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4:16 AM
@BalarkaSen wow. I never thought of it this way. Can you draw me one? (I need some starting point)
Is a sylow-p subgroup isomorphic to every other sylow-p subgroups?
 
Yes.
 
4:33 AM
@Subhasis There's pictures of this in Artin, eg in the section where he classifies groups of order $12$
Any two Sylow $p$-subgroups are conjugate; conjugacy is an isomorphism
Draw a literal flower where the stem is $\Bbb Z_q$, and the petals are $\Bbb Z_p$'s sticking out of $1 \in \Bbb Z_q$, the only place they intersect. That's a good enough schematic.
 
 
3 hours later…
7:25 AM
Mooorning
 
8:22 AM
morning
 
 
1 hour later…
9:38 AM
mroning
morning
need halp
ok i dont need actually
 
 
3 hours later…
12:36 PM
Hello
 
Let $a \in \Bbb{R}$. Let $\Bbb{Z}$ act on $S^1$ via $(n,z) \mapsto ze^{2 \pi i \cdot an}$. Claim: The is action is not free if and only if $a \Bbb{Q}$. Here's an attempt at the forward direction: If the action is not free, there is some nonzero $n$ and $z \in S^1$ such that $ze^{2 \pi i \cdot an} = 1$. Note $z = e^{2 \pi i \theta}$ for some $\theta \in [0,1)$.
Then the equation becomes $e^{2 \pi i(\theta + an)} = 1$, which holds if and only if $2\pi (\theta + an) = 2 \pi k$ for some $k \in \Bbb{Z}$. Solving for $a$ gives $a = \frac{k-\theta}{n}$...
What if $\theta$ is irrational...what did I do wrong?
 
12:56 PM
@BalarkaSen Hey remember that picture I drew a while ago explaining $[X,Y]$?
Unifying the like three different definitions for it
I vaguely remember a near-identical picture for $\nabla_XY$
but I can't remember what it is
Do you know what I could've been thinking of?
(The dotted arrows in that picture are the result of taking the parallel not-dotted arrows and "flowing" them backwards through one of the vector fields)
The catalyst for asking this is this Reddit post:
'cause I understand that second one but I'm having a hard time explaining it in words
(Re: the first one: a matrix transpose "looks" like the equation $Ax\cdot y=x\cdot A^\top y$. Which implies several things, like how $A^\top x$ is perpendicular to $A^{-1}x^\top$ where $x^\top$ is the vector space perpendicular to $x$.)
 
1:16 PM
(In other words $A^\top x$ is $(A^{-1}x^\top)^\top$?)
(though that doesn't tell us the magnitude)
 
1:30 PM
Oh, also, 3B1B released another video on differential equations
It feels stylistically closer to his "Essence of" videos than his usual videos
 
 
2 hours later…
3:30 PM
@Semiclassical Logunov is talking about spherical harmonics for some reason
This seems like very basic undergrad physics stuff
Maybe mathematicians don’t see this ever
 
3:51 PM
Well, it’s possibly also a matter of establishing the conventions being used
 
4:45 PM
Maybe this woom is hunting wabbits...
it's vewy, vewy quiet.
2
 
@LeakyNun it seems possible to me that two different f.p. groups have the same profinite completion
 
@MatheinBoulomenos what are the groups lmao
 
I don't know, but from the abstract of that paper, they seem to exist
@RyanUnger I can confirm, I know roughly that they are a nice basis for $L^2(SO(3))$ or something, but I've never worked with spherical harmonics
 
5:08 PM
@Semiclassical no he spent like 30+ minutes
The end game was to construct eigenfunctions on the sphere for arb large eigenvalue with two nodal domains
 
Wait @BalarkaSen
Or anyone who knows Riemannian geometry
If $f:M\to\Bbb R^n$
Do we have $\nabla_XYf=XYf$
 
Yes
 
Thanks Ryan
@RyanUnger Do you have a formal proof? Right now I only have an intuition
 
3x_{11} + 5x_{12} + 4x_{21} + 3 x_{22} - (10x_{11}x_{22} + 2x_{12}x_{21})

Is this function a convex function?
 
