Mathematics - 2018-03-04 (page 1 of 3)

Lozansky

12:00 AM

What do you mean by $f \circ -$?

Ted Shifrin

$g(x)=f(-x)$.

Adeek

@TedShifrin can I ask a quick question again ?

Lozansky

Ah

Ted Shifrin

I think that should work, @Lozansky.

Karim, just ask. It's more disruptive to keep asking that.

Lozansky

@TedShifrin But that's just what I wrote above

Ted Shifrin

12:01 AM

Where did you write it above? I'm applying to two different test functions.

Lozansky

@TedShifrin What are their domains?

Adeek

Let $\phi(z)$ be analytic function on $\{|z| < 1\}$ if $\phi(0) = 0$ and $|\phi(z)| < 1$ for all z in D, show that $|\phi^{\prime \prime}(0)| \leq 2 - 2|\phi^{\prime \prime}(0)|$. It is hinted to apply Schwartz lemma to $\psi(z) = \frac{h(z) - h(0)}{1 - \overline{h}(0) h(z)}$ where $h(z) = \frac{\phi(z)}{z}$

so I solved it

Lozansky

I guess whole of $\mathbb{R}$ since they are test functions

Adeek

but I am having one major issue

how come we can define $h(0)$

it is undefined at zero

I guess we can still use it as it it is removable zero right ?

Ted Shifrin

@Lozansky: They're compactly supported functions on $\Bbb R$, of course.

Adeek

12:05 AM

@TedShifrin I think we can still use this because of the removable zero issue right ?

Ted Shifrin

Karim: You should be able to answer that, yes. Define it appropriately at $0$.

@anon !!!

Adeek

okay

yeah that makes sense

anon

hello

Adeek

Hi anon

anon

so here's a question. what's a set of three foods such that any two of them together is good, but all three together is bad.

Adeek

12:07 AM

chicken rice and pizza

haha

Ted Shifrin

Isn't "good" and "bad" regarding food a matter of individual taste?

I wouldn't do rice + pizza, Karim.

Lozansky

@TedShifrin Well then $\delta[f] = \delta[g]$?

anon

sure, it's subjective

Adeek

I would @TedShifrin :D

Ted Shifrin

@Lozansky: That's my suggested definition, yes. (For all $f$ and associated $g$.)

anon

12:07 AM

you put chicken or rice on your pizza, but you wouldn't put both?

Balarka Sen

@anon a borromean pasta, red flavored, green flavored and yellow flavored

anon

haha

Ted Shifrin

LOL at borromean pasta

Adeek

I have never had Borromean pasta

Lozansky

@TedShifrin OK... but doesn't that suggest $\delta^+ = \delta$?

Adeek

12:09 AM

I should try Borromean pasta at some point together with Torus pasta

lol

Ted Shifrin

I'm talking general distributions at the moment, @Lozansky. Now what is $\delta^+$?

Adeek

I should work on my jokes lol I was jocking with my wife the other day and was saying something that sounds funny to me

and I was laughing and I only heard crickets on the other side on my wife

haha

Ted Shifrin

@anon: So what is your candidate for an answer?

Lozansky

The even extension of $\delta$?

Adeek

@TedShifrin want to hear something funny ?

anon

12:10 AM

dunno if there's an answer for my tastes

Joe Shmo

@TedShifrin is (rho x c)* = rho* x c* ?

Ted Shifrin

Why is it an extension? I don't get that. What we're saying is that $\delta$, according to my definition, is itself an even distribution.

That doesn't make sense, @JoeShmo, does it?

Adeek

@TedShifrin I was giving a talk last month. So, after the talk people were clapping then I started clapping with them like an idiot.

hahaha

Ted Shifrin

Well, our idiot president does that all the time, Karim.

Lozansky

@TedShifrin I'm not saying it makes any sense, it's straight from my book

Ted Shifrin

12:12 AM

I don't like the x's in there, to start with, @JoeShmo. And you know I have a marginal opinion of your book already.

