Mathematics - 2019-06-08

Rithaniel

00:00

Ah, gotcha.

So this what I'm looking at, then: math.franklin.uga.edu/sites/default/files/inline-files/…

124 pages, fair enough. Might take me two weekends.

Ted Shifrin

Right. Jan 2018 is the last update. If you work exercises (and there are some good ones), that'll slow you down significantly. That's the only way to actually absorb/learn math :P

Rithaniel

I can agree with that. Keep in mind, me claiming that I can do it in two weekends is my form of a joke.

Stan Shunpike

@TedShifrin I guess I'm struggling with grasping the physical intuition for what a bivector is. The idea that, if I take the cartesian prodcut of two vector spaces and treat that as a vector space made sense to me. that's the tensor product, right?

So like, in the simplest case, angular momentum takes in a vector for the rod and then a vector for the force applied to that rod, and spits out a number (the magnitude) and the direction (the bivector). is that more or less right?

So I'm confused because like, with the cross product, it's perpendicular, but this diagram makes the bivector seem like it lies in thie plane of x and p

why would the bivector be an area?

Ted Shifrin

No, the cartesian product is not the tensor product.

You think of a bivector as a signed parallelogram. :)

Stan Shunpike

@TedShifrin I thought the tensor product was the cartesian product with a set of rules to satisfy certain bilinearity conditions

@TedShifrin Is that incorrect?

Ted Shifrin

00:12

No. You wouldn't get enough dimensions that way.

Stan Shunpike

oh

hmmm

Ted Shifrin

You need the vector space generated by all ordered pairs (in this case), with an equivalence relation building in bilinearity.

Don't mess with tensor products. For differential forms and multivectors you can do things much more concretely. See my lectures.

I need to go do some cooking now, so we'll have to continue later ...

Stan Shunpike

ohhhh

ok

i found the passages i've been using

Ted Shifrin

The key thing is that with forms (or multi-vectors), with an inner product on $V$ you have the Hodge star operator which gives you an isomorphism $\Lambda^2 V \cong \Lambda^{n-2}V$, where $\dim V=n$. So this gives bivectors "back" as vectors in $\Bbb R^3$. Physicists call the cross-product a pseudovector. They're actually right.

What you're showing must have been written by a physicist. It's wrong.

Stan Shunpike

damn

ok

Ted Shifrin

00:19

Or at least very misleading.

Stan Shunpike

i'll use a different book

thanks ted

Ted Shifrin

LOL. Thanks for nothing, you mean.

Talk soon!

Stan Shunpike

@TedShifrin ciao! :) always a pleasure my friend

Stan Shunpike

00:49

@TedShifrin wow you're right. that was a terrible way of writing it. no wonder i got stuck. i found a better explanation. i'll have to work it out and then check again with u

Ajay Mishra

03:14

10 hours ago, by Ajay Mishra

What does it mean to integrate $f(x,y,z) = xy$ over a unit sphere centered at origin?

anakhro

@StanShunpike despite Ted's insistence to not do differential forms that way, I have found that the tensor product route is much more understandable, personally. So definitely have a look at both that way and the others and see what makes sense for you.

Don't be stuck looking at it one way in the long run (though of course for learning, feel free to learn it one way, then learn the others much later).

Ajay Mishra

@anakhro Do you know what the author meant from that? ^ ?

anakhro

@AjayMishra I answer it with another question: what troubles you about that statement?

Ryan Unger

It seems like a perfectly valid surface integral

Ajay Mishra

I have done surface integral when their is a surface, and their is a vector field. I have to add up the components of vector field perpendicular to the surface dotted with infinitesmial area element. I have no idea about this, Here I have one surface and there is another surface, I don't know how to make sense of that.

anakhro

03:29

You mean you have done flux with integrals but not just a general function?

Ajay Mishra

No, I was following MIT 18.02 and khan academy. They didn't discussed that.

Though I can find the formula for doing that on paul's notes, but He didn't explained anything about that at all.

@anakhro Exactly.

anakhro

I am not familiar with MIT 18.02, but I found these "supplementary notes": math.mit.edu/~jorloff/suppnotes/suppnotes02/v9.pdf

In the 6th page, you will find an explanation, though somewhat short.

