Mathematics - 2018-06-03 (page 4 of 5)

Poline Sandra

20:00

Someone help me to prove that $(x_n)$ is a cauchy sequence

@AlessandroCodenotti can you help me

Alessandro Codenotti

@MikeMiller Sure, but if I think about the directions to move in I visualize $T_pX$, I can't "see" $TX$ as a whole or even an open subset of it

Akiva Weinberger

@PolineSandra I don't think it necessarily is, under those conditions.

If there's a $c<1$ such that $d(f(x),f(y))\le cd(x,y)$, then it's called a contraction mapping and $x_n$ is necessarily Cauchy

but that's a stronger hypothesis

MatheinBoulomenos

it's not stronger on a compact space

but in general, it is, yeah

Akiva Weinberger

@MatheinBoulomenos Why, actually? I know it should be true but I'm not seeing the proof

Wait, isn't $\frac x{1+x}$ on $[0,1]$ a counterexample? In that there is no $c$, not that the $x_n$ doesn't converge @MatheinBoulomenos

Mike Miller

@AlessandroCodenotti Well you get TU for U a chart

MatheinBoulomenos

20:15

@AkivaWeinberger oh yeah I meant you get a fix point on compact space, but you have to use a different argument

I didn't read everything, I just assumed you want to show that the map has a fix point

Mike Miller

But I agree, I can only "see" TM when M is a subspace of a small Euclidean space

MatheinBoulomenos

because I guessed that's why you'd worry about the sequence $x,f(x),f(f(x)), \dots$

I may wrong though

Alessandro Codenotti

Hi @Ted

Ted Shifrin

Hi @Alessandro. Did you figure out that geometry exercise? :)

DogAteMy: Yes, @Mathein is right. If you're on a compact metric space, you don't need a contraction map.

Mike Miller

Not every self-map of a compact contractibile space has a fixed point

MatheinBoulomenos

20:24

$\operatorname{Im}(f^n)$ is a decreasing sequence of non-empty compact subsets, thus we must have a non-empty intersection $Y:=\cap_{n \geq 0} \operatorname{Im}(f^n) \neq \varnothing$. But using the fact that every continuous map on a compact set attains a maximum applied to the map $\operatorname{Im}(f^n) \times \operatorname{Im}(f^n) \to \Bbb R, (x,y) \mapsto d(x,y)$ and the condition on $f$, we get that the diameter of $\operatorname{Im}(f^n)$ goes to $0$ as $n \to \infty$

Mike Miller

There is some counterexample involving a space that looks like a roll of toilet paper

Ted Shifrin

<-- confuzled

MatheinBoulomenos

@MikeMiller we're still assuming $d(f(x),f(y)) < d(x,y)$

Mike Miller

Yes but I was responding to Akiva

MatheinBoulomenos

but not $d(f(x),f(y)) \leq \kappa d(x,y)$ for $\kappa < 1$

Mike Miller

20:25

Just as a general statement people might care about

I see

I should have caught up with the convo first

MatheinBoulomenos

I just assumed that's what is to be shown from seeing the sequence $x,f(x),f(f(x)), \dots$

Alessandro Codenotti

@TedShifrin Not yet, I've been busy going through past exams papers :(

Ted Shifrin

Find anything interesting, @Alessandro?

Oh, did you guys discuss asymptotic directions, anyhow?

Alessandro Codenotti

Not really, mostly a lot of computations

Yes @Ted

Ted Shifrin

OK, so my exercise is relephant.

Do you want to see one of my final exams?

Alessandro Codenotti

20:27

I'm afraid

Nûr

Let $f : [0,1] \cap \Bbb Q \mapsto \Bbb R$, continuous. Is $f$ bounded?

Alessandro Codenotti

Usually in this course the written exam is computation heavy, while proofs and more interesting stuff are kept for the oral exam

Ted Shifrin

My course was taught for students not nearly in your league. But they had my homeworks, many of which showed up on the exam. But you're not required to look, @Alessandro. Plus, I don't think I have your email.

@Nûr: What do you think?

Nûr

I know the answer

:D

Ted Shifrin

My exams were both, @Alessandro.

Then why are you asking, @Nûr?

