00:00 - 13:0013:00 - 00:00

12:00 AM
just write -iziert on the end of English verbs
and you can speak German

haha

The Spektrum of $\sigma(T)$ ist kompakt und satisfiziert

ze* Spektrum

haaa
@Thorgott sorry to shit on your language

it's kein problem

12:02 AM
are you yourself sure
such an insult to my language would make me foxdevilwild

what do you call a Klein bottle with no L?

Kein Botte

lol
I have failed

Interesting fact: it should actually be a Klein surface but the translator mistook "fläche" for "flasche" so we call it a Klein bottle

Difference between kein bottle and klein bottle: one of them exists, the other doesn't.

12:04 AM
heh

Now I wonder what other identities I can play with that are only one level of nesting to make the error of my approximation zero 🤔
I mean 10^-7 is something I could live with... but not without trying for better, now, if not later.

Am I being an idiot: one defines $K$-rational points in projective space as $\Bbb P^n(K) := \lbrace [x_0, \dots, x_n] \in \Bbb P^n : \text{all}\ x_i \in K\rbrace$. The author then says "Note that if $P = [x_0, \dots, x_n] \in \Bbb P^n(K)$, it does not follow that each $x_i \in K$."
but like

The tantalizing bit is that all the error functions look like some kind of lemniscate.

wot
that doesn't sound right

well
it sounds dumb
hahaha

12:08 AM
is this in Vakil's notes

no this is in Silverman's elliptic curves
So he makes that statement and then says "choosing some $i$ with $x_i \neq 0$ it does follow that $x_j/x_i \in K$ for every $j$ "

perhaps he means to say that $x_i$ is not well-defined as it depends on a choice of representative, whereas $x_j/x_i$ is independent of the choice of representative
(the condition $x_i\neq0$ is of course independent of choice of representative too)

so each $x_j$ is well-defined up to a factor $\lambda \in \overline{K}^\times$ and dividing by some non-zero $x_i$ gets rid of this $\lambda$
this seems like a good choice for a "standard representative" of a point in projective space (because you end up with a $1$ in some position)

yeah

but his definition of $K$-rational points seems wrong then

12:13 AM
this gives you the standard coordinate atlas on projective space when you make it into a smooth manifold
what's a $K$-rational point again

I wrote the definition above

@Clarinetist what topics do you need?

huh, I'm confused
what's projective space for you exactly
cause P^n and P^n(K) are the same to me

affine space^{n+1} modulo $\overline{K}^\times$

$\overline{K}$?
oh

12:15 AM
algebraic closure lol

I see

Right so you have $\overline{K}$ and then a $K$-rational point is apparently just a point of projective n-space whose coordinates lie in $K$

yeah, that makes sense

But then the remark is dumb, whereas the remark makes sense but only if you modify the definition somehow lol

hmm no, it says that just because one representative has all coordinates in $K$, other representatives need not to
Like $[i\colon i]\in\mathbb{P}^2(\mathbb{R})$, but $i\not\in\mathbb{R}$

12:18 AM
hmm the definition is confusing then
I mean, for me
idk
it's suspicious because he writes "all $x_i \in K$" and not "$x_i \in K$ for all $i$"
rofl

probably just sloppy wording

Is there actually a method for converting from a recursive relation to a function or non-recursive relation?
Thing is, I can keep creating error functions that will supposedly continue to increase in accuracy because I have a gudermannian approximation. I just store the difference of a true hyperbolic function and a circular function approximation over and over infinitely many times and you get an infinite precision transcendental function through infinite recursion, or at least in theory :)
Imagine using the same e^x to approximate the function you're trying to approximate by approximating e^x using an approximation that uses e^x...

