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4:01 AM
Is there any easy way to see that any arbitrary union of right open intervals of the form $[a, b)$ in $\mathbb R$ has a minimal element?
@S.D. $\bigcup_{\varepsilon > 0} [\varepsilon, 1)$ ?
4:24 AM
@user76284 Interesting. So that is not actually true. Thanks!
2 hours later…
6:34 AM
@TedShifrin @MikeMiller What I want has been proved by Forster and Rampsott, see first paragraph in the introduction here.
4 hours later…
10:19 AM
Am I right in saying: (Complex Analysis)
If our f is holomorphic in some region D, but we have some closed contour C that goes around some hole somewhere inside (but not included in) D, then we cannot use Cauchy's result to conclude the integral is 0, since f is hence not holomorphic on and inside C. (Since we're only given that f is holomorphic in D and not the 'hole')
10:36 AM
So in this question:
Q: Cauchy's Theorem, Deformation Theorem, similarities and differences - Complex Variable

Euler_Salter1) Let $f$ have an antiderivative $F$ in a region $D$ containing a $\underline{closed \,\,\,contour}$ $C$. Then \begin{equation} \int_Cf(z)dz = 0 \end{equation} 2) Also, Cauchy's theorem says: If f is $\underline{holomorphic}$ inside and on a $\underline{closed \,\,\,contour}$ $C$ then \begin{eq...

The question says 'Is it now obvious? Of course they are, they are zero!'
The error in this reasoning is that there's an assumption that the contour's interiors are in D
Am I correct?
yeah, the integrals aren't necessarily $0$
I get that, but it's because of my argument or something else?
well, the reason why they aren't necessarily $0$ is that counter-examples exist; your argument explains why this doesn't contradict Cauchy's theorem
Ok, thank you :)
10:47 AM
Hi everyone
I actually hail from the physics side of things, but its MP so that could be a compromise
My object of interest is the case where I have a product of many many rational expressions in some variable
So I already covered the case where I have more zeroes in the denominator, I found something like this:
Anyway, I'd also like to find something like this variant of the Heaviside rule but for polynomial long division
Finding reading material on this question seems to be a rather difficult task, I was hoping maybe someone here knew how to divide a general polynomial by another general polynomial, ofc assuming these polynomials don't have the same coefficients
And assuming that the polynomial in the denom P_D has a smaller degree than the one in the one in the numerator
P_N say
11:27 AM
Does anyone know what "two symmetric edges" means in graph theory ?
Two edges between the same two vertices I think (with the same orientation if the graph is oriented maybe)
11:59 AM
@robjohn Hello sir! Can you please give me a proof that $$f(x)= \begin{cases} 0 & if~x~is~irrational \\ 1/q & if ~x~is~of~the ~form~p/q~(it ~in ~simplest~form~and~p,q \in \mathbb Z , q\neq 0 \end{cases}$$ ($0\lt x\lt 1$) have a limit zero at any $0\lt a \lt 1$.
Here is the proof given by Spivak
But I understand your proofs quite easily, so if you have time can you please write up an elegant proof (like you do). It’s a request not a demand.
@Knight The idea is simple: If you fix the denominator, there a discrete amount of rationals so the smallest distance from $a$ to the nearest such one can't be too small
12:24 PM
I'm asking for your advice on the following matter
1 hour later…
1:29 PM
Is there a canonical volume form on the unit tangent bundle?
@Knight For any $q$, there are fewer than $q^2+q$ rational numbers with denominators less than or equal to $q$ within $1$ of $a$. Find the closest of these $\le q^2+q$, say at a distance $\delta$ from $a$, of course, not counting $a$ if $a\in\mathbb{Q}$. For any $|x-a|\lt\delta$, you have $f(x)\lt\frac1q$.