Maybe it will fit in better with the rest of this course, I don't know.

skullpatrol

04:38

@RyanUnger whelp, that was a helpful hint :P

@anakhro wow! "...with the assistance of T.Shifrin..."

Shifrin Math 3500 Day 1: Introduction to Vectors

►

Ryan Unger

04:53

@skullpatrol now he knows what to google.

Stan Shunpike

05:10

@anakhro i'm not giving up on tensor products

but i need a good resource

do you have any recommendations?

Ryan Unger

05:41

@StanShunpike read the wikipedia article

it's one of the good ones

Adam

08:28

Does anyone else get the situation whereby there is no ability to access any learning resources on a subject, but they immediately know themselves to be capable of to a reasonable extent, producing successful outcomes by employing the methods in the subject for which you are not allowed complete access to?

I just find it strange that we are predisposed almost inherently equipped psychologically for such specific things

skillpatrol

when i find myself in that kind of situation; i tend to wait until the subject is formally introduced

i find that educators usually present topics in a logically linked manner

Adam

09:25

well I am sorry as skilful as that advice is,no such educator exists, well any that I would not require to be subject to the methodology first in order to assure his claimed qualification

skillpatrol

09:38

sure, qualifications play a big role in whether one wishes to follow their methodology

Adam

Well if it's the only option left I suppose I need to authorize my self for carrying out extrajudicial processes like everyone else has done in historical circumstance it was required

skillpatrol

if you're confident in your abilities, go for it!

Adam

10:43

Itr doesn't matter I just now decided that I want to become a qualified empath and mother is basically mary poppins except prehistoric and seemily having a constant stroke when ever she looks at me, so the world leading expert in the field arguably

leader*

Adam

11:05

it's surprising how little people have picked up on the fact that a lot of animation we see could have only been produced or produced in collaboration with clandestine intelligence agencies and contain humour that's a real life thing happening right now that people would be appalled by if it was leaked by whistle blowers

it's like real time satire which isn't going to be funny to anyone oc

skillpatrol

11:23

the catalyst to this process is that people can find anything they want to believe in on the internet

(including social media)

Adam

but for shits and giggles is there really a trade course available in professional extrajudicial killings? I mean there is no real way of telling who or why it happens when it does of course because there is no legal accountability for the people that are allowed to do it

skillpatrol

no idea

Adam

well if you did they wouldn't be very professional would they

skillpatrol

nope

Adam

But it is interesting to estimate how quickly the wrong type of person could eliminate all potential threats to rulership over a nation simply by data mining a state controlled social media platform with conditions that order predator drones strikes on those potential threats deemed necessary to neutralize

I guess it depends how many they had in their fleet on automata recon

but if a smart phone only costs $50 surely it isn't that expensive

skillpatrol

11:40

perhaps, we should continue this chat in your room?

Luyw

12:24

Hello, let ABC be a right triangle with BC the hypotenuse, $k^2((AB)^2+(AC)^2)=k^2(BC)^2$ suffice to show that scaling one side by a factor of $k$ will result in all the other sides being scaled also by a factor of $k$? Would this implicitly say that the angles remained constant?

skillpatrol

what do you think?

Luyw

I don't see a problem with it.

skillpatrol

you are right :-)

Luyw

Would it be considered rigorous though since the angles are implicitly said to remain constant?

skillpatrol

yes, it is rigorous

multiply the Pythagorean theorem by k^2 on both sides

anakhro

12:33

@StanShunpike are you good with linear algebra? If so, I have a relatively brief reference for you.

Albas

13:54

Its pretty easy to prove that a smooth manifold can always have a Riemannian metric. Are there any conditions on when can a manifold have a Lorentz metric on it?

Subhasis Biswas

14:06

does there exist any non trivial homomorphism from $\mathbb{Z}/n\mathbb{Z}$ to $\mathbb{Q}$?

My answer will be no. Since $\phi(D)$ has to be cyclic, where $D$ is the domain group.

is it a correct response?

@Albas, what do you think

Albas

Yea it is correct I think, one is cyclic and the other isn't.

Alessandro Codenotti

@SubhasisBiswas What about $\Bbb Z\to\Bbb Q$? Can there be nontrivial homomorphisms?

