Mathematics - 2024-04-30

12:33 AM

NEWTON-RAPHSON FOR THE WIN!!!!!!!

Srini

Not sure what the right way to draw more attention to a post is. Due to low number of views, I wanted to try my luck asking here. Anyone has any suggestions for my question here: math.stackexchange.com/questions/4905659/… ? Thanks

12:53 AM

@Srini ROC is ambiguous. Don't use terms you haven't properly explained. Writing ROC \in Re(p) > 0 is not good at all

Srini

1:15 AM

Region of convergence. Fairly standard term in Laplace transform literature

Re() is the real part of a complex number. p in this case is the complex variable used in the transform integral

1:55 AM

Don't use acronyms / initialisms if you don't have to. The character limit for SE posts is quite permissive. Write out "radius of convergence".

I also hate the \limits command, particularly for inline mathematics. Why not just display that integral?

And logarithms should be typeset with \log.

I would also be curious to know where the integral comes from. Why should anyone care?

In other news, this question is, what I think the children would call, "cringe".

2:35 AM

$$
f(x) = 6(5[\frac{x}{3}] + (x \mod 3) + (x^2 \mod 3)), x= 1..6
$$

12,18,30,42,60,72 covered!

Those are the twin prime averages in order for $x = 1...6$

NJ Wildberger already proved it dang!

😎😎

2:51 AM

I just watched the video. Obviously this guy needs some YT attention so gave a provocative title. But I don't think the content has solved anything.

@Shaun hey!

Been busy lately with work

Srini

3:10 AM

@XanderHenderson, I am still learning Matjax. Will use your suggestions for future posts. As you where the integral comes from, it's part of a larger problem I was solving and this just happens to be one of the pieces. I didn't have any answer for why anyone should care. Laplace transforms of many non-basic functions are good as reference.

PNDas

3:52 AM

Any inequalities on $(1+x)^p$, $p\geq1$? e.g. $(1+x)^p\geq 1+px$ when $x\geq-1$, $(1+x)^p\leq 1+p(1+x)^{p-1}x$ when $x\geq-1$.

pn: there are some you can derive for at least some values of x by taking more than just the linear terms in the general binomial series expansion of (1+x)^p and analyzing the remainder

PNDas

In the last inequality, instead of ${p-1}$ it should be $p+1$.

PNDas

4:19 AM

How to prove the inequality with $p+1$?

6:29 AM

@DanielDonnelly Hi :)

6:45 AM

@Hi

@Shaun

What are you up to these days?

8:10 AM

Not downvoted. But I find the labeling "monoglot, Western, white, and a man" linguist, occidentalist, racist and sexist. If there is an inclusive science, it's math. It does not care where you come from. The so-named Pythogorean theorem has been discovered by Chinese, Indian, Babylonian, Greek, Egyptian and who knows how many other cultures. Binary number systems have been discovered by the Polynesians. The dihedral group structure is used by the Aborigines around Uluru (Australia) for their marriage taboos. Even animals can do some basic math. And machines, who are now getting better at it. — Captain Emacs yesterday

I think I must agree with this comment

What the fuck?

8:41 AM

First of all, the grouping of people into black, white etc. is inherently a racist concept, I believe used for the purposes of the Nazi Germany. While it is quite wide-spread, those are, by nature, racist terms. People come into various shades, they can be of darker or lighter skin color.

What the fuck are you posting here @Jakobian matrix?

LOL

I'm not sure I understand wtf is going on

I see racism :|

Focus on the mathematics pls

Okay maybe for nazi's it was more elaborate than that, but today its used in a similar way as to group people into classes. The problem is that your skin coloration is subjective and depends on opinion of you or other people

Second. The fact that you assign yourself to all of those groupings, and infer that its harder for you to be inclusive because of that, based on your "race" or your sex, is indeed racist and sexist. You are judging, not only yourself, but all of the people that would be categorized like that by others, as somehow inherently bad. This is a very toxic viewpoint

@Shaun this is my feedback for your academia post

Please contribute to society by not spreading dangerous political ideologies online

user 70432

9:08 AM

Flag any racism as such and move on.

