point-set-topology

[gentle osprey]

doubledual

[tepid depot]

$\tilde{f}(1)-\tilde{f}(0) = 1$

[gentle osprey]

doubledual

[tepid depot]

also, 1 is a regular value of f

f^{-1}(1) = {0,1}

but d(0) + d(1) = 2

no good

[nimble cipher]

Ohhh now I see it

[tepid depot]

but, if you use any other regular value, it's fine

like i is a regular value of f

f^{-1}(i) = {pi/2}

degree at pi/2 is 1, it works out

[nimble cipher]

yeah is it because the the derivative vanishes at 1?

[tepid depot]

the derivative doesn't vanish at one

the problem is, endpoints get counted twice if they're in the preimage

but if it's a closed loop, morally it should be counted once

because morally, a closed loop is a map from S^1, not a map from [0,1] with equal endpoints

anyway i would email your prof and mention that example to him

one way to fix it is to just not double count the endpoints

[nimble cipher]

Okay. Thank you. I will ask him before I attempt this one.

[sleek thicket]

If I have a bundle $E \to M$ with compact fibers and a smooth map $f : E \to \R$ is the function $F : M \to \R$ given by $F(p) = \inf_{x \in E_p}f(x)$ smooth?

[gentle osprey]

Shalmorc

[sleek thicket]

in the case that I care about E is the bundle Gr_2(TM) and f is the sectional curvature

[tight agate]

found it

it's in EKMM

[nimble cipher]

Hi going back to my problem here. How do I show that $\tilde{f}(s_{i+1}) - \tilde{f}(s_i) = (d(s_{i+1}) + d(s_i))/2$? Thank you

[gentle osprey]

emphatic_wax

[sleek thicket]

Alright so I figured out that I don't need F to be smooth, just continuous

And smoothness feels unlikely...

So by going local my question is now this:

(ty for the hmm ttera I made a typo)

Let $X, Y$ be topological spaces and suppose $Y$ is compact. Say I have a continuous function $f : X \times Y \to \R$ and define $F(x) = \min_{y \in Y} f(x, y)$. Is $F$ continuous?

[gentle osprey]

Shalmorc

[sleek thicket]

You can assume X, Y are actually smooth manifolds if it somehow simplifies things

hmm this feels less true now

tfw

@gritty widget so I need to find a positive smooth function on M s.t. α(x) <= sec(Π) for all Π a plane in T_x M

It suffices to find a continous one

And it sure would be nice if α(x) = min sec(Π) worked

But I don't see why it would

:(((

[gritty widget]

are there any conditions on M

i'm trying to imagine how such a function would act like on a few different surfaces

[sleek thicket]

Oh shit there totally is

M is complete

I didn't think that would matter but maybe it does??

[gritty widget]

if such a function didn't exist... maybe there's something to be said?

cause like

uh

hm let me think more

[sleek thicket]

lol

So it suffices to define it locally

You can stich via a partition of unity

[gritty widget]

i love partitions of unity so much

[sleek thicket]

???

are you insane

that is

bruh

blocking you

[gritty widget]

i'm joking

pou's are good

[sleek thicket]

some things shouldn't be joked about tterra

I have a strong emotional dependence on partitions of unity

[gritty widget]

[sleek thicket]

Wow

This is a toxic environment

[gritty widget]

is this a lee problem

i wanna see full context

just to make sure i'm not thinking about this in the wrong way ig

[sleek thicket]

It's the last problem in IRM lol

[gritty widget]

[sleek thicket]

Trying to do step 1 of the hint

pls do not spoil the rest of the problem

I've only thought about defining α so far

[tough imp]

Have you tried looking at it on a distinguished open?

[sleek thicket]

yeah

[tough imp]

Rip I have no other tricks left

[sleek thicket]

Wait distinguished open means coordinate chart right?

:smug:

[gritty widget]

[fading vale]

@sleek thicket i used it for u

[gritty widget]

imagine not having nitro

[fading vale]

still dont know when my nitro runs out

[sleek thicket]

I think I proved my lemma

[gritty widget]

right i feel like you could do this in coords and then patch it up

but coords gross

[sleek thicket]

I used to have nitro but then I stopped spending time on discord

but then I became degen again

[fading vale]

good

i missed u

[sleek thicket]

Okay so I have a truly awful proof lol

aww ty moth

[gritty widget]

oh hey this problem is in do carmo (without a hint-solution )

[tough imp]

Reword

Now

[fading vale]

its true

[tough imp]

Now

I will abuse honorable powers if you don’t >:(

[fading vale]

its true

[sleek thicket]

Damn the turn tables

[tough imp]

I give you 30 seconds

[gritty widget]

i need to do more RG exercises

all i do are my hw ones, painfully slowly

[fading vale]

[tough imp]

You made me do it

[fading vale]

smh chmonkey abuse

its true tho

[tough imp]

:(

[fading vale]

who else would buy it for me

[sleek thicket]

So let $X, Y$ be first countable topological spaces with $Y$ compact hausdorff (thus $Y$ is also sequentially compact). Suppose we have a continuous map $f : X \times Y \to \R$. Then the function $F : X \to \R$ given by $F(x) = \min_{y\in Y} f(x, y)$ is continuous.

[gentle osprey]

Shalmorc

[sleek thicket]

Because $X$ is first countable, it suffices to show that for any convergent sequence $x_n \to x$ in $X$ we have $f(x_n) \to f(x)$

[gentle osprey]

Shalmorc

[sleek thicket]

This is equivalent to the claim that for any subsequence ${f(x_{n_k})}_{k=1}^\infty$ there is a further subsequence converging to $f(x)$

[gentle osprey]

Shalmorc

[sleek thicket]

For each $k$ we can choose $y_{k} \in Y$ such that $f(x_{n_k}, y_k) = F(x_{n_k})$, i.e. $y_{k}$ is a minimizer for $x_{n_k}$

[gentle osprey]

Shalmorc

[sleek thicket]

Since $Y$ is sequentially compact we can find a subsequence ${y_{k_j}}{j=1}^\infty$ which converges to some $y \in Y$. Then $F(x{n_{k_j}}) = f(x_{n_{k_j}}, y_{k_j}) \to f(x, y)$

[gentle osprey]

Shalmorc

[sleek thicket]

Okay so it suffices to show that $y$ minimizes for $x$, ie $f(x, y) = F(x)$

[gentle osprey]

Shalmorc

[sleek thicket]

If not there's some $y' \in Y$ with $f(x, y') < f(x, y)$. Choose some $c\in \R$ with $f(x, y') < c < f(x, y)$. By how product topologies work, we can find neighborhoods $U$ of $x$, $V$ of $y$, and $V'$ of $y'$ such that $U \times V' \subseteq f^{-1}((-\infty, c))$ and $U \times V \subseteq f^{-1}((c, \infty))$. Because $x_n \to x$ and $y_k \to y$ we can find some large $j$ such that $x_{n_{k_j}} \in U$ and $y_{k_j} \in V$ (in fact this holds for all large $j$) and so $f(x_{n_{k_j}}, y') < c < f(x_{n_{k_j}}, y_{k_j})$, which contradicts the definition of the ${y_k}$

[gentle osprey]

Shalmorc

[sleek thicket]

This sucks lol, I bet there's a much nicer proof which doesn't involve sequences

just for posterity, this is proving that if $f : X \times Y \to \R$ is a continuous function and $Y$ is compact and $X, Y$ are nice spaces (manifolds) then $F : X \to \R$ defined by $F(x) = \min_y f(x, y)$ is continuous

[gentle osprey]

Shalmorc

[sleek thicket]

okay so it's obvious that $F^{-1}((-\infty, c))$ is open, since $F(x) < c$ iff $f(x, y) < c$ for some $y$

[gentle osprey]

Shalmorc

[sleek thicket]

But $F(x) > c$ iff $f(x, y) > c$ for each $y$, and that's not obviously open...

