Mathematics - 2018-04-22 (page 1 of 2)

Well, that seems reasonable, but wasn't the computation I had in mind.

I'm suggesting that you want to factor out leading behavior and see what's left.

"Ah, I see," said the blind man.

I get that gist, I'm just not sure how that connects to derivative information. I'm sure it does, but I don't exactly know how.

Goodness, I've found the paper! Yay!

Well, what if you factor out $x$ from $x^{1/3}(1-x)^{2/3}$?

Now I'm partying!

PVAL-inactive

12:06 AM

Hi @Ted

Heya @PVAL

Oh my god, I think it's clicking now, lol.

Hey @PVAL!

The reason the name Lehmer seemed familiar is that one of the theorems in the paper is called Lehmer's theorem

He was a reasonably famous number theorist, at Berkeley a number of years ago.

12:08 AM

Did someone say number theory?

Whoops.

Eric Silva

@Daminark lol nerd

This comes as news to you, Eric?

Eric Silva

nah

Makes sense.

PVAL-inactive

12:12 AM

Gotta keep up the mental abuse once you start it.

To whom was that addressed, @PVAL?

The theorem concerns the degree of $\cos(360 k/n)$ and $\sin(360 k/n)$.

PVAL-inactive

at Eric I guess but not seriously obviously.

@TedShifrin I've hit another snag. Going to dinner now though. Rest assured this is going to be the only thing on my mind, lol

Well, don't be more antisocial than usual, @Fargle. I'm leaving shortly, too.

12:13 AM

@Daminark I find Number theory beautiful (although I'm beginner).

@TedShifrin I'll ping you when I either crack it or quit.

Take care chat!

I still have relatives visiting, @Fargle, so won't be around predictably ...

But OK.

See you @Fargle!

@user537566 I too am very new in this subject. I really just know a bit of Weil's "Number theory for beginners" book plus stuff that came up in algebra, but it's definitely something I like quite a bit

At least at this level

@Daminark That's the one that mentions Haar measure in the first page lol.

Nope, that's "Basic Number Theory"

12:25 AM

hello

@Daminark Oh makes sense! I was like, watch out we got a modest expert over here!

I just learned a bit from Niven's book -- although I skipped a lot of problems (because there is way too many of them).

Hey @0celo7, what's up?

Ah yeah I've been told Niven is supposed to be good

@Daminark stuck on the PMT in multiple places, so starting on my summer project

you?

Yeah, but it has too many problems. It sounds a weird thing to complain about, since that can surely only be a good thing, but the point of the exercises is to make one digest the subject matter fully, and if there are too many problem you don't know if the problem you didn't do is one of the more important ones designed to drive the point home, if that makes sense?

How's Weil's book?

Heyo people

12:31 AM

@user537566 Conway's functional analysis book probably has 1,600 exercises (Fermi calculation) and it's really hard to tell which are important

I liked it a good bit. Compact, problems were fun, etc. I did a few chapters at the beginning over the summer, but then when I took algebra I think a lot of later parts in the book were subsumed by ring theory

Stuff like Fermat's theorem on the sum of two squares, $\mathbb{Z}[i]$ being a PID, all that good stuff

@Daminark I am reading some insane italian dude's treatise on level set flow right now

@0celo7 what's PMT?

positive mass theorem

I see

12:34 AM

@0celo7 I thought I was crazy for having this issue with the book! Thank you!

Hmm, let me guess, is this Minicozzi?

@Daminark it's a GMT problem disguised as a Riemannian geometry problem disguised as a physics problem

@Daminark Bill Minicozzi is not Italian as far as I know :P

I mean an Italian Italian

Ah

@Daminark Sounds interesting. I'm gonna check it. The only Weil number theory book I had heard of was the infamous other one. I like older books too. I learned some unusual techniques from Landau's book (but they were not too advanced).

@Daminark This guy (well, not him, but the book is about it) does everything in terms of the distance to the boundary

it's a curious fact that everything in hypersurface geometry can be understood using the distance to the boundary

and this allows for weak formulations because it requires zero regularity

12:38 AM

But yeah on my side @0celo7, I'm finishing up the last bit of my notes on this Lipschitz graphs paper. I think I'll present that part on Monday and finally be done. At which point further efforts in learning GMT will be mostly relegated to actually reading the beginning of Mattila and learning the stuff I should've known a long time ago. Such as, how Hausdorff measure actually works :P

@Semiclassical You can also use \lnot, for "logical not".

