3:01 PM
@ÍgjøgnumMeg what are you trying to do?

I have an equation $\dddot{\mathbf{x}} = -\nabla V(\mathbf{x})$ where $V(\mathbf{x}) = \sin(kxy)$ with $k \geq 0$
oops
should be $m\ddot{x}$
ARGH $m\ddot{\mathbf{x}}$***

lol $\dddot x$
$\ddddot x$

hahahaha

$\dddddot x$
aww
@ÍgjøgnumMeg that's not a PDE, it's a system of ODEs

and I've linearised this (taylor expanded $V$ get $V \approx kxy$)
@0celo7 alright fair, that's fine
but the problem is

3:05 PM
So you have $\ddot x =-ky$ and $\ddot y=-kx$

well yeah but
the eigenvalues of the Hessian matrix are called the principal curvatures right?

Not in this context

How come?

I hear ODEs

hey would someone mind clearing up a little confusion i have about showing convergence in metric spaces :/ ?

3:07 PM
Eh well if the graph is a surface then they probably are. Whatever, why does it matter?

Well I'm doing an exercise and am supposed to display a dependence on the principal curvatures

Eigenvalues of the shape operator are principal curvatures

@BalarkaSen Is the shape operator for a graph the Hessian of the function?
I really should know that but meh

Right, and what are the eigenvalues of the hessian in this case? I get $\pm k$ as the eigenvalues of the hessian and the lecturer told us that the eigenvalues of the hessian are the principal curvatures...

3:09 PM
So... I have spent 5 hours tidied up 400+ unread emails. Now it looks a lot nicer as each sender is placed into categories thus when the folder is read vertically, I have the full timeline on how the sender's interaction with my email changes

It's the derivative of the Gauss map isn't it

(For context, I don't know any differential geometry or whatever, it's just part of my course)

I think it's the Hessian of the function it's the level set of @0celo7

@BalarkaSen I dunno. Studying for an exam and trying to pay attention to fluids lecture
bye

Is it common to talk about "projection matrices" for rectangular matrices? I have only heard of projection matrices as idempotent (square) matrices.
The context where I have heard "projection matrices" for rectangular matrices is when reducing a set of ODEs to a smaller set of ODEs, by multiplying to the left by a rectangular matrix.

3:14 PM
@BalarkaSen It just seems a little counterintuitive; the graph of the function (near the origin) is a saddle so there needs to be a positive and a negative curvature. The eigenvalues of the hessian are $\pm k$ in this case so that works out, but when taylor expanding $V(\mathbf{x})$ around the origin, we get $V \approx kxy$ so that the equations I get are $m\ddot{x} = -ky$ and $m\ddot{y} = -kx$
And now I guess there's a choice to make; which equation "gets" the $-k$ and which equation gets the $+k$?

@ÍgjøgnumMeg Hm, let's see. Say $S$ is locally cut out by $f(x, y, z) = c$. Then the Gauss map $f : S \to \Bbb R^3$ is given by sending $(x, y, z)$ to $\nabla f(x, y, z)/\|\nabla f(x, y, z)\|$
So if I take the derivative of that, upto normalization, that's $D\nabla f(x, y, z)$ isn't it. That's a Hessian
To be more precise, the shape operator is $S(U, V) = Df(U) \cdot V$ at the tangent space $T_p S$ of $S$ at $p = (x, y, z)$

right

If I let $D\nabla f(x, y, z) = H$, which is the Hessian, then $S(U, V) = HU \cdot V = V^T H U$, right?
It really is the Hessian form

Yeah I know, that's not the point
The point is that I don't see the dependence on these curvatures

Oh I know, I was just trying to clarify my own thoughts. Let me read your question.

3:19 PM
Ah okay fair

I don't want to be wrong when helping you :)

Yeah that's fine, it's frustrating because the lecturer hasn#t actually mentioned any of this stuff and now expects us to produce answers...... hahaha

Hello,
Can a local homeomorphism with homeomorphic fibers fail to be a covering map?

@ÍgjøgnumMeg So you have the direct product of two systems of 2nd order ODEs $(x'', y'') = (-kx, -ky)$ (you linearize the "potential function" $V(x, y) = k\sin(xy)$ to get this, which is fine). Er, so, why does this have much to do with the Hessian or the eigenvalues?