This is a $\delta$-$\frac1q$ proof, instead of $\delta$-$\epsilon$
2:11 PM
In complex analysis, can we substitute the z and a in "limit as z approaches a" with their respective conjugates?
not entirely sure what you have in mind, but most likely the answer is yes
Have to show that if f(z) is holomorphic and the domain also contains all its conjugates, then conj(f(conj(z)) is also holomorphic
Basically wrote down the def'n of the derivative for conj(f(conj(z)) and worked my way back to the def'n. of the derivative for f(conj(z))
that sounds like a good way of going about it
what you want to use is that complex conjugation is continuous
2:50 PM
@Thorgott why do I need this?
I hate thinking like this, but this was an exam question
3:09 PM
if this is an exam question, I'm not interested in talking about it
3:32 PM
@Thorgott not one of my exam questions, i'm doing past papers of complex analysis. regardless, of course you're not obliged to talk about anything.
3:51 PM
ah, ok, in that case consider what continuous functions do with limist
well... they'd be equal to the limit
I gave it some thought and used this argument:
First I show that |z-a| = |conj z - conj a|
Then I replace |z-a| in the eps-del definition, and there I have it...?
4:04 PM
sorry, then you have what?
4:36 PM
Q: Show that the ratio of limits converges to the nearest Riemann zeta zero except when the ratio is a singularity

Mats GranvikLet $h(s,n)$ be: $$h(s,n)=\lim_{c\to 1} \, \frac{(-1)^{n-2}}{(n-2)!}\zeta (c)^{n-2} \sum _{k=1}^{n-1} \frac{(-1)^{k-1} \binom{n-2}{k-1}}{\zeta ((c-1) (k-1)+s)}$$ and let $g(s,n)$ be: $$g(s,n)=\lim_{c\to 1} \, \frac{(-1)^{n-1}}{(n-1)!} \zeta (c)^{n-1} \sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1...

Did you see the new ratio? It gives zeta zeros. The limit as such is a bit slow to compute because it requires so much symbol manipulation by the computer. But never the less I think there is actually hope that this could lead somewhere, since by a "global" derivative formula for the Riemann zeta function
(that I told Leaky Nun about earlier in chat), one can cross out all derivatives by nesting recursively the limit for the derivatives to any degree derivative.
$$h(s,n)=\lim_{c\to 1} \, \frac{(-1)^{n-2}}{(n-2)!}\zeta (c)^{n-2} \sum _{k=1}^{n-1} \frac{(-1)^{k-1} \binom{n-2}{k-1}}{\zeta ((c-1) (k-1)+s)}$$
$$g(s,n)=\lim_{c\to 1} \, \frac{(-1)^{n-1}}{(n-1)!} \zeta (c)^{n-1} \sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta ((c-1) (k-1)+s)}$$
$$\rho_s = i s+\lim\limits_{n \rightarrow \infty}\frac{h(i s,n)}{g(i s,n)}$$
For $s=15$ we get: $0.5+14.1347i$
The nested derivatives I posted in the answer. What you see here is the end result.
$$-\frac{\zeta '(s)}{\zeta (s)}=\lim_{c\to 1} \, \left(\frac{\zeta (c) \zeta (s)}{\zeta (c+s-1)}-\zeta (c)\right)$$
$$\zeta '(s)=\lim_{c\to 1} \, \left(\zeta (c) \zeta (s)-\frac{\zeta (c) \zeta (s)^2}{\zeta (c+s-1)}\right)$$
$$\zeta ''(s)=\lim_{c\to 1} \, \left(\zeta (c) \zeta '(s)-\frac{\zeta (c) \zeta '(s)^2}{\zeta '(c+s-1)}\right)$$
$$\zeta ^{(3)}(s)=\lim_{c\to 1} \, \left(\zeta (c) \zeta ''(s)-\frac{\zeta (c) \zeta ''(s)^2}{\zeta ''(c+s-1)}\right)$$
$$\zeta ^{(4)}(s)=\lim_{c\to 1} \, \left(\zeta (c) \zeta ^{(3)}(s)-\frac{\zeta (c) \zeta ^{(3)}(s)^2}{\zeta ^{(3)}(c+s-1)}\right)$$
What we need though in order to understand the series expansion of: $$\frac{1}{\zeta(s)}$$
is the derivatives of the reciprocal above. That is what I tried to explain in the answer.
