point-set-topology

[summer jolt]

Then he defines the interior as

Can anyone explain why zint A ≠ A?

[sleek thicket]

have you thought about what this looks like for the topological space X = R^n with the euclidean (standard) topology?

it is possible to have Int A = A, that's the definition of A being open

[obtuse meteor]

yeah. Imagine a set like {0} in R, and reason out what Int {0} is :)

do this for a few different kinds of sets in R and you'll start to gain more of an intuition

and once you've done that try it for stuff in R^n

[summer jolt]

So interior here would be for example a closed ball minus the edge points. What I am thinking about is how a point can be in X but X not be a neighbourhood of it.

[sleek thicket]

right, so the closed ball is a good example

what does it mean for S to be a neighborhood of x in R^2? Like, what's the definition?

where S is some set containing x

[sleek thicket]

also if "X" is the entire topological space then X will be a neighborhood of each of its point

this follows from N2 and the fact that the collection of neighborhoods of x is nonempty

[obtuse meteor]

hrm, how does N1-N4 show that the collection of neighborhoods of x is nonempty

[sleek thicket]

it does not

[obtuse meteor]

lol. It should have that

[sleek thicket]

no I mean brown includes it above these axioms

they're concerning a system of nonempty sets at each point

[obtuse meteor]

oh good :P

I was a bit concerned

[sleek thicket]

actually snype I can give you a concrete example, instead of looking at A = a closed disk in R^2, what about A = a closed interval [0,1] in R? that example at the bottom of my screenshot tells us what the neighborhoods are in R, so you can show A is not a neighborhood of its boundary points 0,1

(also this is the book I first learned topology from!)

[obtuse meteor]

wow that is hyper based sham

altho

[sleek thicket]

it is not!

[obtuse meteor]

Now I will bully you

ugct

[sleek thicket]

this book sucks lmfao

[obtuse meteor]

I've never read it

¯_(ツ)_/¯

I do think groupoids are more natural though and I'll always stand by that

[sleek thicket]

so in my frosh analysis course we had to write a paper in spring

like on any math thing we wanted

and this course did multivar. and complex analysis so we ended up invoking the jordan curve theorem several times

so i was like

hm, i would like to prove that

but i thought homology seemed scary and categories did not

and Topology and Groupoids proves the jordan curve theorem using seifert van kampen for groupoids

I have the paper on my website still but it's very bad http://untyped.me/groupoids.pdf

anyways i do not have a high opinion of the book on reflection

[summer jolt]

Makes sense now!

[sleek thicket]

nice!

if you think about R^n a set will fail to be a neighborhood of some point if you can travel in some direction and fall outside the set

travel for arbitrarily small distances I should say

[obtuse meteor]

baby nudges

drive you outside

no matter how baby they are

[summer jolt]

Yes I get it now. It makes sense in rn but in other spaces I suppose one just deals with it

[sleek thicket]

well like

you can't really say what happens in arbitrary spaces imo

it depends on the space

like the simplest criterion is just "A is not a neighborhood of x", that's atomic

[obtuse meteor]

yeah the concept of a neighborhood is really atomic

[sleek thicket]

literally in brown's case!

[obtuse meteor]

there are a ton of different ways of giving a definition of a topological space

IDK which is my favorite hrm

[sleek thicket]

kuratowski closure with a single axiom

[summer jolt]

The thing is that I think I found it hard to get definition on neighbourhood in rn from the 4 axioms

[sleek thicket]

for meme points

a topological space is a pair of a set $X$ and a function $\mathbf{c} : 2^X \to 2^X$ satisfying ${\displaystyle A\cup \mathbf {c} (A)\cup \mathbf {c} (\mathbf {c} (B))=\mathbf {c} (A\cup B)\setminus \mathbf {c} (\varnothing )}$ for all $A,B \subseteq X$

[obtuse meteor]

wdym by this?

there are multiple topologies on R^n

so there are multiple definitions of a neighborhood in R^n that are consistent with the axioms

[gentle osprey]

shamroc$\overline{k}$

[summer jolt]

Yeah sorry I meant it didn't immediately follow in my head that the usual definition given in metric spaces is compatible with the axioms

[obtuse meteor]

ahh gotcha :)

that can be a little bit harder to verify

personally I think it's always a good idea to introduce these kinda've together

[summer jolt]

BTW do you recommend this book?

[sleek thicket]

no

[obtuse meteor]

like the metric space or R^n definition along with the general definition for topological spaces

[sleek thicket]

i anti-recommend it

as someone who learned topology from it for the first time

I do not think it is a good introduction

the groupoid stuff is vvv cool and it's worth coming back to

[summer jolt]

Ok would munkres be better?

[sleek thicket]

but I think the exposition, selection of problems, and selection of topics are weird (not talking about groupoids)

I haven't read munkres but I've heard good things ¯_(ツ)_/¯

[obtuse meteor]

munkres is fairly dry imo

a lot of people have been recommending this book

https://topology.mitpress.mit.edu/

and I have been too

but it's a newer book so it's less well-tested than classics like munkres

[sleek thicket]

this looks pretty cool

[obtuse meteor]

my first topology book was in french and I can't remember what it was but it was super good :(

[sleek thicket]

I like the selection of topics better

oh wow you're fluent in french? I didn't know that

[obtuse meteor]

it was pretty terse and covered it more like french style but it was good

I'm not fluent in french

[summer jolt]

I see some category Theory there that's quite appealing

[sleek thicket]

oh okay

[obtuse meteor]

I am fluent in "throw a pdf into google translate and make do"

[sleek thicket]

lmfao

[obtuse meteor]

I think it's a good way to introduce topological and categorical concepts at the same time

and probably builds a lot of mathematical maturity

[sinful pecan]

damn where was that book when i was taking topology

back in my day we had to make do with looking everything up on the nlab

[summer jolt]

Thats nice to hear because I am also reading basic category Theory by Leinster

[sleek thicket]

hot take: i don't understand how people learn from the nlab. everything is all so jumbled and circularly dependent

[sinful pecan]

100%

[sleek thicket]

i say this despite referencing the nlab quite a bit

[sinful pecan]

yeah nlab is good for finding references when learning, not so much as a first exposure

[shut moat]

A moduli stack of elliptic curves is a moduli stack of elliptic curves

[obtuse meteor]

I use the nlab all the time for learning category stuff and as a reference

I don’t find it too circular

[cedar pebble]

nlab is very hit or miss. The stuff there on classical category theory is very good, I think (it used to be a lot worse but it has got better over the years)

[sleek thicket]

maybe circular is the wrong word? I was thinking about how I looked at the dold kan page the other day (which had a lot of useful info/intuition) and the 2nd paragraph jumps into infinity groupoids. this makes sense, the nlab's whole shtick is the npov, but my eyes essentially glaze over when it does this (which is on like half the articles). Part of my motivation for learning about dold-kan is to understand simplicial stuff and eventually infinity cats, but the exposition on nlab relies on already having that understanding

I'm not saying "nlab bad" but large parts of it are written as a reference for people who already know higher category theory, and so the presentation can feel p upside down

[cedar pebble]

yea I think that's fair enough, there are a lot of pages that jump to this point of view too quickly

[red garden]

Let Y be the topological space that looks like the letter Y. Show that Conf(Y,n) is connected.

what does this mean

'Let Y be the space that looks like Y'

how can i know anything from this description

im new to topology

and?

i dont get it

whats the open setsg

XD

im sorry

[gentle osprey]

slimvesus

[obtuse meteor]

in other words, they'll be open intervals on the three line segments, and at the intersection it's a little y shape with open ends on each "stick"

this forms a basis for the topology

[gritty widget]

time dependent vector fields can eat my ass

that is all

[dusk heron]

Consider a complete Riemannian $4$-manifold $(M,g)$ with a nontrivial Killing field $K$ vanishing at a point $p\in M$. Then the flow of $K$ induces a one-parameter group of isometries on the tangent space $T_pM$ and this one-parameter group leaves at least one $2$-plane $V\subseteq T_pM$ invariant. If $\gamma$ is a geodesic with $\gamma(0)=p$ and $\gamma'(0)\in V$, and if $\gamma$ has a conjugate point $q\in M$, does $K$ also vanish at $q$?