Well the exercises says to clearly exhibit the dependence on the principal curvatures

3:29 PM
@Arrow Yes, I believe so. The example that comes to mind is the espace etale $|\mathscr{O}(\Bbb C)|$ of the sheaf of holomorphic functions on $\Bbb C$. Fibers of the projection $\pi : |\mathscr{O}(\Bbb C)| \to \Bbb C$ are germs of holomorphic functions at some $p \in \Bbb C$, all of which are homeomorphic. And this is a local homeomorphism by construction.
This is, however, not a covering map (there exists holomorphic germs of arbitrarily small radius of convergence, so you don't get slices above a fixed neighborhood of a point)

@BalarkaSen And I'm not sure where to see this dependence, I see the $-k$ factor in the ODEs but which of the equations has the negative eigenvalue and which has the positive eigenvalue? Or do you make a choice? Or should one just look at the graph of the function? hahaha

@ÍgjøgnumMeg I am kind of confused too. I don't see what's the Hessian here.

The hessian is the matrix $$\begin{pmatrix} \partial_x^2 V & \partial_{xy}V\\ \partial_{xy} V & \partial{y}^2 V\end{pmatrix}_{(0,0)}= \begin{pmatrix} 0 &k \\ k &0\end{pmatrix}$$
ergh
this is evaluated at $(0,0)$
and then the eigenvalues of this guy are $\pm k$
And my lecturer tells us that these eigenvalues are the principal curvatures of the graph
at $(0, 0)$

They are the principal curvatures of the level sets of $V$ at $(0, 0)$, that should be true because of what I wrote above.

@BalarkaSen I will try to understand your second remark. In the meantime, do you know any nice conditions on a local homeomorphism to make it a covering space? (I am hoping for properties of the map itself, rather than conditions on the domain and codomain such as here.)

3:36 PM
But I don't see what this has to do with the ODE, hm.

@BalarkaSen Right, but how do I see the dependence on these values in the system of ODEs I have?

@Arrow Proper local homeomorphisms are covering maps.
@ÍgjøgnumMeg We are confused about the same things, I am afraid, haha

Ah, so properness enters again.. Great, thanks.

@BalarkaSen Alright no problem :) Thanks for the help anyway

@Arrow Well, no, you actually need more conditions.
If the domain is non-Hausdorff this does not work (think about projection of line with two origins to R)
And upon googling it seems you also need Hausdorffness + local compactness of the base. But that's it.

3:38 PM
@BalarkaSen yes, I googled also. I wonder if there's a nice condition at least for maps between locally connected spaces. I would like to avoid Hausdorff assumptions.
Maybe you could help me with some intuition though
"why" is having homeomorphic fibers not enough for a local homeo to be a covering map?

Unfortunately covering spaces with non-Hausdorff base or total spaces are unnatural.
@Arrow The obstruction, as I said, is properness.

When I e.g glue two copies of a space along an open subset (the best example I know of local homeos which aren't covering maps) it seems the only problem is the fact fibers may change from point to point. I guess my picture isn't general enough.
Do you by any chance have a good mental picture to hint why properness is key?

You can think about it as, if $f : Y \to X$ is not proper, $f$ does not necessarily send "infinity to infinity." (which is what properness guarantees). So say $f^{-1}(p)$ for a $p \in X$ is a non-compact set
The proper which might happen is that, even though $f^{-1}(p)$ is discrete, every $q_i \in f^-1(p)$ has disjoint neighborhoods, but they are arbitrarily small towards the infinity
So if $U$ is a neighborhood of $p$, $f^{-1}(U)$ would not decompose as a disjoint union of neighborhoods of $q_i$.
Because the neighborhoods of $q_i$ as "$i \to \infty$" are arbitrarily small
You should look up the proof that assuming Hausdorffness + local compactness, proper local homeomorphisms are covering spaces. That is very revealing.

I will. Thanks for the help.

3:56 PM
Guys. My book defined the derivative of $f\in R[X]$ (where $R$ is a ring) as $f’=H(X,0)$, where $f(X+Y)-f(X)=Y\cdot H$. Now they want to show the product rule $(fg)’=f’g+fg’$. They start out as follows: $f(X+Y)-f(X)=YH_1$ and $g(X+Y)-g(X)=YH_2$, thus
$$f(X+Y)g(X+Y)-f(X)g(X)=Y(f(X)H_2+g(X)H_1+YH_1H_2).$$
I don’t really see how they arrive at this equality. Could someone give me a hint? I think it must be straightforward, as they omitted the steps.
oh wait.. maybe I should just write it out and not be lazy
yea so never mind, I can't delete it anymore.

