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12:00 AM
@Fyn I'll take that as a sign that I wasn't spouting gibberish
(speaking nonsense)
 
 
1 hour later…
1:08 AM
Gee, I’ve missed so much fun and frolic!
 
party noisemaker sound
 
1:58 AM
Entertainment by Munchkin? (See above for origin of her name.)
@AkivaWeinberger This reminds me of J. Alfred Prufrock.
 
Would you entertain an app to accompany your YouTube course sir?
 
LOL. What would it do?
 
I suppose there's already a YouTube app...
Interesting how LaTeX doesn't italicize capital Greek letters: $L(x)\Gamma(x)$. More consistent would be $L(x)\mathit\Gamma(x)$ (or I guess $\mathrm L(x)\Gamma(x)$)
I suppose one consequence is that, to be consistent with LaTeX's conventions and in light of the fact that the $\Beta$ command doesn't exist, the Beta function should be written $\mathrm B(z_1,z_2)$ instead of $B(z_1,z_2)$.
 
2:17 AM
You have my sympathy @TedShifrin I didn't realize how bad authors had it until now. I just saw a chegg ad where they're basically marketing chegg as a replacement for textbooks. There is no real replacement to a textbook except for another textbook but even then there's always subtleties from textbook to textbook on a subject
My generation unfortunately just wants answers and doesn't want to do any work.
 
Thanks, Obliv. I don’t feel sorry for myself. I wrote books to improve the learning of students, not ultimately to make money. Most of the books out there are disappointing, and so is most of the stuff freely available on the web.
But, yes, learning is hard work and requires mental dedication. You can’t just get a knowledge implant.
That said, I am proud of my diff geo text, which is a totally free pdf.
 
As a student who used to self study to some degree, I appreciate that. We're really blessed to have access to such high quality work such as your own.
By the way, would it make sense to follow your lecture series on multivariable mathematics along with your textbook, or do you jump around?
I haven't seen the series yet, been saving it but I'm guessing if it follows it then you mention where you are in the text each week
something that is hard to work up to but extremely rewarding in studying math is having that moment when everything clicks and studying becomes really fluid.
I have a hard time getting in that state of mind lately, but I know if I try hard enough it'll come back.
then I can hopefully read your diff geo text without asking for help every other minute :P
 
2:39 AM
ted himself engaged in the longstanding differential geometry tradition of ripping off do carmo. that was the chegg of its time.
and do carmo ripped off... who was it, again?
 
We all ripped off Riemann didn't we?
 
curves and surfaces is more gauss and older folks, but broadly, yes.
 
creativity is knowing how to cover your tracks
 
imagine a world where copyright protection in math extended to specific results, and copyright terms were infinite, so you'd have, like, riemann's great-great-great-grandchildren riding around in lambos and doing cocaine and being influencers.
 
But without Euler, Riemann probably wouldn't have made his Zeta function right?
@leslietownes LOL
Hmm.. but Riemann would owe shares of his results to other mathematicians. Maybe the ancient greeks would have the most shares in mathematical results :P
 
2:45 AM
Can you copyright a number?
π®
 
@TedShifrin Also, had to look this reference up and I am very pleased to have discovered this gem. This opening part about living in one place for so long and you go back and it's like a museum rang true youtube.com/watch?v=ZNSI9XemXdo I sometimes visit my childhood town and it's crazzyy how time affects an area.
Leaving some things changed and others unchanged
@user4539917 If so, I want 3.
 
The lectures followed the text (although I did different examples, sometimes different heuristics/proofs). The only exception was that I reversed the order of chapters 6 and 7,
 
Why so?
 
The diff geo is meant for students who’ve had standard multi and linear algebra, not my hard course.
@user4539917 Because I wanted to start the second semester with integrals, not the technicalities of the inverse/implicit function theorem.
 
hmm...
 
2:55 AM
@leslietownes Hell no. We are brothers, but his book is far from my choice.
When Chern taught out of doCarmo, I had to run problem sessions explaining the book. And that was Berkeley students.
Chern’s notes are classical in part, but over-filled with tensor notation and lacking in exercises. I ripped neither of them off. :)
 
i was kidding. and it's not surprising that all curves/surfaces books include some of the same classic problems as examples, absent any ripping off or contractual obligation to do so. there aren't so many things you can nicely compute by hand at 'homework problem' length.
 
DoCarmo insists on treating surfaces with abstract manifold machinery. Chern is the opposite. I am aligned closer to Chern, for sure.
 
