Mathematics - 2024-01-02

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Jakobian

00:03

Differentiable curve

Differential geometry of curves is the branch of geometry that deals with smooth curves in the plane and the Euclidean space by methods of differential and integral calculus. Many specific curves have been thoroughly investigated using the synthetic approach. Differential geometry takes another path: curves are represented in a parametrized form, and their geometric properties and various quantities associated with them, such as the curvature and the arc length, are expressed via derivatives and integrals using vector calculus. One of the most important tools used to analyze a curve is the Frenet...

I'm looking at the natural parametrization here, but why would this be smooth?

Is there another way to make $\|\gamma'\| = 1$?

Ted Shifrin

No. If you start with a $C^k$ regular parametrization, the reparametrization is also $C^k$, I think. What do you mean by smooth?

Jakobian

infinitely differentiable

Ted Shifrin

Then you need to start with smooth.

Jakobian

user image

because I want to do this

Ted Shifrin

He’s starting with smooth.

Jakobian

00:11

ah, yeah I'm starting with it

Ted Shifrin

And nowhere vanishing derivative.

Jakobian

so $t$ will be a diffeomorphism?

Ted Shifrin

What’s $t$?

Jakobian

function $t(s)$ on wikipedia

or $s(t)$ because they're inverses

Ted Shifrin

It’s the inverse function of $s$, which is smooth.

Jakobian

00:14

ah, yes

ah, because its non-zero, the norm is differentiable. I see

thank you

Ted Shifrin

Sure

1 hour later…

copper.hat

01:25

@TedShifrin Happy $\binom{24}{3}$

Ted Shifrin

01:36

I’ll guess what that means. How’s the body?

HNY to you, too.

Jakobian

01:55

happy tuesday

Ted Shifrin

02:17

Noch nicht!

PM 2Ring

02:27

Also $2024=2^{10}+10^3$

We're almost at perihelion.

user image

Times are in UTC, with a 6 hour timestep.

Some Japanese funk to help Joe recover from his operation.

WHAT IS HIP? [ TOKYO GROOVE JYOSHI COVER ]

leslie townes

03:23

ted: 2024 is a tetrahedral number, but we don't yet have 2024 golf balls... :(

Ted Shifrin

04:00

I’m still waiting for 6 to get used :)

3 hours later…

one potato two potato

07:06

@Jakobian I used a pdf translator online but it sucks.

3 hours later…

Alessandro Codenotti

09:55

@Jakobian Does the topologist's sine curve without the last point in the tail still have the FPP?

No it doesn't

I think a modification of your example in which there is a segment and copies of the topologists's since curve sticking out instead of more segments will work though (realized as a subspace of $\Bbb R^3$ say)

one potato two potato

10:19

Suppose $f:\Bbb R\to\Bbb R$ is a continuously differentiable function such that $\lim_{x\to\infty}(f'(x)+f(x)) = 0$. Then $\lim_{x\to\infty}f(x) = 0$.

1 hour later…

Mad

11:23

Regarding the differential of a smooth map between manifolds, one says it is a map between $dF_p :T_p M \mapsto T_{f(p)} N $ given by the definition $ dF_p(v)(f) := v(f \circ F) $ i do not consider this definition mathematically accurate and clean at all! i think it should be written in a way different manner, i am thinking something of sorts:
$ v \mapsto v ( - \circ F)$ which is now notationally more accurate in my opinion, since the element on the right is truly a member of the tangent target space, do you agree, that i defined this correctly, if not why?

In addition, following this one would write \\
$ v ( - \circ F ) : \mathcal{F}_{F(p)} \rightarrow \mathbb{R} \\ g \, \text{some function defined on a neighbourhood of F(p) } \mapsto v (g \circ F) $

And the final form of $ v (g \circ F ) := \sum_i v(x_i) \partial (g \circ F) / \partial x_i $ at p

where are $x_i$ local coordinates, do you agree? how would one find the form of $v(x_i)$

i guess using the same definition but by inserting the form of the local coordinates right?

IE $ v (x_i):= v(\phi_i) := \sum_j \partial \phi_i / \partial u_j $ where as the $u_j$ are standard coordinates and thus standard partial derivatives in R^n and Phi is some chart

1 hour later…

Jakobian

12:46

@AlessandroCodenotti oh yeah, that might work

Thank you

Jakobian

13:19

Oh actually they give me an example in the plane as well

Alessandro Codenotti

Where?