[gentle osprey]

gustavn64

[dusk heron]

In this situation we have a Jacobi field $J$ along $\gamma$ which vanishes at $p$ and $q$, and which is also orthogonal to $\gamma$, right?

[gentle osprey]

gustavn64

[gritty widget]

can confirm the second thing is true, ya. (the second derivative of <J, γ'> vanishes by the jacobi equation, so <J, γ'> is an affine function; if it vanishes at two points it vanishes everywhere.) don't know how to answer the first immediately though, sorry

[placid thorn]

so if I have a topology over a set $X$

[gentle osprey]

Corgwn

[placid thorn]

and $A$ is a subset of $X$ where for all elements $x$ of $A$, you can find a subset $U$ wherein $x$ is an element and $U$ is a subset of $A$

[gentle osprey]

Corgwn

[placid thorn]

how would you show that $A$ is open?

[gentle osprey]

Corgwn

[placid thorn]

does the opennness of these sets only specifically apply to the set of subsets paired with $X$ that we include for our topological space, $T$

[gentle osprey]

Corgwn

[placid thorn]

or are all elligible sets that satisfy the openness property contained in $T$?

[gentle osprey]

Corgwn

[tacit stratus]

let U(x) be some choice of U for a given x \in A

then take the union of all U(x)

this is a subest of A, and contains every element of A, so it must equal A, and it is open

[astral cedar]

U has to be open for this to be true, otherwise you could just take the singleton

And that would be true for any set

[tacit stratus]

oh thats right i missed that U wasn't specififed to be open

[placid thorn]

How do you mean?

Sorry I’m very new to thinking about topology

[gentle osprey]

slimvesus

[placid thorn]

right

so...is that just the definition of a set being open?

[little hemlock]

wdym by "that?" The definition of open set is a subset which is a member of the topology defined on the set

[placid thorn]

right, so $A$ might not necessarily be included in that topology

[gentle osprey]

Corgwn

[little hemlock]

an arbitrary set A might not be open, i.e. contained in the topology, that is correct.

[placid thorn]

right

[placid thorn]

"Recall that a set Z is called countable if it is either finite or it admits a bijection f:N→A. "

is A supposed to be Z here?

[gritty widget]

probably

[placid thorn]

if I had, say, a set [-1,1]

how would a disprove that a set that contains 1 is open on that set?

[ivory dragon]

under what topology?

because a subset containing 1 may be open under the subspace topology

i think i know exactly what munkres exercise this comes from lmao

[placid thorn]

oh, I meant a set over [-1,1]

a set like [x| 1/2 < x < 1]

[ivory dragon]

yes, what topology of [-1, 1]?

or let me rephrase

does this happen to be munkres section 13 problem 4?

maybe it was a 14

one of those

[placid thorn]

if it is, it's by coincidence

[bleak helm]

@placid thorn are you going to answer the question lol? You are trying to prove its not open under what topology?

[placid thorn]

I'm trying to prove that [x| 1/2 < x < 1] isn't open over [-1,1] with 1 as my counterexample

because I know it's on the "edge"

[bleak helm]

Do you know what a topology is?

[placid thorn]

its a collection of subsets of a given bigger space

so, [x| 1/2 < x < 1] might be a member of a topology over [-1,1]

[sweet wing]

wat

like (1/2,1] is open with the usual topology

or if you use the discrete topology {1} is open

[placid thorn]

I should say that the euclidian metric is being used here

[sweet wing]

ye so that's what we usually call the usual topology

[placid thorn]

well why

on the point 1, there's no "neighborhood" around 1

there's no way for me to go an arbitrary positive distance beyond 1 and have space in my set leftover

[bleak helm]

This is what we were trying to pry out of you 😂

If your space is X, then the ball of radius r around a point a is {x in X: |x-a| < r}. So you only look at points from X. So here your space is [-1, 1], so you only look at points inside that interval

So the ball of radius 1/2 around 1 is the interval (1/2, 1]

[placid thorn]

right... so what if it was [x| 1/2 < |x| < 1]?

[ivory dragon]

i dont understand that notation

do you mean the set of x such that 1/2 < |x| < 1?

if so that doesnt contain 1

or does writing using closed brackets [] instead of curly brackets {} imply the closure of the set or something?

because i havent seen that before

[placid thorn]

oh woops

there

the absolute value of $x$ is what I meant. Typo.

[gentle osprey]

Corgwn

[bleak helm]

That's also open. It's just an open interval, (1/2, 1)

Oh, nvm

[ivory dragon]

well (-1, -1/2) U (1/2, 1)

[bleak helm]

I'm dumbo

But, it is open

[ivory dragon]

but in any case its a union of open sets

so its open

and it doesnt contain 1

so i dont really understand your question

[placid thorn]

well, okay, what's an example of a set over $[-1,1]$ on the euclidian metric that isn't open?

[gentle osprey]

Corgwn

[placid thorn]

I thought that a not open set was anything on the "border" of a topology

[ivory dragon]

[1/2, 1) for example

is not open

singleton sets are also not open

i'd recommend you review your definition of a topology - a topology is a collection of subsets (in this case of [-1, 1]), and we call a set "open" if it is in that topology

in the euclidean topology on a set X, a subset U is open if, for all x in U and some r > 0, (x-r, x+r) intersect X is a subset of U

[placid thorn]

right

so, for (X, T) in this case

X would be [-1,1]

and T would be the euclidian norm/distance

well what does being "open" with respect to the euclidian norm mean here?

[ivory dragon]

the "intersect X" is important for this specific question

btw

since it means we'll be intersecting any such intervals with [-1, 1]

when considering whether sets are open

[placid thorn]

Okay, I think I’ve bothered my brain with enough math today

I’ll leave with this question:

So, I’m trying to prove that this weirdly defined set of subsets Is a topology

And to do that I need to show that the intersection of a finite numbers of its elements is also in the topology

[ivory dragon]

right, that's one thing you need to show.