I'll just define $\Bbb P^n(K)$ as the set of points of projective n-space that are invariant under $\operatorname{Gal}(\overline{K}/K)$

Would you say this question is better for meta math.se? :P

Use a generating function or smth idk

12:29 AM
How exactly would that work?

as in
you have your $a_{n+1}$ as a function of $a_0, \dots, a_n$ and then define $f(X) := \sum_{n=0}^\infty a_n X^n$
and then do things

I'm currently defining a natural transformation whose elements are natural transformations
Tautological bookkeeping is seldom this irritating

Bookkeeping is the only English word containing three consecutive pairs of repeated letters

The recursive relation computes a set of infinitely many error functions that approximate a circular function, say cos(x), that then approximates the Gudernmannian function and goes back to the original hyperbolic functions, cos(gd x) of course being sech(x). so you generate the approximated hyperbolic, subtract it from the exact sech(x), and then repeat the cycle infinitely many times.

TIL

12:35 AM
I need to read the proof that the definition of profinite group is equivalent to a Hausdorff compact t.d. group but I remember it being long and boring but I feel like it's something I should know

isn't it just a top group that's a Stone space

yes

see, easy

hahaha
I mean, the thing I want to prove is the thing that it is just

So I guess in a way, I did find an infinite precision closed form of the circular functions that converges more quickly than other infinite series as a recursive function...

12:39 AM
$G = \varprojlim_{N\ \text{finite normal}} G/N \iff$ $G$ top group stone space

Man... but I didn't want an infinite series!

I am gonna blackbox so, so much junk this semester

Hey, is there like some gigantic list of trig composition identities somewhere that I can use? The one on wikipedia is rather sparse.
I need something that makes a known closed form function from a circular function with the constraint that the argument to the circular function must not be another circular function (or its inverse).
If I get that, 'tis game over.
At least in theory

0

Could we consider the statement that the Earth is Cube is the negation of the Earth is round?

well, have you ever seen an object that was neither round nor cubic

12:48 AM
Yes. squircles.
Though if that's 3D, would that be squircloids? lol

Squircle? You mean an $\ell^3$-ball?

A what ball
But you can see my idea here, right? These are "one level of nesting"... if I knew the error function for any single one of these, I could make the error zero: desmos.com/calculator/w4wecmcpws
So I mean... if one of you knows how (without parametric equations) to make a circle using only one level of nesting rather than something like cos(arcsin(x)), I'm all ears.

I've visited this site 1528 out of the 1700-ish days since I joined

Oh no...
He's about to say something significant probably. hides

the significance is that I'd probably be a lot better at maths if I hadn't ever joined this site

12:59 AM
<in old man voice>: Y'know, I've been in this industry for 51 years out of my whole life since I was 1 year old... I've ne'er seen in all those 51 years <some significant object of interest> in all my life!
I thought you were going to go for something like that

Alas, I am only 26

20 going on 21 here o/
Negation of Earth as cube though...
I therefore assert Minecraft as the negation of spherical geometry.
Err I mean spherical lattice geometry
How do you multiply without multiplying?

For instance, in the tropical semiring the multiplication is addition of two elements

Oh, so you're saying that in the tropical semiring, e^i + e^x = cos(x) + isin(x)? Sounds lovely!

In the category of matrices, the product of two natural numbers is their sum

1:12 AM
for finite $\Bbb Z$-modules $\prod = \bigoplus$
heh

Hm, I'm getting answers... so what if I ask "how to compute cosine without computing cosine?"

what does that even mean

Same thing in principle as How to multiply without multiplying

for instance, Taylor expand cosine and then truncate
then you haven't computed cosine but you've got something close to cosine if you just take enough terms
for some suitable definition of enough

1:14 AM
1. Just reverse engineer a closed form from the infinite series. 2. Profit.
If I can convert anything to an infinite series, why is there not an operation to convert back from an infinite series?
(To a closed form on the reals)
Also would also be nice if I could just, y'know... do something like xsinh(1)/(1+x^2) and get a circular function, but the infinite series doesn't yield to that nor does it plot a circular function.
(2k+1)! varies, so I guess it is a matter of finding an infinite series where the denominator/coefficient is constant which is also a valid definition of the circular functions. Same of course obviously goes for (-1)^k.
Now if I could find a suitable function that maps linear and continuous rotation to the perimeter of a square, that would be great. I already have sine and cosine of a square.
Question: can you make the Archimedean spiral using rectangles or square, or no?

