Mathematics

12:03 AM

It's counterproductive to forget about the grouping if you want to combat racism. For example, if suddenly you want to declare that we should not be calling the native americans as exactly that -- this has the effect of showing lip service solidarity with them while making it impossible for any kind of reparation to happen (eg, giving them back their lands)

It's important to remember who committed acts of atrocities towards whom, in the context of combatting racism. Race therefore cannot be forgone.

But even outside of that, I find the premise that race was developed just for racists to perpetrate racial crimes inane. Race is just a condensed term for the ethnic and historical background of a group of people --- that's a good thing! Why would you abandon that?

Jakobian

Okay I give in, I was being simplistic

Balarka Sen

And yes, you did not explicitly say any language was racist towards white people, but your scaremongering language (I paraphrase) "mention of race (and other intersectional identifiers or "groupings" as you call it) itself is racist! This is very dangerous!" in the context of a post made by someone who merely identified as a normative white man in the beginning of his post, is very problematic.

Thorgott

I appreciate the reflection

Balarka Sen

You had no ill will, but realize this is the kind of logic that is actively used by bad-faith actors.

Jakobian

@BalarkaSen I don't think that's dangerous, I had a second point that said something like, you thinking that you're not capable of being as inclusive as other people by being, lets say, normative white man as you said, is what the dangerous ideology was mainly referring to.

Balarka Sen

12:11 AM

There's a historical record of Caucasians not being inclusive yes? :-)

I get your point though

Thorgott

I don't think that was the implication! He is capable of being as inclusive and the post was made in an effort to realize that capability.

The point of the self-identification is not that there is some inherent factor preventing white people from being inclusive, but that white people are typically socialized in ways that makes them not as inclusive (all going back to the society and history again)

John Zimmerman

Is it possible for $f'(x)/(g(x)f(x))$ to be oscillatory if $f'(x)$ nor $g(x)$ nor $f(x)$ is oscillatory?

cool graph!

Jakobian

@Thorgott well I've imagined the situation like this. You're putting those things like, you're white, male, western, monolinguist. And in American media I see this sort of self-shaming for identifying as a white person, or however you want to call it. Of course, stereotypes go in, you can see Balarka's comment. That's how it sounded, after all, what else purpose does it serve to state those things

Monolinguist, western. Okay this makes sense with how you understood it

Lukas Heger

silly topology question: Is there anything that goes wrong if you define characteristic classes for a non-Lie topological group? I know we won't get Chern-Weil theory obviously, but as long as we can compute some cohomology groups, you should get some char classes? In particular for discrete groups this should be related to group cohomology and covering spaces

Balarka Sen

Simply, it serves the purpose of stating "I am aware of the huge amount of atrocities the people who look like me have committed throughout history, but we are trying to be better." I think it's less self-shaming and more a sign of historical maturity, a hopeful signal that the world is changing.

@LukasHeger I think $BG$ for a topological group $G$ has rational homotopy type of a product of some bunch of $K(\Bbb Q, n)$'s.

I forget the exact $n$'s. But you should be able to derive all characteristic classes from here.

Lukas Heger

12:24 AM

intersting thanks!

Jakobian

@BalarkaSen We can grant this the benefit of the doubt and assume this. But I still see it as self-shaming. No need to apologize for crimes you haven't commited

Balarka Sen

I never think of it as an apology (Because, as you said, why apologize for something you haven't done?), but an acknowledgement. A nod to history.

This was very hard to get out of people even in the 50's!

Jakobian

Well okay. What purpose would what you are saying serve in that specific post

Balarka Sen

I think it was a nice, considerate preamble. I wouldn't mind if they didn't put it.

I don't think of every Caucasian male as a racist lol

Jakobian

Its very suspicious. Why would someone be this mindful without stating it as such, on some random academia post

I find it more believable that it was a byproduct of some inner thoughts

Balarka Sen

12:29 AM

Such kinds of preambles certainly should not be enforced

I know that's another loud minority opinion which I find inane.

