Mathematics

2:34 AM

9

To complete Student's answer : $\mathbb Z_p \subset \mathbb Q_p \subset \mathbb C_p$, and since $\mathbb C_p$, the algebraic closure of $\mathbb Q_p$, is isomorphic to $\mathbb C$, the field of the complex numbers, we know that $$ |\mathbb R| = |\mathbb Z_p| \le |\mathbb Q_p| \le |\mathbb C_p| =...

In short, yes.

Balarka Sen

I am here to chew ass and kick bubblegum.

2

EnjoysMath

3:13 AM

0

Q: Would this suffice in a visual type theory to define a list?

See the image. I got that from: wikipedia article. I want to make a visual type theory so that we aren't stuck comprehending pure text for eternity. Also, is the abstract type List a product of some sort? The diagram doesn't indicate this since it's based on a product diagram of $E \times L$...

Do you think that searching for isomorphic subgraphs of a category diagram is a bad idea?

Handmade diagrams don't get super huge

The user can then a the click of a button apply a rule defined by a diagrma

Failed to be a Mathematician

4:29 AM

math.stackexchange.com/users/464147/… @secret Am I correct?

Secret

which?

Failed to be a Mathematician

proof

math.stackexchange.com/users/464147/…

Alpha Romeo

5:27 AM

Hey, can someone please verify/give reference to this: If partial derivates exist at every point in a neighborhood of a $R^2$ -> $R$ function, then the partial derivatives are continuous as well.

Secret

5:51 AM

in The Symposium, 16 mins ago, by Secret

> Come now, I will tell you ... the only routes of inquiry that are for thinking: the one, that it is and that it is not possible for it not to be, is the path of Persuasion (for it attends upon Truth), the other, that it is not and that it is right that it not be, this indeed I declare to you to be a path entirely unable to be investigated: For neither can you know what is not (for it is not to be accomplished) nor can you declare it. [2=B2]
For the same thing is for thinking and for being. [3=B3]

Erik M

6:20 AM

What’s the most surprising math result you know of?

Secret

6:38 AM

@FailedtobeaMathematician there are tons of proof questions, which?

Chandler Watson

@Jagerber48 hello!

Jagerber48

Hello! Thanks for having a look at my question!

Chandler Watson

Of course! Just took an intro course to diff geo and the question looked really well thought out :)

Jagerber48

Thanks :)

It sounds like you're getting the idea of what I'm asking

Chandler Watson

No, totally -- it's strange to me too

The reason stuff like the Hodge star and musical isomorphisms and what have you come up is because to properly represent things in a coordinate-invariant manner, we use what are called differential forms

Have you heard the term before?

Jagerber48

6:40 AM

Yes I've heard of them but my understanding is not great.

I've read about them and self-teach myself but I've never taken a proper differential geometry course.

Chandler Watson

No problemo at all, honestly my own understanding is not too deep yet

Jagerber48

I understand decently well what something like dx is as a covector

and it seems like differential forms are anti-symmetric tensor products of covectors like dx

Chandler Watson

Nice, okay, so you're familiar with the idea of the cotangent space and everything?

Jagerber48

but I can't get my head around why that would be interesting or the properties of the objects

Chandler Watson

Okay, that's actually a great way of putting it

Why it's interesting (afaik, anyone feel free to correct me at any point) is because we have two parallel ways of seeing things

We have the standard calculus way, where we work with fields, and have div, grad, and curl

And then we have working with differential forms

We can kinda switch back and forth between the two using the "musical isomorphism" -- does that sound familiar?

Jagerber48

6:46 AM

I'm in physics so I'm familiar with raising and lowering operators on tensors/vectors using the metric tensor. it seems like the musical isomorphism is capturing exactly this sort of thing. Thinking of it as converting between a "standard calculus" way of working with fields etc. and the way of working with differential forms sounds interesting to me.

Chandler Watson

That's exactly it. So you're pretty much free using that isomorphism to flip back and forth between the tangent space and the cotangent space (i.e. vector field to differential form and back again)

Jagerber48

ok

Are you able to explain how grad and div come into the discussion of forms?

Chandler Watson

Yup, just piecing together the details (took the class in winter, trying to remember most of it! :P)

Jagerber48

ok great! thank you!

Chandler Watson

np!

