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15:00
abelianizes chat Hlelo polepe
I never did think through Ted's "probably total internal reflection in disguise" question
Well played
@Semiclassical it's a god damn algebra haven
15:00
Consider a smooth stack
there's never analysis here
consider a set of finite perimeter $E$ with measure theoretic boundary $\partial_ME$...
It's either topology or algebra
I think it should go like this. Suppose you have a laser light shining through glass onto a glass-air-interface. If the laser is normal to the surface, you'll get no refraction.
@BalarkaSen I don't know
@Akiva How does directional derivative act on dot products?
15:01
@BalarkaSen functorially?
Are we doing connections on sheaves?
On the tangent sheaf, yes
If you now start to sweep the angle of the laser light, you'll find that the beam both strikes the interface farther along and you get refraction of the transmitted ray to a larger angle.
Eventually, that ray will become parallel to the surface and you'll get TIR.
@BalarkaSen $(a\cdot b)'=a\cdot b'+a'\cdot b$?
I'm going to see the classic Riemann-Hilbert correspondence in part two of a for-fun lecture series, part 1 was today and covered connections, although in DG land
15:04
@Narcissusjewel nice
@Akiva Yeah. So $D_Z g(X, Y)$ should be....
I've been interested in R-H problems for a while, though it's never really gotten me anywhere
The same thing but with Ds and gs
($g(X, Y)$ is a scalar function on $M$, so I can directional derivative it along a smooth vector field$Z$)
I mean technically I haven't learned what $g$ is yet
but I guess it's an inner product
15:05
Well maybe I can come back and say something about it. The end of the talk was super rushed and introduced holonomy and monodromy and I didn't get to take the notes
on the tangent plane?
Inner product on each tangent space, yeah. $g_p(\bullet, \bullet)$ is an inner product on $T_p M$. being smooth means $g(X, Y)$ is a smooth function on $M$ for smooth vector fields $X, Y$
@Semiclassical I thought you meant the Riemann Hypothesis.
nope
Riemann-Hilbert
@Narcissusjewel ever seen numerical Riemann-Hilbert stuff?
15:06
@Waiting Hey, J here. How are you? =D
Not yet, and I haven't seen R-H yet ever before
@BalarkaSen jesus, what is with the giant dots
The goal is to do classical R-H correspondence, and then do the D-module version
@Akiva I think you should write down what $D_Z g(X, Y)$ should be for me
Hey @Gasparo how are things these days?
15:07
$•$_•
@0celo7 $\langle \large{\bullet}, \bullet \rangle$
@BalarkaSen $g(X,D_ZY)+g(D_ZX,Y)$?
@Narcissusjewel I don't remember your username. Would you give me a hint who you are?
@Akiva Oof, what does $D_Z X$ mean?
Claim from today: "One can do curvature without a metric" is that legit?
@Gasparo A
15:08
He is the music man, he comes from far away
@Narcissusjewel Yes, curvature of connection makes sense
@Narcissusjewel You mean AC?
@Gasparo Indeed :)
@Narcissusjewel WTF LOL
@BalarkaSen What's $D_Z$ again?
15:10
@AkivaWeinberger If $f$ is a smooth function on $M$, $D_Z f$ is another smooth function on $M$ defined by $(D_Z f)(p) = D_{Z(p)} f$. Directional derivative of $f$ in the direction of $Z$
They wrote $D$ as the connection, and said $D\circ D$ is the curvature, I don't understand the significance
But what's directional derivative of a vector field? Refer to previous conversation
@Gasparo Hi there. How are doing? Working on some results lately.
@BalarkaSen Oh. Hm… maybe $\mathcal L$?
@Waiting Has your book been published? I would like to take a look at it if it has. =D
15:11
@AkivaWeinberger Close. The conversation's about a connection :)
I'm almost certain @Waiting = @Gasparo lmao, always coincide in activity for years
I really want $D_Z g(X, Y) = g(\nabla_Z X, Y) + g(X, \nabla_Z Y)$, not? :)
@Narcissusjewel LOL, someone else thought that too, and I told them you think so too
$\nabla$? @BalarkaSen
@BalarkaSen Don't ask me about it here. I prefer not to talk about my mathematics here (anymore).
