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8:03 PM
@TedShifrin I think the link, which is not so easy to read as it includes a lot of unusual locutions and sentence structures, said the same as yours. @Thorgott
 
@Thorgott $\iota^* \omega_p = \omega$, no? Like you wrote earlier.
 
right, that's a typo (the latter one)
 
But I think I would like Sylvester's writing as he loved language and poetry, which should make his writing embellished with a lot of metaphoric terms.
 
@robjohn have you tried writing the Putnam?
 
oh, I figured out how I was sloppy
those functions $a_I$ were only defined $U\cap S$, not on $U$
so extend them via bump functions
 
8:19 PM
@BalarkaSen not sure if you are around but I had a rather elementary problem with something
 
I am a bit busy but you can say it
 
How do you define bump functions?
 
It might just be an easy answer in any case: the common definition for an integral of a 1-form $\alpha$ along a path $\gamma\colon[a,b]\to M$ is $\int_\gamma \alpha = \int \gamma^*\alpha$. This definition makes sense to me in any case, but it also matches the usual one for integration on manifolds (where $\gamma$ is the inverse of a chart for $\im\gamma$), but only as long as $\gamma$ is injective (otherwise at double points you run into problems).
 
I have further questions about Sylvester's law of inertia. Though the signature of a real quadratic form $A$ is an invariant of $A$, the eigenvalues of $A$ are not an invariant of $A$, right? Because a quadratic form is not a linear transformation.
 
So my question is whether integration of 1-forms can be defined easily for immersions or something like that.
 
8:23 PM
You can pullback by an immersion
So it makes sense, right?
 
The notation and everything makes sense. It's just I don't ever see it in that generality in books where they are integrating forms. So forms integrate more generally than on manifolds?
And that's just another fact of life?
 
Line integral is done all the time over immersed parametrized curves
You have surely seen this in complex analysis
 
urgh, this isn't as easy, right
but surely the function has a smooth extension at least locally
proof by handwaving
 
$U \cap S$ may not be a neighborhood in $U$, right?
 
yeah
 
8:27 PM
I guess there's no getting away from tubular neighborhoods.
 
That's what spawns this question. I am sitting here writing notes about doing complex analysis "right" with geometry, and so I reached this gap in my knowledge where I was like "surely this notation makes sense and works, but I have never seen it before.
 
Just use tubular neighborhood theorem for crying out loud.
 
what if I just go through a chart and then just say such extensions are possible in Euclidean spaces
 
The function $a_I$ has a canonical extension locally, it's a function on $\Bbb R^{n-k} \subset \Bbb R^n$. Just define the same damn function on all of $\Bbb R^n$
Why do you guys spend time on inconsequential details
Infuriating stuff
 
Details are usually needed for comfort.
 
8:29 PM
I know I'm ribbing them
 
that's the same as what I'm saying, no
 
It's impossible to tell what you're saying, it's $\iota_* \rho_p \omega_p$ everywhere
$\rho_p$? Are you serious?
 
Pig
@BalarkaSen so that does work? I was thinking the same but felt a bit uncomfortable about it
 
Yeah it should work, @Pig.
 
do you take issue with my notation
@CaptainBohemian what're Eigenvalues of a quadratic form?
if you mean Eigenvalues of a corresponding Gramian matrix, these are in general not invariant
 
Pig
8:32 PM
ah i see now
No that doesn't work
@BalarkaSen The problem with that argument is assuming the $\mathbb{R}^n$ you mentioned actually covers the ambient manifold $M$
That's false
just take $d\theta$ on $S^1 \subset \mathbb{R}^2$ for example
you can extend that to $\mathbb{R}^2 - 0$ alright but you can't cover 0
 
They're taking little $\Bbb R^n$ charts covering $S$ in $M$
 
locally extending suffices
 
You don't want a full extension to $M$, you want an extension to a neighborhood of $S$
 
a neighborhood (in $M$) of a point in $S$ suffices
 
Pig
neighborhood is fine, but that's why you do need bump function. You can't use the exact same function
 
