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17:08
Ah... okay. This is precisely metric compatibility $X\langle Y, \nu \rangle = \langle \widetilde{\nabla}_X Y, \nu \rangle + \langle \widetilde{\nabla}_X \nu, Y \rangle$. The left hand side is zero because $Y$ is vector field on $M$ so it has no projection along $\nu$.
We get $\langle \widetilde{\nabla}_XY, \nu \rangle = -\langle \widetilde{\nabla}_X\nu, Y \rangle = -h(X,Y)$.
what does the minus sign mean though?
o.9
o.9
hello everyone
how are we doing
17:23
How to type this?
oh god, no
before you ask how, you should ask why
Why?
o.9
o.9
this can probably get you started tex.stackexchange.com/a/308243
yeah, the tex SE is full of people who will lecture you on why not to do something while also telling you 10 non equivalent ways of doing it
sickos, but the harmless kind of sickos
o.9
o.9
17:48
latex is bloat
troff ftw
@feynhat There's a sign difference here, no? Just differentiate $\langle \nu,Y\rangle = 0$.
@leslie Is the munchkin back to wreaking havoc?
wth is that double root
she complained about having to walk by herself. she's a little wobbly and we'll have to pick her up early from day care again. i think it's only 80% a plea for attention.
Well, sounds normal :)
the operator for a number if it is in a flying saucer
17:52
Insanity @shin
everyone's focusing on the double root. what about the horrible spacing around the = in the middle of that.
It's not LaTeX.
the vertical stuff seems off too. you could probably do something like this in a mix of MS equation editor and just typing in word.
But who's writing such garbage?
if you focus you can also notice a secret message written backwards from the sheet's behind
17:54
@Wolgwang For the love of sweet baby jeebus! WHY?
yet another secret message from the devil. i'm ignoring you, satan.
@TedShifrin hey Ted!
Heya @Stan. Long time!
@TedShifrin yeah had a few health issues, but things are better now.
looking for jobs now
That's good. I'm falling apart completely. Every disk in my back is moderately to severely impaired. Amazing I can still walk :P
Jobs!! Wow. The end of an era!
17:56
Ah, but learning never stops :)
in TeX, LaTeX and Friends, 2 mins ago, by David Carlisle
@Wolgwang is don't an answer? :-)
Oh, I quit long ago.
I'm only loyal to learning :'-)
@o.9 My masters advisor thought that LaTeX was a bloated piece of garbage, and used Plain TeX exclusively. I learned on Plain TeX, but switched over to LaTeX when I started my PhD (my masters thesis is written in Plain TeX).
did it long before school and will do it long after
@TedShifrin When I was a kid, I just sat around reading the encyclopedia my parents had available
was a fun way to learn new things!
17:57
Michel is nuts, @Xander. I switched from AMS-TeX to LaTeX when I had to revise my first book (of four!) and realized that every time I reordered exercises or added a lemma I'd have to manually correct numbering throughout the whole document. Insanity.
there was a guy at berkeley who used troff well into the 2000s. i think he finally switched over to some flavor of tex.
I used to read encyclopedias and French dictionaries, @Stan.
I've never heard of Troff (other than in a context we don't wish to discuss here).
@TedShifrin hahahaha omg it figures, right? lol well my spanish was no where near that good, so i never got around to that.
what's nice about encyclopedias is you could have two encyclopedias deal with the same subject in an interesting manner from two separate article author's perspective
wikipedia sort of doesn't do that
switching languages is as close as you get
but then some pages are just translations
@Stan So you're applying in the econ world? Corporate?
18:06
@TedShifrin yeah thanks. I keep forgetting that connection=derivative.