Joe Shmo

@TedShifrin i.. dont ... know .. :S

@TedShifrin you and me both

don't get me started

Lozansky

"The even extension of $\delta_{\epsilon}$ is $\delta_{\epsilon}^+ = \delta_{\epsilon}+\delta_{-\epsilon}$. If we let $\epsilon \to 0$ we get $\delta^+ = 2\delta$"

Joe Shmo

I think the author was thinking in French, writing in English

Ted Shifrin

So we're trying to compute $(\rho c)^*\omega$ for a $1$-form $\omega$ on $\Bbb R^2-\{0\}$. $\rho$ is a scalar function, so it can't pull back anything.

Joe Shmo

AAAAAAHHHHHHH

Ted Shifrin

12:13 AM

@Lozansky: OK ... but I do not like calling it an even extension at all.

It's not an extension.

Lozansky

@TedShifrin I should say that in the problem, we have a condition for $x>0$ involving $\delta$. We then extend the domain to $\mathbb{R}$ to make use of integral transforms

Ted Shifrin

So you're only applying to test functions with compact support in $\{x>0\}\subset\Bbb R$? That's bizarre.

Lozansky

Yes that is correct

More specifically, it is assumed $x\geq 0$. Then we have a condition $u(x,0) = c \delta_{\epsilon}(x)$ where $\epsilon>0$ is assumed to be close to $0$.

Ted Shifrin

Then it won't make sense to apply such a distribution to a function on $\Bbb R$.

Lozansky

It is a physics problem, I should say

Ted Shifrin

12:16 AM

I dunno. I give up.

Lozansky

I ran into an even weirder result before in my book

Ted Shifrin

@JoeShmo: When you pull back, you will get two terms with $\rho'(t)\,dt$ appearing, but they should cancel out.

Joe Shmo

@TedShifrin i figured that's what im working towards

Lozansky

"Let $f=f(x)$ be defined on $x\geq 0$. The even extension of $f$ is $f^+(x) = \cases{f(x), & x>0 \\ f(-x), & x<0}$. Then $(f^+)''= (f'')^+ + 2f'(0)\delta(x)$"

Ted Shifrin

@JoeShmo: And in the other terms, $\rho^2(t)$ in the numerator cancels $\rho^2(t)$ in the denominator.

Joe Shmo

12:18 AM

but i can't fit it all in. in particular, i don't know what rho x c would look like in their individual compoennts?

Ted Shifrin

Stop writing the damn x.

Lozansky

They state it without proof :(

Joe Shmo

one sec let me snap a picture of what i have so you know im not milking answers for free

well if i write * I'll confuse it with the pullback

how's rho \times c

Ted Shifrin

$(\rho c)(t) = \rho(t)(c_1(t),c_2(t)) = (\rho(t)c_1(t),\rho(t)c_2(t))$.

Joe Shmo

thats a lot of backslashes

Ted Shifrin

12:20 AM

Just omit multiplication altogether, just like scalar multiplication of a vector. Yikes.

Joe Shmo

oh yeah

duh duh duh

Lozansky

@TedShifrin It is a function, not distribution

Joe Shmo

duh

idiot

faceplam

Ted Shifrin

@Lozansky: I'm ok with that. That's really an extension (except you need to include $0$ in the domain).

Joe Shmo

because its scalar

Ted Shifrin

12:21 AM

Yup.

Lozansky

@TedShifrin Does the 2nd derivative make sense to you?

I mean, take $f(x) = e^x$. There is no jump discontinuity for $(f^+)''$ at $0$?

Ted Shifrin

Sure there is. Look at the first derivative. You get $1$ from the right and $-1$ from the left. So the first derivative is essentially a Heaviside function. What's its derivative?

Lozansky

$\delta$...

Ted Shifrin

There you go.

You have to think about the definition of the second derivative.

Lozansky

But if $f(x) = e^x, x \leq 0$, $g(x) = f(-x), x<0$ then $f''(x) = e^x$ and $g''(x) = e^-x$ and so $f''(0) = g''(0)$?

Ted Shifrin

12:29 AM

That's not valid.

Lozansky

Why not?