[gentle osprey]

Shalmorc

[nimble cipher]

Hi. So far, I proved that f^{-1}(z) is finite. I'm still confused how to show that the sum is equal to the difference of the values of the lifts

[nimble cipher]

I was also able to show this bit: $\tilde{f}(s_{i+1}) - \tilde{f}(si) = (d(s{i+1}) + d(s_i))/2$

[gentle osprey]

emphatic_wax

[nimble cipher]

I just need the full equality in the problem

[sleek thicket]

What are the geodesics of the cylinder? Are they circles, vertical lines, and helixes?

[gritty widget]

sounds right

if you cut out a vertical line from the cylinder you get something isometric to the plane (or rather an infinite horizontal section thereof)

the geodesics there get mapped to one of those three

well that might not be entirely convincing but i think it forms the basis for a proof

if this is for the second part of that problem in lee then you can also ||take a paraboloid, say, z = x^2 + y^2. the sectional curvature at the vertex is 0, but any meridian is a line|| this is wrong, the sectional curvature of this paraboloid at the vertex is 4 (but does go to 0 as you go further up it!)

[sleek thicket]

No, different problem

[gritty widget]

ah i c

[sleek thicket]

I already had an example for that one

[gritty widget]

i thought "zero curvature" so i thought you were doing the other part lol

[sleek thicket]

I think I have a counterexample to a problem lol

[gritty widget]

post it on MSE and have lee himself roast you

well that's not even necessary

[sleek thicket]

o o f

so like

[gritty widget]

he's your prof!

[sleek thicket]

23a

[gritty widget]

i don't know cut locus / injectivity radius stuff

[sleek thicket]

Take M to be a cylinder

oh no worries I don't either

Here's the relevant section

So the cut locus of a point on the cylinder is the antipodal line

and thus the distance is π (right?)

I'm taking R = 1

So take p = (1,0,0) and q = (1,0,π)

Then d(p, q) = π yeah?

[gritty widget]

mmhm

did u check errata

[sleek thicket]

yes

So the only geodesic going through p and q is γ(t) = (0,0,t), right?

Like any unit speed geodesic going through p looks like γ(t) = (cos(at), sin(at), bt) where a^2+b^2 = 1

And I did some algebra and found that a is forced to be 0

Maybe that's wrong??

oh yeah so like p and q aren't conjugate points on the straight line geodesic because the straight line geodesic is minimizing for all time, I think

[gritty widget]

something like that, i'd think. conjugate points hurt my brain

[sleek thicket]

Yeah so like, I know that if a conjugate point occurs then for all further times the geodesic stops being minimizing

By prop 10.32a

ahhhhhh

Confusing

[gritty widget]

okay maybe i do not understand the geodesics of the cylinder

i am failing to visualize the result of the problem

[sleek thicket]

mood

[gritty widget]

but i see where you're going with it

[sleek thicket]

It seems like this is a counterexample and I'm confused

Maybe I should look at like R^2 \ {0} instead

[gritty widget]

not complete

[sleek thicket]

oh right lol

But the cylinder is right? It's a closed subset of a complete riemannian manifold, R^3

[gritty widget]

yes

[sleek thicket]

Also all the geodesics exist for all time lol

Because I just classified the geodesics

[gritty widget]

ok so if your points p and q aren't conjugate then i'm imagining like

uh let me draw something

gonna do this in paint since i'm too lazy to get my phone

[sleek thicket]

So I have a proof when q is actually in the cut locus

it's kind of gnarly tho

[gritty widget]

okay nevermind i cannot draw

[sleek thicket]

lol

[gritty widget]

i want to say you can have two helical geodesics from p that wrap around and meet q in such a way that their tangent vectors are negatives

but like

i am not sure that's possible

yeah i don't have any ideas right now

i'm gonna go eat, gl on the problem

[sleek thicket]

Yeah but like

Okay that makes sense but I couldn't find them

And also why would they be minimizing??

Like they need to have length < π

Maybe I fucked up computing the lengths...

[gritty widget]

it didn't seem like there were any errors

i'm having trouble imagining a geodesic of the cylinder shorter than just the straight vertical line from p to q

[sleek thicket]

Okay so like, γ(t) = (cos(at), sin(at), bt). To go from p to q you gotta complete a single loop in the circle, so a t0 = 2πn and b t0 = π

[gritty widget]

okay i am actually gonna eat now

[sleek thicket]

Right?

okay go eat

have fun "eating"

[gritty widget]

[sleek thicket]

eating (derisive)

Hmm okay so wlog γ has unit speed and so t0^2 = (a^2 + b^2) t0^2 = (4n^2 + 1)π^2

Which is never < π...

If we choose n = 0 then t0 = π but then b t0 = π forces b = 1 and since a^2 + b^2 = 1 this can only happen if a = 0, so we're looking at a straight line

so in general like b = 1/sqrt(4n^2 + 1) and a = 2n/sqrt(4n^2 + 1)

Okay yeah so for each integer n we get a helical geodesic which is (1,0,0) at time 0 and (1,0,π) at time t = π sqrt(1+4n^2)

And flipping the sign of n gives a geodesic with opposite velocity

[sleek thicket]

I'm 100% convinced there's an error

So you can actually compute the geodesics very easily. S^1 × R is a lie group with a bi invariant metrics, so geodesics starting at (1,0,0) are one parameter subgroups, which are all of the form γ(t) = (e^(i a t), bt)

And I did the algebra several times for when these geodesics pass through (1,0,π) if you start at (1,0,0)

blech

[tight agate]

I think there was some author who would pay you if you found an error in their textbook

[gritty widget]

oh so there actually was a helical geodesic like the one i was thinking of?

i felt like it'd fail because of some uniqueness-of-geodesics thingy (imagining the tangent vectors at q as being tangent to the circle at height pi so by uniqueness they have to be circles...) or something

maybe i shouldn't avoid computations

maybe i'll do this one after i finish this lie group exercise

lie groups are very nice

everything is just

do thing at T_eG
left or right multiply
what could be more nice?

[sleek thicket]

Exactly

Actually maybe they aren't orthogonal...

So let γ(t) = (cos(2nt/sqrt(4n^2 + 1)), sin(2nt/sqrt(4n^2 + 1)), t/sqrt(4n^2 + 1)) and σ(t) = (cos(2nt/sqrt(4n^2 + 1)), -sin(2nt/sqrt(4n^2 + 1)), t/sqrt(4n^2 + 1)). Then γ'(t) = (-2n/sqrt(4n^2 + 1) * sin(2nt/sqrt(4n^2 + 1)), 2n/sqrt(4n^2 + 1) * cos(2nt/sqrt(4n^2 + 1)), 1/sqrt(4n^2 + 1)) and σ'(t) = (-2n/sqrt(4n^2 + 1) * sin(2nt/sqrt(4n^2 + 1)), -2n/sqrt(4n^2 + 1) * cos(2nt/sqrt(4n^2 + 1)), 1/sqrt(4n^2 + 1)), so at time t0 = π sqrt(4n^2+1) we get γ'(t0) = (0, 2n/sqrt(4n^2 + 1), 1/sqrt(4n^2 + 1)) and σ'(t0) = (0,-2n/sqrt(4n^2 + 1), 1/sqrt(4n^2 + 1))

So you don't get γ'(t0) = -σ'(t0)

The z coordinates are the same (and this is geometrically obvious)

I'm glad I'm doing this because it'll prepare me well for the math gre lol

[quaint agate]

Hey

Does anyone here happen to know if the extended (or signed) gauss code of a link diagram can be written in a canonical way? If so, please let me know I’ve been trying to find a resource which shares how to do this but I haven’t found one thus far.