@user70432 the problem is that its a very subtle form of racism. And flagging it won't explain why its racist

I don't think Shaun was aware of this. I think it just came naturally as a byproduct of society they live in

10:38 AM

@DanielDonnelly Studying. I have finally figured out how to use the local HPC facilities. I might have new data soon :)

@Jakobian Thank you for your opinion.

@Shaun

Express the twin primes a group word problem :)

Exercise

That's interesting, @DanielDonnelly. Let me think about it :)

Remember: all twin prime averages > 4 are 6x for some x

prove that first

using modular arithmetic

then consider that modulo $q = 5$ you must have $6x$ that is never $\pm 1 \pmod q$.

So you need to "jump them" to proceed to the next valid position

So you end up with jump sequences for mod 5: $\overline{2,1,2\}$, (starting at 0) then you add 2, then 1 then 2, then repeat

for $7$ it's $2,1,1,1,2$ etc

So there's group theory written all over it

and maybe these $2,1$'s can be expressed as generators in some finitely presented group.

Because if you do this on a graph, you can always make the abstract negatives as going the opposite direction

11:07 AM

I can express this problem as the simultaneous Mall shoppers problem. Imagine a family of $n$ shopers enter a hexagonal mall. However they all split up into individual shopping habits.

Some stores the skipped but what's peculiar about their paths is that it happens to be for shopper $s_1 = \overline{2,1,2}$ around the hexagon meaning they enter at node $0$ then jump to store at node $2$ (skipping store 1) and then they go to store 3 then jump to skip to store $5$ (skipping $-1 \pmod 5$) and so on... $s_2 = \overline{2,1,1,1,2}$ and in general shopper $s_n = $ the same thing with a total of two $2$'s and $p_{n + 2} - 4$ ones.

They each spend 0 minutes at each store, because they're smaht. They take 2 minutes when they skip a store and 1 minute when they simply go to the next store. In how many minutes will they all be at node $0$ again. For $n = 1$ the answer is $12$ that's the next twin prime average

Something Strange Happens When You Follow Einstein's Math

12:31 PM

►

it's about parallel universes

Peter

12:50 PM

Why do people waste their short life with such nonsensical ideas ?

Derivative

1:09 PM

The area of a hypersurface $f^{-1}(r)$ has an interesting formula obtained with Stokes theorem $\int_{\left\{f<r\right\}} \frac{\Delta f}{\left|\nabla f\right|}dVol$

under mild assumptions

Mad

hello,
when we say "E is a family of functions f:X to Y " do we mean, the elements of E are functions, or is E a subset of the power set of "Func(X,Y)" set of all functions from X to Y IE

is an element of E a function or may also be {f,g,h..}this form

because a family is defined as a subset of the powerset, however, it does not seem consistent with what comes after

Elements of E are functions

@Mad I have never heard the idea that "family" is a subset of the powerset.

That's a new one on me. Do you have a reference for that?

In all of the math that I have read, "family", "set", and "collection" (among others) are all synonymous.

A: What is the difference between "family" and "set"?

I thought the same but apparently there is a distinction

30

Strictly speaking, a family is a function $I \to U$, where $I$ is an index set and $U$ is a universe that contains the members of the family. Strictly speaking, a set is not a family indexed by itself: it's either the image of the family, if the members are the elements, or the union of that fam...

@SoumikMukherjee That's not what that answer says.

That answer says that a "family" is a set which is indexed by some other set. But, again, I don't think that most work-a-day mathematicians are going to be too fussed about that distinction. Note that other answers to the same question disagree, and prefer a more information notion, i.e. a family is a set is a collection is a ... .

Even so, the original question here is about "a family of functions $f : X \to Y$", which is just a set of functions $f : X \to Y$.

(So the elements are functions, not sets of functions.)

1:27 PM

@XanderHenderson Yeah ofcourse, I never heard the subset of power set thing before

1:48 PM

Every set is a subset of a power set if you think hard enough about it

@Jakobian Sure, but I'm not sure that is helpful.

While this is forgetting the data of a given set $X$ we take power set of, why do we even need that data in the first place?

Does an $n$th degree relative cohomology of noncompact $n$ manifold vanish?