Compare \land $\land$ and \lor $\lor$

@Daminark If you like this classical stuff, Maggi is good too

I'll be using that a lot this summer

And some insane German guy who rewrote the perimeter stuff to work on R. manifolds

\lnot,\neg $\neg$; \land,\wedge $\wedge$; \lor\vee $\vee$

Hmm, I'll check it out

$\lnot x$ vs $\neg x$

mmm... identical spacing... mmm...

12:44 AM

$\knot x$

$x \land y$ vs $x \wedge y$ --- $x \lor y$ vs $x \vee y$

>:(

the spacing works there, too

$\newcommand{\knot}{\ddot \smile}$ @AkivaWeinberger $\knot$

$\ddot\cup$

You can't put diareses on a capital U!

nor a cup!

12:49 AM

proof?

Ü

$\bar{\ddot\cup}$

you're doing that thing Balarka does

@0celo7 Make a unibrow?

making us worry about you

there's a 95% chance Balarka has a unibrow

I wonder if there's a decent way to find the hypervolume of $S^{n+2}$ from that of $S^n$.

12:59 AM

@AkivaWeinberger Induction

I think I'm asking for the induction step

ManishKumar Singh

why is R[X] as a R-algebra generated by <X> and not <1. X>

You think you are asking for the induction step? or are you asking for the induction step? Because if you aren't sure, there isn't much that I can do to help you...

Finitely generated algebra

It depends on what you meant when you said "induction"

If I had a way to get $S^{n+2}$ from $S^n$, I could use induction to get $S^n$ from $S^1$ or $S^2$.

I don't see how it's helpful for getting $S^{n+2}$ from $S^n$.

@ManishKumarSingh How does your book define a generating set?

So this is cool:

The red square has area 1. The rest of the colors are chosen to have total area 1 as well.

(and also be symmetric about the diagonal)

This uniquely determines each of the areas. Apparently, the shape approaches a circle as you use more colors.

You can see that the circle drawn in black only intersects the purple layer and none of the others.

ManishKumar Singh

@AkivaWeinberger I am using the definition given in wiki en.wikipedia.org/wiki/Finitely_generated_algebra

In mathematics, a finitely generated algebra (also called an algebra of finite type) is an associative algebra A over a field K where there exists a finite set of elements a1,…,an of A such that every element of A can be expressed as a polynomial in a1,…,an, with coefficients in K. If it is necessary to emphasize the field K then the algebra is said to be finitely generated over K . Algebras that are not finitely generated are called infinitely generated. Finitely generated reduced commutative algebras are basic objects of consideration in modern algebraic geometry, where they correspond to affine...

1:10 AM

Right, so everything in R[X] is a polynomial in X with coefficients in R. Remember that 1 is a polynomial.

ManishKumar Singh

so is there not a fixed definition for finitely generated algebra.

or I think I can resolve it, if X is a generator, am I allowed to take X^0??

Yes.

SK19

Hello there!

Woo basically done with GMT

An immeasurable achievement

1:19 AM

:thinking:

1:29 AM

That picture above seems not to be drawn to scale? Or am I missing something

Maybe I should mess around with Geogebra

(Each color is meant to have the same total area)

Oh never mind

It is drawn to scale, I just made an arithmetic error

1:43 AM

@BalarkaSen stupid question: why does chain homotopy need to safisfy $g-f=dP+Pd$ when the $Pd$ term doesn't contribute to the fact that the induced homology morphisms are identical?

I mean, can't we have $g-f=dP+2Pd$ instead?

2:16 AM

DogAteMy: Do you mean the surface areas of the spheres or do you mean the volumes of the balls? The latter is easy to do going up by $2$s. @Akiva

@TedShifrin Volume

Surface area of $S^n$ follows from volume of $S^{n+1}$ anyway so they're kinda equivalent

${\rm SA}(S^{n-1})=n{\rm Vol}(S^n)$

Unit sphere, anyway, otherwise put an $r$ on the left

So you should say $B^n$ or $D^n$ for the ball.

