Mathematics - 2020-04-13 (page 2 of 2)

space

17:00

What can I do in this situation: someone asked a question on Math.SE and received an answer. Someone else posted a comment to that answer asking to clarify something but has received no answer yet. I would like to comment in an attempt to clarify it but I don't have enough reputation points. Is there no other way for me to help that person? (Besides farming reputation...)

Apologies for the silly question.

Tobias Kildetoft

17:17

@space Not sure there is much else you can do, unless the clarifications would make sense as an additional answer to the question.

space

@TobiasKildetoft I don't think it would make sense as an additional answer. The answer itself is good and answers the question since OP accepted it and has no further questions.

Tobias Kildetoft

@space If you point me to the answer, I can make a comment inviting the commenter to this chat in case they are still around

space

https://math.stackexchange.com/questions/103081/taylor-expansion-of-functional

my comment would be an answer to the comment posted at the first answer.

geocalc33

Why is 6 afraid of 7?

space

7 8 9

Balarka Sen

17:23

and 6 9

Tobias Kildetoft

@space Ahh, that comment is more than a year old. It doesn't really make sense to do much about it now

space

Fair enough, thanks for your time though

geocalc33

What happened after 6 9'd 7

7 got ir8

space

Is that what mathematicians do when they don't do math?

3

geocalc33

no

no?

Mike Miller

17:32

@BalarkaSen I need good math to think about

No more bad math

Balarka Sen

@MikeMiller Explain to me why if you have a constant negative curvature surface in R^3 and you parametrize it by $\phi$ so that the asymptotic directions are coordinate curves then the surface normal is parallel to $\partial^2 \phi/\partial x \partial y$

Mike Miller

No way

Balarka Sen

I don't see the picture

It's a computation that uses $(u \times v) \times w = (u \cdot w) v - (v \cdot w) u$

Black magic stuff

Tobias Kildetoft

@BalarkaSen Well, that is part of the definition of a cross product algebra :)

Balarka Sen

It's a crucial lemma in Hilbert's proof that there are no constant negative curvature complete smooth surface in R^3

Lol @Tobias

Tobias Kildetoft

17:38

(I found the review I mentioned)

Balarka Sen

but Hilbert man

what a nutcase

Knight

@Sebastiano I don’t want to say anything about the users of Phyiscs.SE.

Balarka Sen

The basic idea is that any such surface admits a nice parametrization whose coordinate curves are always asymptotic and the rectangles formed by the asymptotic curves are geometric rectangles (opposite sides have equal length)

And then you just compute by taking larger and larger rectangles that the volume must be bounded by $2\pi$

Garbage, hyperbolic plane has infinite volume

geocalc33

there's people in my vicinity trying to calculate the area of a circle

Knight

@geocalc33 What are their methods?

geocalc33

17:41

um

I heard "but why would you multiply the radius by the diameter?"

I hope they succeed

Knight

God help them!

Is Calculus and Analyticla Geometry allowed in that discussion?

geocalc33

How do you analytically continue $f(x)=\sum_{n=1}^\infty n^{\frac{1}{\log_n(x)}}$?

i've gotten stuck because I don't understand how to define split complex numbers within the logarithm. I'm guessing I need to choose the proper branch?

@Knight I'm not entirely sure!

I rewrote the sum as $f(x)=\sum_{n=1}^\infty e^{\frac{\ln^2(n)}{\ln(x)}}.$ from inspection one recognizes this as the sum of (nonlinear) hyperbolas $\ln(x)\ln(y)=\ln^2(n)$ nonlinear in the $x-y$ coordinate plane. transferring to $u=\ln(x)$ and $v=\ln(y)$ yields regular rectangular hyperbolas clearly

and the sum only converges because of the nonlinearity of the hyperbolas. Redacting to u-v coordinates yields a divergent sum

but the series is so difficult

Mike Miller

18:04

I don't see a picture either @BalarkaSen

Balarka Sen

I suspect it might be very complicated to see because in the argument you use constant Gaussian curvature crucially

I only understand asymptotic directions as those two little lines you get when you intersect the tangent plane with the surface

Mike Miller

Too hard

Leaky Nun

@BalarkaSen Evert the exotic 4-sphere

Mike Miller

none embed in $S^5$

Balarka Sen

18:22

Even if I didn't know the fact Mike said I am pretty sure it's a garbage question because isotopy classes of immersions of M in N are determined by bundle embeddings of TM in TN, and tangent bundles of exotic spheres are no different than usual spheres

Mike Miller

Trying to remember why none embed in $S^5$. Can't just use h-cobordism

geocalc33

does analytic continuation even exist for split complex stuff lol

Mike Miller

@BalarkaSen You might like this

Sebastiano

@Knight Why I'm very well in TeX.SE and sometimes in Math.SE? There is too much rigor and there are many who make unnecessary controversy. Have you seen my scores? I don't care about my scores. The negative scores, however, are also due to hater and deprive me of asking questions.

