Mathematics

Ted Shifrin

12:00 AM

You can claim that exhaustion made you slow.

Daminark

Ah perfect

anakhro

12:24 AM

@TedShifrin Am I crazy in thinking this is the proper way to calculate $\iota_X\beta$ for $X = (xy - z)\frac{\partial}{\partial y} + (yz - x)\frac{\partial}{\partial z}$ and $\beta = x\,dy\wedge dz$:
$$\iota_X\beta = \iota_X(x(dy\otimes dz - dz\otimes dy)) = x(dy(X)\otimes dz - dz(X)\otimes dy) = x((xy-z)\,dz - (yz-x)\,dy) = \dotsc$$

Because I have a more extensive example calculation I am trying to work out but it's not getting anything remotely close to the right answer.

Ted Shifrin

You can't do interior product with a general tensor. Leave it alone with the differential form.

So you'll get $x(xy-z)dz - x(yz-x)dy$.

Yeah, that's what you had.

Why is @Eric here when he's supposed to be dining in splendor?

anakhro

Hmmm. So suppose I then took $d$ of that. I get
$$2xy\,dx\wedge dz + x^2\,dy\wedge dz - (yz-1)\,dx\wedge dy - xy\,dz\wedge dy ?$$

Ted Shifrin

What happened to the $z\,dx\wedge dz$?

And where's the $2x\,dx\wedge dy$?

I give up.

anakhro

Ah yes I messed up that.

Ted Shifrin

Where did the $+1 dx\wedge dy$ come from?

anakhro

12:35 AM

$\frac{d(xyz-x^2)}{dx}dx\wedge dy$, right. It should have been a $(yz - 2x)$ not $(yz - 1)$.

Maybe I am just doing this part too quickly and need to take my time and make sure I am not making errors like that.

Ted Shifrin

OK, that addresses my second complaint, then.

anakhro

The general principle I think I have right, though. So that's good to know. Just sloppy.

I will do it after I make my calzone dough and see if it works out better this time. Thanks @TedShifrin

Ted Shifrin

I've spent 45 years using differential forms, so I'm semi-competent.

Send me a calzone.

anakhro

Uncooked, or cooked?

Ted Shifrin

Hmmm ... depends how fast the stagecoach will be.

anakhro

12:39 AM

Heh. I am pretty sure it would spoil before it got there in either case.

Stan Shunpike

1:18 AM

hola

Ted Shifrin

heya @Stan. I seem to recall I answered a ping of yours.

Stan Shunpike

Yes that was very helpful

Ted Shifrin

Cool.

Stan Shunpike

So I tried to read more and from what I understand, a tensor product is a quotient vector space

is that accurate?

Ted Shifrin

Yes, but rather than thinking about the construction, you should understand the universal mapping property it has and just use that most of the time.

So, before I disappear to cook dinner, have you any questions to raise tonight?

Stan Shunpike

1:30 AM

Yeah that was my first, so I guess I should start by reading about that then?

I learned about like equivalence classes and relations and quotient sets

and feel more comfortable with those now

Ted Shifrin

You've never learned modular arithmetic?

Stan Shunpike

uh i did but like idk

thinking of it like in these terms

anakhro

Stan, you are more from a physics background, si?

Stan Shunpike

with basis vectors

and coorodinate free viewpoint

hasn't come easily to me

Ted Shifrin

Hmm ... thinking of $\Bbb Z$ mod $m$ as a ring is all about equivalence relations.

I should send you the appendix from my algebra book.

Stan Shunpike

1:31 AM

yeah that made it clear

but then I wasn't sure like, how that relates to the tensor product being an equivalence relation

didn't quite get that far

Ted Shifrin

It's not per se.

OK, I'm gonna send you the appendix anyhow. Not sure if I have a separate file for that.

Stan Shunpike

@anakhro sort of. I read it on my own but have almost no formal training. I was an econ major

Ted Shifrin

Stan just chose to hang around here and get bullied.

Stan Shunpike

@TedShifrin hahahahahaha basically

willing to hang in there

@TedShifrin so should I read about this universal property first?

the quotient vector space approach left me still confused how wedge products work.

I just got sick of the psuedo vector concept and want like the full version. None of the cool physics books use psuedo vectors

they all use differential geometry

Ted Shifrin

They don't know what diffferential geometry is :P

Stan Shunpike

1:35 AM

im gathering

should I watch your lecture on it? you started off with boundaries

so i wasn't sure if that would lead to the tensor product

Ted Shifrin

On what?