Jakobian

Fixed point properties by Connel but maybe its easier to describe

In the vertical strip of the topologist sine curve take only one point

This has the FPP

Alessandro Codenotti

Ah of course, the same proof as for the topologist's sine curve works for the FPP

Thorgott

13:39

@Mad I don't see the difference

Soumik Mukherjee

@Jakobian Do you mean the $\{0\}\times[-1,1]$ part?

Jakobian

@SoumikMukherjee yes

Soumik Mukherjee

Okay

1 hour later…

Mad

14:46

@Thorgott is not the first one an element of R not the tangent space.

Thorgott

you define a map by specifying its values

$dF_p(v)$ is a tangent vector of $N$ at $f(p)$, so it is a derivation on smooth germs at $F(p)$ and this derivation is given by mapping a germ $f$ to the real number $v(f\circ F)$

Mad

when you write v(f circ F) this here is a member of the reals.

I am talking about notation, if you write this like that, then the set on the right should be the reals , not the tangent space of N at f(p)

Thorgott

@Mad yes

@Mad no, you're mixing things up

$dF_p(v)$ is an element of $T_{f(p)}N$, $dF_p(v)(f)$ is an element of $\mathbb{R}$

Mad

Exactly!

the definition uses f in the definition and explicitly writes out $v(f \circ F)$ where they mention the sets above them being the tangent space, this is why i am saying, i dont think this is accurate notation, hence my suggested notation

Thorgott

they are saying that $dF_p(v)$ is the element of $T_{f(p)}N$ defined by $dF_p(v)(f)=v(f\circ F)$ (for all smooth germs $f$ at $F(p)$). this is completely fine.

there's nothing wrong with your suggested notation either, for the record, but they're both ways of saying the exact same thing

Mad

15:00

Yes but you are wording it differently.

if it was worded like this i wouldnt even bother writing here, since it confused me for a while, now my confusion has been cleared up.

Thanks!

Thorgott

15:13

glad it's cleared up

Jakobian

15:33

@TedShifrin It has a geometric flavour to it, I think you might like it

Thorgott

15:49

I have a hard time believing Ted hasn't seen this proof

In dimension $1$, geodesics parametrized by unit speed unique (up to orientation). This implies a connected $1$-manifold is the image of such a geodesic and the geodesic is either injective or closed. Then it's an analysis by cases.

leslie townes

16:40

i think it's an exercise in spivak? it's an exercise somewhere. surely he's seen it.

it may have fallen out of his memory when he upgraded from punch cards to magnetic media

psie

I have a basic question. I'm studying series of the form $\sum_{k\in\mathbb Z}f_k(x)$. First of all, I'm confused about where we start summing from. With the index set being $\mathbb N$, we start mostly from $0$ or $1$, but here I'm unsure. Should one think of $\sum_{k\in\mathbb Z}f_k(x)$ as $\sum_{k=0}^{\infty}f_k(x)+\sum_{k=1}^{\infty}f_{-k}(x)$? For the Weierstrass M-test, do we need a sequence such that $|f_k(x)|\leq M_k$ for all $k\in\mathbb Z$?

Jakobian

@psie context?

psie

Fourier series.

Jakobian

Then $\lim_{N\to \infty} \sum_{n=-N}^N f_k(x)$ is correct, if I recall

leslie townes

psie if the things being summed are nonnegative, any sensible way of interpreting the sum will coincide with any other way (which could be thought of in an index and order free as "the supremum over all finite sub-sums"

if the things being summed are more general, you will need to actually pay attention to hypotheses of theorems for how people are using that notation

Jakobian

16:48

You think of it as limit of $s_N(x) = \sum_{n=-N}^N f_n(x)$ in this case

leslie townes

if you have something like an M-test (i.e. with those absolute values in it and a sum M_k that is convergent) that dominates everything, it won't matter

Jakobian

its a special thing for Fourier series

psie

ok

leslie townes

as jakobian says in a lot of specific fourier series contexts, the symmetric partial sums are considered for various reasons

including because they correspond to a particularly well studied family of kernels

Jakobian

you want to give it this special interpretation

leslie townes

16:49

but its not like some general rule that it's always what that means, even people doing trigonometric series will consider other kinds of sums, so you have to pay attention

psie

I will pay attention :) the book I'm reading just suddenly introduced $\sum_{k=-\infty}^\infty c_k e^{ikx}$ out of nowhere...

leslie townes

you see a miniature versions of this with sums indexed by nonnegative integers or positive integers, where if its absolutely convergent any rearrangement gives you the same thing, but if it's not, "a + b + .... + z + ...." without parentheses in it can be ambiguous and "sum_{n=1}^{infty} a_n" is with reference to a specific indexing and limiting process

Mad

17:03

for given vector fields when one writes $X_p Y $ is this notation or some kind of algebraic multiplication of vector fields on a manifold?

