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12:05 AM
@Thorgott Ya that's correct
 
nice
but I still haven't figured out why this implies your vector field is smooth
restricting the differential to the subbundle gives a bundle map that is a fiberwise isomorphism, but itself not invertible, so it's still awkward
 
It actually gives a smooth bundle isomorphism $F : (\ker df)^\perp \to M \times \Bbb R$, $F(x, v) = (x, df(v))$
As bundles over $M$
 
ah, that looks like the map I was missing
 
Therefore define a smooth section $s : M \to (\ker df)^\perp$ of the first bundle by $s(p) = (F^{-1} \circ c)(p)$ where $c : M \to M \times \Bbb R$ is the constant section $c(p) = (p, 1)$.
As $(\ker df)^\perp \subset TM$ is a subbundle of the tangent bundle, $s$ defines a section of the tangent bundle too (composing with the subbundle inclusion), which is our vector field $(df)^*(d/dt)$
So in this formalism smoothness of the vector field is automatic.
 
1:07 AM
$F$ is clearly a bundle map and a fiberwise linear isomorphism. I can also see that $F$ is smooth, but I can't see why $F^{-1}$ is smooth (this is the same thing I was struggling with earlier, essentially).
 
Bundle isomorphisms are diffeomorphisms
 
But I don't know it's a bundle iso if I don't know the inverse is smooth in the first place
(I'm taking "bundle isomorphism" to mean bundle map with an inverse bundle map)
 
This is a language issue. Bundle isomorphism to me just means a smooth bundle homomorphism which is a fiberwise isomorphism. This is the same as your definition, small exercise
It's just the fact that inverse of an invertible matrix varies smoothly with the matrix, nothing fancy
These things you can sit down and figure out on your own, they're really basic :P
 
That can't be true, a smooth bundle map that is a fiberwise isomorphism doesn't even need to be bijective
 
Bundle homomorphism fixes the base for me
$E_1, E_2$ two vector bundles over the same base $B$, $f : E_1 \to E_2$ a smooth bundle map over $B$ taking the fiber over $b$ in $E_1$ to the fiber over $b$ in $E_2$ isomorphically for all $b \in B$. Then $f$ is a diffeomorphism
 
1:14 AM
ok, that sounds useful, I'll try it
 
this was the whole point of the construction of $F$
 
yeah, that makes sense
but I wasn't aware of the useful fact and struggled accordingly
 
well now u do
 
1:52 AM
@Balarka Ok, let $\pi_1,\pi_2$ be the projections. Let $p,q\in E_1$ s.t. $f(p)=f(q)$. Then $\pi_1(p)=\pi_2(f(p))=\pi_2(f(q))=\pi_1(q)$, so $p,q$ live in the same fiber of $B$ in $E_1$, but $f$ is bijective when restricted to that fiber, hence $p=q$, i.e. $f$ is injective. Surjectivity of $f$ is obvious since each element of $E_2$ lies in some fiber of $B$.
Choose local trivializations $\varphi_i\colon\pi_i^{-1}(U)\rightarrow U\rightarrow\mathbb{R}^n,\,i=1,2$ for some open $U\subset B$ (same $U$ for both WLOG cause otherwise take intersection). Then $\varphi_2\circ f\vert_{\pi_1^{-1}(U)}\cir
 
Ya sounds good
 
this is neat
now our vector field is just given by the compositions $M\rightarrow M\times\mathbb{R}\rightarrow V\hookrightarrow TM$
 
What topic of math are you guys discussing? I feel like I am a complete stranger to it.
 
differential topology, I'd say
 
Well it's out of my range. It's 3rd yr course.
I am reading combinatorial analysis. It asks from a group of 5 women and 7 men, how many different committees consisting of 2 women and 3 men can be formed? Which is easy but it asks. What If 2 of men are feuding and refuse to serve on the committee together?
What do you think it means by Now suppose that 2 of men refuse to serve together. Because a total of {2\choose 2}{5 \choose 1}=5cout of {7\choose 3}=35 possible groups of men contain both of the feuding men, it follows that there are 35-5=30 possible groups that do not contain both feuding men.
I am confused by the unexplained calculation {2\choose 2} {5\choose 1}.
 