Subhasis Biswas

@AlessandroCodenotti nope, i guess. $\mathbb{Z}$ is an infinite cyclic group generated by $1$ and $-1$

Alessandro Codenotti

Why isn't the inclusion an homomorphism?

Subhasis Biswas

closure will be violated?

user131753

14:17

Yes, I was just going to say this but @AlessandroCodenotti pointed out it well. The fact that the domain of a group homomorphism is cyclic and the codomain of it is non-cyclic doesn't imply that there can't be any non-trivial group homomorphism between them.

Alessandro Codenotti

@SubhasisBiswas Closure?

user131753

@SubhasisBiswas Can you show that the inclusion map is not an homomorphism?

Subhasis Biswas

Let some $\phi(k)=a$, where $ k \in \mathbb{Z} $ and $a \in \mathbb{Q}$.

Repeat $\phi(k)+\phi(k)+...$

@user170039 i am trying... Now, $Z$ is a subset of $Q$

if we map each member of $Z$ to the same members of $Q$...

then we get an homomorphism

Alessandro Codenotti

No, we don't

Subhasis Biswas

@AlessandroCodenotti I am definitely missing something

wait.

will be back

Alessandro Codenotti

14:21

Write down the definition of ring homomorphism

user131753

@AlessandroCodenotti I think he was talking about group homomorphism from $(\mathbb{Z}_n,+)$ to $(\mathbb{Q},+)$.

Albas

Yeah

Alessandro Codenotti

Makes no difference, write down the definition of homomorphism

user131753

I think it does because then the inclusion map from $\mathbb{Z}$ to $\mathbb{Q}$ is a group homomorphism. Isn't it @AlessandroCodenotti?

Alessandro Codenotti

Why wouldn't it be a ring hom?

Subhasis Biswas

14:26

i was talking about a group homomorphism

user131753

@AlessandroCodenotti Of course it will be.

Albas

From a subgroup to the original group the inclusion is always a group homomorphism

Subhasis Biswas

the finite cyclic group $Z_n$ has no non trivial homomorphism (??). Now, I am still thinking about the $\mathbb{Z} \to \mathbb{Q}$. A trivial homomorphism is the one that sends all element to the identity element of the codomain group,right?

Alessandro Codenotti

right

You didn't provide a proof for $Z_n$

user131753

@Albas In fact you can take the following as a definition of "substructure" in a very general sense provided you have a notion of "structure morphism" and "structure": Let $S$ be a structure. A structure $T$ is said to be a substructure of $S$ if the inclusion map from $T$ to $S$ is a structure morphism.

user131753

14:31

(Of course assuming you can define an inclusion map from $T$ to $S$)

user131753

Now replace "structure", "substructure", "structure morphism" respectively by,

Subhasis Biswas

Suppose, for some $a \in Z_n$, $f(a) \neq 0$. Now, let $f(a)=t$ (say).

Evidently, $f(a+a+a+a+...$ n times $)=nf(a)=nt$. The domain being finite, so will be the range. Therefore, by the Archimedean property, for some $n$, $nt>$ any finite number. (if $t$ negative, consider $-t$)

user131753

1. Group, subgroup, group homomorphism,

user131753

2. Ring, subring, ring homomorphism,

Albas

@user170039 Topological space

Continuous mapping

user131753

14:34

3. Category, subcategory, functor

user131753

The key thing to note in this definition is that the notion of a substructure is dependent only on the notion of structure morphism.

Alessandro Codenotti

@SubhasisBiswas $(\Bbb Q,+)$ has no order

Subhasis Biswas

@AlessandroCodenotti okay. Can we say then, for each new $n$, we are getting a new rational number, and after one stage, it is going to get out of the range?

user131753

@AlessandroCodenotti: I was thinking about formulating a definition of structure. Care to chat with me a bit regarding this in a separate room?

Alessandro Codenotti

What is $a+a+a+\cdots+a$, with $n$ additions?

Subhasis Biswas

14:38

@AlessandroCodenotti $+_n$.

Alessandro Codenotti

@user170039 Not right now. I think the model theory definition is the most general you can get

@SubhasisBiswas I'm asking what's the result of that operation

user131753

@AlessandroCodenotti Really? How does the concept of a (large)category fit into it?