[placid thorn]

so the set is specifically, for any given input set $X$

[gentle osprey]

Corgwn

[placid thorn]

the set of all $U$ such that $X-U$ is countable or equal to the set itself

[gentle osprey]

Corgwn

[placid thorn]

it obviously contains the null set

and obviously, a set made up of the intersection of the null set is also gonna be the null set

but what I want to show is if a set $Y$ made up of the intersections of a bunch of different sets $b$ where for $X-b$, $X-b$ isn't equal to $X$ but is countable, $X-Y$ is also countable

[gentle osprey]

Corgwn

[placid thorn]

also, for uncountable $X$, is it possible that there exist $X-b = X$ for b besides the null set?

[gentle osprey]

Corgwn

[ivory dragon]

not if b is a subset of X.

[placid thorn]

right

which all eligible $b$ are assumed to be members of, given the whole topology thing

[gentle osprey]

Corgwn

[placid thorn]

but back to my first question

for any $b$ that $Y$ is intersecting, $X-b$ might be countable, but since $Y \subseteq b$, couldn't the "uncountable" elements of $X$ that $b$ is removing in $X-b$ go away from $X-Y$?

[ivory dragon]

the "root" of this is showing that X - (U cap V) is countable when X - U is and X - V is

[gentle osprey]

Corgwn

[placid thorn]

right

I'm tempted to say

[ivory dragon]

hint: union of two countable sets is countable.

[placid thorn]

"It's trivial to show by induction that any finite $X - U$ intersection such and such are also countably finite"

oh

[gentle osprey]

Corgwn

[placid thorn]

oh I'm big dumbus

[ivory dragon]

and then the rest of the statement follows by induction yeah

[placid thorn]

In what cases could X-u be infinite?

[ivory dragon]

suppose X = R, U = {x | x is irrational or x < 0}

then X - U is exactly the positive rationals and 0

i dont think theres a good, like... characterization of when X-U in this construction will be finite/infinite

[desert bloom]

I have a question about the shape operator

I know that its self adjoint, therefore, the principle curvatures are orthogonal to each other

but is there an intuitive reason why they must be self adjoint?

like, if I forcibly try to construct a local surface by defining two curves that are not orthogonal to each other, can I even do that?

[tacit stratus]

I think one way is to treat the local behavior of a surface as its quadratic approximation

So for any given point on a surface, take a coordinate system where the surface is the graph of the function, and that point is the critical point (basically, the x-y plane corresponding to the tangent space)

Then, since the second derivative is full rank, it can be locally approximated by the graph of a quadratic, and the principal curves of that are always orthogonal

You could imagine defining a surface as a graph that isn’t locally quadratic

Like this ( x^3 - xy^2)

The thing is, since the second derivative at the origin is zero, the curvatures are both zero anyways, so it’s not a contradiction

[tacit stratus]

Essentially, You can draw a surface where it looks like the principle curvatures aren’t orthogonal at one point, but that necessitates that the second derivative vanishes at that point, i.e the curvature is zero in all directions (i think)

[crimson imp]

People, how can I prove that the Gram-Schmidt process which maps GL(n) to O(n) is a deformation retract, in the sense that $i\circ r$ is homotopic to the identity on GL(n)? Here, r(M) is the matrix obtained applying the process on $M\in GL(n)$

[gentle osprey]

pmorelli

[crimson imp]

And i is the inclusion of O(n) on GL(n)

I tried to find a explicit homotopy but it doesn’t seem to be a clever approach

[cedar pebble]

Gram-Schmidt without the normalization of vectors preserves determinants, is continuous, and commutes with scalar multiplication

so you can get the desired deform retract by first scaling everything so that it has determinant 1 and then apply Gram-Schmidt

[crimson imp]

Hmmm it makes sense, I’ll try to write it down

Thanks!

[cedar pebble]

@obtuse meteor https://ncatlab.org/nlab/show/Introduction+to+Stable+Homotopy+Theory

[obtuse meteor]

thanks

[cedar pebble]

bunch of stable homotopy theory nonsense but there are background chapters at the bottom of the table of contents that do homological algebra up through singular homology and cohomology and then basic algebraic topology up through fundamental groups

[obtuse meteor]

neat!

[placid thorn]

"Let $T_\infty$ be the collection of all subsets $U$ of an arbitrary set $X$ such that $X-U$ is infinite, empty, or all of $X$"

[gentle osprey]

Corgwn

[placid thorn]

is $T_\infty$ not a topology? I feel like that's the case

[gentle osprey]

Corgwn

[placid thorn]

since you could have a bunch of $U$ where, individually, $X-U$ is infinite

[gentle osprey]

Corgwn

[placid thorn]

but when you put all of them together, $X-U'$ becomes countably finite

[gentle osprey]

Corgwn

[placid thorn]

so...could a union of infinite finite sets become uncountable?

[obtuse meteor]

Yes

Write R as a union of singletons

[placid thorn]

right...

and for any infinite set, you can just say without explaining yourself that there exists a set $U$ such that $X-U$ is finite

[gentle osprey]

Corgwn

[placid thorn]

by just defining $U$ as something like

[gentle osprey]

Corgwn

[placid thorn]

"Everything in X except for these, three elements"

[placid thorn]

how do you make a continuous function from a circle to a square?

[shut moat]

maybe a drawing could help you?

try and draw what such a function would look like

[gritty widget]

it might be easier to do if you imagine them as concentric

[shut moat]

ya

[gritty widget]

you would know, approximately circle

[placid thorn]

hm...

how could I transform an arbitrary circle $(x-a)^2 + (y-b)^2 = r^2$ to the unit circle $x^2+y^2 = 1$

[gentle osprey]

Corgwn

[gritty widget]

translate and scale

[placid thorn]

in that order?

[gritty widget]

up to you

good catch in that you usually have to be careful about that

but here it shouldn't matter, if you do it right

i'd translate it to the origin, and then scale, since then the scaling should have a particularly simple form

[placid thorn]

right, the scaling is just dividing by the square of r yea?

*root

[gritty widget]

divide the coordinates by r, not sqrt r

it's a circle of radius r

and you wanna go to one of radius 1

like (x, y) -> (x/r, y/r)

[placid thorn]

hm...

[gritty widget]

now i haven't had a coffee today but i'm somewhat confident in that

try it out

[placid thorn]

ya it's good

but then how would you make the translation happen?

would it just be... subtracting every point's x coords by a^2?

[gritty widget]

the center of the circle given by (x - a)^2 + (y - b)^2 = r^2 is (a, b); how do you send that to the origin with a translation?

[placid thorn]

oh

subtracting all the x coords by a and all the y coords by b

right

[gritty widget]

yes

so what's the whole transformation?

(x, y) -> (?, ?)

[placid thorn]

((x-a)/r, (y-b)/r)

[gritty widget]

cool

[placid thorn]

thank u TTerra

sorry that I have a very small brain

[gritty widget]

that's okay, it will grow

just feed it more math

remember to water it too

[placid thorn]

how would I prove that two sets aren't isomorphic over the euclidian metric?

find for any bijective function between them $f$, there's one point $x,y$ where the euclidian distance between $x$ and $y$ isn't equal to $f(x),f(y)$'s euclidian distance, right?

[gritty widget]

what do you mean by isomorphic? isometric?