3
15:12
@Gasparo Actually I think like 4 of you are the same person
Like I said, strange people in this chat.
4
@Akiva The connection, the thing I told you about here
Fun fact: (a) Dragon Ball Z in Japanese is Doragon Bōru Zetto and (b) Why the hell are they naming their own goddamn things in English
The second one is not a fact but still
@Waiting OK then. I won't ask you then.
If I have a function $u(x,t)$ can I make partial Laplace wrt one variable like this $\mathcal{L}_x u_{xx}(s,t) = s^2 \mathcal{L}_x u(s,t)$ or do I need to include initial terms?
15:13
@BalarkaSen Oh, you actually sniped me there
I didn't see you write the answer
@Gasparo Are you better lately?
@Narcissusjewel I can assure you I always only have one account at a time. I do not tell lies.
@Gasparo That's what someone who lies would say :P.
@Waiting I believe I will get well this year.
@Narcissusjewel Everyone lies on their bed at night to sleep.
15:15
@Akiva So anyway a connection which satisfies $D_Z g(X, Y) = g(\nabla_Z X, Y) + g(X, \nabla_Z Y)$ is called a "metric connection"
'Cuz it's compatible with the Riemannian metric
Like directional derivative is compatible with dot product... so dir. derv. is a metric connection on $\Bbb R^n$ with the Euclidean metric
Don't you mean $D_Zg(X,Y)$?
@Gasparo That sounds good. Hope you will.
Good catch
@BalarkaSen why are you switching from $D$ to $\nabla$
Typo. I just fixed it
15:17
@Waiting I will, because I am Will Hunting.
??
Looks the same to me
He changed the LHS
the RHS and LHS don't match
Different domains
What the hell are you talking about
15:18
why are you using $D$ and $\nabla$ for the same connection
$D$'s got Euclidean domain, $\nabla$'s got $M$ domain
I am not?
@Gasparo I'm sure you will. What did you study lately?
$D_Z$ means directional derivative
$g(X, Y)$ is a scalar function
15:19
@AkivaWeinberger the connection extends trivially to the whole tensor bundle.
which I am directional deriving in the direction of $Z$
@0celo7 Oh shut up
@BalarkaSen why not write $Z g(X,Y)$ then
@Waiting Nothing. I need to get well first before doing any mathematics, though that might sound strange to you. But I will probably try to learn some languages this year, just for fun.
Yeah that's the standard notation but it's something that took me time to get used to
So I didn't want to use it
I'm pretty sure do Carmo writes $Z\langle X,Y\rangle$
15:20
He does
@AkivaWeinberger OK, so we have a connection on Riemannian manifold $(M, g)$ now which is compatible with the Riemannian metric. There's one more condition, which brings us back to the question you asked
@Gasparo Yeah, learning some languages is fun.
Namely, $\mathcal{L}_X Y = - \mathcal{L}_Y X$ in case of Lie derivatives. What's the relation between $\nabla_X Y$ and $\nabla_Y X$?
@Narcissusjewel Well, I will email you another time to tell you more.
In general, nothing
R I E M A N N + error
15:23
The right analogue is apparently $\nabla_X Y - \nabla_Y X =\mathcal{L}_Y X$
Dammit I was just about to write that
Is that still a connection
@Akiva Oh you knew this?
Ah yeah, that's true too.
@BalarkaSen No, I was about to guess that it was probably useful
@AkivaWeinberger No, the Lie derivative isn't tensorial, hence not a connection.
15:24
by analogy with $\mathcal L$
@Gasparo I heard lately that some from MSE received some emails about being involved in some kind of experiment.
@Akiva Wow good guess.
So you were about to write "$\nabla_X Y - \nabla_Y X$" I suppose?
@BalarkaSen Oh, I didn't guess that it would equal $\mathcal L$ though
@Waiting Never heard of that. What kind of experiments?