@AkivaWeinberger $\nabla_Xu=Xu$ by definitoin
when $u$ is a function
$Yf$ is a function
 
5:21 PM
I meant $(\nabla_XY)f$
Also I don't actually think that's true anymore, you need a projection there
 
no right
@AkivaWeinberger you need a commtator or something
$X(Yf)-(\nabla_XY)f$ is called the Hessian
 
So I'm looking at the Hessian of the embedding
($f$ being an embedding $f:M\to\Bbb R^n$)
If you have an embedding in $\Bbb R^n$, you also have a natural map $T_p\to\Bbb R^n$
which basically looks like translating the tangent plane to the origin
I think $v\mapsto vf$ is that map
 
5:36 PM
Hessian of the embedding? What?
What are you trying to do
 
mumbles that DogAteMy needs to remember that parentheses are important in mathematics
hi @Ryan, DogAteMy
@robjohn !!!
 
@RyanUnger $f$ is the embedding
 
Hi Ted
 
Normally when you write $Xf$, $~f$ is a function $M\to\Bbb R$
but it can be a function $M\to\Bbb R^n$ also and nothing changes
 
yes so I’m wondering what you’re up to here
 
5:42 PM
Correct.
 
So basically what I want is for the Hessian of the embedding to be perpendicular to the embedding
 
So are you trying to compute the mean curvature vector by taking the laplacian of $f$?
 
I'm trying to remember why $\nabla_XY-\nabla_YX=[X,Y]$
 
Oh, good grief.
 
Lol
 
5:44 PM
That's definition of torsion-free.
 
Yes
I know
I'm remember to see why $\nabla$ is torsion free,
 
So it doesn't hold for general connections.
 
where I'm defining it as the connection inherited from the embedding
(that is, it's the derivative of $Y$ in the direction of $X$ projected onto the surface)
 
Right, I think I know that.
I would do it with moving frames, of course. But if you check that the difference is a tensor (which it is), then you can just check on coordinate vector fields.
 
The difference being $\nabla_XY-\nabla_YX-[X,Y]$?
 
5:46 PM
Yes.
You will get the difference of Christoffel symbols, which are symmetric in the Levi-Civita connection.
 
The catalyst for me asking this is this Reddit post
> Why do we use the Levi-Civita connection as opposed to so other connection? Why are its defining properties desirable, especially the symmetry requirement?
> I kind of get the intuition for Rn. I'm just not sure how the conditions of the Levi-Civita connection correspond to this.
I wanted to collect my thoughts here before responding to him
 
@TedShifrin Hey, Ted!
 
Torsion is a very subtle thing, DogAteMy. But torsion-free + metric compatible is the way to get a canonical connection.
 
@AkivaWeinberger I think it just follows from the fact that the commutator of tangential fields is tangential
So the equation is true for the ambient connection and then you project both sides
 
I'm not sure I follow
$XY-YX$ is tangential, yeah
What's that have to do with $\nabla_XY-\nabla_YX$?
 
5:55 PM
What’s the definition of nabla
 
DogAteMy: I looked at the link. You're writing garbage with regard to the transpose stuff. Why should a linear map from $\Bbb R^n$ to $\Bbb R^m$ have an inverse in the first place? And for goodness sake don't use $x^\top$ to mean the orthogonal complement when it already means something.
 
6:14 PM
Oh right
Also I was assuming square :/
Wait so what's the right notation
$x^\bot$?
In any case
If we have $f:M\to\Bbb R^n$, then $df(v)=vf$, yeah?
Identifying $T_{f(p)}\Bbb R^n$ with $\Bbb R^n$
 
If you’re still trying to compute the torsion you should do what I suggested
 
6:58 PM
OK, this is what I'm gonna write
"Maybe before we see why ∇*_X_*Y−∇*_Y_*X=[X,Y] holds for _subspaces_ of ℝ^(n), maybe we should first check that it works for ℝ^n _itself_.

Hint: show that ∇*_X_*Yι=XYι, where ι:ℝ^(n)→ℝ^n is the identity map. (Notice, by the way, that if we identify T*_p_*ℝ^n with ℝ^(n), we have X=Xι.)"
 
