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04:01
In Analysis I, we took the ideas of calc I and made them rigorous. In Analyis II, we made several ideas from Calc II rigorous as well as some calc 3 topics(learned multivariable differentiation)
But what about the topics of Calc 3: integration of multi-valued functions, and vector calculus, etc? What course are these ideas typiclaly made rigorous?
Sounds kinda like a dumb question: I am really just asking what is the advanced version of calc 3 typically called?
this almost seems to be a question about course titles, which would be specific to a particular school or program of study, and who knows. but some of that is covered in classes with titles like "differential geometry", "differential topology", "differentiable manifolds," "theory of manifolds", etc.
even course titles like "calc I" and "calc II" aren't all that precise, and any general link between course name and course content only gets harder to follow with subsequent classes.
OK. I understand what you are saying, but let me sitll ask: So like a class in measure theory and integration, will that make rigorous the multivariable integration techniques I did in calc 3?
it might, it might not. my guess would be generally not.
but it depends on the school/course/instructor.
Sure. OK. thanks
No. But typically multivariable analysis — which is often not taught — covers such things, including differential forms and Stokes’s Theorem.
And of course the inverse and implicit function theorems first.
04:38
An undergraduate tried to publish a new proof of the Pythagorean theorem in Arxiv and was rejected, I don't know the reason but is there any somewhere that accepts new proof of Pythagorean theorem still?
Do people care about this btw?
05:02
all sorts of places might accept a paper like that, e.g. journals with an emphasis on expository writing, maybe with sections relating specifically to geometry or college math. past copies of a journal are pretty good indicators of what a journal will publish.
i have no idea what it takes to get a paper 'accepted' to the arxiv (although i have my own papers there), but, i don't think that it regards itself as a place to 'publish' anything. it began life as a preprint server and still kind of is one. i can understand not wanting to open the floodgates to being, like, Host Your Mathematics For Free dot com.
it probably doesn't help that this is a proof of the pythagorean theorem. that might be a particularly bad fit for the arxiv, even if it's perfectly publishable in the right place.
the arxiv's "general mathematics" section used to, and maybe still sometimes does, include stuff like that. perhaps to a degree that it is something of a cesspool
maa.org/press/periodicals/mathematics-magazine is an example of a journal (sadly, super paywalled) that might publish stuff like that. it is maybe more rigorously reviewed than most such journals, however. e.g. they might actually assign it to a reviewer who will actually perform something resembling a check that an allegedly "new" proof of the pythagorean theorem is, in whatever sense appropriate to its editors, actually "new."
i don't know many people who care about proofs of the pythagorean theorem. i know one guy who would definitely care, but if he were here, he'd probably say that nobody who hasn't seen at least 50 non-equivalent proofs of the pythagorean theorem is in any position to say whether or not a proof of that is "new."
people who get way into classical geometry tend to be like that.
05:57
Hmm what a big world
06:28
I'm reading a proof and the author write $\mathbb Z_2(x^2,y^2)=QF(\mathbb Z_2[x^2,y^2])$. Where QF stands for quotient field aka fraction field.
But these fields aren't equal, they're isomorphic, right? What am I not getting here?
06:55
uh, depends. it might be a definition? if the author does give separate meaning to Z_2(x^2, y^2), then yes, they mean isomorphism. [depending on how formal you are being, "QF( )" might not define something except up to isomorphism]
$\mathbb Z_2(x^2,y^2)$ is the smallest field containing $x^2,y^2$ and $\mathbb Z_2$
07:55
I found this problem: "Let $f(x)=x+L$, $L\in\mathbb{R}$, and let $g \in C^1(\mathbb{R})$ be such that $f \circ g=g\circ f$. Prove that $g'$ is periodic." I tried this: by the hypothesis $f\circ g=g\circ f$, we have $g(x+L)=g(x)+L$. Since $g$ is differentiable, taking derivatives both sides and using the chain rule lead to $g'(x+L)=g'(x)$, so $g'$ is $L$-periodic. If this is correct, is the continuity of the derivative in the hypotheses superfluous?
08:09
Can anyone tell what changes I need to do in this code to make it work for finite fields $\mathbb{F}_{p^n}, n>1$ math.stackexchange.com/a/3722724/1093844
09:05
Coxeter group seems quite interesting topic
09:51
So the example I said, tiling $\Bbb R^2$ by equilateral triangles (fundamental domain is equilateral triangle) then each loop represents some relation in Coxeter group. So if we do this to triangulated surface...
btw there's a theorem named Torelli theorem. Its name is similar to Tonelli's theorem but completely different
Wiki say it's in AG of RS but if I replace theorem to group (Torelli group), then it's about topology
Are they somehow related? I have no idea
Is it always possible to realize a group (I would assume finitely presented) as a tiling of some space?