Nûr

20:29

see if people will respond well

Ted Shifrin

Oh, so you're testing the denizens of the room?

Nûr

yes :)

@TedShifrin Do you have ideas about this one: Consider 4 lines in $\Bbb R^3$ in the "average situation", how many lines intersect all of them?

I do not have the answer

Alessandro Codenotti

I think you have it, you sent me your exercises for GP once, but that was quite some time ago, my address is [email protected] @Ted

Ted Shifrin

That turns out to be one of my favorite questions, @Nûr, one which I solved in the final chapter of my algebra book (using projective geometry).

Oh @AlessandroCodenotti: Does that mean you want one? :)

MatheinBoulomenos

Is $\beta \Bbb N$ totally disconnected?

I think it is, but I'm not sure

Alessandro Codenotti

20:33

Well I'm curious now!

Ted Shifrin

@AlessandroCodenotti You know that curiosity killed the cat.

Nûr

does this require only basic knowledge?

Mike Miller

@MatheinBoulomenos Can you translate that into a criterion in the algebra of functions?

Ted Shifrin

@Nûr: By the way, it's better to pose your question in $\Bbb C^3$ than in $\Bbb R^3$. The answer in $\Bbb R^3$ is "maybe $a$", "maybe $b$"!

Alessandro Codenotti

@MatheinBoulomenos I think so

Mike Miller

20:36

The fact I want to use is that C(beta X) = L^infty(X)

Nûr

ok

Ted Shifrin

@Nûr: I don't know what "basic knowledge" means. Projective geometry is the proper setting for the question.

Alessandro Codenotti

whoa that's a lot of exercises @Ted, how much time was given for the exam?

MatheinBoulomenos

@MikeMiller huh, that's interesting

Ted Shifrin

3 hours, @Alessandro.

Also remember that I don't expect 90% for an A.

And they've seen almost all of that before, or something much like it.

Akiva Weinberger

20:37

@Nûr Nah

Nûr

I did not study projective geometry, and it was given in oral to someone in the same situation

Akiva Weinberger

$\frac1{2x^2-1}$

Alessandro Codenotti

We have 3 exercises in 3 hours :P (But they are longer and once is expected to do all of them to get 30 out of 30)

MatheinBoulomenos

@MikeMiller ah I think I have it

Akiva Weinberger

@TedShifrin What geometry exercise?

Nûr

20:40

hmmm, actually it seems difficult without background

Daminark

Wow 3 hour exams, I mostly had 2

Except in 2 classes

Nûr

I've taken a 6 hours few weeks ago :p

Ted Shifrin

You don't need projective geometry, but it is the natural setting. Here's an exercise to start: Bare hands, take three "generic" lines like the $x$ axis, the line $z=x$, $y=1$, and the line $z=2x$, $y=2$. What lines meet all three of those? @Nûr

Alessandro Codenotti

I have no idea about 5 and the last point of 11 (because I don't know what holonomy means), but I'm confident I know how to do the rest, except maybe 10 because I don't know the notation $\nabla_{x_u}x_u$

Ted Shifrin

DogAteMy: One of the questions in my diff geo notes.

Daminark

20:41

Ironically, one of the two was not an exceptionally long test and we had 4 hours, the other was an hour and a half and very tough to finish in time

Ted Shifrin

Didn't you guys do covariant derivatives, @Alessandro?

MatheinBoulomenos

if $X$ is a discrete space, then for any family of fields $(F_x)_{x \in X}$ one has that $\operatorname{Spec}( \prod_{x \in X} F_x) \cong \beta X$ where the Zariski topology agrees with the topology on the Stone-Cech compacitifcation. Now $\prod_{x \in X} F_x$ is a zero-dimensional ring and it's well-known that this is equivalent to the Zariski topology on the spectrum being totally disconnected

Alessandro Codenotti

Yes, we use $D_\gamma X(t)$ for the covariant derivative of the vector field $X$ along the curve $\gamma$

Ted Shifrin

Oh, so my $\nabla$ is your $D$.

MatheinBoulomenos

@TedShifrin they only did contravariant derivatives

Ted Shifrin

20:42

I used $D$ for usual Euclidean derivative, then $\nabla$ is the projection onto the tangent space of the surface.

