Mathematics - 2020-08-06 (page 1 of 2)

lmfao

Adam L

that's fine I'll go back to facebook and take my frustration out on high functioning differently abled people

lmfao

7:49 AM

@AdamL didn't you get banned for consistently making sexist and casually racist comments on this chat?

Or was that a different Adam

9:29 AM

Did he proof that $\pi_1(\Sigma_i)$ is hyperbolic for $i > 2$ in the previous session @BalarkaSen ?

I don't recall seeing it

No.

It follows from Svarc-Milnor lemma and quasi-isometry invariance of hyperbolicity

So we get the cayley graph is quasi-isometric to the universal cover of $\Sigma_i$

that's the plane, isn't it ?

That's the hyperbolic plane

$\pi_1(\Sigma)$ acts isometrically on the hyperbolic plane

oh ok. And that's the normal plane for $\Sigma_1$. I see

Genus >= 2 is crucial

9:34 AM

Maybe this is complete unrelated - but is there any way to prove it (that $\pi_1(\Sigma)$ acts isometrically on the hyperbolic plane) with using uniformization theorem, and the fact that $\Sigma_i$ is a Riemann surface ?

ok seems he is back

@Lelouch this is killing a fly with a nuclear bomb

$\pi_1(\Sigma)$ acts isometrically on $\Bbb H^2$ by drawing a picture

@BalarkaSen ok sorry, yes, this is obvious by drawing a suitable $2n$-gon in $\mathbb{H}^2$

my bad

9:55 AM

I thought general topology could get ugly, but topological dynamics is way worse, it's all nonsense

You can have a space $X$ and an automorphism $f$ such that every point has a dense orbit in the associated Z action, but the space is not homogeneous

And $X$ is like compact metrizable, as good as spaces can be

Hardik Rajpal

11:00 AM

guys what's the procedure to start a new stackexchange website?

11:14 AM

@HardikRajpal area51.stackexchange.com

11:48 AM

https://math.stackexchange.com/a/2684915/586524
My notes emphasise that the partials must be continuous. I know that if a real function is differentiable, it is continuous. That being said, is: "If continuous partial derivatives exist at some point a" less redundantly stated as "If partial derivatives exist ..."?

what does it mean for continuous partial derivatives to exist at a point

if you want to talk about continuity, you better have the partials exist in an entire neighborhood

That's how I understand it too...

Wait, let me quote the notes directly:

"Assume that u and v have continuous first-order partial derivatives in D and that they satisfy the Cauchy-Riemann equations at some point a in D. Then f is (complex) differentiable at a"

Is continuous here redundant? If they're (u and v) partially differentiable they must be continuous, or does this not follow for partial derivatives?

https://math.stackexchange.com/questions/3781829/calculate-int-0-infty-frac-sin-xx3x-mathrmdx-find-my-mistake
can someone tell what I do wrong here?

It's not saying $u$ and $v$ are continuous, it's saying $\frac{\partial u}{\partial t}$ and $\frac{\partial v}{\partial t}$ are continuous

I mean I take a pacman path

and everything zeroes out

and I can't understand why

12:00 PM

I'm pretty sure it's not redundant

@Thorgott are you speaking to me?

@Threnody For a more elementary example from real analysis, it's possible to have a function f : R^2 -> R whose partial derivatives exist at a point, but aren't continuous at that point, and this can in fact cause f to fail to be differentiable

@hash no

:(

The condition of the partial derivatives themselves being continuous is stronger than them simply existing

12:04 PM

@Fargle I see, so... for partial derivatives, it does not follow that (they exist => they're continuous)

Indeed

But I'm right in saying that it does follow for one variable function derivatives

no, that isn't true

Derivatives don't have to be continuous even in single variable

x^2 sin (1/x)

Oh..

No no... uhh

12:05 PM

good ol' sin(1/x)

The original function has to be continuous if it's differentiable, but the derivative doesn't

Yes! Ok that's what I was trying to say

here's a nice thing to read for a reality check math.stackexchange.com/questions/112067/…

makes you realize just how bad derivatives can be

Multivariable is a weird case even in real analysis, because even if partial derivatives exist, you don't have to have differentiability---in fact, it's possible for both partial derivatives of f to exist at a point, and yet f is discontinuous at that point

I see...