10:26
The lifting to triangulated surface, now I think does not make sense
10:56
I googled perron family and suddenly the horror movie conjuring comes out
11:34
1
Q: Let $f(x)$ be a polynomial with integer coefficients. Assume that $3$ divides the value $f(n)$ for each integer $n.$ Prove that when $f(x)$ is divided

Fdst ZfsyLet $f(x)$ be a polynomial with integer coefficients. Assume that $3$ divides the value $f(n)$ for each integer $n.$ Prove that when $f(x)$ is divided by $x^3-x, $ the remainder is of the form $3r(x),$ where $r(x)$ is a polynomial with integer coefficients. An Edited Solution is presented below f...

I am so confused with this arguments/solutions of mine. Could anyone explain whether my solution is justified as far as it's validity is concerned ?
11:56
@XanderHenderson elliptic, parabolic and hyperbolic?
12:19
wow dirichlet problem is solvable on any region if its boundary is not very weird. I only know it's solvable on a simply connected region.
12:48
what would be a definition that captures the number of disjoint continuous curves on a graph? for instance, $f(x) = 1/x$ has two such curves, $f(x) = x$ has one such curve, $f(x) = \tan x$ has infinite such curves
$x^{2}y^{2}-x^{2}-y^{2}=0$ has 4 such curves
13:18
@shintuku number of path components?
i'll read up on that, thank you vm
It's worth mentioning that something like the trig functions usually have a restricted domain, though.
This also happens to be the case for 1/x, too. It does not have zero in its domain.
hm, right
So it probably is a little tricky to formulate this without getting too wordy, because you are looking at the inclusion of the graph of a function (which may not be defined on all of R) into the plane.
So it's path components of the image of this inclusion.
I doubt anyone would care though if you resorted to "pieces" or something like that.
hm, i'll look into that. seems that i'd look into topology to give an adequate definition
thanks
13:32
It doesn't really need topology in the general sense since this is all a subset of R^2. So everything is the standard topology.
Define the image of the inclusion as $A$, and then it's just how many equivalence classes are there for the relation $x\sim y \iff$ there is a continuous path $p\colon[0,1]\to A$ in $A$ such that $p(0) = x$ and $p(1)=y$. The most you need to think about then is what does it mean for a path to be continuous in $A$, which technically invokes the (standard) subspace topology on the subset of $A$.
Surprising fact: If $f$ is an entire function, then the map $F(z)= \int_0^{2\pi} f(e^{i\theta} z) d\theta$ need not be continuous.
@Koro what's your reason for believing it would be continuous?
pf: Substitute $e^{i\theta}z= u$. Then if $z\ne 0$, we get $F(z)= \int_z^z -if(u)/zdu=0$ and $F(0)=2\pi f(0)$.
@anak I was expecting continuity because we are in C, where non obvious things are not un-common: bounded entire is constant etc.
But the fact above about continuity seemed obvious to me somehow and it is false.
Does that kind of change of variable be allowed?
Is $f’^\text{v}(x)$ standard notation for fourth derivative?
13:46
@PlaceReporter99 I've never seen that notation. What is the "v" for?
Oh, god, no!
iv
@onepotatotwopotato yeah, why not?
@XanderHenderson edited question, I meant fourth derivative
No, that is not notation that I have ever seen before.
@Koro I don't know. Why allowed?
Why would I interpret a tic mark and a v as the Roman numeral IV?
13:48
@onepotatotwopotato tbh, I used it un-consciously. I'll think about it.
The most common way to extend the primed notation is to use Hindu-Arabic numerals in parentheses, e.g. $f^{(4)}$ for the fourth derivative.
$f^{'^4}$
@anak: is complex change of variable allowed?
$f$ is analytic so we can consider mean value theorem
13:50
$z$ is independent during the integration
@onepotatotwopotato don't we need z to be real there?
$z = re^{i\phi}$
@PlaceReporter99 I would not regard that page as a reliable source.
I mean, they can't even format their trig functions correctly.
I don't trust them.
In any event, I have never seen the notation $f'^{v}$. It is gross, and I don't like it.
Use $f^{(4)}$. Or, better yet, $D^4 f$ or $\frac{\mathrm{d}^4}{{\mathrm{d}x}^4} f$.