Alessandro Codenotti

Ah, I see

Ted Shifrin

@Alessandro: Holonomy means, briefly, the angle through which a vector turns when you parallel translate around a closed curve.

You also probably can't do #7 if you didn't discuss the hyperbolic plane, @Alessandro.

MatheinBoulomenos

@TedShifrin I found a really simple way to compute $\pi_1(S^1)$

Lozansky

@Daminark We have 5 hour exams :P

Daminark

What the hell

Ted Shifrin

20:44

What étale cohomotopy did you compute, @Mathein?

MatheinBoulomenos

overleaf.com/read/qvqpmfzjchss#/63957916

@TedShifrin I actually did use étale cohomology, yes

Alessandro Codenotti

@TedShifrin Makes sense intuitively. We didn't spend a lot of time on $\Bbb H$ but we did define it and saw what its geodesics are so I can improvise something for #7

Ted Shifrin

Yeah, real simple.

Nûr

The 6 hours exam was once, otherwise it is 4 hours

Ted Shifrin

So, @Alessandro, you glad you never got stuck with me as a prof? :D

Alessandro Codenotti

20:46

Hmm, I'm not sure, your exercises seem nicer :D

MatheinBoulomenos

I also used etale fundamental groups, poincaré duality for compactly supported étale cohomology, cellular approximation, Brown representability, Hurewicz and both versions of universal coefficients

Ted Shifrin

As I said, the prereqs for my course are lighter than you guys have. Only multivariable calculus and linear algebra required, although a handful of students had taken my more advanced multivariable math class.

Alessandro Codenotti

I'm a bit annoyed by our past papers, some seem to be more about testing my ability in differentiating some awful function twice without messing up the calculations than my knowledge of diffgeo

Ted Shifrin

Yeah, @Mathein, super elementary.

Yeah, @Alessandro, I think that's stooopid.

Notice, also, that I gave them a ton of formulas ...

I don't do that frequently in courses, but for this course of course it is necessary.

MatheinBoulomenos

you have to do horrible calculatiosn in diff geo, that's what the subject is about (it seems)

Alessandro Codenotti

20:48

We also have a sheet with useful formulas made by the professor, it's impossible to know some of them

Ted Shifrin

No, not really. But for a beginning undergraduate course, doing some calculations is important. Advanced courses the computations are more involving Lie groups and maybe some differential equations.

@Alessandro: With differential forms instead of the "classic" approach, the formulas become super easy.

Alessandro Codenotti

One more reason to learn about them!

Ted Shifrin

See the bottom two lines of my sheet. That's all of Gauss, Codazzi, etc.

MatheinBoulomenos

@TedShifrin yeah the latter was a bit of a problem for me. We don't really learn to learn solve ODEs, depending on which analysis prof you have. All the physics students solved them easily, but although I still remember the proof of Picard-Lindelöf (at least the gist of it), I never really worried about solving ODEs

Alessandro Codenotti

Well that does look appealing

Ted Shifrin

20:51

Plus, @Alessandro, it's the tip of a Lie algebra iceberg :P This is all $SO(2)$ and $\mathfrak{so}(2)$. But $SO(n)$ shows up .... and other Lie groups.

Daminark

Lol I remember a friend of mine taped some Christoffel symbols to my door. We're not friends anymore >:(

People have no restraint these days with April Fools pranks

Ted Shifrin

Did you ever play the word spelling game Hangman?

MatheinBoulomenos

I also remember one case where you had to do about 2 pages of calculations to determine the rank of some rather large matrices to check some transversality condition

Ted Shifrin

You'll have to be more explicit than that ...

Daminark

@Ted yep

Ted Shifrin

20:53

That's what Christoffel symbols are for, Demonark :P

Alessandro Codenotti

lol

Ted Shifrin

Anyhow, @Alessandro, if you decide you want to discuss any of my questions, I might be able to remember how to do some of 'em.

Daminark

Really makes you thonk

Ted Shifrin

@Nûr: You thinking about the three lines I gave you?

Alessandro Codenotti

@Mathei Isn't $\beta X$ totally disconnected regardless of $X$ actually?