I arrived at the Riemann zeta zero ratio limit above by applying Mathematicas FullSimplify command to the nested derivatives.
4:52 PM
@robjohn "For any $q$, there are fewer than $q^2 +q$ rational numbers" I couldn't understand this part.
This is the third derivative:
$$\frac{\partial ^3\frac{1}{\zeta (s)}}{\partial s^3} = \frac{6 \zeta '(s)^3+\zeta ^{(3)}(s) \zeta (s)^2-6 \zeta (s) \zeta '(s) \zeta ''(s)}{\zeta (s)^4} = \lim_{c\to 1} \, \left(\frac{\zeta (c)}{\frac{1}{\frac{\zeta (c)}{\frac{1}{\frac{\zeta (c)}{\zeta (s)}-\frac{\zeta (c)}{\zeta (c+s-1)}}}-\frac{\zeta (c)}{\frac{1}{\frac{\zeta (c)}{\zeta (c+s-1)}-\frac{\zeta (c)}{\zeta (c+c+s-1-1)}}}}}-\frac{\zeta (c)}{\frac{1}{\frac{\zeta (c)}{\frac{1}{\frac{\zeta (c)}{\zeta (c+s-1)}-\frac{\zeta (c)}{\zeta (c+c+s-1-1)}}}-\frac{\zeta (c)}{\frac{1}{\frac{\zeta (c)}{\zeta (c+c+
@Thorgott Yes. For me the easiest way to write it down is using orthonormal frames.
By applying the `FullSimplify` command in Mathematica to right hand side we get:
$$\zeta (c)^3 \left(\frac{1}{\zeta (s)}-\frac{3}{\zeta (c+s-1)}+\frac{3}{\zeta (2 c+s-2)}-\frac{1}{\zeta (3 c+s-3)}\right)$$

Which appear to have alternating binomial coefficients in the numerator.
"Can you do addition?" the White Queen asked. "What's one and one and one and one and one and one and one and one and one and one?" "I don't know," said Alice. "I lost count."
Through the Looking Glass.
5:19 PM
@Ted what does it look like?
@Thorgott then i have the eps-delta definition of the limit of some function g(z) as z approaches some a, written in terms of the conjugates of both z and a instead. So... I suppose that translates to the shorthand syntax of [lim z-> a]{something} being the same as [lim conj z -> conj a]{same thing}
Here you go. Let $e_1,\dots,e_n$ be an orthonormal frame (working on an open set, of course), and let $e_1$ be the chosen element of the unit tangent bundle. Then we have the dual coframe $\omega_1,\dots,\omega_n$ and we have $de_1 = \sum\limits_{j=2}^n \omega_{1j}e_j$, so the volume form on the fiber is $\omega_{12}\wedge\dots\wedge\omega_{1n}$.
Thus, the volume element of the whole bundle is $\omega_1\wedge\dots\wedge\omega_n\wedge\omega_{12}\wedge\dots\wedge\omega_{1n}$.
You might complain that $de_1$ doesn't make sense. But this is how you should understand this intuitively. I can put the canonical connection on the tangent bundle and do $\nabla e_1$, or I can do it all in terms of differentiating the $\omega_i$.
@TedShifrin hey Ted!
Hey @Stan
@TedShifrin I think I found the trapezoid inertia tensor
It's so ugly
5:29 PM
Yes, it's an ugly shape.
Even for a rectangular box with sides of different lengths, it's not going to be pretty.
I ended up having to use mathamica. I did one dimension and verified as I went along that my formulas were correct
thank god for mathematica. it would have taken forever
the solution is like a half page
Well, you are computing nine things.
Just the original one (about the "height" axis of the object) wouldn't be so bad by itself.