Alessandro Codenotti

I guess you can work in NBG if you really want to talk about class sized structures

I'm fine with set sized structures

Subhasis Biswas

$\overline{a}$ (here denoted by $a$) is the class of objects that is of the form $mk+a$. [here talking about $Z_m$]

Addition of such classes is is defined as the remaineder obtained when divided by $n$.

Alessandro Codenotti

I know how it is defined

I want to know the result of that operation

If I ask you what's 3+5 you tell me 8, not the definition of addition

Subhasis Biswas

14:44

@AlessandroCodenotti $na \equiv k (\mod m)$. The result will be $k$.

Alessandro Codenotti

What's $5+5+5+\cdots+5$ with $11$ 5s, in $\Bbb Z_{11}$?

Albas

@SubhasisBiswas No man

He's asking you 1+1 in modulo 2 arithmetic

Subhasis Biswas

it will return the identity $0$

Albas

Yeah

Alessandro Codenotti

So now what's $a+a+a+\cdots+a$ with $n$ $a$, in $\Bbb Z_n$?

Subhasis Biswas

14:46

@SubhasisBiswas correct that to $Z_m$

@AlessandroCodenotti notational mistake

Alessandro Codenotti

Just give me the result ffs

Subhasis Biswas

@AlessandroCodenotti $0$

Alessandro Codenotti

Ok finally

@SubhasisBiswas So look at this again

Subhasis Biswas

@AlessandroCodenotti yes. That is a mistake. The "proof" is wrong

Alessandro Codenotti

But it can be fixed

Subhasis Biswas

14:48

but what if I change it to $Z_m$ with $m \neq n$?

Alessandro Codenotti

Look at it again remembering that $a+a+\cdots+a$ $n$ times is $0$ in $\Bbb Z_n$

Subhasis Biswas

@AlessandroCodenotti identity always goes to identity

Alessandro Codenotti

Agreed

Subhasis Biswas

wow.

$nt \neq 0$

since $t \neq 0$

and $n \neq 0$

Albas

It all hinges on the fact that Z_n is finite cyclic. Should have written that.

Alessandro Codenotti

14:51

That's good

Now write down a complete proper proof

Subhasis Biswas

@AlessandroCodenotti ok.

Suppose there is a nontrivial group homomorphism from $(\mathbb{Z_n},+)$ to $(\mathbb{Q},+)$. So, in order for the homomorphism to be nontrivial, for some $a \in \mathbb{Z_n}$, $f(a) \neq 0$ , $0$ being the identity element of $\mathbb{Q}$.

Suppose $f(a) = t$. We know that with respect to the group operation in $(\mathbb{Z_n},+)$, $na=0$. So, $f(a+a+a+.. n$ times $)=f(0) = f(a)+f(a)+f(a)+... n$ times $=nt$, by the definition of homomorphism.

Again, we note the fact that the identity element of the domain group is always mapped to the identity element of the codomain group. But, in this …

Alessandro Codenotti

Good

Subhasis Biswas

@AlessandroCodenotti can it be made better?

Alessandro Codenotti

So $Z_n$ being cyclic was not enough, you also needed to use that's it is finite

Subhasis Biswas

@AlessandroCodenotti yes. Now let us try for $\mathbb{Z}$

Alessandro Codenotti

14:58

What about $\Bbb Z\to\Bbb Q$

Subhasis Biswas

@AlessandroCodenotti identity mapping?

Alessandro Codenotti

Indeed

Are there group homomorphisms $\Bbb Z\to\Bbb Q$ that are not the inclusion? @SubhasisBiswas

Subhasis Biswas

@AlessandroCodenotti was thinking about it. You mean not the identity?

Or $f$ takes one element outside of $Z$?

Alessandro Codenotti

The identity maps a set in itself, this is an inclusion

Anyway, the map sending $z$ to $z$

Ultradark

"maps a set in itself"

Subhasis Biswas

15:03

so, not an inclusion map means that $f$ assumes at least one element that is not in $Z$?

