$ and then distribute the b??

Doug's finishing up his undergrad this semester; I talk to him daily. I knew Sarah went to MIT, but I haven't really heard from her

Remember that $I$ is closed under outside multiplication @usukidoll

UGhummm

@TheGreatDuck no problem, as said if in future you want talk or we can help us, we can do it. Thanks.

(ra) and all of the B strings

$ra(b_{1}+...+b_{n})$

@FaraadArmwood, I finished up in the summer, but I haven't applied for grad school yet. I'm thinking about it in the near future, but I'm not sure I want to. I've been suffering from severe depression lately. Just taking a break atm from going balls-to-the-wall in undergrad and trying to do some stuff to improve my mental health

~.~

I've had depression for close to a decade now, but it got particularly bad in my last semester of undergrad.

What's the question @usukidoll ?

@KajHansen f(a+b)=f(a)+f(b), if it's linear. can we say therefor f(a-b)=f(a)-f(b)?

tryin to prove the ra in I and b in J

I didn't apply to grad school right away since I felt like I needed a break from problem set grinding. I think perhaps I should've taken it a little easier in undergrad, haha. I kept taking like 3 maths / semester (4 at one point, not counting 4950)

@KajHansen: I've told multiple people about this (depression). I think the math community has to do a better job at handling the mental well-being of its participants. It's something that I think almost every graduate student has dealt with time and time again.

max of maths I've took during undergrad was 2

It's truly a shame @FaraadArmwood. Nearly everyone I know personally in the math community suffers to some degree

I think grad school in general does a piss poor job at it

how so?

@KajHansen: It's great that you took time off, but collect yourself and get back to it! You just need to surround yourself with good people, and pick out a good program.

It's not just a math community thing.

Oh, it suffices to show $ra \in I$ @usukidoll. Think about the axioms for an ideal

That's good advice @FaraadArmwood. I'm studying some representation theory on my own at the moment, and trying to post here more. I was away from here for a long while.

Basically trying to maintain my knowledge from undergrad and pick up some new stuff in the meanwhile

@KajHansen: I run graduate tea and coffee here at NDSU and we talk all the time about stuff like this. Its a really good emotional release. Also what I found that helped me a lot was just to dispose any thoughts of inferiority towards others. When you study mathematics, study it for yourself. Who cares what other people know!

Hey @Faraad

@KajHansen: Also, mathematicians don't have to computers. If something is really on your heart, say it! Don't keep it in! Feelings are not a sign of weakness.

@Cbjork: My man! wassup?

Also, I apologize for those really trying to do mathematics in this forum. It's just really nice to catch up with some friends.

I want a sugary doughnut D:

@KajHansen me 2

our prof Uses Fulton and harris

That last point is perfect advice @FaraadArmwood. I've been talking to Doug very in-depth regarding my feelings and emotions. I don't know what I'd do without a social release

@KajHansen that is good. I think one good thing I am thinking about that I should be doing next semester is wake up early in the morning go to gym and study afterwards.

@KajHansen is the following true? every linear function is injective. for a function $R\to R$ this holds, but im not sure about higher dimensions

That definitely is an easy trap to get stuck in @Faraad. It helps to remind myself that there's always going to be someone better than me. Unless I'm Mochizuki :P

@KajHansen I used to tell myself I should be doing math 24/7 but excerise is very important for mental being and being able to focus.

eats @Kaj ' s finger

False @Null

Sorry @usukidoll, lots of distractions :P

.-.

@Null: For a very easy example, let $f \equiv 0$.

@KajHansen can you give a small example? i can't think of one

what does it mean to get the R\I? ???

@FaraadArmwood ah, i see

ah lol

@Faraad you know just in the middle of apps

then this doesnt hold even for R^^

having a good time and working on problems

How are you @Faraad?

eats

The gym's absolutely wonderful @Adeek. I went daily last two years of undergrad. I came to depend on that endo-opioid release for some relief. It's harder to get to the gym back home, unfortunately, so I do calisthenics instead (not as good, but something)

@usukidoll, $R/I$ will have cardinality $|R| / |I|$. This greatly restricts the possibilities for each

?! so the zero ideal is just 0?

@KajHansen I am depressive too, i wonder wether that's the case, because I am agnostic(mild atheism).

@Cbjork: Noiice!! Where are you applying? I hope that thing I sent you helped.

In particular, the only possibilities will be the trivial ring, $\mathbb{Z}_2$, and $\mathbb{Z}_3$

yeah @KajHansen I will go daily starting from next semester. It is not a waste of time to have some 40 min stress release.

I've been taking antidepressants for almost a year. there we go

(And $\mathbb{Z}_6$ for the trivial ideal)

aren't those the maximals?!

@Null, what does your religion have to do with depression?