12:08 PM

yeah, however the existence of bounded partials (in a neighborhood) implies continuity

classic example is f(x,y) = xy/(x^2+y^2) for (x,y) =/= (0,0), f(x,y) = 0 for (x,y) = (0,0)

Partial derivatives exist at (0,0) but the function isn't even continuous there

Yeah I see it!

@Thorgott lol that's way above my paygrade, i don't even know what a measure is yet

This is why you have notions like the class of C^n functions, which means "(partial) derivatives of nth order and below all exist and are continuous"

That "and are continuous" does a lot of work

(C^0 just means continuous)

Oh, so there's interest in talking about a class of functions?

the point is just that derivatives can be discontinuous on large sets

12:13 PM

Ignore my choice of words there.

Ok :D

C^n(R^k, R^m) (C^n functions from R^k to R^m) is just a set

From wikipedia: "In mathematics, the term large set is sometimes used to refer to any set that is "large" in some sense" lol

But, for example, x^2 sin (1/x) is differentiable (hence continuous aka C^0), but not C^1

yeah, the ways in which large can be made precise are what's listed in the answer :P

12:17 PM

@Fargle I see.. we haven't done much about this, probably because I chose combinatorics and graphs instead of more analysis and set theory ._.

Can't blame you, I was never very strong at analysis myself

I like it, I just don't like being an automatic theorem prover

But it's helpful to carry around a bucket of counterexamples, just to have something to sharpen concepts against

Thank you all, have a good one

@Balarka @Alessandro open.spotify.com/album/…

12:21 PM

@EdwardEvans fanks

gonna listen soon

fookin bangers on that album

also

someone tell me what Hodge theory is

what do u need it for

idk there's just a course on p-adic Hodge theory this semester (I probs won't take it, looks like I don't have the prereqs) but idek what Hodge theory is

lol

eek

Gonna start a band called Matrix specifically so that I can form a supergroup with Vektor and call it Tensor

12:24 PM

I can tell you what Hodge theory on Riemannian manifolds is, but I think that's not what you want

@Lelouch Fuchsian groups were studied in the late 1800s, certainly before uniformization was proved!! (It was conjectured around the same time, though)

Grothendieck didn't know $\pi_1 \Sigma$ is a Fuchsian group

it's somehow related to the algebro-geometric Hodge theory, but I don't know any of that

fair

I'll probs just stare at the wikipedia page until I start crying

therefore we conclude Lelouch is Grothendieck

12:25 PM

Somehow I don't think p-adic hodge theory uses the Laplacian

which makes one wonder: the fuck do these guys think hodge theory is

@BalarkaSen It's easy to forget such concrete and obvious things when you work at a high level of abstraction

reminds me of how if you say the fourier transform to an algebraic geometer they think you mean...

Lmao yeah wtf is Fourier-Mukai

oh god wtf

lol idk, which MSC does Hodge theory fall under?

12:27 PM

Algebraic geometry is cancelled due to hurting my brain and heart

DiffTop?

classical Hodge theory is DiffGeo

(I think)

I think I have to learn ultralimits

welp

Thanks @Ed I'll check it out

Also Gojira dropped a new song

npnp

orly

12:41 PM

damn

listening

@BalarkaSen noice, from which source?

dunno

just read the definition from wikipedia

A btw @Balarka @Thorgott I asked the topologically/point transitive question on MSE earliee

did u check if that weird example works

12:42 PM

Hopefully it'll get an answer math.stackexchange.com/questions/3781838/…

@BalarkaSen I thought about it, but I couldn't convince myself that the action is topologically transitive

let me see

it should work

nice

@Alessandro well it was certainly a Gojira song

Most definitely

say $f : X \to X$ is topologically transitive, look at the $\Bbb R$-action on $T_f$. Pick any two open subsets $U, V \subset T_f$ intersecting the meridian nontrivially say, so $U \cap X, V \cap X$ are open subsets of $X$, we know by transitivity there exists some $n$ such that $f^n(U \cap X) \cap (V \cap X) \neq \emptyset$.

