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12:00 AM
In more generality a 2 dimensional surface is something you can locally parametrize by 2 coordinates
 
12:10 AM
For continuity, there's two definitions usually. A metric one and one with open sets. They're equivalent.

This boils down to the metric topology is generated by open balls so it suffices to check continuity on the open balls which is the definition of continuity.

Why is the equivalence of definitions important/what problems does it solve? It makes one's mind set up connections between metric space concepts and topology concepts so that one can work with both better. In particular abstract results about topology can walk through the bridge to give us results about metric spaces. But they g
 
I met an economics major who knew about the banach tarski paradox
unrelated:
What if all ocean water left the planet at the speed of light?
but simultaneously all sharks turned into black holes
 
"What if xyz" isn't a question
 
12:29 AM
What if "What if xyz" were a question
 
What even is a Calabi–Yau manifold
@Rithaniel What if this question weren't a question?
 
What if humans had never developed the word "what"
 
¿Qué?
Alternative gag: Scottish accent Whit?
 
That is the abstract notion signified by the word "what," not specifically the English sound made to represent the notion
 
12:50 AM
Consider a sketch of a flow in the upper half plane. In fact there are two identical copies of the same flow, with one reflected across the middle.

If we let the flows interact then this amounts to adding the vectors in the box portion. There should be generally a vertical flow after this adding process.

I see how to add the vectors in the box but not the vectors in the top right corner and the top left corner.

[![enter image description here][1]][1]


[1]: https://i.stack.imgur.com/mY6la.jpg
 
Yet Another Sphere Eversion Video
 
How do you add vectors that don't overlap?
 
What's in the blank spaces? Zero vectors?
 
or maybe I should say how do you add
flows
yeah!
zero vectors!
maybe I'm not understanding something, but does that just mean you add the vectors in the box and the rest stays the same?
 
12:56 AM
Now that is a smooth sphere eversion
 
that's what I call
 
testing:
 
I love sphere eversions because it was such a complete accidental discovery
 
fail
nope, didn't work
 
It's a start
$\triangleright$
 
1:02 AM
it's supposed to be an animation
though right now it's cycling too fast
slowed-down example: i.stack.imgur.com/SxoME.gif
that's better
@AkivaWeinberger this is an animation of your trisection construction from earlier
evolving from a triangle to a hexagon towards something smooth
 
Neat
I still haven't tried to find an equation for it, but I think two possible approaches are:
1) write out coordinates for the vertices of P_n in barycentric coordinates (probably in base 3) and see if I find a pattern
 
one nice thing about doing it in mathematica is that I can extract the areas of the progressive pictures
 
2) For each $\theta$, see if there's an equation for the tangent line at angle $\theta$
 
and, relative to the area of the initial triangle, I get the following so far:
 
(like a sort of dual polar coordinates)
 
1:09 AM
1, 2/3, 16/27, 140/243, 1252/2187, 11252/19683, 101236/177147, 911060/1594323, 8199412/14348907, 73794452/129140163, 664149556/1162261467
 
Decimals?
Ah
4/7
 
1., 0.666667, 0.592593, 0.576132, 0.572474, 0.571661, 0.57148, 0.57144, 0.571431, 0.571429, 0.571429
oh, nice
 
4/7 is indeed the correct answer (but I don't know why)
(I previously said I knew the area but nothing else)
 
well, I am a physicist, so obtaining it experimentally works for me :P
 
This also means 6/7 of the hexagon I think
 
1:11 AM
hexagon is 2/3 of the triangle, so that checks out
I'm more interested in the sequence of fractions at this point
 
Do you have a static image of the final shape?
 