8:34 PM
On a neighborhood you don't need bump function to extend
Outside you do
 
Pig
sure, fair enough
 
$a_I$ extends perfectly fine from $U \cap S$ to $U$, right?
That's what I was arguing for, sorry for being unclear
 
yeah, I buy that
 
Pig
No it's okay, I was thinking along the same lines but I thought it directly extends to $M$, which confused me
 
and then we stitch them together with partitions of unity and we're done
tubular neighborhoods = avoided
 
8:36 PM
angery
 
@aperspicaciouslycuriousmind writing the Putnam? I have not.
 
Pig
that's almost tubular neighborhood to be honest
 
Yeah
 
Pig
you just need one more compactness argument to argue that the width of your local neighborhood can't go to 0
so effectively you proved tubular neighborhood theorem along the way
 
8:48 PM
ok, so a way to recast the ambient coordinates thing is as follows: Let $S\subset M$ be a smooth submanifold, $\iota\colon S\rightarrow M$ the inclusion and $\Phi\colon S\rightarrow\mathbb{R}$ a smooth function. Assume it possesses a smooth extension $\tilde{\Phi}\colon M\rightarrow\mathbb{R}$. Fix $p\in S$ and a chart $(U,x^1,...,x^n)$ of $M$ (not necessarily adapted) around $p$.
Then $d\tilde{\Phi}=\sum_ia_idx^i$ on $U$ for some smooth functions $a_i$. It follows that $d\Phi=d\iota^{\ast}\tilde{\Phi}=\iota^{\ast}d\Phi=\iota^{\ast}(\sum_ia_idx^i)=\sum_i\iota^{\ast}a_i\iota^{\ast}dx^i=\sum_
 
i'm reading a book and the author sais: Let $M$ be an n-dim topological manifold with boundary, A chart for M is a pair $(U,\phi)$ where $\phi: U\rightarrow \mathbb{R}^n$ is a map such that $\phi$ is a homeomorphism onto an open subset of $\mathbb{H}^n$ or $\mathbb{R}^n$. If $p\in M$ is not a boundary point, in which case, restricting the coordinate map to $\phi^{-1}{(int\mathbb{H}^n)}$ is an interior chart whose domain contains p.
isn't he assuming that $\phi$ is continuous?
 
last time I checked homeomorphisms were continuous
 
notice, he said its a map $\phi : U\rightarrow \mathbb{R}^n$ that is a homeomorphism $\textbf{onto an}$ open subset....
 
Yes that necessitates continuity.
 
@Thorgott What are you doing here?
 
8:55 PM
That just means if you restrict it to its image, it has a continuous inverse.
 
I'm trying to formalize what it means to write down the exterior derivative of a $0$-form in ambient coordinates
 
ohk. I thought you were still on $\iota^*$ is surjective.
 
Like $df_p(v_p) = v_p(f)$ in local coordinates, @Thorgott?
Or what do you mean by "in ambient coordinates"?
 
what that means is what I'm trying to figure out
 
mf trying to formalize $df = \sum \partial_i f \, dx_i$
 
9:00 PM
lol
I love you, Balarka
You will make it.
But Thorgott might not. :((((((
 
thorgott is good mane
he just also likes to formalize nahmean
he'll probably end up interpreting differential geometry higher category theoretically
 
@Thorgott after you are done with that, write it formally in Coq for me.
thx
 
no man, local coordinates =/= local ambient coordinates
im not just being redundant i swear
 
@anakhro: Just to emphasize. You can pull back forms by any smooth mapping, of maximum rank or not.
 
I actually did read some of the nlab page on synthetic differential geometry recently
 
9:04 PM
uh oh
 
@TedShifrin yeah my problem seems to lie with the fact that anywhere I see integration of forms in a textbook, we are pulling back by a chart (a diffeomorphism), but since we only care about things locally, pulling back by an immersion seems to work with my head alright.
 