Think of $B^n$ as the $n$-dimensional "equator" of $B^{n+2}$. Then fixing a point of $B^n$ you get a $2$-dimensional disk as the cross-section.

So, actually, I lied. Better to think of a $2$D equator and take $B^n$'s of appropriate radii as the cross-sections.

2:46 AM

So I think it's something like $\int_{B_n}\pi(1-|r|^2)dA$?

Well, this is why I admitted I'd lied. Better to do a 2D integral in polar coordinates.

Whereas if I wanted to only find the volume of $B^{n+1}$ rather than $B^{n+2}$, it'd be $\int_{B_n}2\sqrt{1-|r|^2}dA$ or some such, which looks much more annoying

But, yes, you have the gist of the idea.

The $B^n$ integral you wrote down will be a pain, because you'll need $n$-dimensional spherical coordinates and it won't work out well.

Nah, it looks like a cylinder minus a double cone actually

but in higher dimensions

But it won't give you the recursive relationship nicely, whereas doing the integral over a 2D disk in polar coordinates gives it immediately.

2:49 AM

Hm, actually, I remember that geometric construction, for finding the volume of a sphere. I wonder if I can find pictures

I didn't realize it generalizes to higher dimensions

That goes back to Archimedes ... the volume of a 3D ball.

Anyhow, try my corrected approach and it'll work easily.

Oh, wait, not a cone…

It's not going to scale correctly in higher dimensions, DogAteMy.

@TedShifrin So, $\int_{B^2}{\rm Vol}(B^n)(1-r^2)dA$ or something? Or do I need an exponent somewhere

$^{n/2}$

Right.

But this will work out nicely, cuz polar coordinates.

2:53 AM

Right without the exponent?

No, right with the exponent.

${\rm Vol}(B^n)\int_0^{2\pi}d\theta\int_0^1(1-r^2)^{n/2}rdr$?

them swift edits though

Looks right.

heya @Fargle

Heya @Ted, @Akiva, @anyone-else-who-is-active

2:58 AM

What would this be if I wanted $B^{n+1}$ instead of $B^{n+2}$? ${\rm Vol}(B^n)\int_{-1}^1(1-x^2)^{n/2}dx$? That doesn't look right.

That's just the same thing as before but without the $2\pi$, so that can't be right…

No, it's not the same. You're missing an $x$.

Oh, you're right. I see.

And $\int x(1-x^2)^{n/2}dx$ is much easier with the $x$ than without, 'cause I can do substitution.

That was the whole point :)

It's $\dfrac1{n+2}(1-x^2)^{(n+2)/2}$.

Er, negative that.

And then from $0$ to $1$, gives $\dfrac1{n+2}$.

And with the $2\pi$ and the ${\rm Vol}(B^n)$, that's

${\rm Vol}(B^{n+2})={\rm Vol}(B^n)\dfrac{2\pi}{n+2}$

Can we check for $n=1$ and $2$?

3:04 AM

$n=0$ and $1$, maybe, I don't know the volume of $B^4$ off-hand.

${\rm Vol}(B^0)=1$ seems to make sense.

$\pi\stackrel?=1\dfrac{2\pi}2$

$\dfrac43\pi\stackrel?=\dfrac{2\pi}3$

Seems to work! And it predicts that ${\rm Vol}(B^4)$ is $\dfrac12\pi^2$.

Which Google confirms.

Yup, which be correct.

All right! I wonder if there's a more geometric way to do this.

Why the hell is it four?

There's also an interesting question, DogAteMy. Plot this function of $n$ and does the result surprise you?

Don't know how to plot it, but I know it starts to decrease to $0$ after dimension $5$ I think.

3:11 AM

@AkivaWeinberger do you know it? or did someone else tell you that?

I knew it before

There's also some bizarre question about whether $n$-dimensional unit balls fit inside $m$-dimensional unit balls for $m>n$.

From a probability perspective, this means: if we choose $n$ numbers randomly from $[-1,1]$, the odds that the sum of their squares is less than $1$ is much less than the odds that they're all positive.

@TedShifrin I thought the equator of S^(n+1) is S^n

I didn't state this right.

