it might be the correct answer, but it is not useful for me if I dont get it 100%

so far what he did was, gave a group structure

Oh, I was thinking $xzv$. Let me think about that.

that is fine but , too many questions I have left to accept that answer

@TedShifrin as admin , do you have the power to lift that question up so more ppl can ponder on it ? :D

I'm not an admin. I have only very modest power in this room.

okay =p

can u declare a state of emergency

@ÉricoMeloSilva -.-

i did ask him about that product, but he did not reply

i might have been a typo but still ._.

what is the identity element of this group G

How come (0,0,0,0,0,X) is the Kernell

To what Map ?

Many things i dont quite get

Hey guys

Hi Demonark.

sup nerd

I'm busy doing something IRL but will get back to this.

How's it going?

hi dami

@Daminark math.stackexchange.com/questions/3069049/…

that is the question we trying to make sense of =p

Quick question, is $\sum_l \frac{\partial^2 f(x)^l}{\partial x^i \partial x^j} \frac{\partial x^k}{\partial f(x)^l} dx^i \otimes dx^j \otimes \frac{\partial}{\partial x^k}$ just $(\nabla \nabla f) \cdot (\nabla f)^{-1}$?

@Kasmir: The $y$ has to be a typo, as $y=0$ is allowed in the matrices to start with.

And the identity is $(1,0,1,0,1,1)$.

@TedShifrin thanks! that is logical :D

@TedShifrin why the last entry is 1 ?

@user76284: I have no idea what $\partial x^k/\partial f^l$ means.

Because $D=1/\det$, @Kasmir.

@Ted just so you know, I just sent you an email

Oh.

I've been a bit distracted by real life drama/misery. I'll check.

@TedShifrin is that D , the same D he put in the matrix?

@Daminark is this a good time to email Ted ? -.- we are in the middle of something :D

He never put a $D$ in a matrix?

@Kasmir: Stop that.

sorry

You do not have any business saying something like that.

it was a joke Ted

hence my smilyes

It doesn't read like a joke. And I'm in a shitty mood.

okay sorry !

@TedShifrin It shows up in the expression for the Schwarzian derivative: mathoverflow.net/questions/320590/…

I don't think it makes any sense. I suspect it's supposed to stand for an entry of the inverse matrix, but as it stands it's garbage.

i know right

With some assumptions on $f$, one can make sense of it.

the answer is not complete

@Kasmir: That was to @user76284, not to you.

I've answered your question completely.

aha oups

oh, but why does that determinant enters there?

if that group came from how we multiply matrices

So that $DD'$ will be the entry in the product.

why is the 6 th entry differ from all others?

Just write everything down without $D$. Work with $5$-tuples instead.

He was trying to give you a way to pick a representative of the coset in the original matrix group, I suspect.

Because the quotient group is not a matrix group.

in that case all what he did is described matrix product via algebraic way

and that does not solve anything

It gave you a group to map to with the right kernel.

There isn't a "pretty" answer.

There is no obvious subgroup of matrices.

what are we doing?

hmm

i think i get it

what he did was exactly when someone tries to open a door with a wrong key, and modified that key untill it opned the door

because that whole contruction was kinda "forced" not natural

Right. That's a good analogy.

all right thanks Ted!

FYI Ted, the question that I started with , was the set of upper triangular matrices with diagonal entries = 1 .

That was the subgroup, not the big group.

that was the quotient i wanted to have

Now things are totally muddled.

You told us upper triangular elements of $GL(3)$ was the big group and we wanted the kernel to be diagonal $1$ with arbitrary $13$-entry.

So we start with something 6-dimensional and should end up with something 5-dimensional.

upper triangular with diagonal entries = 1 mod identity matrix with only 1,3 entry = anything

I swear you never said all diagonal entries 1 in the big group.

I never claimed I said this!

you are right

I just posted that question like that, then he understood it that way, and then we all understood it that way

You posted the question here and on main as I just stated.

looking back now at the question , it was diagonal entries = 1

Yes but you know me by now ,i missed to type that part

Doesn't change much, actually.

well it kinda does

becaue if we mutiply two such matrices

only the entry 1,3 gets effected

and we add those entries

A_1,3 + B_1,3 , rest stays the same

wait wait

am multiplying elements of the kernel with each other

in that case what i wrote is true

You can restrict to the case $x=z=v=1$.

yeah :)

quotient groups are magical

But don't blame us if you gave us all the wrong question.

am not !

it was a very good question that came out of that mistake

I think the questions are equally difficult, actually.

Just a 2-dimensional answer instead of a 5-dimensional answer.

but still Ted, I still dont know how one deals with cosets if they infinite

You know lots of examples of infinite cosets.

theta +2PI

for angles :D

and paralell lines for v.s

Take $\Bbb R^2$ and mod out by a $1$-dimensional subspace.

Right, you just said that.

null space in general

x_0 + x_p are cosets

but when I view them in such exercices like you solved now, it does not look nice

ican take elemnt of H and multiply it by K

but I dont feel the answer

Sure.

Well, you won't always.

hmm that it why I asked the question

to understand the quotient group without going to cosets!

For example, if you take $SO(n)/SO(n-1)$ you can't picture it algebraically. You have to picture it in terms of group actions and you get the $(n-1)$-sphere.