I hope this is a joke. @Gasparo
15:25
@BalarkaSen Yeah
Gotcha
@AkivaWeinberger Yeah that's the fun part
@Gasparo I don't know the details. Rumours.
I mean it's an assumption
It's called being a "torsionless connection", or "symmetric connection"
Not all connections are like that
It's a hypothesis
But nobody really understand the meaning of this hypothesis
15:26
Oh. So the ones for which the thingy equals the L thingy are an interesting subset
There's an MO post about it
@Akiva Yeah it's interesting because of this theorem:
@Waiting I still have the same email address I gave you the last time. If you would like to talk to me, go ahead. If not, never mind.
@BalarkaSen Uh, I would say torsionless is defined by $\nabla$ commuting with itself on functions.
Theorem: There is a unique connection $\nabla$ on a Riemannian manifold $(M, g)$ which is metric ($\nabla_Z g(X, Y) = g(\nabla_Z X, Y) + g(X, \nabla_Z Y)$) and torsionless ($\nabla_X Y - \nabla_Y X = \mathcal{L}_Y X$).
That means any geometry that arises from such a connection is intrinsic to the Riemannian geometry
Hmm, apparently do Carmo does it with the commutator.
15:28
Oh cool
It's unique and always exists?
It's a magic theorem
Waiting's and Jasper's return might mean another era is incoming...
5
@0celo7 Hmm, what does that mean again?
15:29
Considering that now at least more than 5 of the weird now knew about The Plan, something is going to happen very sooon
@Secret Perhaps 2018 will be the year of miracles, where many people will get their miracles.
We'll see
@Gasparo OK. Sure.
2018 so far is ok, even the casting of the "dark magic" is less frequent than the past year
@BalarkaSen Sorry I was thinking about the physicist definition. It means that $\nabla_i\nabla_jf=\nabla_j\nabla_i f$ for any smooth function $f$
15:30
Ahh yes
'Cuz $[\partial_i, \partial_j] = 0$
Yeah, so it's all equivalent.
@AkivaWeinberger you can write down a formula for the connection given those axioms
called the Koszul formula
So can I get an explicit example?
so it exists and is unique
really appears from the proof of the theorem
On a plane it's probably a directional derivative. What is it on a sphere?
15:32
Considering that tobias will not be interested in pondering about semirings and other weak structures for the moment being, and that Mathien and co. will keep the algebra going, this transition is expected to be smooth
But then, I have no idea what the next era will be, since we already have an integral era back in 2014
@Akiva If X and Y are two vector fields on the 2-sphere in R^3, the connection is directional derivative of Y in the direction of X, projected to the tangent plane of the sphere
$(\nabla_Y X)(p) = \text{proj}_{T_pS^2} (D_Y X)(p)$
Embedded manifolds are really convenient
@Secret Maybe a series era, one with many hypergeometric series.
Check Ted's notes to see how the geometry looks for the sphere
Ted does it very concretely
@Waiting hmm, that will be good cause semi is good at special functions, and I have been recently investigating limit and epsilon delta proofs at the 1st principle level
but we still have a lot of variables, for we have an even distribution of regulars for all fields of maths right now
@Akiva In particular, definition: If $\gamma$ is a curve in $M$, it's said to be a geodesic if $\nabla_{\gamma'} \gamma' = 0$. $\gamma'$ means the tangent vector field to $\gamma$ (extended to all of $M$ by a bump function)
More or less, it means a particle falls off along $\gamma$ with no acceleration
The geodesics in the sphere turns out to be exactly the (arcs of the) great circles with the connection I wrote down
15:37
Right. You can't define it that way with $\mathcal L$, because the pullback thingy means that $\mathcal L_{\gamma'}\gamma'$ is always zero.
@Secret I almost have a whole notebook with hypergeometric series. Working slowly lately since I also have to finish other tasks.
Too bad we are not going to see it based on your few comments above
27 mins ago, by Waiting
@BalarkaSen Don't ask me about it here. I prefer not to talk about my mathematics here (anymore).