Ultradark

do you embed something "in" or "on" a manifold

Subhasis Biswas

@Ultradark plis. no

Ultradark

@SubhasisBiswas you plis no

Alessandro Codenotti

Not the inclusion means not this map, it doesn't have to land outside of $\Bbb Z$

@Ultradark in

Subhasis Biswas

@AlessandroCodenotti got it

@Ultradark never mind

Ultradark

15:06

And if the manifold is a just surface then it would be "on" I guess?

anakhro

Depends on the context.

It would still be "in" technically.

for example, R^2 is a "surface" in obvious ways

But without the context of it already being a surface embedded in another manifold, it makes no sense to use "on".

Ultradark

seven months ago I didn't know what a manifold was

Subhasis Biswas

@Ultradark that's really motivating for me

@AlessandroCodenotti yes

$\phi(x)=2x$

Alessandro Codenotti

Yep

Subhasis Biswas

@AlessandroCodenotti what if.... it is not even an inclusion map?

i mean, landing outside Z?

Alessandro Codenotti

15:16

What do you think?

Subhasis Biswas

suppose, $\phi(x)=\frac{3}{2} x$

anakhro

To be honest, the definition of a manifold is just a mouthful.

Subhasis Biswas

@Ultradark how long did it take you to learn?

anakhro

It can be...streamlined...let's say.

Ultradark

@SubhasisBiswas I mean I just read the definition seven months ago and thought about it a while. Doesn't mean I understand everything there is to know about manifolds lol

user131753

15:20

@anakhro How?

Subhasis Biswas

@AlessandroCodenotti can you provide me an insightful exercise on some topic? Just a single exercise.

anakhro

@user170039 How to streamline it? Remove the Hausdorff & second countable condition.

Albas

@anakhro Just say it's a subspace of R^n

anakhro

@Albas that would be called lying, though. :(

user131753

@anakhro But then you would be generalizing the usual notion of topological manifold, right?

anakhro

15:26

@user170039 it's one of the more general definitions, yes.

But for someone who you don't think is ready to discuss Hausdorff or second countable, the relevant "counterexamples" aren't really a concern.

user131753

@anakhro I thought you were talking about a really compact way to represent the usual definition. My mistake.

anakhro

@user170039 Well like I mean, if you want to teach someone the definition of a manifold, they will need to know (a) the definition of a topological space, (b) the definition of a continuous function, and (c) what locally Euclidan means.

Subhasis Biswas

@AlessandroCodenotti, on linear algebra and group theory. My linear algebra is even worse. Maybe an exercise regarding linear transformations

anakhro

@user170039 (a) and (b) are easy to define, but would take longer to motivate, and locally Euclidean is basically the thing that makes a manifold a manifold.

Albas

Yea @anakhro especially physicists. They don't like the other stuff that comes with the definition of manifold.

user131753

15:30

@anakhro: I am never going to teach someone manifold theory. I hate geometry.

anakhro

Well I wouldn't say physicists don't like the other stuff, but rather it's not too much of a concern when you are practically always working in local coordinates.

user131753

@anakhro Yep. It was the only interesting thing in the definition.

anakhro

@user170039 oh, it's unfortunate you don't like geometry. What made you hate it?

Ultradark

I'm making a soundtrack (single track) for a theoretical car racing video game

Albas

I had to teach manifolds to a bunch of physics majors. It was hell.

Ultradark

15:33

what questions did they ask?

user131753

@anakhro In no particular order of importance: algebraic topology, differential topology, the definition and construction of smooth structure, preference of geometric intuition over rigour are some of the reasons.

anakhro

@Albas I would have thought that would have been fun.

Ultradark

one person's fun is another person's not fun

anakhro

@user170039 What about those makes you "hate" geometry, though? Like certainly maybe they don't enthuse you, but where is the hate coming from? As for the geometric intuition, I think it's a charm of geometry that you can actually apply geometric intuition semi-reliably, but I agree that it is a problem when people think it substitutes for actual rigour.

Albas

@anakhro I thought so too, but no it was not. They didn't want any definitions. They just wanted examples. They didn't care about things like partitions of unity, etc. They wanted me to teach them Riemannian geometry before even knowing the definiton of a manifold.

user131753

15:36

@anakhro "I think it's a charm of geometry that you can actually apply geometric intuition semi-reliably" - that is one of the reasons.

anakhro

@user170039 what about that makes you "hate" it?

user131753

Plus the fact that I don't really understand diagrammatic reasoning.

anakhro

Is it that you are frustrated because you don't understand the geometric intuition?