Jakobian

Moreover, the author was here, and didn't comment to even hint at an explanation of this type

Thorgott

this self-shaming aka "white guilt" is also one of those thing crypto-Nazis (and actual Nazis, for the record) really love to emphasize, but it's largely a boogeyman

Jakobian

This suggests that my explanation with self-shaming, if the author gone through my insane ramblings and actually understood them, might be more plausible

Thorgott

no significant amount of people is apologizing for crimes they haven't committed

there is simply an awareness of how history has shaped our society and ourselves

Lukas Heger

@BalarkaSen As a German I have to say that is something we're reasonably good at: accounting for the past, admitting that our history is really dark and our ancestors committed unfathomable attrocities. We learn a lot about Nazi Germany in school and this kind of "Erinnerungskultur"(remembrance culture) is quite important imo.

Jakobian

12:33 AM

well I did see it in media, maybe they picked those people

Thorgott

media has a tendency to depict extremes and be divisive

especially American media

cause it's all privately owned over there

Jakobian

But the fact that its, probably widespread, in media can be a factor

because we're talking about an individual here and not a statistic

Balarka Sen

Hah yeah the US 24 hour news cycles are probably the one thing decidedly ripping the country apart

Lukas Heger

@BalarkaSen couldn't there intersting say $\Bbb Z/2\Bbb Z$-valued char classes, too?

Balarka Sen

@LukasHeger this is a good question honestly

Try looking in Sullivan's "Localization, Periodicity & Galois Symmetry", he might have something to say about the p-localized cohomology ring of BG

Thorgott

12:37 AM

@Jakobian well, I don't think he did anything but acknowledging your response, but I don't wanna speculate on his motivations, none of us should really

one potato two potato

@Thorgott seems it's not a consequence that an ordinary cohomology vanishes

Thorgott

my point is that the introduction to his post does not necessitate some warped worldview

Lukas Heger

@BalarkaSen I'll look into it, thanks

Jakobian

if I don't accept the explanation I've given already then I don't have a good one

Thorgott

it could also just mean to indicate that his background and how he was socialized was comparatively non-inclusive and he's trying to change that

Lukas Heger

12:38 AM

I think I heard that paper mentioned before

Xander Henderson

@Thorgott No, but I do think that it is symptomatic of (1) a real misunderstanding of what "diversity" and "inclusion" mean, and (2) a fear of "stepping out of line" and being subjected to a struggle session.

Thorgott

@Lukas @Balarka wasn't there a huge industry of trying to make topological Pontryagin classes work

Balarka Sen

yeah lol

Xander Henderson

@Thorgott Honestly, a lot of that is garbage. Yes, the Aztecs had a somewhat more sophisticated numeration system than their European counterparts, but the Aztec approach to mathematics has had essentially zero influence on the Western practice of mathematics. What the Aztec were doing is interesting, but from an anthropological/historical point of view, rather than a mathematical point of view.

Balarka Sen

ended with someone becoming a very successful US senator

Thorgott

12:43 AM

@XanderHenderson it could be, I don't really have a particular opinion on the post as such, I just didn't like the way Jakobian was taking issue with it first and foremost

@XanderHenderson yeah

Jakobian

haven't we learned by experience already that I have bad way of handling issues, especially those political ones

Thorgott

@LukasHeger this is an interesting question btw, it's true when we take profinite instead of residually finite groups

Odestheory12

if the hypotenuse of a triangle is a multiple of the square root of two, does it necessarily imply that the two legs of the triangle are equal?

Xander Henderson

@Thorgott I think something to understand about @Jakobian is that he comes from a place where there is very little diversity. Everyone around him is Polish. It makes it very hard to understand the struggles in places like the US, where there are groups who have been historically systematically oppressed.

Odestheory12

i.e if $x^2+y^2 = 2k^2$ does it mean $x=y$? And what if $x,y,k \in \mathbb{N}$?

Lukas Heger

12:45 AM

@Thorgott the answer is no, because there are residually finite groups with the same profinite completion. But at least one can say that two res. finite groups have the same profinite completion iff their restricted Hom functors are isomorphic

Balarka Sen

ooh nice

Xander Henderson

It is also worth keeping in mind, @Jakobian, that "white" and "black" actually have very little to do with skin tone. Indeed, there was a well-researched book published a decade or two ago which made a pretty strong case that both the Irish and Italians were considered "colored" in the US for a period of time.