I can get one
If I drop the triangle and rescale to the hexagon, then the sequence of fractions is 1, 8/9, 70/81, 626/729, 5626/6561, 50618/59049, 455530/531441, 4099706/4782969, 36897226/43046721, 332074778/387420489
converging to 6/7 evidently
the denominators are powers of 9
oeis doesn't recognize the sequence of numerators
ah, but I can instead write those as 1,1-1/9, 1-11/81, 1-103/729, 1-935/6561
and oeis does recognize 1,11,103,935...
specifically, as the coefficients in the expansion of 1/((1-2*x)*(1-9*x))
which suggests some recurrence relation
oh. better observation: the differences between successive fractions (i.e. the area lost at each step) are 1/9, 2/9^2, 2^2/9^3, 2^3/9^4, ...
so the area lost at each step decreases by 2/9
which is also true if I start from the triangle, so I guess that should go back
 
1:35 AM
So you end up with $1-\sum 2^n/9^{n+1}$? @Semiclassical
 
looks right, yeah
 
Neat pic (Waterman polyhedron, convex hull of lattice points contained within a sphere)
 
1:52 AM
can the addition of two hyperbolic flows be hyperbolic?
 
Do you even know the definitions of the words you are using?
5
 
Are you serious?
 
2:11 AM
I don't know the answer to that question
 
god that's so w i d e
 
Well - I don't fully understand the definition, but wouldn't a hyperbolic flow added to itself be hyperbolic?
But yeah try Math SE
 
okay
 
What subject would that be? Differential geometry?
 
dynamical systems
 
2:12 AM
Aha. I should learn that someday
So this is connected to double pendulums and Lorenz attractors somehow?
 
probably those systems have some kinda configuration spaces which are manifolds and then you ... study how a point on said manifold moves
and that forms whatever a "flow" is
 
akiva, I like your question. what about an Anosov flow added to a reflected copy of itself
 
Prof Ghrist (creator of the Calculus Blue series on YouTube) has some dynamical systems/chaos animation
 
then there would be a flow path that essentially remains invariant over the course of the evolution
the invariant would be the central vertical line dividing the flow into symmetric parts
the flow below splits everywhere except one path
 
for reference, this Tao's blog post on the paper: terrytao.wordpress.com/2019/09/10/…
2
 
vzn
3:08 AM
anyone around here looked at the proof? :)
 
I continue to have mixed feelings about Quanta
 
vzn
have not met many experts who are fans of pop sci, quite to the contrary... :|
 
3:24 AM
@AkivaWeinberger it's a necessary evil
 
3:48 AM
Hello
If $R$ is a division ring, then the only ideals of $R$ are $R$ and $\left\{0 \right\}$
Suppose $I \subseteq R$ is an ideal of $R$. Assume $I \ne \left\{0 \right\}$. Suppose $a \ne 0 \in I$ and $b \in R$. Then there's $x \in R$ s.t. $ax =xa= 1$. Thus $1 \in I$. Hence $1 \cdot b = b \cdot 1 = b\in I$. So $R \subseteq I$. Therefore $I = R$.
This is what the book said: suppose $I \subseteq R$ is an ideal of $R$. Assume $I \ne \left\{0 \right\}$. Let $a \ne 0 \in I$ and $b \in R$. Then $ax=b$ is solved in $R$, hence $b \in I$, thus $I=R$.
Could someone explain what "$ax = b$ is solved in $R$" means?
 
4:16 AM
the book means that there exists $x$ such that $ax=b$
 
4:32 AM
Thanks.
 
5:08 AM
Put together a sphere eversion playlist
Any video it's missing?
 
5:26 AM
Balarka probably tried to explain this to me at one point. I still understand nothing
> A jet is really nothing more than the collection of all of the low-order
derivatives of a function up to a certain point. For instance, the
two-jet of f(x), a function of one variable, can be represented
by the triple (f, df/dx, d^2f/dx^2)
Hm OK
(from here)
 
5:45 AM
does it really "matter" whether "x is a limit point of A" is defined to mean "every open contains a point of A" with or without the "other than x" condition?
Hatcher in his general topology notes omits the "other than x", and calls attention to it: p7 here
isn't his definition what goes by (iirc) "adherent point"?
and like i said does it matter at all for any applications (say, you know, if i'm doing algebraic topology with one definition vs another)
 
They differ at isolated points, if I remember right
I think one important sequence (which probably has a name) is $A,A',A'',\dots$
where $'$ is the set of limit points
 
huh wonder if that always stabilises
 
and you can define $A^{(\omega)}=A\cap A'\cap A''\dotsb$ and continue up the ordinal hierarchy
Dunno if that's the right notation but then $A^{(\omega)} '=A^{(\omega+1)}$
 
to me the simplest difference is whether a the range of a(n eventually) constant sequence has the constant it tends to as a limit point, which it does with Hatcher's definition and not otherwise
 
(dunno why the LaTeX isn't working)
 
5:51 AM
0
Q: Algorithm to get the list of triangles resulting from clipping a triangle against a convex polygon.