@Null: There is some correlation between religion and the mental state. It may sound easy when I say this but, If you want to be happy, just be happy! There's nothing to it.

@Faraad I'm applying all over the place. Biggest two are UC Davis and UT Austin. But I'm focusing on a lot of lower schools, too. Yea, your list gave me a lot to think about. It will definitely go into my consideration

@KajHansen i don't know, maybe hinduistic people are happier because they belief in their stuff.

I actually became convinced of Christianity over the past year after a lifetime of atheism. That's one hell of a discussion that can't easily fit in this chat format though.

(i picked hinduism exactly for that reason, to not offend christians lol)

(and to not start a discussion about that )

@KajHansen I am atheist.

cardinality is the size of the set

k can we like move to math stuffs?

@usukidoll, you can find the ideals in $\mathbb{Z}_6$ fairly easily by picking generating sets

First, see what is generated by $1$, $2$, $3$, etc individually

Next, see what's generated by sets of two elements

Oh cool

yeah the notation of the second part of the question threw me off

I was like wha?

Wait, what do you mean by "2"?

4 is 2 because 4 is already a part of the ideal generated by 2

Then it won't be distinct from 0, 2, 4

@Cbjork: Yeah, really look into the schools and not just their rank. I have friends in top 5,10, etc schools and it's not really about that. A good school will help you get a job and that's for sure. But as you can see from this chat, there are other things to think about than just working at the best school. You need to be happy where you are.

and then 5 is 1 because 5 is already of a part of the ideal generated by 1

@FaraadArmwood, I think a common trap is looking at literal National Review rank for grad school. I think it's more productive to think about a specific area you're interested in and see which professors have research interests most aligned with yours.

@usukidoll: Not to butt in, but if you would like a really good exercise try showing that for any integer $n$, one of $n,n+1,n+2$ is divisible by 3.

after I eat a big dinner

There might be a great professor at a so-so school. Like Pomerance at UGA.

cuz I came home tired and hungry and I'm still at that stage

@FaraadArmwood for sure. I'm doing my due diligence for the schools

@KajHansen: Definitely! One of my advisors is the student of a field's medalist. So yeah, there are tons of other things to consider.

If you have all the ideals @usukidoll, then note the cardinality of $R/I$ will be $|R| / |I|$.

For primes $p$, there is a unique ring of cardinality $p$

@KajHansen: The funniest thing about this all is that we post so much in tex we use $ in chat and it doesn't even show hahaha

It actually does @FaraadArmwood. There's a plugin that makes it show up if you click the links on the top right

Oh man. Send it to me! I didn't know.

@FaraadArmwood math.ucla.edu/~robjohn/math/mathjax.html

Yeah @usukidoll, all the possible factor rings have prime cardinality

k. I'm just waiting for whiteout to dry

the cardinal sin for math papers....PEN!

xD

haha

@KajHansen: Noicee!! thanks a lot!

all I have left is 1, 4b,4c, and 5b which has something to do with first isomorphism theorem. actually I did this but my mapping was backwards

Yep!

I think J+I would be the kernel

I found it a bit easier to have my map backwards

but that's like the converse... so how will I do direct?

Oh I see

(a+I) + (b + I) = a+b + I

@Cbjork: do u have a field of interest yet? (just asking out of curiosity)

And this maps to $\phi(a+b)$ based on how the map is defined

@Faraad Low-dimensional topology

Note also $a+I \mapsto \phi(a)$ and likewise for $b$. So you can see that this is additive @usukidoll

@Cbjork: Oh nice!!!!

A similar argument for the multiplicativity of the map. It's pretty similar to what you have above

The thing left to show would be well-definedness

@Faraad thanks. I never knew, what are you studying?>

what does it mean to be well-defined? I forgot D:

A map is well-defined if all the things in a given equivalence class map to the same place

So for a well-defined map out of the integers mod $5$, we should have the image of $3$ and the image of $8$ the same

Since $\overline{3} = \overline{8}$

@Cbjork: With my advisor, 3-manifolds & Geometric Group Theory, but I really love diff geo so I study that on the side. ODE has also become of interest to be due to my liking of Poincare.

ugh I want to switch the order it was easier that way

well defined for a sequence means that no term is undefined? (0/0 for example)

@Cbjork: You need to talk with Dr.Gay asap! He knows almost everyone there is to know in the field.

Yes @usukidoll

snickers

so I let a and b be in R/I

To show well-definedness, you need to select $c, d \in I$ and show that $\overline{\varphi}(a+c) = \overline{\varphi}(a + d)$ for a given $a$

(It's important to remember that "a+I" is a set of values that are all "equal" under the "mod I" equivalence relation)

It's pretty straightforward in this case. Just a formality really

@Cbjork: Don't leave me hanging. What's the course?

