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12:38 AM
@Thorgott Make it be vertical in terms our local product structure.
@Rithaniel don't ask me!
1:15 AM
Fair enough, I shall refrain from doing so
1:54 AM
A local trivialization $SU\cong U\times S^{n-1}$ gives $TSU\cong TU\times TS^{n-1}$, the latter factor being the vertical component. The map $\pi\circ s$ becomes a map $U\rightarrow S^{n-1}$ under this identification (ignoring the first component, where it is the identity), but the map isn't any more explicit.
I guess we can also try to trivialize $SO(TU)\cong U\times SO(n)$. The frame then becomes a map $U\rightarrow SO(n)$ (ignoring the component where it is the identity) and we can bull pack the connection forms (which vanish on the other component) along this. But how do we relate the c
3 hours later…
4:34 AM
@EdwardEvans no that wasn't me care to reference those exact quotes please?\
4:55 AM
Hi @Ted
@Thorgott I actually do not like working with the trivialization. I'd rather work with the local frame as a moving frame on $U$, and then $s^*\tilde\omega^j_i = \omega^j_i$ is the connection matrix for that moving frame (as forms on $U$).
Hi, a @Balarka.
My roommate is reading Chern's proof of CGB and he had some questions so I told him he should come around here and ask you stuff. He'll probably join soon with some questions.
@Thor: Then it is precisely the equation I wrote down earlier. For that moving frame $e_1,\dots,e_n$ on $U$, I have $\nabla e_i = \omega^j_i\otimes e_j$ as a tensor equation on $U$.
He should talk to @Thor, @Balarka. That proof is not one I've ever presented. Chern was such an amazing mind that he just pulled the families of forms out of his *** (or brain) and knew they would work. I really prefer the proof I usually gave in class. The disadvantage of the Grassmannian proof is that you don't get the boundary theorem. I.e., you don't get the form on the bundle whose exterior derivative is the Pfaffian.
@Thor (summation symbol omitted, of course).
Why is everyone doing CGB these days, a @Balarka?
Ah OK yeah he was talking about some boundary term. I dipped out because I don't know CGB at all
You and Thor can talk to him
@TedShifrin I don't know why it's hype suddenly haha
If he's literally reading Chern's proof, Thor won't be of any use.
5:00 AM
I might read some proof of CGB while everyone's at it as well
Raghunathan has a proof using Morse theory lol
Well, you should start with mine (with the universal bundle on the Grassmannian and the necessary Schubert cycle).
Ah yeah I could do that
I guess I don't immediately see how the Pfaffian comes in through Morse theory.
And I presume that does the boundaryless theorem only.
I don't know any details, I just know it exists. Here, if you want to have a quick look.
Within the first 30 seconds, I saw a typo in it. :P
5:05 AM
It is Principle vs Principal
I'll save it to examine. He's reducing the structure group using Morse Theory. And uses the fact that the integral is independent of choice of connection.
No, it was "structure" being misspelled.
Haha I see
And then there is "eigen values." Ugh.
Classic Raghunathan
I see. It's localizing at the zeroes of the gradient vector field to pick up the Euler characteristic, so it's totally in the spirit of Chern's proof. I'll have to look more carefully.
5:08 AM
Thought you might be able to appreciate it better than I
Downloaded, thanks.
I think I like my Grassmannian proof, because one sees from the particular Poincaré dual why you're getting the sum of the indices of a particular Morse function (actually).
Ah that's interesting
I'll read it in detail, I should have some off-time this week.
I sort of understand why a lot of it is about the cohomology of the Grassmannian (which is, in the limit, a polynomial algebra on the Chern classes) but don't know the geometry of it all
Well, in one set of notes, it's just a sketch in the last page, but the technical details will be understanding the Schubert cycles and Poincaré duality argument earlier.
My proof is much more classical.
Right, it's high time I learn Schubert cycles.
I did the complex stuff pretty carefully, but did most of the details for the real oriented Grassmannian case.