@BalarkaSen Topological spaces are basically $\infty$-groupoids anyway, so you only need to put a smooth structure on those
 
apparently differential equations can be treated "naturally" in toposes
 
In mathematics, synthetic differential geometry is a formalization of the theory of differential geometry in the language of topos theory.
F
 
@anakhro: But if it's not one-to-one you may be "over-counting" when you integrate.
(For example, think about $S^1\to S^1$ given by $z\rightsquigarrow z^2$.
 
9:06 PM
@TedShifrin when does that occur exactly? Not every time you integrate a non-injective immersion, surely, otherwise complex analysis is dead?
 
Well, in complex analysis, you count "with multiplicities" all the time.
 
Perhaps I am not sure what you mean by over-counting, then.
 
basically it goes back and forth over the image
 
You're no longer computing the integral over the image if the image is covered up more than once and/or unevenly.
 
@Ted You can always nudge the immersion a little smoothly so that it becomes an immersion with transverse double points only, and that only changes the value of the integral only a little bit, though. So the "overcounting" might be illusory.
 
9:09 PM
But there is the degree formula, which I'm sure you know.
$$\int_M f^*\omega = \deg(f)\int_N \omega \quad\text{when } f\colon M\to N.$$
 
In the context of curves in $\Bbb R^2$
 
I was worrying not about immersiveness, Balarka, as that really doesn't matter globally. I was worrying about multiple-sheetedness or even not covering situations.
 
Right, I understood the point you're trying to make (namely, the formula you wrote down).
 
But in non-compact situations, we may have total un-evenness.
 
Oh I see what you mean. So for something like Balarka says where you have transverse double points, there is zero contribution to the integral, but like in your case, multiple sheets can wreck the idea of integrating over the image itself.
 
9:14 PM
is there a natural example of a 3-category
 
So in a definition for $\int_\gamma \alpha$ for a path $\gamma$, should we include in our assumption that $\gamma$ only has transverse double points?
Otherwise the notation $\int_\gamma$ doesn't quite make sense all the time.
 
No, the parametrized path can be as weird as you want. You define the integral by pullback, not as an integral over the image "manifold" which isn't a manifold.
 
So you are saying even writing that notation is bad?
 
You are integrating over a singular chain, not over the image.
It's standard in complex analysis, for example, for the curve to be piecewise-smooth, so no big deal.
 
How should I notate it?
 
9:22 PM
The notation is fine. You just need to understand what you're talking about.
 
The notation (with its definition) seems to suggest I am integrating over the image of $\gamma$.
At least, to me. So I should not be making this connection?
 
@Thorgott I mean the entries you get when you diagonize a quadratic form (choose a basis so that the matrix associated with the quadratic form becomes diagonal).
 
If $\gamma$ is a mapping, then this is defined to mean $\int_I \gamma^*\omega$ where $I$ is the domain of the mapping. The image may be a singular thing.
Back to my singular chains comment. If the image is a piecewise-smooth submanifold (without multiplicities), then no problem.
 
Piece-wise smooth submanifold would include what Balarka was talking about where we only have transverse double points?
 
Piecewise-smooth means it is a finite union of smooth things. They do not need to meet transversely, but certainly includes that case.
 
9:27 PM
So it also includes the case where you would have multiplicities like your example with z^2?
 
No.
 
I ordered a book: Serge Lang's Algebra
31 USD total
 
That is a very tough book to learn algebra from.
 
I alread know lots of algebra, and I did "look inside" on amazon. Looks great
 
But in your example, $im\gamma = S^1$ which is a smooth submanifold.
 
9:29 PM
for me
Also working on BananaCats again
when it's done we'll have another way to add commutative diagrams to posts
 
That's the trouble with your notation, @anakhro. Clearly just the image does not suffice. So you need a one-to-one mapping or you need to interpret it as a singular chain and pull back.
 