Albas

I mean they don't need to go in any depth as mathematics and slight rigour is not their top priority but atleast know the stuff that you are using properly.

Ultradark

yes^

anakhro

15:38

@Albas Well admittedly, partitions of unity are one of those things you learn once then put in the back of your mind and forget about until someone asks you "are you sure?".

user131753

@anakhro Nope. I don't like geometric intuition.

anakhro

@user170039 why not?

user131753

Maybe because it's not really fully reliable.

Albas

I would say I find algbera and number theory pretty dry and lack of any intuition. I can get algebra as a tool to study other things.

Ultradark

diversity is the spice of life

anakhro

15:40

@user170039 Reliable for what? "intuition" seems to presuppose that it's not an actual formal argument.

@Albas Algebra is neat because of how well organized it is. It's absolutely fantastic. It's like the field of organization.

I agree that without a good background on why they care about some of these things in algebra, it might seem like just a large concrete building with no purpose.

user131753

@anakhro: We will talk about that later. For now I have to go.

anakhro

Enjoy!

Albas

@anakhro Yea but like there were other things such as when is something a submanifold. Some basics about lie groups without even wanting to know what's the definiton of a group.

I shouldn't say lack of any intuition for algebra. Number theory no intuition. But algebra feels very dry.

Ultradark

I haven't taken a strict geometry class since 9th grade

will be taking geometry again this fall

anakhro

@Albas well it's admittedly very tough to teach a class of aspiring mathematicians about manifolds and have the majority of them understand. It follows it is very very tough to teach a class of aspiring physicists who have little formal training in mathematics.

No doubt it sounds like a lot of them were suffering from impatience, but teaching requires you to sometimes take some rather drastic changes to otherwise standard approaches in order to adjust for a different audience.

Ultradark

15:47

well said^

anakhro

The good part is now you know what apparently doesn't work for physicists! So next time you have the opportunity, you can try a new approach!

Albas

@anakhro True. Well first try in teaching. Will take a note of that

@anakhro Yea I kind of will be interacting with them a lot more as time goes.

anakhro

Teaching is notoriously difficult. As evidenced by the plethora of bad teachers.

3

It's undervalued, as well.

Ultradark

yeah^

Albas

Yup

anakhro

15:49

So good on you for even attempting.

Albas

@anakhro Well I do like it.

anakhro

Me too. :)

What was the physics level of these students?

Albas

They had taken courses in things like QFT, QM, Classical Mechanics, Electrodynamics. I was asked to give a series of talks on manifolds and Lie groups as they were taking General Relativity in their next semesters.

anakhro

What is your physics background like?

Albas

I also think that things like Symplectic Geometry should be introduced while talking about Classical Mechanics and atleast hinted upon while teaching QM in Quantisation.

anakhro

15:54

Yeah, my initial suggestion would be introducing symplectic geometry first via Hamiltonian/Lagrangian formalisms.

Ultradark

yeah I agree @Albas

Albas

@anakhro Well since I am a mathematics major, I haven't taken any official courses but I have studied Classical Mechanics, Electrodynamics, Quantum Mechanics, very introductory general relativity. Currently upto some QFT.

I would really like to do General Relativity properly someday.

anakhro

Nice. So you have a robust knowledge of how manifolds can apply to physics. That's a great start.

Albas

@anakhro Yea even on the mathematics side of things they tend to forget things like the Hamiltonian mechanics when trying to introduce symplectic geometry. The entire poisson bracket thing and hamiltonian vector fields put symplectic geometry in a lot of perspective

Ultradark

@Albas I asked a question on hamiltonian vector fields and symplectic geometry on overflow. If you want to look at it and give me some constructive feedback feel free :)

It's a pretty novice question

Albas

15:59

@Ultradark I can try, I am still quite early into symplectic geometry so I can't guarantee

Ultradark

no problem, I'll take any feedback lol

Ill post the link here

anakhro

The problem you will probably see in a series of lectures on diff. geometry for physicsts is that you will be inclined to cover the basics in the first lecture in full generality so that you can do more interesting things later. The problem is that this will probably be boring to the general physics student. I'd probably try to come up with well thought out examples that operate as mnemonics for physics students.