I hate real Grassmannians :P
Anyhow, it's cool that everyone's submersed in CGB. :P
5:17 AM
Funnily I think the cohomology of the real Grassmannian is easier than the real oriented Grassmannian, in the limit
$H^*(BO(n); \Bbb Z_2)$ is a polynomial algebra on the Stiefel-Whitney classes whereas for $H^*(BSO(n); \Bbb Z_2)$ you get Pontryagin classes as well, which are beasts I don't get
One of my honors advisees from years ago (who never was brave enough to take a class from me) just sent me an email that he's doing a Zoom defense of his Ph.D. thesis in applied geometry on Tuesday. He was a math/German/music triple major, if I remember correctly.
As you can imagine, a @Balarka, I don't like doing the $\Bbb Z_2$ stuff.
It doesn't fit forms too well.
I may have to send you different notes where I did the Pontryagin classes a bit more.
@TedShifrin Oh cool!
In the notes I think I sent you I did complex and Euler form only.
Yeah you never sent me anything on Pontryagin classes
I never learnt anything about them either
I just know they're a thing
5:23 AM
if you say so, @Edward.
lol, [insert time-zone appropriate greeting], @Ted
yo @EdwardEvans
@BalarkaSen You only see Pontryagin classes in integral cohomology, and they appear in the cohomology of BO(n); they're chern classes of complexifications, which makes sense for any real VB, not just oriented. In Z/2 cohomology H^*(BSO(n)) is the polynomial algebra on $w_2, \cdots, w_n$.
Ah ok
5:29 AM
Yeah, you need oriented only for the Euler class (Pfaffian).
@Balarka: If you have a folder with all the stuff I've sent you (whether you wanted it or not), see if you have MATH897.pdf in there.
Yeah of course, I am not sure what I was smoking, BZ/2 -> BO(n) -> BSO(n) so I guess in cohomology you just kill H^*(BZ/2; Z/2) from H^*(BO(n); Z/2) which makes w_1 die
@TedShifrin Hmm, I am not sure I have this folder, let me check
Ah yes I do have MATH897.pdf
OK, so that has a tiny bit about Pontryagin in there. Plus the end is the cute stuff on counting cusps of Gauss map of a projective surface with Chern classes.
Ah great. Yes, I remember reading it a little, but never got to Pontryagin classes.
There's more differential geometry in the 8260 notes I sent from the year before I retired. And the CGB proof is slightly more sketched there.
Yep, I have those
5:35 AM
Yup, I knew that. So I should expect a question or two perhaps ...
I learnt moving frames in higher dimensions from 8260 of course, we discussed it sometime
CGB is the thing I never learnt
I'll just read it
There's a new person answering all sorts of questions on main who seems to know lots of different things. Angina Seng. Don't know if you've run into her/him.
Nope, I don't know them
Does seem very active
He/she has 5000-ish answers
6:41 AM
hi i know i am in the wrong room
7:36 AM
Who just starred everything?
This is the shittiest starboard. Nvm, the room owner can probably undo the stars.
8:23 AM
Is there anything I can do to increase the chances of getting an answer to this question? Or just making a bounty? math.stackexchange.com/questions/3777437/…
8:51 AM
hi chat
hi Astyx
@northerner adding a bounty is the usual way. Linking the question on social media will bring in viewers, but that does not often bring in someone who can answer the question.
9:39 AM
hello world!
"Show that $\operatorname{Spec} R \to \operatorname{Spec} R[X], \mathfrak{p} \mapsto \mathfrak{p}R[X]$ is well-defined" seems like a weird exercise to me..
10:17 AM
Does anybody know what happened to knotilus.math.uwo.ca? I can't access it since few days.
10:33 AM
@EdwardEvans Lol
@Balarka it is literally as simple as I think it is right?
they mean its a morphism of schemes
well there's no context, it just says to show that that map is well-defined
so.. free point I guess
I'll just imagine it to mean they want you to observe it's obtained from taking Spec of R[x] -> R[x]/(x)
because only then can you say its a morphism of schemes, i.e., well defined in the category of objects the domain and target are understood to be
stupid exercise anyway
lol, that will come next semester I guess, this is just commutative algebra, AlgGeo will start in October
10:39 AM
hm ok
i cant imagine what else well defined means though
Same hahaha
there's no weird equivalence relation going on
the psets in this course have just been weird
its fine if they want you to observe its coming from a ring hom
should phrase it differently is all
Hey everyone, I need to clarify some linear algebra concepts;
Row vectors [1,0,0,0],[0,1,0,0] \in R^4 span a subspace of dimension 2.