[gentle osprey]

Corgwn

[placid thorn]

yea sorry

that's what I meant

[gritty widget]

i see what you're getting at (negate the definition of "isometric" for metric spaces), but one maybe easier thing you can try is to find isometry invariants of the space that aren't the same for your two spaces

by "isometry invariants" i mean properties of the spaces that are preserved by isometries, like boundedness, diameter, completeness (cauchy sequences converge), etc.

so like a sphere of radius 2 wouldn't be isometric to a sphere of radius 3, both with the euclidean metric, since as metric spaces they have a different diameter (even though they are homeomorphic)

a general tip: if you know your spaces are homeomorphic, the property you find should be defined in terms of the metric

[astral cedar]

find for any bijective function between them $f$, there's one point $x,y$ where the euclidian distance between $x$ and $y$ isn't equal to $f(x),f(y)$'s euclidian distance, right?
@placid thorn that’s less of a strategy and more of a restatement 🙂

(A correct restatement, if that was the question)

[obtuse meteor]

Invariants good. Definitions bad

[placid thorn]

so the way I did it for R and (0,1) was

"let x be arbitrary and let y = x +50"

so there's a euclidian distance of 50 between x and y

but since f(x) and f(y) need to map to (0,1), the maximum distance between f(x) and f(y) is 1

well, that's the upper bound

and so since 1 \le 50, the euclidian distance between f(x) and f(y) isn't 50, and so regardless of if a valid f exists for R and (0,1), it's never isometric"

[gritty widget]

that's one way to do it

you can essentially rephrase this as a 1-liner though

||one of 'em is unbounded, other isn't||

usually when we wanna prove two spaces aren't homemorphic/diffeomorphic/isometric/whatever, we do it like that, showing that one space has a property that the other doesn't that would usually be preserved if they were actually homemorphic/...

as cohomo says, definitions bad, invariants good

[placid thorn]

really

I don't think we've learned that bound equivalence is cause for isometry in the course I'm in yet

is (0,\infty) technically "bounded"?

yes of course it is

but I don't want to use that one liner to show that R and (0, infty) aren't isometric

[gritty widget]

why is (0, \infty) bounded?

let's say, with the euclidean metric

[placid thorn]

yea

[gritty widget]

uh

you typed (0, 1) at the start

[placid thorn]

because there's a value, 0, such that all elements within that set are greater than it

well I figured out (0,1)

now I'm trying to figure out (0,\infty)

[gritty widget]

ah, you're on to another exercise

mb

[placid thorn]

yea

maybe...thinking about x=x and y = -x

so, over $\RR$, their distance is 2x

[gentle osprey]

Corgwn
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[placid thorn]

euclidianistically

or maybe showing that for every x and y = -x pair, assuming that the distance between $f(x)$ and $f(-x)$ is equal to $2x$ leads to contradictions

[gentle osprey]

Corgwn

[placid thorn]

and so there's gotta be one that doesn't have a distance of 2x

maybe?

[gritty widget]

answer, using an isometry invariant: ||(0, \infty) is not a complete metric space, whereas R is||

[placid thorn]

but we're talking about $(0,\infty)$ here

[gentle osprey]

Corgwn

[gritty widget]

fixed

[placid thorn]

we haven't talked about complete metric spaces in the course though

[gritty widget]

sadistic prof lol

[placid thorn]

so I feel like using that here is cheating

[gritty widget]

so you're saying, suppose we have an isometry f: R -> (0, \infty), then for any x in R, |x - (-x)| = 2|x|, so |f(x) - f(-x)| = 2|x| as well, then you want to derive a contradiction?

hmm

well then f(x) - f(-x) = 2|x| or f(x) - f(-x) = -2|x|

and f > 0

i don't immediately see it, but you can try playing with that a bit

just noticing that f will always be positive might be enough to pull out a contradiction

[placid thorn]

I dunno, I've got no better idea

*ideas

[gritty widget]

you could try rephrasing the complete metric space bs in terms of actual cauchy sequences and use that to get your contradiction

like, ||1/n is cauchy, its image in R under an isometry f: (0, \infty) -> R is also cauchy, so its image converges in R, but...||

tbh i don't really see how to do this without just nuking it using completeness

[placid thorn]

oh shit I know!!!!

[gritty widget]

👀

[placid thorn]

we just proved that R and (0,1) aren't isometric

could we show that (0,1) and (0,\infty)...

no, we couldn't

[gritty widget]

😔

[placid thorn]

well wait, if the argument works for (0,1), it will work for (0,a) right?

and if we make a arbitrarily large...isn't that the same as showing that (0,\infty) isn't isometric with R?

[gritty widget]

you mean like your earlier argument will apply to show non-isometry of (0, a) and R? it should, yeah

but i'd be very wary about taking a large like that

[placid thorn]

yeah it feels too easy

[gritty widget]

because now you're asking about the (non)isometry of metric spaces that depend on a and whether that is preserved as a gets large

seems hard

[placid thorn]

yea

[gritty widget]

i got to go but good luck on the question, hopefully you find an answer that isn't a completeness nuke

[placid thorn]

yeah me too

[placid thorn]

" In order to disprove isometry, given the euclidean norm, we must find $x$ and $y$ contained in the real numbers such that $x - y \not= f(x) - f(y)$. Suppose we fixed $y$ as 0, here, obtaining that $x = f(x) - f(0)$, or that $x - f(0) = f(x)$. Consider that $f(0)$ is a strictly positive value, since $f$'s output is over all positive reals, so suppose we then set $x$'s value as $-f(0)$, valid, since $x$ is over all the reals, including the negatives. Thus, our expression would expand to $-2f(0) = f(f(0))$, or that $f(0) = f(f(0))/-2$. This is plainly absurd, however, as it implies that there exists an $x$, $f(0)$, such that $-f(x)/2$ is equal to a positive value, and therefore that $f(x)$ returns a negative output. Thus, we must conclude that $x - y \not= f(x) - f(y)$, when setting $x$ to $f(0)$, and $y$ to $0$.
"

@gritty widget

[gentle osprey]

Corgwn

[placid thorn]

how would you show that a triangle and a square in R2 aren't isometric?

leveraging the fact that one only has 3 vertices whereas the other has 4?

[placid thorn]

so, for the metric space ([-1,1], euclidean), {x | 1/2 < |x| < 1} is an open set, yea?

[little hemlock]

@placid thorn are you still stuck on any of these?

[placid thorn]

The last one, yes.

Sorry

[little hemlock]

yes for the last one. Proving that is just a matter of showing that for each x in the set, you can find a sufficiently small open ball containing x contained in the set.

[placid thorn]

Right, where the ball (or line, in this case) is just a “set slider” of sorts.

That brushes up against a bunch of points

Hm...so for ([-1,1], euclidean), {x | 1/2 < |x| /ge 1} isn’t open, because we can’t make a ball that catches 1, right?

Or can we? We could just set our ball’s center at .99 with a radius of .1, right?

[little hemlock]

you have the wrong idea here. An open ball in [-1, 1] is the intersection of an open ball in R with [-1, 1]

We could just set our ball’s center at .99 with a radius of .1, right?
think about the definition of open ball. This open ball would not contain 1.

[placid thorn]

well I'm not quite sure what an open ball means here, I'm still having trouble wrapping my head around this.

is it like...a circle and everything in the circle?

or is it just the boundary of a circle?