 
6 hours later…
9:16 AM
so in finite group theory, a group action on a set $X$ partitions $X$. Is the same true when considering the action of an infinite group on a set $X$?
 
what partition are you thinking of there? into orbits of the action? that certainly works in any group, whether finite or infinite
 
yes, ah okay
 
it's mostly the existence of inverses that you use there, for semigroups that aren't groups, you might not get a partition
even for actions of very nice semigroups, like cancellative abelian semigroups
 
9:37 AM
is there a way to determine some form of order (as in |G|) of a set containing all the points on a hemisphere? is that a sensible question xD
 
10:08 AM
uh, not sure there. the set of points on a hemisphere is infinite, unless 'hemisphere' means something weird.
there's the orbit-stabilizer theorem, which in finite groups tells you something about the size of orbits. that generalizes, although once a topology on the group gets involved i think you need to make assumptions about the topology or the action to get a nice theorem.
but it would say for example that if you have the circle acting on a sphere by rotations, then it's no accident that the orbits of points (other than the poles) are homeomorphic to circles (i.e. they "look like" copies of the group that is doing the acting)
but thats more about the space as a topological space, not just what its order is. in at least a lot of infinite settings, 'order' alone won't tell you much.
 
 
2 hours later…
12:40 PM
Either Calc 3 is pretty easy or I just haven't yet gotten to the part where it gets challenging
 
@ペガサスSeiya You know that when you say things like that, you kind of sound like an ass, right? Braggadocio is not a good look...
 
@XanderHenderson "or I just haven't yet gotten to the part where it gets challenging" I wrote this specifically to make sure I don't come off as arrogant
@XanderHenderson Other than that, you've said stuff even worse than that and I haven't called it out. Please, next time, assume good faith instead of assuming that everyone who says "X seems easy" is being boastful
 
@ペガサスSeiya And I am telling you that the tag didn't help. From my perspective, what you wrote came across badly. You can listen to what I am saying, and perhaps make an attempt to improve, or you can ignore me.
It is entirely up to you.
 
12:55 PM
I will choose to ignore. Someone misinterpreting my words on multiple occasions is not my problem
 
@ペガサスSeiya One person "misinterpreting" your words on one occasion might not be your problem. But multiple people have commented on your approach in chat on multiple occasions. This starts to look like a pattern, which you might want to do something about.
 
1:22 PM
Good afternoon 🥱
 
2:01 PM
@SineoftheTime Morning here. Just past 7am.
I have recorded one lecture today. Two more to go.
 
2:29 PM
Can anyone please verify my proof for 'orbit space of action SO(n) on R^n is $[0, \infty)$'?
 
@XanderHenderson it 4 pm here. This morning I did the exam of abstract algebra 😵
 
I don't think this is true, Koro
ah no, perhaps it is
but you're missing a step in any case, namely that if two elements have the same image under $f$, they lie in the same orbit. this will be equivalent to injectivity of $\mathbb{R}^n/SO(n)\rightarrow[0,\infty)$.
 
2:46 PM
ohh, I first show that f is constant on fibers of q: this ensures that we have a map that makes the diagram commute.
But, if I show f to be a quotient map, then by property of quotient space, I get the homeomorphism.
So the question is: how to prove that f is a quotient map? For that, I define a map j from [0, infty) to R^n. Noting that f is onto and continuous, and that foj = identity of [0, infty), it follows that f is a quotient map.
so we are done.
 
that's all fine and dandy, but it doesn't suffice
 
why not?
 
why would it?
 
because I have the homeomorphism!
 
hi
 
2:49 PM
why do you think it is a homeomorphism?
 
like I said: because f is a quotient map.
 
so any quotient map is a homeomorphism?
 
no!
f induces a map from R^n/ SO(n) to [0, infty) so that the diagram commutes.
this induced map is homeomorphism iff f is a quotient map (this is property of quotient spaces).
^the universal property
 
no, the universal property of the quotient space just says that the map at the bottom is continuous
 
yes, but that's not the only thing it says. It also says that: the map at the bottom is a homeomorphism iff f is a quotient map :-).
 
2:55 PM
so what's your claim? if I have a triangle like X -p-> X/~ -g-> Y, then g is a homeomorphism iff gp is a quotient map?
 
not necessarily. But add one more condition: given an onto continuous map f from X to Y, such that f is constant on fibers of p
f=gp
 
ok, so that is your claim if we add that gp be onto?
 
g is continuous
 
sorry, mistyped
 
gp onto and constant on fibers of p, then g is a homeomorphism.
 