The Great DuckWe have a 2D triangle. It has 3 points. We have a 2D convex polygon. It has N points. Normally when clipping, one wants to produce the polygon or list of triangles inside the convex polygon. I want to do the opposite. I want to get the list of triangles that do not intersect aside from borderin...

 
i don't see what's wrong with the latex wtf
 
And if I remember right this was Cantor's original motivation for defining the ordinal hierarchy in the first place
 
@AkivaWeinberger do you have the plugin enabled in your web browser?
 
interesting
 
I don't remember the exact statement this is used to prove
 
5:52 AM
@SohamChowdhury Well atleast in functional analysis it does matter. To show that l^infinity is not metrizable or seperable, you need to use that the neighborhood intersects the set at points other than center of the neighborhood
 
@TheGreatDuck Long time no see. It's a link on the top right, remember?
 
@AkivaWeinberger doesnt answer my question
 
@SohamChowdhury A limit point of a sequence is different from the limit point of a set
In a sequence you can have the same point multiple (or infinitely many) times
@TheGreatDuck I have the LaTeX renderer thingy yeah
 
anyway
please help
 
yeah i'm talking about the set definition
 
5:54 AM
been banging my head all night
uuurgh
 
range of the sequence not the sequence itself
 
I dunno specific examples where it matters
 
All my point-set topology knowledge is used in service of other-topology
LaTeX test
 
it doesn't really matter for other-topology is what i'm hoping
 
5:56 AM
$A^{(\omega)} '$
 
yeah that's double superscript
' is actually a superscript
 
${a^b}'$
$a^{b\prime}$
Huh
 
how about $(a^b)'$
 
5:57 AM
$A^{(\omega)\prime}$
${A^{(\omega)}}'$
I guess I have a choice
$A^{(\omega)}\prime$
 
this sequence can be used to show that every closed set in $\Bbb R$ is a countable set union a perfect set
 
Perfect
@LeakyNun That's what it was
 
the last one looks hideous
 
which is used to prove continuum hypothesis for closed subsets of $\Bbb R$
 
Perfect is like, Cantor set-y things and interval-y things?
 
5:58 AM
every point is a limit point
 
@SohamChowdhury ${A^{(\omega)}}_{\huge\prime}$
 
ahh my eyes
 
@AkivaWeinberger yeah
every non-empty perfect set contains a Cantor set
so that's $2^{\aleph_0}$
 
$\stackrel\prime{A^{(\omega)}}$
 
looks like a diacritic
 
6:00 AM
$Á^{(\omega)}$
 
$\stackrel{\prime}\prime$
$\stackrel\prime{\prime^\prime_\prime}$
${\stackrel\prime\prime}^\prime_\prime$
 
^It's raining
upside-down rain
(Raindrops are actually shaped like pancakes and parachutes but whatever)
 
6:46 AM
Guess what?
Online interactive sphere eversion demo! homepages.loria.fr/BLevy/GEOGRAM/geogram_demo_Evert.html
 
7:03 AM
Theoretically this works too (www3.math.tu-berlin.de/geometrie/lab/vr.shtml#sphereEversion) but I haven't been able to get it to work
 
 
2 hours later…
8:39 AM
Good morning, I have found a funky definition in a text book and on wikipedia that a complex square matrix is always unitary
Am I misunderstanding this ? that doesn't seem right
Unitary would mean A*A = I, but I can think of so many complex square matrices where this isn't the case
 
9:13 AM
that can't be right
for instance, for a unitary matrix clearly det(A)^2 = 1
er, i mean |det(A)| = 1
 