[mystic geyser]

I feel dumb but could anyone give me a clue as to how he gets this?

I mean, the indices between the LHS and RHS don't even match...right?

This is in the context of flows, sigma is the flow generated by X.

[elder yew]

Chain rule?

$$ \varepsilon X^{\nu}(x) = \varepsilon X^{\mu}(x) \partial_{\nu} Y^{\mu} (x) $$

[gentle osprey]

MoonBears-C-

[elder yew]

It's either that or

$$ \partial_{\nu} Y^{\mu} = \delta_{\mu \nu} $$

[gentle osprey]

MoonBears-C-

[mystic geyser]

Here $X^\mu$ and $Y^\mu$ are the vector components while $\partial_\nu =\partial/\partial x^\nu$, so I don't think it would give a delta

[mystic geyser]

RIght?

I also gets multiplied by Y^\mu so yeah I got no idea

[uncut surge]

These indices are a disaster

I think the X^\mu in the second line should be an X^\nu, then this is an okay Taylor expansion I believe

@mystic geyser

So it's prolly just a typo. It looks like the X and the \partial have the same index as he continues with his calculation, too

[mystic geyser]

ohhh lmao I see it now

thanks a bunch

[desert bloom]

I just learnt about the shape operator and I have some questions

On a surface, which direction are the eigenvectors of the operator?

Is there a way to visualize the principal curvatures and its directions?

[sleek thicket]

I don't really have a good general rule for you, sorry (because I don't really understand the shape operator lol)

but I think it's helpful to look at some examples

eg the cylinder, the hyperbolic paraboloid

also general surfaces of revolution are doable and you can write down the principal directions easily

oh yeah this is sort of what I was trying to get at with my examples but I didn't know how to communicate it lol

also spectral theorem stuff tells you that the principal directions will be the vectors $v$ of norm $1$ which maximize $\langle v, sv\rangle$

[gentle osprey]

Shalmorc

[desert bloom]

Oh, I see

The taylor expansion thing is new. They haven't brought that up at all in class

Most extreme curvatures makes sense

they taught me that the principal curvatures are the eigenvalues of the shape operator

and that the eigenvectors are the corresponding directions

Its still kind of hard to visualise what exactly the shape operator is doing

They say in spirit, its the derivative of n (sigma)^-1

[marsh forge]

Hi so my gf has this exercise

It’s really similar to one from harsthorne 2.7 specialized to X being Spec A

And I think her version might be misstated? I don’t know a lot about this stuff but I can’t think or any reason why it should be true

Whereas the harstorne restatement is a lot simpler to prove I think

(Hartshornes version says given A->L we get a bijection between points in Spec A and maps A_p/m ->L)

@tough imp @sleek thicket it’s ur time 2 shine

[tough imp]

This is true I think

you know the kernel is prime as it's an integral domain

then you can embed that integral domain in different ways I think

[sleek thicket]

i agree

[tough imp]

That map is injective so it induces a map on the field of fractions of the integral domain

and by commutative algebra stuff that's k(p)

[sleek thicket]

so to be clear, maps A -> L are determined by a prime p (their kernel) and an injection A/p -> L. If B is an integral domain then injections B -> L are the same as injection from the field of fractions of B into L, and fof(A/p) = k(p)

[marsh forge]

Oh I see I see

[tough imp]

Fuck dude

this problem was so ahrd for me originally

but now it's like obvious

[marsh forge]

I feel silly now

Thank you

[tough imp]

[marsh forge]

This stuff is fun honestly maybe I should be a geometer

Jk I’ll just pretend to understand A1 htpy

[sleek thicket]

Max you should help me with geometry rn

I have a smooth positive function α : R -> R

[marsh forge]

I’ve seen the cursed shit ur doing sham

[gritty widget]

daily reminder that geometry is painful

[marsh forge]

I want no part of it

That’s not the geometry I mean ttera

That’s cursed as hell

[gritty widget]

(i know)

[sleek thicket]

Why does the initial value problem u'' + α u = 0 with u(0) = 1, u'(0) = 0 have a solution defined for all time

pls

I need geometry

My children are starving

[gritty widget]

rauch stuff?

just handwave the ode results

[sleek thicket]

lmao

[marsh forge]

Have you considered homotoping it

[sleek thicket]

You are both very helpful

[gritty widget]

just solve the ode

[sleek thicket]

I tried

But I was too weak

It turns out any 2nd order ODE y'' + qy' + ry = 0 is of this form up to a substitution (but maybe α is not positive anymore)

That's pretty cool

[marsh forge]

It’s cursed

[gritty widget]

for u'' + α u = 0 i think this one depends on the sign of α

but like

three cases

so it's not that bad

[sleek thicket]

Hmmmst

[gritty widget]

like

α = 0 gives you a constant solution, done
α < 0 gives you something with exponentials, done
α > 0 gives you something with sin and cos, done

[sleek thicket]

I mean I know how to solve it when α is constant lol

With that sign stuff

[gritty widget]

oh alpha is function of time?

frig

[sleek thicket]

Yes

Lmao

[gritty widget]

[sleek thicket]

I'm not that bad at ODE

[gritty widget]

ODEs belong in wolframalpha

or for that one guy in #multivariable-calculus to solve

oh wait you posted in there

maybe he'll show up

[sleek thicket]

Lmfao

W|A didn't help

What if you could like

Make this into a geodesic equation or smth

Idk lol

[gritty widget]

just rewrite the proof of existence and uniqueness

[sleek thicket]

Some second order equation we already know has solutions

[gritty widget]

[sleek thicket]

[cloud owl]

wolfram given alpha

[gritty widget]

okay off the top of my head i don't know how to show this thing has a solution for all time

usually if you gave me a 2nd order ode id like

try to turn it into a linear one using v = u'

but :(

[sleek thicket]

Yeah...

This is just part of a hint in IRM

[gritty widget]

so it's probably got a full worked solution in do carmo

[sleek thicket]

He says "let u : R -> R be the solution to the IVP..."

With no further explanation

lmao

[gritty widget]

i vaguely remember like

if your ODE satisfies some super specific lipschitz condition then you can extend it for all time

idr the details

something like if you can find a lipschitz constant that works everywhere (independent of time) then you can extend to all time (sounds false but it's the best i remember)

but i see no reason why that should be the case for this problem

that's all i got

[sleek thicket]

oh wait you do actually get that solutions exist by introducing v = u' right?

[gritty widget]

how do you deal with the u

if you do that

[sleek thicket]

we have the linear system
u' - v = 0
v' + αu = 0

Solutions for this exist by Picard lindelof

Locally I mean

[gritty widget]

hmmmmmmmm

[sleek thicket]

I'm looking at the proof of existence and uniqueness of geodesics lol

This is how you prove that

[gritty widget]

ODEs bad i'll go back to my lie groups

[sleek thicket]

Okay yeah Lee actually proves a linear ode thing which gives existence for all time!