I'm trying to find a napoleon quote that I very much liked but have forgotten regarding the difference between the scholar and the non scholar
 
(The parentheses in ℝ^(n) are a Reddit formatting thing)
 
he based much of his success on principles like this I cant believe ive forgotten it
it's basically saying that it's a waste of time to throw a parade for a scholar or win he or she over with compliments and awards etc but this is the biggest source of sense of purpose in the non scholar
 
@RyanUnger hmm. Hard to avoid spherical harmonics when doing eigenfunctions on the sphere, but seems like it it shouldn’t require much time to cover that
Separation of variables for Laplace's equation blah blah blah
I guess if one wants to derive them rather than merely recall them, it requires a bit of board time
 
7:36 PM
\o @robjohn
 
I'm in a period in my life that I can't go to college, does someone know what I can do to feel most closed to it and free? I mean, I like very much to study and I want to study math
 
Have you thought about self-studying from textbooks? @PichiWuana
 
@skullpatrol Yes!! But I always wonder which are the best books (or even in which order to study) and then I encounter few problems or problems without answers (so I feel I can't check my self-study)
 
@PichiWuana that's part of the reason why Math.SE exists :-)
 
yeah there is this thing called the internet and well yes there are better books than others you can study from provided they are not stolen from you by drug dealers you should buy a text book that they base university courses on if you can save for one
I was working from "Problems in Analytic Number Theory" Second Edition, by M.Ram Murty prior to the idiots robbing me and taking that with them which was a fantastic book to self learn from one of the best ive had actually
 
7:51 PM
I'm sorry for that
Maybe a book for university course is not the best for self-study... better for courses with a teacher
 
@AkivaWeinberger You're thinking of the interpretation for Riemann curvature tensor, likely.
$R(X, Y)Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X, Y]} Z$
It's parallel transporting $Z$ about that truncated square you drew, so $R(X, Y)Z$ would measure how much the angle deviation is after it comes back
i.e., "Riemann curvature tensor is infinitisimal holonomy"
 
Yeah I wasn't happy about it either it was more than $200 usd actually well look if you want my honest opinion self study doesn't exist, you are still being taught something by Euclid if you read his works despite him having died a few thousand years ago but he is as much a teacher as you'll get, and if you don't plan on reading the works of others, to maintain some sort of purity in the word self study, well, no you have failed in life and should give up entirely. but that is a very good book
regardless of you attending Princeton university or not
 
> "The only geometry one needs is that every set is a plane and its subsets are potatoids."
 
@Adam Of course I wouldn't read only Euclid
 
The connection induced from the Euclidean connection is torsion free simply because mixed partial derivatives commute, @Akiva
 
8:01 PM
how do I change the sound it makes when someone tags me in my god its like a spring pops lose from my head and shakes my whole body every freakin time
 
lol
 
lol
 
l
o
l
@Adam there's a speaker icon just above the room title in the upper right hand corner, click on it.
Mathematics
 
oh right well thankyou sir you have prevented all further spasms of shock and spilling of coca cola
 
8:19 PM
also I didn't say only Euclid in fact the whole point of my comment is lost if you stick only in there buddy
 
\o @Daminark
 
Hello
 
how's it going?
 
Hi @Dami
 
Yo Balarka! And everything's alright, how about you?
 
8:23 PM
chillin
 
@BalarkaSen still on my math forum? lol that was before I learnt to speak like a normal human and relatively act like one to the extent communication is at least semi worth while
 
i remember being on a couple math forums, which one is yours
i dont visit any of them now though
 
yeah me neither you are the only one I remember talking to on it but I have been well and truly banned from this IP address for that forum now, which, which was as you might have guessed for being too polite and sensitive to delicate religious sensibilities
 
waddup
 
i dont remember being in any notorious forums
Hi @ÍgjøgnumMeg
 
8:29 PM
but no it's not my forum I just remembered it was one of the first I started talking math on, and it was a long road for someone like me being receptive to constructive criticism, especially from a kid a third my age which according to your profile at the time you were
i have a chronological disability that prevents me from accurately recalling exactly when this was, don't worry about it
 
i was pretty young when i started roaming around in random math forums on the internet yeah
might be me
 
well yeah it said you were 10, so it was a troubling thought to be getting advice from a ten year old at the time i think i was still holding on to some sort of hopes of a career in non stupidity related fields which was at some point abandoned
 
@skullpatrol o/
 
how's it going? @robjohn
 
9:05 PM
@skullpatrol things have been rough for a while, but I'm getting a bit more time to do things here.
 
9:16 PM
my computer is acting plain weird again
kind of like a math chat room filled with spam about an extradition bill but still extrajudicial is my new fav word so I guess I should be happy there
 
9:37 PM
@PichiWuana I don't know where you are in your math studies, but you might find my lectures on linear algebra and multivariable calculus/analysis interesting. See the link in my profile.
 