[little hemlock]

an open ball in $\bR^n$ would be something like $B_r(a) = {x \in \bR^n : |x| < r}$. Like a filled in sphere w/o its boundary.

[gentle osprey]

kxrider

[placid thorn]

right, okay

[little hemlock]

the relevant fact is that an open set in euclidean space is a union of these guys

[placid thorn]

so let's just take this open ball, and make it's radius .2. Now it contains 1 and 1.05

or did I violate some rule by trying to do that?

[little hemlock]

closer to the right idea. The space we are working in is [-1, 1]. An open ball in [-1, 1] will never contain the point 1.05. But there is an open ball of radius .2 centered at .99

[placid thorn]

right

so our "big space" is like a filter on the bounds our balls can take

[little hemlock]

for example, the open ball centered at .99 of radius .2 would just be the set (.97, 1]

[placid thorn]

for this particular metric space

[little hemlock]

yea. btw typo from earlier, $B_r(a) = {x \in \bR^n : |x - a| < r}$ oops

[gentle osprey]

kxrider

[placid thorn]

right. So we don't need to worry about the fact that 1 is on the "edge" of this particular subset of the big set this metric space is over

[little hemlock]

yup

[placid thorn]

and since 1 U the set from earlier is just this set, and we know all the other points from the earlier set are open, we're good here, yea?

[little hemlock]

all the other points from the earlier set are open
i assume you mean that there are open balls containing each point contained in the set. You're right that each point of the old set is an interior point of the new set for the same reason they were interior points of the old set, and that you only have to check that 1 is in the interior of the new set to show that the whole thing is open.

[placid thorn]

right, so then what about ([-1,1], euclidean), {x | 1/2 /ge |x| /ge 1}?

I heard from someone that this set isn't open, but I'm not sure why

what about the $/ge$ being after the 1/2 would cause this set to stop being open?

[gentle osprey]

Corgwn

[little hemlock]

the idea is to show that 1/2 is not in the interior of that set (i.e. that there is no open ball containing 1/2 contained in the set).

[placid thorn]

right..., well, why not the ball centered at .51 with radius .1?

wouldn't it go to the left of our number line and snatch 1/2?

the distance from .51 to .5 is .01, which is less than .1

[little hemlock]

because that ball is not contained in {x | 1/2 /ge |x| /ge 1}.
you have that .41 is in the ball of radius .1 centered at .51 while .41 is not in {x | 1/2 /ge |x| /ge 1}.

[placid thorn]

okay, fine, why not a ball centered there with radius .01?

...that wouldn't contain 1/2 itself, because 1/2 would be just on the border

and if bump up our radius of the ball even a teensy but such that it's now able to capture 1/2, it'll also be containing stuff outside of {x | 1/2 /ge |x| /ge 1} as well.

is that right?

[little hemlock]

yep

[placid thorn]

so hold the phone

why were we able to get that to work for ([-1,1], euclidean), {x | 1/2 < |x| /ge 1}?

shouldn't the same argument apply to any ball that's trying to-

no

it doesn't!

because the ball doesn't intersect with anything outside of that set!

since we're restricted to [-1,1]

okay, I think that's where my confusion with this definition was

so, by that, (R, euclidean), {x | 1/2 < |x| /ge 1} isn't open

[little hemlock]

right, {x | 1/2 < |x| /ge 1} is open in [-1, 1] but not in R under their usual metrics

[placid thorn]

okay

then I want clarity with one last set...

{x | 0 < |x| < 1 and 1/x \not\in NN}

I think this is open

[little hemlock]

what is NN?

[placid thorn]

because even though those restrictions against x where 1/x is natural numbers is a bit of a minefield here for the positive members of x,

natural numbers

so, x couldn't be 1/2, 1/3, 1/4

any member of the harmonic series

as long as you aren't a member of the harmonic series itself, you're gonna have a finite distance between yourself and the next harmonic down, right?

so you can just make your bubble big enough to avoid the next harmonic down

or the next harmonic up

but you'll always have a buffer of points between either

at least I think so

[little hemlock]

right, this set is open in $\bR$. You have that an open interval is open in $\bR$, and the union of open sets is open. So this set can just be written as $$ \bigcup_{n=1}^\infty (1/(n+1), 1/n)$$

[gentle osprey]

kxrider

[obtuse meteor]

You also need (-1, 0)

To be ultra pedantic

[little hemlock]

oops forgot about the absolute values

[placid thorn]

Yea

And that set also fits everything in [-1,1] right?

I don’t see why it wouldn’t

It seems like shrinking the set that spawns the topology or whatever is better here

[little hemlock]

sure, its also open in [-1, 1] for basically the same reason.

It seems like shrinking the set that spawns the topology or whatever is better here
there isn't really a convention about this, but i think i know what you mean. You can right this set as a union of open sets in R which are small enough to be open sets in [-1, 1] on their own

[tidal forge]

Are there any techniques people know of to determine a maximal polytope containment? I.e., given two d-polytopes, and have the second one fixed, determine whether the first polytope is contained within the fixed polytope given some rotations and/or translations

[red garden]

are normal spaces

like hausdorff spaces but for closed sets?

and is tietzes important?

[tepid depot]

yes

[tepid depot]

i've seen it applied in functional analysis like once. I don't think the proof is important. it's nice to be aware it exists

[red garden]

what about uyrson

uryshon lemma

[tepid depot]

so like

point set topology is of limited importance

there are basic things you gotta know

you gotta know the language

it's very helpful in that way

but pure point set topology theorems are not very helpful

nobody studies this stuff

https://pi.math.cornell.edu/~hatcher/Top/TopNotes.pdf

these notes are the best way to learn point set in my opinion

they're good, and only focused on things people actually use

people studied pure point set a little bit in the 40s-60s, but everything interesting quickly got solved

and what's left is actually set theory/logic

so unless you're interested in that, i wouldn't spend a lot of effort on things outside the doc i linked

you can always learn a random point set theorem when you need it

I was googling and i came across this which explains it really well

From Moschovakis's Notes on Set Theory:

General (pointset) topology is to set theory like parsley to Greek food: some of it gets in almost every dish, but there are no great “parsley recipes” that the good Greek cook needs to know.

the thing is, when more point set comes up beyond that, the book you're reading usually explains it

or gives you enough info to get going

like if i've got the name of the theorem and the buzz words involved, i can probably look it up

and make sense of it

ah i see

that's annoying

you definitely hit diminishing returns after the stuff in hatcher's notes. I don't disagree that knowing these slightly more general theorems exist is worth

basically i agree with you lol

[obtuse meteor]

The person who says this scares me

[rugged swan]

Based

[sleek thicket]

i wanna be the guy

this person sounds awesome

[dim meadow]

Ultra 5 years ago

That sounds like a lot of fun actually

A computability theorists point set topology course would just be a semester on the baire category theorem

I wonder if anyone's ever given a course exclusively on applications of BCT

[shy moss]

in a topological space, How do you show that boundary of boundary is equal boundary

[marsh forge]

?

do you mean boundary of a boundary is empty?

or 0 (in a chain complex, say)?