Balarka Sen

Suppose $M$ is the exotic $4$-sphere, and $M \to \Bbb R^5$ be an embedding. I claim this is not isotopic through immersions to $M \to \Bbb R^5 \to \Bbb R^5$ where the last map is the antipodal map. By Smale-Hirsch theorem, $\text{Imm}(M, \Bbb R^5) \cong \text{Emb}(TM, T\Bbb R^5)$ which is classified by $\pi_4(SO(5)) \cong \Bbb Z_2$, and if you chase the map you'll see those two maps give the two distinct elements

So regardless of if $M$ embeds in $S^5$ or not the answer is the same as the usual 4-sphere

@MikeMiller !

Looking

Mike Miller

18:32

I can't get access

Balarka Sen

Oh strange I can

I want to learn the proof of quarter pinched sphere theorem eventually

the topological version is a simple corollary of Rauch comparison theorem i believe

@MikeMiller Oh wow ok: $M$ is a Riemannian manifold such that cut locus of $p$ is at a constant distance away from $p$ if and only if $M$ is a disk glued to a closed disk bundle over a manifold with spherical boundary

$p$ is just some point I am sure

not every

Mike Miller

What

Balarka Sen

That should totally be Morse theory with the distance^2 function

Mike Miller

Link

Balarka Sen

it's in the thing you linked

Mike Miller

18:43

send me a pdf elsewhere

Balarka Sen

yup ok

Sent

Mike Miller

I will obtain it at some point

Balarka Sen

you can flow along radial geodesics man im telling you

im pretty sure gromov should have noticed this let me google

AHahahah

Curvature, diameter, and Betti numbers

M Gromov

Mike Miller

There are very few examples of disc bundles whose boundary is a sphere

Balarka Sen

I know the tautological bundle - not much else comes to mind

Mike Miller

18:52

Sure, both real, complex, quaternionic, and octonionic over $S^8$

I think other than some things ht. eq but not diffeo to these, that's it

Balarka Sen

Wow.

Mike Miller

Surely we can prove this.

Balarka Sen

Yeah OK we want spheres bundles with total spaces spheres

Mike Miller

Suppose $S^d \to S^k \to M$ is a fibration. The Gysin sequence gives us $e: H^*(M) \to H^{*+d+1}(M)$, cup product with a specific class $e(F)$. Gysin should say that $e$ is an isomorphism except in top degree, I think

That would give $H^*(M;R) = R[e]/e^c$, where $|e| = d+1$ and $c = (k+1)/(d+1)$

(I guess there is some fiddling here with whether or not this is an oriented bundle. If it's not oriented, then $\pi_1 M \neq 0$, so we must have $k=1$ or $d=0$; the $k=1$ case is so small it's silly, and the $d=0$ case can only give things homotopy equivalent to $\Bbb{RP}^n$.)

Then our base has cohomology $\Bbb Z[x]/x^c$ for some $c$. I'm pretty sure this is only possible when $|x| = 1,2,4,8$ and when $|x|=8$, for $c=2,3$

And all such spaces are ht.eq to projective spaces

Balarka Sen

Sounds like Adam's theorem

Mike Miller

19:01

I think it's a Steenrod square computation

Ah ok

Balarka Sen

too hard

Mike Miller

Let me say it anyawy

The point I think is the Adem relation

Oops no time now

Ted Shifrin

Howdy, a @Balarka, MikeM.

Balarka Sen

Hi @Ted

Stan Shunpike

19:16

hello

Simone

Hello

Ted Shifrin

heya @Stan

Alessandro Codenotti

Hi Ted

Ted Shifrin

Hi, demonic @Alessandro

Masterphile

i am again refreshing some basics of real analysis, but i am not sure should i now first observe the whole set-theoretic construction of R starting from N, or that again I choose axioms for R for granted to be a starting point, the whole construction, although needed, is a bit tedious

Ted Shifrin

19:28

Good grief. If you go back to the set-theoretic beginning, you'll never get to analysis. Just grant the real numbers with the least upper bound property.

Masterphile

yes, i usually do that

Ted Shifrin

Once you know about Cauchy sequences, you can equivalent think of $\Bbb R$ as the completion of $\Bbb Q$ (the set of equivalence classes of Cauchy sequences).

Masterphile

i know about them, they are "used" also to complete some more general spaces if i remember correctly

Ted Shifrin

Sure. Any metric space.