[gritty widget]

nice

what page?

i'm curious to see

[sleek thicket]

If you do the substitution trick I think this applies

Page 4.31

[gritty widget]

smh he could have stuck that in the appendix

makes sense

very nice

[sleek thicket]

Oh yeah Page 4.31 does not exist

What am I saying lmao

tfw you haven't eaten

[gritty widget]

smh go eat

[sleek thicket]

Page is 106

[sleek thicket]

The sectional curvature at the vertex is 4, right?

Like sectional curvature = gaussian curvature for a surface

And I computed that the gaussian curvature there was 4 last week lol

By surface of revolution stuff

Now I'm confused as to why the meridian isn't minimizing tho...

[gritty widget]

it shouldn't be 4 at the vertex

if it were you'd contradict bonnet myers or something like t hat

i haven't actually computed it

uhhhhhhh

okay i might be wrong

let me be a bit more careful to avoid saying anything dum

no, you are right, it should be 4

i misspoke, sorry

[sleek thicket]

See here

For curvature computation

Kk

Anyways I think you can look at like S^2 × R?

This is a product of manifolds with nonnegative sectional curvature so it had nonnegative sectional curvature too

Maybe R^2 so sectional curvature isn't degenerate lol

I think you can compute that vertical lines are lines

Oh wait actually

We did this problem last week

If G is any nonabelian compact connected lie group I can choose a bi invariant metric

[gritty widget]

oh hey i just did that exercise

[sleek thicket]

Oh nice lol

Anyways just like

If G is a nonabelian compact connected lie group

Look at G × R

This will have nonnegative sectional curvature since it's a lie group with a bi invariant metric

But it's nonabelian still so not flat

I should've known, every exercise is better with lie groups

[gritty widget]

how do you visualize a line/vector bundle over a scheme? I can't seem to find any visualization so is there like a reason for this?

[meager python]

Can you visualize it for a variety?

Grassmannian bundles should give you plenty of visualization for the easier cases

[gritty widget]

@sleek thicket stackexchange has a nicer proof for X locally compact Hausdorff

[sleek thicket]

sick

For the minimizing a function along fibers thing?

[gritty widget]

https://math.stackexchange.com/questions/1604210/when-is-the-image-of-a-proper-map-closed

[meager python]

Also you can try to visualize O(1) @gritty widget

[sleek thicket]

Wait what

[gritty widget]

So if you have an open set U in XY and some aY in U

Then you can find open set V in X s.t. V*Y is in U bc Y is compact

It follows that X*Y to X is a closed map

[sleek thicket]

I don't see why this is relevant to me, sorry

[gritty widget]

@meager python I can’t visualize O(1) on projective line oops

Somehow it’s different from O(-1) idk how visualize these

[meager python]

Take the projective line and remove a point. Take a line disjoint from the point and project away from the line. This is a realization of O(1)

[gritty widget]

I just need to think of it as glueing together the trivial bundles on open subsets

[meager python]

I think O(-1) is similar but opposite

[gritty widget]

oops sorry i still don’t see it

[sleek thicket]

What was this about btw?

[gritty widget]

yes

[sleek thicket]

I don't see how the projection being closed proves that the function F(x) = min_y f(x, y) is continous

[gritty widget]

@meager python is this visualization of O1 correct?

Oh I think Finverse((-infty, c]) is closed @sleek thicket

[meager python]

You should get a twist, I think it’s called the serre twist

[sleek thicket]

Oh I see, it's the image of f^-1((-infty,c])

[meager python]

Similarly to moebius strip

[sleek thicket]

Thanks!

[gritty widget]

Yeah except it’s like an algebraic twist

But is this like good intuition though? Basically I’m trying to glue together two trivial line bundles.

[little hemlock]

im not seeing why the underlined part is true

agh wait a second hm

yea, so my understanding is that every sawtooth has to "touch" 1, so why can't we just take a basic open set of the form $$U = \prod_{x \in [0,1]} [0, 1/2).$$ Then, if $f : [0,1] \to [0,1]$ is any sawtooth, there is $y \in [0,1]$ such that $f(y) = 1$ so $f \notin U$.

[gentle osprey]

kxrider

[little hemlock]

i.e. U is an open neighborhood of the zero function which doesn't intersect the set of sawtooths. ig i'm misunderstanding something tho?

[uncut surge]

Is it really clear that this U is an open set in the product topology? Since this is an infinite product of sets, I don't think this is clear

Wiki says that if you have a product like this, but almost all factors in the product are equal to the whole space (i.e. [0,1]), then that's defined to be open, but this one doesn't fit that description

[little hemlock]

agh oops that's exactly right thanks lol

[uncut surge]

Ye, and if you have open sets like that, where almost all factors are [0,1], you don't run into this problem

Naisu, I feel like I learnt something from this

[digital glacier]

What book is this?

[sleek thicket]

Introduction to Riemannian Manifolds by Jack Lee

[digital glacier]

Thanks. I remember considering reading that

[sleek thicket]

It's good

[quartz rover]

@chrome dew i think what the authors want to say is this actually

${(A^T)_\alpha}^\mu$

but they wrote the indices backwards

[gentle osprey]

Floccinaucinihilipilificator

[chrome dew]

yeah I agree

[quartz rover]

cuz if u dont do this then the dimensions dont match up

ok thanks for giving it a look over and reminding me we are dealing with scalars

that was really hurting my head

[chrome dew]

yeah, no problem

[rugged swan]

Hi, is there a non trivial continuous function from A^1 to R ?

where A^1 is the affine space on R, its closed sets are the finite subsets

[gentle osprey]

slimvesus

[rugged swan]

oh yes fk

thx

do you have an example of topological spaces X and Y and a function f : [0,1]xX -> Y which is continuous on both variables but not continuous ?

i.e. f(t, .) continuous for all t in [0,1] and f(.,x) continuous for all x in X

ty

wtf

ok then what do we get if we study homotopy but with the definition that an homotopy between f and g is a function H : [0,1] x X -> Y continuous in each variable. Thus we can see H exactly as a function H : [0,1] -> C(X,Y) where the topology of C(X,Y) is the topology of the simple convergence ?

I'm sure we can define some groups as the fundamental group

[sleek thicket]

iirc the book topology and groupoids has an exercise on this and most things are degenerate

Let me see if I can find it

So this should show that the "fundamental group" of the circle is trivial

[meager python]

What are two schemes X,Y such that Hom(X, Y) is empty?

[rugged swan]

yeah thx @sleek thicket ! the homotopy H(t,e^2ipix) = e^2ipitx works well too imo

[sleek thicket]

@meager python Spec F2 and Spec F3, yeah ?

Any two affine schemes with incompatible characteristics should work

[meager python]

Ah truuu, no ring morphisms

[cold vine]

How do I go about showing that suspension is functorial. Im showing that continuous maps induce continuous maps. So I have f:X->Y and then clearly fxid:XxI->YxI or even p*fxid:XxI->SY but how do I show that there is a continuous map from SX->SY? Is there some universal property that helps here?

[rugged swan]

bc p*f ((x,0)) = point

and p*f((x,1)) = point

you can factorise by Xx[0,1]/~

(it's the universal property of the quotient space)

[cold vine]

Ah yeah true, thanks! For some reason I thought that wouldn't work.

[cold vine]

So now I am to show that since I've shown that suspension S is a functor that now if f:S^n->S^n has degree m then Sf:SS^n->SS^n has degree m. Well I know that SS^n=S^n+1 and that S and H_n are functors. And that by definition of degree Hn(f) is multiplication by m and now i need to show that Hn+1(Sf) is also multiplication by m but not sure how to continue

[rugged swan]

You have a natural isomorphism H_n+1 (S X) -> H_n(X)

use naturality

[cold vine]

hmm ok. how does it follow that that is an isomorphism?