Good point.
It's a wonder that this notion of geodesic agrees with what you'd think ("locally length minimizing curves" - that definition involves NO connections. You can define length using the Riemannian metric only: $\int \sqrt{g(\gamma', \gamma')}$)
Riemannian geometry is a fucking magic
15:41
@Waiting hypergeometric series identities are a weird magic
@Secret All the publishable stuff in there will be published one day, not sure when though.
@Waiting You must be Ramanujan reborn, since he also uses notebooks.
@Semiclassical True.
about the only identities there I claim to understand in any depth are those which come from the differential equation they satisfy
all hail the Wronksian blah blah blah
The only identity I truly understand is 1+1=2.
15:42
@Gasparo In mathematics you have to be like you, not like others. To bring your unique contribution.
3
OK I have to go back to chemistry now
3
ugh
@Waiting OK. I was thinking maybe you are the goddess who came to Ramanujan's dreams instead.
@Gasparo haha
also this chat is getting too fucking weird
6
imma shitpost somewhere else
I am still best known here in math chat for all sort of weird maths from division by zero algebras, to a real counterpart of a cauchy integral like formula, to tensor visualisation, infinite sets investigations and so on
15:44
@Gasparo A goddess talking on the MSE chat. The place where the people and gods come together.
@BalarkaSen You are already a weird ever since that chemistry incident
and it is because of that fact that the true form is not shown
|This is because my rule of thumb is as follows:
> The probability of my shit flood post increases with the decrease in weirdness
@Secret That sounds fine to me.
@Secret If thumbs are also fingers, and we have 2 thumbs and 10 fingers, then we have 8 non-thumb fingers, interesting...
and here reveals:
This is one reason math chat flood a lot more serverely when Dattier gone into question leech mode, and when Mike Miller, Eric Silva is on
because they are some of the non-weird people
I never saw robjohn for years.
15:48
@Waiting Same here, think he is very busy.
and whenever chat get too non weird, I want to DDoS it out of existence
Having said that, I expect Eric will become weird soon after I have enough real analysis background to communicate with him
@Gasparo It's possible the mods to be here using some kind of invisible option.
Mike Miller is much harder, fo it will take me at least a decade to have enough knowledge on algbraic topology
Lemma: Non weird people are often just tolerating me, instead of being genuinely interested
5
Proof: In a thousand words
I have a full list of people who are non weird, and whenever they make the chat room non weird, shit postings will skyrocket
This is rare nowadays as ever since we got most of the topology bunch to become weird, it is very unlikely for the weirdness level of the chat t drop below critical levels
@XanderHenderson speak of the devil
Balarka used to be a non-weird and together with Slereah frequently flipped into a neutrino, the two combined with 0celo to form the Phase Lock Trio, which result in huge drop in the weirdness level in h bar, thus result in a lot of shitposting floods
Ever since the chemistry incident, Balarka becomes weird, and for unknown reason, Slereah becomes more unlikely to become a neutrino, thus the Trio rarely Phase Lock now. In addition
Balarka helped me to grasp differential geometry, thus making it more comprehensible and hence more potential to generate weirdness, this is why h bar does not have much shitposting lately
We are not sure what exactly is the chemistry incident, but one guess might be Balarka start taking chemistry
16:02
@Secret what's your favourite book on the calculation of integrals?
I have not read much books yet, due to my attention spread between chemistry, physics, mathematics, fine arts and other things
so I don't know the answer to that question
I see.
So far all my integral attempts are based on trying to approach them from first principle, with the most recent attempt being trying to turn epsilon delta into an algorithm that can give me value of a given limit
Details in the Mathworks room
OK. I have to get back to some very hard work. I'll return here once in a while.