So, the way of taking the gradient is to take your function from a 0-vector to a 0-covector (pretty much the same thing), take the exterior derivative to get a 1-covector, then convert it back from a 1-covector to a 1-vector. And that's it :)

The gradient of a vector field is just a nicely-wrapped exterior derivative

When you try curl, things get a tiny bit tricker, but not a whole lot worse.

Jagerber48

6:59 AM

ok, I can kind of follow for grad but I'm curious to hear about div. I've struggled with that one and definitely don't understand the Hodge star, other than understanding that like, 3-2 = 1 and 3-1 = 2 has something to do with it...

Chandler Watson

Yeah, exactly. At first, the Hodge star seems a little arbitrary and strange.

I'll do curl first, because it also uses the Hodge star, but it takes us "internally" from 1-covectors to 2-covectors

Jagerber48

ok

Chandler Watson

So curl, we start with a 1-vector and convert it to a 1-covector

We then take the exterior derivative as usual, but then we're stuck with a 2-covector

And we need some way of getting out a 1-vector when we're all done

Jagerber48

How do you think about the exterior derivative here?

Chandler Watson

So we use the Hodge star operator. Since we're working in 3D, the Hodge star takes us from a k-covector to a (3 - k) covector

The exterior derivative here gets a little tricky, since we have to worry about all those wedge products now

So intuition understandably gets a little trickier.

Wikipedia's way of thinking about it talks about thinking about the differential forms as measuring flux through small parallelpipeds, and taking the exterior derivative just adds another dimension to that parallelpiped.

I've heard good things about the approach taken in this document tho, and it discusses intuition on Hodge star as well (tbh I might read that later tonight :)

In any case, we use the Hodge star to get us back from a 2-covector to a 1-covector, and then we can just go back to a 1-vector as normal

Jagerber48

7:07 AM

ok

Chandler Watson

Intuitively, the Hodge star swaps out everything that's "there" for everything that's "not there", i.e. the hodge star of dx^dy is dz

Jagerber48

hmm ok

Chandler Watson

That's up to sign though, cyclic permutations preserve the sign but doing something like dy^dx yields -dz

Jagerber48

ok

Chandler Watson

So you can see that it clearly takes you from 2-covectors to 1-covectors, tho of course it does seem a little funny at first.

For div (the grand finale) you go from 1-vector to 1-covector, take a hodge star first this time to get to a 2-covector (we'd already done 0-covectors and 1-covectors, so this is the intuitive next step), then take the exterior derivative to get a 3-covector.

We want out a 1-vector tho, so we hodge star again to get a 1-covector and then just send it back.

So grad is pretty much a fancy exterior derivative on 0-forms, curl is a fancy exterior derivative on 1-forms, and div is a fancy exterior derivative on 2-forms, but they're all pushed and hodged around so that they play nice with 3D vector fields

What's fun is that you can actually do out those computations on a generic vector field in Euclidean space, and get out the good old expressions for grad, div, and curl :)

Jagerber48

7:14 AM

Yeah I think that is what I would like to try to do. That would probably help with my question

Your explanation of Hodge star seems pretty straightforward and helpful

The thing I wouldn't know how to calculate is the exterior derivative of a 1-form or 2-form

Chandler Watson

Oh, gotcha -- the rules are pretty nice, so I can just give a quick example

so if you have something like x dx + y dz, taking the exterior derivative just means that you wedge things together, i.e. you get dx^dx + dy^dz = dy^dz, because anything wedge itself is zero by antisymmetry

Jagerber48

ok

So I guess it is important that there is a the presence of both a function and a covector

because you need to take the differential of the function to get the second covector in the product

yeah that's not too bad

Chandler Watson

it is indeed :) and the wiki page has some nice extra details if you need them

like if you have xy dz, it's the sum of all the partial derivatives, i.e. y dx^dz + x dy^dz

also, d(df) = 0 for any function f is useful

Jagerber48

right

cool thanks! Yeah I think I'll be able to unravel these definitions a bit now.

I guess my original question still stands about the $\nabla$ operator standing on its own

Chandler Watson

of course, I've starred your question because I'm pretty curious about that too

Jagerber48

7:23 AM

It seems like the answer must be wrapped in this differential geometry stuff somewhere!

