Morning

I have a graph theory question

@EdwardEvans I'm stuck with some annoying analysis

I need to upgrade a cone field to a pair of vector fields

How do I

fuck knows

but if you wanna hear a generic example of Galois theory failing for infinite extensions then hmu

lolol

actually quick noob topology question: is there a difference between system of neighbourhoods and fundamental system of neighbourhoods of a point?

or just .. a different name

i imagine both means a neighborhood basis

Ah okay, that makes sense

@BalarkaSen A cone field is really a $(\Bbb Z/2)^2$-bundle

An oriented cone field is a $\Bbb Z/2$ bundle

Guys, donno if I asked this here before, but does anyone know how I can show that we always need to exclude at least one curve starting at the origin that goes to infinity (not intersecting itself), if we want to have a continuous log function in the plane?

@Astyx You there?

Yes

I think all I need is to show that if we don't have such a curve, then somehow it's possible to construct a loop around zero

(which would then contradict continuity, so that's good)

@ShaVuklia Let $B$ be the set you delete, so $\Bbb C\setminus B$ is the domain of the logarithm. If $B$ does not stretch to infinity, then this set contains a non-contractible loop around 0

Proof: take your loop to the boundary of the ball of radius n for n large

Since B is bounded (does not stretch to infinity), one of these loops will never enter B

wait

@MikeMiller Yeah

if could still be that B stretches to infinity, right

@Astyx Can you please help me here

?

I'm only assuming that there is no (continuous) curve that stretches to infinity

But since B contains 0 and that loop isn't zero in R^2 \ 0, it couldn't have been zero in R^2 \ B either

Not right now sorry

I'm busy

there could still be arbitrary far points in B, no?

@ShaVuklia I mean then you're wrong

Let B be the straight line through the origin near the origin

And then at some point to the right, turn that into a topologist's sine curve

And then extend that to infinity by another line to the right

That isn't a curve but there won't be a loop around 0 in C \ B

The problem is that B is taken completely arbitrarily

right, that's a good point

I must say, I don't have clearly in mind what I want to show exactly

the thing is

I want to somehow say what would be the greatest domain of a continuous logarithm

and my idea was: one where only a continuous curve has been taking out of the plane, starting at zero and going to infinity

if I could show that this is at least the condition we need for continuity

Sure, but the topologist's sine curve example shows that you can't just "improve" B by replacing it with a curve

If you want B to be a curve you need to replace it entirely, not just pass to a subset

right, maybe I don't fully understand your example yet, so I'll look at that first, and then I might reformulate my question

I'll draw a picture

If you don’t see what I’m doing with B- look up the topologist’s sine curve

The claims require some nontrivisl t

nontrivial work

right thanks, I'm going to have a look at this (am in a meeting rn, so can't concentrate fully)

@MikeMiller Good example

I mean horrible example, that is

If $(\psi_n) \in H^1(\mathbb R^3)$ is bounded, has $L^2$ norm 1 and converges weakly to 0 in $L^2$, how can I prove $\int_{\mathbb R^3}{|\psi_n|^2\over|x|}dx$ converges to 0 ?

I have a question and I would really appreciate any help/suggestions. I'll attach the screenshot of the problem below.

The usual way eq. 1 is proved is by using an infinitesimal gaussian pill-box. But, I wish to show eq. 1 from eq. 2.

@MikeMiller I'm having problems with asymptotic curves which comes back and intersects itself

To avoid these things I want to upgrade the cone field given by kernel of the second fundamental form to a pair of well-defined vector fields

When can I do that?

Let $V$ be a 2-dimensional real vector space and $Q$ a symmetric bilinear nondegenerate indefinite quadratic form. What are the transformations $T : V \to V$ which do not switch the two null lines in $\ker Q$?