And rows [0,1,0,0] [0,0,0,1] as well, and I think they do not span the same subspace, but have same dimension, and I'm not sure
why is that?
When span of set of vectors is equal to span of standard basis with the same dimension
11:17 AM
@Balarka man just read a proof of CGB, it really shouldn't be difficult for you
I'll see man
I have 80 different things to read
I should be reading Atiyah-MacDonald to prepare for my algebra exam
but instead I'm trying to figure out covariant derivatives smh
the latter is better than the former
admit it
idk, I can't say I feel too stoked about any of this after spending like 3 days trying to figure out something that should allegedly be obvious
I am currently thinking about some kind of definition of natural numbers such that you cannot single out any member of them
a "continuum" of natural numbers so to speak
But then, by peano arithmetic, having a zero and then the successor function seemed to thwart that
11:54 AM
@BalarkaSen and 75 of those are written by Gromov?
I have only one thing to read (well not really, I have 80 too but one that I want to read now) but it's hurting my mind
In top. space $X$, for universal net $S$ is it true that $B,C\subset X$ are cluster points of $S \iff B\cap C \neq \emptyset$ ?
i.e $S$ is frequently in $B,C$.
def Net $S$ is said to be universal iff for each subset of the space $A\subset X$ the net $S$ is eventually either in $A$ or in $X\setminus A$.
Also does anyone know a good book where uniformities/uniform spaces and their main properties are explained?
12:25 PM
does $lim_t\to\infty e^{-2k^2t(k^2x_0^{-2})-1}=0$ fir $0<x_0<k$
there's a section on that in Engelking, but I assume you know that already
ok i am asking stupid question
also someone's messing with the starboard again, sigh
@AlessandroCodenotti What do you have to read
yeah this starboard thing is getting old
@BalarkaSen is tomorrow's lecture happening ?
12:27 PM
you got the zoom link for htis week ?
yeah, you didn't get?
no ?
neither did other from cmi
I mean, not the same as last week, right ?
12:28 PM
no, new links
yeah sure
@BalarkaSen your mail is 13 right ?
@BalarkaSen sent it
Got it
@BalarkaSen some dynamical systems stuff
12:34 PM
@Lelouch Replied back
I'm stuck on an exercise, I'm pretty sure both that I have the right solution, and that I can't show it is the correct solution because I'm missing something stupid
@BalarkaSen thanks !
Should I let him know that y'all didn't get the mailing list, or you can say that during class?
No problem, I'll send him a mail. I think it's better if I say it
Right, do that
@AlessandroCodenotti topological dynamics again? :P
why dont you read something not crazy
like Brin-Stuck
12:38 PM
@BalarkaSen I want to learn about universal minimal flows mostly
@BalarkaSen yep, the exercise is to show that a minimal flow with connected phase space is totally minimal, and some other properties of minimal but not totally minimal flows
@BalarkaSen Yes, I told him and he replied rn that he forwarded it to a CMI prof. He also forwarded me the links, but thanks nevertheless
Is intersection of any two cluster points always non-empty for universal net $S$ in top space?
what's the intersection of points
@BalarkaSen the first three chapters look quite interesting, I'll check it out! I don't really care about ergodic stuff right now though
12:43 PM
of cluster points.
Suppose $B,C$ are subsets of space $X$, the intersection is $B \cap C$.
@AlessandroCodenotti minimality is the smallest invariant set business, right?
points =/= subsets
Can't cluster point be a subset?