@onepotatotwopotato relative to what

a submanifold

2:05 PM

$H^2(T^2, S^1) = \mathbb{Z}^2$ according to this

where by $S^1$ we mean some contractible copy of $S^1$ in $T^2$

oh, non-compact

sorry I missed that part

A bit pedantic but the authors should have required $U$ nonempty here right?

@onepotatotwopotato If you remove a point from $T^2$ and consider a contractible copy of $S^1$ in it, then the second cohomology will stay the same, right?

or does that not work

@EE18 They did...?

Oh I guess I meant in the second part of the if and only if

But I see, you mean from the first part

OK, I can see how it should be thought of as "carrying over"

It is all part of the same example.

$U$ is non-empty.

2:11 PM

True, that's fair. Thnks Xander

it's known that the top dimensional homology or cohomology of a noncompact manifold vanishes

2:21 PM

what is a name for a knot that is on a given surface?

first thing that comes to mind is a "torus knot"

but in general?

Umm shouldn't it be a simple closed curve eventually?

@JohnZimmerman according to wikipedia there is a concept of a submanifold of a manifold being "knotted"

I remember there's a seifert surface associated to a general knot

hmm

Do you know the answer @Thorgott to the cohomology question?

2:29 PM

@Peter it is not nonsensical. it comes out of Einstein's equations

2

straight out of them. no trickery

the prediction is on par with the prediction of a black hole

@RyderRude Granted, we know that Einstein's equations aren't the final product so the prediction could be nonsensical, no?

any prediction of a model could be nonsensical

YourLordJoyBoy

Ugh.....made a friggin 53 on the final. And a 66.06 overall for the class

@EE18 could be, as in, anything could be non sensical. but given what we know, it is a natural prediction

as in, one doesnt need to put it in. it just comes out

note that black holes were considered non sensical once

Another question for folks. There is the usual homomorphism from the set of polynomials in $m$ indeterminates $K[X_1,...,X_m]$ into the set of functions $K^{K^m}$. We have further that this homomorphism is an injection iff $K$ infinite

2:33 PM

isn't this pretty well known physics?

the fact itself, I mean

black holes were considered non sensical becuz they were kinda pathological solutions

@Jakobian multiverses arent accepted as fact

@Jakobian but yes, it is well known physics

oh. Multiverses or whatever. Sure. I meant Einstein equations and black holes. Forgot we what we were talking about

Now my question is as regards the discussion below. The fact that the identification of polynomials in $K^{K^m}$ is a subspace follows from the homomorphism above being, in fact, a linear map between the two larger vector spaces (when we view them as such). But what can we say in the $K$ finite case?

yes.. that much is well tested physics

That is, is the image of this homomorphism no longer a subspace (why, would it fail to be a linear map in the case of it not being injective?)

@RyderRude I guess what I mean (and i am far from an expert, though I imagine Sean Carroll might say something similar) is that with quantum mechanics we have at this point zero reason to doubt its veracity, whereas we already know that GR is not the final word when it comes to spacetime (for various reasons). Thus we might reasonably be more skeptical of certain conclusions from GR than from conclusions that fall out of QM

2:36 PM

@EE18 by identification of polynomials you mean image of this map?

Correct :)

The authors say nothing in the finite case so I wanted to check

As far as I can tell the map is linear whether $K$ finite or not. If so, then the image is a subspace from basic vector space theory

Yes

@EE18 i would say a reasonable opinion is that neither GR nor QM is a correct theory

You are writing way too much, I'm not reading all that

@Jakobian No prob. It sound like you agree with my more succinct final statement that "As far as I can tell the map is linear whether $K$ finite or not. If so, then the image is a subspace from basic vector space theory" so I'm all good

2:39 PM

Polynomial functions with $m$ variables form a subspace of $K^{K^m}$

Whether or not $K$ finite right?

