Biot-Savart rule, Ampere's circuital law, etc etc

@Balarka I remember you saying diagonalizable matrices are the roots to some polynomial in their coefficients

and I'm forgetting why

Neat. Any Faraday stuff yet?

Both integral and differential forms, @Balarka'?

Or, rather, nondiagonalizable matrices

Huh?

@AkivaWeinberger Is this Cayley-Hamilton business? I am not entirely sure what I said

Man, how drunk were you

None of Cayley-Hamilton has anything to do with diagonalizability.

@arctictern sorry to bug you again, but I need to show that $[(x,y)] \mapsto \frac{x}{y}$ is a ring homomorphism as well, right?

I am high on yellow bugpowder

a matrix satisfies its own characteristic polynomial

yes

Hi again, tern :)

@arctictern So, how do I define multiplication and addition on the equivalence classes $[(x,y)]$?

hello

@ALannister guess / look it up

@Akiva I just remember talking to you about characteristic polynomial of nilpotent matrices

Which is, of course, X^k

(I don't consider electromagnetism to properly start until Faraday's law of induction comes up)

@Semiclassical Nope.

(Until then it's electrostatics and magnetoststics, with no interaction between them)

@TedShifrin Well... one of the proofs can

@arctictern For $[(x_{1}, y_{1})]$, $[x_{2},y_{2}]$, $[(x_{1},y_{1})]+[(x_{2},y_{2})] = [(x_{1}y_{2} + x_{2}y_{1}, y_{1}y_{2})]$?

Yes, Demonark, if you allow a little analysis.

@TedShifrin Or a little algebraic geometry

Permanence of identities or something?

@Balarka Zariski?

Artin uses that phrase.

Though you can infer the necessity of Faraday's law of induction if you pick the right example

You just do the analytic argument for the Zariski topology on GL_n :)

@ALannister yes

Well, that's "analysis," Balarka'!

Hehe, fair enough

cool. So it's pretty straightforward to show this part then.

I can give that example, but maybe it's better once you see induction

@Daminark I have never used a book for it, only a combination of videos, articles, and courses. I often learn higher level things and then infill down.

Okay so you're not the one I battle to the death with

heya robjohn

I demand trial by combat.

Hey @robjohn long time no see.

@TedShifrin Hey, Ted. what's up here?

nothing deep, @robjohn :)

@TedShifrin Good; I'd hate to get in too deep!

@BalarkaSen My question above was essentially equivalent t to "why are they dense in Zariski"

I'm planning another drive up to and back from the Bay Area, @robjohn, but this time no stopping in LA ... trying to get across as fast as possible ;)

@TedShifrin Oh, well. It's too hot here anyway. Best pass us by and get to cooler climes.

well, Palo Alto is always hot, too ... but I don't relish the drive back on I5.

@TedShifrin the Coast Highway is closed, but I think that is farther north. That takes longer, but is more pleasant.

I did that once ... and yes, it's still landslided. I'm gonna take 101 up. Won't add tooo much time.

And I've never done it.

@AkivaWeinberger Because the set of all matrices is irreducible, and the set of those with all eigenvalues distinct is an open set

Doing the coast makes it 2 days.

@TedShifrin really? I drove to Cupertino along the coast in 1 day.

@Daminark You are looking for the other one @BenjaminR

chat.stackexchange.com/users/90992/benjamin-r

Must have been a 12-hour day or so, robjohn.

@TedShifrin don't remember

Oh, well, I have to add time to get from SD to LA and you're starting on the north side of LA ... so that could be a 3-hour difference.

sorry went afk

How dare you @Balarka

"You went away from keyboard? GULAG"

Random math fact of the day: For reasons, I'm presently interested in the function $e^{xe^x}$

I looked at the first few derivatives, found they were integers, and plugged those into OEIS

@Semiclassical derivatives evaluated where?

x = 0 I hope

zero, yeah

@Balarka actually at 7

And apparently that sequence counts the number of forests with n nodes and height at most one

@Ted what'd you say between complex, probability, or geometry?

Huh?

No no no, the derivatives were all constant integers

Hi chat

Hi @Eric

Hi Eric and chat

Hey @Eric

Sup

Hi, Eric (and @Astyx). Eric, you saw my ping yesterday that I heard from Spivak?

@Astyx Hi Eric

@Ted wrt lectures, I'm thinking of which two to pick

I did not but that is good

@Balarka Hi Balarka

For a rare instance, Eric, I pinged you by full name.