Well, when it differentiates a scalar function, it is just plain old derivative :P
shintuku interesting point. another thing about paper encyclopedias is that at least some of the time, there is an effort to make cross-referenced material somewhat consistent. e.g. an entry on a math topic that cites another math topic might not point to something using completely different terms or notation. often because a small group of people would be responsible for editing multiple entries on a topic.
individual wikipedia pages in math can be good, but you click over to read more about some piece, and it could be a completely different set of people working on it, and mutual inconsistency in notation/definitions, etc.
o.9
o.9
@XanderHenderson ur master's advisor sounds wise
@TedShifrin deep learning researcher is i think my preferred area
specializing in healthcare
still figuring it out
just found out that terminology this morning by talking to an expert in the field
18:23
XD
at least latex gets the spacing right. note it also puts the bar of the fraction a little lower, which is clearly The Right Answer.
just goes to show that even if you have bad ideas and want to choose chaos, latex makes it more effective to typeset chaos.
what was satan using that for? arithmetic-geometric mean?
@leslietownes :-o How do you know?
i was thinking about what i might use it for, and then figured that's probably what satan would do.
@leslietownes AM lol
so some demon photoshopped that in there. OK.
was that demon you?
18:37
Nope... a typo
19:03
@Wolgwang math shrug
latest math conundrum: if I average $f_1(\hat{n}\cdot \hat{a}) f_2(\hat{n}\cdot \hat{b})$ over the sphere, must the result be some function of $\hat{a}\cdot\hat{b}$?
i think the answer is yes, and yet my computation (for a specific special case) keeps laughing in my face
(in my case i'm getting $\cos(\theta/2)$ when I think i should be getting $\cos\theta=\hat{a}\cdot \hat{b}$, and it is driving me mental)
19:22
@TedShifrin Michel doesn't use either Tex or LaTeX. He has an administrative assistant who converts all of his handwriting into LaTeX. My masters advisor, on the other hand...
@o.9 He is, though I disagree with him on this one. LaTeX is a horrible pile of bloat and incompatible libraries and whatnot, but it has the advantage of being designed to converge semantics into pretty documents. TeX doesn't really understand semantics---the user is meant to have to think about layout themselves.
It is like the difference between html and css. css tells the browser what a page should look like (i.e. how it is formatted), while the html gives the semantic meaning (e.g. "this is a paragraph", "this is a header", "this is a table", etc).
Since I am neither an expert on typesetting, nor do I want to be, I choose to trust that the choices made by Leslie Lamport (and others) reflect good typesetting practice, so I use LaTeX.
On the other hand, my masters advisor also believed that C++ was a terrible, bloated piece of garbage, and insisted that I code in C when I had to code. I took an undergraduate C++ class, but I am much more comfortable in plain C than C++ (OO is nonsense).
@Wolgwang That is a war crime.
7
Q: Example of a bounded linear functional $T:L^ \infty \to \mathbb{R}$ which cannot be expressed by an integral $\int_{[a,b]}fg$ for an integrable $f$

Pascal's WagerLet $m$ be the Lebesgue measure on $[a,b]$. What would be an example of a bounded linear functional $T:L^ \infty \to \mathbb{R}$ which cannot be expressed by an integral $\int_{[a,b]}fg$ for an integrable $f$? I know that, if $1 \leq p < \infty$, then any bounded linear functional $T:L^p \to \m...

I am working on a similar problem, but I am trying to show that the linear functional defined by $L(f) = f(2/3)$ cannot be expressed as such an integral. Is it possible to modify the example used by Quoka to work with the linear functional I am dealing with?
The fact that his sequence works is because $f_n(0) = 1$, i.e., it is nonzero.
But for large enough $n$, $f_n(2/3) = 0$, which won't get me the same contradiction that Quoka obtained.
Basically, I just need to redefine $f_n$ such that $f_n(2/3) =0$ for all $n$ or for sufficiently large $n$, so that the limit in the end is nonzero---which is precisely where the contradiction arises.
And I still need the functions to converge to $0$.
I guess a spike at $2/3$ should work...
So, continuous functions approximating the indicator function $1_{\{2/3\}$...not sure if there is a simpler example. Let me write down the functions.
Oh, wait...that won't converge pointwise to zero, but to $1_{\{2/3\}}$
But that function is $0$ almost every...Does the dominating convergence theorem work when the pointwise convergence is almost everywhere?