@XanderHenderson the page is using roman numerals, which is fairly common (though not the most common). PlaceReporter just didn't transcribe is properly.
13:57
Wikipedia says that Roman numerals are due to Lagrange: https://en.wikipedia.org/wiki/Notation_for_differentiation#Lagrange's_notation .
And chat butchers the link...
In any event, that implies that it is French. And we all know about those French mathematicians. :P
I've never seen it before, and I would guess that most American mathematicians are not familiar with it. As it is a French notation, it is more likely that Continental mathematicians (and former colonies of France, maybe) would be familiar with the notation.
It gets used in physics more often in my experiece.
@Koro I think there are special considerations to be made.
@onepotatotwopotato It works. Thanks a lot. :-) F indeed is continuous (infact constant).
So the intuition was correct.
@anak It isn't event that bad. Mostly, you just have to avoid singularities.
@anak I see. One red flag that I noted after @onepotatotwopotato's comment was that the variable of integration was changing from $\theta$ (real variable) to $u$ (a complex variable) after using the complex valued substitution.
@XanderHenderson Yeah, I was thinking about poles. As long as it is holomorphic, I was feeling it should be fine. However, I don't trust my complex analysis skills very much.
14:07
Suppose that we are given an antipodal map $f:S^1\to \mathbb R$. What this means is that $f(u)=f(-u)$.
$S^1\subset \mathbb C$ and suppose that f is continuous.
I want to extend $f$ holomorphically throughout $D=\{z: |z|\le 1\}$.
How about the following?: For any $a\in D$, join a to centre of D by a straight line and suppose that this line meets periphery at $a'$, then define $F(a)=f(a')$.
This F clearly extends f. Is it holomorphic?
@anak Just think about how the proof goes. In one complex variable, you can argue via the fundamental theorem of calculus, and the chain rule. You need differentiability for those, but not much more than that.
hmm, with this F, the F is real valued and real valued is not holomorphic unless it is constant.
Suppose f is non constant.
then this F won't give me the desired extension. Is there any other extension?
3
Q: Closed-form solution to the transcendental equation

PiotrCould you give me advice on how to find a closed-form solution $t>0$ to the following transcendental equation: $$(t+1)^a - t^a = g$$ where $a>1$ and $g>1$. An accurate closed-form approximation for $t>0$ to see how $t$ depends on $a$ and $g$ could be also fine. I guess the above equation defines ...

I don't understand this. What is the variable here?
14:23
@geocalc33 What do you mean?
The question clearly says "how do I solve this for $t$?"
@XanderHenderson Oh I get now
@Koro If it can be extended, we can use reflection principle to obtain a meromorphic function with finitely many poles on $\Bbb C$.
I was thinking about that but I'm having difficulty.
@Koro Life is difficulty.
Or, perhaps more appropriately, жизн трудно.
It sounds better with a Slavic accent...
14:40
So suppose that such an extension F is possible. I want to further extend it to C (here I wish I had Schwarz reflection for D). Considering that F is real on boundary of D, |F| must attain maximum there. So the extension must be bounded hence F must be constant, contradiction.
But I'm not sure how to extend beyond D. Say, $z\not\in cl D$, then I defined $\tilde F(z)= \overline {F(1/z})$
If I show this tilde F to be holom. then the result follows by Schwarz's lemma.
But I'm not sure how to show this tilde F to be holo.
@XanderHenderson that looks like 'ochen priyatna'.
:P
Similarly, I saw in various videos, on Russian planes , POKKA is written.
(i.e., when one tries to transliterates it into english.)
I don't understand slavic letters so I try to connect them with english alphabets and the words such as above come out as result.
😅
why for which assertion of mine?
both two statements
14:55
becase S^1 is compact, F is non constant continuous function (given) so |F| must attain maximum on S^1.
and constantness at the end follows by Liouville's. To conclude boundedness, I use maximum modulus principal.
@Koro Nope. That would be more like очень приятно.
@Koro that's for entire function
yes. So the question boils down to 'Does there exist a Schwarz reflection principal for a disk?' If it does, then using this I make my F entire.
This way the business is reduced to entire functions. This is the stage I applied the aforementioned theorems.
4
Q: Schwarz Reflection Principle on a unit disk

timetosleepSuppose $f$ is a analytic function defined on $\bar{D}(0;1)$ and has real value on the boundary. I'm trying to show $f$ can be extended to entire plane by $$g(z) = \begin{cases}f(z) &, \lvert z\rvert \leqslant 1\\ \frac{1}{\overline{f(\overline{z}^{-1})}}, &, \lvert z\rvert > 1\end{cases} $$ I t...

