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7:02 PM
@user2103480 You have lots of macros that don't work here :P
 
@user2103480 E(X'AX) = E(tr(X'AX)) = E(tr(AXX')) = tr(E(AXX')) = tr(AE(XX')) = tr(A*Sigma)
 
howdy, a @Balarka
 
@TedShifrin oh :D
shoot, totally forgot
 
I have the same problem because I have so many macros I use in my math documents/books.
 
Hi, @Ted
 
7:03 PM
Hi @TedShifrin. About to talk Fary-Milnor with the kiddo.
 
@BalarkaSen ah yes I could've worked with the trace without substitution
 
Oh nice :)
 
I'm getting the impression that discrete approximation is an important tool in integral geometry
 
That's Milnor's proof, not the one I gave.
Nah, I did lots of integral geometry. No discrete approximations.
 
Ted's uses Crofton's formula
You can also do it by hand using Gauss Bonnet
 
7:05 PM
I don't call that by hand. That proof is in doCarmo, using the tube.
 
yeah thats what i was remembering
 
He was looking at these GTech notes where the author uses the wrong names for everything.
 
I actually like the proof I have (which I learned from Griffiths — who knows who "did" it first) because it has Morse flavor.
And of course I love Crofton.
 
Ghomi. He calls Crofton's formula Cauchy's formula and he calls Fary-Milnor Fox-Milnor.
???
 
Lol
 
7:08 PM
Shame on him.
Cauchy did a Crofton-like thing for convex bodies.
 
@TedShifrin I have to get around to reading your little lecture notes titled "Geometry through the ages".
 
Nah, you don't, a @Balarka. That was lectures for a low-level audience.
 
I can't recall what caught my eye but it had many Morse like things
 
I did a variety of classic things, but did end with a bit of Gauss-Bonnet and the local integro-geometric version.
 
Hmm yeah that must be it
 
7:10 PM
hi chat
 
I should make a list of things I can do using Morse theory
 
hi Astyx
You can answer the question on main about when all $f^{-1}(c)$ are diffeomorphic for $c\ge 0$. :P
 
Haha there we go
I proved Lefschetz hyperplane theorem today
Someone asked my how much we know about topology of Stein manifolds so I had to cook up a proof for him
The basic idea is the squared distance to a generic point is a harmonic function along 1D complex submanifolds of the embedded Stein manifold which I find very cool.
 
Lefschetz hyperplane is a beautiful theorem. I know (knew) the complex geometry proof using Kodaira vanishing.
 
Ahh, ok.
But your version is for submanifolds of $\Bbb{CP}^n$?
 
7:16 PM
No. Just compact complex manifold.
 
I was thinking the variant for $\Bbb C^n$; affine analytic subvarieties deformation retract to half-dimensional CW complexes.
Ah OK.
 
(Kähler follows from the hypothesis that there's a positive smooth hypersurface.) I guess Bott has the general version.
 
What's a hyperplane section for an abstract complex manifold? Some zero locus of a bundle?
 
Right, you assume you have a smooth hypersurface that's positive (in cohomology, if you like).
So, it's not just any old hypersurface (e.g., the exceptional divisor when you blow up won't do).
 
That makes sense. Nice result!
 
7:19 PM
See Griffiths/Harris :)
 
@TedShifrin Yeah, good example
 
Interestingly, Ramanujam showed that Kodaira vanishing and the Lefschetz theorem are equivalent.
 
Have you seen The Man Who Knew Infinity?
 
Different Indian mathematician :P
 
Yeah C P Ramanujam
 
7:27 PM
I suppose they all look alike and have the same names. :)
 
lol
Ramanujam is different from Ramanujan though
 
Yes, that was sorta my point :P
Although, to be honest, this is the only time I've run across his name.
 
He and his circle of friends at TIFR are semi-famous in the country at least
All algebraic geometers of various flavors
You might have heard of Narasimhan (there are two, one is M. S., one is Raghavan :P)
 
I met the Narasimhan who was at U Chicago and who wrote a book or more on complex variables.
 
Ahhh
That's Raghavan yeah
 
7:36 PM
And then there's Narasimham, I believe, who posts a lot of geometry junk on main, lots of it wrong.
 
Lol
I think I have seen them
 
I try mostly to ignore him.
Oh oh, it's @Edward.
It gets really annoying when people delete their questions as soon as they have an answer. I am starting to flag all such and ask for the OP to be instructed in proper etiquette.
 
What do you call an integral that has the same area when you change the parameter?
 
What does it mean for an integral to have area?
 
The area A is invariant under perturbations on $n$ here $$ A=\int_0^\infty \frac{e^{-\frac{1}{tn}}(e^{-tn}(tn+1))}{t}~dt $$
 
7:49 PM
Your use of language is just so bad. The integral remains the same when you change the value of $n$. There is no word for this other than saying it is independent of $n$.
 
@TedShifrin hullo!
 
Finally, @Edward.
 
My sincerest apologies
 
I never trust you to be sincere.
 
does this occur a lot?
 
7:50 PM
@TedShifrin woaaah woah, why not?
 
sin seer
 
woah, never thought of the etymology of sincere
without smth
what's "wrong" in Spanish or French or smth
 
faux
 
One of my favorite French wines is Sancerre!
 
7:52 PM
Lol
 
sans cere
what are the integrals called?
 
No word.
 
Hey @Lukas and @Balarka btw
 
hey @Edward
 
7:53 PM
I'm dying with assignments this week
Still a physics assignment remaining
 
are these integrals useful?
 
same tho
big big big assignment for p-adic Hodge theory, and a fairly easy one for ANT
 
Just be glad you no longer have Ted homeworks, @Balarka.
 