I've never done this before, so its all new to me :'). Mathematica is unbelievably useful for engineering. It's really great for long derivations where you want to know at least what ur shooting for
Oh, Mathematica is highly useful for all of us. Since I was retired and couldn't legally get departmental copies anymore, I had to give in and buy it for myself.
@TedShifrin wow! it's that useful??? I had no idea
What do u use it for?
5:44 PM
Computations (often for answering MSE questions when I don't want to do the whole thing out by hand), but I've always used it for graphics (especially for differential geometry). Most of the pictures in my geometry notes are from Mathematica, and I have all sorts of animations written and programs to solve DE that can't be solved other than numerically and then plot them (e.g., geodesics on surfaces).
I have a Mathematica primer I wrote for my diff geo students. I can send it to you if you're interested.
That would be terrific! I'd love that. Please send it.
@Stan: Sent.
Hi, FakeMod.
@TedShifrin thanks! It's interesting because, with the problems you've assigned me in the past, I don't think any of them involved mathematica. What role do you think programming plays in math education?
@Threnody yeah, because a sequence $(z_n)_n$ converges to $a$ iff $(\overline{z_n})_n$ converges to $\overline{a}$
5:49 PM
Let $\mathbf r$ be a function of $q_1,q_2,\dots,q_n,t$, where $t$ is time. I will be referring to total time derivatives by a dot over the symbol. Now, how do I prove that$$\frac{\partial \dot{\mathbf r}}{\partial \dot{q}_j}=\frac{\partial \mathbf r}{\partial q_j}$$ where $j\in\{1,2,\dots, n\}$?
@TedShifrin Hi, how do I prove the above identity? :-)
@Ted Hmm, I'm not seeing why that is the volume form on the fiber. The volume form can be expressed in the frame as $\sum_{i=1}^n(-1)^ie_i\omega_1\wedge...\wedge\widehat{\omega_i}\wedge...\wedge\omega_n$, but I don't see why that is $\omega_{12}\wedge...\wedge\omega_{1n}$.
@Stan: You weren't doing things that lend themselves to graphics or complex computations.
@Thorgott: What you wrote doesn't actually make sense in general; it's a vector-valued form, not a form. Think about the unit sphere. $e_1$ is your position vector, and $e_2,\dots,e_n$ give an orthonormal basis for the tangent space at $e_1$.
@FakeMod: If you're not doing Hamilton's equations, this looks very suspicious.
@TedShifrin I encountered this while learning Lagrangian mechanics.
Yeah, it looks remarkably like Euler-Lagrange, but it's not.
5:53 PM
So yeah, it was someway in the path of reaching Hamiltonian formulation.
So I don't care about $\mathbf r$.
@TedShifrin What is that supposed to mean? 0_o
You have $f(q,t)$, where $q$ is a (vector) function of $t$. So when you write $\dot f$, this is $\sum \partial f/\partial q_j\, \dot q_j + \partial f/\partial t$.
Look at that formula. What is $\partial \dot f/\partial\dot q_j$?
5:55 PM
Ooh! Got it, thanks :D
You're welcome. :)
@TedShifrin I suppose the most i've had to use wolfram products is double checking my math to make sure i don't have a sloppy algebra error or something when doing derivations
Well, if you run my notebook you'll see all sorts of things that you wouldn't know how to do without Mathematica :P
I used it some in both multivariable and diff geo classes, too, especially for animations.
@TedShifrin do you program in any other languages?
Nah. Not really. I'm a woefully inadequate programmer.
6:06 PM
@Knight Think of the rationals with denominator $q$ in an interval of length $1$ (mod $1$): $\frac0q,\frac1q,\frac2q,\ldots,\frac{q-1}q$. There are $q$ of them. Summing for all denominators up to $q$ gives $\frac{q^2+q}2$. Adding those on the other side, gives $q^2+q$. Of course, there will be overlaps where the numerators and denominators have common factors, but that only reduces the number.
G'morning, @robjohn :)
Mathematica can be used pretty much in the same way Excel spreadsheet formulas work. The hard part is to learn how to use speed ups with operators such as the hashtag operator #. Michael De Vlieger in the OEIS is skilled at using those Mathematica operators.