[shy moss]

Let A be a subset of a topological space, then ∂(∂A)=∂A

[shy moss]

is that true?

[shut moat]

I could be wrong but I think they're sticking with the same ambient space instead of putting subset topology on the boundary

[marsh forge]

Oh by $\partial A$ do you mean $A-int(A)$?

[gentle osprey]

doja max

[shut moat]

Closure\interior I think

[marsh forge]

Ah right.

[shut moat]

[REDACTED]

[shy moss]

yes

[shut moat]

ok R is the boundary of Q so I take back most of what i said

if empty interior, boundary of boundary is boundary

by defn

[tough imp]

But boundary = boundary isn’t true. Take Q in R, the boundary is R, boundary of that is empty

[gritty widget]

what we should take away from this is that the notion of boundary transcends mathematics and enters the field of sociology

[shut moat]

lmao

the rational numbers are such a weird set

R ftw

[gritty widget]

one might say that the notion of boundary is at the boundary of mathematics

[ivory dragon]

i wouldn't.

but one might, i suppose.

[shy moss]

if A is closed then interior of the boundary of A is empty set

[shut moat]

my guess is that for super nice subsets like submanifolds, it's true

[shy moss]

ok, thanks

[shut moat]

like ∂S^n = S^n I think

[tough imp]

The closure of the boundary is always the boundary I believe

Since it’s closed

[tepid depot]

it's funny we were talking tietze today, because I think i can use it to solve a problem my (likely) advisor gave me

[tough imp]

are you in grad school?

[tepid depot]

yeah

[tough imp]

oh huh I thought you were still an UG

I have no clue why

lol

[tepid depot]

🤷‍♂️

still a baby grad student tho

idk i feel like i'm crushing a fly with a hammer here, i'm going to post the problem and my solution in #advanced-analysis to see if anyone has any better ways to do what i did

[little hemlock]

my question is about this last part. How does this map send B_n'' to 0?

[sleek thicket]

probably: chain maps preserve B's so the construction of the map B_n'' into B_{n-1}', which is killed in the quotient

also this is more appropriate for #groups-rings-fields, it's a homological algebra question

here's a real answer: if $z$ were in $B_n''$ we could write $z = \partial'' x$ for $x \in S''{n+1}$. then surjectivity of the map $p : S \to S''$ implies that we can lift $x$ to $y \in S{n+1}$. we can then choose $s_n = \partial y$ since $p(\partial y) = \partial''(p y) = \partial'' x = z$, as $p$ is a chain map. then $\partial s_n = \partial(\partial y) = 0$, so $s'_{n-1} = 0$

[gentle osprey]

shamroc$\overline{k}$

[sleek thicket]

and then the class of that in homology is zero too

[little hemlock]

ah okay i think i get it

also, yea, forgot this question was purely algebraic

[sleek thicket]

@gritty widget sphere quiz

What are the n such that the tangent bundle of S^n can be given a complex structure

[gritty widget]

[sleek thicket]

Ie we can make every tangent space into a complex vector space compatible with the existing real vector space structure

[gritty widget]

,w What are the n such that the tangent bundle of S^n can be given a complex structure

[gentle osprey]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by TTerra#5291

[sleek thicket]

if you were a real geometer you would find this intuitive

And know automatically

[gritty widget]

they're all even dimensional so all of them

[sleek thicket]

Hmm

[gritty widget]

[sleek thicket]

X

[gritty widget]

[sleek thicket]

are you saying all S^n or all S^(2n)?

[gritty widget]

all the tangent bundles are even dimensional

[sleek thicket]

X

[tight agate]

isnt this an open problem?

for n=6

[sleek thicket]

it is not

[gritty widget]

[sleek thicket]

It's true for n = 6

at least that's what my textbook says

[gritty widget]

you're n = 6

[sleek thicket]

the answer I was looking for is n = 2, n = 6

[gritty widget]

n = even prime

[tight agate]

there's an almost complex structure on 6

I dont know if there's a complex structure

[sleek thicket]

what do you mean by an almost complex structure?

[tight agate]

an endomorphism that squares to -1

[sleek thicket]

That is what I mean by a complex structure

[gritty widget]

[tough imp]

oh well you didn't specify that

TTerra would've known the answer if you did

lol

[sleek thicket]

That's what a complex structure on a vector bundle is defined as in this book

[tight agate]

there's an integrability condition for it to be complex

[tough imp]

shhhhhhhhhhhhhhhhhhhhh

[tight agate]

it's an open problem

apparently some physics dude claims to have found one

[sleek thicket]

yeah, if you want the underlying manifold to a complex manifold then you have an integrability conditiom

[gritty widget]

yeah, what, did he smash some fucking particles together and find one?

[sleek thicket]

But what I said above was just a complex multiplication on the tangent spaces

[tough imp]

who here thinks Hartshorne is too easy hahaha

I can't be alone right

hahahaha

[gritty widget]

yellow chat

[tight agate]

[tough imp]

Rip brofibration

[gritty widget]

brofibration for honorable?

[sleek thicket]

simply mute brofibration

[tough imp]

Piss color?

[sleek thicket]

Then we have yellow chat

[tight agate]

[gritty widget]

piss chat

please

if you want piss chat, check out #help-0

through #help-9

[tough imp]

#❓how-to-get-help

[sleek thicket]

#❓how-to-get-help

[gritty widget]

👀

@tight agate #❓how-to-get-help

[sleek thicket]

therapist moment

[tough imp]

did you know if you spell therapist backwards it says
tsipareht

[tight agate]

huh

it actually seems pretty easy to rule out a lot of the spheres

by just using Hirzebruch Riemann Roch

[gritty burrow]

Hi everyone,
I was wondering if a parametric surface in 3D $p(s,t): D \subset \mathbb R^2 \rightarrow I \subset \mathbb R$ can have a one-to-many mapping between its domain and image?

[gentle osprey]

avneesh

[gritty burrow]

Surface is smooth and non-intersecting

[gritty burrow]

By one-to-many I meant that multiple elements from the domain map to a single element in the image

[marsh forge]

thats many-to-one

i think

one-to-many is not even a function unless im being dumb

[gritty burrow]

@gritty widget I am sorry I am very new to the subject. It might imply injectivity. I just wanted to confirm

[marsh forge]

an 'intersection' is exactly a point of non-injectivity

by definition, really

[gritty burrow]

@marsh forge Oh yes. I meant one element from domain to many in image

@marsh forge Okay. Thanks for clarifying

[marsh forge]

wait

one element in the domain can never go to multiple in the image

as long as you are working w functions

[gritty burrow]

@marsh forge but doesn't that apply to only functions? Or does it apply to parameteric curves/surfaces too? Again, I am sorry if my questions are trivial

[marsh forge]

all of this stuff is normally defined as functions

it's possible that you might be wrong about what the domain is, however

[gritty burrow]

@gritty widget yes but a circle cannot be a graph of a function (because it has one to many relationships). But it's representated as a parameteric curve

[marsh forge]

a circle is a function

take [0,2pi] under e^ix

the image is (homeomorphic to) a circle

[gritty burrow]

Okay. Thanks

[desert bloom]

A slightly dumb question

I am trying to proof that the gauss map is invariant under reparameterisation

But I got stuck

I swear I knew how to do it 9 hours ago

I probably even wrote it somewhere underneath the heap of paper I unceremoniously decide to throw on the floor

Now it’s lost in the sea of scribbles

And my brain is kinda fried

Someone help

[desert bloom]

<@&286206848099549185>

[desert bloom]

I’ll try and figure it out

Update, I think I got it

[little hemlock]

pretty hyped for my first class in manifolds/DG today

[gritty widget]

have fun

[little hemlock]

thanks

[gritty widget]

what's the syllabus

[little hemlock]

basic knowledge about smooth manifolds,including the basic concepts, tensors and differential forms, de Rham cohomology, vectorbundles and covariant derivatives, and (possibly) Riemannian metrics.