Get far enough into probability that you start doing generating functions. Then represent those coefficients as complex contour integrals :P

Weird I actually didn't get notified

Hmm, go check, Eric :)

@Daminark I just can't stop listening to Rick Astley now

Demonark? I don't know enough. Are the teachers all equally good?

Complex is graduate? What are the other two?

No I mean like, I'd be lecturing on two of those topics, like this summer

Ohhh

@TobiasKildetoft Oh, OK. So I could take the discriminant of the characteristic polynomial

(I'm actually running into that function in the opposite direction of that, i.e. starting from the integral )

@AkivaWeinberger Exactly, yeah

You should do one geometry, for sure, Demonark. I dunno what you guys are doing in probability or complex.

@TedShifrin Taking the 101 is only 30 minutes longer and better than the 5.

@Daminark do a geometry and probability

Well, in complex, today we were supposed to do Cauchy's bound, Liouville, FTA, Morera, some stuff on zeroes of a holomorphic function

Instead we did Cauchy integral formula and the fact that holomorphic functions are analytic

@robjohn: Really only 30 minutes? Going to Palo Alto I figured it wouldn't be too different, but from Berkeley the I5 is more of a straight shot.

@daminark I vote for devoting some time to learning something you might not see in a course here

Those are important things, @Demonark. Magical proof ... switching series and integral :)

@TedShifrin At least according to Google from my house to Apple.

Eric makes a good point.

Does anyone know of any good, simple references on stochastic dominance? Beyond Wikipedia?

I ran into a couple of questions in a professor's notes about FOSD, SOSD, monotone likelihood ratios, etc. and the relationships between them

Yeah, for sure. Part of why I'm thinking of complex is that we've had 2 underprepared lectures, so I could probably get us caught up in about a day

And yeah true

and I'm trying to find reference to maybe shed some light on them

If I went back in time id have done a probability lecture in addition to the ones I did

I've commented numerous times about what a mistake I made not taking/teaching probability earlier in my life.

@Ted found your message, very relieving news on spivak AND on his books

Which I want to acquire at some point

Glad you found it, Eric.

Probability is extremely cool at like every level

From basic stuff to fancy stuff

@Ted I know the sectional curvature of Lie groups (with biinvariant metrics) now.

I don't know the fancy stuff. I think every math major should know the basic stuff (including continuous distributions, law of large numbers, etc.) — but not the grad level which is just measure theory and all theory.

Do we do it more like measure theory, wombo combo, what?

That fórmula Is very nice @Balarka

I assigned that for homework, @Balarka'.

Have you worked out my exercise yet on the meaning of sectional curvature?

I don't remember the formula

Oh my Portuguese keyboard added an accent lol

We saw, Eric.

Some triple Lie bracket I guess

Aha, cute

It's like $1/4 |[X,Y]|^{2}$ @Mike

Or something like that

On the probability side of things, I'm running into large deviations theory of late

oh sectional not riemannian

which is neat

Well the Riemannian one is a triple lie bracket

Still cute

yeah I calculated

Gotta run chat, have a good day ppl

See you!

i like the triple one better cuz it naturally pops up from thinkin about what curvature is and what identities it satisfies

Goldbach's conjecture is "obviously" a $\Pi^0_1$ statement. The Riemann hypothesis is also equivalent to a known $\Pi^0_1$ statement.

But this isn't obvious, of course.

How about the Collatz conjecture? It's "obviously" a $\Pi^0_2$ statement, but is it equivalent to a known $\Pi^0_1$ statement?

Sorry, had to go afk again

@MikeMiller $R(X, Y)Z = 1/2 [[X, Y], Z]$

Just to prove I remember it :)

":)"

:)'

@Ted Your exercise being, sectional curvature is Gaussian curvature of the exp of the 2-plane the vectors span in TpM?

Looks like there's no MathJax here? I saw my LaTeX code come through as LaTeX code instead of as formatted math, and it took me a minute to realize that that wasn't what I wanted. :D

right.

@TobiasKildetoft So if we call the discriminant of the characteristic polynomial $d$, and we call the characteristic polynomial $p$, then we know $d(A)\cdot p(A)=0$. (If $A$ is diagonalizable, this is the easy case of C–H, so $p(A)=0$. If $A$ isn't diagonal, its eigenvalues aren't all distinct, so $d(A)=0$.) If the product of two polynomials is identically zero, at least one is (essentially, look at the term with the highest power); since $d$ is not identically zero, $p$ must be.