Mathematica has so many more powers than an Excel spreadsheet!!!
@TedShifrin yes agree
I am not particularly interested in random sequences.
But, yes, I've used the # in some of my Mathematica programs.
I'm behind the times, though. Most of the programs I wrote I wrote 15 years ago or more.
6:21 PM
@TedShifrin Hey, Ted!
I would say that the intersection of tasks that a spreadsheet is suited for and those Mathematica (or other similar tools) are suited for is pretty close to just the basic calculator
Well, you can certainly make Table work like you were using Excel. But often the things I put in the Table are far more intricate than Excel could handle.
Not that I have used anything like Mathematica for a very long time.
I assume you used Magma a lot earlier on.
No, I used GAP actually
6:25 PM
And for quite a lot of stuff I used version 3 which was supplanted in 1997 (because the Chevie package for dealing with Hecke algebras had not been ported to version 4)
Only after doing a ton of coding and computations did I learn about the PyCox package for SAGE which would have been much more suitable since it was actually kept up to date
It should be $\sum_{i=1}^n(-1)^ide_i\vert_p\omega_1\vert_p\wedge...\wedge\widehat{\omega_i\vert_p}\wedge...\wedge\omega_n$ (which makes sense after identifying the tangent spaces as subspaces of $T_pM$) as volume form on the unit sphere in $T_pM$.
You should try to understand what I gave you, as it is correct.
Now I am just very happy that I don't have to maintain any of the code I wrote as part of that.
I think most of it is not even properly indented
Academic code has quite the reputation anyway
6:34 PM
At least the output is nice (once you run it through LaTeX. You do not want to try reading the actual .tex file)
6:49 PM
Thorgott trying to understand volume forms on the unit tangent bundle?
a @Balarka, yes. He seems not to like my answer.
@Thor: You keep going back to the formula for the volume form for $S^{n-1}\subset\Bbb R^n$ in terms of coordinates. But when you're working with a manifold, you have no such coordinates. That's why I'm doing everything in terms of frames for the tangent space.
hey chat
Hi, Lucas.
@Ted, would you help me out with a general topology question?
I'm supposed to solve it using real analysis, but I know the result holds for any topological space.
The frame does give us coordinates. $de_1\vert_p,...,de_n\vert_p$ is a global chart for $T_pM$ when we view it as a manifold in its own right and then contraction with a unit normal field can be made to work. I am trying to understand how you arrive at your volume form, but I neither get that on its own nor do I get why it's the same as my volume form.
7:02 PM
No, what you're writing down doesn't make sense on an abstract manifold. I don't even understand how you're getting a chart for $T_pM$. It's the $e_i$, not their derivatives, that give a basis for the tangent space.
To differentiate, you need a connection.
Do you agree that if we take $S^{n-1}\subset\Bbb R^n$ and choose an orthonormal frame for the tangent space locally, namely $e_2,\dots,e_n$, then wedge product of the dual coframe elements gives the volume form?
@TedShifrin Nice volume form.
If we write $x$ for the position vector on the sphere, then we're writing $dx = \sum\limits_{j=2}^n \eta_j\otimes e_j$ if $\eta_j$ is the dual coframe. I'm using $e_1=x$ in the setting we were doing, and then writing $\eta_j = \omega_{1j}$, because when you're used to the moving frame set up, this is telling you how $e_1$ twists toward $e_j$.
Oh oh, @MikeM is here to make trouble.
No I'm not
Just here
Hey, Mike. Have you heard anything from my old friend and your old roommate?
hi chat
7:08 PM
Salut, @Astyx!
Comment va ?
We haven't talked much since we graduated, and only briefly after he moved, so I've got nothing for you.
If we dualize to the unit cotangent bundle instead, we can cook up a $1$-form which eats a tangent vector at a covector and spits out the value of the covector at the underlying tangent vector (after projecting to base). This is the tautological $1$-form $\alpha$, and my volume form would be $\alpha \wedge (d\alpha)^{n-1}$, the one coming from the contact structure on $T^*_1 M$.