@AkivaWeinberger Someone's been thinking about sphere eversion I see
I dunno if I ever gave you this puzzle but: Can you find a deformation of the graph of $z = x^2 + y^2$ over $\Bbb D^2 \setminus (0, 0)$ to the graph of $z = -(x^2 + y^2)$ over $\Bbb D^2 \setminus (0, 0)$ such that at no step in the deformation the surface is horizontal, i.e., some point of the surface has unit normal vertically pointing up.
Essentially everting a punctured hemisphere without ever making any part of the surface horizontal
@Akiva The "simple sphere eversion" is great. I always thought there should be some way to just to it by doing calculus on the singularity but never managed to figure out what's happening.
Somehow you have introduce a circle's worth of double points, and apparently you can do that by some magic on the cancelling pair of fold lines meeting at two cusps
 
9:53 AM
Wonder what I should talk about in an upcoming student colloquium
 
10:33 AM
Is the stereoraphic projection orientation reversing or preserving (from the Riemann sphere to the complex plane)... I have two different sources ( Gamelin "Complex analysis" and my course) contradicting each other on this.
 
That depends on what your choice of orientation on the Riemann sphere and the complex plane are, respectively.
 
Gamelin: "s.p. is orientation reversing, as a map from the sphere with orientation determined by the outer unit normal vector to the complex plane with the usual orientation"
 
That's correct
 
So if the inner unit normal is taken the s.p. is orientation preserving?
 
Yup
 
10:38 AM
Btw is there a convention in / standard choice of outer/inner unit normal or is this entirely text dependent...?
Thanks
 
The standard orientation on the sphere is understood to be determined by the outward pointing normals.
 
Thanks!
 
Huh, I never observed that stereographic projection is orientation reversing. I see it now though, if I draw a little circle just east of the North Pole and project it, it flips upside down.
It's because 1/r is orientation reversing on R^+ I suppose
 
Yeah, it goes the other way wrt the usual chart maps of CP^1
 
@SohamChowdhury yes. But it says that any complex square matrix is always unitary. So that's not possible right?
 
10:53 AM
As Soham pointed out, "of course not". Who says that?
 
I think wikipedia
If U is a square, complex matrix, then the following conditions are equivalent:

U is unitary.
it really confuses me
In mathematics, a complex square matrix U is unitary if its conjugate transpose U∗ is also its inverse—that is, if U ∗ U = U U ∗ = I , {\displaystyle U^{*}U=UU^{*}=I,} where I is the identity matrix. In physics, especially in quantum mechanics, the Hermitian conjugate of a matrix is denoted by a dagger (†) and the equation above becomes U...
 
You're misinterpreting what "the following conditions are equivalent" means
 
You knew the problem is not simple when it shows pseudosymmetry in its graph
 
There is a textbook where I found the same phrase
 
10:54 AM
Yes, the phrase is fine.
 
oh @balar
@BalarkaSen thank you
I was thinking I'm misunderstanding this phrase
what does it mean?
ahhh
that if it's unitary it's got to be that ....
?
 
The following conditions are equivalent doesn't mean the condition is equivalent to the hypothesis, it means the conditions (a), (b), (c), ... as stated are equivalent
 
ahhhhh
OMG thank you
I thought I was completely off somewhere
 
Also your gravatar is broken
 
so just means that if it's complex and square to say a, is the same as b and c
not sure how to fix it
guess I just have to upload a new one
 
10:58 AM
I am reading about regular curves, and definition of that is: A differentiable curve is said to be regular if its derivative never vanishes. Wikipedia has this explanation about it: "In words, a regular curve never slows to a stop or backtracks on itself." I can't understand what 'backtracks on itself' means.
 
@MikeMiller Sent you a message elsewhere
@Silent Consider $f : [-1, 1] \to \Bbb R^3$, $f(t) = (t^2, 0, 0)$. This is trajectory of a particle travelling from $(1, 0, 0)$ to $(0, 0, 0)$ and then back again to $(1, 0, 0)$.
This is what backtracking on itself means; note how $f'(0) = (0, 0, 0)$.
 
@BalarkaSen I suspect that backwarding will always give a point where the derivative of curve vanishes, because of Darboux theorem, am i right?
Thanks for that example!
 
No, nothing to do with Darboux.
This is just Rolle's theorem
 
11:25 AM
Let the relation R = {(n, m) | n, m ∈ ℤ, ⌊n/4⌋ = ⌊m/4⌋}. is any of the set below a correct equivalence class of R? A) {2, 4, 6, 8}. B) {1, 3, 5, 7}. C) {4, 5, 6, 7}. D) {1, 2, 3, 4}.
I think answer is C but can somebody explain it?
 
