Mathematics

Thorgott

00:00

matter in which way?

Xander Henderson

@Thorgott Something something Higgs boson?

Jakobian

I'm just not sure why its important here, or is it important?

Thorgott

important to what end?

Jakobian

I'm basically asking why you mentioned that its maximal

Thorgott

cause together with locality this implies any non-zero element is a product of a unit and a power of $x$

hence inverting $x$ suffices to yield the quotient field

differently put, $k[[X]]\setminus\{0\}$ is the saturation of the multiplicative subset generated by $x$

Jakobian

00:07

that sounds so complicated

Thorgott

it's not saying anything more complicated than that a formal power series is invertible iff constant term is non-zero

which is an observation you definitely need to make

Jakobian

ah yeah. Yeah I see now

I was just thinking the viewpoint of $k[[X]]$ being a local field with maximal ideal $(x)$ is a bit too much for me, but I remember doing similar argument in Liu

and I get why its true

Balarka Sen

its needlessly fancy but it gels well with how a holomorphic function $f(z)$ near $0$ can always be written as $f(z) = z^n g(z)$ where $g(0) \neq 0$ (and therefore, $g$ is locally invertible around $0$)

Xander Henderson

None of you have mentioned "sheaves" yet. For shame.

Balarka Sen

00:22

Given a topological space $X$, call a "relative Riemann surface over $X$" a map $f : \Sigma \to X$ from a Riemann surface $\Sigma$. Let $\pi : M \to \Sigma$ be a complex submersion. The moduli stack $\mathrm{Hol}(M/\Sigma/X)$ is a sheaf on $\mathrm{Top}$ such that...

Xander Henderson

@BalarkaSen THANK you.

Thorgott

01:06

@BalarkaSen bruh

Random Variable

01:38

@robjohn It seems to only be an issue in Windows and Linux. I thought a clean install of Firefox would at least fix it temporarily, but it didn't. As you stated previously, it's like it's getting stuck before the final render (although some stuff doesn't render at all.) It's been broken for over 2 months now.

EE18

02:56

Unfortunately I got sidetracked so just catching up on the discussion

thanks to all for the comments

My final question of the evening...is the reason we demand that the ring R here be commutative so that the homomorphism is indeed a homomorphism?

That is, there's no problem defining the ring of formal power series in multiple indeterminates whether or not R is commutative, right?

(Actually I should probably strike that question from the record, I know Leslie will correctly point out I shouldn't ask about noncommutative rings at this point!)

leslie townes

03:13

oh you can totally look at noncommutative rings, but literally everything gets more complicated. what you're not allowed to do is ask questions that cause people to wonder whether a "ring" is definitionally required to have a multiplicative identity or not.

so "ring with 1," FOH, but, noncommutative ring, fine. but note if you stare long enough at the "usual formula" for multiplying sum c_k x^k and sum d_l x^l, i.e. what you'd want to think of as "multiplication in R[x]," you are implicitly using, or at least wanting ot use, that the c's and d's commute with x.

which does indeed complicate what 'evaluation' is going to be, but also what maybe implicates what "R[x]" is and how to define a multiplication on it so that it's a ring.

i will decline the invitation that is always out there from the universe to race immediately to the most general abstract structure i can think of that generalizes this situation and how to surmount all of the difficulties that arise in general.

but basically that is why that hypothesis is there.

EE18

03:42

Makes sense, thank you as always Leslie!

Noted re no rngs!

Ryder Rude

08:47

if $\sigma _r$ is an oriented r-simplex, what is the geometric interpretation of $2\sigma _r$ (which is a r-chain)?

i mean an interpretation of a general r-chain

@EE18 hi. what is this subject called

Jakobian

10:11

@RyderRude ring theory

Ryder Rude

10:48

thanks

Thorgott

12:39

@EE18 yes

the bigger question, as leslie indicates, is what you want the formal variables to commute with

and the evaluation morphisms will require different hypotheses to be well-defined

@RyderRude in this generality, none

Ryder Rude

13:16

@Thorgott oh. and sometimes it can be interpreted as a union surface of simplexes, right?

Sha Vuklia

13:42

@leslietownes hahahahah leslie, thanks for the laugh

that was some good joke mastery (i expected something stale since this template is used a lot)

Ryder Rude

14:20

can all mathematical isomorphisms be seen as different notations to do the same math

Balarka Sen

14:50

@RyderRude You should think of it as a map $f : \Delta^r \cup \Delta^r \to X$ where $f$ is given by $\sigma_r$ on both factors.

Any general $r$-chain can be thought as map from a Delta-complex to your space.