There is no geometry in the question you asked.

okay :D

At least, none that I see.

but what did he mean by , operation defined on polynomials?

in terms of *

Algebraic group structure.

All the pieces of the group law are polynomial expressions.

aha neat ._.

I want to learn more of these

what course would one take ? :D

I don't know.

@TedShifrin All of the relevant papers I found have exactly the same expression. But I'm not entirely sure how to interpret it either.

I never like it when people write $dx/df$.

I keep forgetting that you like analysis and geometry more =p

well Ted thanks alot for your patience and help ! <3

@user76284: If $f$ is a diffeomorphism, then it's just a change of coordinates and one can make sense of it. But it's not the entry of the inverse matrix of the jacobian, of course. That only works in one dimension.

Found this: math.stackexchange.com/a/1744211/76284

That seems to suggest it is? Or am I reading it wrong.

Is infinity undefined?

No, it specifically says it is NOT when you have more than one variable, @user76284.

I meant it's an entry of the inverse matrix, just not the same one?

i.e. $\frac{\partial x^k}{\partial f^l} = ((\nabla f)^{-1})_k^l$ rather than $((\nabla f)_k^l)^{-1}$

Infinity has a definition, so it's not undefined. Perhaps you're being confused by the limit of $\frac{1}{x}$ as $x$ approaches $0^+$.

Which is infinity, but $\frac{1}{0}$ is undefined.

It's neither.

What is it?

Is there an extra transpose?

No, it's way more complicated than that.

You have matrices that are inverses. Entries are not related to entries ... other than by a giant mess with cofactors.

I know, but I'm indexing outside the inverse (i.e. assuming the matrix inverse has already been taken), right?

I should've written $\frac{\partial x^k}{\partial f^l} = ((\nabla f)^{-1})_l^k$ instead (had the indices switched).

But it's never that except in one dimension.

Maybe these papers all have some convention that I don't know about.

What is the answer I linked to saying, then, in terms of coordinates?

This part: "Instead of the relationship (1) holding, we have instead..."

It's saying what I've been saying. The two matrices are inverse matrices.

Let $\rho = f^1, \phi = f^2, x = x^1, y = x^2$. Then the matrix being inverted on the RHS is the Jacobian, correct?

Yeah, and this is the case of (locally anyhow) a diffeomorphism.

So it's saying that (to take the top right element of the matrix on the LHS as an example) $\frac{\partial x}{\partial \phi} = \frac{\partial x^1}{\partial f^2} = $ the entry at row 1, column 2 of the RHS.

@TedShifrin I finally figured out what you were trying to say a few days ago about the component functions $x^i$ in charts

I typed up the answer there

But the RHS is just the inverse of the Jacobian. So it's saying $\frac{\partial x^1}{\partial f^2} = (J^{-1})^1_2$, right?

Hey, Ted!

How's the weather there? We are supposed to have 3/4" of rain here tonight.

@user76284: I'm busy with something very important IRL, so I can't keep looking bacak and forth. Sorry.

We're supposed to get rain here tonight, too, @robjohn.

We are meeting friends from your area tomorrow for dinner, about half-way in between.

It's supposed to be mostly over by morning, last time I checked.

Yeah, same here. However, it picks up again from Monday through Thursday.

Wet week.

Nothing in the forecast for here ...

I posted the question at math.stackexchange.com/questions/3070450/…. Let's see if I'm right.

Yeah, that is right, assuming $f$ is a diffeomorphism so things make sense.

@robjohn: Oh, we're supposed to get some showers Wed, it says.

@user76284 I would think that is the case.

but the author should really make that clear.

why df = f_x dx + f_y dy?

@robjohn: I really hate $\partial x/\partial f$ notations.

Why not, @Leaky? See my lectures :P

Okay here's a fun problem. So let's say you have two groups of n people, A and B, and each of them has a preference listing for members in the other group. The Gale-Shapley algorithm gives a stable matching, meaning it pairs any two people up and you don't have a person in group A and in group B which prefer each other to their partners

What does the LHS mean? $df(p)(v)$ means the directional derivative at $p$ in direction $v$.

I'm trying to translate the Schwarzian derivative given in mathoverflow.net/questions/320590/… to index-free notation. I got the first term down as $(\nabla f)^{-1} \cdot \nabla \nabla f$ but I'm having trouble with the second term. It looks like $-\frac{1}{n+1}g(\nabla \log \det \nabla f)$ but I'm not sure about $g$.

The algorithm works by having people in the first group propose down their preference lists, and people in the second group will accept any offer if they're unmatched, and any offer made by someone they prefer to their current partner. Now, if every person in the first group (those proposing) has a different top choice, they will get their top choices regardless of the preferences of people in group B.

@user76284: You might have wanted to ping robjohn on that

@TedShifrin sure, one could prove it using the limit definition or whatnot

or am I just choosing a basis

No, not limit definition. But it's expanding the $1$-form $df$ (which I just defined for you) in terms of the basis $dx$, $dy$.

In particular, every person in group B can end up with the person they prefer least. Question: is it possible that everyone in group A can end up with their worst choice?

so {dx, dy} is a basis of R^2*

and then define f_x to be the coefficient of dx?

Remember that the directional derivative (for $C^1$ functions) is $\nabla f(p)\cdot v$, and that's what you get when you evaluate the RHS on $v$.

No, $f_x$ is the partial.