Polynomials with $m$ variables form a subspace of $K^{K^m}$ when $K$ is infinite

If $K$ is finite, then $K[x_1, ..., x_m]$ is infinite, but $K^{K^m}$ is finite, so it doesn't form a subspace

Oh woah

OK, I will have to think on why then

Simply because $x_1, x_1^2, ...$ is an infinite sequence of polynomials

Polynomial is an element of $K[x_1, ..., x_m]$. Polynomial function is a function $f:K^m\to K$ given by some polynomial

@EE18 look up Penrose's comment on Many Worlds. he uses the hypothesis that QM is incomplete to say that Many worlds is false

many worlds relies on taking QM literally to the word

while General Relativity multiverses are a relatively more comfortable prediction

2:49 PM

I will look it up for sure :) though IIRC, when it comes to QM foundations Penrose is not taken as seriously as he once was?

yeah... because he hypothesises some connections to consciousness (which i think is reasonable)

but regardless of that, his comment on many worlds is objectively reasonable

it's in the reception section en.m.wikipedia.org/wiki/Many-worlds_interpretation

@Jakobian I'm not sure I follow why this is enough to establish things. Certainly the image of $K[X_1,...X_m]$ cannot be infinite given that its a subset of a finite set, but why does that stop it from being a subspace of that finite set (which is a vector space)?

@EE18 what is it?

a finite vector space cannot have an infinite subspace

I'm not restricting to any assumptions. Take this at face value

and if I am, then I'm stating it. The only assumption is that $K$ is a field

I agree, but where are we getting that the subspace is infinite? Why does the image of an infinite set (the polynomials) have to be infinite? (We know it’s not, so I’m just not seeing how we’re getting a contradiction)

subspace?

what subspace

Listen. You've filled the gaps wrong

21 mins ago, by Jakobian

If $K$ is finite, then $K[x_1, ..., x_m]$ is infinite, but $K^{K^m}$ is finite, so it doesn't form a subspace

how do you intepret this sentence, and why do you think I'm talking about polynomial functions?

start using the correct terminology instead of saying vague expressions like "it" or "the subspace" or "things"

If I'm saying "it" in a sentence, then surely I must be referring to something in that sentence

Polynomials don't form a subspace

AND

I've already said that polynomial functions do form a subspace

why would I be contradicting myself right now

geez

3:37 PM

I will look back on our Convo. Thank you as always for the help Jakobian.

Peter

parallel universes are nonsense , full stop.

Derivative

is the set of points in $\Bbb R^3$ with exactly one rational coordinate connected?

2

Koro

3:57 PM

@Peter Perhaps. But what intrigues me more is if it is finite or infinite.

@Derivative interesting. I know $\mathbb{Q}^2\cup (\mathbb{R}\setminus\mathbb{Q})^2$ is connected but I'm not sure about your example

Koro

4:13 PM

@Jakobian can we say something in R^2?

Projection onto the first two coordinates of this set is $\mathbb{R}^2\setminus\mathbb{Q}^2$ which is known to be connected

Lucky Chouhan

4:42 PM

@Jakobian would you suggest me a book on topology?

Engelking Sieklucki that Soumik posted image of recently is very good

standard recommendation on topology, but a lot more stale, is Munkres

@Jakobian That book is awesome

Lucky Chouhan

4:58 PM

@Jakobian Thank you :)

@SoumikMukherjee would you like to show me table of contents?

5:31 PM

The quotient $\Bbb R^n/\Bbb Z^n$ gives a torus, $\Bbb T^n$. Take the quotient $\Bbb R^n/\Bbb J^n$ where we have the relation $\exp(\Bbb Z^n)=\Bbb J^n$.

The notation $\exp$ is shorthand for each point in the coordinate pairs of the $\Bbb Z^n$ lattice being exponentiated. Ex. $(3,2) \mapsto (e^3,e^2).$

What is $\Bbb R^n/\Bbb J^n$ isometric to?

Show that $\Bbb R^n/\Bbb J^n$ is noncompact

@EE18 one way of maybe detangling this would be to avoid using the word "polynomial" in more than one context at once (e.g. while it is formally clear that a "polynomial" is not the same thing as a "polynomial function," it may not be helpful to keep phrasing that discussion in those terms until you understand it).

in abstract terms, if you have spaces U and V and a linear map T from U to V, it is helpful to distinguish (in notation and speech) between U and T(U), and between a given u in U and its image T(u) under T, even in those cases where T is injective (when T is actually an isomorphism from U to T(U), where it is possible to think of T(u) as 'the same thing as' u, or to 'identify' T(u) with u, or whatever). i realize that your source material does not do this. but it might help if you did this.