3:13 AM

You want cylinders maybe?

I know the bizzare result with a cube and $2^n$ and one spheres

I'll have to reconstruct the weird fact sometime.

$2^n$ spheres in the corners and one "small" (but actually big) sphere in the center

I'm remembering something that should fit inside is actually too big ...

The red sphere becomes arbitrarily large (larger than the cube!) as the dimension increases.

That's the fact I was thinking of. Yes.

3:19 AM

Finally got through this darn GMT paper

Are you giving a lecture or throwing a party?

While you were doing geometric measure theory, we were measuring some geometric things

I thought I was done earlier but I very much wasn't

Yeah I've been giving some lectures over the past few days but with classes I've barely been able to prepare just a few pages ahead of where I'd reach in lecture

Tomorrow's the last day, I finish off the paper

Well, not tomorrow, methinks.

Yeah, day after. Few years ago I got in the habit of "tomorrow" meaning "next class" and that just stuck

@AkivaWeinberger "geometric" and "measur(e/ing)" commute, don't they?

3:29 AM

Ted helped me prove that ${\rm Vol}(B^{n+2})={\rm Vol}(B^n)\dfrac{2\pi}{n+2}$ with integrals, but I'm wondering if there's a more geometric way.

Hello chat!

Wait that's strange, huh

Note, by the way, that it goes to zero, while the unit cube $[0,1]^n$ stays at volume $1$. (Note also that the cube containing the sphere is $[-1,1]^n$, so I'm comparing the sphere to something that's $1/2^n$ the size of its bounding box.)

Can any1 please suggest a good reference of the proof for tangent space at identity is isomorphic to the set of all left invariant vector fields of the Lie group $G$ ?, I am trying but the notations are blocking me away :'(

Can any1 guide me through the proof

Given a vector at the identity, translate the space to get a vector at any other point

3:32 AM

DogAteMy: I don't know one, so I'd be curious if you come up with something.

By translate I mean the map $x\mapsto gx$ for $g\in G$

@Baymax: If you define $X_g = L_{g*} V$ then $X$ is left-invariant.

Well, you need to differentiate that, DogAteMy.

Right, fine, the map $T_eG\to T_gG$ induced by that map

Correctomundo.

($T_gG$ meaning the tangent space at $g$)

3:35 AM

Thank you so much but Can we discuss the notations at first please?

like

$X_{g}$ is vector field of $G$ at $g$

Have you studied basic manifolds things?

yes

Right. I meant $X(g)$, if you prefer.

But most people use the subscript because you want things like $Xf$ where $X$ is a function.

What book are you doing this from?

3:38 AM

$L_{g} (X_{h}) = X_{gh}$ ?

Strange phenomenon when a guy express the pascel triangle in terms of digital roots

You need to differentiate $L_g$. That's what the lower star means.

Also, you can think of $S^3$ as a (trippy) specific example of a Lie group

@AkivaWeinberger actually I found these notes

I dunno if there's another easier to visualize

3:39 AM

I wonder if the residential number theorists here have some idea on what is happening

@Secret Digital root is just mod 9, right?

You get lots of weird patterns when you take Pascal's triangle mod something

@AkivaWeinberger@TedShifrin - bose.res.in/~amitabha/diffgeom/chap21.pdf

Mod 2 gives you the Sierpiński triangle

Not quite: $dr(n) = n- 9 \lfloor \frac{n-1}{9}\rfloor$

it has a floor component

…That's the same thing, no?

Only difference is you write "9" instead of "0"

3:42 AM

Not all free notes you find are good, Baymax.

But I'm not going to look at it.

Let me check how it equals to the form $n+9k$...

(so somehow, the floor function as taking the remainder, hmm...)

Well, it makes sense, in that $n-\floor n$ is almost like "mod 1" @Secret

Okay

let me try

what is $V$ Ted?

Where do you see a $V$?

ah right, it gives a similar looking graph too

so the search should be narrowed down to pascel triangle mod 9, hmmm...

3:48 AM

14 mins ago, by Ted Shifrin

@Baymax: If you define $X_g = L_{g*} V$ then $X$ is left-invariant.

The Sierpinski gasket lives in Pascal's triangle. There is a really lovely construction there that I understood at one point in my life.