it suggests that there is such reflection principal on unit disk but I'm not sure how to show holomorphicity of g when |z|>1.
and another confusion that I have is why they specifically chose $\frac{1}{\overline{f(1/\bar z)}}$. What's wrong with more straightforward and intuitive $f(1/z)$?
15:10
@Koro You haven't understood the proof. Why we take \bar f(1/\bar{z}) ? Do you know 1/bar{z} reflects any point about the unit circle?
is it at all possible to graph a cubic equation without a computer?
@shintuku Depending on how accurately you want to graph it, sure.
You can find the zeros using Tartaglia's formula, and use Fermat's theorem on fixed points to find extreme values. The derivative will let you know where the function is increasing and decreasing, too.
hm, but i'm talking about a graph of a cubic that's not necessarily a function
@shintuku Rotate it first.
Oh, no. I see what you are saying.
for instance, $x^{3}y-x^{2}+y^{2}=0$
15:14
Yeah, that's harder. But you can still make some progress using tools from calculus. It should be doable.
and algebra
@Koro Another way to think about the problem: Find a Mobius map that map conformally real axis to unit circle.
@user977780 I don't understand this 'about unit circle'. What do you mean by that?
at Xander: how would you tackle this at first? obviously it has a solution at the origin, and then we see that an increase or decrease in y must somehow have the opposite motion in x
@shintuku I mean, I probably wouldn't, because graphing stuff by hand is not necessarily a skill that I think is valuable in that much generality.
I found this problem: "Let $f(x)=x+L$, $L\in\mathbb{R}$, and let $g \in C^1(\mathbb{R})$ be such that $f \circ g=g\circ f$. Prove that $g'$ is periodic." I tried this: by the hypothesis $f\circ g=g\circ f$, we have $g(x+L)=g(x)+L$. Since $g$ is differentiable, taking derivatives both sides and using the chain rule lead to $g'(x+L)=g'(x)$, so $g'$ is $L$-periodic. If this is correct, is the continuity of the derivative in the hypotheses superfluous?
15:17
But you could solve that equation for $y$ (as you can do with any cubic), and graph each "branch" separately using standard tools for graphing functions.
oh, there's a way to decompose the graphs of cubics?
reflection principle on the unit circle... I actually asked this question last year in this chatroom
@shintuku I mean, you can solve for $y$, n'est-ce pas?
@user977780 hmm, sounds cool. I'm thinking about this part.
(i-z)/(i+z) should do it.
@XanderHenderson i'll tinker
15:22
Regardless, I don't understand what 'reflection in the unit circle' is.
FTC implies $F$ can't be a rational function
why not?
F is originally a continuous non constant function on S^1 that takes real values.
It should be rational times $e^{h(z)}$ for some entire function $h$.
then we are extending it.
because D is simply connected?
but for this I think you need F (i.e. extension of F) to be non vanishing.
@Koro Can you give me your original question?
15:26
1 hour ago, by Koro
Suppose that we are given an antipodal map $f:S^1\to \mathbb R$. What this means is that $f(u)=f(-u)$.
@user977780 can you please elaborate on what 'reflection in the unit circle' means?
the said reflection can happen in two directions atleast so I'm not sure which one to take.
15:39
@user977780 it doesn't seem to be working.
I shifted everything to $\mathbb C$ but the problem is in showing holomorphicity using Schwarz's lemma.
@Koro What do you mean by "z_1, z_2" are symmetric about real axis?
Take the mobius map Tz=(z-i)/(z+i) , then T maps real axis conformally to the unit circle.
z_1, z_2 are symmetric about real axis then Tz_1, Tz_2 are symmetric about unit circle.
@Koro I haven't understood your question. f: {|z|=1}\to\mathbb{R} .
A real valued function of a complex variable either not differentiable or derivative is zero.
15:55
Consider the space of smooth maps $f: [0,1]^3 \to [0,1]^3$ s.t. $f$ preserves the convexity of the class of all closed surfaces satisfying $f(0,1)=1$ and $f(1,0)=0.$ What do I need to add here to only allow surfaces with genus $0$
Zero derivative on domain implies constant.
@user977780 Ohh, it means their imaginary parts are of opposite signs.
Because as it stands you could have a convex surface which is something of a sliced-in- half zonoid
oh wait
Forget about the linked part. Let me write it again: Given non constant continuous $f: S^1\to R$ such that $f(z)= f(-z)$, I want to extend it throughout the unit disk holomorphically.