5:47 PM

@LuckyChouhan here

@AlessandroCodenotti Is the Nobeling space $N_n^{2n+1}$ connected for $n > 0$?

Ah no that should easily be path connected

$N_n^k\subseteq \mathbb{R}^k$ is a subspace of points where at most $n$ are rational

your source has a linear map T from U to V, and in the last sentence of your excerpt, the author says that in a situation where T is injective and an "identification" is made, this "means that U is also a subspace of V." the author is not saying that U is literally a subspace of V. the author is saying that in this situation U is isomorphic to a subspace of V via the map T.

in abstract terms, in some of the above discussion you are wrestling with, well, how could T(U) not be a subspace of V. the significant point in the author's context is not that T(U) is a subspace of V (it always is, whenever U and V are vector spaces and T is a linear map from one to the other). the point is that in context, T(U) is a subspace of V that is isomorphic to U, so this subspace T(U) of V can be "identified" [via T] with U.

@Derivative can I know the origin of this question?

6:13 PM

I agree it's path connected

any idea about $A = \{x\in \mathbb{R}^3 : \exists!_i x_i\in\mathbb{Q}\}$?

What about it?

is it connected

sorry I made a typo

Isn't it still path connected?

why?

what paths do you have in mind

6:19 PM

By "drawing" it in $\Bbb R^2$

what do you mean by $\mathbb{R}^2$

@leslietownes Ah OK, so this is what I am missing. I am fortunate after reading enough of enderton that I do follow the distinction between one structure and its isomorphic embedding in the other, so after reading what you've said I think I follow why I was confused this morning: I was taking the "identification" too loosely to just mean the underlying homomorphism. So I was confused why Amann Escher remarked only that the image of this homomorphism was a subspace of...

..the space of functions $K^{K^m}$ when $K$ being finite was irrelevant to that end. But you point out here what I missed: that while being a subspace simply follows from the set of polynomials to set of polynomial functions being a linear map, for the two subspaces to be isomorphic requires that there be a bijection (enough to be an injection here between the two). And that's where being infinite is necessary

I see now that Jakobian was trying to point that out too

6:37 PM

@AlessandroCodenotti commenting on $\mathbb{Q}\times(\mathbb{R}\setminus\mathbb{Q})\cup (\mathbb{R}\setminus\mathbb{Q})\times \mathbb{Q}$?

EE18: this issue pops up almost every time someone talks about "identifying" something with something else (or similar phrasing). in the background there is often some U, V and a map T: U to V whose properties allow one to regard U and T(U) as "the same" for the purpose of the discussion, and in "identifying" one thing with the other, the author is saying "i am going to begin doing this"

@Jakobian the same kind of subspace but in the plane

Oh wait, you want exactly one rational coordinate?

yes

EE18: note that it's pretty easy to talk this way without expressly introducing notation for what "U" and "V" and "T" are (your author's "U" and "V" are clearly identified, but the map "T" assigning to a polynomial its function on k^m is not given a symbolic name, only described in words). and indeed often the point of "identifying" things with other things is to avoid having to introduce a ton of notation that will end up not mattering very much for the issue the author is trying to discuss

Derivative

7:11 PM

@Jakobian someone at impa told me about it

I feel that the space is totally disconnected

7:39 PM

@DanielDonnelly This might interest you: leanprover-community.github.io/blog/posts/FLT-announcement

YourLordJoyBoy

Meant to mention this prior....my grandma passed.

hey chat, dumb question

i've seen a limit that uses this trick: since $n! \sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$ by Stirling's, take the $n$-th root in both sides and you get an asymptotic approximation to $(n!)^{1/n}$

however, this kind of trick doesn't work for the n-th power - for instance, $1 + 1/n \sim 1$ but taking the n-power gives $e = 1$, which is false

how does one argue that this process is valid? if we know that $f(x) \sim g(x)$ and $g \ge f$ for all x, then $g/f \ge 1$ and thus $(g/f) \ge (g/f)^{1/n} \ge 1$ and our trick works

8:00 PM

Well taken on all fronts as always @leslietownes. Thanks for the help :) Will mark this down in my notes.