Ted Shifrin

Because the first derivative doesn't exist at $0$. And the second derivative is the derivative of the first derivative.

Lozansky

I'm looking at the 2nd derivative from left and right and they are equal?

Oh

Ted Shifrin

NO.

Lozansky

That's a good point

Ted Shifrin

12:31 AM

You can't just do one-sided second derivatives. It's garbage.

Joe Shmo

@TedShifrin got it!

Ted Shifrin

If I give you $f(x) = \begin{cases} x^2, & x\ge 0 \\ -x^2, & x<0\end{cases}$, how do you decide what $f'(0)$ is? @Lozansky

Yippee @JoeShmo.

Joe Shmo

rho^2 / rho^2

Lozansky

@TedShifrin $0$?

Ted Shifrin

But you can't just do $2x$ and $-2x$, @Lozansky. You have to use the limit definition $$\lim_{h\to 0} \frac{f(h)-f(0)}h.$$

Knowing $f'(x)=2x$ for $x>0$ and $f'(x)=-2x$ for $x<0$ doesn't tell me anything about $f'(0)$ UNLESS I know that the first derivative exists and is continuous.

This is an important mistake to understand :P

Lozansky

12:35 AM

@TedShifrin Then I can't say anything

Ted Shifrin

Of course you can.

You do the limit as I said.

Lozansky

Yeah well $(h^2-0)/h=0$ in the limit

Ted Shifrin

You have to do two-sided there!

$\pm h^2$.

So that's why that works. But the same principle applies for your second derivative (but it's more subtle because of the different one-sided derivatives).

Eric Silva

sup

Semiclassical

hi

Ted Shifrin

12:39 AM

heya Eric

Semiclassical

So, here's a question I'm trying to figure out how to get my head around.

Ted Shifrin

Just met a really smart high school kid on Thursday who's doing independent study out of my multivariable book, but U Chicago is high on his list. He thinks he's interested in computer science and maybe math. I gave him a warning and told him that I could put him in touch with you guys if he wanted me to. He seems to know some kids from his school there so he probably won't ask.

Easier to remove your head, @Semiclassic.

Semiclassical

lol

Lozansky

@TedShifrin But I still don't see how they arrive at $(f^+)'' = (f'')^++2f'(0)\delta(x)$?

Semiclassical

First, an easier one which I know an answer to. Consider the map $U\mapsto U^T U$, and let $S$ be the set of 3-by-3 real matrices with unit columns.

Balarka Sen

12:41 AM

@EricSilva did you see my ping

Ted Shifrin

@Lozansky: For $x\ne 0$, you get $(f'')^+(x)$. And our Heaviside discussion gives the $2f'(0)\delta$.

Semiclassical

(Why those names? I dunno, had to pick something)

Ted Shifrin

@EricSilva: I forgot to ping you with that long paragraph.

Semiclassical

What's the image of $S$ with respect to that mapping?

Ted Shifrin

not "with respect to" ! You mean, "under"?

Semiclassical

12:43 AM

blah

yeah

skullpatrol

When does your AoPS class end Professor?

Ted Shifrin

First guess: all symmetric $3\times 3$ matrices with $1$'s on the diagonals and numbers with absolute value $\le 1$ on the off-diagonals.

Not 'til mid June, skull.

We're doing linear algebra for the rest of the course.

skullpatrol

cool

Semiclassical

That's definitely the codomain, but the image is actually smaller than that.

Ted Shifrin

I meant image.

Semiclassical

12:45 AM

The characterization I know goes like this. Pick any real 3-vector $x$; then $x^T U^T U x=\|U x\|^2\geq 0$, so $U^T U$ is positive semidefinite.

Ted Shifrin

Yeah, because the mutual dot products aren't entirely independent.

Semiclassical

Right.

Ted Shifrin

Yeah, I meant to say semidefinite, but forgot to type it.

Semiclassical

okay

My understanding is that that's enough to characterizes the image.

Ted Shifrin

Hmm.

Semiclassical

12:46 AM

i.e. if $M$ is positive-semidefinite and satisfies the rest of the requirements you gave, then $M=U^T U$ for some appropriate $U$.