Yeah, we should be able to replace pointwise convergence with pointwise convergence almost everywhere---that usually works.
19:52
"f(2/3)" isn't defined for an element of L^oo
it's defined on C[0,1] or other function spaces whose elements have point evaluation
Yes, I am defining a linear functional on $C[0,1]$, then using Hahn-Banach to extend it to $L^{\infty}[0,1]$, and then arguing it doesn't arise as an integral for some $g \in L^{1}[0,1]$.
And then from there I am supposed to conclude that $L^{\infty}[0,1]$ is not reflexive.
o.9
o.9
what is OO ?
I like c++ because I'm a noob
oh, please don't do that to me again
@o.9 Object Oriented
o.9
o.9
oh
19:57
@o.9 No. Oh Oh.
$\frac{1}{n(n+a)} = \frac{1}{an} - \frac{1}{a(a+n))}$ is a useful identity that is not immediately obvious. the idea might be silly, but is there any sort of compilation of algebraic identities like these somewhere?
o.9
o.9
-.-
user, seems you have the right idea in that you're trying to show that an integrable function, if it implements that functional, has to be zero a.e. (and thus can't implement that functional). how you formalize may be up to you.
o.9
o.9
I think I only know how to use objecto oriented
shin: partial fraction decomposition?
19:58
oh, it's partial fraction decomposition
heh
thanks
one of the funny things about partial fraction decomposition is that it is sometimes taught before calculus, when there's very little use for it. it's just presented as something you can do. at least, i got it that way.
the problem with pre-calculus is that, if the only reason to learn it is "it'll be useful in calculus", then you don't have any motivation for it at that moment
that said, the thing which does come with partial fractions is the concept of "if two polynomials are equal for all inputs, then they're the same polynomial and you can identify coefficients"
and i do think that's not a bad lesson
i tossed this in chat earlier, but i do want a second opinion on this:
1 hour ago, by Semiclassical
latest math conundrum: if I average $f_1(\hat{n}\cdot \hat{a}) f_2(\hat{n}\cdot \hat{b})$ over the sphere, must the result be some function of $\hat{a}\cdot\hat{b}$?
so $\hat{n}\sim \text{Unif}(S^2)$
@Semiclassical Which is why the pre-calc curriculum is terrible, and why I kind of hate teaching the class sometimes.
i think you can prove that without reference to roots or factorization (e.g. by induction), but whatever. i'm not objecting to partial fractions.
semi did your geometric description change since before or is it the same? i was trying to think about that
same
oh
are we talking about what i just linked
or before that
my practical version now does look different
i was misinterpreting my source a bit
20:10
oh hrm
i'm about to get on a call but i will jump back to whatever the newest version is afterward
the practical version of this is now that $f_1(n\cdot a)=1$ if $(a+b)\cdot n<0$ and 0 otherwise. (yes, $f_1$ has implicit $b$ dependence as well. i think that can't be avoided in my problem of interest)
while $f_2(n\cdot b)=\frac12(1+n\cdot b)$
maybe the implicit $b$ dependence in $f_1(n\cdot a)$ is the issue
its presence does confuse me
20:41
@leslietownes I think f_n(x) = exp(-n(x-2/3)^2) will work.
20:55
every subset of an affine independent set is affine independent?
@XanderHenderson The memory management in C++, RAII, is easier than the manual management in C, I think.
21:09
@leslietownes the perplexing point is that, if my calculation is right, it seems like $f_1(n\cdot a)=1$ if $n\cdot a<0$ should work for my purpose
but my source definitely doesn't do this
21:33
ahhhhhh, now i get it. it really is what I just said
no $(a+b)\cdot n$, just $a\cdot n$
by golly that paper was tough to decrypt
@Semiclassic Hermann Weyl's Theorem on Vector Invariants says that any $O(n)$-invariant function of a collection of vectors is a function of their mutual dot products. With $SO(n)$ one throws in determinants, as well (which wouldn't apply in your case, anyhow).
I'll steal a quote of the theorem from my thesis for you.