Orientability is not the only obstruction; imagine $Q(x, y) = x^2 - y^2$ and a $\pi/2$ rotation

$v_1, v_2$ be an orthogonal basis of $Q$ with norm $\lambda_1, \lambda_2$. The canonical null vectors are $v_1/\sqrt{|\lambda_1|} \pm v_2/\sqrt{|\lambda_2|}$. I want $T$ to not switch the two

yeah this is unnecessary ill just choose basis such that $Q(x, y) = x^2 - y^2$ exactly and ask $T$ to not switch the lines spanned by $(1, 1)$ and $(1, -1)$

yeah this is just $SO^+(1, 1)$; fine

@BalarkaSen The two asymptotic lines are weighted sums of the +ve and -ve eigenvalues so you can tell them apart; you get two line fields

I suspect that's nonsense

if $M$ is my surface then the cone field gives a classifying map $M \to BO(1, 1)$; since $\pi_1 M = 0$ this lifts to $M \to BSO^+(1, 1)$, I think, is how I want to argue

I crucially want simple connectedness because I think asymptotic lines can wrap back and intersect in catenoids or something

this is the same as what you suggested in the beginning, right? $\Bbb Z_2^2 \to BSO^+(1, 1) \to BO(1, 1)$ is a universal cover, so if $\pi_1 M = 0$, the classifying map of the tangent bundle $M \to BO(1, 1)$ lifts up to $BSO^+(1, 1)$

That gives a pair of vector fields

This is much better than the bullshit do Carmo does

he keeps writing man

Is the only thing that could go wrong when considering a branch of the logarithm that we can find a loop around zero? or should our set satisfy some other condition as well? (the domain of a branch being defined as a open, connected subset of $\mathbb C^\times$ such that the log function is continuous on it)

can we prove that if we can't find a loop that goes around zero, then it always goes well?

"it always goes well" meaning: there exists a (continuous) branch

polytope.net/hedrondude/spaces.htm

Ok, reading this guy and comparing it with ordinal arithmetic gives me some idea on how on earth to visualise:

$$\sum_{k=0}^{\infty}a_k$$

@Thorgott How does one check if a two-variable function is improperly Riemann integrable over a region? I was looking at HK integrability and remembered that you said any improper Riemann integrable is HK integrable, but how would I check if f(x,y) was improperly Riemann integrable in the first place?

Just as in googology the growth rate of ordinals can be understood as the growth of its nth term e.g. $\omega^2$ is the same as $n^2$ in terms of growth, we can generalise this further into infinite series by noting how it is a growth but with many different variables e.g. $mnp$.

And this give us an alternate way to write out partial sums of this infinite series:

$$\sum_{k=0}^{\infty} a_k= \lim_{n\to \infty} \sum_{k=0}^{n}f_n(a_k)$$

How to get the latex symbol for "belongs to" commonly used with sets...

\in

For example $\sum_{k=1}^{\infty}\frac{1}{k^2} = \lim_{n\to \infty} \frac{1}{k f_n(k)}$

where $f_n$ is defined as: $k \mapsto 1$ for $n=1$, $ \mapsto 2$ for $n=2$ and not the first term

Thus in general $f_n(k) = n$ for $k \geq n$ and $f_{n-1}(k)$ for $k \geq n-1$ and so on

Thus the result is a partial sum of the form:

each of these infinite sums are easier to manage, and so you end up with a sequence of numbers and then you take the limit of that sequence to get the original infinite sum

Dumb question: given an infinite extension $\Omega/k$ and Galois groups $G(K/k)$ with $K/k$ the finite Galois subextensions of $\Omega$, Neukirch mentions that $U := \prod_{K\neq K_0} G(K/k) \times \lbrace \sigma \rbrace$ is a subbasis of open sets for $\prod_K G(K/k)$ where $K_0/k$ run over ALL finite subextensions of $\Omega$ and $\sigma \in G(K_0/k)$.

How should I interpret $U$? As $(\prod_{K\neq K_0} G(K/k)) \times \lbrace \sigma \rbrace$ or as $\prod_{K\neq K_0} (G(K/k) \times \lbrace \sigma \rbrace)$

I think the second choice lol

oops

$G(K/k)$ are assumed to have the discrete topology

No, no, @Edward, the former

orly

The "slices" define the Krull topology - or any product topology on product of discrete spaces in general

Ah nice :) thanks

@Secret Are you free? (I mean free enough to solve one of my doubts )

Don't feel like helping out someone's homework today

Hahahahaha it’s not homework but a conceptual problem.