Balarka Sen

i think its an important question to understand if profinite completion of 3-manifold fundamental groups distinguish the 3-manifolds

when and when not

heard a talk by Alan Reid on this a while ago

Lukas Heger

@Balarka intersting. Does every residually finite group arise as the fundamental group of a 3-manifold? I know that (somehow through geometrization magic) the $\pi_1$ of 3-folds is always res. finite

Balarka Sen

No :) $\Bbb Z^2$ is not fundamental group of a closed $3$-manifold.

Lukas Heger

12:49 AM

oh I see

Thorgott

@LukasHeger ah, that's cool

@BalarkaSen end reasons?

@XanderHenderson to be fair, I would not assume my background is all too dissimilar

Xander Henderson

@Thorgott I have no idea about your background. I just know that I have discussed these issues before with Jakobian, and have found that his perspective is very much one of an outsider.

Balarka Sen

(1) Universal cover Z^2-sheeted cover so noncompact
(2) pi_2 = 0 otherwise by sphere theorem the manifold would be a connect sum and Z^2 is not free product of two groups in a nontrivial way
(3) so pi_1 = pi_2 = 0 of the universal cover and pi_3 = H_3 = 0 by noncompactness. ad infinitum, pi_n = 0 for all n.
(4) universal cover contractible so manifold is homotopy equivalent to K(Z^2, 1) = T^2.
(5) there is no such 3-manifold as H_3(T^2; Z/2) = 0

Jakobian

@Thorgott Yeah I heard the cause of our lack of diversity is the genocide caused by Adolf Hitler

Lukas Heger

@BalarkaSen nice

Jakobian

12:53 AM

pre-war there was lots of diversity in our country, including Jews, Romas, and so on.

Balarka Sen

Ah the glory days of Kakania

This reminds me I need to finish Man Without Qualities by Robert Musil

Lukas Heger

I don't quite get the sphere theorem step though

Jakobian

This is why they attacked us in the first place after all, Hitler was aiming at Jews and Poland had the most population of them from what I recall from history

Thorgott

@BalarkaSen ah nice, the middle part is similar to the argument that knot complements are Eilenberg-McLane

Xander Henderson

@Jakobian It is a factor, but Poland has never really had the kind of diversity that US has. It is mostly all a bunch of Slavs, as opposed to forced labor from Africa, poor itinerant workers from China, immigrants from Ireland and Italy, as well as all of the early European settlers and First Nations people.

Balarka Sen

12:55 AM

@LukasHeger It states that if $\pi_2(M) \neq 0$ for a 3-manifold $M$, then there must be an embedded sphere in $M$ representing a nontrivial element of $\pi_2(M)$.

Now if you cut along that sphere the manifold separates out into a connect sum

Lukas Heger

oh I see

Balarka Sen

(Unless it's $S^1 \times S^2$ or some weird Klein bottle variant, but there are the only exceptions)

Xander Henderson

So, while Hitler killing of a bunch of folk certainly didn't help the diversity on the continent, the state of low-diversity existed before he came along.

Honestly, I suspect that is part of what allowed him to be successful---he could scapegoat small minority populations, and win over a populist crowd.

Jakobian

But hey, I did see like 5 people with black color of skin in our country (in my entire life)

Balarka Sen

@Thorgott yeah.

Thorgott

12:57 AM

@Jakobian Polish Jews were obviously a target, but Hitler also committed genocide on the indigenous Poles if that's what you can call them

one potato two potato

math.stackexchange.com/a/3271966/668308 @BalarkaSen during the proof of the knot complement is a K(G,1) space, why can I assume PL embedding for $S^2\hookrightarrow M$ not just a topological embedding?

Thorgott

their intent was to ethnically cleanse most of Eastern Europe under the name "Generalplan Ost"

psie

I have a basic question.

Jakobian

Yeah Slavs in general were also targeted, definitely

psie

Consider the sets of half open intervals in $\mathbb R$, that is $(a,b]$, $(a,\infty)$ and $\varnothing$, where $-\infty\leq a<b<\infty$. The collection $\mathcal A$ of finite disjoint unions of h-intervals forms an algebra. Now consider any sequence of disjoint h-intervals $(I_n)_{n\in\mathbb N}$ with $\bigcup_1^\infty I_n\in\mathcal A$. Then $\bigcup_1^\infty I_n$ is a finite disjoint union of h-intervals, call them $\Delta_1,\ldots,\Delta_n$.