@BalarkaSen yes, minimal means that it has no proper closed invariant subsets
12:45 PM
Which is easy to see is equivalent to every orbit being dense
a cluster point is a point, by definition
i think about it more in terms of foliations than invariant sets
I don't know foliations :P
you should read Candel-Conlon at some point
why are you doing topological dynamics on Polish spaces man
just do something not crazy
mathematics is all about descent into madness
12:48 PM
I'm going to talk about pairs $(X,f)$ of a space and an homeo instead of Z actions because it's nicer notationally. Totally minimal means that $(X,f^n)$ is minimal for all $n$
@BalarkaSen but then there's no descriptive set theory :(
Let me rephrase:
Let $S$ be a universal net which is frequently in subsets $B,C\subset X$ of the topological space. Is it true that $B\cap C \neq \emptyset$ always?
@AlessandroCodenotti Good
@BalarkaSen that's a huge book
polish topologists make a career out of cross referencing "well known results" from old 1000 page tomes
so thats not a problem for you
"follows easily from a result in Fremlin's measure theory"
Without even specifying which volume
12:54 PM
You probably can say that about any measure theory fact you'll ever need to be fair
@flowian well, it's either eventually in $B$ or eventually in $X\setminus B$ since it's universal, but it's frequently in $B$, so necessarily it has to be eventually in $B$, same for $C$ and so it's eventually in both $B$ and $C$, which implies $B\cap C\neq\emptyset$
@Thorgott Thanks for confirming. I was doing an exercise which asked to prove that if universal net frequently in a subset, then is eventually either in it or in its' complement. This led me thinking what if there are 2 such subsets...
hi all
Q: About analytic continuation of the function $f_r(s) + v = (\sum_{n=1}^{\infty} \frac{1}{n + n^s + r}) + v $ with respect to $s$.

mickConsider the function $f_r(s) + v = (\sum_{n=1}^{\infty} \frac{1}{n + n^s + r}) + v $ for real $r,v$ and complex $s$. ( with respect to variable $s$ and fixed $r,v$) I wonder how this function looks like and behaves with respect to $s$. Where are the poles and zero's ? Is there a critical line l...

Q: analytic continuation of $f(s) = \sum_{n=1}^{\infty} \frac{1}{a_i^s} $?

mickConsider a nondecreasing infinite sequence of positive integers that starts with $1$ : $a_n$ ( so $a_1 = 1$ ) Example $a_1 = 1 , a_2 = 3 , a_3 = 3 , a_4 = 7 , a_5 = 12,... $ Now consider $f(s) = \sum_{n=1}^{\infty} \frac{1}{a_i^s} $ How do I know if I can do analytic continuation so that $f(s)$ w...

and a bounty here :
Q: Analytic continuation for $\sum_{n=0}^{\infty}(\sqrt n+1/3)^{-s}$

sigmaDefine a function $F(s)$ by: $$F(s)=\sum_{n=0}^{\infty}(\sqrt n+1/3)^{-s}$$ Is there a closed form expression for the analytic continuation of $F(s)$ to $F(-s)$?

any ideas ?
1:15 PM
@flowian A universal net is eventually in any set of its complement, because that is the definition of universal. If you know it is frequently in that set, it necessarily has to be eventually in it, because being frequently in a set and eventually in its complement would be a contradiction.
@Thorgott Precisely so!
how does the tensor product of a differential form and a vector field work?
ah nvm
1:51 PM
Hi everyone! A small question : can we "partially differentiate" a equality on both sides?
You don't differentiate equations. You differentiate functions. If two functions are equal, then so will be their partial derivatives.
@Thorgott How does this work though?
@feynhat Right Thanks!
Hello all, I am currently struggling to formulate a scenario in mathematics.

For .e.g There are 90 cases of COVID in location A and 10 cases of COVID in location B. I would like to create a simulation that would suggest me how many vaccines should I distribute in location A and location B based on different weight factors assigned to the location and predict the number of actives cases after X number of days given I have only 30 vaccines available? Which mathematical problem is it?
More generally suppose that $E_1$ and $E_2$ are two vector bundles over the same base $B$. And suppose $s_1 : B \to E_1$ and $s_2 : B \to E_2$ are two sections. How do you define $s_1 \otimes s_2$?
Hmm. I guess it will be a section of $E_1 \otimes E_2$, obtained by taking pointwise tensor product.