The definition actually works perfectly
16:00
0
Q: Does Schwarz reflection principle exist for a disk?

KoroSchwarz's lemma: Suppose that $K\subset \mathbb C$ is an open connected set that is symmetric w.r.t. the real line. Let $K^+$ denote the part of $K$ above the real line and $K^-$; the part below the line. If $f_1: K^+\to \mathbb C$ and $f_2: K^-\to \mathbb C$ are holomorphic and extend continuo...

because i specified "closed surface"
@onepotatotwopotato I have posted this.
i.e. compact without boundary
I do like the idea of Mobius map but I got stuck while using it.
@user977780
3 mins ago, by Koro
Forget about the linked part. Let me write it again: Given non constant continuous $f: S^1\to R$ such that $f(z)= f(-z)$, I want to extend it throughout the unit disk holomorphically.
I remember I checked that principle by hand. Use mobius transformation that maps upper half plane to a circle. You already know reflection principle for upper half plane
16:04
yeah, I tried but a problem occured in the way.
Let me explain that:
I just received a notification that some comment of mine was flagged as offensive or spam or something. I'm not sure which comment exactly as the notification didn't say that.
So we have H conformally equivalent with D (open unit disk).
The map is $f:z\mapsto (i-z)/(i+z)$ from H to D. The same map is conformal map $S^1\to R$.
Suppose that the desired extension is $F$ (defined on circle and its interior). Then I define g on H as follows: g(z)= F(f(z)).
this is real valued on R.
so it should extend holomorphically to C.
For $z$ in $L$, $g(z)= \overline{F(f(\bar z))}$
how to get the corresponding F ?
I can define it using $F(f(\bar z))= \overline{g(z)}$.
f is a bijection so well defined. Fine, so F is extended to all of C.
How to show that F is holomorphic outside disk?
Holomorphic iff analytic.
16:19
how does that answer my question?
Power series.
So set $F(u)= g(f^{-1}(u))$ when u is in D, and F(u)=$ \overline {g(\overline{f^{-1}(u)}})$
otherwise
nope, all this seems wrong.
F is not yet extended.
F is defined on H only.
Is there any book which discusses Schwarz on unit circle?
16:36
$\Bbb D$ to $\Bbb H$, take reflection, then map back to $\Bbb D$.
16:49
@onepotatotwopotato but that's what I did.
Anyone have any idea about what the "Y+iZ" method is, while solving, the differerential operator equation $\frac{1}{f(D^2)}\sin ax$ such that $f(-a^2)=0$ ?
17:11
I don't know what you're doing. Do some calculations via explicit maps
@onepotatotwopotato Explicit content is a violation of the Code of Conduct. Be careful!
3
17:30
In the first equation of this answer: https://math.stackexchange.com/a/361431/987127
what does d floor(t) mean and n+ in the limits of integration mean?
can you graph $y^{4}x+x^{2}-y=0$ without a computer?
or, estimate it's graph without a computer
17:45
you can: solve for $x$
more difficult question, what about: $y^{4}x^{3}+x^{4}-y=0$?
18:24
Strange that some people used the word 'reflection in circle' today but none of them answered what it means, how it is defined when asked about it.
seems this is not well defined and is a way to handwave stuff.
So many posts asking about Schwarz's lemma but all of these seem to be lacking a good proper answer.
Koro, you tend to criticize everyone else way too fast. Reflection in the unit circle (or the unit sphere in any dimension) is defined by $R(x)=x/\|x\|^2$. The image is a point on the same ray through the origin, and the product of the distances is $1$.
This is a totally classical notion. Inversive geometry is in all sorts of classic books.
18:39
Good evening (or good morning :P)
Good night? @Mr.Feynman?
Can that be considered a salutation, though? It's 9pm here so I don't know what to say. I should have settled with "hello"
@TedShifrin May I ask you a differential geometry question about Stokes theorem and hypersurfaces (which is actually the reason why I've just joined the chat)?
Hello is so risk-free. What's your question? :)
My question is the following:
2 days ago, by Mr. Feynman
Given a Riemannian manifold $(M,g)$ with the Riemannian volume form (in coordinates) $\sqrt{|g|}dx^1\wedge...\wedge dx^n$ hypersurface. Given a hypersurface $S$ with induced metric (pullback) $\gamma$, can I conclude that the induced volume form is (in coordinates) $\sqrt{|\gamma|}du^1\wedge...\wedge du^{n-1}$?