Put $$A_i=\{j\in\mathbb{N}: I_j\subset \Delta_i\}, \ i=1,\dots,n.$$ Is it true that $$\bigcup_{j\in A_i} I_j=\Delta_i?$$ I have big problems with the $\supset$ inclusion...

Balarka Sen

1:01 AM

@onepotatotwopotato Hah, good question. This is part of the sphere theorem, comes out of the way it's proved.

one potato two potato

Hmm... because the base manifold has a PL structure, the statement of the sphere theorem can be improved to a PL embedding you mean

Balarka Sen

Yes, during the proof you start with a triangulation of M

Then try to resolve the singularities of a self intersecting PL sphere in M

one potato two potato

Ok, Alexander horned sphere does not count because $\Bbb R^3$ is simply connected

Xander Henderson

@BalarkaSen That makes it a "porism". :)

Like a corollary, but it follows from the proof, not the statement.

Balarka Sen

Oh, good to know

That's a good word for it.

Lukas Heger

1:06 AM

regarding the fact that fundamental groups are residually finite in dimension $\leq 3$, but not in dim $\geq 4$: I find it amusing that this implies together with a bunch of other stuff that in dimension $\leq 3$ a closed manifold is simply connected iff all bundles admitting a flat connection are trivial, but this equivalence is no longer true in dim $\geq 4$

Balarka Sen

Alas not many simply connected closed 3-manifolds!

Lukas Heger

true

the purely algebraic part of that statement is nontrivial. It's not clear that there's a finitely presented group with no nontrivial finite-dimensional representation over $\Bbb R$

Jakobian

@psie its not true

Imagine you are dissecting $I = (0, 1]$ into a bunch of smaller intervals $\Delta_i$

the issue is that $\Delta_i$ need to be assumed maximal in some sense

psie

It's inspired from Folland's, but maybe there's something missing, hmm

Jakobian

Instead of taking arbitrary such half-intervals $\Delta_i$, take convex components or however you call them

Lukas Heger

1:12 AM

But one can show that finitely generated linear groups are residually finite, thus a f.g. group has trivial profinite completion iff all finite-dimensional representations over all fields are trivial

so that reduces the search to a f.p. group with trivial profinite completion

psie

@Jakobian here, but it's a bit out of context, so you need to probably have the book at hand

Jakobian

@psie what are you trying to tell me

psie

that is where I got it from

Jakobian

okay. I didn't need to know that, but thanks

1) This is still false as was stated. 2) To fix it you consider components whetever you want to think of it as topological or convex as in order theory

I.e. for a point $x\in A$ you consider the union of all intervals $I\subseteq A$ containing $x$

This now will have the property that every $I_i$ is contained in exactly one $\Delta_j$ (basically from definition)

Jakobian

1:40 AM

I've been bullied to oblivion today *sigh*
I'm not sure if I'm tired because I'm tired or I'm tired because I've been arguing at this ungodly hour

both

and now I'm talking to myself as if I'm some kind of teenager that can't trigger their inner monologue otherwise

Daniel Donnelly

2:12 AM

@Shaun Yes, I came up with that simultaneous Hexagon Mall shoppers problem.

John Zimmerman

4:39 AM

Is it okay to vote to reopen a question of one's own given that it was closed?

leslie townes

4:51 AM

if the system allows it, there can be nothing generally wrong with it (it would even make perfect sense to allow that behavior, to let an OP draw attention to an improvement to a closed question). i would just apply to the decision the same criteria you would to any vote to reopen (i.e., have a reason for voting to reopen that relates to the usual reasons for voting to reopen, and not just because it's your question and the system allows it)

John Zimmerman

5:26 AM

sanity check: $f,g,f'$ are strictly decreasing analytic functions on the real line, can f'/fg exhibit oscillatory behavior? Osciallotrying behavior I will define as neither strictly increasing nor strictly decreasing, but rather non strictly increasing monotonically.

nvm

John Zimmerman

6:56 AM

0

Q: Show $M$ is a compact surface diffeomorphic to $\Bbb T^2$

Take $H = w_1^{\mathbb Z} \odot w_2^{\mathbb Z},$ where $w_1=(a,1)$ and $w_2=(1,b)$ for some $a,b>0; a,b\neq 1,$ and $(x_1,y_1)\odot(x_2,y_2)=(x_1 x_2, y_1 y_2)$ and $w^{\mathbb Z } = \{ w^n \mid n \in \mathbb{Z} \}.$ Consider the quotient space $ M=\Bbb R^2_{\gt 0}/H $. I want to show that $M$ i...

thoughts?