OK, I don't even know what city he is in now.
Sorry, it's the $\omega_i\vert_p$ that give a global chart on $T_pM$
7:10 PM
This is probably the same as your moving frames volume form
It looks similar
I'm not sure, @Balarka. How is yours built out of the Riemannian metric?
@TedShifrin Yes (if the frame is oriented)
@TedShifrin Yeah, it's not immediately clear, it's just somehow baked in the definition of the unit cotangent bundle.
I could write it down
I'm not convinced, a @Balarka.
Doing the sphere, where you work globally with $SO(n)$ is a good test.
@TedShifrin I might be able to get academic versions, but I think that the Personal version is about as expensive and so I haven't bothered getting the academic versions.
7:14 PM
@robjohn: I think I wangled a 50% deal because of my "ex"-academic status.
Then the license is still valid after I retire.
Yeah, mine should last my lifetime now.
Which, of course, could just be a few months the way our country is going.
I have a copy of 12.1 that I can't use until I upgrade my OS
However, I have found out how to overcome a lot of the problems I was having with 11.3
Mine's 12.0. I did upgrade and when I did, I lost all Microsoft and Adobe products I had purchased ten years ago. So now if I have to update the pictures in any of my books (for which I used Illustrator) I'm up the creek.
@Thor: The point is that to get a coframe on the tangent bundle (rather than on the base manifold) you need to know how the frame is twisting as you move around.
@TedShifrin when I got a new laptop, I was unable to use Photoshop and I refused to get the subscription license. Pay every year instead of once.
7:17 PM
@robjohn: It's MONTHLY. I was livid.
(That was before every day of living made me livid.)
@TedShifrin I guess I forgot about that. I was just unwilling to do that.
LOL, me too.
Especially since I may never need to use it again. I did find Inkscape, which is free. It seems to have most of the tools that Illustrator did, if only I can figure out how to use them.
If there were another program that handled all the plug ins I purchased, I would be happy. I used it a lot to process my astrophotos.
I've never once used Photoshop, so I can't help.
I was already angry at Adobe for telling Apple that it had to stop working on QuickDraw GX or Adobe would stop producing Mac applications. That cost me my job at Apple.
7:21 PM
Why did they do that? Stepping on their proprietary toes somehow?
QuickDraw GX made it extremely easy for third party developers to mimic things that only PS or Illustrator could do. So rather than trying to do something great using QDGX, they said "stop it!"
Ah, can't have healthy competition.
That seems to be Adobe's position
Are they whispering in Tromp's ear, just like Facebook and Google?
@Thor: I don't mean to ignore you. Are you sorting things out so that what I said makes sense?
who knows? They may not be big enough for him to notice.
7:25 PM
I am not sure if there has been healthy competition on anything involving any of the major software companies for quite a while (maybe apart from offering cloud solutions)
@Thor: The approach I'm taking, of course, appears in Cartan and especially in Chern, when he did Gauss-Bonnet using the sphere bundle.
@TedShifrin I really, really like "Tromp"
I can't explain why
@Balarka: It's my more polite editing for chat.
@TobiasKildetoft and I don't trust the cloud. Don't want all my things on someone else's server.
Yeah, I have limited things in the cloud (I refuse to pay for extra storage up there).
7:27 PM
Yeah, I'm trying to make something work, albeit with very limited success
and yeah, CGB is the context in which I'm encountering this
@TedShifrin I often say that the "T" is silent.
@robjohn Well, you could set up your own cloud of course :)
@robjohn: I appended a -olini to what I typed here.
@Tobias: That requires more knowledge than I have.
@TobiasKildetoft I may do that soon.
OK, @Thor. Officially, I should have used $\pi^*\omega_j$ to put the base coframe up on the unit tangent bundle, but I didn't want to complicate things.