[rugged swan]

uh

it follows from the excision theorem

let CX be the cone of X, you have a s.e.c. 0->C(X)->C(CX)->C(CX,X)->0

which induce a long exact sequence

and we have a natural quasi-isomorphism C(CX,X) -> (CX/X,{X}) by excision

[cold vine]

ahh alright i think i get it! thanks!

Oh I actually have proved that before for reduced homology😅 and it works here

[rugged swan]

yup

H_n+1 (S X) -> H_n(X) here we have an iso in reduced homology but since it's the classical homology in degrees > 1 I didn' t mention it

[cold vine]

yup!

thanks

[tough imp]

Alternatively let Y be Spec 0 haha

[cold vine]

Any hint appriciated on how to proceed: i have f:S^n->S^n with d(f)=0 and I am to show that there are distinct x,y with f(x)=x and f(y)=-y

the degree of the map

so f*:Hn(S^n)->Hn(S^n) is multiplication by d(f)

yeah that i just figured out

Ok so now i assume that no point maps to its antipode. And i think this might then be homotopic tobidentity

-f maps x to -x?

Ahhh. So -f has a fixed point as well since d(-f)=0 and now this fixed point is y

Damn. Nice 😅 Gotta get used to these

Thanks a lot

[quiet pilot]

The zero set of an ideal I is the union of the zero set of all minimal primes containing I, right?

As a set without structure

In a polynomial ring k[x1,...,xn]

[sleek thicket]

Yes. A maximal ideal m (eg of the form (x-a1,...,x-an)) contains I iff it contains some minimal prime containing I

because if m contains I then the ideal m/I of k[X]/I must contain some minimal prime via zorn's

[gritty widget]

[sleek thicket]

why

why am I being d

[gritty widget]

[sleek thicket]

[pseudo crane]

[keen elm]

I thought zorn's lemma was obviously false

[sleek thicket]

very cool, mathew

[marsh forge]

isnt zorn the one no one knows about

in that joke

well ordering is obviously false

[sleek thicket]

yeah lol

[keen elm]

rip

I have failed

[elder yew]

How would you guys argue that x^2 - y^4 + z^6 = 1 is not compact

@sleek thicket ?

[sleek thicket]

hmm

Oh I think I see

Umm no sorry

[gritty widget]

[elder yew]

If you substitute

[sleek thicket]

Was lying

[elder yew]

you get an odd degree

[sleek thicket]

Like clearly you want to get size to blow up

[elder yew]

I know odd degrees are always fucky

[sleek thicket]

Umm this admits a map from a hyperbolic paraboloid

Right

Err maybe that's the wrong term

[tough imp]

not compact

[sleek thicket]

But like let u = y^2, v = z^3

[tough imp]

is what he had said

I think

[gritty widget]

oh

[sleek thicket]

Then x^2 - u^2 + v^3 = 1 is a hyperbolic space

Oh yeah so there's a surjection from this onto hyperbolic space

Right?

f(x, y, z) = (x, z^3, y^2)

[elder yew]

that's what I was thinking

but dunno how to word

[sleek thicket]

And the hyperbolic plane is homeomorphic to an open disk

Well that function gives a bijection with the pringle model

And it's continuous

Sorry not bijection

Surjection

yeah?

let H = { (u, v, t) : u^2 + v^2 - t^2 = 1 } and let X = { (x, y, z) : x^2 - y^4 + z^6 = 1 }

Define f : X -> H by f(x, y, z) = (x, z^3, y^2)

This is clearly continuous

And it's surjective because you can just take cube/square roots

Does that make sense moonbears?

[elder yew]

yeah makes sense to me

Not my lower division student I'm helping right now

[sleek thicket]

Well H isn't conpact

Oh oof

Umm

We should be able to engineer an explicit sequence going to infinity

Okay I see

So for each y I believe we can find x, z such that x^2 + z^6 = 1 + y^4

Just take x = sqrt(y)

Err sqrt(1+y^4)

So define p(t) = (sqrt(1+t^4), 0, t)

This is a curve in our set yeah?

And |p(t)| > |t|

[gritty widget]

pringle model

[sleek thicket]

The rhs goes to infinity so p becomes unbounded, so X isn't compact

@elder yew did that make sense?

[elder yew]

That's good

That works

This is due tomorrow and he decides to come in for tutoring on this today

[sleek thicket]

Ooof

I'm scrambling to finish my analysis hw

Very very stuck

It's also due tomorrow

[elder yew]

I can try to help sham

after I help this guy

If you need it

[sleek thicket]

Nah it's okay

[elder yew]

I'm so fukin' done with the past few days, PhD apps, job apps, writing, massive amounts of tutoring, spouse is stressed

:/

I had a cool problem at Tterra, to show if there's a vector parallel to the tangent plane on a simple surface at some point, then you can find a curve whose derivative is parallel to that

To that tangent vector

@gritty widget

[sleek thicket]

Oh I also had a cool problem for ttera but then I didn't finish solving it

[chilly rune]

So I was working on a problem related to the inverse limits of spheres with surjective monotone bonding maps and needed to show a lemma, namely that for a locally connected separable continuum $X$ if for any $x,y\in X$ and any open connected sets $U_x,U_y\subset X$ we have $X\setminus (U_x\cup U_y)$ disconnected then so is $X\setminus {x,y}$ (this is not true if $X$ is not lc for example take the closure of $sin(1/x)$ in the plane).

I'm pretty sure that this can be generalised, to be precise let $X$ be a locally connected, separable continuum and $B\subset X$ be a subset with an associated family ${B_{i}}_{i\in I}$ of subsets such that:

1. the collection is closed under union and $B=\cap_{i} B_{i}$

2. for all $i,j$ there exists $k$ with $B_k\subset B_i\cap B_j$

3. for any disjoint open connected sets $U_{\alpha}\supset B_{\alpha}$, where $B_{\alpha}$ are connected components of $B$ we may write $U_{\alpha}=\cup_{r\in R_{\alpha}} B_r$ for some $R_{\alpha}\subset I$ and all $\alpha$

4. for any open set $U\supset B$ there is an element of the family $B_{i}\subset U$ s.t. $X\setminus B_{i}$ is disconnected

5. for all $A\subset X$ with non empty interior there is some $B'\subset A$ homeomorphic to $B$ with an associated family ${B'j}{j\in J}$ which satisfies 1)-4)

if all these conditions are met $X\setminus B'$ is disconnected for any subset $B'$ homeomorphic to $B$.

My attempted proof of this is found here in lemma 1

\url{https://pnkaddaj.files.wordpress.com/2020/12/inverse_limits-21.pdf }

I am still quite new at these things so I am not sure if there is a mistake in the proof, I would be really grateful for any comments

[gentle osprey]

pnkadd421
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[tight agate]

Just learned something super cool

Let R* be an even periodic cohomology theory

with a kunneth formula

Then R* (CP -infinity) is iso to R* (pt)[[X]]

and as CP-infinity is a K(Z, 2), it is a topological group

and the multiplication map induces a map R*(CP-infinity) to R*(CP-infinity) \otimes R*(CP-infinty) (kunneth)

and a choice of iso from R* (CP -infinity) to R* (pt)[[X]] gives a formal group law

and a theorem of Quillen states that the complex cobordism spectrum carries the universal formal group law

[marsh forge]

Are you familiar w chromatic homotopy theory?