Ultradark

1

Q: Counting fixed points for Hamiltonian symplectomorphisms on $T^{2}$

This question is motivated by the Lorenz curve used in economic analysis and also the Penrose diagram used in general relativity, used by physicists in order to visualise causal relationships in compactified Minkowski space time models. It is also motivated deeply by Hamiltonian mechanics, symp...

anakhro

So that if they wonder what $x$ is, then they can remember "oh yeah, $x$ is like example $y$ from physics!".

skullpetrol

and vice versa :-)

minus the generality, of course

Albas

16:08

Yeah that is a very good idea. That would make it seem like less of unnecessary jargon. That would be a difficult thing to do in general I guess, like how do you motivate a physicist to care about something like say submanifolds, things like the level set theorem or transversality?

anakhro

Well level set theorem and transversality can be treated more or less like one the same.

For submanifolds I am thinking Lagrangians.

Or Legendrians.

But I think you can keep submanifolds as a very naive concept and get away with a lot.

naive concept like "subsets of a manifold which are a manifold in their own right".

Albas

Yeah

Secret

59 mins ago, by anakhro

@user170039 How to streamline it? Remove the Hausdorff & second countable condition.

It is unthinkable just how huge $\omega_1$ is, it is the second ordinal that it takes a net as long as itself to converge to it

anakhro

@Albas another fancy concept in teaching is that if you interest students enough, they will let you get away with some unmotivated material sometimes, too.

So you can save the concept of submanifolds for later.

Or transversality would be a good one to save for later, motivating it just in the sense that it is a generic property (I'd definitely skip out on Sard's theorem, though).

@Secret you quoted me but you comment doesn't seem to be about the quote?

Secret

16:24

Ok, that's because I omitted too much chain of thoughts in the middle. The full thing is like this:

See "second countable condition" -> Respond "I am perfectly fine dropping even first countable" -> counterexamples that are not first nor second countable are often involving $\omega_1$ -> start thinking about the huge size of $\omega_1$ again -> wondering about infinite manifolds that are uncountable dimensions

Having said that, compared to Slereah, I knew almost nothing about non hausedoff manifolds, other than some cool examples like the line of n origins

(That the chain of thoughts is so convoluted is one reason I don't ping you in the reply, for pings are be reserved for urgent things only)

Albas

Lol @anakhro didn't know that.

Secret

If anakhro want to know, he will notice it and knows

This is partly why some people think I am spamming because my posts are designed such that it gives maximum freedom for the intended receipient to scroll past it if they want

and thus to people seeing from the outside, it seems as if my posts directed at no one

Those [Random]s are the worst examples of my communication style. Since Ted seemed to be near breaking point of being annoyed by them, I have now decided to cut them off entirely, which should reduce the (perceived) spams to zero

Aneesh

17:03

could someone help with this? math.stackexchange.com/questions/3255257/…, even after the comments i'm not exactly sure

2

anakhro

@Secret I think he was responding to a message I sent him before that.

anakhro

17:32

@Secret you should definitely try making a blog to categorize all your thoughts and ideas. That allows you to archive them in a systematic way, as well as share them without spamming people (if you feel you are spamming at times).

Leaky Nun

18:05

hi @Ted

Ted Shifrin

hi Leaky

@Aneesh: Of course it's little-oh of $t^2$, not big-oh. But what do you not understand?

Aneesh

@Ted right, its little-oh of $t^2$ (the ...), the bit im having trouble understanding is how he goes from that line, to the result of (2.27)

@Ted but actually, doesn't the remainder being little oh of $t^2$ in that interval also mean its big o of $t^2$ in that interval? the remainder is a continuous

Ted Shifrin

Of course you can get $s$ is (one of two) function(s) of $t$. Factor out $t^2$ from the power series on the left, and take square root.

You want little-oh or else you don't have the right leading term.

Aneesh

ah okay, so it really is just as simple as factoring out $t^2$ and then it reduces to what the book says it does?

(with your little o correction as well)

Ted Shifrin

Yes.