1:41 AM
hey wait a secant...
Can't I get "hyperbolic square functions", map from hyperbolic functions to the hyperbolic square functions, then map from hyperbolic square functions to square functions and then to the circumference of a circle to get the circular functions?
Could that actually work?

Let $\mathcal{C}$ be a category and $I\in\mathcal{C}$. What do the monomorphisms in $\mathcal{C}/I$ look like? Obviously, if a morphism in $\mathcal{C}/I$ is a monomorphism in $\mathcal{C}$, it is a monomorphism in $\mathcal{C}/I$, but I don't see whether the reverse implication holds.
I don't really see a way of tautologically conjuring up the necessary commutative triangles

Right, well I'll be heading off to bed now. I bet you there's some way to map things to the Euclidean plane using the square's sine and cosine from other geometries... even from complex, if only in part, by defining a square in that plane, doing whatever mappings you want, and then putting it on a square in the Euclidean plane.
Makes sense, right?
Anyways, good night!
(You know, like grafting...)
(good night)

cya

2 hours later…
3:57 AM
Classic problem solving technique

4:20 AM
still a w a k e

5:03 AM
> “There was a seminar for advanced students in Zürich that I was teaching and von Neumann was in the class. I came to a certain theorem, and I said it is not proved and it may be difficult. Von Neumann didn’t say anything but after five minutes he raised his hand. When I called on him he went to the blackboard and proceeded to write down the proof. After that I was afraid of von Neumann.”
― George Pólya

2 hours later…
7:04 AM
@EdwardEvans no more

7:53 AM
still a w a k e

lol wtf sleep

I‘m in the city center
I just learned that the additive group of a local field is locally profinite and the multiplicative group too
Cool

8:13 AM
@EdwardEvans proof by cases
hmm but the multiplicative group of the function field is very weird
wait but I suppose it's still the limit of the units
like $\mathcal O^\times = \lim (\mathcal O/\varpi^n)^\times$ right
is this the proof

8:44 AM
@Leaky the groups $1 + \mathfrak{p}^n$ are compact open subgroups of $F^\times$ and they're a system of open neighbourhoods of $1$
so by definition $F^\times$ is locally profinite lol

wait when is that the definition
@MartinR wolfram alpha?

Locally profinite means every neighbourhood of the identity contains a compact open subgroup; closed subgroups of locally profinite groups are locally profinite and open subgroups are clopen, and locally profinite groups that are compact are just profinite groups, so by this definition, every open neighbourhood of the identity contains a profinite group
so.. the terminology is apt
lol

no but profinite means a projective limit of finite
that's what it means by definition right

sure and this is equivalent to Hausdorff, compact and totally disconnected
locally profinite is just Hausdorff, locally compact, totally disconnected
lol

I don't accept them as definitions :P

8:52 AM
locally profinite is to profinite as locally compact is to compact
just alter the definition to "open neighbourhoods contain ____"

yes, so I showed that $\mathcal O^\times$ is profinite

yes but $\mathcal{O}^\times \neq F^\times$

but it's an open neighbourhood of $1$
$\mathcal O^\times$ is profinite so $F^\times$ is locally profinite
ah right, there are two definitions of "locally compact" as well
I'm using the "weaker" one

I see lol
anyway

right, anyway

8:55 AM
I need to sleep for like 16 hours

ja, das brauchst du

I'll shout more things about smooth reps of locally profinite groups into chat periodically over the next few months
stay tuned
und ja das brauch ich
hahaha

how would you translate das brauch ich?