Representation theorists, invariant theorists, and integral geometers know this. I wouldn't be surprised if a certain cadre of physicists know it, along with the unitary analogue.
right
something something Schur's lemma probably
No, it's far deeper than that.
LSS
LSS
21:36
It is rght to say that |a/b| = |a|/|b| for complex numbers, right?
Ah, I only have the unitary statement in my thesis.
If you want to read about it (and see a proof), see Spivak Volume 5 pp. 317 ff.
@LSS yes
LSS
LSS
Thank you Ted
Use polar form to prove $|ab|=|a||b|$ or your variant of it (which is immediate).
Incidentally, @Semiclassic, Weyl's Main Theorem was fundamental in Chern's original definition of Chern classes, the Riemannian tube formula, etc.
Weyl's The Classical Groups is an impossibility to read, however.
LSS
LSS
Yes, it was exactly what i did. But not being a mathematician makes me fear sometimes that i am assuming things that should be proved.
Are you a retired professor Ted?
Yup, @LSS. I'm a bum.
LSS
LSS
21:40
Not a english speaker native so i am not sure what does it means. I will search it, wait a minute.
LOL, don't worry.
LSS
LSS
Oh i see now hahah. I am a bum too, but an idiot bum. At least you are smart
It is only a rumor.
LSS
LSS
I see you have experience enough to answer one of the main question that has perseived me thoughout my life, can i ask you?
Don't expect too much. Go ahead.
LSS
LSS
21:44
Do you think "to be genius" is more about DNA/biological process, that is, you need to have the biological advantage to be it. Or do you think it is more about persistence and train? Or maybe both?
Since you were a professor, you probably saw a lot of type of students.
Oh, this is not a science question. This is a philosophy and emotion question.
LSS
LSS
Yes
Or maybe biological, but not so exact
"Genius" is not well-defined unless you use IQ, which we know is dubious.
@TedShifrin neat
I have seen a lot of innate talent in students and a lot of others in mathematics who succeeded quite well with very hard work and dedication. But I don't throw around the word genius. Even the mathematicians I know who were astonishing in their accomplishments, I never think about "genius."
Anyhow, @Semiclassic, because you're averaging, your result is of course orthogonally invariant and so your conjecture must be true :)
21:47
@TedShifrin oh, really? I usually suspect I should be saying something about Schur’s lemme when it comes to stuff like this
I think it's immediate from Weyl's main theorem. You have an $O(3)$ action on the unit sphere and your average value is invariant under that action. If $g\in O(3)$, it's clear (by moving the $g$ to $\vec n$) that $F(g\vec a,g\vec b) = F(\vec a,\vec b)$, no?
That seems reasonable. I don’t think I’ve seen Weyls main theorem
His theorem holds for any number of vectors, of course.
I stated it for you.
As I said, you can find a lot of discussion in Spivak, Volume 5.
There are ways of doing characteristic classes directly without appealing to that, so most modern treatments do.
21:52
I’ll have to look that up
Did your physicist colleague answer the email question?
Nope. But my having decrypted the paper means we have an answer
LOL, aha.
I should have mentioned Weyl's result last night. Sorry.
I mean, I’ve seen that concept before and was definitely taking for granted that it should only be a function of the dot product
But my flawed results from earlier were undermining my confidence in it
(I get why it wasn’t working now)
It's always nice to know you have theorems backing up your physicist's intuition.
21:57
What’s left is to figure out if there’s a cute way to compute the following without spherical coordinates
What’s the average value of $a\cdot n$ over a hemisphere?
WLOG we can assume the northern hemisphere of course
Doing that in spherical coordinates is trivial of course but it seems a waste
The only other method I know comes from the trick for integrating any polynomial over the sphere, which is to multiply by the Gaussian and use homogeneity. I haven't thought about whether that works for a hemisphere only.
The easy way is probably to use Weyl to conclude that it must be a function of $a_z$ alone, and deduce the value when $a_z=1$
I'm not sure it's a "waste." :D
Nah, you don't want to use a sledge-hammer on this.