X_p is a tangent vector at P. we define a VF X or Y as assignment of points P to tangents.

Thorgott

context?

Mad

Vectorfields :DD i think its notation. (defining the space of Vectorfields with the anti commutator as a Lie algebra)

Thorgott

ah ok, then they mean by $X_pY$ the function that takes a smooth function $f$ to $X_p(Yf)$ (here, $Yf$ is the smooth function $q\mapsto Y_qf$)

Mad

i see yes. you are right, well versedi n this subject i see.

Ted Shifrin

17:57

Warning: $XY$ is not again a vector field when $X$ and $Y$ are.

Only the commutator is.

@Jakobian I read through it when I was an undergraduate, but I've never taught it. Like Sard's Theorem, I just quoted the result as I didn't find the proof that interesting. Milnor uses arclength; Guillemin and Pollack use Morse theory.

1 hour later…

Jakobian

19:27

@TedShifrin I find it more interesting than verifying that $f^{-1}(y)$ is a smooth manifold with boundary for a regular value $y$, for example. Its less technical and more geometric (but formalizing the details seems to be a little annoying here).

definitely something that would be nice to sketch the argument of, in my opinion

maybe not necessarily the whole proof in detail

copper.hat

@PM2Ring thanks :-) just saw it by accident while scrolling back!

robjohn

19:56

@PM2Ring tomorrow

PM 2Ring

@copper.hat No worries. :) I suppose I should've pinged you. You might get a chuckle from this skit: chat.stackexchange.com/transcript/36?m=64919701#64919701

copper.hat

@PM2Ring brilliant!

PM 2Ring

@robjohn In about 4 hours. But I tend to consider the perihelion of the Earth-Moon barycentre as more significant, for the reasons I give here: astronomy.stackexchange.com/a/49546/16685

@copper.hat Eleanor is from Scotland. One of her recurring characters is Craig the unenthusiastic tour guide.

Scottish Tourguide for the Museum of Childhood Doesn't Give a F***

Sha Vuklia

20:36

Let $G$ be a compact topological group, let $M(G)$ be the space of Radon measures on $G$, and for any $f:G\to \mathbb C$, let ${}_x f(y)=f(xy)$. We define $L_x:M(G)\to M(G)$ by $L_x(\mu)(f)=\int{}_x f d\mu$ (invoking the Riesz representation thm). My book now claims that $L_x L_y=L_{yx}$. To me it seems it goes the other way around: $L_x L_y=L_{xy}$, because $L_x(\mu)=(L_x)_*\mu$ (the push-forward measure) by the following theorem:

user image

And taking the push-forward measure is a covariant functor

Sine of the Time

$\{e_n\}_{n\in \Bbb N}$ is an orthonormal basis of a Hilbert space $H$, consider the map $T(x)=\sum_{n\ge1}\frac{\langle x, e_n \rangle}ne_n$. Prove that $T(H)$ is dense in $H$ but it does not coincide with $H$.

To prove the second part, I was trying to show $T(H)$ is not closed

any ideas?

1 hour later…

Xander Henderson

21:44

So, like, the Taylor series for the exponential function converges pretty quickly, right? With that happy little $n!$ in the denominator?

Would you call that a Taylor Swift series?

PM 2Ring

21:55

Of course. The primorial reciprocals converge even faster. math.stackexchange.com/q/4817726/207316

Jakobian

@SineoftheTime Let's take $y = \sum (1/n^{1/2+r})e_n$, $0 < r < 1/2$

PM 2Ring

Taylor Swift PROVES that the square root of 2 is irrational

Sine of the Time

I'm trying to compute $T(y)$

Jakobian

No, try to solve $T(x) = y$

Sine of the Time

ok

$\frac1n\langle x,e_n\rangle=\frac1{n^{1/2+r}}$

copper.hat

22:05

@PM2Ring it's a bit sad, but my ability to discriminate accents has diminished considerably

PM 2Ring

22:23

@copper.hat Don't worry. Eleanor does a variety of accents, and freely admits that her natural accent is a bit strange. Dave Huxtable is a polyglot and accent expert. This skit gives a glimpse of his talent.

Hilarious English Accent Spoof Pharma Ad

1 hour later…

leslie townes

23:23

@ShaVuklia i always get stuff like this wrong (and even books/notes seem to be a coin flip on this) but your analysis feels right. at least, a perhaps more common convention for x acting on a function f(y) in this situation would be with f(x^{-1} y)

because it makes the multiplication law work out for what you'd expect for a left action

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