2:05 AM
I have a quick question: It is easy to see, using the Lagrangian for length in rectangular coordinates in the plane to see that straight lines minimize the length functional. I ran across an exercise which converted the Lagrangian to polar coordinates, and then asked me to write the Euler-Lagrange system in those coordinates.

Let me say the result is horrific. Is that probably because they wanted to make the point that choice of coordinates will dramatically affect the difficulty of solving the Euler-Lagrange equations, and they just wanted you to see two ends of the spectrum?
 
What does it actually means?
 
@rschwieb Haha sounds like physicists being physicists
 
${2\choose 2}$ are the ways of choosing the two feuding men from the two feuding men (of course, there's only one way to do so), and ${5\choose 1}$ are the ways to choose the remaining third man (there are $3$ men total in a committee) from the $5$ men that are not the two feuding men
 
@Thorgott I thought the same but I think remaining men should be {3\choose 1} since you may choose wamen.
 
we are not considering the women right now, only the number of groups of $3$ men containing both two feuding men
 
2:13 AM
Oh now it's okay to say Ohhhhhhhh....... I understand. It's so easy to see.
@Thorgott you have fericious mathematical instinct!
 
@Balarka horrible question time: I have a smooth manifold $M$, then, for each $p\in M$, I can consider the tangent space $T_pM$ and $I_p$, the set of all inner products on $T_pM$ and $I=\coprod_{p\in M}I_p$. This isn't quite a vector bundle, as each $I_p$ is only a positive cone. Regardless, can we put a differentiable structure on $I$ in such a way that the "sections" of $I$ are precisely the Riemannian metrics?
 
Are you all like 3rd yr undergraduate student? And by looking at profile of your guys it looks like you guys are professors in math department.
Or top student in university.
May be geniuses.
 
@Thorgott Yeah :) First let's look at a space which is a vector bundle, namely, space of all smoothly varying pointwise symmetric bilinear forms
 
@WhyWhatWhereWhenHow this will come naturally to you once you've done a bit more combinatorics
 
A bilinear form on a finite dimensional vector space $V$ is just an element of $(V \otimes V)^* = V^* \otimes V^*$, where $\otimes$ denotes the tensor product. You know this interpretation, yeah?
 
2:28 AM
a bilinear map on $V$ uniquely factors as linear map through $V\otimes V$, so the first part is clear
 
yeah forget about the second part, the isomorphism $V^* \otimes W^* \to (V \otimes W)^*$ is given by $\phi \otimes \psi \mapsto (\alpha \otimes \beta \mapsto \phi(\alpha)\psi(\beta))$ which you can check is injective and since we're in finite dimensions it's an iso
 
@WhyWhatWhereWhenHow Nah I suck at maths, I just try really really hard and get lame, adequate grades
 
That's formal nonsense
 
@Thorgott Everything comes naturally but sometimes I go to wrong direction. My be everyone is like same .They do do mistake. Anyway you can have busy conversation may be I am bit being a distraction in highly mathematical conversation.
 
we all do a lot of mistakes
 
2:30 AM
this is a rant room just hang around and rant nobody will be bothered by it
thats what we all do
 
Mike said this a couple days ago in here, but it's by doing mistakes that you learn the most
and yeah, I'm buying that formal nonsense
 
@EdwardEvans It's normal. But I think about student having perfect grades which quiet scares the crap out of me.
 
(we will start talking about tensor products in my commutative algebra lecture next week, so I'll do enough of that then)
 
@WhyWhatWhereWhenHow lol tbf you can't go through life comparing yourself to others else you'll never be happy
cuz Peter Scholze will always be better than you
 
lmfaooo
 
2:32 AM
rofl
 
or will he indeed
 
hahaha
 
couldnt have put it better
such a great quote i will use and abuse this
 
Spraying my wisdom around
 
Gauss will always be greater than peter scholze.
 