@TedShifrin oh thanks a lot. Now that the definition is clear. How does it match the intuitive definition of 'reflection'?
i.e. reflection in a straight line.
18:46
The context is that of this answer of yours. I wondered if that same volume form could be written in such a form
I tried working out the calculations but I got lost with the pullback metric determinant $\gamma$
I don't want to disturb your diff. geometry discussion. I'll look for some more sources now other than Stein and Shakarchi's to get answers.
@Koro It matches, as onepotato said, if you consider the conformal isomorphism between the disk and the upper half plane. Then reflection in the boundary corresponds to reflection in the boundary (which is reflection across a line).
Mr.Feynman, have you worked this out explicitly in the case of a surface in $\Bbb R^3$? It's all going to boil down to the Gram determinant.
Yes, I worked it out for the $\mathbb{R}^3$ case with euclidean metric tensor, getting the "common" divergence theorem
Ted, how do i become a master at parametrizing algebraic varieties
Most cannot be parametrized :)
18:54
the path of mathematics is a path of sorrow
8
I didn't know the term "Gram matrix", checking the wiki article it should be what I call "matrix of the (components of) metric tensor", go the Gram determinant is my $g$. The problem I have is in relating $g$ (Gram determinant in the manifold) to $\gamma$ (Induced Gram determinant on the hypersurface), actually
I don't understand your question, Mr.Feynman. There's no issue with substituting $\gamma$ for $g$; it is the induced metric. Please clarify your precise question.
Are you trying to understand why restricting $\sum (-1)^{i-1} n_i dx^1\wedge\dots\widehat{dx^i}\dots\wedge dx^n$ gives $\sqrt{|\gamma|}du^1\wedge\dots\wedge du^{n-1}$ if you choose a parametrization by $u$?
what about algebraic curves?
Most cannot be parametrized. What do you mean by parametrization?
19:00
given the equation of an algebraic curve, i'd like to be able to think about it's shape without graphing it on a computer
i was thinking that decomposing an equation into parametrized equations of a single variable could make it easier to think of the geometry of plane curves on the basis of the equations
Yeah, @Mr.Feynman, you want to think about this in terms of the geometry rather than the calculus. Take a parallelelepiped spanned by $v_1,\dots,v_{n-1}$ in the tangent plane and then the unit normal $\nu$. Then $dV(v_1,\dots,v_{n-1},\nu)$ is the volume of the parallelepiped, which you then get by the Gram determinant.
@shintuku I disagree, but this is a personal opinion. I take that path for discovery and enjoyment.
@shintuku It's essentially impossible. What you mean by parametrization would ordinarily mean parametrization by rational functions of $t$, which means we're talking only about rational curves. Those all have genus $0$. The standard example is the general cubic $y^2=x(x-1)(x-2)$. It cannot be parametrized.
@Mr.Feynman, to be one step more explicit: You will apply this with $v_1,\dots,v_{n-1}$ given by $\partial/\partial u^1,\dots,\partial/\partial u^{n-1}$ for your parametrization.
hm i see
By the way, if I give you the parametrization $\left(\frac{2u}{1+u^2},\frac{1-u^2}{1+u^2}\right)$ of a curve in the plane, does that help you see the curve?
19:06
no but it makes it painful in my eyes
If I'm not mistaken, this is the construction of the induced volume form taking the interior product with the $dV$ volume form of the manifold and then pulling it back (in fact you evaluated it on $\partial/\partial u^i$)
That's just the unit circle. How did a parametrization "help" you?
@Mr.Feynman Yes, that's right. But that's where my silly formula for the hypersurface area form came from.
but how do people who do algebraic geometry get an intuition of the graphs of equations of varieties? do they fully rely on computers to get a sense of the shape?
If that's what you did, I understand that is a volume form on the hypersurface. What I don't understand (intuitively I do) is why the induced volume form is the Riemannian volume form with the induced metric of all possible volume forms
19:08
Classic algebraic geometry did all sorts of amazing things long before there were computers. You're not going to understand why there are 27 lines on a (smooth) cubic surface in $\Bbb P^3$ by drawing pictures.
Because precisely of how I computed that volume. I'm getting the induced "hyperarea" of that parallelepiped by turning it into a complete box and using "area of base times height."
i have no doubt about it
In general, they don't.
i'm having a hard time thinking of how people learn to think of the geometry of equations that can't easily produce an isolated $x$ or $y$
@TedShifrin I'm a bit confused about why the "area of base times height" thing is specific to the Riemannian volume form and not to any good volume form
Huh? The induced volume of the parallelepiped in the tangent plane is precisely what I just did. That's the induced metric, eh?