[tight agate]

not really

we're learning a bit of it

[marsh forge]

ah yeah, this is the starting point which is why i ask

[tough imp]

I like your funny words

[tight agate]

oh, and there's more

the homotopy groups give us the Lazard ring

[marsh forge]

if you want a cool thing to look at, every complex oriented cohomology theory gives a FGL and its natural to ask when an FGL gives a cohomology theory. There's something called the landweber exact functor theorem which tells you when this is possible

[gritty widget]

lizard ring

[tight agate]

yeah we did that too

[marsh forge]

nice

[tight agate]

flat maps Spec R to Mfg give even persiodic cohomology theories with R0 iso to R

pretty neat stuff

[marsh forge]

ideally i think i want to do my thesis related to this stuff eventually

[tight agate]

chromatic homotopy?

[marsh forge]

yeah

[tight agate]

yeah it looks dope

[marsh forge]

if i can compute one homotopy group of S^0 i will die happy

[tight agate]

heh

[tight agate]

You can do the whole thing algebraically. There's a functor from nil Augmented R-algebras to sets which sends a map A --> R to its kernel

this functor is "representable" by the power series ring in one variable over R

with the adic topology

and we're looking for lifts of this functor to groups

which is in 1-1 correspondence with co-group structures on R[[x]]

which is the same data as a one dimensional commutative formal group law

as the comultiplication map R[[x]] to R[[y,z]] is determined by where you send x

which is some power series F(y,z)

and the cogroup axioms force all the formal group law conditions

[wanton marsh]

what is an "even periodic cohomology theory" ?

[tight agate]

and then if you look at the functor which assigns to every comm ring R the set of all formal group laws over it, it is representable, and represented by the Lazard ring

[tight agate]

even if R^n is zero when n is odd

and periodic if there's an element in R^2 such that multiplication by that element induces isos R^n to R^n+2

for example complex K-theory is even periodic

[wanton marsh]

what's a cohomology theory then

[tight agate]

uuh

something that satisfies all the eilenberg maclane axioms

except the dimension axiom

[tight agate]

and instead, you can ask for a functor that goes from comm rings to the groupoid of formal group laws

and if you "force" descent for etale covers

you end up with the moduli stack of formal group laws

the moduli stack of elliptic curves sits inside the moduli stack of formal group laws

i am not sure how, but I assume for each elliptic curve you just take the formal completion at the identity

[wanton marsh]

can you do all that in ZFC ?

[tight agate]

all 'normal math' is done in ZFC no?

[tough imp]

Most normal math is done in ZFC + "I don't really pay much attention to the nitty gritty"-axiom I think

Inb4 Ultra

[wanton marsh]

if a cohomology theory is a sequence of functors from rings to set, you run into some problem cuz of how that's not a small category

[tight agate]

size doesnt matter

and I do not see the problem

but i do not know anything about set-theoretic issues

[wanton marsh]

I don't remember such an axiom in the ZFC axioms

what's a nice book explaining all that stuff ?

an LCA ?

wait are you doing topological modular forms

[tight agate]

this stuff does lead up to tmf, yes

but we arent going to learn anything about them

as there's only 1 lec left

[wanton marsh]

I mean books about those K(Z,2) CP-infinity formal group laws things

[tight agate]

Lurie spectral AG

a survey of elliptic cohomology

EKMM

[wanton marsh]

what is this sorcery lol

[tight agate]

and any book/article which does stuff with the adams spectral sequence ig

[wanton marsh]

ah that's the authors' initials

[tight agate]

yes

there's a lot of stuff

in ekmm

you probably dont need all of it

[wanton marsh]

alright, thanks I'll try to read them

[tight agate]

oh and there's 2 EKMMs

the bigger one is probably the one you want

[wanton marsh]

that link was 268 pages

[tight agate]

yeah that's the bigger one

[slender plank]

Hello I need to show that sin(n) is dense in [0,1]

can someone do a demonstration for me ??

same fos cos

[tough imp]

IIRC this boils down to rational estimates of pi

That’s all I’m willing to say cuz I forget how to do this and really don’t wanna work it out haha

[slender plank]

do you have any precise demo zritten ?

[gritty widget]

also, did you mean dense in [-1.1]?

[slender plank]

yeah yeah

[slender plank]

y

[pseudo crane]

maybe they meant [-1,1]

[gritty widget]

blocked

[pseudo crane]

🥺

[gritty widget]

@slender plank it is related to the question you asked in #advanced-analysis

||sin(n)=sin(n+2πm), π is irrational||

[slender plank]

ok

i dont get the link

one is constant

the other is not

[gritty widget]

?

sqrt(3)?

[slender plank]

no the function

[gritty widget]

elaborate

ok lemme give you this final hint

|| if a is irrational then { n+am | n,m in Z} is dense in R ||

[elder yew]

CityHunter, how is that proved rigorously

I've seen it all the time

[gritty widget]

do you know that fractional part of an is dense in [0,1]?

@elder yew

[elder yew]

No

I mean I'd believe it

but in general no

[gritty widget]

so... I guess you mean you haven't seen its proof

https://math.stackexchange.com/questions/903142/for-an-irrational-number-a-the-fractional-part-of-na-for-n-in-mathbb-n-is

Rest is pretty straight forward

you can adjust the integral part of an+m as you please with m

and fractional part is dense independent of m

[tight agate]

this is an instance of a more general fact: any additive subgroup of R is either cyclic or dense in R

and the argument is pretty much: look at the infimum of all positive elements in the subgroup

if it is in the subgroup, then you can show that it generates it

[elder yew]

that the infimum generates it

or the subgroup generates it?

[tight agate]

as for any other positive element, if it is not a multiple of the inf, then by subtracting you can get a smaller element in the subgroup (not exactly, but i think you can make this idea work )

if the inf is outside the subgroup

you can show that the subgroup is dense

by just "translating"

yes

same idea

[elder yew]

Yeah I've seen it hand wavy for the circle

But never seen a full proof. I've seen it come up on quals

[sleek thicket]

On my RG homework I was supposed to produce a counterexample to something

My construction started with "let G be a nonabelian connected compact lie group", and the argument works

Except I also said "eg G = SO(2)"

🤔

Oh yeah ultra I ended up not needing any of that

[quiet pilot]

I'm trying to prove that a line in a smooth surface lie in the tangent plane of any of the points the line contains. I'm probably missing something obvious but I'm not sure how to proceed

For simplicity we can restrict to P^4 such that the line is the intersection of two planes

Clearly the line lies in the tangent space of the two planes defining it

[marsh forge]

just email the ta and say you typo'ed

[gritty widget]

so if $\alpha$ is any curve in $SL(n, \bC)$ with $\alpha(0) = I$, then you want to show that $\alpha'(0)$ has trace zero (after which comparing dimensions finishes the proof). try differentiating $$\det(\alpha(t)) = 1$$ at $t = 0$ and using the fact that $\det(e^A) = e^{tr(A)}$ to simplify. (i.e. find $d(\det)_I$)

[gentle osprey]

parallel TTransport

[gritty widget]

any element of the tangent space is given by the velocity vector (at the identity) of some curve

what i outlined only shows one inclusion (tangent space \subset trace-zero matrices), and the other one follows from basic linear algebra

anyways a big hint for finding the differential of the determinant at the identity is || e^{tA} is a curve through I at t = 0 with velocity vector A, so by the definition of the differential, d(det)_I(A) is the derivative of det(e^{tA}) at t = 0 ||

which i've spoilered since it's basically the main idea of the proof

[gritty widget]

well you will get that $tr(\alpha'(0)) = 0$

[gentle osprey]

parallel TTransport

[gritty widget]

differentiating $\det(\alpha(t)) = 1$ at $t = 0$ gives $$ 0 = (\det\circ\alpha)'(0) = d(\det)_{\alpha(0)}(\alpha'(0)) = d(\det)_I(\alpha'(0))=tr(\alpha'(0)) $$

[gentle osprey]

parallel TTransport

[gritty widget]

the second equality sign is the chain rule, the third because alpha(0) = I, and the fourth by the fact that d(det)_I = tr, which follows from the identity det(e^A) = e^{tr(A)}

[blissful dune]

I started do Carmo!

it's a great textbook so far

[gritty widget]

curves and surfaces or rg?