I need that

8:57 AM
also I need that

how would you translate them differently

Hmm

this is needed for me ?

hmm

No it's a subtle thing but idk how to articulate it

8:58 AM
maybe das brauch ich = I need that

also hi

As a "stand alone" statement, "Das brauch ich" und "Ich brauch das" are basically the same, but I think "das brauch ich" is more common and sounds more natural
On the other hand, after something like "ja" you have to say das brauch ich
You can't say "Ja ich brauch das" or smth
idk

yes, that is what you need

Ja, das ist das, was ich brauche

yeah ok they're different
I don't know, I don't speak German

9:01 AM
Yeah I've reached a level where I don't know how to explain why things are correct

yes, that you need

9:42 AM
Guys I have some problem with these so called naming and isomorphism topic
The statement "Isomorphic groups are essentially same"
I think this is only true when you are renaming the elements of group and there is isomorphism between group and the constructed group by renaming.
And it is not true for arbitrary group.
Also in second page it says "${\bar{F}$ has same elements as $G$"

the isomorphism is the renaming

I don't know why "${\bar{F}$ has same elements as $G$"?
@Astyx I hope you have read whole page but this is what I thought first.

But it doesn't mention this isomorphism is renaming so I assume for arbitrary
@Astyx you can mention this but I think let's not care about minor error
I don't have a lot of time so I will write my precompiled question from my brain
"Isomorphic groupoids are essentially same" I think if a group $\bar{G}$ constructed by renaming elements then they are Isomorphic and are essentially same. But above statement means arbitrary Isomorphic groupoid are essentially same. Or may be I am wrong since I couldn't see two arbitrary groupoid are same. The property of Isomorphic groupoid IMO says it isn't necessary for Isomorphic groupoid to have same elements and binary operation thus renaming is the option remaining.

the keyword is "essentially"
means that the names of the elements are irrelevant
they have the same properties
one is discrete iff the other is
one is finite iff the other is

9:56 AM
"Groupoid isomorphic with with a groupoid F is indistinguishable from a suitable renaming of $F$ as far as the elementsand the way they multiply are concerned" I think the Isomorphic used in the groupoid Isomorphic with a groupoid $F$ and it's renaming should be same.
And that isomorphism is renaming is renaming one. Or else I don't understand how can arbitrary isomorphism make the above statement true.

Your sentences are unreadable. An isomorphism is renaming elements in a way that the structure is preserved, like Astyx is said; so you can identify isomorphic mathematical objects. Be it groupoids, or whatever.

Intuitively I think above statement is true but by how book says I think description is not proving the same thing as the statement.
I got only 2 min.
Also why F bar has the same elements as G?

Come back when you have more time so that your questions are not written incoherently :P

@BalarkaSen ok ;_; sir
@BalarkaSen How you know it's just renaming? By the description it just says it need to be homo and bijective
I know renaming will make isomorphism but I don't understand why isomorphism is just renaming
damn I need to go offline

1 hour later…
11:25 AM
Good morning!

11:38 AM
@BalarkaSen I have more time now.
So my what my brain is saying is "I know the renaming is isomorphism but I don't know why is isomorphism renaming?"
What if there are some isomorphism not renaming?

An isomorphism doesn't in it's technical definition make any reference to names

Now if you know "An isomorphism is renaming elements in a way that the structure is preserved" then prove it.
@Charlie That's what I am trying to say.

How does one prove that, that's what an isomorphism is

Out of my mind lol
What leaky nun said is something I already know.

You cannot prove that an isomorphism is a structure preserving bijective because that's what it is by definition. You could find specific examples of isomorphisms and prove that they satisfy those properties

11:42 AM
@Charlie Yes I know this too.

Then I'm not sure what you're confused about

An isomorphism is by definition a "renaming" of elements.

My problem is why does groupoid $G$ has same elements as $\bar{F}$?
And now old problem is why "An isomorphism is by definition a "renaming" of elements."?

It doesn't. You can relabel the elements of $G$ by those of $\bar{F}$ in a specific manner, compatible with the binary operator (so that (label of a*b) = (label of a)*(label of b))

Or is book defining this way,

11:45 AM
It's not a definition. It's a way of thinking about the term "isomorphism". Work out some examples.