Sha Vuklia

@leslietownes Nice observation! I didn't expect $R$ to turn out so nicely. Thanks :)

Jakobian

@RyderRude uh... no?

Balarka Sen

Only if all triangles of isomorphisms commute

Jakobian

I'm not exactly sure what you mean. But isomorphism often is something that preserves structure so that the objects are the same up to relabeling

But isomorphism in itself is also a very abstract notion

It doesn't have to obey this rule, as, in my opinion, category theorists butchered the beauty of in some part

Ryder Rude

@BalarkaSen oh

@Jakobian yes..this is how i think about it

@BalarkaSen i havent studied delta complex yet tho

Balarka Sen

15:07

It's advisable to study delta/simplicial complexes before embarking on singular homology.

Ryder Rude

Nakahara's book introduces simplexes and then homology group

sorry simplical complexes

and then homology groups

ive studied simplical complex

Balarka Sen

simplicial complexes are more or less the same as delta complexes for the purposes of this discussion

Ryder Rude

@BalarkaSen what is X

Balarka Sen

Whatever your space is

Ryder Rude

let's say we mean the r-chain group of a simplical complex K. then is X=K?

Balarka Sen

15:13

Oh, you're doing simplicial homology

Ryder Rude

yeah

Balarka Sen

Yes. You should just think of $2\sigma_r$ as a particular simplex "counted twice".

In $K$

Ryder Rude

okay. and i can think of $\sigma _{1r}+\sigma_{2r}$ as the union, right?

Balarka Sen

Weighted union.

Ryder Rude

so when the coefficient is 1, it's just the union space

Balarka Sen

15:15

Sure.

Ryder Rude

and when the coefficient is other things, i shud think of it as a stack of simplexes

Balarka Sen

Yes

Precisely

Ryder Rude

thanks

this addition algebra thing is basically the algebraification of spaces and unions, so we can compute boundaries using an algebraic computation

Balarka Sen

The key point is to allow weights, which can be negative.

This is not present in the naive union

Ryder Rude

yeah. things cancel out in the boundary computation

it's genius

Pizza

16:45

Consider a relation $R \subseteq A \times B$, then:
$$R = G(f) \Leftrightarrow \forall x \in A, \exists !y \in B : xRy$$$$PROOF$$
with $f : A \to B$ application and $G(f)$ graph of $f$.
$\Rightarrow$) The application $f : A \to B$, by definition, associates with each element $x \in A$ and only one element $y \in B : f(x) = y$. Then it follows that:
$$(x,y) \in G(f) \Rightarrow (x,y) \in R \Rightarrow xRy.$$
$\Leftarrow$) We define a map $f : A \to B$ which makes every $x \in A$ correspond to the unique element $y \in B$ such that $xRy$. Therefore $y = f(x)$ is equivalent to $xRy$ and theref…

Maybe I understood the first part but I don't understand the proof :(

the proof start from where is $\Rightarrow$ , but i cant edit the message :)

EE18

This is more or less the definition of a function as a special case of relation, no?

(That is, how has your booked defined "function")

Thorgott

likely not properly

Pizza

@EE18 the definition is the normal

i dont understand the proof

Thorgott

that is not a helpful response

Pizza

then I'm reading these things from a power point, because they aren't there in the book... the book starts immediately with the matrices...

Sine of the Time

16:59

who teaches maths using power points ? :(

EE18

I suspect "function" is defined as a "rule associating every input in the domain to a unique output"

Pizza

@SineoftheTime its a 399 slides ppt ...

EE18

Pizza, without a clear statement of how your book defined function it's gonna be tough for me at least to answer you. Maybe someone more experienced can though

Pizza

no wait

nickbros123

ive never seen $\exists !$, does that mean "there does not exist"?

EE18

17:00

"there exists a unique"

can be made rigorous using just "there exists" and some stuff inside the formula to which it applies

Sine of the Time

there is one and only

Pizza

A function or application $f$ consists of a set $A$ called the domain of $f$ , of a set $B$
called the codomain of $f$ and of a law that associates to each element $x \in A$ and
a single element $f (x) \in B$.

nickbros123

I see, then it falls straight from the defn

Jakobian

cabbage rolls are just so delicious

Thorgott

what is a "law"

that's not a definition

Pizza

17:12

@Thorgott would be the expression of the function

Thorgott

that's not a formal concept

Pizza

this is what is written on the slide..

Mad

i dont get the concept of how you can construct positive roots in an abstract root systems, where i am looking, the same thing appears "Construct a hyperplane that does not contain any root, and call elements on one side positive the other negative" wth does this even mean. And how can i construct a hyperplane that does not contain any root? why does it even exist.