4:09 PM
@ShaVuklia what is H?
oh, $H(X,Y)$ is $h \in R[X]$ such that $f(X+Y) - f(X) = Yh$
@GFauxPas hi

hello sir

@MathematicsAminPhysics Seeing that you are currently working on some general topology, I have sent you invite to general topology chat room, in case you are interested. Of course, other users with topological interests are welcom there too.

@LeakyNun lol yea

4:39 PM
Oh then take a look at this question .... ROTFL !

Obama was probably well underage when that came out.

Rick Astley is an outdated meme. What is this, 2007?

Obama is an outdated meme.
What is this, 2008?

lmao
Well, the person who replaced Obama's meme status is way funnier, in a scary way.
So I am not sure how I feel about that

The questions related to 80's pop culture are different.
Was the meglomaniac businessman character based on Donald Trump?
The answer is usually yes.

4:45 PM
Hahah

e.g. Wall Street, Back to the Future/

5:08 PM
Fun topology question : is this right (should be, but still, old language):

Random challenge, probably easy:
A number is in base-b simplified Goodstein form if it is written as
b + b + ... + b + c, c ≤ b
The simplified Goodstein sequence of a number starts with writing the number in base-1 simplified Goodstein form, then replacing all 1's with 2's and subtracting 1. Rewrite the result in base-2 simplified Goodstein form, then replace all 2's with 3's and subtract 1, etc. until you reach 0. For example: If you start with 3, it looks like this.
1 + 1 + 1                | 3 = 1 + 1 + 1
2 + 2 + 1                | Change 1's to 2's, then subtract 1. (2 + 2 + 2 - 1 = 2 + 2 + 1)
3 + 3                    | 3 + 3 + 1 - 1 = 3 + 3
4 + 3                    | 4 + 4 - 1 = 4 + 3
5 + 2                    | 5 + 3 - 1 = 5 + 2
6 + 1                    | 6 + 2 - 1 = 6 + 1
7                        | 7 + 1 - 1 = 7
7                        | 8 - 1 = 7
6                        | Numbers are now lower than the base, so just keep subtracting 1.
5                        |
Let G(n) be the length of this sequence starting with n. From the above example, we see that G(3) = 15.
Can you derive the closed form for G(n)?

IDK if I misread something or not, but isn't it a necessary and sufficient condition for any real function to be tending to a limit (not just a bounded function) such that given any $\epsilon$, one can give a real $g(\epsilon)$ such that for all $x,y > g(\epsilon)$, $|f(x) - f(y)| < \epsilon$ ?
Hardy's book says its just for obunded functions.

@AlexKChen That clearly doesn't cover $x\mapsto x^2$ blowing up at infinity
that limit "exists" in a sense, but not using the condition you wrote

Oh, but if the function has a finite limit, then this condition is ture, right ?
Oh darn then the function is bounded :/

Yeah if it's continuous and has a finite limit at infinty it's bounded.

5:17 PM
Sorry I am just so stupid... in just the next section unbounded functions are covered...should have read a bit further before asking.

5:33 PM
@TedShifrin Hey Ted :)

Hi Kasmir

You logged in like 2 seconds after me
like mutial agreement ><

hi @Ted

But dont worry I wont bother you no more with stupid Q's :D

Hi @Alessandro

5:34 PM
Hi @Ted

OK, Kasmir ... depends on your definition of "stupid Q" :P
@Alessandro: How many more people have you mowed down?
Hi Balarka

Not a lot
I'm doing much more math than driving now that the semester begun

Guess you're becoming a better judge of position and velocity, Alessandro :)

any questions about the reading matrial i meant :D

Ohhh ... That's why I'm safe. thinks of more math questions for Alessandro

5:35 PM
@BalarkaSen Did you see this?

Very cool that you're officially in the bibliography for asking the question, Mike :)
I skimmed the paper. I'm not going to mess with B and BO. :)

@MikeMiller Ah no I did not
I remember your question though

Me too.

I agree. The argument is pretty clean, just some calculations.

I'd make a joke about the Heisenberg uncertainty principle but uh

5:36 PM
With regard to Alessandro's driving, Semiclassic?

i need to have lunch before i'll have the brain power to make puns
2

Were we punning?
It seems that whenever I pun, Demonark shows up, as if on cue, so I'm trying to be careful.
6

@TedShifrin sure, if you have nice abstract algebra questions

On what sort of topics, Alessandro?

@TedShifrin 80% of the time that's me calling me on discord for comedic effect

5:39 PM
I still don't think of you as having stand-up comic ability, Balarka.