@Faraad 4th dim. euclidean geometry

(a+c)+I = (a+d) +I

disguised as a 5210 class

pukes

@Cbjork: Hmm, probably related to his work on tri-sections.

That's really cool though.

Not with +I anymore @usukidoll. Just a + c and a+d with c, d \in I

(a+c) = (a+d)

Exactly, modulo $I$

but do I keep my homomorphisms with the +I?

You still send elements of $I$ to zero @usukidoll

so c and d is in I and they are sent to 0

@faraad yea I thought so. It runs at 8am so we'll see how dedicated I am to the subject

leaving me with a=a

Basically lol

This is the first time I've ever heard of a major-level math class starting at 8 AM

Earliest are usually 905

@Cbjork: Thanks for saying something about that course. I actually need to read Four Pillars of Geometry. That seals the deal for my two projects over christmas hahaha

meh, how do i grow up to a meaningful member of science :/

i am tired to always ask retarded questions

@Null: If you do what you love and keep your heart in the right place you'll be remembered.

in fact so tired i forget "?"

the earliest math class I had was 8:30 am

and I lived far so I had to wake up at 5am

@usukidoll: If I ever have to take a math class that early, expect all my hw's to be submitted by email.

partial differential equations :/

LOL @FaraadArmwood

I was like Imma go get an A so I won't have to deal with this early time sxxxxxxx ever again

@usukidoll: In that case, I never signed up to begin with ahaha

but I needed 2 400 level courses for my ba so ;p

@FaraadArmwood exactly what i think haha (homeworks per mail)

@KajHansen: hahaha

@FaraadArmwood i just feel i ask "how is 1+1=2", where other talk about multidimensional knots in a hyperplane

@Null: Rome wasn't built in a day and even those who built it stare in amazement.

oh crud I just had to show well-defined on this one xD

I went too far;p

@Null: In other words, those talking about multi-dimensional knots are still baffled by problems in calculus.

no wait back up that was for a

well damn I've done a ;p

@FaraadArmwood i think a big problem of me is that i am even too lazy to read the definitions up :/

yeah I went too far on b

@Null: What do you mean?

@FaraadArmwood i go on my homeworks and try to solve them without actually knowing what they are about. so i end up stuck without reason. Because i am not too dumb to understand the definitions, they are just one giant pile of uninteresting stuff for me :/

@Null: Come on..how can you expect to do a problem and not know the definitions? Why are the definitions are uninteresting?

An even bigger question, why are you approaching mathematics the way you do?

@FaraadArmwood because I don't understand the NEED for them, as why has anyone invented them. what was the problem he tried to solve?

for the beginning stuff you get all those nice examples

Hey @FaraadArmwood can you look at this question? math.stackexchange.com/questions/2047574/…

weeeeeeeee without the definitions or understanding of them , what's the point? the proof will turn to garbage fast

@Cbjork: Point-set is not mu forte at all. I could think about that problem, maybe solve it, but you'd be out of graduate school by then.

@faraad haha thanks anyway

@Cbjork: Haha. However, if you ever have something diff geo/diff top, manifold related just let me know. I like to think about that stuff.

but I should use a different letter like s = w

@Null: I'm trying really hard to understand what you are saying. Are you asking why was mathematics invented?

Those letters are fine @usukidoll

but yeah

woo

$phi(r) \in J$ and $\phi(s) \in J$, so $\phi(rs) \in J$, and pull it back

Kind of a lame question

yeah

@FaraadArmwood not quite. I'm just wondering around why certain things got a definition. But at least our conversation has revealed my bad approach. Gotta work on that instead of my math-problems

@Null: Also it is to my understanding that the european approach to mathematics is much different than mine, namely the former is one in which I've heard believes in doing nothing with extreme rigor and proof. That's not really how I feel so I'm probably not the best person to talk to.

@FaraadArmwood is unrigorousness really a european thing? :D

@Null: Don't beat yourself up about something like that! Just enjoy it, really. No rigor is the european staple (what I've heard at least). What you mentioned is what I suffer from hahaha

@Null: Also, take nothing I say here to be absolute truth. I can only give my opinion and how I internalize things. You have to come up with a way that works for you.

another problem is, that I use stuff that I don't understand. From today on I will not do so.

that is good @Null

for example partial integrating was a pure remembering thing for me. it made no sense, i just was good at applying it to my calc(if needed)

that is unsatisfieing

@Null: I've told you all I can. You just have to go get it! If you are unsatisfied with your understanding of something, get off the chat, read some mathematics and whatever you don't understand, ask here or just on stackexchange. It's literally that easy. I see all the time people posting about how to get better at math. Well, do more math! That simple, come on.