John Zimmerman

7:08 AM

I have some thoughts on this

quotienting R^2 by purely transcendentally valued ordered pairs which get sparser tending to infinity

I don't think this $M$ can be related to a torus actually... 🤔

Clearly $\{(a^m,b^n)\mid m,n\in\Bbb Z\}$ is isomorphic to $\Bbb Z^2$

via $(a^m,b^n)\leftrightarrow (m,n)$

however this easibility is destroyed via the quotient

I think this boils down to

leslie townes

given a and b satisfying your conditions, is H intended to be {(a^m, b^n): m, n in Z}? your question (and other questions like this) might get a wider audience if you presented the relevant definitions in shorter ways, instead of in terms of nested defined concepts that are not used more than once

or even fully defined at all (e.g. you purport to define "w^Z" for an element w of R^2 in terms of "w^n" for integers n, but you do not expressly say what "w^n" means, when w is in R^2 and n is an integer; similarly you purport to define cdot on R^2, but then write w_1^Z "cdot" w_2^Z where "w_1^Z" and "w_2^Z" themselves appear to be some kind of sets, and not elements of R^2)

John Zimmerman

Hmm

@leslietownes I guess I will change it to $\{(a^m,b^n)\mid m,n\in\Bbb Z\}$ I think it's more concise you're right

leslie townes

7:24 AM

it might also be kind to define what R^2_{>0} is, i mean, i can guess because you're always asking questions about this, but a random MSE reader stumbling across the post shouldn't have to guess

my last bit of feedback would be to say more about what you mean by the quotient of R^2_{>0} by H

and by "more" i don't mean "everything you can think of that might be relevant to the problem," just, "enough so that someone knows what this object is"

John Zimmerman

true!!

ok

leslie townes

(e.g. if it's some standard textbook quotient construction or something with a wikipedia link, just say that, and don't unravel every definition that might go into that)

and the PSQ police would probably add: having indicated how the quotient is defined, it might help to give the reader some indication of why you might expect the quotient to be whatever you think it is

John Zimmerman

edited

thanks so much

0

Q: Show $M$ is a compact surface diffeomorphic to $\Bbb T^2$

Consider the lattice $$H:=\{(e^m,e^n)\mid m,n\in\Bbb Z\}$$ which is isomorphic to $\Bbb Z^2$ via $(e^m,e^n)\leftrightarrow (m,n).$ Consider the quotient space $ M=\Bbb R^2_{\gt 0}/H $. I want to show that $M$ is a compact surface diffeomorphic to $\Bbb T^2$. Here $\Bbb R^2_{\gt 0}:=\big\lbrace (x...

general-topology manifolds differential-topology smooth-manifolds integer-lattices

I think it's a lot better :)

possibly related: isomorphic vector spaces admit isomorphic complexifications.

maybe isomorphic lattices admit something similar

Tinkeringbell

7:56 AM

@LuckyChouhan That could be called doxxing. Please don't do that!

7

Dannyu NDos

Is the notion of Fourier transformation extendable to dual number-valued functions?

Wait, let me get rid of the XY problem.

What are the primitive roots of unity in the finite field $\mathbb{Z}_2$?

$x^2 + 1 = (x+1)^2$ is reducible in $\mathbb{Z}[x]$, so

leslie townes

dannyu: are you sure that's what you want? there's only one root of unity in Z_2, whether primitive or not, it's 1. (there are only two things in Z_2 whatsoever, 0 and 1)

Dannyu NDos

I mean... the second primitive root of unity in $\mathbb{Q}$ is $-1$, but the third primitive root of unity in $\mathbb{Q}$ is not in $\mathbb{Q}$. But in $\mathbb{Z}_2$, $-1 = 1$. Do I need some kind of split-complex numbers?

Nevermind; I guess what I wanted to define is impractical compared to NTT.

Soumik Mukherjee

8:35 AM

What is NTT?

Number theory something?

John Zimmerman

anyone know complex analysis?