My notation being $\int_\gamma$?
 
Yes. Unless $\gamma$ is known to be a piecewise-smooth curve, you need to define this as the integral over a singular chain.
 
Smooth + transverse double points implies piecewise smooth curve, though?
I don't quite see why $z\mapsto z^2$ was not piecewise smooth.
 
The transverse double point stuff really has no business here. It is irrelevant for integration what the singular point looks like.
It's globally smooth, but it is multiply covering. So you do not integrate over the image, which is just a single circle.
I'm saying the same thing five times now.
 
9:36 PM
I understand because it covers the image twice you get a factor of 2, but I asked if piece-wise smooth submanifold includes that case, and you said no.
 
Just looking at the image you have no idea how it is mapped to.
 
Okay, thanks for your help.
 
10:10 PM
Hey, am I correct in assuming that there is no known closed form for $\Gamma(\frac{1}{3})$?
 
hi anyone here ???
 
user462942
hey
 
@Joanna I need help with some modulo arithmetic. I have posted a question here: math.stackexchange.com/questions/3698830/…
I just can't figure out why this is not true for non-multiples
 
user462942
@TedShifrin Do you like Michael Artin's Algebra book more than an advanced linear algebra book, if we are comparing strictly the linear-algebraic topics?
 
Yes, I like Artin's book supremely.
He shows you how linear algebra is part of mathematics, not just an isolated topic.
 
user462942
10:18 PM
Wow ... @TedShifrin
 
I read part of it one time, but it doesn't compare to the breadth of topics covered in Dummit & Foote
 
Dummit and Foote is a standard graduate text. Comprehensive. Nothing inspirational.
Artin's book has all sorts of non-standard material for a (hard) undergraduate text.
 
Pig
what's nonstandard in Artin?
 
user462942
Yes @TedShifrin I very much dislike the idea that linear algebra leads us to scientific computing and numerical analysis (like in our math department) -- I never liked this route and would've preferred to see the deeper, broader abstract structures developed.
 
The Hopf map is discussed. Representation theory is discussed. The beginnings of algebraic geometry and algebraic number theory are discussed.
 
Pig
10:20 PM
hm I see
 
user462942
@TedShifrin I read the preface by Artin, and I think he said something like it was time to recognize that Algebra had modernized, and thus many of his chapters were now linear-algebraic topics. I might buy the book?
 
He didn't change his book significantly from first edition to second. Unless there's a third I don't know about.
 
user462942
@TedShifrin second edition, I think
 
user462942
@TedShifrin Hmm ... how many problems should I do for each chapter?
 
user462942
Wondering if I could finish it by end of summer :)
 
user462942
10:24 PM
3 months ...
 
There are some super-challenging problems in there, as well as plenty of standard ones. See what you can do.
 
user462942
I see
 
user462942
What are your thoughts on reading it cover-to-cover? Worth the time?
 
I would say that if you know half pretty well, read it through but don't spend the time on that as much as on things you don't know.
 
user462942
I see, so, move on to new material within his book.
 
user462942
10:29 PM
Thanks @TedShifrin :)
 
Have fun!
 
user462942
I can't wait; Artin was my professor's advisor, btw.
 
Oh, who is that?
 
user462942
I'm gonna stay anonymous @TedShifrin ;)
 
LOL, that's not divulging your identity.
 
user462942
10:30 PM
But he's a great guy; wrote my rec letters.
 
user462942
true haha :)
 
I might know him, which is why I'm curious.
 
user462942
You probably do ...
 
user462942
He tells riveting stories during lecture @TedShifrin
 
That isn't exactly a useful clue.
 
user462942
10:32 PM
hahaha
 
user462942
@TedShifrin As you know, since yesterday, I'm feeling regretful having moved to such applied math work. I feel lost. Thanks for chatting -- I really appreciate your time :)
 
user462942
I did it because of all the hype, ya know?
 