Mike Miller

For $i < 2j$ we have $$\text{Sq}^i \text{Sq}^j = \sum_{k=0}^{i/2} \binom{j-k-1}{i-2k} \text{Sq}^{i+j-k} \text{Sq}^k.$$ What can I do with this? I claim that Steenrod operations are generated by the powers $\text{Sq}^{2^n}$. To see this, take an arbitrary natural $x$ which is not a power of $2$. Write $x = 2^m + i$ where $i < 2^m = j$. Then applying the Adem relation and re-arranging (using crucially that $i < j$ and that $j-1 = 2^m -1$ to see that the binomial coefficient is nonzero), we find...

Ted Shifrin

19:31

has heart palpitations

Mike Miller

$$\text{Sq}^x = \text{Sq}^i \text{Sq}^j - \sum_{k=1}^{i/2} \binom{j-k-1}{i-2k} \text{Sq}^{x-k} \text{Sq}^k.$$

In particular this inductively proves the claim above

Wil continue in a bit

Leaky Nun

Tom Scott just did an abso-bloody-lutely fan-something-tastic video

Ted Shifrin

Leaky, you're sounding particularly British.

Stan Shunpike

who is tom scott?

Balarka Sen

@TedShifrin lol same

Leaky Nun

19:33

Abso-b████y-lutely: Expletive Infixation

►

Ted Shifrin

Although I took a third-quarter alg. top. course my first year of grad school, I never really learned Steenrod squares.

CaptainAmerica16

Hi

Balarka Sen

they look scary and too hard

Ted Shifrin

Just lots of axioms, @Balarka :P

Balarka Sen

i have come to realize i just cant compute anything in algebraic topology

Ted Shifrin

19:35

hi @Captain

Balarka Sen

yeah

Leaky Nun

@BalarkaSen switch sides to number theory, you'll be able to compute everything

Masterphile

@TedShifrin with a burden of incompleteness, i am going to skip again the whole construction

Balarka Sen

and it doesn't help to understand where the axioms come from

Ted Shifrin

I sat through a month of a first analysis course my first year in college in which the professor did first decimals (to convince us how hard it is to define operations) and then Dedekind cuts. I was far more interested in the rest of the course.

Leaky Nun

19:38

@BalarkaSen compute $\pi_4(S^3)$

CaptainAmerica16

Why was he trying to convince you "how hard it is"?

Balarka Sen

Z

Leaky Nun

why?

Ted Shifrin

@Captain: He was an anal-retentive probabilist/analyst.

By far not one of the best teachers I had in college.

Balarka Sen

Z/2 actually hah.

Hopf on S^1 -> S^3 -> S^2

Ted Shifrin

19:39

We can probably do that one with Thom-Pontryagin, @Balarka. I think I did that in one of the diff top courses I taught.

Balarka Sen

pi_4 S^3 = pi_4 S^2 = Z/2 because that's the stable range

CaptainAmerica16

@TedShifrin Blech.

Balarka Sen

@TedShifrin Yeah, the stable 2nd homotopy group of S^2 is pi_2 SO = Z/2

Ted Shifrin

I don't know (any more) this stable stuff.

Balarka Sen

But it requires some work to show why every 2-manifold is framed cobordant to S^2

Ted Shifrin

19:41

I just remember doing the framing discussion directly for the undergrads.

No, aren't we framing $1$-manifolds?

Leaky Nun

@BalarkaSen what's the highest homotopy group of sphere you've computed

CaptainAmerica16

If anyone here is studying intro Real Analysis and/or Proof-based Linear Algebra, let me know if you want to form a study group.

Balarka Sen

@TedShifrin Homotopy classes of maps $S^{n+2} \to S^n$ corresponds to framed cobordism classes of codimension $n$ ie dimension $2$ framed submanifolds of $S^{n+2}$ upto framed cobordism, right?

Masterphile

@TedShifrin i would like that start with decimals, or more generally, with all the bases, i like when reals are so concretely described

Balarka Sen

Let $n \to \infty$ and you're classifying framed cobordism classes of framed 2-manifolds

Oh, I see.

You want to compute $\pi_4 S^3$ and not $\pi_4 S^2$

Nice, yes.

Framed 1-manifolds work

Ted Shifrin

19:45

Whew. You scared me.

Balarka Sen

Oh then I did something wrong. $\pi_4 S^2$ is not the stable range of $\pi_2^{st}$.

$\pi_2 SO = 0$, I miscalculated

2nd stable homotopy group of spheres is $0$

Ted Shifrin

$\pi_2$ of any Lie group is $0$.