I think it is some way to think but now what isomorphism is but I will try to work out with example to make sense out of it.

You could for instance change the symbol representing the integers 0-9 to random scribbles, but as long as they still have the same relationships re addition and multiplication you haven't created a new object, you've just relabelled the points

@BalarkaSen Wait so it was just relabeling those elements? I was thinking about that since they used $\theta$ in both
then is has same elements

I feel like the phrase "structure preserving bijection" leaves little room for ambiguity but fair enough :P

@Stupidquestioninc Look at simple cases of groups like S_3, D_3, GL_2(F_2) and others. You are studying a structure and you are saying two structures are the same if they behave the same way. It doesn't matter if I call one structure hell and the other heaven

11:48 AM
$\theta$ is a map, not an element of $G$ or $\overline{F}$. It's the isomorphism actually.

I think book has some ambiguity i have thought about that earlier but now I don't remember

Which book are you reading from anyway? Seems like a terrible book.

@BalarkaSen I know that lol

Then what does the question "I was thinking about that since they used θ in both" mean?

11:49 AM

@BalarkaSen isomorphic map

Oh no no

I bought it too late
by infamous mcgrawhill
even rudin has lots of terrible error in book that I corrected which left me in terrible confusion

@Stupidquestioninc But what is your question? "isomorphic map" is 2 words, it's not a question.

You need to understand what a relabelling is formally
It's a function that takes labels and give other ones, in an injective fashion
If you want to preserve the group structure, you ask the relabelling to be compatible with your group operation

11:51 AM
@BalarkaSen my question was just why is isomorphism renaming and why does $\bar{F}$ have same element as G?

And what you get is exactly an isomorphism

@Stupid (1) Write down what the groupoid $\overline{F}$ is (2) Write down what the groupoid $G$ is
Tell me when you're done.

@Astyx I know this exactly but you guys were telling me isomorphism is relabelling not relabelling is one type of isomorphism

I'm telling you relabelling compatible with group structure IS isomorphism
In both directions

what do you mean by both directions?

11:56 AM
It's an equality
Not an implication

Dude do (1) and (2)

hey, whaddya know... I actually have a good question for once within my question about the mapping from complex to reals: math.stackexchange.com/questions/3829746/…

That's a long comment chain

You don't say lol
All these years later, and SE still hasn't implemented a "side chat" feature for questions.
There's a reason why moderators still end up deleting the comment section and moving everything to a private chat room leaving behind only a reminder: "The comment section is not a chatroom..." etc.
ez fix

$\overline{F}$ is group constructed by relabelling elements of $F$ and by this we mean $f\theta=\overline{f}$ where f is element of F and f bar is element of F bar . Where theta is mapping which maps F to F bar obeying bijection and homomorphism.
1
and 2

12:01 PM
By the way I would still like to mandatorily state that this is not an xy-problem even though I'm asking for a function of x and y :P
hides

groupoid G is just isomorphism with F with same mapping theta
@BalarkaSen I write that in informal way
so both have same mapping that means both elements are same

@Stupidquestioninc What you said cannot be correct, because according to your first comment $\theta$ is a map $\theta :F \to \overline{F}$ whereas according to your second comment $\theta$ is a map $\theta : F \to G$.

this is what book is sayin
;_;

Do you understand why what you are saying is incorrect? I don't care about the book.
I want you to understand
A single map cannot have different ranges.

Finally, I've met a teacher that actually cares. If only I had such a maths teacher in school...

12:05 PM
@BalarkaSen can be if both are same i mean both are equal
range
so I thought they mean G and F bar are equal

@Stupidquestioninc Right, so what you said makes sense only if $\overline{F} = G$, equal as sets. Elements of $\overline{F}$ are the same as the elements of $G$, on the nose. No relabelling required.
You're correct.
But that doesn't give any insight into what's happening, we just walked in a circle.

what i was confused was one is defined relabeling and another is just mapping to arbitrary so book makes no sense lol

Yup.

so it means book is wrong then what am I learning?