Finish quick
@TedShifrin Those were nice though
 
OK, we enjoyed Milnor's paper.
 
7:55 PM
@EdwardEvans hi
 
The proof that crookedness 1 implies unknot was fun to do by hand.
 
Hey @Leaky and @MillMiker
 
@geocalc: Here's an old problem from the Putnam exam. Evaluate $$\int_0^{\pi/2}\frac{d\theta}{1+(\tan\theta)^{\pi}}.$$
 
-1/12
 
I've actually only glanced at Milnor's paper. Shame on Ted.
 
7:57 PM
haha that's a good integral
 
So how're you going to do it?
 
Darn it I don't understand any of the physics for this assignment I have to do
Time to read I guess
Can't wing it
 
Good evening everyone
 
buonasera
 
@BalarkaSen this but replace physics with literally anything
@Alessandro Grüß Gott
right I'm out, cy'all later
 
8:00 PM
hi, Demonic
what are you learning in physics, @Balarka?
 
Right now I'll read a section on the multipole expansion
 
So you're doing electricity and magnetism?
 
Yeah, electrostatics for now.
 
Gotcha.
 
A lot of electrostatics you can wing because it's just harmonic functions
But I need to know their terminologies
 
8:03 PM
Well, there's a lot more to physics than just the mathematics, honestly.
But, yes, the mean value property of harmonic functions is relephant.
 
100%, I'm just too lazy hah.
 
Hey'all
 
Eg a capacitor is nothing but a solution of $\Delta \Phi = 0$ on $\Bbb R^2 \times [0, d]$, with appropriate boundary conditions on the plates $z = 0, z = d$.
You can say stuff like this and get away lol
 
You should try to develop some physics intuition, mr math geek.
heya @Krijn
 
But this homework is on dielectrics which I don't understand
Yeah agreed.
 
8:06 PM
I remember studying this in 1971, but I don't remember the content.
 
@BalarkaSen hegelian dielectrics
6
 
LOL
 
lol
 
łøł
 
@TedShifrin Crookedness in v direction = number of local maxima of v * r(t). Crookedness = min crookedness in v direction (letting v vary).
Crookedness = 1 means that you can rotate space so that the curve's height has exactly one local maximum.
So pick a knot diagram so that height has exactly one local maximum, and crossings all have different height.
Exactly one local maximum => portion above the top crossing lies inside the knot, and forms a simple closed polygon in the plane
(left side of crossing goes up to that local max, as does right side of crossing)
 
8:09 PM
Right, so there's the same Morse theoretic flavor as finding a direction with precisely two critical points and filling in the disk.
 
By JCT you can then isotope that polygon to nothing, removing a crossing.
 
I still like the Griffiths proof I stole.
 
Yes, I then explained the proof of PL JCT by height functions to see the yoga here.
 
Sorta similar to thin positions of knots
 
Gotcha.
 
8:11 PM
Eg a knot with a projection to a line in R^3 with exactly two critical points is an unknot
 
Right, that's what this proof is saying :)
 
Mjolnir, Norse Theory
4
(@Krijn)
 
Floor homology
 
(¿Que?)
 
@MikeMiller IIRC thin position of a knot is a knot with minimum number of critical points or something of this sort
 
8:16 PM
Gotcha
 
On a related note (knote), waist of a knot is the smallest number $n$ such you can isotope the knot so that it intersects each $z = c$ in at most $n$ points
 
Hej, is there some established notation like Re() and Im() for complex numbers to get real and imaginary part for quaterions? I.e. how would I denote that we only take vector part of the quaterion for further calculations. so given q=(u,v) and i only want to get v :)
 
It was a conjecture of Gromov if the $(p, q)$-torus knots have waist $2p$
 
@smihael Yes, the imaginary part of the quaternion contains all the $i,j,k$ part.
 
@TedShifrin thanks :) so i can write Im(..)
 
8:40 PM
Any clever way to show that $x^6-3$ is irreducible over Q(\sqrt{-3})?
Only direct computation? I think there is a way to do this without that
 
Ring of integers of $\Bbb Q(\sqrt{-3})$ is $\Bbb Z[\omega]$ and $\sqrt{-3} = 1 + 2\omega$, so $|\sqrt{-3}| = 1^2 - 2 + 2^2 = 3$ which is a prime so it seems to me that $\sqrt{-3}$ is irreducible. Do Eisenstein's criterion over $\Bbb Z[\omega]$?
And then Gauss's lemma
I could be wrong, @Thorgott would be able to check if this works
 
is $\mathbb{Z}[\omega]$ a UFD
 
Yeah man
 
prime elements in UFDs give rise to discrete valuations on their quotient fields
and Eisenstein's works with any discrete valuation
but you want the constant term to have valuation 1 for Eisenstein, so I'm not sure what your idea is here
 
8:56 PM
I don't understand this language. $\sqrt{-3}$ divides the constant term, but it's square doesn't
It's an irreducible over $\Bbb Z[\omega]$, a fine Euclidean domain with a norm
So Eisenstein goes through no?
 
its square is the consatnt term
$x^6-3$
$-3=\sqrt{-3}^2$
yes? no?
 
Oh right I forgot the polynomial or something
 
@BalarkaSen Who proved it
 
John Pardon
lol
 
sure
 
8:59 PM
Actually I don't know if x^6-3 is irreducible over Q(\sqrt{-3}) but it seems irreducible
 

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