Stan Shunpike

differential forms

Ted Shifrin

No, no, I never did tensor product, purposely.

Stan Shunpike

ok

Ted Shifrin

Boundaries?

I defined wedge products just by determinants, no fancy stuff.

Stan Shunpike

1:36 AM

ok, i'll watch it and come back with questions

Ted Shifrin

Do you want the appendix on equivalence relations, or are you cool?

Stan Shunpike

yes definitely

thanks ted!

i think i have too many books written by physicists

Ted Shifrin

Make sure you do the exercises :)

Stan Shunpike

claiming to be mathematicians

hahaha i will and then come here when i inevitably get stuck

anakhro

@StanShunpike for the coordinate free def. of the tensor product $V\otimes W$ as the "free-est" vector space where bilinear maps on $V\times W$ are viewed as linear maps on $V\otimes W$.

Ted Shifrin

1:40 AM

Stan: sent.

Of course, this requires double-duality to untangle.

anakhro

Or in other words, the universal property just says that the tensor product factors bilinear maps into linear maps.

Stan Shunpike

@TedShifrin you're the best!

Ted Shifrin

Yup, anakhro tells the truth (for once).

Stan Shunpike

@anakhro so does "free" mean like 'coordinate free'? i saw this notion of a free module, but i didn't really get what "free" meant

Ted Shifrin

no, nothing to do with coordinates.

Just means the things have no relations.

Stan Shunpike

1:43 AM

is the tensor product in your differential geometry book?

Ted Shifrin

Noooo.

I think of tensor product as definitely graduate stuff.

Stan Shunpike

oh really? I assumed that was undergrad

wow ok

idk why. its just everywhere

so i assumed you had to know it

Ted Shifrin

Well, all sorts of crazy things are undergrad at UC, but not elsewhere.

Daminark

Lmao

Ted Shifrin

I did learn tensor product as an undergrad, but it's far from typical.

Well, Demonark stirred to laugh, rather than issue another horrid pun.

Stan Shunpike

1:45 AM

and u ended up being a pro differential geometer

anakhro

@StanShunpike take a look at Gauge Fields, Knots and Gravity by Munian & Baez.

It might be more up your alley if you are a physicist.

Stan Shunpike

@anakhro oooo sounds very intriguing

anakhro

pg. 169 for tensor products

Ted Shifrin

Stan is not a physicist.

Stan Shunpike

i will definitely look

Daminark

1:45 AM

Yeah here the honors algebra class talks about tensor products toward the end but regular doesn't

Stan Shunpike

I tell jokes and make puns

anakhro

@TedShifrin maybe he wants to be one

Ted Shifrin

It does show up in Spivak's Calculus on Manifolds, so fancy honors calculus courses sometimes do it.

Stan Shunpike

and drive the Knight Bus

Ted Shifrin

Stan, you and Demonark should get together for a drink.

anakhro

1:46 AM

Tensor product is done in the second linear algebra class at my university

The module theory one.

Daminark

Should be soon if we do, I won't be around Chicago much longer

Ted Shifrin

Artin's algebra book doesn't do tensor product, for example, and it's damn sophisticated.

anakhro

Does Artin do modules?

Ted Shifrin

Good point.

Of course.

anakhro

Weird.

Daminark

1:47 AM

Does he do STFGMPID?

Ted Shifrin

But tensor product normally is the purview of commutative algebra, not first algebra.

anakhro

I like Berberian for linear algebra.

Ted Shifrin

Yes, of course, Demonark.

I've never heard of it, anakhro.

But lots of people haven't heard of my books, so it's fair.

Stan Shunpike

Are tensor products a requirement to understand wedge products?

Ted Shifrin

OK, time for me to disappear and go cook dinner. Bye.

Daminark

1:48 AM

Same. I used and liked Hoffman-Kunze though it's fairly old school

Stan Shunpike

@TedShifrin ciao

Ted Shifrin

@Stan: Of course not. My course/book/videos don't use tensor product to do wedge products.

Daminark

I hear good things about Linear Algebra Done Wrong

Stan Shunpike

@TedShifrin yeah i'm gonna try to learn it your way then

Ted Shifrin

LOL.

OK, have fun :)

anakhro

1:52 AM

@TedShifrin this is how Berberian does it.

Which roughly follows what my university did, except my university introduces modules in the second course.