7:29 PM
I am still hoping that the project I work on will end up being based on an upcoming Danish government cloud
@Thor: What I'm talking about also works naturally on Grassmannians/Stiefel manifolds. And, with universality, you can do CGB just by reducing to the tautological bundle on the Grassmannian.
@Tobias @robjohn: How would I make my own cloud? I guess I have storage space somewhere due to my internet subscription. But I certainly don't have any 'server' other than my own desktop.
Mainly because the alternative is most likely that it gets hosted on a couple of dedicated servers with not much scaling potential and which will be horribly tricky to get access to for stuff like CI/CD
@TedShifrin I am not deep enough into the details that I would know how to get started
Oh, OK. Well, this is well beyond me.
@TedShifrin I have a server that I am going to use for that purpose. A private cloud for saving things. One thing I hate about clouds is that "synching" is mostly a black box and you don't always have control over what gets deleted when you simply want to remove it from a particular device, or having it come back when you've deleted it.
Right. One of my friends, who keeps lots of photos and videos on the cloud, worries about such things constantly. I'm oblivious.
7:34 PM
(also, we might not quite mean the same thing when we say "cloud")
I will simply have a repository of things I want on the server. That way I can free up space on my phone, etc, and still have access if I should want it later.
Well, given that robjohn has his own dedicated servers, this should be a no-brainer for him.
@TobiasKildetoft "synch" and "cloud" are too weakly defined for me to trust them.
@robjohn wants a Sobolev embedding theorem for his.
When I hear "cloud" I think "distributed"
7:36 PM
BTW, @robjohn, when you were talking about your friend and his adviser at Princeton, I had originally assumed it was math. But when I read your link, I discovered it obviously was not.
No, he was an architecture grad student. We had dorm rooms next to each other for a few years.
@EdwardEvans @Alessandro John Petrucci from Dream Theater dropped a new song
7:51 PM
@Balarka is it as good as this
the "percussionist" in that video is the real mvp
8:09 PM
Oh neat let's listen
I'm not a big dream theater fan tbh
let me pull you under then
Metropolis Pt. 2 is a fantastic album
Most of the rest is alright
I mean technically speaking they're incredible of course, but nothing gets even close to metropolis as far as lyrics etc. are concerned
Also the singer in live shows... I'd rather not comment
aight imma head to bed
see yall on the other side
8:16 PM
I'll hurt my head with dynamical systems for a bit longer before sleeping
hmm, how do I show that $E \subseteq \mathbb R$ consists of only is of isolated points $\implies |E| \le |\mathbb N|$?
By using that $\Bbb R$ is second countable
second countable?
It has a countable basis
I don't know sufficient topology to know what that means.
8:25 PM
Ok so suppose $A\subseteq\Bbb R$ is uncountable, let $A'$ denote the set of limit points of $A$, we want to show that $A\setminus A'$ is countable, right?
isn't $A\backslash A' \ne \emptyset$ sufficient?
No, $A\setminus A'$ is the set of isolated points of $A$, I want to show that if $A$ is uncountable, then uncountably many of its points are not isolated, but for that I need $A\setminus A'$ to be small (a priori we might have $A'=\varnothing$ otherwise)
\neq was fine, we are trying to show that some point of $A$ is a limit point of $A$
(We are going to get uncountably many for free)
consider $\{1/n\colon n\in\mathbb{N}\}$
this set consists of isolated points, but possesses a limit point
I wanted to make this point, but I forgot why I wanted to make it
8:38 PM
Lucas wants to prove that if a subspace of the reals is made of isolated points then it is at most countable. I want to prove that if a subspace of the reals is uncountable then many of its points are limit points, {1/n} is not problematic
yeah, I agree
I need to leave for a while, but I'm sure Thorgott or somebody else can help with this problem
ok, this may not be what Alessandro was going for, but you can argue as follows by contradiction:
a) if there were an uncountable set consisting of isolated points, you would be able to find uncountably many disjoint open intervals in $\mathbb{R}$ (why?)
b) this is impossible (why?)