Balarka Sen

I suck at this, like I said.

Yeah

Ted Shifrin

I'm shocked I remembered doing this back in 1980.

Balarka Sen

@LeakyNun I don't remember. I did some nonsense with the Serre spectral sequence.

Every time I do it I forget it

I don't see the point

It's whatever

@TedShifrin Yes, you were absolutely correct with everything you said.

Leaky Nun

19:49

what is $\pi_1(\operatorname{DIFF}(\Bbb R^n))$?

Balarka Sen

Diff(R^n) is homotopy equivalent to O(n)

Leaky Nun

that's disappointing

Ted Shifrin

So the framing of the normal bundle is coming from $\pi_1(SO(3)) = \Bbb Z/2$. Hmm, how did I get away with that in that course?

Balarka Sen

Given any diffeomorphism f : R^n -> R^n fixing 0 it is isotopic to the linear diffeomorphism df_0 : R^n -> R^n in a canonical way

Ted Shifrin

@Leaky: That's a lemma in Milnor's tiny book on Diff Top. Or in G&P.

Balarka Sen

19:53

Anyway this stuff sucks

Screw homotopy theory

Screw algebraic topology

CaptainAmerica16

lol, dang

Leaky Nun

then which side are you going

Balarka Sen

im gonna do what i want

Ted Shifrin

(he says petulantly)

Alessandro Codenotti

@BalarkaSen I'm afraid to ask what that is

Leaky Nun

19:56

petulantly
adverb
UK /ˈpetʃ.ə.lənt.li/ US /ˈpetʃ.ə.lənt.li/

in a way that is petulant (= easily annoyed and rude, like a child):
"Well, he didn't invite me to his party so I'm certainly not inviting him to mine!" she said petulantly.
He stamped a foot petulantly.

Balarka Sen

Lol

Ted Shifrin

Was my parenthetic remark not sufficient, @Leaky?

Leaky Nun

no

Mike Miller

@BalarkaSen I respect the computational guys

Their ideas are not totally out there i think

Balarka Sen

I cant do it

I respect it but I just find it extremely bad lol

Leaky Nun

19:59

what is $\pi_n$ of a closed orientable connected $n$-manifold?

Mike Miller

try some examples

Balarka Sen

dead end

►

Mike Miller

Anyway I know you hate taking it for granted, but @BalarkaSen, the fact about truncated polynomial rings I think follows quickly from what I said above --- that all Steenrod operations can be decomposed, except for those in degree $2^n$

Balarka Sen

I said it more like this

Leaky Nun

$\pi_2(S^1 \times S^1) = \pi_2(S^1)^2 = 0$?

Balarka Sen

20:01

@MikeMiller Hm OK

Mike Miller

yes

So if your space has $H^*(X;\Bbb F_2) = \Bbb F_2[e]/e^c$ for some $c$, then we know that $\text{Sq}^{|e|}(e) = e^2$ --- again, one of the axioms

Ted Shifrin

@Leaky: Where did you make up that rule from?

Balarka Sen

That's true

Homotopy group of product is product of homotopy groups

Mike Miller

It just follows from maps to a product being maps to each component

But since $\text{Sq}^{|e|}$ is decomposable unless $|e| = 2^n$, and there is no cohomology in degree between $|e|$ and $2|e|$, we see that necessarily $|e| = 2^n$ for some $n$

Ted Shifrin

@Leaky: I did say above that $\pi_2(G) = 0$ for any (connected) Lie group, so ....

Alessandro Codenotti

20:03

@BalarkaSen see homotopy is so much easier than homology to compute

Leaky Nun

oh lol

Ted Shifrin

You might be interested in the Hurewicz Theorem.

Mike Miller

@BalarkaSen So that gives us something mod 2. What do we get mod p?

Leaky Nun

but it isn't simply connected

oh I should think about universal covers

Mike Miller

Mod p, instead of Steenrod squares, we have the Steenrod powers $P^i$ with $|P^i| = 2pi$. And again you have a rule saying $P^i$ can be decomposed unless $i = p^n$

Balarka Sen

20:05

Hmmk

Mike Miller

The other relevant axiom is that $P^{|x|/2}(x) = x^p$ for even-degree classes.

Leaky Nun

what does universal cover of T^2 # T^2 look like?