[blissful dune]

shoutout to whoever sent me it

Curves and Surfaces

then I'll do RG

[gritty widget]

have fun

[blissful dune]

so far it's fun

[gritty widget]

i just finished an RG course that did nearly all of do carmo's book

so i'll say that if you're planning to do his RG book after curves, you'll see a lot of stuff repeated

[blissful dune]

oh

[gritty widget]

just in more generality

[blissful dune]

well ig i'll just skim those parts

[gritty widget]

skimming do carmo's rg is how you die

[blissful dune]

that's great because i want to die

[woeful oasis]

i was so happy when dover released do carmo's curves & surfaces 2nd ed

[tepid depot]

No need to worry too much about content repeating. Historically curves and surfaces stuff came first and was developed by Gauss. Riemann came up with RG as a generalization. So it's gonna help a lot to do curves and surfaces first

Gives a lot of context and motivation

[blissful dune]

ah dope

[marsh forge]

My gf was wondering if anyone had a hint for how to do this

I’ll be honest I’ve thought about this for exactly 0 seconds

[sleek thicket]

So give A^4 coordinates (u, v, s, t). Then im φ is contained in Z(vs - ut, v^2 - uv - u^2 s, t^2 - ts - s^3). You can prove im φ = Z(vs - ut, v^2 - uv - u^2 s, t^2 - ts - s^3) by cases on whether u = 0 and s = 0 or whether one is nonzero

this shows closedenss

what's her definition of normality of a curve?

I don't really want to check if it's irreducible but assuming it is, the coordinate ring isn't integrally closed in the field of fractions

because (t/s)^2 - t/s = s

(basically here I solved for y)

I'm not sure how to show t/s isn't in the rign easily

Left as an exercise to the Max's girlfriend

Oh I think I see. Consider the point φ(0,0) = (0,0,0,0) = φ(0,1). You should be able to use like, continuity to argue that y isn't a function on the variety

I think you can say that on the dense open set D(x) of A^2 we have f ° φ = y where f is the function t/s = v/u, so this is true globally, which causes a contradiction

[meager python]

Did you use some computer algebra to get the zero set or did you just compute it that fast ;3

[gritty widget]

sham's just big brained

[sleek thicket]

I just thought about relations

like the first one is sort of obvious

But im φ ≠ Z(vs - ut) because of dimesion reasons

So maybe you notice xy - x = x(y-1), and then xy^2 - xy = xy(y-1), so v^2 - uv = x^2 y^2 - x^2 y = x^2 y(y-1) = u^2 s

But then finally you see that the image doesn't contain (0,0,0,0), but that satisfies the relations vs = ut and v^2 - uv = u^2 s. The trick is that you need to say "if s = 0 then t = 0"

That one was harder but I sort of just focused on s and t exclusively

[meager python]

Yeah its basically relations of x, y, y-1

Was wondering if you had some algorithm or just exhausted most possibilities

[sleek thicket]

nope, just instinct & guesswork

and like, having done similar problems before

[elder yew]

I computed the second fundamental form correctly the first time with my student

Holy shit

Usually I mess up some cross product or something

[tight agate]

I just noticed that moonbears' pfp is a frog lmao

I used to think it was a weird crab and monkey hybrid with a visor/ski goggles

[elder yew]

Just a happy frog lookin' for a fly

[gritty widget]

second fundamental form

[elder yew]

yA

you take some parameterization x(u,v)

Then ya like ya know

$$ N = x_u \times x_v $$

[gentle osprey]

MoonBears-C-

[elder yew]

(normalized)

And then ya set

$$ L = - x_u \cdot N_u, M = -1/2(x_u \cdot N_v + x_v \cdot N_u), M = -x_v \cdot N_v $$

[gentle osprey]

MoonBears-C-

[elder yew]

So finally

$$ II = L \ du^2 + 2 M \ du dv + N \ dv^2 $$

[gentle osprey]

MoonBears-C-

[sleek thicket]

Lmao I thought moonbears' pfp was Kermit specifically

didn't realize it was a real frog

[elder yew]

I'm getting bullied

@gritty widget @sleek thicket All the fancier curvatures reduce down to gaussian curvature

I think it's like

II/I for the fundamental forms

Surface area is given by

$$ \int_{U} \sqrt{EG - F^2} $$

[gentle osprey]

MoonBears-C-

[sleek thicket]

I mean kind of

[elder yew]

That's what I saw in my old notes anyway

[sleek thicket]

so like, sectional curvature determines the curvature tensor, and all the other curvatures are derived from the curvature tensor

[elder yew]

Good enough for me

[sleek thicket]

and like, if I have a plane Π in the tangent space at p I can push it forward via the exponential map to get a surface

and the sectional curvature of Π is sort of the gaussian curvature of the surface

but gaussian curvature only makes sense if you have an embedding into R^3

you can define an abstract gaussian curvature, but it's defined in terms of the scalar curvature, which is derived from the riemann curvature tensor

I would not say that everything reduces down to gaussian curvature

[elder yew]

yA

It was just a specific case

[sleek thicket]

oh lol

sorry

[elder yew]

I was referring to an earlier discussion where I had asked if it reduced down to classical DG curvature

Riemannian/Sectional etc.

But it didn't seem like it did

But it seemed the notion that did end up getting generalized was gaussian curvature

Even if it doesn't all boil down to that

[woeful oasis]

If $f:R^1 \to R^n$ is a smooth map, $\omega$ a 1-form on $R^n$. Is it true that the pullback $f^{*}\omega$ is exact?

[gentle osprey]

Apopheniac

[woeful oasis]

This is just FTC right?

[gritty widget]

ya

any 1 form on R^1 is closed hence exact (proof uses ftc!)

[woeful oasis]

Good thing I computed a bunch of pullback anyway, just to decide not to integrate at the end. lol

thanks 🙂

[cursive flume]

how to define trace of a tensor field coordinate free?

i.e. riemann tensor field

[gritty widget]

from appendix b of lee's riemannian manifolds book

[elder yew]

jesus

they actually write it out

[cursive flume]

but this is in coordinates

[gritty widget]

🤨

[cursive flume]

i would like without tensor components

defined as a map from a tensor field to a lower rank tensor field without any reference to components

[gritty widget]

i'm not sure what you're asking for, there's a coordinate-free definition right there

lee just gives the coordinate definition right after

[cursive flume]

i cant see which is the coord free def

im confusee

i see he uses components to write it out

[gritty widget]

If $$ F \colon V^* \times \cdots \times V^* \times V \times \cdots \times V \to \bR $$ is a $(k+1,l+1)$-tensor (eating $k$ covectors and $l$ vectors) then the trace of $F$ is the $(k,l)$-tensor defined by $$ (\mathrm{tr}(F))(\omega^1,\dots,\omega^k, v_1,\dots,v_l)=\mathrm{tr}(F(\omega^1,\dots,\omega^k, \cdot, v_1,\dots,v_l, \cdot)), $$ where the trace on the right is the trace of a $(1,1)$-tensor.