8:26 PM

@LucasHenrique Because $a_n^{1/n}\to 1$ if $a_n\to 1$

because $x^y$ is continuous for $x > 0$ and $y\in\mathbb{R}$

@Jakobian wouldn't this argument still apply to $a_n^n \to 1$ if $a_n \to 1$? (which is false)

8:45 PM

lucas, while the sequences (a_n, 1/n) and (a_n, n) are both sequences in the subset of R^2 that jakobian mentions, only one of them is convergent in that subset (the former sequence converges to (1,0), the latter sequence does not converge)

@leslietownes ah. this is embarassing

to make things more fun, maybe someone can find (or write) a book where the continuity of x^y is proved in terms of the fact that a_n^{1/n} goes to 1 if a_n goes to 1

@leslietownes I'm writing one right now! (not)

I'm grappling with a graph

it's behaving strangely I think it might be the way it's rendering

9:01 PM

Imagine you're writing a book because Leslie told you to

that would be awful

Well that depends if you're having fun while doing so

the whole premise sounds funny

you're right

it may very well be fun

leslie what kind of math book should i write

9:46 PM

10:53 PM

This comment is boring and vain. Just the tendency to immediately insinuate hypocrisy rather than engaging with a position tells as much.
This self-fellating about how inclusive of a science math is is incredibly cringe. Yes, we all know the contributions of the Aztecs, the Islamic Golden Age, etc. to early mathematics. However, starting with the advent of calculus, the history of mathematical developments that lie at the foundation of what one would reasonably call "modern mathematics" was predominantly European and (later) American. Of course, there's nothing inherently non-inclusive abou…

This is an entirely unproductive attitude. If you want to act like grouping people into categories is inherently racist/etc, this is a slippery slope to just rejecting anything descriptive (and if one smugly acts like describing someone as a monoglot is linguist, one is way past slipping that slope), then you can feel free to do so.
However, politics matter because it actually affects people. It is a fact that people have different skin colors, that this difference has historically lead to massive amounts of discrimination (colonialism, slavery, etc.) and that these injustices still have re…

Also, that reference to Nazi Germany (presumably made in a poor attempt at shock factor) is nonsensical. The differentiation of people based on skin color obviously predates Nazi Germany.
If you wanna talk about Nazi rhetoric regarding black people, then it's not something as simple as categorizing people into black and white, it's saying shit like "The Jews have been and are the ones bringing the <German equivalent of the N-word> to the Rhine, always with the same intention and goal of destroying the white race they hate through the necessarily occurring bastardization, bring it down from …

@Jakobian Mirror.

@leslietownes maybe one can do it by showing that $a_n^{1/n}$ is Cauchy. then pick a monotone subsequence and then use my trick

@onepotatotwopotato Let's say $M$ connected for simplicity. If $M$ is orientable, Poincaré duality yields $H^n(M,S)=H_0^{lf}(M-S)$, where this is locally finite homology. This group vanishes, because you can choose a proper ray starting at any point and subdivide it into intervals, yielding a locally finite chain with only that point as boundary, making it trivial in $H_0^{lf}$. I don't have the willpower to think about the non-orientable case.

Q: $N$ simultaneous Hexagon Mall shoppers skip some stores in a certain regular pattern. Minutes before they all meet up at node 0? Fundamental group?

@YourLordJoyBoy You have my condolences.

@DanielDonnelly That's the first I've heard of that problem.

Did you come up with it yourself? I can't find it, other than . . .

1

Let the $k$th shopper $k = 1..N$ skip the $v$th store they visit if and only if $v \neq \pm 1 \pmod {p_{k + 2}}$. In other words shopper $k =1$ will start at node $0$ (blue in picture) which is the Hexagon Mall entrance together with shopper $k=2$ (orange in picture). Shopper $1$ moves in patt...

Regarding . . .