When you start doing ring stuff

Stan Shunpike

@anakhro that book looks so great

anakhro

Berberian's linear algebra book. @StanShunpike it is quite cheap through Dover if I recall correctly.

Daminark

It feels similar to H-K but more geometric

anakhro

However, the biggest part of math is to not worry too much about books, but more about doing exercises and getting appropriate help when needed.

Daminark

Maybe covers a bit more actually, H-K doesn't really do much tensor products or PIDs

~~It's the instructor's job to worry about books~~

anakhro

1:58 AM

I have one professor who always teaches brand new courses and does the lecture notes ad hoc, compiling a usually 100pg textbook at the end.

If I could be as half a good teacher as him, I would be happy, I think.

Math geek

0

Q: Continuity of the improper integral $\int_{0}^{x}y^{-1/2}dy.$

The improper integral $\int_{0}^{x}y^{-1/2}dy$ is $1.$ Continuous on $[0,\infty).$ $2.$ Continuous on $(0,\infty).$ $3.$ Continuous only on $[1/2,\infty)$ It is cleat that option $3$rd is wrong one. I am confused about $1$st and $2$nd option. I tried it as $$\int_{0}^{x}y^{-1/2}dy=2\sqrt{x}$...

How it is continuous at $0$?

in the negative real axis. the function doesn't exist in real.

Stan Shunpike

@anakhro so what's ur specialty?

/ area of interest

anakhro

@StanShunpike I like contact and symplectic geometry.

But I am pretty newbie.

I am a masters student, though, if that indicates any level of expertise to you.

Stan Shunpike

i still don't understand what symplectic geometry is

mainly because

i still haven't grasped the wedge product

anakhro

Well it's differential geometry, but with a tool for measuring area.

Stan Shunpike

2:06 AM

cuz i thought i needed the tensor product for that

OHH

really?

wait how is rimenanian geometry different then?

anakhro

Oversimplified, but basically.

Riemannian geometry measures angles.

Like an inner product space.

Stan Shunpike

hmmm

anakhro

angles/distances are to Riemannian, as areas/volumes are to symplectic.

Stan Shunpike

so why does this topic come up in relation to hamiltonians?

in physics

anakhro

The area/volume idea I highlight is not so impressive with respect to its origins in Hamiltonian mechanics AFAIK.

Do you know anything about Hamiltonian mechanics?

Like the main thing to keep in mind is you want to determine the dynamics from the Hamiltonian (a function on your phase space).

Stan Shunpike

2:15 AM

From what i've read

and i don't understand a ton, symplectic geometry is one of the benefits of using hamiltonians vs lagrangians

i have done some simple problesm with hamiltonians in classical mechanics and a few standard ones for quantum

but not a lot of experience beyond that

anakhro

I am not very good at physics, but as far as I know, the benefits of using the Hamiltonian vs. the Lagrangian is based on the symmetries you know about your problem.

Physics effectively is about solving physical problems using symmetries (analogously, math is about studying symmetries, despite no apparent application to physical problems).

So some symmetries are easier to take advantage of when dealing with $H = K + U$ and some are easier to take advantage of when dealing with $L = K - U$.

@StanShunpike the main point is you want to take a Hamiltonian $H$ to a vector field $v$ whose dynamics (the flow lines of $v$) describe the dynamics/physics of your Hamiltonian.

Does that sort of make sense?

Simone

8:40 AM

I'm having a tough time with improper integrals, I own a copy of Spivak's calculus but he only dedicates a couple of exercises to them.
Is there a textbook that deals with them more comprehensively?

Simone

8:57 AM

For instance, I see people using tests that I'd use for infinite series, like Abel's or Dirichlet's tests for these improper integrals. I understand it makes sense due to the definition of the integral, but I've never done this myself and I need practice which Spivak doesn't provide.

northerner

9:27 AM

Does anyone have a reference for how to do the Lucas Probable Prime test? I find the wikipedia page hard to follow en.wikipedia.org/wiki/…

Micrified

9:38 AM

This might be a noob-ish question, but I'm trying to understand the following statement: In general, if $q=p^{m}$ where $p$ is prime and $m \geq 2$, the element $p$ would not have a multiplicative inverse, since there exists no $b$ such that $b \times p = 1 mod p^{m}$.

I guess that since multiplying a prime by any number other than 1 means it is no longer prime, that I find it hard to understand why you couldn't get $1$ when modulo'ing the result by $p^{m}$ ...

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$Mathematics$ Mathematics