I can't quite remember why it worked, but it is still very nice.

An element of the tangent space at the identity, or in other words, a vector at the identity @BAYMAX

Found a paper and an illustration

matwbn.icm.edu.pl/ksiazki/aa/aa78/aa7843.pdf

pappubahry.com/misc/pascal_tri/index.html

So we need to show isomorphism between $T_{e}G \mapsto T_{g}G$

@AkivaWeinberger

If we denote $L(G) $ the set of ll invariant vector fields of $G$

then we need to show the isomorphism b/w $L(G)$ and $T_{e}G$

hmm strange, it seems as n runs from 2 to 100, there's some repeated pattern that seemed to behave like the inverse of doubling

dymocks.com.au/book/…

arxiv.org/abs/chao-dyn/9610005

3:54 AM

Given a vector at $e$, you can get a vector at $g$. By doing this for all $g\in G$, you get a vector at every point, i.e. a vector field.

is it possible not only the Sierpinski gasket is a discrete chaotic system, but also the function $\text{Pascel Triangle }\mod n$? (to be investigated)

@AkivaWeinberger yes like taking $h=g$ tghen we can map from identity to any element by $e \mapsto eg$ ?

Yeah

And then by taking the differential of that map, you can take any vector at $e$ to a vector at $g$

how can we take the differential of the above map?

otherwise you can suggest a book and we can discuss the proof from that ?

4:12 AM

If $f:M\to N$ is any diffeomorphism between two differentiable manifolds, and $p\in M$, then you get a map $df:T_pM\to T_{f(p )}N$ from the tangent space at $p$ to the tangent space at $f(p )$.

Do Carmo's book Riemannian Geometry covers this in Chapter 0

(Chapter 0 is on differential geometry, which is needed for Riemannian geometry)

The way you construct that map is,

take a curve through $p$ whose tangent vector at $p$ is $v$.

In other words, you have $c:(-\epsilon,\epsilon)\to M$ with $c(0)=p\in M$ and $c'(0)=v\in T_pM$.

What you then do is take the image of this curve under $f$.

That gives you a curve $f\circ c$ that goes through $f(p)$. It's tangent vector there is $df(v)$.

You can also think of it as $\frac{\partial f}{\partial v}$.

${}\in T_{f(p )}M$.

I don't know a better reference

@BalarkaSen, what reference would you recommend for the above material? (Differential topology)

Let me understand this Akiva

Also

4 hours ago, by 0celo7

you're doing that thing Balarka does

4:31 AM

Hmm nice@AkivaWeinberger I am getting smthng now, can we proceed?

Sure

So given a point and a ccurve passing through it and a tangent to the curve at that point we found what will be the behavior when it will be under a map $f$, like its tangent vector will be $df(v)$ at the point $f(p)$

Now how will this help us?

So in our Lie group, we have the identity element $e$ and a vector at $e$ called $v$

Let $L_g:G\to G$ be the map $x\mapsto gx$.

This is a diffeomorphism. It's also bijective.

We can look at $dL_g(v)$. This will be a vector at $L_g(e)=g$.

So now let's make our vector field. We'll call it $X$.

I'll write $X(g)$ to mean the value of the vector field at the point $g$, so $X(g)$ will be a vector at $g$. That is, $X(g)\in T_gG$.

Define $X$ such that $X(g)=dL_g(v)$.

That will be your left-invariant vector field corresponding to $v$.

To go the other way: given a left-invariant vector field $X$, take $X(e)$ to get its corresponding element of the tangent space at $e$.

So now we have a correspondence between elements of the tangent space at $e$, and left-invariant vector fields.

We need to check that $X$ is actually left-invariant. We also need to check that $X$ is the only left-invariant vector field such that $X(e)=v$.

4:48 AM

@AkivaWeinberger $dL_{g}(v) : T_{e}G \mapsto T_{L_{g}(e)}G$

or $T_{e}G \mapsto T_{ge} G$

or $T_{e}G \mapsto T_{g}G$

am i correct?

Yeah

@AkivaWeinberger Till this we have shown that, given " ??" we have the left invariant vector field corresponding to $v$ ?

Hm?