13 hours ago, by Shaun

@DanielDonnelly Studying. I have finally figured out how to use the local HPC facilities. I might have new data soon :)

I did get new data! :)

11:15 PM

@Thorgott I'm talking about origins of concept of race itself here. I'm sorry if the fact that this was propagated by nazi germany offends you

Let $\mathbf{Grp}_{RF}$ be the category of residually finite groups and $\mathbf{Grp}_{fin}$ be the category of finite groups. Is the restricted Yoneda embedding $\mathbf{Grp}_{RF} \to [\mathbf{Grp}_{fin}^{op},\mathbf{Set}]$ fully faithful?

I feel like there should be an easy argument

@Thorgott Its very convinient to only read part of what I said and completely ignore the rest e.g.

15 hours ago, by Jakobian

Okay maybe for nazi's it was more elaborate than that, but today its used in a similar way as to group people into classes. The problem is that your skin coloration is subjective and depends on opinion of you or other people

@Jakobian if we abandoned the talk about race or skincolor, this would not make racism disappear, in fact, it would make it impossible to address racism

@LukasHeger I'm not trying to abandon skin color or anything. Just race really

And how would you abandon race

11:29 PM

Well you can talk about people with dark skin coloration and bright skin coloration

instead of white/black people

Why would I? If you want to do it go ahead

because its subjective to say if someone is white/black

Many people think their race/ethnicity is something worth preserving

Yes, everything is subjective, dear

Welcome to the world

@Jakobian That is not my point. My point is that this has nothing concrete to do with Nazi Germany and its rhetoric. If you wanna talk about the concept of race, do that, but if you bring Nazi Germany into it, either do it justice or you will be guilty of (unintentionally) relativizing Nazi Germany and its rhetoric and charged as such. The neither-here-nor-there walking back with saying "maybe it was more elaborate than that" does not absolve you of getting called out on this.

Saying we shouldn't talk about it is not the same as abandoning the concept. And as I said, how are you going to talk about actual concrete issues relating to racism without mentioning race/ethnicity?

11:31 PM

You clearly misunderstood something

@Jakobian white/black is not necessarily about skin color

@LukasHeger I don't have that power and if someone wants to call people black/white then I'm not going to stop them. Probably in a paper it should be defined precisely what those terms mean

Then try to communicate better. "Abandon race" makes it sound like you want everyone to forgo their racial and ethnic background and heritage

this is not about subjective interpretation of skin color, it's about how the cultural status of certain groups of people has evolved over time

@Jakobian there is plenty of academic literature employing these terms and discussing their meaning, for the record

@Jakobian my point is this: you said that it was inherently racist to even call someone e.g. a white person. I'm saying that is necessary to talk about this kind of sttuff if we want to understand e.g. how race/ethnicity affects stuff like socioeconomic status etc.
If there's a correlation between race/ethnicity and socioeconomic status, that's potentially problematic, as it can hint at structural racism. But we cannot even formulate this problem if we do not, as you suggested, talk about race. So I'm saying that there are situations where it is important to talk about race to combat racism…

11:37 PM

@LukasHeger I don't think that calling someone white is racist, but this grouping can lead to racism

@LukasHeger Yeah, that is also what I was trying to get at with my earlier comment.

@Jakobian The existence of races (which, to be clear, is not a biological but a sociopolitical concept when applied to humans) can lead to racism. The potential is unavoidable and not addressing it neither makes it go away nor helps us in overcoming the racism that did historically happen and continues to happen and have repercussions to this day.

why did you all came at me at 2 am to discuss this anyway

I was out all day

Anyway arguing on the internet about if certain language is racist towards white people is always kind of sus, because (a) there's no historical evidence of racism towards Caucasians (its usually the other way) and (b) the people who make these kind of arguments are exactly the internet crypto-nazis

> (b) the people who make these kind of arguments are exactly the internet crypto-nazis

11:45 PM

nah, it's definitely a thing

to be clear, that's not to say I think you're a crypto-Nazi, because I don't

What Thorgott said

There's a very prominent loud minority of such people out there who use EXACTLY the same argument.

I wasn't saying (any) language was racist towards white people anyway

Ryan M

@Jakobian handling this flagged message to avoid a hopefully avoidable escalation to a ban: mind the language, please

@RyanM alright. I was trying to say the argument is bad. I phrased it bad. Sorry for that