$\omega\otimes X$ maps $Y$ to $\omega(Y)X$
1:58 PM
I see. I was wondering if anything interesting happens in your case, $E_1 = \Lambda^k(T^*M)$ and $E_2 = TM$. I guess that's what ^.
but my tensor product isn't a section of the tensor product bundle, it's an operator that takes sections of $TM$ to sections of $TM$
$\omega$ is a 1-form?
and $X,Y$ are vector fields
@Thorgott Isn't there an isomorphism between $V^*\otimes V$ and $\mathcal{L}(V, V)$.
ah of course
so we can actually interpret it that way
2:08 PM
(Lol... Sorry If I am digressing from what you're doing. I am sure you're doing some useful geometry stuff. Just ignore this).
and this actually makes sure this is the natural interpretation
nah man, this is a very good remark
2:39 PM
just think of what variable the operator is tensorial in
you can contract the other stuff and then consider that tensor
Yo Bal.
Yes feign?
Do you know Toru Dutt?
She was a 19th century poet, who wrote in English and French.
2:47 PM
19th century British India poets is equivalent to the old English Romantics from 80 million years back
She also translated bunch of French poems to English. One of her volumes of translated French poems was titled, A Sheaf Gleaned in French Fields.
...and she died at 21.
Name a more iconic trio: French, Fields, dying in early 20s.
That's a great punchline
I thought you were going to make me read that crap
lmao that's great
I'm sure there's a joke about sheaves lurking as well
2:51 PM
Sheaves are only gleaned in French fields unfortunately
Quick off topic Q. I want a quick opinion on whether a question could/should be closed as a duplicate of another if the questions are clearly different but an answer on the linked question directly applies.

More specifically, https://math.stackexchange.com/questions/3786256/a-problem-on-basic-arithmetic I'm thinking could/should be closed as https://math.stackexchange.com/questions/733754/visually-stunning-math-concepts-which-are-easy-to-explain/733805#733805 but I'm second guessing myself and since I've a gold badge in combinatorics, it wouldn't go up for vote if I did
The rest of us drink vodka and gamble and take limit of spaces
Stewart's multivariable chapters are so painfully sloppy
he recommends maximizing certain functions relative to some constraints by substitutuing and maximizing unconstrainted --- eg, find the points on $x^2 - y^2 - z^2 = 1$ closest to the origin by writing the distance squared to the origin as $d^2(y,z) = 1+2y^2+2z^2$, taking the gradient, and seeing that $y=0, z=0$
@robjohn Sir, I got that message as soon as connected my Mac to the hotspot created by my phone. I don't understand it fully.
This invariably misses critical points because this "substitution" is done at the cost of not noticing that sometimes the gradient projects of $x^2 + y^2 + z^2$ projects as zero to the yz-plane
2:55 PM
Does it mean that someone is using my IP address for doing something not legit?
Eg, take the level curve $x^2 + y^2 = 1$ and maximize $f(x,y) = y^2$. Stewart would just have you ignore the $x$ variable, thus missing two critical points!
@JMoravitz Perhaps try asking in the CURED chatroom, I think that's where people discuss these things
Imagine being robjohn and one day realizing you're now 10 people's tech support for every computer issue they have
Oh my good! I realized now that everyone here can see my IP address
@robjohn being a mod can see anything.
Thank you Zanna
3:17 PM
@Knight It means that someone has probably connected to your hotspot. You may need to change the hot spot password.
Thanks robbie sir.
@robjohn Is your breakfast done? With chai-latte :) ?
3:52 PM
@Knight That address is private (generally allocated by the router, i.e. your phone). Do you get your IP addresses dynamically (generated by the phone) or statically (set in your computer). The message seems to indicate the latter. If statically, try changing the address of your computer.
How do the results of partial fractions change in complex analysis? In the real case we have things like 'irreducible quadratic factors'. These have their own case, such as (Ax + B)/(x^2 + c). But factors like these can be easily reduced as linear factors of complex numbers. E.g. (x^2 + 1) = (x - i)(x + i)
So in the context of complex numbers, does the 'irreducible' case no longer exist?
@Knight One thing is sure; no one will get any information from having seen that address. It is local between your router and computer.