2:34 AM
screams in perfectoid spaces
 
to gauss everyone is a smoothbrain so
 
cough Tilted cough
 
Well, when you are stuck in research and getting nowhere, the only reason to keep going is because you love the subject. But if you really love math, then at least personally, comparing myself to others doesn't come up at all
 
this
 
You just want to figure stuff out, you know?
 
2:35 AM
So $(V \otimes V)^*$ has a decomposition $\text{Sym}^2(V^*) \oplus \Lambda^2(V^*)$ into the subspace of symmetric bilinear forms and alternating bilinear forms, by sending a bilinear form $\langle -, - \rangle$ to the average after permuting coordinates or the difference after permuting coordinates ("symmetrization" and "alternatization")
I don't know why I am doing this to you but since I am doing it anyway I will keep doing it
 
fair one
 
Does latex not compile automatically here?
 
of course, this is just the Eigenspace decomposition with respect to swapping coordinates
 
yep
 
I don't know how can human being reach intelligence of those freakin geniuses like Gauss, Ramanujan, Terry ....
 
2:36 AM
right I just spent like 6 hours doing functional analysis and now I actually wanna die, so I'm gonna go to bed
night y'all
 
night
 
I am reading everything here in plain latex since the tinyur thing doesn't work in android version of chrome.
 
so no i will define a notion of fiberwise tensor product and dualization in bundles so i can push this formalism through
or do u know this already
 
nah
 
@EdwardEvans That hit hard rolfmao.
 
2:42 AM
Let $E$ be a vector bundle over $B$, choose trivializing neighborhoods $U_\alpha$ of $B$ such that over each $U_\alpha$ you have a trivialization map $\varphi_\alpha : U_\alpha \times \Bbb R^k \to \pi^{-1}(U_\alpha)$. For any pair of indices $\alpha, \beta$, there are these transition functions $\varphi_\beta^{-1} \varphi_\alpha : U_{\alpha\beta} \times \Bbb R^k \to U_{\alpha \beta} \times \Bbb R^k$ defined over the overlaps $U_{\alpha\beta} = U_\alpha \cap U_\beta$
These guys are identity on the first component, so the only interesting this to look at what is going on in the second component. Keeping track of that I get a map $\phi_{\alpha\beta} : U_{\alpha\beta} \to \text{GL}_k(\Bbb R)$, where $\varphi_\beta^{-1}\varphi_\alpha(x, v) = (x, \phi_{\alpha\beta}(x) v)$.
These are called the transition functions for the vector bundle, and they satisfy the identities $\phi_{\alpha\beta}\phi_{\beta\alpha}^{-1} = I$ and $\phi_{\alpha\beta} \phi_{\beta\gamma} \phi_{\gamma \alpha} = I$ on every triple intersection $U_{\alpha\beta\gamma}$, where $I$ means the map which sends everything to the identity matrix, and multiplication means multiplication pointwise in $\text{GL}_k(\Bbb R)$
 
I think everyone who viewed my question in stackmath is self harming by looking at my question. Or just gave up lol. XD
 
Such family of functions are also called Cech 1-cocycles (you can prolly guess why, given our earlier interaction on sheaves)
 
oh god
I've been tricked into sheaves again
 
sheaves are great! don't be scared!
 