I haven't actually. it's not entirely obvious to me how to do that.

^Proof of C–H without using (Euclidean) density

It's essentially the same as the proof using Zariski density

You need to think about the Gauss equations for a submanifold, @Balarka'.

@AkivaWeinberger Woops, read your argument wrong

I don't know the Gauss equations. Are those the Gauss-Codazzi things in R^3?

I suppose I should have said "the roots of the characteristic polynomial aren't all distinct"

'cause of algebraic multiplicity vs geometric multiplicity or something

They hold in a general Riemannian manifold for a submanifold.

Oh, also, I meant *diagonalizable rather than diagonal at one point

@AkivaWeinberger I think you are cheating with that argument. The discriminant is not zero as a polynomial just because the roots are not distinct

@MikeMiller The sectional curvature formula is interesting because it says $\text{sec}(X, Y) = 0$ if $X$ and $Y$ commute. Which is sort of how it should be, given the parallel transport along "$XYX^{-1}Y^{-1}$" interpretation of the Riemann curvature tensor.

@TobiasKildetoft What do you mean? At the end, I said the discriminant isn't identically zero

(because there are plenty of matrices for which the char. poly. has distinct roots)

@AkivaWeinberger Ahh, but you are evaluating the discriminant at a matrix

@TedShifrin Hm. Should I be able to prove it if I think hard about the Gauss-Codazzi formulas in your notes?

It's Gauss specifically, Balarka. But you want it in a more general case than $\Bbb R^3$.

We know $\forall A,d(A)p(A)=0$. Algebraic stuffs ensures this implies $(\forall A,d(A)=0)\lor(\forall A,p(A)=0)$. @TobiasKildetoft

@AkivaWeinberger I guess I am not sure why $d(A) = 0$ when $A$ has repeated eigenvalues

Let's say, I try to prove for 3-manifolds.

It comes back to that question we were discussing about whether $\exp_p(\Lambda)$ is totally-geodesic or not. And it isn't, but ...

OK, @Balarka'.

@TobiasKildetoft As I said, I should have said repeated "roots of the characteristic polynomial" instaid

and that's just the basic property of how the discriminant works

@AkivaWeinberger Same thing. I still don't see why.

Isn't that the defining trait of a discriminant? The discriminant of a polynomial is zero iff it has repeated roots

Wait, how do you even evaluate the discriminant at a matrix?

@AkivaWeinberger Yes, but that is the discriminant evaluated at the coefficients of the polynomial

The characteristic polynomial is a polynomial in $\lambda$, whose coefficients are polynomials in terms of the entries of the matrix

So the discriminant is a polynomial in terms of the entries of the matrix.

So I'm really evaluating it at the tuple defined by its entries, I guess

boop

dooq

@AkivaWeinberger There might be an argument along these lines, but it needs to be made more precise

beep

Assuming you accept this (which essentially says the ring of multivariate polynomials has no zero divisors), I don't see what else needs to be made precise

Well

@AkivaWeinberger Well, I am still not sure which values you are putting into the discriminant polynomial

I guess the way I described $d$ was confusing

[crazy modem noises in Talking Heads song]

@TobiasKildetoft It's a multivariate polynomial in $n^2$ variables

so I'm plugging in the $n^2$ entries of the matrix

@AkivaWeinberger But those are not the coefficients of the characteristic polynomial

What's the characteristic polynomial of $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ look like

It's something in $\Bbb R[a,b,c,d,\lambda]$

${}=\Bbb R[a,b,c,d][\lambda]$

Gotta run

@AkivaWeinberger No, it is a polynomial in one variable

@TedShifrin I think I can guess how to derive it from Gauss' equation for 3-manifolds. As a baby case I can pick $X = \partial/\partial x, Y = \partial/\partial y$ in a coordinate chart and then $R(X, Y) = \nabla_{\partial/\partial x} \nabla_{\partial/\partial y} - \nabla_{\partial /\partial y}\nabla_{\partial/\partial x}$. If I expand this in terms of Christoffel symbols, I should get one of those Gauss formulas.

But this seems computation-heavy

I don't see that that gives anything to do with exp ?

He's talking about deriving general Gauss-Codazzi

I'm sticking to moving frames ... Bah.

Hmm. I think I'm basically reconciling sectional curvature of a 2-manifold with the Gaussian curvature.

Well, I grant that.

So the rest should argue that sectional curvature of $\exp(\Lambda)$ is sectional curvature of $M$ (our ambient 3-manifold) at $\Lambda$?