[gritty widget]

[shut moat]

hot

a question of my own- do you think classical diff geo (like do carmo's diff geo of curves and surfaces) is a prerequisite for abstract diffgeo?

is there much harm done in doing lee's trilogy first

[gritty widget]

abstract diffgeo as in what kxrider just posted? nah

maybe for riemannian though, yeah

tbh

i just went through a smooth manifolds course and a riemannian geometry course without knowing the classical theory

it was a little bumpy in the RG class, but not knowing the classic stuff didn't hold me back too much

also like, i think lee goes over the classic stuff here and there

not as a focus, but as a motivation, or series of examples

you're reading hubbard and hubbard's vector calc book, yeah? that should be enough to jump into lee

[shut moat]

Great! I gave do carmo a peek and i was bored out of my mind lmao

glad I don't need to go through that

ty!

[gritty widget]

do carmo has an RG book that does a lot of the things he does in his curves and surfaces book but in more generality

but do carmo's rg book is also...

[shut moat]

ye probs won't touch it on my own

Old books suck

was traumatized by rudin and i don't want to go through that again

[gritty widget]

it's not a bad book, it's just rather confusing to read through imo

garbage notation

but the last few chapters of do carmo's rg are great in my opinion

lee on the other hand is very kind and doesn't write confusing things

[shut moat]

yeah I read some of LTM and like the first bits of LSM, and they were super nicely written

is a shame that there are no published solutions to exercises tho:(

[gritty widget]

just read my exercise solutions

[desert bloom]

By the way

Can someone check my solution?

I’m not really used to frechet derivatives

[shut moat]

wait how're you defining gauss map? Isn't it a map from S--> S^2? by defn it doesn't care about the parametrization you use

[desert bloom]

Yeah

That’s why I wanna proof that it’s invariant under reparameterisation

It’s just 1 line in my notes

Like so

It’s not a question, I just wanna make sure I get the chain rule right

And I feel kinda dumb for not being able to this quickly tbh

I learnt my entire course but I still mess up with trivial stuff...

[gritty widget]

what is sigma_u and sigma_v?

[desert bloom]

Partial derivative

[gritty widget]

ah the coordinate vectors given by the parametrization

ok let me think about it for a moment

[desert bloom]

Thanks man

[gritty widget]

n(p) is one of the unit normals to the surface at p, and i guess you would want to show that that is independent of parametrization, up to a sign

ok so

[desert bloom]

I wrote my answer down up there somewhere

I’m not sure if it’s right though

[gritty widget]

i am having a brain fart

chain rule hard

what should happen if $\tilde{\sigma} = \sigma \circ \varphi$ is in $$\frac{\tilde{\sigma}_u \times \tilde{\sigma}_v}{|\tilde{\sigma}_u \times \tilde{\sigma}_v|},$$ the numerator will pick up a scalar involving $\det D\varphi$ and the denominator the same scalar with absolute values, so then the above thing is $$\pm \frac{\sigma_u \times \sigma_v}{|\sigma_u \times \sigma_v|}$$

[gentle osprey]

TTerra

[gritty widget]

now let me take a careful look at yours

ugh i hate cross products

i'm not sure what happened here

[gentle osprey]

TTerra

TTerra

[gritty widget]

and then if you compute cross products, hopefully things cancel

[gentle osprey]

TTerra

[gritty widget]

and then that becomes plus or minus 1, which verifies what you wrote

actually, it should be pretty easy to show that, just using the properties of the cross product

@desert bloom

sorry that took so long

[desert bloom]

Np

[gritty widget]

what do u think S^1

[shut moat]

I ws just wondering why this is a thing that needs to be proved

cuz like, the gauss map definitely isn't unique, there are two normal vector fields for each connected component of S (if it's orientable at all)

[gritty widget]

maybe if i phrased it as "prove that the unit normal is, up to a sign, independent of chart" it would make a bit more sense

since that's essentially what this is

[desert bloom]

Intuitively, it makes sense, I need it to make sense mathematically too

Plus practise

[gritty widget]

(at least, that's what i'm thinking)

[desert bloom]

Yeah, that makes sense

[shut moat]

makes sense, ig I'm just used to thinking of vector fields as inherently chart independent

[gentle osprey]

TTerra

[gritty widget]

and there you have it

the entire thing is a bit disjointed but it's just a computation, so it might be good to rewrite it all together and make sure it makes sense

too much of DG is checking that things are chart independent

for example

here's an excerpt from my manifolds course notes

[shut moat]

mamma mia

wait so why's this necessary? we know that dx^i is a 1-form, and da_i is a differential which we know is well defined and is a 1-form, so the wedge should automatically be a well defined form, right?

[gritty widget]

wanna make sure it's equal in the overlap

since it's defined wrt coordinates

then you can extend to all of M

like, it's well-defined in the coordinate chart, yes, but we want something on all of M

[shut moat]

spooky

[shut moat]

I'm not particularly sure why this^ fails though

[gritty widget]

that's absolutely true, it's just not the point

let me post the full thing

[shut moat]

ooh, so it's that the form is the same

heh pun unintended

[gritty widget]

choice of representative <-> choice of coordinate chart, in this case

if we're comparing w algebra

[shut moat]

I meant form as in like the exterior derivative takes the same form in each chart, accidental crash of terminology lol

I wish I knew algebra tho

[gritty widget]

idk what those are lol

rings are neat but i don't know what groups are

[shut moat]

miss me with that category shit

[tough imp]

I like magma

[obtuse meteor]

me: looking in chat and seeing partials

[gritty widget]

i guess i should have approached marco polo's question categorically huh

[tough imp]

I use partials a lot

in my homological algebra course

[obtuse meteor]

[obtuse meteor]

based

the only reason I post sully at partials

is I don't like them and I can't do those computations

this will hopefully change since I'm doin analysis on manifolds this seeester

so invariably we will be doing these kinds of proofs

and I will be sad

[sleek thicket]

let the total derivative into your heart

it will warm you when you are cold and bring you peace when you are troubled

computations are good 🤤

anyways, I gotta use gram schmidt to get an orthonormal basis in stereographic coordinates for my homework 🙃

[gritty widget]

🤢

[sleek thicket]

ttterrrra please help me with bundles

[gritty widget]

prolly 0% chance i can

tbh

[sleek thicket]

[gritty widget]

[tough imp]

[desert bloom]

Heya

Can I ask if anyone have a more qualitative way of thinking about the second fundamental form?