22:00
I guess
Do you understand my Gaussian trick?
Not off the top of my head
You convert from an integral over the sphere (hemisphere?) to an integral over all Euclidean space.
Let me just scribble for a moment so I don't say nonsense.
I think I've written this down in some class notes or paper, but I don't remember where.
So I'm using spherical coordinates (inexplicitly) $(t,\sigma)$ on $\Bbb R^n-\{0\}$. You want to integrate $\int_{S^{n-1}} P(\sigma)\,d\sigma$. $P$ is a homogeneous polynomial of some degree here.
Rather than work it all out, I'll cheat and link you to this.
22:05
Ooo, Bargmann
I can't read past the first page.
But the trick (which someone taught me in grad school, I think) is to multiply by the Gaussian and then it's easy magic.
I’ll see if I have access when I get home
You get convergence on all of $\Bbb R^n$ and things break up into standard factorial thingies.
This is reminding me of something I know I don’t know
Namely, the relation between spherical harmonics and harmonic polynomials
By harmonic polynomials, you mean harmonic on all of Euclidean space?
22:11
Yeah
And then something something restrict to the sphere
You suffer through spherical harmonics in physics grad school
Right, so it's just a matter of understanding the laplacian on the sphere versus on space. Look at the laplacian in spherical coordinates, e.g.
OK, I have my thing figured out.
I have vague recollections of $Y_{lm}{\theta,.phi)$
And the misery of associated Legendre polynomials
Say $P$ is homogeneous of degree $m$. Then we take spherical coordinates $(t,\sigma)$ on $\Bbb R^n$, with $t>0$, $\sigma\in S^{n-1}$. We want $\int_{S^{n-1}} P(\sigma)\,d\sigma$.
OK, hush and follow what I'm doing :)
22:15
Sure
So we consider $\int_{\Bbb R^n} P(t\sigma) e^{-\|t\sigma\|^2}dV =\int_0^\infty\int_{S^{n-1}} P(t\sigma)e^{-t^2\|\sigma\|^2}\,t^{n-1}\,dt\,d\sigma.$
Damn I hate typing hard math on here.
OK, I think that's right.
Thinking
By homogeneity, this will be $\int_0^\infty t^{m+n-1}e^{-t^2}\,dt \int_{S^{n-1}} P(\sigma)\,d\sigma$.
I switched in my brain.
Right. So for this to be easy I think we’d want $m+n$ to be even
But if you write $P(x) = x_1^{\alpha_1}x_2^{\alpha_2} \dots x_n^{\alpha_n}$, you can now write this as a product of $n$ $1$-dimensional integrals, breaking the Gaussian up as the product of $n$ Gaussians.
Regardless, these integrals are well known in terms of Gamma.
22:21
True
I’d only worry about them being error functions but presumably we’re not that unfortunate
Details, tho
So you end up with something like $$\prod_{i=1}^n \int_0^\infty e^{-x_i^2}x_i^{\alpha_i}dx_i.$$
Since we know all these, we solve for $\int_{S^{n-1}} P(\sigma)\,d\sigma$.
Cool, eh?
I’ll need to study that a bit, probably pick an example
I should have found a way to sneak this into my multivariable math book. I don't think I did.
The question is — can we reduce to a hemisphere? I'm not sure.
22:25
Won’t that just mean $P(t|\sigma)=0$ if we’re on the wrong side of the sphere? That doesn’t seem too bad
It looks like you're integrating over a half-space instead, so one variable is different. Oh, these integrals should have been from $-\infty$ to $\infty$.
And then if $x_n\ge 0$, for the $x_n$ integral you just have $0$ to $\infty$.
Ahh, no wonder you were confident about them being Gamma’s
Of course, if $\alpha_i$ is odd, the integral should be $0$, so I needed to have $-\infty$ to $\infty$ above.
Robjohn will no doubt come in later and show us the sneakiest way to do this, but I'm confident I've reconstructed the method. Every physicist should know this :P
I've earned my double martini for tonight later.
heh
does $\sigma=(\cos\phi\sin\theta,\sin\phi\sin\theta,\cos\theta)$ here? i'd assumed so but i haven't actually seen that notation before
i do like it tho
I don't need to write coordinates on the sphere, ever.