@BalarkaSen I quoted Darboux because it seems like backwarding will reverse the direction of tangent, hence, component-wise, we can apply Darboux. Where am I wrong? (Thanks for quoting Rolle's btw)
 
Another dumb question if I may, how can a complex matrix be hermitian if it isn't square?
 
@oliver As far as i know, Hermitian matrix is square matrix!
 
11:41 AM
I thought so, I'm asked to show that an mxn hermitian matrix has rank n
i found some references
to mxn hermitian matrices where m != n
@silen
@Silent
i'm probably misunderstanding the question, happens to me a lot :-p
 
@oliver oh! then i dont know
 
it's ok I'll figure it out
the journey to AHA is always nice ! :-D
 
yeah! :)
 
12:13 PM
@Silent That's too complicated. Backwarding means $f(a) = f(b)$ happens, the curve is not injective.
In particular, that is. Backtracking is stronger than being non-injective of course
 
@BalarkaSen I was typing exactly this :)
 
But that's precisely why you can project and apply Rolle's theorem.
@LeakyNun @ÍgjøgnumMeg Cyclotomic extension are too complicated for me.
 
@BalarkaSen What do you mean by 'project'?
 
@Balarka sem. I'm planning on reading Washington's book in full over the next 2 years lol
 
@Silent Formulate the question appropriately and try to think about it. I am not going to write it down
 
12:21 PM
ok
 
@ÍgjøgnumMeg So much number theory, I can't begin to start
It hurts my eyes
 
hehe I'm interested in learning Iwasawa theory and I think Washington's book is the most basic reference for that
very very cool
 
@ÍgjøgnumMeg have you heard back from the professor you asked to supervise your master's thesis?
 
@Alessandro not yet :/
 
How do I even begin to understand $\Bbb Q(\zeta_n)/\Bbb Q$? Start by understanding what the minimal polynomial for $\zeta_n$, maybe? The $n$-th cyclotomic polynomial of whatever it is called.
All the automorphisms should send $\zeta_n$ to $\zeta_n^m$ where $(n, m) = 1$, preservation of order, etc
 
12:28 PM
ye
 
Doesn't it directly imply $\Phi_n(x) = \prod_{(m, n) = 1} (x - \zeta_n^m)$ is irreducible? It's a minimal polynomial fam
Why does Dummit-Foote rant about it so much
 
no idea lel
cyclotomic extensions are the setting for a pretty cool partial proof of FLT
 
You'll end up asking when $\Bbb Z[\zeta_p] = \mathcal{O}_{\Bbb Q(\zeta_p)}$ is class number 1, I guess.
Kummer primes etc
 
well, in that proof all that matters is that there's no $p$-torsion in the class group of the $p$-th cyclotomic field
 
Also the proof of that $\Bbb Z[\zeta_p] = \mathcal{O}_{\Bbb Q(\zeta_p)}$ thing is a series of magic tricks from what I gathered by attending a lecture on an ANT class
How do number theorists do it
 
12:32 PM
I haven't seen it rofl
 
Its soul crushing
 
i think Milne's notes have a proof
but I don't remember it
 
i 'followed' the proof when i was attending the class but thats about it
number theory is seriously corrupting
 
hahaha it's pretty dutty
 
i hate number theory, fuck y'all
 
12:35 PM
Go back to topology @Balarka you walked the dark path but you can still redeem yourself
 
but i have no topology friends
 
Time to find new friends I guess
 
lol Milne shows that $\mathcal{O}_K = \Bbb Z[\zeta] + p^m\mathcal{O}_K$ and that for large enough $m$, $p^m\mathcal{O}_K \subset \Bbb Z[\zeta]$
 
i dont want to be the lonely wolf
@ÍgjøgnumMeg yeah something like that, it's horrendous
 
It is expected that 61% of primes are
regular and 39% are irregular
fun
 
12:41 PM
@BalarkaSen, it seems to me that the issue is that even though Rolle guarantees that there is a point at which derivative of each component vanishes, but they may be distinct points for different components.
 