Pizza

I can find something on the internet

Balarka Sen

@Thorgott What do you mean by "formal concept"?

Pizza

17:16

a function is a relation between two sets, called the domain and codomain of the function, which associates one and only one element of the codomain to each element of the domain.

its this

If the domain and codomain of the function $f$ are respectively indicated by $A$ and $B$, the relation is indicated by $f : A \to B$

EE18

Leslie I was about to ask a question about rings without unity

I resisted the temptation successfully

Thorgott

yes, this is a definition and it is the same thing as what you wanted to show to be equivalent earlier

the bottom line is: it makes no sense to try and show two things are equivalent when you can't define one of them

EE18

Pizza are you quoting from lecture slides or a book?

If a book, please advise as to which one

Balarka Sen

@Thorgott You still have not defined a formal concept for me

Thorgott

the actual definition of what a function is will be in terms of a relation

it is worthwhile to explore why that definition captures which intuition

but that is not the subject of a proof

@Balarka I don't have the time to take your bait :P

Balarka Sen

17:21

@Thorgott What do you mean by bait?

Jakobian

standard definition at least, I'm sure some people in foundations have like 5 different definitions for a function

Thorgott

@BalarkaSen exercise to the reader

Pizza

i mean this proof: $\Rightarrow$) The application $f : A \to B$, by definition, associates with each element $x \in A$ and only one element $y \in B : f(x) = y$. Then it follows that:
$$(x,y) \in G(f) \Rightarrow (x,y) \in R \Rightarrow xRy.$$
$\Leftarrow$) We define a map $f : A \to B$ which makes every $x \in A$ correspond to the unique element $y \in B$ such that $xRy$. Therefore $y = f(x)$ is equivalent to $xRy$ and therefore $R = G(f).$

EE18

When one speaks of an ideal of a field, is that ideal still a ring or a subfield?

Jakobian

no because rings/subfields have to have identity element

Pizza

17:23

@EE18 im reading the slides from power point

Balarka Sen

@EE18 The answer is one does not speak of an ideal of a field, because there is no such interesting thing

Pizza

i also have a book , but it start with matrix

Jakobian

The only way an ideal can be a subring is for it to be the whole space

EE18

Balarka, that's what I'm trying to prove :)

Balarka Sen

@EE18 What is an ideal?

EE18

17:24

Let R be a ring with unity. A subring I is called an ideal of R if RI=IR=I. An
ideal is proper if it is a proper subset of R.

I'm trying to prove that the only ideals of a field are the trivial ones

Thorgott

I could mention a really funny fact right now, but I'll refrain for the sake of pedagogy

Jakobian

Prove that $I = R$ if and only if $I$ contains a unit

EE18

But my current proof idea demands that $a \in I \implies a^{-1} \in I$

Did that, that was part (a)

Balarka Sen

@EE18 This is a nonsensical definition

Jakobian

since all non-zero elements of a field are units, you have your proof

EE18

17:26

huh? I don't follow. Only 1 is the unit?

Jakobian

No, any invertible element is called a unit

Thorgott

"a unit" =/= "the multiplicative unit"

EE18

Oh ok. So then in (a) I merely proved "An idealI is proper if and only if 1∈I."

Will work on the above then. I guess it relates to my proof idea

Jakobian

yes, as you see, after you show that your question becomes trivial

EE18

Oy, I see now

Balarka Sen

17:28

The reason you're having a hard time proving what you want to prove is because your definition of ideal is wrong

Pizza

I don't understand what you don't understand about what I wrote...

Balarka Sen

An ideal of $R$ is kernel of a ring homomorphism $\varphi :R \to R'$ to some other ring

EE18

Same proof idea as I had, I don't need $a^{-1}$ in I because it's in $R$

@BalarkaSen Oh dang really?

is this not a correct definition?

Jakobian

Don't listen to Balarka, he's giving you misleading information

its not a standard definition of an ideal, what he proposes

Balarka Sen

jakobian is of course the arbiter of standard in mathematics

@EE18 it is but it is meaningless as such

Jakobian

17:30

your definition is completely fine

almost fine... actually it isn't fine

$I$ is not a subring but merely a subgroup closed under multiplication, it doesn't have to contain $1$

EE18

Anyway, this is what I ended up doing: It remains to show that these are the only ideals. Suppose $I$ is a proper ideal of $K$ (if not then $I = K$), whence $1 \notin K$ from (a). Now suppose $a \neq 0 \in I$. Since $a^{-1} \in K$ we have $aa^{-1} = 1 \in IK = I$, a contradiction.