what a nightmare

give me a moment to fix this shit

[cursive flume]

we have trace on lhs and rhs

[gritty widget]

yeah wait

[cursive flume]

how

[gentle osprey]

TTerra

[gritty widget]

the trace on the right is the trace of the tensor F

the trace on the right is the trace of a linear map (in the picture, you can see lee mentions there is an identification of (1,1)-tensors and linear endomorphisms V-> V, the latter of which certainly has a well-defined trace)

[cursive flume]

ok then i need to understand the trace of a linear map

i havent done that without matrices coordinate free :/

[gritty widget]

there's definitely a coordinate-free definition of the trace of a linear map somewhere out there

[cursive flume]

and how is F a linear map

its a tensor

[gritty widget]

okay then let me post the part where he identifies (1,1)-tensors and linear maps V -> V

the thing on the right inside the argument of trace is a (1,1)-tensor

(finite-dimensional is important)

So when we write $$ F(\omega^1, \dots, \omega^k, \cdot, v_1, \dots, v_l, \cdot), $$ we are referring to a $(1,1)$-tensor on $V$. (The empty arguments mean you put things in there.) Under the identification given in Proposition B.1, we can think of this $(1,1)$-tensor as a linear endomorphism $V \to V$, which certainly has a well-defined trace.

[gentle osprey]

TTerra

[gritty widget]

and that's why the definition makes sense

[cursive flume]

can you pls write out explicitly the trace of an endomorphism v->v?

[gritty widget]

[cursive flume]

then I think I got it

[gritty widget]

i mean there are a few ways to do so

most straightforward is to just pick a basis and take it to be the trace of the matrix

coordinate-free would be as the sum of the eigenvalues

[cursive flume]

how can one define eigenvalues basis free

[gritty widget]

🤨

$T(v) = \lambda v$ is very basis-free.

[gentle osprey]

TTerra

[cursive flume]

ah right i was overthinking it

so this works for tensors

[gritty widget]

ya, it's all in appendix B of lee's IRM

and i mean all of this

[cursive flume]

is there no restriction for tensor fields?

cause riemann tensor isnt rly a tensor

[gritty widget]

the trace of a (k+1, l+1) tensor field will just be a (k, l) tensor field whose value at each point is the trace of the tensor at the point

[cursive flume]

its a section on the module gamma tm

[gritty widget]

i was just reading about this stuff from lee's book the other day

[cursive flume]

i’m trying to derive einstein eqns coordinate free

[gritty widget]

all of the multilinear algebra that gets developed on a vector space transfers over to tensor fields pointwise

just a general-philosophy thing that i think is good to keep in mind

[cursive flume]

my diffgeo class is so yikes

prof follows janich’s vectoranalysis book

i wanna kms

super old approach of dg not even coord free

does lee define grad curl div all coordinate free?

[gritty widget]

ya

lee always like, gives the coordinate-free definition of something, then immediately gives its coordinate representation (usually as an exercise)

grad, div, and curl are all in chapter 2 and its exercises of irm

[cursive flume]

wish my prof did lee

[gritty widget]

i took RG out of do carmo's book and there's basically no mention of differential forms or tensors outside of like, one tiny section that never comes up again

one of my only gripes with the book

(also do carmo's notation is shit)

[cursive flume]

where can one read about spinor bundles and cartan’s approach to diffgeo?

in a self contained way

seems like spinor fields we use in qft are a highly nontrivial mathematical structure

sections of spinor bundles but idk how to define spinor bundles

it has to do with associated and principal bundles somehow but idk precisely how 😦

[gritty widget]

i don't know where you could find stuff about spinors, but i know that tu has a book on principal bundles that may be related

[runic sun]

Stupid question, just starting learning about topology, isn’t any space under the topology (X,P(X)) a Hausdorff topological space? If so what’s the point of a Hausdorff space then? I also have the same issue with the definition of a topological space, with this topology or (X,{Ø,X}) isn’t any set a topological space? What‘s the point of these definitions of any set satisfies them?

[cloud owl]

the set isn't the important part, as i understand it

the topology is

[runic sun]

Oh so the point is to be taking about different topologies? It‘s just that with things like manifolds I see that in the definition there‘s conditions for the existence of a topology which is Hausdorff, what’s the point of that part, if every set has a topology that‘s Hausdorff?

[cloud owl]

every set has a topology which is hausdorff, but not every topology is hausdorff, and the things only want the ones which behave nicely?

[runic sun]

Yeah I got that

[cloud owl]

i don't see the problem

[runic sun]

What I’m asking is what‘s the point of including conditions on the existence of a Hausdorff topology in a definition

If there always is one

[cloud owl]

oh right

[runic sun]

So what is the point of it?

[cloud owl]

uhhh i don't actually know

[runic sun]

Huh alright, still thanks for clearing up some things

If someone else knows please ping me

[runic sun]

Ohhh, I get it

[tepid totem]

It‘s just that with things like manifolds I see that in the definition there‘s conditions for the existence of a topology which is Hausdorff, what’s the point of that part, if every set has a topology that‘s Hausdorff?

This is not true. There is no condition on the 'existence' of a Hausdorff topology,since as you say, this always exists so its a pretty useless condition. Instead, a manifold is a set, together with a topology that is Hausdorff and satisfies some other conditions. In particular, when asking if something is a manifold, you are already given a set and a topology on that set, and you wanna check if the topology satisfies your conditions. You don't just take a set and ask 'is there a topology that makes it a manifold', the topology is already given.

[runic sun]

Yes thanks

I get it now

[coarse kestrel]

For the <= direction I don’t understand the sentence “alter the C_i by adding to each C_{i+1} a finite union of compact sets”

What is the finite union of compact sets

What are we even adding it to? C_i or C_{i+1}

[gritty widget]

Hi. Can someone tell me if I solved this correctly?

I have the set A = B(0_2, 1) \ {0_2} ⊆ R^2 where B is a closed ball
and I need to find the int, fr of A
and find if A is a closed or open set
is it correct to say that int(A) = A\fr(a)?
and that fr(A) = {(x_1, x_2 ∈ R^2 | (x_1)^2 + (x_2)^2 = 1 }
and A is a closed set?

[coarse kestrel]

Not quite

fr(A) also has one extra point

[gritty widget]

o_2, no?

[coarse kestrel]

Yes

[gritty widget]

then it isn't a closed set

[coarse kestrel]

No it isn’t

[gritty widget]

thanks!

and one more question

if I have the set A = QxQ ⊆ R^2

what is int of A?

oh, it is {0}

[gritty widget]

yup

[marsh forge]

add to C_{i+1} a finite union of compact sets covering C_i

[coarse kestrel]

Why is it possible to find a finite union of compact sets whose interior cover C_i

[marsh forge]

X is locally compact. Around every point in C_i take a compact neighborhood. This is an open cover of C_i, which is compact. Take a finite subcover.

Technically the interiors are an open cover but whatever

@coarse kestrel

[coarse kestrel]

Oh

Right

Lmao that was obvious

[tough imp]

For anyone that has a copy of Hartshorne, can you check the statment of exercise III.2.5 on page 213? I think my copy must have a typo as it defines F_p to be j^*F but that isn't defined

I suspect it maybe meant j_*F or j^-1F but I'm not sure

[wanton timber]

For what it's worth, Springer's online copy also has j^*F

[tight agate]

I think it should be j^-1 tho

as F is just a sheaf