Formalism aside, $\phi_{\alpha\beta}$ completely determines the bundle in the following sense. Consider the topological space $\coprod_\alpha U_\alpha \times \Bbb R^k /\sim$ where $\sim$ is the equivalence relation defined by $(x, v) \sim (x, \phi_{\alpha\beta}(x) v)$ whenever $x \in U_{\alpha\beta}$ and $v \in \Bbb R^k$
This is homeomorphic to $E$, and the projection to $U_\alpha$ defines the bundle projection $E \to B$
So all information in the bundles are contained in the transition cocycles
I am doing this topologically but to get the smooth structure etc back you want $\phi_{\alpha\beta}$ to be smooth maps to $\text{GL}_k(\Bbb R)$, just to get that nonsense out of the way
Let $E, F$ be two vector bundles over the same base $B$ of rank $n$ and $m$ repectively. Choose a common trivializing cover $U_\alpha$ on $B$ for both $E$ and $F$, and let the corresponding transition cocycles by $\phi_{\alpha\beta}, \psi_{\alpha\beta}$.
(1) $E \oplus F$ is defined by the transition cocycles $\phi_{\alpha\beta} \oplus \psi_{\alpha\beta}$, maps to $\text{GL}_n(\Bbb R) \times \text{GL}_m(\Bbb R) \subset \text{GL}_{n+m}(\Bbb R)$ as a block matrix. This is the fiberwise direct sum.
(2) $E \otimes F$ is defined by the transition cocyles $\phi_{\alpha\beta} \otimes \psi_{\alpha \beta}$, where tensor means pointwise tensor product of matrices, maps to $\text{GL}_n(\Bbb R) \times \text{GL}_m(\Bbb R) \to \text{GL}_{nm}(\Bbb R)$. This is the fiberwise tensor product
(3) $E^*$ is defined by the transition cocycle $\phi_{\alpha\beta}^*$, where dualizing a matrix pointwise means just taking transpose $\text{GL}_n(\Bbb R) \to \text{GL}_n(\Bbb R)$. This is the fiberwise dual
The space of bilinear forms (without adjecitives) on $M$ is then $T^* M \otimes T^* M$ because that is the correct notion fiberwise. A smooth section of the bundle $T^*M \otimes T^* M$ gives a smoothly varying bilinear form on the tangent spaces of $M$
You can convince yourself that you can similarly define the symmetrization $\text{Sym}^2(E)$ of $E \otimes E$ for a vector bundle $E$, and that defines $\text{Sym}^2(T^*M)$, the space of symmetric bilinear forms on $M$, a section of which defines a smoothly varying symmetric bilinear form on $M$
The space of Riemannian metric is a subspace of $\text{Sym}^2(T^* M)$ which you rightly pointed out is the fiberwise open cone of the positive definite guys. This is not a vector bundle, but a bundle nonetheless, with cone fibers.
A smooth section of $\text{Sym}^2(T^*M)$ whose image lies in this cone bundle is a Riemannian metric on $M$
Sorry I didn't pause for breath there
 
2:59 AM
does it make a difference to construct $\mathrm{Sym}^2(T^{\ast}M)$? that is, why don't we immediately consider sections of $T^{\ast}M\otimes T^{\ast}M$ whose image lies in the cone?
 
yeah it doesnt really i was just being really specific
i mean it makes a difference in the sense that the subspace is not an open subspace of T^*M o T^*M anymore
what the formalism tells you eg is that riemannian metrics are open in the space of all symmetric 2-forms on a smooth manifold, because if you wiggle a section of Sym^2 which is inside the open cone it'll always be inside the open cone because its open
 
ah, that sounds reasonable
how do we recover the smooth structure from the cocycles?
 
also its a fiberwise convex open subset, with the "boundary" being what i would call "degenerate metrics", which are positive semidefinite and vanish at some places
 
Riemannian pseudometrics?
 
sounds right, but i think you also call indefinite guys are pseudo-Riemannian
detour: this convexity and openness is useful because there's a theorem of gromov which says fiberwise convex subsets of bundles are special, you can approximate sections which lie strictly inside the cone by sections which lie on the boundary of the cones
 
3:09 AM
no opportunity to mention gromov was missed
 
lmfao yeah totally just for that
@Thorgott ehh to get the smooth structure you also want to the transition functions for $U_\alpha$ considered as manifold charts
then $U_\alpha \times \Bbb R^k$ has a smooth structure, you have transition functions of the manifold and the cocycles of the bundle which glue these to a smooth structure of $E$
well $U_\alpha$ is just placeholder for $\Bbb R^n$, the thing $U_\alpha$ gets mapped to by the chart diffeo, in that case
 