3:03 PM
gp is automatically constant on fibers of p
cause p already is constant on fibers of p
 
then I think that g should be a homeomorphism.
 
ok, so take ~ to be the identity relation and suddenly you're saying every quotient map is a homeomorphism
 
no, I am not saying that.
You are considering the diagram differently. We want to consider a map F from X to Y, then this induces and map at the bottom.
not that given a map at the bottom, and then we are getting a map on the side
 
there's no real difference, you get the same diagram
 
@Thorgott I understand now that considering the diagram this way has such problems.
can you please look at theorem 22.2 in Munkres?
 
3:15 PM
@XanderHenderson All I claimed is that calc 3, as far as we've done it, is rather easy. That's all there is to it. If saying something like this is against the rules, I'll stop
 
sure
 
@Thorgott Here is the universal property: Suppose X, Y, Z are spaces. Suppose that q: X-->Y is a quotient map. Given any continuous map g from X to Z which is constant on fibres of q, a map from Y to Z is induced (let's call it h)
We have the following: 1) h is continuous, 2) h if homeomorphism iff f is a quotient map.
 
I'm looking at the theorem and that's not what it says
 
In your example: you didn't induce any map.
 
sure I do, g is induced from gp
 
3:18 PM
please look 22.3 (corollary)
 
the fact I gave it a name first doesn't make a difference
yes, I see Corollary 22.3, the scenario there is different and that difference means the world
 
And I misstated it in my last message: Y should be X/~
@Thorgott yes, and in my solution above, that scenario is met.
 
perhaps it is, but you haven't shown that
 
I did show that!!
'f is a quotient map' was shown.
 
yes, but that's not the crux
 
3:20 PM
why not?
 
corollary 22.3 does not just take about any quotient of $X$, it talks about a specific quotient of $X$ that depends on the map $g$ (in his notation)
 
constant on fibers of ... is just a restatement of that. Isn't it?
fibers terminology helps me remember that corollary. That's why I like it and use it :-).
 
no, constant of fibers just means the relation is at least as fine as the one of corollary 22.3
in corollary 22.3, the fibers of $g$ (in his notation) are precisely the equivalence classes of the relation
constant of fibers just means the fibers are unions of equivalence classes
in your attempt above, this means you don't just have to show that $f$ is constant on $SO(n)$-orbits, you have to show that the fibers of $f$ are precisely the $SO(n)$-orbits
which is missing a reverse direction: if $f(x)=f(y)$, there is an $A\in SO(n)$ such that $Ax=y$
 
@Thorgott Ohh that's a valid point.
I should show that q is also constant on fibers of f.
I showed that f is constant on fibers of q but not the other way round.
Thanks a lot @Thorgott. I understand it now.
 
a quotient map is a homeomorphism if and only if is injective, so in the scenario of theorem 22.2, the map $f$ is a homeomorphism if and only if it is injective if and only if $g(x)=g(y)$ implies $p(x)=p(y)$, i.e. iff the fibers of $p$ and the fibers of $q$ define the same partition of $X$
(just a different way to word the same observations)
 
3:31 PM
and I suppose that we can't really say that if ||x||=||y||, then there exists A in SO(n) such that x=Ay
 
also, perhaps this helps for intuition, you can compare it to an analogous situation in linear algebra: a surjective linear map $f\colon V\rightarrow W$ of vector spaces induces a linear map $\overline{f}\colon V/U\rightarrow W$ if $f$ vanishes on $U$, but that map will be an isomorphism iff it is injective iff $U=\ker(f)$
 
Avv
Thank you. Can you please refer me to a source where I can find proof for cosine interpolation formula $ x(t) = x_1 + t (x_2 - x_1) , y(t) = y_1 + t (y_2 - y_1) $?
 
3:51 PM
@Thorgott: I did the following patch now- Suppose that the induced map in the image is g: R^n/SO(n)-->[0,infty). Then this g is onto continuous and in fact a quotient map. I show this g to be one to one as follows: Take any [x], [y] in R^n/SO(n) such that g([x])=g([y]). This implies that f(x)=f(y), i.e., ||x||=||y||. If x=0, then y=x=0. If x is non zero, then y/x lies on $S^{n-1}$.
This means that y and x differ only by a rotation, i.e., there exists an A in SO(n) such that y= Ax, which is same as saying that [x]=[y].
This shows that g is one to one. g is onto (as it is a quotient map). So g is a bijection.
g is bijection+quotient map, so g is a homeomorphism.
probably the only part confusing here is 'y and x differ only by a rotation'. I wrote so because it feels true to me intuitively. I don't yet know how to prove it.
Oops, there is a mistake in it. y/x makes no sense!
I should say that y/||x|| lies on the sphere.
ohh but that's okay because: SO(n) 's action on S^{n-1} is transitive. This proves the 'y and x differ by a rotation' part.
 