Not the end of the world.
 
user462942
NSF grants, meetings, seminars -- it's all exploding at our dept.
 
user462942
Yeah ... true
 
10:36 PM
@TedShifrin so many people use ur books, it's really amazing
 
Not that many, @Stan.
 
I mentioned your book on differential geometry to someone in this reading group and he immediately recognized it
really cool
well, maybe i just run into them all then
:')
 
yeah, maybe
 
@TedShifrin anything fun on your summer reading list?
 
Nothing particular.
 
user462942
10:38 PM
@TedShifrin is differential geometry synonymous with "geometric analysis"?
 
Not synonymous. But that's the NSF grouping for people doing the analytic side of differential geometry. There are other aspects for sure.
 
user462942
I see
 
@TedShifrin do you have anything fun planned for the summer?
 
Thanks to the world, I have absolutely nothing planned.
 
Yes, it has eliminated any ability to travel or go anywhere, for even basic things
 
10:45 PM
LOL, I can't even do normal food shopping.
 
are you doing online?
 
some
 
user462942
@TedShifrin What was your most fun place to travel to?
 
Don't have a good answer. I always love France, but Croatia was amazing.
 
@Pig It's just a beautiful book; nobody writes algebra books like that.
 
10:53 PM
Yes, Balarka was an early convert.
 
I understood Galois theory from reading Artin; he spends a chapter on drawing Riemann surfaces, tinkering around with monodromy, branch cuts, ...
 
Pig
I think my maturity was just too low when I first read the book, so I think it's okay but didn't like it that much
 
user462942
@TedShifrin I love France too
 
Pig
At the time I was more interested in olympiad style problem solving, so Herstein's Topics in Algebra was a good fit for me, despite its terrible notations
 
user462942
Croatia? Interesting -- I've never considered going there
 
10:58 PM
Herstein was a classic, but it was all about symbolic manipulations. Tricks. Never integrated linear algebra into algebra.
I'm not even talking about functions on the right.
 
Pig
i knew some basic linear algebra before that, so i was mostly using it to familiarize myself with group/rings/field
But yeah, I think on the whole it took me a long time to go beyond symbolic manipulations in mathematics
that's why i said my maturity was too low back then :)
 
user462942
How do you pronounce Galois? Is it Gal-lo-Wah or Gal-Wah?
 
Gal-Wah
 
user462942
@Rithaniel thanks!
 
Where "a" is like the "u" of "umbrella"
 
11:02 PM
accent on the ultimate
hey, A. Someone wants into the other room. For some reason it's telling me.
 
Sorry about that. It's because you're an owner
I'll give write access
 
How did I ever become an owner?
 
I added you as one :)
Foundational senior member
 
Very suspicious.
 
11:36 PM
My manifolds homework: Show $\left\{(x,y,z)\in\mathbb{R}^3\colon\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\le1\right\}$ is a smooth manifold with boundary.
My topology homework: Show $\left\{(x,y,z)\in\mathbb{R}^3\colon\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\right\}$ is a smooth manifold.
 
seems a little repetitious, @Thorgott
 
Manifolds homework: See [Topology Homework]
Topology homework: See [Manifolds Homework]
Can you get away with that?
 
lol
it's not even redundant, funnily enough
at least without knowing that the boundary of a manifold with boundary is a manifold
well, it's redundant in the sense that a chart for the former gives rise to a chart for the latter, of course
 
Doing the topology homework is easier, since the regular value theorem is standard. The regular inequality theorem takes a bit more :P
Do I have to do charts, daddy?
I refuse.
 
we haven't even done regular value theorem in topology
however, I can just note it's diffeomorphic to the sphere
 
11:42 PM
Please do.
 
Blah.
Isn't the implicit function theorem a prerequisite?
That's all you need.
 
He has done it in his manifolds course I'm sure
why is topology course doing smooth manifolds?
it should do topological manifolds
 
Who knows what the course is.
 