Aneesh

18:15

okay, so I put $t^2(h''(0)\frac{1}{2} + \frac{\mathcal{o}(t^2)}{t^2}) = -s^2$

and I want to write $s$ in terms of $t$

Ted Shifrin

So you'll need a power series expansion for $\sqrt{c+\epsilon(t)}$ ....

Best way to do that is factor out a $\sqrt c$.

Aneesh

sorry one sec, just going to write this all out on paper so I make sure I'm understanding all this

Ted Shifrin

Oh, absorb the $-$ sign into the LHS so you can take real square root.

Aneesh

ah okay I think I'm almost there

I have gotten up to $t = \frac{-2}{h

$t = \frac{(\frac{-2}{h''(0)})^{\frac{1}{2}})s}{\sqrt(1 + o(1))}$

Ted Shifrin

18:30

That doesn't quite make sense, because the square root you divided by has $t$ in it.

Aneesh

that o(1) term is with respect to t->0

its really o(t^2)/t^2 = o(1)

Ted Shifrin

You want a (formal) power series for the original LHS, writing $s=\sqrt{c}\sqrt{1+\epsilon(t)}$ and using the power series for $\sqrt{1+x}$. Then you can invert to get $t$ as a function of $s$. But you don't need to go very far. Actually, I really don't like the book's writing $g(t)$ equal to a function of $s$. Yuck.

Aneesh

or your \epsilon(t) term

Ted Shifrin

Right, I said it became $o(1)$ way up there ^^^ somewhere.

Aneesh

yup

Ted Shifrin

18:34

But $\sqrt{1+x} = 1+\frac12x+O(x^2)$ and that's all you need.

Alessandro Codenotti

Hi @Ted

Ted Shifrin

Then you can find a power series for the inverse function formally.

heya demonic @Alessandro

anakhro

Hi

Ted Shifrin

hi, @anakhro

anakhro

What's up, Ted?

Ted Shifrin

18:37

Nada.

anakhro

How...uneventful.

Ted Shifrin

I leave the events up to you young'uns.

anakhro

Heh.

I am just working on polishing my thesis but it's ending up with me adding more and more new stuff that I will have to edit.

I might just say "ENOUGH" and go full-time into editing it.

Aneesh

okay so you are saying we write s = const. t (sqrt( 1 + eps(t))), then expand
sqrt(1 + eps(t)) as a power series, and invert the power series to find t in terms of s, as far as inverting the power series goes, is there some formal procedure for this?

F.White

Am I correct that:
$[\frac{\partial}{\partial x},-y\frac{\partial}{\partial x} + x\frac{\partial }{\partial y}]=\partial_y$?

Since this is equal to $\partial_x(-y)\partial_x+\partial_x(x)\partial_y - (-y\partial_x(1)\partial_x + x\partial_y(1)\partial_x)$

Aneesh

18:42

is there a way to process mathjax on chat

F.White

(First time I've actually computed one of these is why I want to verify)

Ted Shifrin

Sure. You do it recursively, term by term, @Aneesh. For example, you want $t=a_0+a_1s+a_2s^2+\dots$ to satisfy $s=1+\frac12t+3t^2$. Plug in and expand, solving first for $a_0$, then for $a_1$, etc.

skullpatrol

@Aneesh math.ucla.edu/~robjohn/math/mathjax.html

Ted Shifrin

Yes, @F.White, you're correct.

Aneesh

ahh okay that makes sense

F.White

18:43

@TedShifrin Thanks Ted

Aneesh

@skullpatrol thanks

skullpatrol

np

Aneesh

@Ted

@Ted for the O(s^2) as the remainder term at the end , will this fall out of this procedure at the end?

Ted Shifrin

Sure, because you already have the linear term from the constant term of the square root.

Ted Shifrin

18:59

@skull: with respect to your mentioning my name about those MIT notes, I wrote a set of notes for multivariable calculus when I was a senior at MIT and teaching the stuff (my first official class). They used a lot of them as supplementary notes for years (apparently, in revised form, still).

skullpatrol

cool

anakhro

Ancient scrolls passed down from Shifrin to Shifrin.

Ted Shifrin

LOL

Well, I did not keep a copy of those when I retired and emptied out my office. Oh well.

Bye for now.

anakhro

Bye!

skullpatrol

you should rename your lecture series to something more descriptive rather than the course number

cya

anakhro