9:53 PM
@TedShifrin thanks for that in bookmarking all of these under 3500, is there a 101 i should start with and find my way into four digits? what level of expertise is required for all of these is a more clear way of asking
 
Well, there are various math sources all over the web, including Khan Academy, etc. My particular course was intended for people seriously interested in mathematics (i.e., proofs as well as computations and applications). The students in there were about half first-year students who had taken BC AP calculus in high school and gotten the top score, about half second-year students who'd taken various first-year calculus paths in college.
 
yeah I did at least the first two years of linear algebra and calculus at university but was not math focused at the time but in watching one of them now enough is familiar to try to keep up i think
 
Yeah, you'll need to remember basics of single-variable calculus, but otherwise it's pretty self-contained if you're motivated.
 
long time ago tho even the credits have expired not the student debt though so i think they are trying to hint i should go back a start from first year and double said debt but im a terrible student it really wasn't worth while the first time round considering my rate of attendance then and how unlikely that would be different going back now
 
Hey everyone, is there a way to derive the Cauchy estimate for power series coefficients without use of complex analysis?
 
10:07 PM
these "virtual" youtube varieties of learning is certainly a less risk financially speaking
 
For what it's worth when I started watching Ted's lectures in the process of understanding multivariable calculus I started from like lecture 2 or something. It's great
 
@Thorgott: No, it depends completely on the Cauchy Integral Formula.
I mean you can turn that into an application of Green's/Stokes's Theorem and "pretend" you don't know Cauchy, but ...
Thanks, a @Balarka. Lecture 1 was defining vectors and vector addition (subtraction, too, very hard).
 
Cauchy's theorem is natural in the Stokes setup but not Cauchy integral formula I don't think.
@TedShifrin There are some good problems even in that section of your book :)
 
Sure, Cauchy's integral formula follows just like you prove Gauss's law using Stokes's Theorem.
 
It feels weird to turn the complex quotient $f(z)/(z - a)$ into something real
 
10:09 PM
The key thing you have to use, though, is that $f$ holomorphic makes $f(z)\,dz$ a closed $1$-form. We need something from complex analysis.
You don't need to turn it into something real.
You'd need to turn the whole $1$-form into real things, not just the function.
 
OK, sure. You'll argue it has an antiderivative on the indented contour by Cauchy's theorem.
 
@BalarkaSen yeah from the number theory i got into in my most recent years it's bizarre how i almost became allergic to calculus i loved it back then and for some reason not quite so when i began focusing on prime numbers
 
The key point to me is still that it's picking up a residue at the pole of order one at $z = a$.
 
In fact, I'm fond of the generalized CIF that has $\iint \partial f/\partial\bar z \,dz\wedge d\bar z$ in it.
I've never been much of a fan of number theory, full disclosure.
Only elementary stuff for pedagogical reasons.
 
@Adam There are some serious analytic flavors to number theory, actually, which might be to your taste.
May 16 at 22:28, by Balarka Sen
What do you all think of this theorem: The number of ways to write $n$ as a sum of four squares is equal to $8$ times the sum of divisors of $n$ if $n$ is odd and $24$ times sum of odd divisors of $n$ if $n$ is even
A proof of this uses (basically) Fourier analysis
Even though it looks rather innocuous albeit surprising result in pure number theory
 
10:14 PM
So it's somehow a question about factorization in the quaternions :P
Hmm, maybe not.
oh, it's nerd Demonark.
 
Hello there nerds
 
Yeah but that's not how you would approach the counting problems usually
 
Lmao I got sniped
How's everything going?
 
@BalarkaSen well because it was what Wikipedia deemed my interests to be categorized as i have simply told myself that is what i am studying, it really starting with me horsing around not even knowing what category of math you call it. actually, ill show you the exact subject you and i discussed on mmf that reminds me you were actually right, i don't know if i would have taken it well at the time tho
 
The number of ways to write $n$ as a sum of four squares, $r_4(n)$, are coefficients in the Fourier expansion of the theta function raised to the power of 4.
 
10:16 PM
I see, thanks.
 
Tbh I think not categorizing oneself is best. There's some cool stuff out there, do some cool stuff
 
So it boils down to an estimation problem for those coefficients. That follows from surprisingly geometric results about cusp forms on $\Bbb H^2/\Gamma(2)$, etc.
 
Obviously a lot of one's work might fall under something or other but even if you aren't as fond of some parts of analysis, you might find others fun/useful
 
Interesting.
 