Under my book's definition a ring need not contain 1 so it agrees with you Jakobian

Thorgott

@BalarkaSen I don't like this as a definition

it's putting the cart before the horse

EE18

What do you know Balarka! The next question... "If φ : R → R′ is a ring homomorphism, then ker(φ) is an ideal of R."

It's allcoming together :)

Balarka Sen

@EE18 Prove the converse too.

@Thorgott I disagree

I would also define a normal subgroup as kernel of a group homomorphism. There's no reason to care otherwise.

Jakobian

@EE18 No, after you have a) you just claim that if $I$ is not $(0)$ it must contain unit so be the whole field

EE18

17:33

@BalarkaSen oh goodness!

Thorgott

there is, because you want a definition that can actually be computed

EE18

Why does I not being $\{0\}$ imply it contains 1?

Balarka Sen

what do you mean by computing a definition wtf lmao

is that even english

Jakobian

unit is an invertible element

Thorgott

checked algorithmically, whatever

Jakobian

17:34

thats the definition of a unit

Balarka Sen

checking a definition algorithmically??

Thorgott

my point is: I have a subset $I$ and wanna know I can quotient by $I$

your definition amounts to: construct the quotient by $I$ to know you can quotient by $I$

which is not helpful

EE18

I still don't see how it's immediate without what I did

Balarka Sen

@Thorgott No, this is nonsense. I define an ideal as kernel of a ring homomorphism. Then I give a proposition which gives necessary and sufficient condition to find out which sets are ideals.

This is perfectly fine.

Jakobian

What you want to prove is that $I = R$ iff $I$ contains a unit i.e. an invertible element

I believe this is what they were asking you to prove in a)

Pizza

17:35

Internal operation: it is an operation that takes 2 elements of the set and produces one more element of the set.

Thorgott

that's reasonable

I still think the order is wrong

Pizza

is this correct?

EE18

I think the disagreement is that in (a) I was asked to prove something regarding the multiplicative identity

Not an arbitrary nonzero element

Thorgott

I care more about quotients than kernels, but this can be a matter of preference

Jakobian

I see. Well, in that case, prove the more general statement

it will be helpful in the future anyway

Balarka Sen

17:36

@Thorgott sure.

Jakobian

I say more general, but both results, when written out, are a bit underwhelming

An example where its helpful is: in a local ring $(R, \mathfrak{m})$, $\mathfrak{m}$ consists of the elements that are not invertible

Balarka Sen

it is only natural that you'd prove trivialities when you keep doing set theory without knowing the real meaning

Jakobian

@EE18 Not NON-ZERO element but INVERTIBLE element

robjohn

@EE18 did you mean "$1\not\in I$"?

nickbros123

do u guys have any suggestions for a book on logic that could bring me from ground to "competent" , I cant believe ive somehow survived thus far without a formal "going thru" of logic

Jakobian

17:51

@nickbros123 I can give you what I've been recommended but didn't actually read

assuming you mean mathematical logic on a decent level

if you just mean something very basic and very introductory, then I don't have a recommendation

nickbros123

@Jakobian fire away, ill check it out and see if it fits me

Jakobian

link.springer.com/book/10.1007/978-3-030-73839-6

nickbros123

pff thats too advanced for me

i was looking for some of the basic stuff tho

Jakobian

I did ask what you are looking for

Pizza

Consider a relation $R \subseteq A \times B$, then:
$$R = G(f) \Leftrightarrow \forall x \in A, \exists !y \in B : xRy$$
with $f : A \to B$ application and $G(f)$ graph of $f$. \ $$PROOF$$
$\Rightarrow$) The application $f : A \to B$, by definition, associates with each element $x \in A$ and only one element $y \in B : f(x) = y$. Then it follows that:
$$(x,y) \in G(f) \Rightarrow (x,y) \in R \Rightarrow xRy.$$
$\Leftarrow$) We define a map $f : A \to B$ which makes every $x \in A$ correspond to the unique element $y \in B$ such that $xRy$. Therefore $y = f(x)$ is equivalent to $xRy$ and the…

I dont understand this proof...

nickbros123

18:00

my bad. I jumped straight into Bartle and Sherbert Intro to analysis after HS calculus, and Hoffman Kunze LA, and I found that I did not have much of a problem with anything "understanding" wise, but i have though come across an embarassing and earth shattering gap in my understanding of formal logic

after this I could never trust books that use english words in their propositions

Jakobian

♦ $Mathematics$ Logic

This room is meant for discussion about logic, including found...

1

6

there is this channel they don't seem to be discriminatory when it comes to basic questions, you just need to be allowed to type

Pizza

@Jakobian what is your pfp?