19:15
@XanderHenderson that was pretty explicit.
@TedShifrin Maybe I'm starting to see it. I'll express it in plain words to see if my geometric intuition is working (as I'm quite weak on geometric intuition). Since we used the Riemannian volume form $dV$ of the manifold to calculate the volume of the parallelepiped in the manifold, then the area of its base (parallelepiped in the hypersurface) can only come from the induced Riemannian volume form
@robjohn and aggressive in nature :P
19:34
@TedShifrin Thanks for helping with my doubt. :)

Good night everyone.
19:50
@robjohn OH NO! I might have to suspend myself!
@XanderHenderson hands Xander bungee cord
::makes popcorn::
How to show reciprocal of zeta function is analytic at s=1?
20:06
@TedShifrin Boy... that got dark fast.
Just a small clarification without which I couldn't sleep. I didn't know that the Riemannian volume form is defined as the unique volume form such that $dV_g(\text{ positive orthonormal basis})=1$. I only knew its coordinate definition. Now everything adds up
Don’t do too much with coordinates.
I try as much as I can to work coordinate free but being a Physics student it's more common to have coordinate expressions all over the place...
@SoumikMukherjee Are you specifically asking about elliptic curves over finite fields? Or do you just want to know how to work with a simple Galois finite field of $p^n$ in SageMath? Sage has extensive support for number fields. doc.sagemath.org/html/en/reference/number_fields/sage/rings/…
20:26
@XanderHenderson No suspended animation.
@Mr.Feynman Physicists also hate differential forms, but they are super-powerful and much more conceptual than long tensor formulas.
@shintuku The terms of Ted's parametrization of a (semi) circle should be at least vaguely familiar. Expressing $\sin(\theta)$ and $\cos(\theta)$ in terms of $\tan(\theta/2)$ has many uses.
oh, yeah i recognize it now
Oh, good. :)
didn't see it until you pointed it out
Still, Ted's point stands. Parameterizations can certainly be useful, but they don't necessarily give you much insight into the nature of the curve. You kind of need to see the whole procedure of how the parametrization was derived.
20:41
i'm still disappointed i can't get an intuition of a variety from its equation
the geometry part of algebraic geometry is a lie
Given a sheaf of rings $\mathcal O_X$ on a top space $X$, $\sigma\in\mathcal O_X(X)$ a global section, is $\mathcal O_X\cdot\sigma$ then a sheaf, or should I sheafify? I'm particularly interested whether given a covering $\{U_i\}$ of $X$ and $\tau_i\in\mathcal O_X(U_i)$ such that $\tau_i\cdot\sigma\vert_{U_i\cap U_j}=\tau_j\cdot\sigma\vert_{U_i\cap U_j}$, the unique element $\eta\in\mathcal O_X(X)$ such that $\eta\vert_{U_i}=\tau_i\cdot\sigma\vert_{U_i}$ belongs to $\mathcal O_X(X)\cdot\sigma$
If a function f(z) is analytic around a point $z_0$, and f(z_0)=0, then is the function analytic at $z_0$?
Slightly NSFW (nothing terrible, just some naughty language), but definitely on-topic:
@PM2Ring yes, I am asking about elliptic curves over finite fields $\mathbb{F}_{p^n}$. The answer there gives the algorithm for $n=1$ case, I am curious about $n>1$ cases.
Is there anybody out there
20:56
@SoumikMukherjee Ah, ok. Sorry, I don't know enough about that topic to give you good advice. There are experts on elliptic curves on finite fields on Cryptography.SE, and they permit Sage questions there. But ECC usually just uses prime fields, not Galois p^n fields.
ECC = Elliptic Curve Cryptography
OTOH, crypto people certainly know about finite field arithmetic, since AES uses GF(2^8) arithmetic. So it's quite possible that someone on that site is happy to discuss elliptic curves on general Galois fields.
@MinecraftPlayer69 What does "analytic around a point" mean?
@XanderHenderson love that one
@ShaVuklia I'm not quite sure what $\mathcal{O}_X\sigma$ is supposed to be?
Howdy, @Thor.
$(\mathcal O_X\sigma)(U)=\{\tau\cdot\sigma\vert_U:\tau\in\mathcal O_X(U)\}$
@TedShifrin a neighbourhood around z_0. Some radius R>0 such that the function is analytic for |z_0+r| where 0<r<R.
21:09
@XanderHenderson what naughty language?