But I got what you meant :-)

hmm... so maybe I can finally tackle this guy...?

$$\sum_{k=1}^{\infty} \frac{H_k}{k} =$$

What does it mean to say a collection of projections on a Hilbert space are pairwise orthogonal?

the inner product between any two of them is zero

What's the inner product on $B(\mathcal{H})$?

Do you mean that there ranges are orthogonal subspaces?

yeah, the subspaces are all orthogonal to each other

what operator is $B$, bounded operators?

a hilbert space is always equipped with an inner product, but what it is depends on what hilbert space you are dealing with

$B(\mathcal{H})$ is the collection of all bounded linear operators on the hilbert space $\mathcal{H}$

ok, then yeah, pairwise orthogonal means any two of the subspaces in the collection are orthogonal to each other, so that the inner product of any pair of vectors taken from each will give zero

Thanks!

Is it true that if $X$ is a topological space and $A\subset X$ is closed and $K\subset X$ is compact then $A\cap K$ is compact?

No, there are non-Hausdorff counterexamples

What do you think?

@MikeMiller I disagree

In compact Hausdorff spaces closed subspaces are compact

Both hypothesis are crucial

@Alessandro Is that a meme

@BalarkaSen I don't see why you need Hausdorff. Let $X$ be compact and $S\subseteq X$ closed. Given an open cover $U_\alpha$ of $S$, $\{U_\alpha\}\cup\{X\setminus S\}$ is an open cover of $X$, pick a finite subcover by compactness, this also covers $S$

Oh, we were thinking of compact subspaces being closed, or something

Yeah that's it

Right, in Hausdorff spaces compact subsets are closed

Lol duh

Also in Hausdorff spaces finite intersections of compact sets are compact but it's false in general

I was thinking line with two origins but each "half-line" is not closed

@MikeMiller Did you see my argument above

I am trying to think of a general counterexample so I can understand how $A \cap K$ extends the finite cover of $K$ into an infinite one

I didn't Balarka

the cone field gives a classifying map $M \to BO(1, 1)$ for the tangent bundle of $M$, and if $\pi_1 M = 0$ this lifts to $M \to \widetilde{BO(1, 1)} = BSO^+(1, 1)$, which gives me a pair of null vector fields out of the cone

Fine, right?

Oh yeah

This is essentially what you said in the beginning, of course

I said something sloppy about eigenvalues that I deleted, but you're simply connected so it all works

$(\Bbb Z_2)^2 \to BSO^+(1, 1) \to BO(1, 1)$ is the universal cover

@MikeMiller ok this may work: $A$ is closed but it contains noncompact sets, and these lie inside $K$ despite $K$ is compact. Thus $A \cap K$ may contain an infinite cover that does not cover $K$ nor $A$

@MikeMiller Yeah I was a bit puzzled. I can draw self-intersecting asymptotic curves in non-simply connected negatively curved surfaces in R^3, I think. I'll "hyperbolize" the Mobius strip by making it saddle-like at every point

there should be examples on the hyperboloid as well but I am not going to work it out.

do Carmo spends 2.5 pages ranting about this particular thing so I like that this is much more concise

@Secret Alessandro proved above that $A \cap K$ is always compact for $A$ closed

@BalarkaSen I don't think you're done though

You've shown you get a pair of vector fields

How do you show the integral curves give a chart?

/ that there is a global foliated chart

Ah yeah; I had already constructed a candidate global chart $\mathbf{x} : \Bbb R^2 \to M$ whose coordinate directions are always tangent to the cone field before

So by uniqueness of ODEs I get that this must be the foliated chart

I don't know how to avoid that jugglery

I'm writing down the details at some place, I'll send you a link when I am done

@MikeMiller I think @Alessendro's proof deal with the special case where $K \subset A$, while Jaakko's example seemed to not even require $K$ to be a proper subset of $A$ I think... unless I am missing something

Oh my god what's the metric for Klein's disk model

@Balarka check discord (private messages)

Thus what Alessndro proved is "In compact spaces closed subspaces are compact"

@Alessandro Ah yeah I saw some notifications earlier this morning; I'll have a closer look tonight.