4:15 PM
sounds good to me
 
5:06 PM
How do I show that $CP^n=D^{2n}/$~, where ~ denotes identification of antipodal points?
For me $CP^n= (C^{n+1}-0)$/lines through origin
 
What you wrote is wrong.
It's not antipodal points. What you described is $\Bbb RP^{2n-1}$.
 
antipodal points of boundary of $D^{2n}$
ohh
yeah, it was wrong.
we have x~y if x=y or x,y are in $\partial D^n$ and q(x)=q(y), where $q: S^{2n+1}\to CP^n$ is the quotient map.
I'll try with this now. I had misunderstood ~ earlier.
 
Right. You're modding out by the $S^1$ action. And you need $S^{2n-1}\to\Bbb CP^{n-1}$.
 
5:24 PM
ah yes, the sphere bundle of the tautological bundle :P
 
Is there some other mystical way of finding a unitary matrix to make two matrices unitarily equivalent besides eigenvectors?
 
Nope.
 
So why make the definition or statement so vague to leave it open to the interpretation that such a thing is possible?
 
Perhaps if you have a geometric characterization of the underlying linear transformations?
 
i.e the definition states: "A and B are unitarily equivalent iff there exists a unitary matirx $P$ s.t $A = P^*BP$"
@TedShifrin ooooh.... well the generalization of orthogonal operators is in a chapter coming up....so I guess it is "to be continued"
 
5:41 PM
How does $S^1$ act on $S^{2n+1}$?
 
Complex multiplication.
 
Suppose w is in S^1. $(z_0,z_1,...,z_n)\in S^{2n+1}$, then $w.(z_0,z_1,...,z_n)=(wz_0,wz_1,...,wz_n)$
hmm
 
5:58 PM
How do you bookmark conversations again?
nevernubd, I found it...glaring right at my face
 
6:08 PM
I have no idea how to prove that one.
Is there any book on 'problems in algebraic topology' with solutions?
 
which part are you trying to prove?
 
6:25 PM
I'm trying to prove that CP^n is D^{2n}/~, where ~ is as above.
 
do you know the antipodal case that Ted mentioned
 
oh, I think there is no antipodal characterization for CP^n.
For RP^n, yes.
 
then just try doing something analogous
 
but the problem in this case is precisely the lack of antipodal representation. I define $f: S^{2n+1}\to D^{2n}: (z_0,z_1,...,z_n)\mapsto (z_0,z_1,...,z_{n-1})$. But this f does not seem to be constant on fibers of $q: S^{2n+1}\to CP^n$.
 
well, forgetting a coordinate certainly won't do you any favors in constructing any homeomorphism
tell me your preferred proof for the RP^n case and then we can discuss how it can be adapted to the complex scenario
 
6:41 PM
Here I show X and Y to be homeomorphic.
This proof.
Two steps: 1) Consider the triangle S^n XY: this gives a map from X to Y. 2) Consider the triangle X(R^{n+1}-0) Y, this gives a map from Y to X.
 
that's not what I mean, sorry, perhaps I wasn't clear
 
These two maps (both continuous) obtained in 1) and 2) turn out to be inverse of each other thus giving the homeomorphism.
 
I'm asking about $D^n/\sim\cong\mathbb{RP}^n$, where $\sim$ identifies antipodal points on the boundary
 
So can we modify this for CP^n?
ohh okay.
 
what you just gave is analogous to $\mathbb{CP}^n=S^{2n+1}/S^1$, which is true, but not the statement we're trying to prove
(to preemptively avoid confusion, I mean the orbit space, not collapsing $S^1$ to a point)
 
6:46 PM
So first D^n is identified (homeomorphically) with the upper hemisphere of S^{n+1}
 
is it weird that I find it more natural to write how many times a goes into b as $\frac{a}{b}$ instead of $\frac{b}{a}$
I just feel like it reads better the other way.
kind of too late to change that notation though I guess.
 
so D^n can be em-bed in S^n, then we consider the quotient map from S^n to S^n with antipodal points identified.
 
well, if you multiply the number of times a goes into b with a, you should get b. I'd say that $\frac{a}{b}a=b$ looks a lot weirder mnemonically than $\frac{b}{a}a=b$. cancelling fractions is very intuitive notation.
 
then we use the universal properties
 
Koro, that sounds good
 
6:53 PM
@Thorgott to each their own. I'll admit that does look weird but I'd read that as the number of times a goes into b * a is b.
Only because we're used to it that way though.
Consider multiplication as done in elementary school, it's just as queer as the line method of multiplication where you draw lines to represent digits and find intersections
and count the intersection points on diagonals
 
you'd read both the same way, my point is that one is visually helpful and the other is not
 
i'm just used to left to right, up to down
I'd argue they'd be equally helpful visually if you have the same amount of practice in both ways.
just preference really
 
7:30 PM
The math above me looks nothing like the math I'm used to. I wonder if/when I'll get to that point
 
I'm sure you'll get to fractions sooner rather than later
 
That looks like far more than mere fractions
 
@thorgott is the surface area of a sphere divided by a circle made by a disc of its diameter also a transcendental value like pi
the area of the circle*
I'm assuming it is
 
that ratio is precisely the diameter itself
 
oh that's neat.
 