Ted Shifrin

Yeah, that's true.

Semiclassical

okay

Ted Shifrin

Usual proof is using the spectral theorem.

Semiclassical

So that's something I understand pretty well. Incidentally, the main requirement for the matrix to be PSD in that scenario is that $\det M\geq 0$.

(It's not the only one, but the others follow from the requirements you listed.)

Ted Shifrin

You need the 2x2 principal minors to have the same.

Semiclassical

12:49 AM

Right. But the principal minors are of the form $\begin{pmatrix} 1 & x \\ x & 1\end{pmatrix}$

Joe Shmo

@TedShifrin don't use x

:P

Semiclassical

So $|x|\leq 1$ is enough to guarantee that the principal minors are nonnegative etc

Ted Shifrin

GRR.

Right, Semiclassic.

So ... next?

Semiclassical

Last point before moving on: If you view $M$ as a function of the matrix elements $m_{12},m_{23},m_{13}$ (the rest are either 1 or set by symmetry)

easy enough to change

Ted Shifrin

Those are your direction cosines, of course.

Semiclassical

12:52 AM

then $\det M = 1+ 2 m_{12}m_{23}m_{13}-m_{12}^2-m_{23}^2-m_{13}^2$.

Ted Shifrin

Whatever.

Semiclassical

Which corresponds to the cubic polynomial $f(x,y,z)=1+2xyz-x^2-y^2-z^2$

and, wouldn't you know it, $f(x,y,z)=0$ is Cayley's cubic nodal surface :>

Akiva Weinberger

What's positive semidefinite again?

Semiclassical

$x^T M x\geq 0$ for all $x$

Ted Shifrin

or all eigenvalues $\ge 0$ ($M$ is symmetric here)

Akiva Weinberger

12:53 AM

What makes it semi, Semi?

Semiclassical

the possibility that $x^T M x=0$

Akiva Weinberger

I guess definite means it's positive for nonzero $x$?

Ted Shifrin

otherwise $>0$ for all $x\ne 0$

Semiclassical

yeah, I should've said $x\neq 0$

Akiva Weinberger

Mhm

Semiclassical

12:55 AM

Anyways. It's just cute that the boundary of the set of PSD matrices of this form happens to be such a nice surface.

So, that was the case I knew well.

For my present case, I instead have the bilinear mapping $A,B\mapsto f(A,B)=A^T B$.

Ted Shifrin

ugh, don't use inner product brackets for a matrix-valued thing

Semiclassical

hmm, fair

Ted Shifrin

OK, now what?

Semiclassical

I still have $A,B$ being matrices with unit columns, but in the context of interest I now have them as 3-by-2 matrices.

Ted Shifrin

Oh, right. This is where I showed you the Gram determinant stuff.

Semiclassical

12:58 AM

Right. There, though, I was doing the determinant

Ted Shifrin

Yeah, yeah.

Semiclassical

and I'm pretty well convinced that's not going to help me much here.

Ted Shifrin

So you want the image of the mapping again?

Akiva Weinberger

So $A^TB$ is $2\times2$ (as opposed to $AB^T$ which is $3\times3$), right?

Semiclassical

Right.

Ted Shifrin

12:58 AM

You both should use ^\top for transpose :P

Semiclassical

ugh

four times as many characters to say the same thing :P

Ted Shifrin

Well, when you actually typeset for posterity, use it.

It really isn't a T.

Semiclassical

$A^\top$ vs. $A^T$...

Ted Shifrin

So we still have unit columns.

Semiclassical

:/

yeah

Ted Shifrin

1:00 AM

I really dislike $A^T$ after years.

Akiva Weinberger

And of course, $A^\top A=A^{\top+1}$

Semiclassical

$S=\{A\in \mathbb{R}^{3\times 2}\mid \forall i(\|Ae_i\|=1)\}$

...bleh, that looks awful.

I am not proud of that formatting.

Ted Shifrin

It looks beyond awful.

Akiva Weinberger

\mid for the vertical bar

It fixes the spacing

0celo7

@Semiclassical Nice

hello people

Semiclassical

1:02 AM

Is there a smarter way to do the for-all bit?