I'm very worried about quotienting
 
and you glue these multiple copies of $\Bbb R^n \times \Bbb R^k$ using chart transition functions and bundle transition functions
 
at the very least the equivalence relation seems open, because the transition maps are homeos
and the graph of the relation should be closed by their continuity
 
sure topologically it's a-ok
 
3:13 AM
so I'm at least buying this gives a top. manifold
 
its a good exercise to write down the smooth structure as well
i'd explain it but i am nearly asleep now having like my 19th smoke
i cant write formulas, only talk in wobbles
 
yikes
ima go sleep over this
 
cool
night
 
after sleep, I'll also have to finally read that last paragraph and figure out why we put so much effort into pulling back a plain vector field
a yway, night
 
its because if u have a vector field where nothing happens for a long stretch of time the topology on the region is trivial
:)
 
3:25 AM
How many user are using stack exchange?
Does anyone know meaning of because nothing here now? In the comment of math.stackexchange.com/questions/3675476/…
Or does he mean... Everything I assumed is right.
 
3:50 AM
@Knight Okay, I won't pretend to be you/
 
I think if no one answers my question and just leaves me with 4 upvotes then I have to say I have mastered real analysis book. (Hurrrrrrayyyyyyy!) Now I can move to Group theory, Onion ring and galois vegetable then Metric unit with spicy measured quantity theory then Topology of Carrot then Quantum Which was hated einstien then general and special relativity then n
Anyway there are endless then
Self studying undergraduate course is really time consuming ! Woosh. I wish there was professor just taught me but they are tooo overpriced.
 
I cost at least a million bucks, @WhyWhatWhereWhenHow. However, if you're interested, my 112 lectures on YouTube are available for free.
@BalarkaSen I told you to quit smoking. Now you're past wobbling!
 
@TedShifrin Yes I want to watch your million bucks lecture. Show me your demonstration. Wait... You are a professor.Yikes.
I finished multivariable calculus using Khan academy looks like Salman Khan didn't cover whole multivariable calculus. I don't even know wth is fubini's theorem.
Khan academy multivariable calculus is cringy but has intutive approach.
@TedShifrin I searched you in youtube looks like you provide detailed lecture in each of the contents on multivariable calculus so it is 30+ minutes long. BTW I like the mustache on your face it's like Einstien's mustache. It does fit on every face.
I want to grow that mustache on my face too.
 
4:14 AM
LOL, right now I also am working on Einstein's hair!! :P
My course incorporates linear algebra and is has all the proofs (as well as plenty of computations).
 
Hey and you got lecture on lagrange interpolation. I was looking for this when I was middle schooler all the time! None of the website satisfied me I will watch you lecture now.
 
It's an application of stuff from linear algebra. If you don't know linear algebra yet, you should be patient and work your way to it.
 
@TedShifrin God bless professor like you who provides Lecture on internet.
2
 
Well, it was thanks to my hard-working students who taped them. But they may not fit your personality or learning style.
 
@TedShifrin Yep I haven't completed linear algebra yet. I was not satisfied with easy khan academy video so I am gonna buy Gilbert strang linear algebra.
@TedShifrin @TedShifrin I can learn anything given appropriate time. I am sure any style fits except learning rote memorization and blind rules.
Sorry for tagging too much. Hope it doesn't annoy you.
@skullpatrol I know the news.
But isn't lagrange interpolatio learned in numerical analysis? Instead of Multivariable calculus or Linear Algebra?
 
4:25 AM
@WhyWhatWhereWhenHow I love Strang, but his books are not great if you're trying to learn things step by step with proofs. But he does an excellent job with applications.
Just start at the beginning of my course :P
 
Mr or Prof Shifrin I would like to read your books rather than watching video. Since I can't do learn in very fast with that 480 ppi resolution. I mostly skip video and watch blackboard .
@TedShifrin I swear I can learn that in a week. XD
But I think book will be faster
Oops it's 110 videos.
May be 2 weeks.
 
LOL, no. This is a very challenging, year-long course taken only by the best college students. :P
But you're welcome to find the books. Because of the world we live in, they're unfortunately not very cheap.
 