9:01 PM
Re: unit tangent bundle. I don't even see why this viewpoint is consistent with the more abstract one. Let $SM$ be the unit tangent bundle and $SO(TM)$ be the oriented orthonormal frame bundle. We have projections $\pi\colon SO(TM)\rightarrow SM$ and $p\colon SM\rightarrow M$ and our frame gives a section $s\colon U\rightarrow SO(TM)$. There are forms $\tilde{\omega}^i_j$ on $SO(TM)$ such that $s^{\ast}(\tilde{\omega}^i_j)=\omega^i_j$ (not only for this, but for any frame). It can be checked that $\tilde{\omega}_1^2\wedge...\wedge\tilde{\omega}_1^n$ is vertical and $SO(n-1)$-invariant, so p
@Thor: Pulling back along $p$? No, the $\omega_1^j$ are not forms on $M$. They're forms on $SM$.
They "reside" on the fibers of $p$.
I'm about to go out for a half hour or so, but shall return.
My $\omega_1^j$ are the connection forms relative to the moving frame. Surely those are forms on $M$? $e_1,...,e_n$ is a frame on $U\subseteq M$, so the coframe $\omega_1,...,\omega_n$ are $1$-forms on $U$ and then $\omega_1^j$ are defined by $d\omega_1=\sum_{j=2}^n\omega_1^j\wedge\omega_j$.
9:18 PM
@Thorgott a) definition of isolated point b) idk :p
think about the rationals
i really don't have a clue
@Thorgott ah right, that's nicer
The rationals are countable and dense @Lucas
yeah, this uses separability rather than second-countability
basically just observing that a separable discrete space is at most countable
9:36 PM
@Thorgott true, even though they are the same for metrizable spaces
hmm, how does separable => second-countable go?
for metric spaces, that is
The basis is balls of rational radius centered on the points of the countable dense set
ah, of course
@AlessandroCodenotti ok, I should use that somehow..
ok, so if that happened, there would always be a rational inside these intervals
so there's a subset of the rationals in that indexes the set of intervals
but the set of intervals is in bijection if our set supposed to be uncountable
a contradiction
is that right?
@Thorgott If you're trying to give a volume form for $SM$, this obviously won't work. You need the volume form for the fiber.
The frame is not on $M$. Remember that $e_1$ is giving us the point in the fiber we're at.
So you need the volume form for $e_1$'s varying over the unit sphere at $p$.
10:10 PM
oh, so you wanted $e_1,...,e_n$ to be a frame of the fiber?
Those are a frame for $T_pM$, of course.
So it's best to think of everything up on the complete frame bundle. I'm choosing a local section of the frame bundle of $M$ as a bundle over $SM$. The $(n-1)$-form I wrote down depends on $e_1$ but is independent of the choice of $e_2,\dots,e_n$.
10:32 PM
Isn't that similar to what I described above? You have connection forms on $SO(TM)$ and then $\omega_1^2\wedge...\wedge\omega_1^n$ is $SO(n-1)$-invariant and pushes forward to a form on $SM$
OK, you had stuff with $p$ going all the way down to the base. That's what was wrong.
Yes, this is fine. This is what I was doing at the outset.
Perhaps I should have made it clearer by going to the full frame bundle.
yeah, what I was wondering was if whether pushing forward the connection forms one has on the frame bundle to $SM$ gives the same forms as when pulling back the connection forms one constructs on an open set via moving frames to $SM$
but I now realize that obviously won't work, cause the former are intrinsic and the latter depend on a choice of moving frame
You can only do the latter if you choose (locally) a fixed moving frame. We can't do that in this situation. You want the Haar measure of the gauge group in there :P (to mix metaphors).
10:51 PM
ok, so now that we agree on our form on $SM$, why does it restrict to the volume form on each fiber?
You mean the non-base part? I explained that three times already. This is how you intrinsically construct the volume form on the unit sphere. Or indeed on any Riemannian manifold. You take the wedge of the orthonormal basis covectors.

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