Sorry, I need to go to bed now

See you in some number of hours

sure akiva, thank you very much for your time andd explanation!

Nicholas Roberts

5:16 AM

Hey can someone clear something up for me quickly? The question is to compute the galois group of $x^n - t$ over the field C(t) where t is an independent variable

C denotes complex numbers. I dont understand the field C(t).

Since C is already algebraically closed, i dont understand why one would compute the galois group over C(t)

@NicholasRoberts C(t) is the field of rational functions in t with complex coefficients

Nicholas Roberts

Oh wow, ok. that helps lol

5:33 AM

@NicholasRoberts indeed, C is algebraically closed means that every finite extension of C is C, but one can still have infinite extensions

Nicholas Roberts

Oh ok, so this is an infinite exentension?

yes

Swapnil Das

5:52 AM

Hi,

I just studied that any odd degree polynomial has range of R. Can anyone provide/link any proof of the statement?

Q: Polynomial functions of odd degree are surjective

So do you buy that if a polynomial $p$ has odd degree (assume without loss of generality that the leading coefficient is positive, otherwise flip the signs and nothing changes), then $\lim_{x\to\infty} p(x) = \infty$ and $\lim_{x\to -\infty} p(x) = -\infty$?

@SwapnilDas

Swapnil Das

Yup, I got an answer from an MSE post. Thanks for responding :)

2

Prove if the function $f: \mathbb{R} \to \mathbb{R}$ is a polynomial function of odd degree, then $f(\mathbb{R}) = \mathbb{R}.$ We know a polynomial, $f(x)=a_nx^n +a_{n−1}x^{n−1} ...a_1x+a_0$ with real coefficients is continuous. Also, $\mathbb{R}$ is connected now since $\mathbb{R}$ is connecte...

It's a shame that we were asked to memorize this :(

DarkVampiric AbstractArtist

6:55 AM

Anyone have a good grasp on the symmetry of a pentagon? Some work relating to Abstract Algebra by the way.

Perturbative

7:53 AM

Hey everyone

There was this cool question on my algebra test yesterday that I couldn't quite get

It was to find all the ring homomorphisms from $(\mathbb{Z}, +, \cdot)$ to $(\mathbb{Z}, +, \cdot)$ where $+$ and $\cdot$ are the usual addition and multiplication on the integers

Silent

@LeakyNun, is it rue that if $p(t)$ is characteristic polynomial for matrix n by n $A$, then $-p(t)$ is characteristic polynomial for $A$, too?

Shobhit

Hello. I am reading product topology from wiki, and somewhere in there it says,$ \{(x,y) : xy=1 \}$ is closed, now there the definition of open set in a product space is also given, but i am not able to show that the above mentioned set is closed. Help please.

Perturbative

8:17 AM

@Shobhit There must be some error, if you have two topological spaces $X$ and $Y$ and you consider the cartesian product $X \times Y$ endowed with the product topology, and you pick an element $(x, y) \in X \times Y$ since there's no algebraic structure on it $xy= 1$ doesn't make any sense

Shobhit

@Perturbative sorry, $x,y \in R^2$

Perturbative

@Shobhit Maybe take a look here : math.stackexchange.com/questions/367155/…

$\{(x,y ) \in \mathbb{R}^2 \ | \ xy = 1\}$ is the graph of the function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \frac{1}{x}$ which is continuous on $\mathbb{R} \setminus \{0\}$

Balarka Sen

@AkivaWeinberger Guillemin-Pollack is a good book

@0celo7 I wish I did, though

Shobhit

@Perturbative thanks :)

Balarka Sen

its 2 indian

8:38 AM

@BalarkaSen did you see my question?

@Silent not really. $p(t)$ is defined to be $\det(tI-A)$ or $\det(A-tI)$ (depending on convention), so it is a specific polynomial

in the first convention the leading coefficient is always $1$, in the second convention $(-1)^n$

@DarkVampiricAbstractArtist yes?

DarkVampiric AbstractArtist

8:51 AM

@balarka imagine a world where there's a book on diff top that's TeX'd

Mary Star

9:06 AM

I want to show that $\|x\|_p\rightarrow \|x\|_{\infty}$ for $p\rightarrow \infty$.