Ted Shifrin

So I guess we want to think about the two $2$-planes (whereas you were looking at the normal vectors).

0celo7

write it out

Akiva Weinberger

$S=\{A\in\Bbb R^{3\times}\mid\forall i,\|Ae_i\|=1\}$, maybe

Semiclassical

yeah, in this case I might as well

Akiva Weinberger

And I guess you want the image of $S\times S$ under $f$

Ted Shifrin

1:03 AM

I am not fond of overuse of symbolic quantifiers in the first place.

Semiclassical

$S=\{A\in \mathbb{R}^{3\times 2}\mid \|Ae_1\|=\|Ae_2\|=1\}$

I might as well just list them since there's only two columns in this case

And yeah, I want $f(S\times S)$

I'm having a hard time finding any hand-hold here, though.

Ted Shifrin

So we're really working on a variant of the Stiefel manifold of frames for $2$-planes in $\Bbb R^3$ ... you're using unit frames rather than orthonormal frames.

Unfortunately, your function doesn't give a well-defined function on pairs of planes.

Semiclassical

I mean, the characterization of the matrix elements as dot products means that they're between -1 and 1 here.

Balarka Sen

@AkivaWeinberger So, why do you believe in $\nabla_X Y - \nabla_Y X = [X, Y]$?

Akiva Weinberger

So this is really things of the form $\begin{bmatrix}A_1\cdot B_1&A_1\cdot B_2\\A_2\cdot B_1&A_2\cdot B_2\end{bmatrix}$

Semiclassical

1:05 AM

Right.

Ted Shifrin

Yup, DogAteMy.

Akiva Weinberger

@BalarkaSen If we choose $X$ and $Y$ such that their coordinates are parallel to the axes, $[X,Y]=0$, right?

Balarka Sen

Correct

Ted Shifrin

So, pursuing my 2-planes, we can generically choose them to have one of those unit vectors in common.

DogAteMy: You mean they're coordinate vector fields.

Not what you said.

Akiva Weinberger

Yes, those

Wait, what?

Ted Shifrin

1:07 AM

Balarka was hasty.

Balarka Sen

He means in a parameterization

The axes are the coordinates

Ted Shifrin

No, you could multiply by functions and still be in those directions.

Akiva Weinberger

Ah yeah fair

Ted Shifrin

General functions won't wash out.

Balarka Sen

Ok, unit ones.

Ted Shifrin

1:08 AM

We don't even know what "unit" means.

Semiclassical

One thing I was playing around with was to consider the mapping $A^T B\mapsto \text{tr}(A^T B R)$ where $R$ is some constant matrix, since that'll correspond some linear function on the matrix elements of $A^T B$.

Balarka Sen

Pushforward of unit vector fields parallel to coordinates.

Semiclassical

I swear I"m trying to will \tr into existence

Akiva Weinberger

Point is

Ted Shifrin

I don't know why LaTeX didn't create trace.

Semiclassical

1:08 AM

yeah, idfk

Akiva Weinberger

For those guys we're just requiring $\nabla_XY=\nabla_YX$, which is true in Euclidean space and everything inherits its $\nabla$ from Euclidean space

so it's a reasonable assumption

Balarka Sen

It's an O-K justification but it doesn't quite leave me satisfied

Semiclassical

And you can get some sorta nice stuff by taking $R$ to be an orthogonal matrix. But I'm not happy with it.

Akiva Weinberger

And that's the same as saying $\Gamma_{ij}^k=\Gamma_{ji}^k$, because guess what, Christoffel symbols are something I know now

Balarka Sen

I have an alternative way to parse what you said, by the way

Antonios-Alexandros Robotis

1:10 AM

Helloo

Semiclassical

@TedShifrin I'm going to have to trust that those words make sense :)

Akiva Weinberger

And then if you just shove the rest into the equation for $\nabla$ in terms of the Christoffel symbols, if you do $\nabla_XY-\nabla_YX$ the Christoffel symbol stuff cancels out and what you're left with happens to equal $[X,Y]$

which, yeah, isn't the best justification, it's just saying "the algebra works out"

Balarka Sen

True

Akiva Weinberger

but still

Semiclassical

Main thing is that I'm having trouble finding any hand-hold here. With the other case, the determinant ended up the natural choice: Any matrix of the desired form with nonnegative determinant was definitely in the image.