Sounds "obvious"

It would be obvious if the surface were totally geodesic. But it's not.

Ah.

@Ted Pick a small exponential chart for the surface in normal coordinates. I would expect that as you linearize this surface to the tangent plane you can show that the derivative of this process (the third order contribution to curvature) is nil.

Oh but that's false.

My bad.

I didn't follow that.

It was nonsense.

If the surface was totally geodesic, we could use normal coordinates at the point to show this, right? Because Riemann curvature tensor becomes the Euclidean tensor at that point.

That's definitely wrong.

The metric is Euclidean to first order.

(i.e., in normal coordinates, Christoffel symbols vanish at the point, but their derivatives don't)

Hm, you're right

That happens twice a year. You caught me on a good day.

I'm still saying you have to pay attention very carefully to the particular surface, namely $\exp_p(\Lambda)$.

But I don't want to.

Wah wah wah ....

I suck at geometry so I am excused if I fail

Bull.

But this is kinda interesting

I guess this is related to the fact that the gradient on a submanifold isn't the projection of the gradient upstairs.

It isn't?

It is in Euclidean space.

I haven't thought about it otherwise.

what is the gradient?

It's the vector field dual to $df$.

(using the Riemannian metric to give duality)

Oh, that guy

My internet dislikes me.

I told it to.

>:'(

Expert evidence says the suicide rate of nerds is higher than the general populace because of their involvement with a dangling internet 24/7.

@Fargle's internet

But I have books, @Balarka.

What are those?

@Fargle Tru

they're like the paper version of pdfs

Wow, @Alessandro. What a novel idea.

I am not taking my laptop with me for the week I'll be gone to the diff geo workshop

@TedShifrin rimshot

instead I will take with myself a notebook, TS Eliot, and Notes from THE UNDERGROUND

Hey guys

You'll have a breakdown, @Balarka'.

@TedShifrin I'm not sold on it yet, it's much harder to search for stuff in them

@Dodsy!!!

Hey, Ted!

:)

how were your travels, my friend.

@Alessandro: I remember there was something called an index when I was a child.

I've been back almost 3 weeks, Nate :P

much harder than ctrl+f :P

@TedShifrin I think I have planned myself super well this time to go totally unhinged

haha, true. :P

The first lash comes from not understanding any of the diff geo talks with probability 1

LOL, Balarka.

Why are you going?

And then, as I have no laptop or internet, I will inevitably read The Waste Land and Notes

that would make me more unhinged and depressed

and the cycle goes on till I go crazy or I die

So your whole goal is to be maximally depressed?

Maximal crisis to the point of derangement, yes

(I shouldn't rephrase sentences in the middle of typing.)

I don't think I like that, Balarka

On the other hand I'm taking Milnor's Morse theory and do Carmo first four chapters with me

so it's going to be a guinea pig experiment. I will either go mad or end up extremely productive after the week

When is this slated to occur?

Depression is no joke, Balarka, make sure you take care of yourself.

Starts from 23rd this month, ends at 31st

Perhaps you could bring some music to aid in your mental health.

@Dodsy Ok.

I could take this

rolls twelve eyes

I suggest this

Ugh. It's too hot out in Minnesota

95 F and humid. I'm sure others have it worse, but yuck

Fake facts.

(100 C may be the boiling point of water, but 100 F is my boiling point)

What's the boiling point of mercury (without googling).

Okay that game sucked, you guys are right.

Dunno

Google says 90 F and 54% humidity

it's 75 here.

I like my weather in the 30 degree Celsius range.

Same

I much prefer 60s Fahrenheit

Hey Daminark.

75-80 is when I'm just like "this needs to stop now"

@Dodsy How's uni

Not there yet!

Ah

How's it going @Dodsy?

September 8th classes start.

Oh, just dandy.

Relaxing, enjoying the sun. :)

Oh, close enough

but our apartment lease starts August first.

Which is exciting.

@TedS I'm killing fields now

what do you mean balarka?

you're reading about pol pot?

the khmer rouge?

or that you are literally killing fields.

oh, he's gone. I'm sure he would have enjoyed that pun

Killing field is a mathematical object (a vector field with special properties)

Oh haha.

Named after a mathematician with the last name Killing, I presume?

yeah Willhelm Killing

Cool.

I actually didn't know that killing field was an actual incident

Yeah, very sad.

that makes me feel like I shouldn't have made that pun

I don't think many people know about it.

I really hope that's not true but uh

Probably is :/