@Silent but you can "zero in" on a common point where derivatives of all three component vanishes because the value of $f$ matches arbitrarily close to the point where the curve turns and starts backtracking on itself. that's why i told you to make it mathematically precise what you want "backtracking" to mean
 
ok.
 
@Alessandro
 
dsm
12:58 PM
can you guys make sense out of that problem?
I mean, it seems like $\tilde{g}^{\mu\nu} \equiv \tilde{g}(e^\mu,e^\nu) = g(e_\mu,e_\nu) = g_{\mu\nu}$, but that's not right.
self-study, not homework
 
$\phi:[a,b]\to\Bbb R^n$ with $c,d\in(a,b)$ and $\phi(d)\in\phi[a,c)$. So $\phi(d)=\phi(x)$ for some $x\in [a,c)$, and $\phi$ has this property that $\phi(t(s))=\phi(c-s)$ for $s\in[c,d]$ where $t:[c,d]\to\Bbb R$ is a continuous function. Is that correct way of defining backtracking?
@BalarkaSen
 
1:27 PM
@TedE no I'm not offended, I just prefer criticism to be constructive, and yours was not
if it does have substance, then I apologise and will expect you to answer one of the questions you deem basic
 
you sound kinda salty tbf
 
what does salty mean?
 
Almost contemptuous.
But to be fair, TedE was constructive. He did make a suggestion for how you could improve.
 
No it's just that if any of the questions ive posted that have not been answered are seen as basic by someone then obviously I want that person to answer them, that's only fair,right?
 
I don't know if that is "fair".
When a student asks me a question which I think they have not really thought about, I usually try to encourage them to look into answering it themselves.
I think that is fairest in those particular cases.
It would be unfair to deprive them of the learning experience that comes with struggling with their own questions.
 
1:43 PM
Based on your logic no question posted on stack exchange should ever be answered
 
I think you misunderstand if you think that's the logic that follows.
But I do agree with the statement that majority of answers on stackexchange exhibit poor pedagogy.
Often times users merely spoonfeed answers to people.
Instead, they should be giving hints and motivation for the questioners to continue working on it on their own.
 
you suggested that Ted is able to easily answer my questions, but does not do so because he doesn't want to deprive me of the learning experience, yes?
 
No, I am saying that it doesn't follow that spoonfeeding answers is the "fair" thing to do.
Whether TedE did or did not do this, I am not sure. I only scrolled up so far.
What TedE might want to do is more clearly exhibit this fact (if he agrees with it) by giving you appropriate hints.
 
"I am salty because the conservatives won a huge majority in the election" is a possible use
 
And to compliment this process, you should perhaps ask for hints if you know the question has a known solution, or ask how you could approach a question that you are unsure has an answer.
This way you exhibit a willingness to learn properly (through working yourself, perhaps with some help from others) and TedE would exhibit a willingness to teach properly (through encouraging independence and helping when you actually need it).
 
1:49 PM
@Silent Looks a little clunky to me. Would it be more accurate to say it's a curve $\phi : I \to \Bbb R^n$ such that you can restrict to a subinterval $[a, b] \subset I$ with a point $c \in (a, b)$ such that exist reparametrizations $\gamma_1, \gamma_2 : [0, 1] \to [a, c], [c, b]$ such that $\phi(\gamma_1(t)) = \phi(\gamma_2(1-t))$?
@ÍgjøgnumMeg "I feel salty because I had too much salt with my morning sandwich"
I'm So Meta, Even This Acronym
(read the first letter in each word)
 
yis
I am going to be salty soon because I just ordered a fucking giant döner to deal with my horrendous hangover
 
I don't get them
 
@BalarkaSen I am going to start looking at the Mishachev + Eliashberg book
 
Awyeah
 
I just put it on my phone, and I was going to head out for my 1 hour commute to school.
ARE YOU IN IT TO WIN IT?
 
1:58 PM
Let me know if you stumble upon cool stuff
I am in my man
 
Will do. Likely you will see any cool stuff before me.
How much of a treat am I in for, not having done a lot of homotopy theory before?
 
The good thing about Eliashberg Mishachev is there's unlimited cool stuff in it
 
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