Ted Shifrin

1:12 AM

Hi @Antonios

Antonios-Alexandros Robotis

how's it goign

Akiva Weinberger

And, besides, it's true in Euclidean space

(and thus in any subspace of Euclidean space)

Semiclassical

Here, though, it's not at all clear to me what the right analogue is.

Ted Shifrin

@Semiclassic: I'm thinking about fixing the $2$-plane the first matrix corresponds to, and similarly the one the second does. Then think about varying the columns but keeping the planes the same.

Balarka Sen

@AkivaWeinberger Suppose $M$ is a smooth manifold and $f$ is a smooth real-valued function on $M$. Assume $x\in M$ is a critical point, i.e., $Df_x = 0$. You can define the Hessian, or the second derivative, of $f$ at this point as a billinear function $Hf : T_x M \times T_x M \to \Bbb R$ such that $Hf(X, Y) = X_x(Yf)$ where $Y$ is extended from $Y_x \in T_x M$ to a neighborhood around $x$ in $M$.

Akiva Weinberger

1:13 AM

OK I need to go now sorry

Ted Shifrin

@Balarka: I think I've told you this before. You can do that for any vector bundle and talk about the intrinsic derivative of a section at a zero.

Bye, DogAteMy.

Semiclassical

That phrase actually went by a bit fast for me, actually: What's a 2-plane?

Balarka Sen

This is, like the Hessian should be, symmetric because $Hf(X, Y) - Hf(Y, X) = X_x(Yf) - Y_x(Xf) = [X, Y]_x(f) = 0$ because $x$ is a critical point of $f$.

Ted Shifrin

@Semiclassic: A 2-dimensional subspace of $\Bbb R^3$? :)

Semiclassical

Hmmm

Yeah, okay

Ted Shifrin

1:14 AM

I'm assuming your column vectors are linearly independent. Things are quite degenerate when they're not.

Balarka Sen

@TedShifrin This is true. But my point is if you have a Riemannian metric, you can define a Hessian everywhere and the symmetricity of the Hessian is fundamentally based on $T\nabla = 0$

$T$ being the torsion

Semiclassical

agreed.

Ted Shifrin

I agree, @Balarka. I just don't stay glued to the Riemannian world.

Balarka Sen

I think it's what I am most comfortable with :)

PVAL-inactive

I use 2-plane for vector spaces that aren't R^3

when I say 2-plane field all the fibers are subspaces

Ted Shifrin

1:16 AM

@Balarka: Torsion only makes sense for the tangent bundle. In some sense, that makes it artificial. But I spent a lot of time trying to understand E. Cartan's interpretation in terms of holonomy of the affine connection.

anakhronizein

Hola.

Antonios-Alexandros Robotis

You know that feeling where you wonder if you're doing a problem the wrong way...

Semiclassical

So I guess the 2-plane corresponding to a 3-by-2 matrix A would just be the set of $Ax$ for arbitrary $x\in \mathbb{R}^2$

Antonios-Alexandros Robotis

hi @anakhronizein

Ted Shifrin

@Semiclassic: Sure, the column space.

anakhronizein

1:17 AM

I know exactly that feeling.

Semiclassical

...yeah, that's easier to say :/

Balarka Sen

@TedShifrin Mm, I see. So when do you have geodesic coordinates on a general vector bundle (with a bundle metric and a metric connection) in general?

Ted Shifrin

So think about two generic planes. They'll intersect in a line. You could pick a unit vector along that line for a column of each of your matrices. (Or you could not.)

Semiclassical

Sure.

hmm, for each of them?

hmm. I think I buy that, but I need to convince myself.

Ted Shifrin

@Balarka: In the usual sense, I think normal coordinates only make sense when we're doing the tangent bundle. There are analogues for projective connections, I guess.