COVID-19 is changing the world we live in professor
 
Multivariable Mathematics: Linear Algebra, Multivariable Calculus, and Manifolds is the book. Ok I will accept the challenge! I will try to finish it in 2 weeks. 500 pages.
 
@skull: definitively ...
 
4:39 AM
:-)
 
@WhyWhatWhereWhenHow ... Working on a book should mean working (and getting correct) at least half the exercises. That alone will take most of a year.
I suspect you've never dealt with a serious college mathematics book before.
 
@TedShifrin is it bad practice to write $\forall$ and $\exists$ in a proof?
are u supposed to use words?
or are the symbols accepted too
 
@TedShifrin @TedShifrin I have seriously dealt with college Mathematics. I have finished whole exercise of analysis book Numbers and functions. Steps into analysis. And read springer undergraduate books of real analysis. It looks me weeks to solve it without looking answer from books. And the logic were correct. But sometimes I go to wrong direction but I always correct my mistake.
I know it takes lots of time.
 
@Stan: I was taught (by Munkres) never to write symbols (other than things like $\implies$ and abbreviations like s.t.). Most mathematical journals do not accept things full of formal symbols. They're very hard to read.
 
@TedShifrin I have not only be correct on half but full. I believe true mathematician can solve everything that has been Discovered.
 
4:45 AM
@WhyWhatWhereWhenHow: I'm not putting you down. I'm just trying to be realistic. But let's see what you can do.
 
@TedShifrin good to know. didn't know that. i agree, i find them illegible. I thought it was the preferred way
since i see it in class all the time
i didn't realize it was shorthand
 
I had almost no professors that wrote symbols, @Stan.
 
@TedShifrin I know you are not trying to put me down.I am just too confident XD.
 
Some write $\forall$ and $\exists$ but plenty of words.
@WhyWhatWhereWhenHow: Well, you can show me a few finished/polished proofs of exercises as you go along, and we'll see :)
 
I believe part of the reason why most students can't handle serious college level textbooks is because there are so few serious high school level textbooks @TedShifrin?
 
4:47 AM
You can ask @Stan: He's submitted a bunch of linear algebra to me and had to sweat.
@skull: All over Europe, Russia, and Asia, the expectations in the US are way more stringent.
 
@TedShifrin Ok I will do your book exercise after I finish enough course that will make me satisfied.
 
But in the US we sort of expect too many students to get isomorphic educations.
 
After finishing undergraduate course I can finally start to daydream Mathematics lol.
 
@TedShifrin to be fair, I sweat with almost all math, but at least your book was fun to sweat over
One of my friends majored in math in college and he says he can barely understand the math we are doing cuz of the notation
that we are using this quarter
very strange experience
i'm looking forward to doing real math again this summer
 
@Stan: So do you feel the little bit we did together helped prepare you?
 
4:51 AM
@TedShifrin i feel like it not only helped, but was frankly the only decent linear algebra training ive gotten and i think it's essential i do some more
It has definitely made my courses much easier
and made finding information much easier
cuz i know what to google if i get stuck
 
@abhas_RewCie You can, after all we are same person using two different accounts :-)
 
In retrospect, I regret using my Caltech "multivariable and linear algebra" combo course to excuse myself from taking linear algebra at uchicago
because i think I did myself a disservice by not taking a full course devoted entirely to it
so i'm trying to rectify that by doing exercises from your book, which i plan on resuming as soon as this quarter is over
 
What I have observed is if your first course or first introduction to Multivariable Calculus is ill-taught then you won’t ever be able to get it right.
 
It's just not sensible to avoid taking a course in linear algebra when it is so prevalent
but i didn't know that when i started college lol
just trying to control my GPA
 
@TedShifrin Is you book Multivariable Mathematics: Linear Algebra, Multivariable Calculus, and Manifolds a standard,Special or Optional course in university?
 
4:54 AM
@Stan: I'm glad it's helping.
 
I think it's year 1 course.
 
Are you saying that all over Europe, Russia, and Asia there are serious high school level textbooks professor @TedShifrin?
 