I've done it in analysis 2 even, but we haven't explicitly discussed it in the topology course yet, so I'm not gonna quote it
it's "topology from the differentiable viewpoint"
 
Oh, Milnor or Guillemin & Pollack?
 
11:44 PM
let's be real, no one wants to do topological manifolds
 
So you're going to do transversality and intersection theory?
 
the prof recommended Milnor as supplementary reading, but the actual contents are rather hodgepodge so far
 
@Thorgott why not
 
The only time I've done anything with topological manifolds was in algebraic topology (Poincaré duality).
 
we've done some wild stuff like proving that manifolds in Euclidean spaces are local differentiable neighborhood retracts and we actually needed that in order to prove the tangent bundle is a manifold
 
11:48 PM
wait — what's a tangent bundle of a topological manifold?
microbundles?
 
@Balarka smooth invariance of domain = easy, topological invariance of domain = no clue
 
I rest my case
 
you need homology, @Thorgott
then it's easy
 
@Ted no, smooth manifold*
 
11:48 PM
well, it used to be ... I don't berember now.
 
what the shit
 
then who needs ANR to get the fact that the tangent bundle is a manifold? that's nuts
 
lmao
 
I don't see an easier way with the Milnor definitions tbh
 
btw smooth submanifolds are much better than ANR/NDR, by the theorem you hate to use
 
11:50 PM
What are Milnor definitions?
I've taught this stuff dozens of times and have never needed to know what a LDNR is
 
Let $X\subseteq\mathbb{R}^m$ and $Y\subseteq\mathbb{R}^n$. A function $f\colon X\rightarrow Y$ is differentiable if, for each $p\in X$, there is a neighborhood $U$ (in $\mathbb{R}^m$) of $p$ and a differentiable (in the classical sense) function $\tilde{f}\colon U\rightarrow\mathbb{R}^n$, such that $\tilde{f}\vert_{U\cap X}=f\vert_{U\cap X}$
 
OK, that's standard for embedded things. G&P do that too.
 
$X\subseteq\mathbb{R}^N$ is called a smooth manifold of dimension $n$ if each point has a neighborhood diffeomorphic to an open subset of $\mathbb{R}^n$
 
The equivalence to the usual one follows once you know the local immersion theorem.
This is all standard, @Thorgott, if you're working with submanifolds of Euclidean space (as opposed to abstract charts).
It's trivial to get the tangent bundle is a manifold.
 
i guess LDNR just means smooth manifolds have adapted charts lmao
 
11:53 PM
Which is the local immersion theorem in this case.
Just the inverse function theorem. No biggie.
But I don't need any of this for the tangent bundle.
Just define it as $\{(x,v): x\in X, v\in T_xX\}\subset M\times \Bbb R^m$, and then produce charts.
 
Now if $M\subseteq\mathbb{R}^N$ is a smooth manifold of dimension $n$, we have the tangent bundle $TM=\{(x,v)\in M\times\mathbb{R}^N\vert v\in T_xM\}\subseteq\mathbb{R}^{2N}$. Let $\phi\colon U\rightarrow\mathbb{R}^n$ be a chart for $U\subseteq M$ open.
For $y\in U$, $d\varphi_y\colon T_yU\rightarrow T_{\varphi(y)}\varphi(U)=\mathbb{R}^n$ is an isomorphism. Let $p\colon TM\rightarrow M$ be the canonical projection. Let $\psi\colon p^{-1}(U)\rightarrow U\times\mathbb{R}^n,\,(y,v)\mapsto(y,d\varphi_y(v))$. This ought to be a chart for $TM$.
 
It's easier to use parametrizations rather than charts.
Particularly with submanifolds of Euclidean space, but it's logically equivalent.
 
isn't parametrization just the inverse of a chart?
 
If I use parametrizations $\psi$, then I can just use $d\psi$ directly to parametrize the subset of the tangent bundle.
 
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