@Semiclassic: The two planes intersect in a line.

Semiclassical

1:20 AM

I guess it should amount to considering $R_1 A^\top B R_2$, and picking $R_1,R_2$ so that the first column of $A,B$ in each case is the same.

I should probably have that as $R_1^\top$ but I don't really want to.

Ted Shifrin

I think that makes it worse, not better.

Semiclassical

I'm not sure it's the right mindset anyways

Ted Shifrin

@Semiclassic: What I'm proposing you think about (cuz I'm not gonna do it) is study what happens to your function if you fix the column spaces (the two planes). Then you can think about varying the angle between the two planes and see how the image varies.

Semiclassical

But I'm okay with considering $A=(e_1,a)$, $B=(e_1,b)$ for the time being.

That's fair.

Ted Shifrin

There's also a symmetry to the problem if we swap columns in one or both matrices.

Semiclassical

1:24 AM

Right.

Ted Shifrin

You should sort of mod out by that symmetry and not worry about it :)

Semiclassical

hah

Ted Shifrin

Well, I mean, figure out what it is but then stop worrying about taking it into account every time.

Semiclassical

Something I could consider on the other side

Ted Shifrin

My intuition expects there to be two extremes — the two planes coincide (degenerate situation), ranging to ... the two planes are orthogonal.

Semiclassical

1:26 AM

Can I find examples of $M$ such that no $A,B$ could possibly exist? (I guess that's just restating the question, though.)

Balarka Sen

@TedShifrin So if $(E, M, \pi, g, \nabla)$ is my total apparatus, doesn't it make sense to ask, for any $p \in M$, a trivializing chart $U$ around it so that $g$ maps to a bundle metric on $U \times \Bbb R^k$ by the trivialization $\pi^{-1}(U) \to U \times \Bbb R^k$ such that it's the identity metric upto first order at the fiber over $p \in U$?

Ted Shifrin

Well, you no longer have symmetry or positive semi-definiteness to go on.

Semiclassical

Right.

There's still some symmetry, of course, but that's not saying much.

Ted Shifrin

@Balarka: No, I don't think so. Interestingly, for complex manifolds and the tangent bundle, you need Kähler to get that.

Balarka Sen

Wow.

Ted Shifrin

1:28 AM

That's an equivalent way of defining Kähler, in fact.

Balarka Sen

That's a fascinating fact

Ted Shifrin

@Semiclassic: Your $M$ needs to at least satisfy the determinant bound you already got from your cross product thing.

Balarka Sen

Can you give me a place to read up about it?

Semiclassical

Yeah.

Ted Shifrin

It's in all the standard complex manifolds books, @Balarka ... certainly G/H, Wells, etc.

Semiclassical

1:30 AM

I guess I do know the following? Suppose $M=A^\top B$. Then for orthogonal 2-by-2 matrices $R_1,R_2$ I have $R_1^\top M R_2 = (AR_1)^\top (B R_2)$.

Balarka Sen

Gotcha. Thanks!

Semiclassical

Though, does that actually help me...hrm.

Ted Shifrin

I'm not sure what orthogonal $2\times 2$ matrices are doing, Semiclassic.

Semiclassical

Well, they're rotations on the column spaces

But now that I write it I'm not sure it makes sense. Is it even obvious that $A$ having unit columns implies that $AR_1$ does so as well?

Ted Shifrin

No, I was about to complain about that.

Because you're rotating a non-orthogonal frame.

Semiclassical

1:32 AM

Right.

I guess it'd work if I did something like $A^\top RB$ for 3-by-3 orthogonal R, but bleh

I don't really like that.

Ted Shifrin

I'm going to ask you to think about what I suggested for a while before you write down more matrix formulas.

Semiclassical

hrm.

Ted Shifrin

Anyhow, I'm going to cook. Bye!

Semiclassical

What I was mainly trying to do above is find some way to make concrete your point regarding the symmetries of this problem.