"What's a sound foundation for integrals" vs. "what are some fun integrals I can do"
16:05
I am more the former, because I like to turn things into algorithms and I prefer general cases
The limit stuff is still in progress, for so far I failed to use it to show that the harmonic series is divergent and the p series are convergent, so something's amiss
I guess the problem is that $x \mapsto \sum_{x=1}^{\infty}$ as a linear functional, is too nonlocal for its value to be determined by epsilon deltas
in Mathworks (Not the main chat!), Jan 28 at 11:13, by Secret
The map: $(x > M) \to (a < x :\sum_{k=a}^x () ? \sum_{k=a}^M ())$ is notrivial and requires knowing whether the stuff in $f()$ is nonegative or nonpositive in the interval $(a,M)$
anyway, it's time to sleep, 3:11 might explain why I am sounding a bit rambling
I have a question, im making a program in C++ and i need some math advice. The problem i have is the following: i have a function that reads 512 blocks of bytes from a file and stores them in memory. Assuming that i dont know how much bytes i need to read (lets call it X), i do know that X > 512 and that ceil(X/512) - floor(X/512) != 0. (ceil(...) will round things up, for example 2.78 will be 3, and floor will round down, for example 4.8 would be 4).
I got this idea to make a cicle that would read 512 bytes at a time from the file, floor(X/5120) times, and in the end i would do (X/512) - floor(X/512) , so i would get the 0,.. part and i would multiply that to 512 so i would get the missing bytes : 512 * 0,... . The problem is that i dont know if the 0,... would be finite or infinite, for example it could be 0,234 or 0,333333(3).
The problem with this is that if it was for example 0,333333(3), the computer would assume it was 0,33333333334 , and it could change the result of the multiplication by 512. Anny ideas on a better way to do this?
When someone start talking about maths in the next hour, the chat should go back to normal, which is much better than to stay fully weird and cut off visitors
Argh. I just went through part of Chapter 0 of do Carmo and I got a headache
Annoying technical proofs
Look I trust you on most of this stuff do Carmo
I mean I think I barely tolerated Hatcher Chapter 0 as well
Chapter 0s are just annoying in general probably
which is why they're optional I guess
It's like an appendix but at the front
16:31
Hi @Antonios
hi @MatheinBoulomenos
How's it going?
not bad
on a train
fiddling with some algebra stuff
sounds good
mostly just filling in some details in a few things
need to get back to those szamuely problems later haha
16:36
@0celo7 UPSILON!
just a quick question I don't remember this
so you can use Borsuk Ulam theorem to prove invariance of topological dimension right ?
@AkivaWeinberger I don't know if you know this
@AkivaWeinberger the issue that I have learned with this technical stuff is you could just skim it and look for the main ideas
then it will get easier when reading such stuff
@TedShifrin Are you there?
16:52
@XanderHenderson in a paper by Fefferman, I promise I didn’t make it up ;)
@0celo7 do you know this geometer ?
I am not sure if this fact is correct or not
@Adeek the proof of topological invariance of dimension that I'm aware of uses de Rham cohomology, but not sure about B-U.
I see @Antonios-AlexandrosRobotis
17:19
@0celo7 I have no doubt about it. Upsilon is awesome!
Claim: If $R$ is a finite boolean ring with identity, then $R \simeq \Bbb{Z}_2 \times ... \times \Bbb{Z}_2$......I'm trying to figure out how to use the Chinese Remainder theorem, but I'm not having much luck. I could use a hint.
17:50
could someone help me with a bit of group theory
Let G be a finite simple group with |G| ≥ 3. Suppose that G
acts on the set X = {1, 2, . . . , n}. Show that every element of
G corresponds to an even permutation of X.
prove that $(n!)!$ is divisible by $(n!)^{(n-1)!}$
It's easy using induction. I am looking for a rigorous proof.
Hi, I have question in Binary
@Abcd Induction isn't rigorous?
00000000000000000000000000000101 (5)
11111111111111111111111111111010 (~5 = -6)
@Semiclassical It might be but my teacher told me to refrain from using it...
17:58
:/
Because in our exam we have Multiple Choice Questions generally.
So we obviously can't use induction that time.
Why ~5 = -6 ?
@Stamimail what's 5+~5 in there?
And what happens when you add 1 to that?

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