Sure, if I get bored of the exercises I already have to do!

Ted Shifrin

20:57

But mine are more interesting :P

Alessandro Codenotti

Fair, but less similar to those which will be on the exam, which is a pretty important point to me at the moment

Nûr

We could use the parametric equations of the lines in that particular case but it looks tedious

MatheinBoulomenos

@AlessandroCodenotti but subspaces of totally disconnected space are totally disconnected

Ted Shifrin

@Nûr: Well, you have to do some calculations, unless you're going to learn Schubert calculus and the cohomology of Grassmannians.

Nûr

: )

MatheinBoulomenos

20:59

And if $X$ is $T_{3.5}$, then the canonical map $X \mapsto \beta X$ is an embedding

Ted Shifrin

Think about it geometrically. "How many" lines meet two skew lines? How many lines will you expect to meet three lines in general position?

user193319

If $T$ is a bounded linear operator on a Hilbert space, why does $\sup \{\langle T v,Tw\rangle | \: \Vert v\Vert = 1, \Vert w\Vert = 1 \} = \Vert T \Vert^2$ hold?

Ted Shifrin

Where are you stuck?

Nûr

for two it is an infinity, but i am not sure about three

Alessandro Codenotti

Ah, wait, I was thinking about a different but related construction

Ted Shifrin

21:04

Think about one line, another line, and a point on the third line.

Poline Sandra

@AkivaWeinberger is a map $f$ which satisfy $d(f(x),f(y))<d(x,y), x\neq y$ is continuous ?

MatheinBoulomenos

uniformly even

Poline Sandra

@MatheinBoulomenos you are with me?

Akiva Weinberger

Yeah, $\lim_{x\to y}d(f(x),f(y))<\lim_{x\to y}d(x,y)=0$ so $\lim_{x\to y}f(x)=f(y)$ so it's continuous

Poline Sandra

thank you

Ted Shifrin

21:07

What you wrote is obviously garbage, DogAteMy, because you have a nonnegative number < 0.

One of many examples of sloppy application of the squeeze principle.

Alessandro Codenotti

I'm missing something @Mathei, doesn't $\beta X$ have a basis of clopen sets?

MatheinBoulomenos

it's a straightforward application of the $\varepsilon-\delta$ definition and you even get uniform continuity. Just always choose $\delta=\varepsilon$

Mike Miller

@MatheinBoulomenos Nice. I imagine one could take your last statement and pull the essence of the proof out to make this entirely explicit

Poline Sandra

@AkivaWeinberger can i say $x_{n+1}=f^{n+1}(x_0)$ is a sequence in a compact metric space so it has a convegent subsequence $x_{\varphi(n)}\to x$ then $x=\lim_{n\to\infty}f(x_{\varphi(n)})\to f(x)$ so $f(x)=x$

ramose

is it considered rude to answer my own question?

Nûr

21:12

the end is not correct

Poline Sandra

@Nûr you are with me?

Akiva Weinberger

^

@MikeMiller You mentioned something earlier about a contractible space with an endomorphism without a fixed point, by the way?

What is it?

Ted Shifrin

DogAteMy: Did you notice my chiding of you above?

Nûr

What is the exercise @PolineSandra

Mike Miller

@Akiva I believe it is constructed as follows but my memory is hazy - I am getting old

Nûr

21:15

Any hint for: Consider an involution in $S_n$. What is the cardinality of its commutant?

Ted Shifrin

rolls 11 + $\sqrt 2/7$ eyes

Akiva Weinberger

@TedShifrin Dammit. $\le$, I mean

Mike Miller

Consider a spiral from the origin that accumulates towards the unit circle, and include the unit circle. Call this S. It is compact and of course not contractible. It shouldn't even be path connected.