Mike Miller

you know the answer to that, i would hope

Balarka Sen

Leaky just go read some book or something

Mike Miller

This gets messy I think

Balarka Sen

20:12

I am not following any of it, so I wouldn't know :)

Mike Miller

Fine, I'll stop bothering

geocalc33

that's enough out of me too

ypercubeᵀᴹ

A great one has left us

4

Balarka Sen

Lmao

skullpatrol

his struggle in the game only lasted 3 days :'(

Balarka Sen

20:21

Oh I literally thought he was joking about people leaving the chat

But that was a message to Conway

Gotit

Forgot game of life is by Conway

geocalc33

What did Jesus get when he did 2+5

Alessandro Codenotti

@BalarkaSen It's kinda sad that most people know him because of the game of life while he did so much cool stuff

4

Masterphile

12?

geocalc33

5,000

Masterphile

there were, according to story, some leftovers

geocalc33

20:25

did he not use banach tarski?

Masterphile

of course, he had full understanding of that theorem, and knew to apply it in full

:D

geocalc33

he knew how to apply it to energy? wtf?

Alessandro Codenotti

I attended a public lecture by Sergey Fomin in November where he was talking about how Conway noticed some things playing around with Frieze patterns that were only understood completely much later through cluster algebras and we were all like "that looks like such a Conway thing"

geocalc33

@Masterphile do you know mathphile

Masterphile

105

Q: Conway's lesser-known results

John Horton Conway is known for many achievements: Life, the three sporadic groups in the "Conway constellation," surreal numbers, his "Look-and-Say" sequence analysis, the Conway-Schneeberger $15$-theorem, the Free-Will theorem—the list goes on and on. But he was so prolific that I bet he estab...

3

@geocalc33 not personally

geocalc33

20:30

ahh

the free will theorem

Masterphile

i still do not understand how for some $n=\displaystyle \prod_{i=1}^r {p_i}^{a_i}$

the number $$\dfrac{1}{\prod_{i=1}^r \dfrac {{p_i}^{a_i+1}-{p_i}^{a_i}}{{p_i}^{a_i+1}-1}+ \prod_{i=1}^r \dfrac {{p_i}^{a_i+1}-2{p_i}^{a_i}+{p_i}^{a_i-1}}{{p_i}^{a_i+1}-1}}$$ can be a natural number

NoName

I reckon Jesus would be a geometer.

Masterphile

inter-universal

NoName

20:48

Moses: topologist; Abraham: algebraist (obviously).

Thorgott

21:09

$n=2$ produces an integer

Masterphile

yes, and some other natural numbers also produce natural numbers

some solutions:

in Hilbert's hotel, 12 hours ago, by Peter

? for(j=1,length(v),n=v[j];print(j," ",n," ",length(digits(n))," ",factor(n)," ",sigma(n)/(eulerphi(n)+n)))
1 2 1 Mat([2, 1]) 1
2 456 3 [2, 3; 3, 1; 19, 1] 2
3 828 3 [2, 2; 3, 2; 23, 1] 2
4 7584 4 [2, 5; 3, 1; 79, 1] 2
5 33462 5 [2, 1; 3, 2; 11, 1; 13, 2] 2
6 1357440 7 [2, 7; 3, 1; 5, 1; 7, 1; 101, 1] 3
7 1596048 7 [2, 4; 3, 1; 41, 1; 811, 1] 2
8 1964544 7 [2, 9; 3, 1; 1279, 1] 2
9 19800384 8 [2, 6; 3, 1; 281, 1; 367, 1] 2

me and Peter are almost exclusively tackling this problem for, i do not know, maybe even two weeks

the sequence is even not in OEIS

Masterphile

21:30

"When the sky is clear we will remember
Through thick and thin we won't surrender
As the years pass, the memories still last
When the sky is clear we will remember"

:D

Masterphile

22:00

Inscribed square problem

The inscribed square problem, also known as the square peg problem or the Toeplitz' conjecture, is an unsolved question in geometry: Does every plane simple closed curve contain all four vertices of some square? This is true if the curve is convex or piecewise smooth and in other special cases. The problem was proposed by Otto Toeplitz in 1911. Some early positive results were obtained by Arnold Emch and Lev Schnirelmann. As of 2017, the general case remains open. == Problem statement == Let C be a Jordan curve. A polygon P is inscribed in C if all vertices of P belong to C. The inscribed square...

TechnoKnight

22:23

Guys

can anyone please help me here?

please?

0

Q: Why these two inequalities are not the same even though they use the same equation?

I just don't get it, like at all. $Un$ is an iteration defined on $\mathbb{N}$, btw. The question was: $\underset{n+1}{U}$ = $\tfrac{8Un - 8}{Un + 2}$ = $8 - \tfrac{24}{Un + 2}$ $U0 = 3$ "Prove that $3 \leqslant Un \leqslant 4$ by using Mathematical Induction" Step 1: Test n = 0, yes it's c...

calculus inequality

Transcript for

$Mathematics$ Mathematics