Modular forms seem like witchcraft
 
10:19 PM
@Adam Oh OK we know each other from MMF. Got it. No need to pull out conversations from there tho :P
Would rather not be reminded of my younger version
 
I still remember Balarka 1.2 a lot.
I think that's Balarka 0.5.
I'm feeling old.
 
I wish they just realized the model is fundamentally defective and stopped producing newer versions
 
LOL ... they?
 
yeah looks like i deleted the stack exchange question on it anyway i had found a discrete Fourier transform for $\lfloor \frac{n}{m} \rfloor$ and you attempted to explain to me that is what it was that's all i remember lol @BalarkaSen
oh and when it comes to transcripts involving me on the internet, don't worry, the younger version of you most definitely will be seen in a positive light, and just contemplating all the possibilities of things said by someone as insane as me, agree that pulling up said past conversations isn't productive
 
Especially since most of us have no idea who you are/were, Adam.
 
10:27 PM
On the internet nobody knows you're a sloth
3
 
Or maybe everyone knows.
 
Oh wait
 
I was half-joking, the point was I was a very pretentious know-it-all back then.
LOL @Dami
 
Ironically, you know more now, a @Balarka.
 
yeah thanks that's what i tell myself @Daminark having said that thinking that way is one of the key ingredients for giving myself enough rope for the danger zone
 
10:28 PM
The weird thing is the more I know the more the number of things I don't know exponentially increases.
2
 
Yup, that's one of the wonders of life, Balarka.
 
absolutely me too but would we have it any other way? i mean i know im like a dog chasing a car as far as any real "purpose" in learning is concerned i think id be terrified if something didnt unfold into a myriad of new things I'm clueless about
 
@Daminark They key thing if I remember correctly was that if you look at the subgroup $\Gamma$ of $\text{PSL}_2(\Bbb Z)$ generated by (1, 2|0, 1) and (0, -1|1, 0), then any holomorphic function $f : \Bbb H^2 \to \Bbb C$ invariant under $\Gamma$ (in the sense that $f(z + 2) = f(z)$ and $f(-1/z) = z^{2k} f(z)$, $2k$ is called the weight) such that the Fourier expansion of $f$ at infinity and $-1$ having no constant coefficients is called a cusp form (on $\Bbb H^2/\Gamma$).
The $r_4(n)$ thing follows as an immediate corollary of the fact that the only weight $2$ cusp form is identically zero.
I can try to recall more if you're interested.
It's insightful to look at the picture of $\Bbb H^2/\Gamma$... it's like, take the line $\Re[z] = 1$, the semicircle $|z| = 1, z > 0$, and the line $\Re[z] = -1$. This gives a certain region in the upper half plane
Paste those two lines, and paste half of the semicircle (from -1 to i, and then from i to 1) to the other half by folding along i
 
Ah so it would probably then follow from some finite-dimensionality type result?
 
Yup, that $E_4$ and $E_6$ generates the space of modular forms, that type of things
I think in general if you start thinking about modular forms as eigenfunctions of a Laplacian, the space generated by the Eisenstein series are orthogonal to the space of cusp forms - there's a general story I don't quite know
Cusp forms vanish at the cusp (those are the $-1$ and $\infty$ points in the quotient $\Bbb H^2/\Gamma$ picture I described above, where the hyperbolic metric gets coned off), whereas given any values on the cusps you can make a linear combination of Eisenstein series which takes those specific values on the cusps
So it sort of makes sense
Regarding that particular result, saying it's a weight 2 cusp form is like specifying a strong decay rate of the cusp form towards the cusp. Indeed, one basically argues like the maximum value theorem in complex analysis
I can give you a reference actually
@Daminark Sent on Washington
 
10:52 PM
@BalarkaSen no you didn't come across as pretentious at all, i can only imagine being so young and having the mind you have would have resulted in many accusing you of such, but really, my experience in life is diverse to say the least, and I've met know it all types that are in everyway detestable, you shouldn't be so hard on your character you are very humble considering your calibre
 
Thanks. I suppose I have grown more humble over the years rather
 
You probably don't realise how low the bar drops when it comes to integrity of character is concerned, trust me, you wouldn't have come as far as you clearly have if you were a know it all
it was actually the best thing for me to have met a 10 year old at the age of 30 that was well beyond what ill ever realistically become as far as math is concerned someone like you is going to be accused of arrogance simply because you intimidate many ignore the good majority of that mate
 

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