MatheinBoulomenos

yeah. The standard basis $D(f)$ on the Zariski topology is actually a basis of clopen sets if the ring is zero-dimensional and you can translate that easily to a basis on $\beta X$ (using the construction as the set of ultrafilters).
The homeomorphism $\operatorname{Spec}(\prod_{x \in X} F_x) \to \beta X$ works by sending a prime ideal $\mathfrak{p}$ to the ultrafilter given by $\mathcal{F}(\mathfrak{p}) = \{ Y \subset X \mid \exists (a_x)_{x \in X} \in \mathfrak{p}: \forall x \in X: x \in Y \Leftrightarrow a_x = 0\}$

Mike Miller

Consider S x I, where the I coordinate is a positive z-coordinate, sitting in R^3

That is S x [0,1] I mean, the first coordinate xy and the second z

Akiva Weinberger

21:18

Sure

Mike Miller

Now take the union with the unit disc in the xy-plane

Nûr

In the Akiva answer about limits, how do we know it exists?

Mike Miller

Apparently it's this fella. I am trying to remember why.

Ted Shifrin

Yes, @Nûr, that's part of my complaint about the sloppiness of squeeze. It's easy to fix here, of course.

Nûr

yes

Ted Shifrin

21:19

But I took off points when my students wrote stuff like that.

Poline Sandra

@AkivaWeinberger?

Akiva Weinberger

@Nûr Sandywichy? Lemon squeezy?

I'm hungry, don't know if you can tell

Mike Miller

I do not see how to make a fixed point free map on this

The statement of the space is somewhere in Hatcher ch2 exercises

Akiva Weinberger

@PolineSandra The end didn't look right

Nûr

So there is no answer for my challenge question? :D

Akiva Weinberger

21:22

I don't think you can get that from just a subspace

@Nûr Which one?

Nûr

Let $f : [0,1] \cap \Bbb Q \mapsto \Bbb R$, continuous. Is $f$ bounded?

MatheinBoulomenos

@MikeMiller a possible counterexample can't be triangulable due to Lefschetz

so it has to be a bit pathological

Mike Miller

@Mathei Indeed, but my space is certainly pathological in that way

Nûr

It was suddenly asked at the very end of an oral exam :S

Mike Miller

It is not locally contractible

Nûr

21:23

how to push to fail under pressure

Ted Shifrin

@Nûr: Akiva already answered you.

Nûr

Ah

Ted Shifrin

It's a perfectly fair question, I think.

Nûr

sorry

Akiva Weinberger

46 mins ago, by Akiva Weinberger

$\frac1{2x^2-1}$

It's${}\to\Bbb Q$ even

Nûr

21:25

The situation gives it too much importance I think.

good

Ted Shifrin

Depends what the exam was about. If it's why we need to have completeness to get the maximum value theorem for continuous functions on compact sets, it's a great question.

Nûr

the main exercise was completely different

Poline Sandra

@AkivaWeinberger so i can't wok with sequences

Nûr

It was: Show $\lim \limits_{x \to 1^-} \sum\limits_{n=0}^\infty (-1)^nx^{n²} = \frac{1}{2} \ $

Ted Shifrin

Hmm, and what is the context for that question?

Nûr

21:29

What do you mean by context?

Ted Shifrin

What techniques are you being tested on?

Nûr

the curriculum is equivalent to the one of a 2 years undergraduate I suppose. We have basic knowledge ( :D ) on series, integral (Riemann) ; actually a 'simple' answer used convexity

to get closer to altenate series

Ted Shifrin

I guess I don't know how to do that problem.

MatheinBoulomenos

I don't know either

But I also don't find it very interesting

Nûr

I have posted this question on MSE ; the answers seems really good but far too knowledgeable for me

Ted Shifrin

21:35

What's the post?

Nûr

I didn't understood them at all

11

Q: How to prove $\lim \limits_{x \to 1^-} \sum\limits_{n=0}^\infty (-1)^nx^{n²} = \frac{1}{2} \ $?

$\lim \limits_{x \to 1^-} \displaystyle \sum_{n=0}^\infty (-1)^nx^{n²} = \frac{1}{2}$ The power $n^2$ is problematic. Can we bring this back to the study of usual power series? I do not really have any idea for the moment.

Fortunately, my teacher has choosen this exercise and has corrected it in a very simple way

Ted Shifrin

I think the question you started with is far more reasonable than this question. I thought about Cesaro summability, but I don't know all the trickiness in these solutions.

