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

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

@TedShifrin can I ask a quick question again ?

Ah

I think that should work, @Lozansky.

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

@TedShifrin But that's just what I wrote above

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

@TedShifrin What are their domains?

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

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

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 ?

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

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

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

@anon !!!

okay

yeah that makes sense

hello

Hi 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.

chicken rice and pizza

haha

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

I wouldn't do rice + pizza, Karim.

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

sure, it's subjective

I would @TedShifrin :D

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

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

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

haha

LOL at borromean pasta

I have never had Borromean pasta

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

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

lol

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

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

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

The even extension of $\delta$?

@TedShifrin want to hear something funny ?

dunno if there's an answer for my tastes

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

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?

@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

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

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

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.

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

@TedShifrin you and me both

don't get me started

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

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

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.

AAAAAAHHHHHHH

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

It's not an extension.

@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

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

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 $0$.

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

It is a physics problem, I should say

I dunno. I give up.

I ran into an even weirder result before in my book

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

@TedShifrin i figured that's what im working towards

"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 $(f^+)''= (f'')^+ + 2f'(0)\delta(x)$"

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

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?

Stop writing the damn x.

They state it without proof :(

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

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

thats a lot of backslashes

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

oh yeah

duh duh duh

@TedShifrin It is a function, not distribution

duh

idiot

faceplam

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

because its scalar

Yup.

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

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

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?

$\delta$...

There you go.

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

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 $f''(0) = g''(0)$?

That's not valid.

Why not?

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

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

Oh

NO.

That's a good point

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

@TedShifrin got it!

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.

rho^2 / rho^2

@TedShifrin $0$?

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

@TedShifrin Then I can't say anything

Of course you can.

You do the limit as I said.

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

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).

sup

hi

heya Eric

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

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.

lol

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

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.

@EricSilva did you see my ping

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

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

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

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

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

blah

yeah

When does your AoPS class end Professor?

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.

cool

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

I meant image.

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.

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

Right.

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

okay

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

Hmm.

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

Yeah, that's true.

okay

Usual proof is using the spectral theorem.

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.)

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

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

@TedShifrin don't use x

:P

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

GRR.

Right, Semiclassic.

So ... next?

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

Those are your direction cosines, of course.

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

Whatever.

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 :>

What's positive semidefinite again?

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

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

What makes it semi, Semi?

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

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

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

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

Mhm

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 $A,B\mapsto f(A,B)=A^T B$.

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

hmm, fair

OK, now what?

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.

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

Right. There, though, I was doing the determinant

Yeah, yeah.

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

So you want the image of the mapping again?

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

Right.

You both should use ^\top for transpose :P

ugh

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

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

It really isn't a T.

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

So we still have unit columns.

:/

yeah

I really dislike $A^T$ after years.

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

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

...bleh, that looks awful.

I am not proud of that formatting.

It looks beyond awful.

\mid for the vertical bar

It fixes the spacing

@Semiclassical Nice

hello people

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

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

write it out

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

yeah, in this case I might as well

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

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

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

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.

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.

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

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

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

Right.

Yup, DogAteMy.

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

Correct

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.

Yes, those

Wait, what?

Balarka was hasty.

He means in a parameterization

The axes are the coordinates

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

Ah yeah fair

General functions won't wash out.

Ok, unit ones.

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

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 $A^T B$.

Pushforward of unit vector fields parallel to coordinates.

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

Point is

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

yeah, idfk

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

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

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

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

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

Helloo

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

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"

True

but still

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.

Hi @Antonios

how's it goign

And, besides, it's true in Euclidean space

(and thus in any subspace of Euclidean space)

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

@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.

@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 $M$.

OK I need to go now sorry

@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.

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

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 $f$.

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

Hmmm

Yeah, okay

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

@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

agreed.

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

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

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

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

@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.

Hola.

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

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

hi @anakhronizein

@Semiclassic: Sure, the column space.

I know exactly that feeling.

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

@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?

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.)

Sure.

hmm, for each of them?

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

@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.

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.

I think that makes it worse, not better.

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

@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.

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

That's fair.

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

Right.

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

hah

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

Something I could consider on the other side

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

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

@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 $p \in U$?

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

Right.

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

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

Wow.

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

That's a fascinating fact

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

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

Yeah.

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

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 $R_1^\top M R_2 = (AR_1)^\top (B R_2)$.

Gotcha. Thanks!

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

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

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?

No, I was about to complain about that.

Because you're rotating a non-orthogonal frame.

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.

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

hrm.

Anyhow, I'm going to cook. Bye!

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

Later.

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

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.

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

@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

@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

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

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)

hi @MatheinBoulomenos

can we discuss something in complex analysis ?

hi @Adeek

okay, I can try

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

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 $1$

I see

I see okay that makes sense thanks

thanks a lot @MatheinBoulomenos

np

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

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?

@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

\o @MikeMiller