I'd like to apply this to showing that $1/\zeta(s)$ is analytic at $s=1$, but it doesnt quite fit that criteria. Just interested
That doesn't really make sense to me, @Minecraft. Neighborhoods generally contain the point in question.
@TedShifrin ok, I hope you understand my question though.
If you are assuming it's analytic on the punctured neighborhood and bounded at the point, then the point is a removable singularity and the function is/can be made analytic on the whole neighborhood.
Right, I proved that in an earlier assignment and I think you helped me lol. Is it considered a singularity though?
21:12
If the function is not defined there, then it's a singularity. But removable means ...
You're talking about a function with a pole at $z=1$, so its reciprocal will in fact have a removable singularity, yes.
Think about Laurent series.
I guess my best best at a counter example is to consider $\operatorname{Spec}(R)$ with its structure sheaf
I'm off to bed tho, and will think about this tomorrow
Ok, not sure if I need laurent series, I think I can use the fact that $\zeta(s)$ is non zero for $Re(s) \ge 1$, and it's analytic continuation for $Re(s) > 0$. Hopefully there aren't any zero's near s=1 which would break my argument. I will check, thank you!
I just enabled desktop notifications
waits for notification
@PM2Ring Thanks, then I may look at Cryptography.SE now
Well I could just assume the Reimann Hypothesis
21:21
If you're going to toss around such weighty things, at least spell Riemann correctly.
:(
Sry Bernhard
the rye man hypothesis
The Zimmerman hypothesis
@shintuku Does it come with pastrami?
Don't forget the Russian dressing!~
21:27
The Gucci Main hypothesis
@TedShifrin And kraut.
Gewiß!
hey @Ted
$\mathbb{Q}(x,e^x)$ can someone elucidate this?
let me give me thought process
before jumping in
I'm thinking...Transcendence degree
I'm thinking...rationals
transcendence degree of those guys over rationals
I'm thinking linear function. and now I'm thinking nonlinear function
The wine is oaky...
22:00
is there a way to compute the number of closed branches of a cubic curve
or whether or not there is an isolated point
Real algebraic geometry is harder than algebraic geometry working over an algebraically closed field.
ok, so say, it is vigorous exercise and difficult, but we can do it?
@PM2Ring The stereographic projection is great for projecting the sky on a screen.
22:17
I doubt it.
woah, you'd think we'd have cubics figured out by now
At least one of Hilbert’s famous problem asks how many connected components a real algebraic curve has.
Or maybe how many the complement has.
lol i was wondering about that today and asked a question on mathSE about it, no wonder no one answered
22:35
shin: re the question math.stackexchange.com/questions/4687575/…, the notion you're asking about (if not clear from the above) is the number of 'connected components' of a topological space. it is a much more general notion than algebraic curves, e.g. you could meaningfully ask what it is for any subset of R^n whatsoever.
oh anak had mentioned something about that too
thanks, appreciated
so it's not really making that concept rigorous that's difficult. it's computing it.
well, depending on what kind of data you're given to determine the subset of interest.
do you typically study that in topology?
can be very hard. even for stuff given by algebraic equations with integer coefficients, this might be (as ted indicated) roughly equivalent to "hey, this subject 'algebraic geometry,' is there some way of doing it?"
it comes up a lot in topology. e.g. it is captured by the "zeroth homology group" at least for nice spaces.
very cool, thanks a lot for the search term
22:38
which isn't to say that that's a particularly useful observation for computing the thing. it's basically just restating the definition.
i was hoping there were exercises related to the idea somewhere i could at least practice or begin to treat the subject
so the search term helps
if you know about equivalence relations and the notion of a 'path' between points in a space (a continuous map from [0,1] into the space that maps 0 to one point and 1 to the other), you can define an equivalence relation on any subset of a topological space by saying two points are related if there's a path from one to the other. the number of connected components is, for nice spaces, the number of equivalence classes that this breaks the space into.
and you can actually compute with that definition, somewhat, and rigorously, if you have tools like the intermediate value theorem. it gives you 1 for {(x,y) in R^2: y = x} and 2 for {(x,y) in R^2: y = 1/x}.
@leslietownes For certain definitions of "nice". :P
of course mr. fractal man has to be fractious about it.
@leslietownes I could give a rat's tuchus for your semi-locally, quasi-path, almost-connected spaces, or whatever the right string of adjectives is.
Give me pathology, or give me death!
22:57
at leslie: noted
23:25
connected, locally path-connected and semi-locally simply connected

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