Let $e_1,...,e_n$ be an oriented, orthonormal frame as before, $\theta^1,...,\theta^n$ the coframe and $\omega_1^1,....,\omega_n^n$ the connection forms. We have $d\theta^i=\sum_{k=1}^n\theta^k\omega_k^i$ for $i=1,...,n$. On one hand $d\theta^i(e_j,e_k)=e_j(\theta^ie_k)-e_k(\theta^ie_j)-\theta^i([e_j,e_k])=-\theta^i([e_j,e_k])$, on the other hand $(\sum_{l=1}^n\theta^l\wedge\omega_l^i)(e_j,e_k)=\sum_{l=1}^n(\delta_{lj}\omega_l^i(e_k)-\delta_{lk}\omega_l^i(e_j))=\omega_j^i(e_k)-\omega_k^i(e_j)$. Since the frame is orthonormal, $\theta^i=\langle e_i,-\rangle$, so we obtain $\langle e_i,[e_j,e
ok, now I do have to ask why $de_i=\nabla e_i$
@Threnody any polynomial factors into linear factors, so you will only get linear denominators in the partial fraction decomposition
4:17 PM
4:33 PM
@Thorgott oh that's interesting and makes integration much easier
what I said is obviously wrong
you will get powers of linear terms as denominators, of course
@Thorgott Yes, in the case of repeated factors, correct?
@Thorgott Why do I never see you talking about real analysis? Did you already complete it many years ago?
TedShifrin Do you know some video lectures on Real Analysis?
@Thorgott This identification is ubiquitous in differential geometry.
4:47 PM
I do talk about real analysis when others talk about it, I don't bring it up myself cause it's not what I'm doing these days
No, @Knight. I don't know anything about video lectures except my own. There is a lot of real analysis in my own course that's posted, but it's mostly in the multivariable setting.
yeah, it makes sense, it's a natural identification to make
I just hadn't seen it in this context before
I was looking for introductory course in real analysis.
Thorgott didn't your university require you to have real analysis on first and second year of undergrad?
first and second semester, but yes
4:52 PM
@Thor: Why are even wanting to consider $de_i$? Even when your manifold is sitting inside $\Bbb R^N$ and you can differentiate a vector-valued function, you still are looking at $\nabla e_i$, as you're projecting $de_i$ back onto the tangent bundle of your submanifold. So this just isn't a natural creature. What is the context?
The point is, I guess, that your $de_i$ is not a $1$-form in any sense. You need to work with things as vector-bundle valued $1$-forms, and that's what the covariant derivative is.
I mean, where does your $(de_i)_p(v)$ actually live?
The $de_i$ is notational abuse. The context is still the same as yesterday. The first component of the frame gives a section $s\colon U\rightarrow SM$ of the sphere bundle and I'm trying to figure out why the vertical part of $ds\vert_p(e_i(p))$ is given by $\nabla_{e_i}e_1(p)$ after identifying the vertical part of tangent space of $SM$ with a subspace of tangent space of $M$ (the one spanned by $e_2,...,e_n$)
Yeah, I think you want to be thinking of $\nabla e_i$. In Chern's early papers, following E. Cartan, he wrote $de_i$, but they were both thinking of the connection, not regular $d$.
So I don't see what your question actually is. That seems totally obvious to me.
Just think of $SM\subset TM$ and take the covariant derivative of the section of $TM$. The fact that we're restricting to the sphere says that we have to end up orthogonal to $e_1$.
@TedShifrin hey ted!
Hi, @Stan.
$ds\vert_p(e_i(p))$ is an element of $T_{(p,e_1(p))}SM$. The vertical part of that lies in $V_{(p,e_1(p))}$, which we can identify with $T_{e_1(p)}S_pM$ ($S_pM$ being the fiber at $p$), which we can in turn identify with a subspace of $T_pM$ (name the one spanned by $e_2,...,e_n$). Why is the element we end up with $\nabla_{e_i}e_1(p)$?
5:08 PM
I keep asking. Why do you even want to think of the derivative of the mapping here? What is the specific context you're getting this from?
When we differentiate sections of bundles we never want to think this way.
We want to get a bundle-valued $1$-form. That's not what you're doing.