Okay I got a somewhat decent question to ask

I don't need to stand up to do comedy
I am that good at it

Why is G /stab (x) in bijective correspandance with orbit (x)

I'm taking three abstract algebra courses this semester, from the one I like the most to the one I like the least: commutative algebra, Galois theory and group theory

I found the map but , more intuitivly

5:40 PM
Kasmir: That's not a good question. It's shown in every book.
That proof is totally intuitive, so understand it.

I know Ted , i mean more like intuitivly

I wish I knew K-theory
or pretty much anything that goes beyond the basics of anything

Clearly the stabilizer subgroup of $x$ fixes $x$. Something not in the stabilizer subgroup moves it, say to $y$. What are all the group elements that move it to $y$?
Balarka: Stop it. You're beyond basics in lots of stuff.
TOO MUCH WHINING in this room.

lol
It's just that once I know something well enough it immediately becomes basic in my eyes and the stuff I don't know looks really cool
4

5:43 PM
@TedShifrin what are the elements that moves it to y ? g stab(x) no ?

So, isn't that QED?
(Of course you have to argue surjectivity. But that's nothing.)

@BalarkaSen I do that too. But then when I say I "don't know any algebraic geometry" people pick on me.
I think that's a fair price to pay.

surjectivity is by def

by definition of what?
runs out to buy extra eyes to roll at Mike and Balarka

that each element of X be in some orbit
orbit does a partition on the set X

5:45 PM
No, that's not relevant here.

@MikeMiller Haha. Well, your Voisin side project seemed to be a pretty good approach to fix that.

hmm
G /stab (x) , i think about them "like " cosets

You're losing track of the question.

Time, nonetheless, stays to be a merciless creature...

so either the elemnt is fixed or in some orbit
it cant be both

5:47 PM
You're not paying attention, Kasmir.
What set are we trying to get the bijection to?

@BalarkaSen No, you misunderstand. The reason people pick on me is because that's not true in the stated sense.
I'm just saying that it seems fair for Ted to pick on you when you say you don't know anything about K-theory.

Oh, Mike, people are gonna pick on you no matter what. We short people have to suffer that. :)

G/ stab (x) and orbit (x) , are in bijective correspandance as sets

@Balarka: Mike and I complained years ago about how you would say things were super hard until you learned them and then they automatically became a triviality.

I can define a fn, f : gstab (x) ---> g.x

5:48 PM
So pay attention to that, Kasmir.
So you have to check well-defined, one-to-one, and onto.

@MikeMiller Oh, so you meant that people pick on you because you know a fair proportion of algebraic geometry even though you deny it?

well the onto part i thought by defintion you said no

@BalarkaSen Yes, as with you and "----".

Well. I really don't know shit about K-theory lol. Not beyond the definitions anyway

@TedShifrin It's a really easy mistake to make! It's one that causes me and I suspect many others undue anxiety about their own knowledge.

5:50 PM
Because you didn't give the correct argument, Kasmir. Reread what you typed.
I think you can be proud of your breadth of knowledge/understanding, Mike, and it's OK to say that certain things you know basically nothing about.

@TedShifrin Indeed, I have grown a little more confidence in myself after those incidents and do not immediately give up on things if I don't understand it the first few times

if I pick an elemnt in this form , g'x , it comes from g' stab(x)

puts up the PSYCHIATRIST IS IN — 5 CENTS sign

lmao
We're all mad here

Do you get the allusion, Balarka?

5:52 PM
Nope, but I wonder if you get mine

@Kasmir: And why is every element of that form?
We're even, Balarka. I'll tell you mine if you tell me yours.

Haha
Alice in the Wonderland

Oh, right ... Mad Hatter says it?
Mine's from Peanuts, decades ago.

The Cheshire Cat said it

the map is 1-1 and since G/stab (x) has the same order as G(x) , then the map must be onto

5:53 PM
@TedShifrin Ahhh

ah, right, Balarka.
What? @Kasmir

we are doing a map between sets

That is certainly begging the question.
Where do you know how to count the orbit not using this formula?
So my point is — "intuitive" or not — you haven't really understood the proofs. You need to completely understand the basic proofs.

ill prove it right here and now -.-

If you assume $\lvert G/\operatorname{Stab}(x)\rvert = \lvert G(x)\rvert$ in order to show that there is a bijection $f : G/\operatorname{Stab}(x) \to G(x)$ then you aren't proving anything

5:56 PM
Oh, I missed Alessandro's answer earlier. Too much algebra in one term!

f : g stab(x) --> g.x
f is 1-1 , g stab(x) = g' stab (x) iff g'g^-1 is in stab (x)

Go figure out the correct answer, Kasmir.

okay -.-
thanks Ted -.-
:)