Jakobian

@Pizza a back alley

Pizza

looks like an image from an anime

Soumik Mukherjee

@Jakobian Is that hand-drawn?

nickbros123

18:05

@Jakobian i feel like your "basic" and my "basic" are wildly different

Jakobian

@SoumikMukherjee yes

Soumik Mukherjee

By you?

Jakobian

not by me

Sahaj

hi jakobian

I forgot to ping you today

Jakobian

its a watercolor painting, found on the internet

@Sahaj hi

Soumik Mukherjee

18:07

The picture is pretty realistic though. Kudos to the artist.

Sahaj

pretty cool picture

Jakobian

@nickbros123 I meant it as in, asking for a recommendation about a basic book on logic

Sahaj

jakobian are you still a student? Or have you completed education?

nickbros123

ive only now noticed that a lot of things i do for granted are actually axiom of choice application

Jakobian

@Sahaj why are you asking me that

Peter

18:17

Noticed a lot of nonsense in the chatroom "Math & Science archives."

Pizza

@TedShifrin How is it going?

Jakobian

@Peter for instance?

Peter

the consequences of the Prisoner dilemma for life , the universe and everything.

Soumik Mukherjee

I'd expect a lot of nonsense in the category theory room.

Peter

@SoumikMukherjee Why there ?

@Jakobian The trillion dollar equation is another one.

Soumik Mukherjee

18:26

@Peter Just joking about category theory being called abstract nonsense.

Jakobian

@Peter why is that nonsense? Prisoner dilemma is an example of a non-cooperative game where either one person risks it and gains benefit from it, both don't risk and obtain neutral outcome, or both risk it and both are punished for it. A similar non-cooperative game can arise in different contexts, so its reasonable it would appear all over the place. Why is this video in particular nonsense?

Of course, Veritasium doesn't have good fame when it comes to discussing mathematical phenomena

Sahaj

19:01

@Jakobian I was just curious lol

you're pretty smart

Jakobian

19:19

I wouldn't call myself smart, but I'm certainly smarter than some people... or at least more knowledgeable

I believe that everyone can appear as smart if they follow the same principles as me

I think that I very rarely proved my intelligence in this chatroom, for one

mostly I discuss things I've already learnt

If the ability to read and learn is seen as a sign of being smart, then sure. But I don't really think it is

Soumik Mukherjee

19:35

@Jakobian What are those principles?

Jakobian

I often have only a vague understanding of what I mean

and I'm certainly not a person to tell you how to lead your life

what I mean here is things that you learn you should always do in certain situations

Soumik Mukherjee

@Jakobian I get it, I just sometime feel curious to know about other people's views on various aspects of life.

Sahaj

20:52

@Jakobian it was just a friendly compliment

Semiclassical

21:44

@Jakobian he's got a decent rep in terms of popularizing things. in terms of being correct about the things he popularizes...less so

i do want to build his setup for Schlieren imaging tho: youtube.com/watch?v=K7pQsR8WFSo

mostly b/c Schlieren imaging is awesome, and colored Schlieren imaging is even more so

Jakobian

@Semiclassical mostly for physics from what I know. Not for math

lorenzo

23:29

Hello, while I was reading a proof of Weierstrass theorem in a real analysis book the author mentioned that if one has an open ball $B(x_0,R)$ and two points $a,b\in B(x_0,R)$ in it then all the points in the segment of endpoints $a$ and $b$ must belong to the open ball $B(x_0,R)$.

Now, geometrically this is intuitive, but how does one prove this? I have tried using the triangle inequality to show that for any point $c$ in the segment of endpoints $a$ and $b$ we have $|c-x_0|<R$ but I got $|c-x_0|=|c-a+a-x_0|\leq |c-a|+|a-x_0|\leq |c-a|+R$ which is not useful.

EE18

@robjohn I did indeed, thanks for catching that robjohn :)

@nickbros123 Are you talking about a mathematical logic book, a set theory book, or a proofs book?

Jakobian

@EE18 I think all of the above

EE18

I would recommend Enderton, Enderton, and Hammack respectively (have read only the last two, and actually to be honest I'm only in the middle reading the middle one)

Jakobian

@lorenzo yes, we say that $B(x_0, R)$ is convex

How can prove it is as follows, a point on the segment from $a$ to $b$ is of the form $c = a+t(b-a)$ for some $t\in [0, 1]$

now plug it in $|c-x_0|$ and use triangle inequality

group the terms with $a$ and with $b$, use that $1 = t+(1-t)$

♦ $Mathematics$ Logic

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$Mathematics$ Mathematics

♦ Logic

Transcript for

♦ $Mathematics$ Logic

$Mathematics$ Mathematics