But yeah otherwise it just seems like the operator can be used and manipulated successfully in too many ways to be more than "just" a convenience.

Iza_lazet

Just an aside: the "blood-related" generalization of vector calculus is geometric calculus (Clifford algebra) and working with multi-vectors (where grad, div, curl are defined on multivectors). Though for diff geo purposes, working with multi-covectors (differential forms) is sometimes more convenient thanks to pullbacks (we can't pullback a vector field, but we can pullback a differential form). The correspondence between multi-vectors and multi-covectors is not always canonical, unfortunately.

Chandler Watson

Right? It's true that because we have coordinate free definitions of each "nabla operator" your observations must hold, but I'd love to see if there's some clean intuitive way to see why that does.

@Iza_lazet whoa, huh -- any source recommendations for geometric calculus?

Iza_lazet

What was the question about nabla again?

Jagerber48

4

Q: Coordinate free definition of $\nabla$ operator

There are a number of posts on this site asking similar questions and some of them have been answered (to my taste) at least partially but none give a complete answer that I am satisfied with. See links at the bottom of this question for a small selection of posts asking related (or even the same...

differential-geometry notation vectors

Thanks for the input on geometric calculus/Clifford algebra. I've distantly heard about that in the context of spinors but maybe I'll have to look into more detail

Iza_lazet

that is easy. In fact, it is exactly analogous to how the exterior derivative is defined: by first agreeing on the 0-dimensional case, then by listing the algebraic properties one wants the operator to have, so that we get a unique operator. Although obviously, students are usually taught just the formula.

I learned geometric calculus from Alan Macdonald's books and notes. It is a very immature field, so there is some disagreement on certain notations (such as contraction) and what properties people want the operators to have.

his books, unfortunately, do not contain the BAC-CAB family of formulas, but somehow his online notes do.

Chandler Watson

7:35 AM

Interesting--thanks so much!

I'm going to nod off for the night here, take care you two

Jagerber48

Have a good night! Thank you for all of the information and taking the time to discuss with me!

Chandler Watson

Ofc! It was interesting and really informative for me too :))

Jagerber48

@Iza_lazet are you able to elaborate on what you mean by the 0-dimensional case we have to agree on and what algebraic properties we would want the nabla operator to have? Or perhaps you could provide an answer to the linked question whenever you have the time?

Iza_lazet

I think either Herstenes or Alan provided the derivation in their books. Basically you want things like associativity, Leibniz's rule, change in grading, etc. The 0-dimensional just means for a function f (0-dimensional), $\nabla f$ is the usual gradient of the function.

also multilinearity of course

Failed to be a Mathematician

0

Q: If $A$ is an ordered set has the Least upper bound property iff it has the greatest lower bound property.

Mentioned as an easy exercise in 'Topology', James R. Munkres, 2e, Pearson[page No:25] If $A$ is an ordered set has the Least upper bound property iff it has the greatest lower bound property. An ordered set $A$ is said to have the least upper bound property if every nonempty subset $A_0$...

proof-verification order-theory supremum-and-infimum

I got the help :)

Secret

7:46 AM

I see

That theorem is very important, so study it carefully.

Iza_lazet

an honest curious question, when do people use this theorem outside the exercise?

Secret

If I recall, you need that to prove nested intervals theorem

and that will allow you to prove Bolzano–Weierstrass theorem, which is needed to show every sequence in a closed real interval has a convergent subsequence, which is used in proving things related to limits

Iza_lazet

8:01 AM

Oh ok. So that is yet another strategy, among many equivalent ones, to approach the fundamental property of real numbers.

Secret

yeah, and it is because the reals obey lub that you can say so much about continuous functions on the reals. But again I am not an expert, as I mostly learn all my real analysis from this chat as I never get a chance to fit a real analysis course in my undergrad

Iza_lazet

personally, I have always preferred the Cauchy-completeness approach, which can be generalized to general metric spaces. Well, most professors like to teach LUB and Dedekind cuts though.

Secret

I also like the dedekind cuts approach more, because it generalises easily to the surreals

also I really don't like to see more than 3 inequality signs in a single line, which is common for the cauchy approach

though I have some understanding of both approaches and Iam actually planning to restudy those constructions and proofs later because I need them to understand dense linear orderings in general