Dannyu NDos

@SoumikMukherjee Number-Theoretic Transform

Soumik Mukherjee

Okay

Dannyu NDos

Basically DFT under a finite field.

leslie townes

@DannyuNDos as your comments seem to illustrate, the expression "primitive nth root of unity in F" may or may not make sense, depending on what the field F is and what n is. a better way of phrasing your observation is that Q does not contain any primitive third roots of unity. not that "the primitive third root of unity in Q" is "not in Q."

it might be helpful to note that Z_2 is not the only finite field of characteristic 2, and that there are larger finite fields of characteristic 2 that do contain roots of unity other than 1. i.e. if you are willing to work with general finite fields, you are not stuck with just Z_2, even if you want characteristic 2

[this is just an attempt to help you organize your terminology more according to the way you might find it used in references, i realize that you have not even stated the Y in your XY problem]

Jakobian

8:54 AM

@JohnZimmerman I think Xander or Leslie might be the closest we have to a complex analysis expert. But probably Xander more

John Zimmerman

@Jakobian okay thanks :)

Jakobian

@leslietownes can confirm/deny the accusation

leslie townes

i have forgotten most of what i know. i'm now the kind of person who feels at home in the open unit disc, and yet alarmed and despondent and out of my depth in the upper half plane.

Jakobian

Oh. Robjohn might know quite a bit about complex analysis

I have no idea who might be good at complex analysis and what subject of it to be perfectly honest

Lukas Heger

@JohnZimmerman I don't like these kind of meta questions. If one answers "yes", there's an immediate expectation that one is able to answer the actual question. If you have a complex analysis question, just ask it and if someone is able and willing to answer it, they shall do so

John Zimmerman

9:02 AM

I've basically been able to grind out a holomorphic (entire) function $K(s)$ that is on the order of $e^{-z}$ however knowing that the inverse mellin transform of $\Gamma(s)$ by examining residues yields $e^{-z}$, I think I must expect that $K(s)$ has many poles like gamma - right?

Wait

i mixed that up

Tryna show that this dude is holomorphic and understand *why* on earth that could be:

$$K(s)=\int_{1/2}^1 \zeta\bigg(-\frac{1}{\log x}\bigg)~x^{-s}~dx $$

Lukas Heger

@JohnZimmerman strictly speaking, $\Gamma(s)$ as a Mellin transform of $e^{-z}$ is only something that is defined for positive real part, as else the integral in the Mellin transform doesn't converge

John Zimmerman

@LukasHeger yes I know

Lukas Heger

ofc you can argue through analytic continuation to extend it to a larger domain

John Zimmerman

The integral converges for $-\infty <\Re(s)<\infty$ it seems 🤔

Lukas Heger

isn't the only question why the integral converges? if it does, holomorphic is automatic as $x^{-s}$ is holomorphic in $s$

John Zimmerman

9:11 AM

implying that $K(s)$ is entire I feel like there is something I'm missing or some contradiction

oh i see @LukasHeger This means you could take the inverse mellin transform on $K(s)$ and use it to extend the domain of $\zeta(-1/\log x)$

via analytic continuation

It is known to me that $\zeta(-1/\log x)$ is a composition s.t. it admits a meromorphic extension to all of $\Bbb C$ aside from a pole at $1/e$ and a branch cut due to the logarithm

I wonder if the former extension would coincide with the latter extension.

Lukas Heger

@JohnZimmerman yes! due to the identity theorem

John Zimmerman

@LukasHeger ah yes this makes so much more sense now

Bml

9:43 AM

Hi everyone. How to calculate the fourth roots of a complex number like $z = e^{-i \frac{\pi}{3}}$?

Lukas Heger

@Bml think geometrically. Put it in polar form first, your example already is in polar form. Note that taking the fourth power corresponds in polar form to taking the fourth power of the radius and multiplying the argument by 4. This should give you an idea for how to solve such questions

Bml

@LukasHeger z^4 = 1^4 e^{-4 i \frac{\pi}{3}}?

Lukas Heger

10:01 AM

@Bml no, you want $w^4=e^{-i\pi/3}$

so write $w=re^{i\theta}$, then solve $w^4=e^{-i\pi/3}$ for $r$ and $\theta$

Bml

10:19 AM

@LukasHeger OK, so $w = 1 e^{-i \frac{\pi}{3 \cdot 4} + \frac{2 k \pi}{4}}$, right?

Lukas Heger

@Bml correct

Bml