Irrelevant, since $K \cap A \subset K$ is closed in the subspace topology (which is what's meant by the topology on $K$), so you can apply that result to this pair to see that $K \cap A$ is compact

Nah I'm writing you a thing now that I prefer not writing here

Oh ok

Maybe it is

Ah ok

Dude fuck this

I am not writing down the Klein disk model metric

It is obvious that $\Bbb H^2 \to \Bbb R^3$ given by the Klein model is a short embedding

So by Nash h-principle I get a $C^1$-isometric embedding of H^2 in R^3

the Klein metric is $(1 - y)^2/(1 - x^2 - y^2)^2 dx^2 + xy/(1 - x^2 - y^2) dxdy + (1 - x)^2/(1 - x^2 - y^2) dy^2$

Ridiculous

rest in peace John Conway

If $X={1,2,3}$ and $Y=\{4,5,6\}$ are topological spacses with topologies $\{\emptyset,X,\{1,2\},\{2,3\},\{2\}\}$ and $\{\emptyset,Y,\{4\},\{5\},\{4,5\}\}$, is it true that $X$ and $Y$ are not homeomorphic. How can I show it?

The union of 4,5 has no counterpart in $X$, so there are no morphism that preserves the topology

Ah. You are right.

How can I show the quotient space isomorphic to an interval? Namely, in we define and equivalence relation $\sim$ is $\mathbb R^2$ by setting $(x_1,y_1)\sim (x_2,y_2)\Leftrightarrow $\sqrt{x_1^2+y_1^2}=\sqrt{x_2^2+y_2^2}$ then the corresponding quotient space is isomorphic to $[0,\infty[$.

Ok, got an upper bound to this thing:

@JaakkoSeppälä Think of it this way: What real number are you essentially identifying a given point with?

@BalarkaSen Is the embedding proper

$$\sum_{k=1}^{\infty} \frac{H_k}{k}< \sum_{k=1}^{\infty} \frac{1}{n^3} = \zeta (3)$$

@MikeMiller It's a C^0 small perturbation of the Klein disk so nah

It's like the crochet model of H^2 crinkled up in space a lot

nice secret :) looks interesting

It'll be some epic nonsense like this

The above sequence is then bounded by:

$< \lim_{n\to \infty} \int_{1}^{\infty}\frac{1}{x}dx + \frac{1}{2}\int_2^{\infty} \frac{1}{x} dx + \frac{1}{3}\int_3^{\infty} \frac{1}{x} dx + \cdots + \frac{1}{n}\int_n^{\infty} \frac{1}{x} dx$

which evaluates to $\zeta (3)$

so now I have a handy way to check the convergence of a sequence by extending the usual integral test

To go any further I need to figure out how to go from an area under curve back to countable function values summed under curve, and that will require a reversal of the direction from integral back to riemannian sum or lebesgue sum

I am not sure how to do that yet

I think you use norms instead, which is nonegative by defintion

@BalarkaSen Is there a proper $C^1$ embedding

@InFlames $|x+y|^2 = |x|^2+|y|^2 + 2x \cdot y$, while $|x + y|^2 = |x|^2+|y|^2 + 2|x||y|$.

Thus the triangle inequality is equivalent to the Cauchy-Schwarz inequality $x \cdot y \leq |x||y|$, and the inequality will be closer to equality the closer those two terms are to equality

Because $x \cdot y = |x||y|\cos(\theta)$, where $\theta$ is the angle between the two vectors, you see that the formula is more true the smaller $\theta$ is

You can turn this into formulas if you want.

o wait nvm

clearly my linear algebra had gone rusty as I have not touched it for 5 years

captain america

Jack Ohara

welcome

@MikeMiller That's a good question

I am not sure, doesn't seem possible

Geocalc: Ok, some more thoughts...:

wait... what are you trying to do again?