@WhyWhatWhereWhenHow: It's used in the toughest Honors courses typically taken by first and second year students. It's been used at Stanford, Vanderbilt, Yale, Georgia, and other places.
Absolutely, @skull. If you paid attention in this room, you'd see that the guys from Germany, France, China ... all had an analysis type course even in high school, but universally first year of university.
 
@skullpatrol Pal! The focus of Asians is more on being fast and calculative, they pay less attention on “hard thinking”
 
Hmm
 
4:56 AM
@WhyWhatWhereWhenHow i think what distinguishes Ted's book from others books is it's thoroughness combined with its intuitiveness. He doesn't leave any stones unturned. I wish I had found his book earlier so I highly recommend it. Many of my peers in grad school all used his book
 
@WhyWhatWhereWhenHow: At these places it's 10-20 students who complete the course, typically. Oh yeah, they used it at UCSD two years ago. 100 or so started, and I think 30 finished the third quarter.
 
China and Korea are the best examples of ^
Ted sir I’m still unable to find the volume contained by two intersecting cylinders
 
I swear I used intutive approach on Real Analysis. I see a lots of words turning into picture so I am sure I can complete it since I believe I am persistent enough to do that.
 
All I need is a visualisation
 
@Knight Wrong.
 
5:02 AM
@TedShifrin Baklara really helped me understand a few things earlier.
"The Gaussian curvature is defined for surfaces; in an n-dim manifold there are n(n−1)/2 "coordinate surfaces" at any point given by exponentiating coordinate planes as I mentioned. You can think of the Riemann curvature tensor as the tuple of all these Gaussian curvatures"
i didn't realize this
back then i didn't know enough physics to understand what role differential geometry plays in physics
 
Professor Shifrin your Multivariable Mathematics: Linear Algebra, Multivariable Calculus, and Manifolds is expensive than my weekly salary lol.
 
@WhyWhatWhereWhenHow Right. Try the JEE
 
@WhyWhatWhereWhenHow Okay, what are your views?
I personally find the courses of Germany to be very very nice.
They are very on-topic. They don’t pretend to teach us “practical science”
 
@Knight They do have intutive approach. They are not just only fast and calculative they are Smart. You can see indian guys lol.
 
Yes, @Stan. As long as you understand Gaussian curvature of surfaces (abstractly). That's why I said starting with curves/surfaces is a significant step to understanding.
 
5:07 AM
@WhyWhatWhereWhenHow I decline the statement that Asians follow more intuitive approach than Americans or Europeans
 
@WhyWhatWhereWhenHow: Notice that I was pushing the free lectures, not the book.
 
Sir is your dinner done ? :-)
 
@Knight: If you're talking to me, it's after 10 PM.
 
You're back home?
 
@TedShifrin Okay! So, are you planning to sleep?
@skullpatrol Did you see the movie Casino?
 
5:09 AM
@skull: I've been back a month now.
 
@Knight I decline your statement. I have my friends in Taiwan who have far more intutive approach than here in America.
 
@Knight yup
@TedShifrin nice to hear
 
@skullpatrol What was “back home” referred to in that movie?
 
nope
 
????
 
5:11 AM
@TedShifrin ohhhhhhhh
you're book makes so much sense now
wow that took me a while
 
Now I have no idea what you're talking about, @Stan :P
 
@Knight The good professor had to leave his home due to COVID-19
 
Guess it was almost 4 years ago I made my trek to Chicago, @Stan.
 
@skullpatrol Okay. I asked what was referred to as “back home” in Casino is same as Ted’s home?
 
Well, that's not literally correct, @skull. I just chose to do so ... but then realized I'd have to leave for years.
 
5:13 AM
@TedShifrin How long did it took to make that book.
 
@TedShifrin that's crazy. it feels so short, but i guess hw makes it feel that way
@TedShifrin I didn't get why you started off your differential geometry book with 2d and 3d concepts when I thought you would jump right into higher dimensions. I didn't realize the tools with Gaussian curvature were used to build up notions of higher dimensional curvature
so I found that confusing, but now it makes sense why you put emphasis on that
 
@Stan: The diff geo course was for undergraduates. It's the right level for that. To do $n$-dimensional Riemannian stuff you really need serious topology, manifolds, bundles, and a lot more sophistication.
 