7:40 PM
@Thorgott can you elaborate a bit? I'm confused on that
 
Actually I think it's just 4
$\frac{4\pi r^2}{\pi r^2}$
 
That's why I said I'm confused
 
I understood you to be talking about the circle, not the disk
if you want the denominator to be the area of the disk, then sure, the ratio is 4
 
oh yeah I meant the area of the enclosed disk
is there a definition of pi other than circ/diameter
like how euler's number can expressed as an infinite sum or a limit of a power of an expression
 
sure, there'll be plenty
off the top of my head, there's $\pi=\sqrt{\sum_{n\ge1}\frac{6}{n^2}}$
 
7:46 PM
Gaussian integral has a nice result with a pi in it
 
Wikipedia probably has an article titled "formulas involving $\pi$" or something
@ペガサスSeiya oh yeah, I second this
 
is that summed from 1 to some n greater than or equal to 1?
 
that one, unlike most others, is actually useful to know
@Obliv summed over all positive integers
 
@Thorgott we were proving it a few days ago using polar coordinates. I didn't like that method, there's gotta be something simpler
 
@Koro I corrected you earlier. You are off in your dimensions.
@ペガサスSeiya No, that's the simplest way to compute the integral. There are other approaches using differentiation under the integral sign (this approach appears as an exercise in my multivariable math book) and Fourier transforms.
 
7:51 PM
polar coordinates are simple and, most importantly, intuitive
you can knock yourself out with math.stackexchange.com/questions/9286/…
 
@TedShifrin so, any polar-free approaches are more complicated?
 
either they use more machinery or they're at an equal level of complication, but more obtuse
 
I say they are more sophisticated for sure.
 
My favorite part about that integral is squaring the inside and then putting a square root over it and then saying "Look, I just did a whole bunch of nothing!"
 
And require more theory (like proving differentiation under the integral sign is valid, which is nontrivial)
The polar coordinates proof still requires justification if you're being rigorous.
 
7:55 PM
We're engineers. You can probably guess how rigorous we are with math
@TedShifrin "this approach appears as an exercise in my multivariable math book" I'll believe you IFF you tell me where to buy the legal copy
 
Enough of this, Seiya. If you want to buy it, it's easily found on Amazon or on the publisher's website. Drop it.
 
@TedShifrin I cannot find it. I'm not kidding
 
it's also possible to compute the gaussian integral only using single variable calc
 
8:12 PM
might have delivery technicalities
 
You mean delivery charges?
 
8:24 PM
What exactly is an integration factor, like why are we taking $e^{\int P(x) dx}$
 
@ペガサスSeiya that and may not ship to your country
 
also given $(a + \frac{b}{c})^d$ we have $a^d + \frac{abd}{c} + (\frac{b}{c})^d$ for $a,b = 1, \lim_{d\to \infty}$ we have it equal to $e$ but what if I change the numbers around and leave d, do i get some permutation of e?
 
@Obliv won't ship to Japan?
 
actually I think I did that algebra wrong
 
@ペガサスSeiya When you are back in the States, one way is to borrow. When I lived in Texas, I borrowed from local university libraries using TexShare library card free for Texas residents. In some cases I can do Inter library loan through a local public library too, IIRC. In Dallas area, for example, the Ted's Multivariable Calculus textbook is available in UNT (Denton) and Texas A&M.
 
8:37 PM
I actually don't even know what terms a general polynomial with order $d$ would have
or if you dont mind having an electronic copy just send ted some cash and get an online copy off the millions of pdf rips
or don't and keep it to yourself :P
 
@GratefulDisciple That's great, and I'm from Texas so that's even better
 
Does someone know how to show the second part in b)? I believe I have to show that $\rho(\omega)=0$, which I can only argue for the trivial representation.
(I don't know which notation might be uncommon, but I can clarify of course)
 
Reviewing calc 3 at 6 in the morning
Can life get any better?
 
9:10 PM
never mind btw
 
I regret not sleeping
 
The common refrain of all who stay up late with responsibilities looming the next day
 
9:34 PM
I'm gonna walk, not safe for me to drive while being drowsy
 

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