Nûr

simple = Using only epsilon, convexity, and alternating series test

MatheinBoulomenos

oh the fact that this about a theta function makes it more interesting

Nûr

But the other question was just for the last minute of the exam :D

geocalc33

21:38

Did you say it was bounded?

Nûr

It was not given to me

some said yes

Leaky Nun

oh god how long have i slept for

Ted Shifrin

Twenty days and twenty nights.

Leaky Nun

indeed

@MatheinBoulomenos so what is the one-parameter subgroup associated to $\Bbb G_m := X+Y+XY$?

Nûr

Another last minute question : What can be said about a real sequence whose all sub-sequences have a sub-sequence that converges towards 0 ?

MatheinBoulomenos

21:42

@LeakyNun some exponential map on the multiplicative group

Leaky Nun

@Nûr it's real

interesting question

my first hypothesis is that it converges to 0 (am I allowed to make wrong guesses?)

Nûr

I think so

Leaky Nun

assuming otherwise, unfolding epsilon-delta, we get that there exists an epsilon such that for all N there exists n such that |a_n| > epsilon

Nûr

Yes

Ted Shifrin

$n>N$

Leaky Nun

21:45

so we can build inductively a sequence that is bounded away from epsilon

thanks @TedShifrin

MatheinBoulomenos

@@LeakyNun the Lie algebra associated to that group law is just $\mathfrak{gl}(2)$

Leaky Nun

and then contradiction

done

@MatheinBoulomenos Lie algebra associated?

Nûr

What can be said about a real sequence whose all sub-sequences have a sub-sequence that converges ?

Leaky Nun

it doesn't have to converge, e.g. (-1)^n

Nûr

I don't see how to change the end so as the sub-sub sequence is usefull

Leaky Nun

21:48

I think it needs to be bounded (otherwise pick a sequence whose magnitude goes to infinity)

Nûr

yes

Leaky Nun

and then it's sufficient

so i'm done

I can go take your oral exams :D

Nûr

the last minute yes :D

MatheinBoulomenos

@LeakyNun ah nevermind

that's only interesting for higher-dimensional group laws

Leaky Nun

@MatheinBoulomenos what is the Lie algebra associated to a formal group law?

MatheinBoulomenos

21:52

@LeakyNun it's always trivial for 1-dimensional group laws we're considering

Leaky Nun

what does the 2D version of X+Y+XY look like?

mercio

@Nûr where do the tangents of slope +-i to a real ellipse intersect ?

Leaky Nun

@mercio after a base change?

Mike Miller

@mercio you really like this question

mercio

yes

Mike Miller

21:53

did there turn out to be a very clean answer?

Nûr

Actually the real problem is that Bourbaki has eliminated geometry at school in France :D

mercio

evil bourbaki !

the answer turns out to be very clean but noone has said it

so far

Leaky Nun

@MatheinBoulomenos so you're saying $\Bbb R \to \Bbb R^\times : t \mapsto e^t$ is where I could see that formal group law?

Mike Miller

@mercio Will you tell me what it is even if I am too lazy to do the work?

MatheinBoulomenos

@LeakyNun the rule X+Y+XY comes from multiplication in your ring/field, where you do a change of transformation to make $0$ the identity

mercio

21:55

no >:c

Mike Miller

If no, what if I said "too busy" instead of "too lazy"?

mercio

then you could take a wild guess

for starters

where are the points from where the tangents have a nonreal slope ?

MatheinBoulomenos

(1+X)(1+Y)=X+Y+XY

Nûr

That's a tragic reality: ellipse has never be defined in math class.

Mike Miller

Hmmm I see clearly that you are trying to make me do work

mercio

21:56

lol

MatheinBoulomenos

the exponential isn't necessary

Leaky Nun

what is the 1-parameter subgroup then

Mike Miller

I guess the stupid wild guess is "on the imaginary plane"

mercio

the real points

some real points have two real tangents to the ellipse

some don't have real tangents

Mike Miller

that confuses me, an ellipse should be smooth

MatheinBoulomenos

21:58

@LeakyNun $t \mapsto e^{t}-1$ over any ring such that the coefficients of $e^t$ are defined is a morphism from the additive group law to the multiplicative group law

Leaky Nun

hmm

that's interesting

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$Mathematics$ Mathematics