I have shown that $\|x\|_{\infty}\leq \|x\|_p\leq n^{\frac{1}{p}}\|x\|_{\infty}$.

To show the above limit I have done the following:
$$\lim_{p\rightarrow \infty}\|x\|_{\infty}\leq \lim_{p\rightarrow \infty}\|x\|_p\leq \lim_{p\rightarrow \infty}n^{\frac{1}{p}}\|x\|_{\infty} \\ \Rightarrow \|x\|_{\infty}\leq \lim_{p\rightarrow \infty}\|x\|_p\leq \|x\|_{\infty} \\ \Rightarrow \lim_{p\rightarrow \infty}\|x\|_p=\|x\|_{\infty}$$

9:23 AM

@LeakyNun Being that a pentagon has ten symmetries. Could I propose these symmetries as e (identity), m1 to m5 (reflections), and what else could i make up?

DarkVampiric AbstractArtist

rotations

I guess I'll try r1=36, r2=72.

+ r3=108, r4=144, and e=360 (identity). Thanks, Leaky Nun.

I'm not weary of functional analysis and measure theory, sorry Mary. I tried solving it, but it's too much for my level.

Balarka Sen

9:57 AM

@LeakyNun nope

@Daminark Lee

well Lee is really a smooth manifolds text than a differential topology text, but close enough

10:12 AM

I didn't look at it much but the vibe that I got from Ted was that Lee was very different in spirit from, say, Guillemin-Pollack and Hirsch

Stuff like transversality and degree apparently just don't happen there

9 hours ago, by Leaky Nun

@BalarkaSen stupid question: why does chain homotopy need to safisfy $g-f=dP+Pd$ when the $Pd$ term doesn't contribute to the fact that the induced homology morphisms are identical?

9 hours ago, by Leaky Nun

I mean, can't we have $g-f=dP+2Pd$ instead?

@BalarkaSen ^

Mary Star

10:31 AM

Let $X = \{f \in C (\mathbb{R}): f (t) = f (t + T) \ \forall t \in \mathbb {R} \} $ be the space of the $ T $ periodic continuous with the maximum norm $ \displaystyle {\| f \|_{\infty} = \max_{t \in \mathbb {R}} | f (t) |} $.

I want to show that $X$ is a Banach space.

That means that $X$ must have a norm, which is satisfied, and that it is complete, right?

How could we show that?

@MaryStar well you already have a norm :P

right, then you just need to show that it is complete :P

(wt am i saying

I would say that it is a subspace of $C(\mathbb{R})$. This is because periodicity $T$ (I hope this is the right word) is preserved under scalar multiplication and addition.

just identify it with a subspace of $C([0,T])$ already

Yeah, now it remains to show that Cauchy sequences converge within this subspace.

Mary Star

I got stuck right now. Why is it a subspace of $C([0,T])$ ? @LeakyNun @lattice

10:47 AM

I believe you can prove this straightforward. Let $\{f_n\}$ be some Cauchy sequence in $X$, then the sequence converges to some $f\in C(\mathbb{R})$ (since this is a Banach space). Now assume that $f$ was not $T$-periodic, take some $t\in\mathbb{R}$ for which $f(t)\neq f(t+T)$ and show that this yields a contradiction to $f$ being the limit of $T$-periodic functions.

Well, it is not really a subspace of $C([0,T])$, but you can identify it as such. A $T$-periodic function is uniquely determined by its values on the interval $[0,T]$, so $X$ is isomorphic to the space obtained by restriction to $[0,T]$.

But either way (identifying $X$ with $C([0,T])$ or just considering it as a subspace of $C(\mathbb{R})$), the important thing is to show that limits of $T$-periodic functions are $T$-periodic. And this can be done straightforward as above, I think.

Silent

@LeakyNun thank u!

@lattice it isn't the entirety of $C([0,T])$

It is $C([0,T])/\Bbb R$ as a vector space

@LeakyNun Right! I just meant it to be a subspace^^ So $X$ is not isomorphic to $C([0,T])$ but to $C([0,T])/\mathbb{R}$ as you say (because we need to guarantee continuity).

But isn't this too much trouble for this question?^^ Why not just consider it as a subspace of $C(\mathbb{R})$?