Later.

jrh

I kind of had a strange thought when working with images; an image can be thought either as a function z = f(x,y) where z is the intensity of the pixel, or as one long function z = f(i) where the image is "flattened" into a single long signal. That made me think, does there exist a transformation that can convert any concave... (hemispherical?) (not sure if that's the right term, but a surface where a straight line on the z axis only intersects once) 3d surface into a function z = f(i)?

seems like it would only work for surfaces that can flatten to a rectangle now that I think of it, a hemisphere wouldn't make much sense

jrh

2:00 AM

I'd guess it would be something like a series of piecewise functions, each one the curve of a cross section of the 3d shape, though I wonder if there's a more efficient method.

anakhronizein

3:17 AM

@BalarkaSen not to bug, but did you have a look yet? :P

Eric Silva

3:35 AM

@TedShifrin I just saw this because i went out but if you think someone could benefit from talking to me feel free to hand out my email liberally

MatheinBoulomenos

@BalarkaSen it's a monomorphism and an epimorphism, yeah

(in the category of (commutative) rings)

the map $k[x] \to k[x,y]/(xy-1)$ is the same as the localization homomorphism $k[x] \to k[x]_x$ (you can just check the universal property) localizations are always epimorphisms and we know exactly what the kernel of the localization homomorphism $R \to S^{-1}R$ is: anything from $R$ that is annihalted by something in $S$. In particular if $R$ is an integral domain and $S$ doesn't contain $0$, then $R \to S^{-1}R$ is injective

Alex

Buzzwords 'Birational equivalence' and 'non-complete varieties'

MatheinBoulomenos

Since $\operatorname{Spec}(k[x]_x)$ is homeomorphic to the basic open $D(x)$, which is the image under the map $\operatorname{Spec}(k[x]_x) \to \operatorname{Spec}(k[x])$ and since double localizations may be replaced by single localizations in a certain way, this map is an open immersion (which works for any localization at a single point)

Adeek

3:53 AM

hi @MatheinBoulomenos

can we discuss something in complex analysis ?

MatheinBoulomenos

hi @Adeek

okay, I can try

Adeek

I am just trying to understand 3 steps in this proof

why z = 0 --> f(z) is a polynomial of degree N. Why all roots are the same ?

also why N = 1?

I understand all the other argument

MatheinBoulomenos

I don't know the previous lemma, but the point is that you can do Taylor expansion of the function around $0$ and that will also be a Laurent expansion at $\infty$. If $f(z)= \sum_k^\infty a_k z^k$ is the Taylor expansion, then $f(1/z)=\sum_k^\infty a_k z^{-k}$. But if $f$ doesn't have an essential singularity at $\infty$ (i.e. $f(1/z)$ doesn't have an essential singularity at $0$), then there can only be finitely many coefficients $a_k$ which are not $0$

Then the Taylor expansion shows that $f$ is a polynomial

All roots are the same because $f$ is injective

$N=1$, because if $N>1$, then $f(z_0+\zeta_N)=a \zeta_N^N=a=f(z_0+1)$, where $\zeta_N$ is any $N$-th root of unity except for $1$

Adeek

I see

I see okay that makes sense thanks

thanks a lot @MatheinBoulomenos

MatheinBoulomenos

np

Joe Shmo

4:19 AM

anyone know how to find the boundary of a k-cube?

user131753

Very recently, I have been accepted to the course Masters in Pure and Applied Logic in University of Barcelona. However, I don't know whether it is a good idea to go to Spain for this degree. Can anyone tell me (1) is this degree highly valued in the academia (in Logic at least) like one obtained from, say, ILLC (here I am specifically talking about this degree) and (2) how is the faculty? Thanks.

Xander Henderson

4:40 AM

One of my colleagues got is masters in Barcelona (I think---he is Catalonian, so I am making some assumptions, I guess)

As to "highly valued", by whom and for what purpose?

Balarka Sen

@MatheinBoulomenos Yup, thanks. We decided this earlier

@anakhronizein Not the bit I didn't comment on, no.

I'll have a look sometime soon

Hi @MikeMiller

skullpatrol

\o @MikeMiller

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$Mathematics$ Mathematics