That's the beauty of this. It's just an abstract point of norm $1$.
22:30
fair
I think I may have written this all up as an exercise for a grad geometry course once. I'm going to look for that.
Yeah, you do not have to do anything with specific spherical coordinates. You only need the volume element on $\Bbb R^n$ in terms of surface area of the sphere and $t^{n-1}dt$.
mostly i was looking at $t\sigma$ and wondering what it meant
A general point $x\ne 0$ in $\Bbb R^n$ is $t\sigma$ for $t=\|x\|$ and $\sigma = x/\|x\|$.
right
it's a point on the sphere and that's all you need
(and it makes good notation)
Well, I used $\sigma$ because it's analogous to surface integral notation you use, I think.
22:32
probably
in physics we tend to use $\Omega$ for solid angle a lot
I'm not doing solid angle, though.
Oh, I suppose I am.
Yeah.
Well, whatever.
heh
we don't use $\Omega$ as representing a point on the sphere, but it's not a bad convention
Oh, the exercise I wrote didn't have polynomials, but it used this method to find the volume of the sphere in arbitrary dimensions.
Just apply this set-up with $P=1$ :P
Anyhow, I'm now confident it works for a hemisphere, too. Or a quarter-sphere, etc.
Interesting.
22:35
right
just different restrictions on $P(\sigma)$
Nah. Just restriction on the integrals!
I want homogeneity. That's essential.
Oh, I suppose you could bump to $0$ and still be homogeneous. I think it's an identical computation.
It's just easier for me to look at the hemisphere and half-space, etc.
It's nice to know my brain still functions a little bit.
this method probably will help me if i ever try to understand the higher-dimensional aspects of this paper. i only needed the $d=2$ case but it's developed for arbitrary $d\geq 2$
Ah, cool :)
22:56
looking for what's popular lately in mathematical modeling for social sciences in the scientific literature. is there any search engine of mathematics journals?
not to play word games, but i doubt that 'mathematics journals' as many would define them would contain that stuff.
hm right, maybe applied mathematics might give better results
a lot of social sciences, even really quantitative stuff, is done by people who don't have any mathematical understanding. people who describe themselves as focused on 'methods' sometimes develop new models, but they are a tiny minority. maybe a few dozen people in the US.
everyone else just copies them with different data and less understanding.
applied mathematics might be better. something that's one step removed from particular data sets.
@leslietownes so what you're saying is that most social scientists give the field a bad name?
not necessarily. but i don't think they leverage mathematical understanding, or mathematical models very much, in what they do. even if their papers are full of what appears to be the output of R or stata or whatever.
so if someone with mathematical background were looking to model something in the social sciences, they might do well not to start with whatever they see, even in "top journals"
there's a tendency for people to duplicate the methods that other people use. editors who don't understand methods can't independently evaluate them, so feel better when people say they are using stuff they can cite as having been used in other papers.
i don't mean this as a diss on the social sciences. i wouldn't look to pure math journals to decide whether a physics paper was physically appropriate. pure math people don't have the language for discussing that.
same kind of thing.
23:14
Greetings @robjohn and slave-of-munchkin.
@TedShifrin howdy
I expected you to have a slicker way than my Gaussian method of integrating polynomials over the unit sphere. :)
I forget who pointed me in this direction many decades ago.
she's still refusing to walk, but will sometimes stand on her own. when she forgets that she claims not to be able to do that.
Some day you’ll regret raising a little you! :)
She does need to start mobilizing. Maybe Olivia can conspire :)
23:56
she's begun backseat driving a little bit, but only with my wife. apparently the light will turn green and she'll shout "GO! IT'S GREEN!" even if there are cars in front of them.
i'm getting a flu vaccine tomorrow, so if anyone wonders why i'm in a mood, or missing something, that's my preemptive excuse for that. anyone know if it interferes with 5G?
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