@TedShifrin I was trying to find the equations for an oblique cylinder. To do this, I used a shear map in the $xz$ plane and then solved for the volume and moment of inertia tensors. However, I just realized that I might want the shear in the $zy$ plane or some plane that involves the $z$ and a combination of $x$ and $y$. What kind of map would I need for that?
Any shear is still a linear map, just a matrix with more nonzero entries.
You're replacing $e_3$ by $e_3+v$ where $v$ is in the $xy$-plane.
oh i see. cool, that makes sense. im gonna try to rewrite it then.
Personally, I would rotate coordinates in the $xy$-plane and reduce to the case you've already done. Then you just need to figure out how a linear change of coordinates affects the moment of inertia tensor. The point of its being a tensor is that this is simple to do.
@TedShifrin that's clever. would i have to do that prior to computing the moment of inertia or would I do the rotation after computing it?
just confirming. i tried rotating the tensor
but it didn't seem to work
but i could have made a calculation error
5:15 PM
How did you "rotate" the tensor?
I took the inertia tensor and then tried to spin it about the z axis with a 3x3 rotational matrix
You need to multiply twice by the change-of-basis matrix because it's a 2-tensor.
It's the transformation rule for a bilinear form (as opposed to a linear transformation), so you should have $P^\top AP$, where $P$ is the change of basis.
And $A$ is the inertia tensor in this case?
@TedShifrin ok i'm gonna go try that out then. im gonna find a simple example and try to rotate it before i try doing it with my cylinder.
very helpful! thanks ted :)
5:19 PM
Sure thing.
The context is this. Let $\pi\colon SO(TM)\rightarrow SM$ (projection onto first element) and $\tilde{\pi}\colon SM\rightarrow M$. We have a natural connection $T(SO(TM))=V(SO(TM))\oplus H(SO(TM))$. Since $d(\tilde{\pi}\circ\pi)=d\tilde{\pi}\circ d\pi$, $d\pi$ maps $V(SO(TM))$ into $V(SM)$ and, conversely, any preimage of an element in $V(SM)$ necessarily lies in $V(SO(TM))$. Since $d\pi$ is surjective, we obtain a connection on $SM$ via $T(SM)=V(SM)\oplus d\pi(H(SO(TM)))$. Now pick an orthonormal frame as before. At each $p\in U$, it gives a diffeomorphism $T_pM\cong\mathbb{R}^n$ that, by
I think you're making this too computation-intensive.
I would love for that to be the case, but I haven't been able to think of anything better
So that form is constructed as I said when we first started talking about this. That's the orthogonally-invariant volume form on the fiber because it's the wedge of the orthogonal coframe for the fiber. The point is that when you pull back the "tautological" forms on the full frame bundle to $SM$ by the obvious section, this equation is a tautology. $s^*\tilde\omega_1^j = \omega_1^j$.
This is the same game one plays pulling universally-defined connection forms on the frame bundle back to the base manifold by a section — which gives the forms on the base manifold for that particular moving frame.
We're just doing it at the level of the sphere bundle, where $e_1$ is the position vector in the manifold.
I guess my $s$ in the equation above is your $\pi\circ s$.
But that equation is tautological.
The notation is horrible, by the way. We should make tildes more compatible, so if we're using tilde for the universal forms on the frame bundle, then we shouldn't have $\tilde\pi\colon SM\to M$. Where are you getting this?
5:45 PM
I made this notation up on the spot, don't expect it to be well-adjusted :P
Ah. I figured you were getting all this from your professor's lectures.
Anyhow, I recommend thinking things through in a slightly less complicated setting (i.e., omitting the sphere bundle and just going from frame bundle to manifold). Once you are convinced there, you're done.
what's the obvious section from the sphere bundle to the frame bundle?
You had a section $s(x)=(e_1(x),e_2(x),\dots,e_n(x))$ to start with.
So it's just $s\circ\tilde\pi$ in your horrid notation :P
but $s(\tilde{\pi}(x,v))=s(x)=(x,e_1(x),...,e_n(x))$ isn't a section
Sure it is. Because $\pi$ of that is $(x,e_1(x))$.
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