Computing the Waiting (a past user who loves summing weird infinite series) Sum:

$$\sum_{k=0}^{\infty} \frac{H_k}{k}$$

oh okay

Previously we have:

$$\sum_{k=1}^{\infty} \frac{H_k}{k} < \zeta(3)$$

We can actually do a little bit better by partition the area under the curve into countably many rectangles, and then shrink them towards the integer points

Doing that give us:

I have a question on "Find a basis for the subspace of $\Bbb{R}^3$ spanned by the set of vectors $(x,y,z)$ such that $x^2+y^2+z^2=1$."

@MikeMiller $(\Bbb R^2, g = dx^2 + \cosh^2(x) dy^2)$ is a model for the hyperbolic plane. Consider the coordinate embedding $f : \Bbb R^2 \to \Bbb R^3$. Then $f^* g_{Euc} = dx^2 + dy^2 < dx^2 + \cosh^2(x) dy^2 = g$, so $f$ is a short embedding, isn't it? By Nash there is a $C^1$-isometric embedding $f_1 : (\Bbb R^2, g) \to \Bbb R^3$ which is $C^0$-close to $f$

$f$ is proper so $f_1$ is as well

@geocalc Hey

Since the set of vectors the set of vectors $(x,y,z)$ such that $x^2+y^2+z^2=1$ doesn't satisfy scalar multiplication, they don't form a subspace, and therefore there is no basis. Correct?

@CaptainAmerica16 hi what are you learning these days?

Intro real anal and Linear algebra

$$\sum_{k=1}^{\infty} \frac{H_k}{k} = \lim_{\epsilon \to 0} (1\int_{1-\epsilon}^{1+\epsilon} \frac{1}{x} dx + \frac{1}{2} \int_{2-\epsilon}^{2+\epsilon} \frac{1}{x} dx + \frac{1}{3} \int_{3-\epsilon}^{3+\epsilon} \frac{1}{x} dx + \cdots) = (\frac{1}{1+\epsilon} + \frac{1}{2} \frac {1}{2+\epsilon} + \frac{1}{3} \frac{1}{3+\epsilon} + \cdots) - (\frac{1}{1-\epsilon} + \frac{1}{2} \frac {1}{2-\epsilon} + \frac{1}{3} \frac{1}{3-\epsilon} + \cdots)$$

@BalarkaSen Oh I see.

Unusual, immediately you'd think there's some problem because you can't knock off $\partial H^2$ to infinity; that's collapsing nontrivial geometric info

@CaptainAmerica16 nice I'm taking real analysis next semester (hopefully)

but somehow

@geocalc33 Cool!

Anyway my model of the hyperbolic plane up there is basically the hyperboloid model; so it's wrinkling the end of the hyperboloid very much I suppose

distance as you get farther and farther is becoming exponentially larger

The latter two sums are of the form $\sum_{k=1}^{\infty} \frac{1}{k(k\pm m)}$ which evaluates to:

Makes perfect sense

I like this better than the Klein idea; thanks for the question!

time to blog lmao

@TobiasKildetoft I think the problem is about identifying circle point to a half line.

What is the difference between generatrix and cross section?

$\sum_{k=1}^{\infty} \frac{1}{k (k+m)} = \frac{1}{m}(\sum_{k=1}^{\infty} \frac{1}{k} - \sum_{k=1}^{\infty} \frac{1}{(k+m)}) = \frac{H_{k+m-1}}{m}$

$\sum_{k}^{\infty} \frac{1}{k(k-m)} = \frac{1}{m} (\sum_{k=1}^{\infty} \frac{1}{k-m} - \sum_{k=1}^{\infty} \frac{1}{k}) = -\frac{H_{k-m}}{m}$

Hi everyoe.

Combing these, we have:

$$\sum_{k=1}^{\infty} \frac{H_k}{k} = \lim_{\epsilon \to 0} \sum_{k=1}^{\infty}\frac{H_{k+\epsilon - 1}-H_{k-\epsilon}}{\epsilon}$$ which gives...

@anakhro good morning

Hi geocalc, how are you?

I am good

how are you?