so there is some $p \in U \cap X$ and $q \in V \cap X$ such that $f^n(p) = q$. That means the $\Bbb R$-orbit of $p$ in $T_f$ contains $q$ as well

so the $\Bbb R$-orbit of $U$ intersects $V$, no? what am I missing

why isnt this trivial

12:52 PM

Wait which direction is the meridian? The one with slices like $X$ or the other one?

the one which slices like $X$

the first return map of the $\Bbb R$-action wrt a meridian is $f$

this is for any two open subsets $U$, $V$ which intersects a common meridian, but you can always flow $U$ a little bit so that they do

Hm wait I don't think I understand the construction anymore

The construction is $T_f := X \times \Bbb R/(f(x), r) \sim (x, r+1)$ if you want me to write it like this

$\Bbb R$ acts on $T_f$ by $s \cdot (x, r) = (x, r + s)$

No ok but how do you flow backwards from points with many preimages through f

yeah, its still a semigroup flow if $f$ is not a homeo

thats annoying

but it's definitely topologically transitive

1:05 PM

Ok I agree with that now

Those semigroups are annoying...

Want to think about ultralimits instead?

Troll

Blocked

Lmao

Dude how do you make an action

Balarka said he wants to learn about them lol

Why am I making semigroup actions

What's really nice is when you don't have forward or backward flow either

1:09 PM

thats me

that's right... it's $e^{-tD}$ for $D$ dirac time

Can you make $\Bbb R_+$ act on $T_f$ by $r \cdot (x, t) = (x, rt)$? lol

Nah

This is too hard

1:37 PM

Is an infinite ``descend" of radicals possible in a nontrivial number field? To be more clear: If $\alpha$ is algebraic (and not equal to $0, 1$) over $\mathbb{Q}$, then does there exists infinitely many integers $n$ for which there's $\beta$ algebraic with $\beta^n =\alpha$, and $\mathbb{Q}(\beta) = \mathbb{Q}(\alpha)$ ?

nevermind, of course not.

@Alessandro Take the $\Bbb Z$-action on $\{0, 1\}^{\Bbb Z}$ by shift, which is topologically transitive, and consider the subspace of all functions $\Bbb Z \to \{0, 1\}$ which is $n\Bbb Z$-invariant for some $n$. This is dense in $\{0, 1\}^{\Bbb Z}$ so the restriction is also topologically transitive

But every orbit is periodic

Is there any direct way to see the above without using unit theorem ?

@BalarkaSen what do you mean? Sequences that have some $n$ values at the "beginning" and then repeat those?

there is no beginning, it's 2-sided. yes, $n$-periodic

look at all periodic sequences

that just works, fullstop, right?

Hmm let's see

1:48 PM

its the same as the doubling map on the circle, that one's semiconjugate to the shift operator on $\{0, 1\}^{\Bbb N}$

and the dyadic rationals are similarly periodic

but it's a fucking $\Bbb N$

What's the problem @BalarkaSen ?

It's obviously dense in the full space

@Lelouch find example of topologically transitive group action which has no dense orbit

I think that works. I need to leave for a moment but I'll think about it soon. Seems good to me though

Yeah I think this is ok

1:50 PM

@BalarkaSen topologically transitive means for any two open sets $U$, $V$, there's an elemeng $g$ such that $gU \cap V$ is nonempty, right ?

lmgtfy but yes

2:00 PM

yeah, why you can't take the integers with cofinite topology and the identity map

Hausdorff man

when did you say Hausdorr ?

*Hausdorff

I didn't

oh

Alessandro wants Hausdorff + separable examples

2:01 PM

oh shit

Hausdorff+separable+Baire forces it to be true

so he wants to understand if something weaker works

Who the fuck does topological dynamics on the cofinite topology dude

Are you nuts

Well that too with the trivial group acting on it:P

lmao

Lelouch = Grothendieck confirmed

lol

Ok, I guess this works, not too sure. Consider the interval $[0,1]$ with the dyadic rationals removed.