Like I understand the first fundamental form as the dot product on the tangent space

But the only way I understand the second fundamental form is through the shape operator

It seems to be the dot product of something on the tangent plane of the gauss map and the tangent plane

[shut moat]

A smooth surface is locally the graph of a smooth function. If you write this function using adapted coordinates (pulling down the surface and rotating it so that the derivative vanishes at the point you're looking at), then the second fundamental form is the quadratic terms in the taylor polynomial of the function

[gritty widget]

https://math.stackexchange.com/questions/757392/first-and-second-fundamental-form-intuition

may also help

[desert bloom]

Ah

[gritty widget]

"the rate of change of the tangent planes taken in various directions within the surface" is cool

[desert bloom]

Ok

That's how I understand the shape operator

[gritty widget]

well now it depends on what you mean by second fundamental form

in the single week that i studied this in RG, i learned that the second fundamental form is uh... many different things

[desert bloom]

Top bit highlighted in green

How can I interpret a dot product?

For now, I interpret it as some sort of preservation of angles

that's how I see it in the first fundamental form I suppose

[cloud owl]

https://en.wikipedia.org/wiki/Dot_product#Geometric_definition

[desert bloom]

Yeah thanks

[cloud owl]

shamroc don't you sully at me

the geometric and scalar projection things actually make sense

[desert bloom]

I think I kind of got it now

I am a little dumb, so if its purely mathematical, I will get a little bit swamped

I hope that some day I could understand what -S^1 said

[shut moat]

i just meant use these coordinates

[sleek thicket]

kai I was sullying you because it seemed like Marco was looking for geometric intuition more specific to this scenario

[shut moat]

then taylor expand the surface locally

[sleek thicket]

But it seems like they found it helpful

so I am the fool

[desert bloom]

Nah

the thanks was sarcastic

Of course I knew what the dot product is...

[sleek thicket]

oh lol my sully was justified then

Yeah haha that was my thinking

I don't know this classical diff geo stuff very well, otherwise I would try to help

Sorry

[desert bloom]

np

[sleek thicket]

I only know this stuff in the general riemannian setting

[desert bloom]

I tried thinking through it myself

I basically tried using Euler's theorem and parameterise the tangent plane so that it coincides with the directions of curvature

[sleek thicket]

Okay ttera's reaction is fair, I don't really know it in that setting either

But I'm supposed to know it

[gritty widget]

that wasn't a "not like you know it there either" reaction lol

[desert bloom]

Lol

[gritty widget]

it was a

[desert bloom]

don't know anything about that

[sleek thicket]

It was

I know what you really meant

[shut moat]

hubbard will explain it better than I did I guess, let me grab a screenshot

[gritty widget]

[sleek thicket]

[gritty widget]

while we're talking dg

i just computed some christoffel symbols

[desert bloom]

I'm taking an intro to topology next semester

[desert bloom]

so maybe I'll know what you guys are talking about

sooon....

[sleek thicket]

Exciting!

[desert bloom]

Are you guys all masters and post grads?

[gritty widget]

LOL

[shut moat]

lmao

[sleek thicket]

nope, I'm a 3rd year undergrad

[gritty widget]

same

[sleek thicket]

I think ttttteeeerrrrrraaaa is too

[gritty widget]

ya, im third year

[shut moat]

I'm a lowly freshman

[desert bloom]

Me too...

There is quite a huge gap between me and you guys it seems

[gritty widget]

you shouldn't compare yourself to others, especially in something like math

[dim meadow]

^^

[desert bloom]

It's a good yardstick

plus I'm not like, super depressed about it or anything

[shut moat]

well then terry tao would make everyone feel bad

[desert bloom]

Its more like I know I need to work a bit more on it

And plus, I want mathematics to be more like a hobby, so I would wanna do it in my free time as well

as of now, I procrastinate too much!

[dim meadow]

It's more important to decide if it's something you want to work for tbh

[desert bloom]

Its more like, when I do marathons like this and try and understand my course, I can see glimpses to how fun it could be

I do enjoy it

I wanna enjoy it more though

but your maths courses sounds intense

Mine is just real analysis and differential geometry

I wanna throw in group theory as well tbh

but I do have to choose physics modules as well

[shut moat]

physics good

[gritty widget]

giving you about a minute before people come in here to yell at you for liking something with even an inkling of physics in it

[sleek thicket]

[gritty widget]

nooo noooo noooooooooooo it's not pure enough nooooooo

[sleek thicket]

liking physics?

go fuck yourself

[gritty widget]

[sleek thicket]

find a cannon

shoot yourself into the sun

[desert bloom]

The more I do physics, the more I like maths

[sleek thicket]

ritually disembowel yourself

[desert bloom]

So can I unshoot myself into the sun?

[gritty widget]

[shut moat]

best thing do do is both

[gritty widget]

you shouldnt have edited that marco polo

.

[sleek thicket]

[gritty widget]

[desert bloom]

My course is literally called maths and physics

[shut moat]

fuck yeah

[desert bloom]

@shut moat high five

[gritty widget]

what?

[desert bloom]

What?

[gritty widget]

someone solve this system of odes for me

[desert bloom]

No

[sleek thicket]

?

[shut moat]

just do it numerically 4head

[sleek thicket]

this is inappropriate for this channel

[gritty widget]

[sleek thicket]

Please post in #multivariable-calculus

[gritty widget]

they're geodesic equations

[sleek thicket]

thank you

[sleek thicket]

#❓how-to-get-help

[gritty widget]

lee problem 5-4

[sleek thicket]

Oh I think I did this one

Is it a surface of revolution?

[gritty widget]

yeah

i already did the equivalent exercise in do carmo

[sleek thicket]

hmm I don't remember computing geodesics

[gritty widget]

but im just typing it up for completeness

one of the parts is to show that meridians are geodesics

so naturally

i try to write geodesic eqns

[sleek thicket]

yeah but you don't need to find all geodesics for that

[gritty widget]

of course lol

im just memeing

those are hard af to solve

[sleek thicket]

[gritty widget]

[sleek thicket]

hmm

X

[desert bloom]

... is it even possible to solve by hand?

[gritty widget]

this question was nice

[gritty widget]

probably not lol

[gritty widget]

if you write out the connection, it turns out it is ||the standard euclidean connection plus the cross product||

[sleek thicket]

Huh I don't recognize that one

I'm curious what it is

It's obviously not symmetric lol

[gritty widget]

yeah lmao

showing that it's metric-compatible is nice

[sleek thicket]

unless we work in char 2...

[gritty widget]

🤨

it says R^3

[sleek thicket]

[gritty widget]

glad to know the real numbers have char 2

[summer jolt]

Suppse a covector field $\alpha$ is given by
$$
\alpha=a_{i} d x^{i}
$$
Then if we hare a smooth vector field $X=X^{j} \frac{\partial}{\partial x^{j}},$ why do we get
$\left(a_{i} d x^{i}\right)\left(X^{j} \frac{\partial}{\partial x^{j}}\right)=a_{i} x^{i}$
I'm aware that $d x^{i} \frac{\partial}{\partial x^{j}}=\delta_{j}^{i}$

[sleek thicket]

distribute the sums out

[gentle osprey]

snypehype46