So let's forget about the book and do an actual example.

12:08 PM
what is reason for this topic to exist ? Just to kick my arse ?
another ambiguity they have is in second paragraph they didn't say what g is element of
@BalarkaSen as you wish mlord!

I don't think the book is wrong, just badly worded

@Astyx so what does book mean :))))))
just want to know what book actually means.

@Stupidquestioninc Do you happen to know any modular arithmetic at all?

I usually find author use their lousy explaination so it gets pretty confusing at the end of the day

The book is trying to explain how relabelling is the same thing as an isomorphism
Specifically, if you have an isomorphism $\theta : F\to G$, then you can relabel every element of $F$ by an element of $G$ by taking the image through $\theta$
This preserves group structure and shows than, up to a relabelling of the elements, $F$ is the same as $G$

12:17 PM

I will define the following groupoid. $V_4 := \{1, a, b, c\}$, multiplication defined by $1 * a = a * 1 = a, 1 * b = b * 1 = b , 1 * c = c * 1 = c$, $a * a = b * b = c * c = 1$, $a * b = b * a = c$.

(1) Verify this defines a groupoid. In particular, check for three things: $V_4$ has an identity, an inverse and a partially defined multiplication. Draw the multiplication table for $V_4$. Bonus: In fact, demonstrate $V_4$ is a group

@BalarkaSen  You mean demonstrate $V_4$ is a groupoid?

No, you did that on the way to the last sentence already. B wants you to show that it is in fact a group

That's pretty easy exercise lol

I will also define the following groupoid. $\Bbb Z_2 \times \Bbb Z_2 := \{(0, 0), (1, 0), (0, 1), (1, 1)\}$, where multiplication is defined by adding componentwise, but you always set $2 = 0$. So $0 + 0 = 0, 0 + 1 = 1 + 0 = 1, 1 + 1 = 0$. This is called addition modulo 2. Computer scientists do this all the time.

(2) Ditto for $\Bbb Z_2 \times \Bbb Z_2$. Verify everything about $V_4$ in exercise (1) for this guy.
(3) Construct an isomorphism $V_4 \to \Bbb Z_2 \times \Bbb Z_2$.
Thanks!

12:20 PM
:)

ok I will do it when I get out of bike

12:46 PM
Fun fact: You do not understand the history of division by zero by just looking at cranks or just looking at academics. You need to read both
Everything in the academic section of this list do publish in some reputable journals
Division by zero may be the subject where cranks and professionals intersect without much disagreement

hey chat

@BalarkaSen hehe no identity no inverse no is groupoid multiplication table i will show you wait for a mom v_4 is not group by along the fact no identity and inverse
1abc
1_abc
aa1c_
bbc1_
cc__1
now for second exercise you gave
beb after a meal

question: assume that $u: \mathbb R^2 \to \mathbb R$ is differentiable at $(x_0, y_0)$, with continuous partial derivatives. how do I prove that there's a linear order approximation of $u$ in terms of $u_x$ and $u_y$? i.e., how do I prove that there are positive reals $\Delta x$, $\Delta y$, $\epsilon_1$, $\epsilon_2$ s.t. $u(x_0+\Delta x, y_0+\Delta y) = \Delta x \, u_x(x_0, y_0) + \Delta y \, u_y(x_0, y_0) + \epsilon_1 \Delta x + \epsilon_2 \Delta_y$?
should be an easy analysis exercise but it's just a step in a "Cauchy-Riemann + $C^1 \implies$ differentiable" proof, so I don't think I should be wasting a lot of time on that

Use the limit definition of the existence of the derivative, and the explicit form of the Jacobian

@Secret I remember being in high school having my own ideas about division by zero. My idea was rather simple, although I haven't actually thought about it much since then: let x/0 be a number in the set of "mirrored" reals such that (x/0)/0 = x, and 0/0 = 0.

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