The basis for the topology is $(\frac{l}{2^m}, \frac{l+1}{2^m})$

And the functions which are acting are the permutations of length $2^{-m}$ open intervals $(0, \frac{1}{2^m}), (\frac{1}{2^m}, \frac{2}{2^m}), ...$

each element has finite orbit ig

What the hell is this do you mean $2x \mod 1$ or what

2:11 PM

@BalarkaSen what ?

you mean your function is $f(x) = 2x \pmod{1}$

you're writing it in a stupid way

no !

Then what is your function lol

for example, you take the four intervals $(0, .25), (.25, .5), (.5, .75), (.75,1)$ and any permutation of them is an valid fucntion

A: Is a topologically transitive action on a second countable space point transitive?

This is godawful I dont understand your example

@Alessandro posted

0

Take the natural $\Bbb Z$-action on $\{0, 1\}^{\Bbb Z}$ by shift, and consider the subspace $X$ of periodic $2$-sided sequences of some unspecified period length. Clearly the original action is topologically transitive, and $X \subset \{0, 1\}^{\Bbb Z}$ is dense. Thus, the $\Bbb Z$-action on $X$ ...

2:13 PM

@BalarkaSen uhh.... Let me be clear: Fix any permutation $\pi$ of $\{0, \cdots, 2^m-1 \}$, and for $x \in (\frac{l}{2^m}, \frac{l+1}{2^m})$, send it to $x - \frac{l + \pi(l)}{2^m}$

that's one function. Now, do it for all permutations and all $m$

these functions forms a group

Ok, it's some awful group, like direct limit over $S_m$'s or some garbage. What is your claim?

The action is topologically transitive, but hte orbit of any element is finite, and thus not dense

Maybe that works

Seems like a pain

both are clear to see, unless I'm mistaken

A shift operator is easier to me :P

2:35 PM

@Lelouch After some thought I agree what you say should work in principle. Your group is something horrifying like the rooted automorphism group of the binary $2$-tree, but it should be topologically transitive + finite orbit.

The point is you can do it with just 1 self-homeomorphism of a space, which is what a $\Bbb Z$-action is.

Anyway good example, if it works.

user777

2:57 PM

small question: lim n/n = 1 as n approaches \infty, and lim (n/n)= lim (1/n + 1/n + .... + 1/n) = 0 + 0 + 0 + .... + 0 as n approaches \infty. Is this possible?

@BalarkaSen accepted, I'm convinced

Now I have another awful dynamical system to think about lol

3:16 PM

@BalarkaSen ok, nice; I see. thanks.

@user777 you can't swap limits and infinite sum without getting weird results.

essentially what you're doing is writing $1 = \sum_{i \geq 1} a_{j_i}$, where $a_{j_i} = \frac{1}{j}$ if $i \leq j$, otherwise $0$, and taking the limit as $j$ goes to infnity. Swapping inifinite sum and limit can be done in some cases, but not always.

@AlessandroCodenotti Uh oh

4:08 PM

@BalarkaSen lol

I'm trying to understand this homeomorphism of a Cantor like space

fun fact: René Descartes didn't call pullback squares Cartesian diagrams

4:28 PM

@LeakyNun If you don't mind, I want to ask an irrelevant question : Are you a 2nd year undergrad?

no

I probably forgot to update something

So, can you please share your current level of academic education?

I just finished year 3

Nice and congratulations

thanks

4:39 PM

:)

GUYS

I have a question

is anyone here?

TechnoKnight? :(

lol same name

Anyway, my question is

There is a plane, let's call it Q

and there's another plance called P

and there is a line formed by their intersecting, let's call it D

My question is, if we put the equation of Q = equation of P. Will we find the equation for the line D?

@TechnoKnight What is the equation of a place?