Yeah, and that's probably why I find do Carmo to be hard to digest. I get stuck on bundles
 
@Knight pardon my misinterpretation pal
 
In part, I like the course because it makes students actually use linear algebra and multivariable calculus. But I like the concreteness and the depth of the material, without requiring a bunch of prerequisites.
DoCarmo doesn't do bundles, actually, other than tangent bundle.
 
5:16 AM
Welp, there goes my excuse
 
@skullpatrol No problem pal
 
@WhyWhatWhereWhenHow Probably three years to get it polished.
 
@TedShifrin my high school physics teacher retired last september to write a book on group theory
 
@TedShifrin indeed sir, this could go on for years.
 
one of my best friends in Athens is a physicist who occasionally teaches a graduate course on group theory for physics. @Stan
 
5:17 AM
@TedShifrin So you single handedly made a book? o_o
 
Great, great teacher.
@WhyWhatWhereWhenHow, four, in fact (one with a coauthor).
 
@TedShifrin what made him a great teacher? Anything in particular?
 
Have you ever thought of writing a number theory book, sir?
 
I actually never saw him lecture, but I know him well, and many of my students raved about him and wrote letters of recommendation for him for teaching awards. He is well-prepared, clear, motivates well, cares deeply about students.
No, @skull. My only use for number theory is to use it to teach the concepts of algebra (so I incorporated some of it into my algebra book).
I actually have never taken a number theory course.
 
@TedShifrin Have you meet geniuses? What does it makes them genius.
 
5:20 AM
@TedShifrin :O
 
@TedShifrin those are really great qualities
 
I don't know what makes a genius. I certainly have taught students who I knew were extremely talented and gifted. Some of them accomplished far more than I ever have. Some perhaps not.
 
@TedShifrin I also think a good ability to inspire people to want to accomplish things is a great quality too
 
@TedShifrin Did they became great mathematician?
I am really curious about people tagged genius.
 
what do u define as a genius?
 
5:24 AM
I don't tag anyone genius, @WhyWhatWhereWhenHow. Some of them have, yes. Others in other fields.
A few of my most incredibly gifted students, sadly, died young.
 
I define them who are able to think outside the box.
Something extraordinary.
It is too broad to define.
 
I don't see that as "genius."
I don't know what "extraordinary" actually means. But I'm happy to talk about talent and gift.
I know that a lot of students I've taught (and a lot of people I interact with in this chat) think they're a lot better than they actually are. By a significant margin.
 
They do give very unexpected answers (Now seriously don't try to be sarcastic lol).
 
Ted is never sarcastic
ever
 
Well, at least I know what it means, unlike our narcissistic emperor.
 
5:28 AM
I'd say genius is someone who sees something that lots of people don't see. Now I am saying this in a sloppy manner.
 
@TedShifrin omg he's the worst. I think he's taken narcissism to new levels
@WhyWhatWhereWhenHow it's a question psychologists have tried to answer. people have tried to come up with ways to measure "genius" and "intelligence" and, in general, it can be difficult to ascribe a single word like genius much meaning since there are so many forms of ability.
Was Mozart more of a genius than Arnold Schoenberg?
 
@StanShunpike I can't compare them but I can compare them to normal people.
Now don't tell me to define normal people lol.
@StanShunpike From my perspective this is kinda like asking is multiplication better than addition.
We as human beings do have nature of comparing things . But comparing human intelligence is complicated. There is pattern for sure. Someday human intelligence can be rigorous compared.
 
5:53 AM
@CalvinKhor were the formulas that valuable that the lectures wanted them back?
 
@skullpatrol sorry what fomulas? and hello
 
Hello.
May 11 at 7:29, by Calvin Khor
i returned all my number theory to my lecturers so don't know, sorry. I know $Li_s$ is notation for polylogarithms but they don't seem related
:-)
What does "i returned all my number theory to my lecturers" mean?
 

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