I mean plane

There is two planes in space, P and Q

4:47 PM

and now, how can you set two equations equal to each other?

what do you mean?

the equation of Q is something like

x + y + z + c = 0

And the equation of P is something like

x +y + z + c = 0

too

I mean, a linear equation

then

we put

x + y + z + c = x + y + z + c

and we find the equation of the line D

Like x' + y' + z' + c = 0

Is that correct or I'm wrong in doing something?

Did you try a few examples?

what do you mean?

Try putting in some numbers and see what the result is

how that's gonna help me?

4:53 PM

The same way doing examples always helps think about math problems

dude

how the heck that's gonna help me?

Do you know what my problem is?

I'm telling you if I set Q = P

will I find the equation of D?

And I am telling you to try an example and see what you get

what the

are you kidding?

is there anyone else here?

chill

I don't have much time.

4:58 PM

In the time it took you to complain you could have written down an example

Lozansky

$x^p-x$ has p distinct zeros in $\mathbb{Z}_p$ for p prime. I know this follows easily from Fermat's, but I want to show it using the fact that the ideal generated by a polynomial $p(x)$ in a field is maximal iff $p(x)$ is irreducible. This should be possible, yes?

What exactly do you have in mind?

@Lozansky Hmm, I don't see any direct way using that

I mean, $(x^p-x)$ clearly isn't maximal, because it is contained in $(x)$, but that only gives us reducibility and says nothing about the roots

Lozansky

5:15 PM

@Thorgott Good point

I can think of an abstract argument that doesn't rely on Fermat to prove this, but not using Fermat honestly just seems needlessly obtuse

We have to prove that
$$
f(x) = \begin{cases}
0 & if~x~is~irrational \\
1/q & if~x~is~of~the~form ~p/q (p , q \in \mathbb Z, q\neq 0)
\end{cases}
$$

$0 \lt x \lt 1$

have a limit zero at any $0 \lt a \lt 1$.

I'm not able to understand the green boxed thing.

(p/q is in the lowest form)

Which part do you not understand?

I don't know why he took $n$? what is it's role?

I understand for all irrational $x$, $f(x)$ will be zero and hence the inequality $|f(x)| \lt \epsilon$ will be satisfied. It's only the rational number that will cause the problems

And he is representing the rational numbers in the form of $p/q$ as
$$
1/2, 1/3, 2/3, 1/4, 3/4 ....
$$
But what's the role of $n$ here?

5:33 PM

Yes, and for a rational number $p/q$ (in lowest terms), $f(p/q)=1/q$, hence $|f(x)-0|=1/q$ and thus we want to know for which natural numbers $q$, $1/q<\epsilon$

Yes, clear that far.

No, $1/q\ge\epsilon$.

okay, maybe thats what the book (Mr. Spivak) is doing.

I knew that page looked 1000% familiar.

What he does next? Why to introduce $n$?

5:36 PM

Because fractions are built out of natural numbers.

@TedShifrin Do you know Michael in real life?

Yes.

That's how my name (and contributions) show up in his book

I cannot find acknowledgements in the book

Ok, so the point now is that $1/q<\epsilon$ for all large enough $q$, hence there are only finitely many $q$ with $1/q\ge\epsilon$ and consequently only finitely many $x$ in the interval $[0,1]$ for which $|f(x)-0|\ge\epsilon$. So, in any small enough neighborhood of $a$, there won't be any of these $x$ except possibly $a$ itself.

Thats a very clear explanation, thanks.

@TedShifrin Now, I realize how cold it is to say "Calculus by Spivak" when people like you (whom I know) have enough contributions in it.

5:41 PM

Not at all. It's perfectly correct to say it's his book.

If you read the preface to the third or fourth editions, I'm in there. I contributed hundreds of problems and (to the fourth edition) some substantial changes to the text.

Did you get to know about Mr. Gilbert Strang?

Yes, I know Strang quite well. Interestingly, when I taught at MIT for two years, I lectured the 350-person multivariable class two times, and one time he was a TA for me :)

I also credited him in my linear algebra book for influencing the way I taught the material and wrote the book.

Wow!

@robjohn Sir did you have any conversation or something like that with Steve Jobs?