Mathematics - 2018-11-11 (page 1 of 3)

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12:02 AM

Basically the transition functions for $s\times_{S^1} L$ have as image equivalence classes $[a,z]$ where a in Spin^c(n) and z in S^1, but we quotient out the relation [a,z]=[az,1] which then when passing to the adjoint vector bundle gives me the transition function for the tensor product

Yeah

And the fact that the scalars are central means we can commute all the things we need to

yeah

do you have time for a follow up question?

(I'm following Spin geometry by Lawson and Michelson), this is Example D.10 in their appendix. assume now that M is spin, so we have a principal Spin bundle P. by taking balanced products P \times_{Z/2} L we can form Spin^c structures for M. The claim is similar, If W is my spinor bundle for P, then the adjoint vector bundle to my Spin^c structure is $W\otimes L$

I think it should again follow by thinking about transition functions, perhaps now with the additional fact (is it a fact?) that the complex Spinor representation is the complexification of the real Spinor representation

I forget my spin geometry, so I am not sure if that is true

Seems like it should be from your statement though

yeah that's what I though but if i try to write down things I'm not anymore allowed to move the entire $S^1$-factor to the left as before, i.e. $[a,z] \neq [az,1]$ i think, since I'm only allowed to move around $\pm 1$

I think you should first start with L trivial - the spin^c structure induced by the spin structure - and then show that the rest of the statement follows by twisting spin^c structures as above

12:12 AM

Ok, which should follows from the fact that the canonical Spin^c representation restrict to the canonical spin-one right?

Yeah

ok I think I got it

thanks a lot

for sure

Heya MikeM

hey

12:49 AM

hi chat

deleted

Hi deleted Shmo.

I am asked to compute $\int_{C_R(0)} \dfrac{dz}{(z^2+4)(z-1)^3}$ in terms of $R$. Just to make sure I'm not doing this wrong. For $R<1$, the integral $=0$. For $1 \le R < 2$, apply the residue theorem, and for $R \ge 2$ I get the integral is $0$ again

for the latter interval, i have a handy dandy exercise that states that $\deg q \ge \deg p + 2 \implies \int_{C_R} \frac{p}{q} = 0$ if $C_R$ contains all the poles

I'd restrict that to $1<R<2$ and $R>2$, since for $R=1,2$ you'd end up with your contour running through a singularity

Érico Melo Silva

hlo chat

ok. there's also $R < 1$, which doesn't run through any singularities, hence the integral is $0$

12:56 AM

Agreed. What I would do for $R>2$ is compute the residue at $\infty$. It's very powerful.

heya @Eric !

yeah, that's another good approach. (good to compute it both ways imo)

Érico Melo Silva

@Ted about to eat at a michelin starred danish place

@Semiclassic: I'd rather not compute 3 different residues if I can do just one :P

lol

well, one of those residues is pretty trivial :P

OK, agreed.

12:59 AM

but yeah, doing the other two involves a small computation

I'm a big believer in the residue at infinity, 'cuz then I get to talk about how you really take residues of meromorphic $1$-forms, not of meromorphic functions.

ah, true

I mostly like it because it gives me a reason to think about the Riemann sphere :)

Sometimes branched coverings of said ...

Érico Melo Silva

@TedShifrin wut do u mean by this

yeah

1:02 AM

@ÉricoMeloSilva What u mean?

Érico Melo Silva

why does the residue at infinity let u talk about 1-forms

Oh, I'm claiming that it's the residue of a $1$-form that's actually well-defined.

This becomes relephant when you do the Residue Theorem on compact Riemann surfaces.

You should get there soon in G/H.

Érico Melo Silva

i see i see

gotta bounce

Oh, have a great dinner!

@TedShifrin Still always read this as an orbit

1:13 AM

LOL ... I can't help your proclivities, Mike.

Let $\mathscr S$ be a subbasis for a topology $\mathscr T'$ on $X$. Let $\mathscr S$ be a basis for a topology $\mathscr T$ on $X$. How do I prove $\mathscr T=\mathscr T'$. Let $X\in \mathscr T \implies X=\bigcup_{S\in \mathscr S}S$. Since, each $S\in \mathscr S, S=S$ So, $X\in \mathscr T'$.

For reverse inclusion, Let $X\in \mathscr T' \implies X=\bigcup_{\alpha \in \Lambda} \bigcap_{\beta \in \Gamma,|\Gamma|<\infty}S_{\alpha \beta}, S_{\alpha\beta}\in \mathscr S$ . If $\bigcap_{\beta \in \Gamma,|\Gamma|<\infty}S_{\alpha \beta}\neq \emptyset$. Let $x\in \bigcap_{\beta \in \Gamma,|\Gamma|<\infty}S_{\alpha \beta}\implies \exists x\in S_x \in \mathscr S: S_x\subset\bigcap_{\beta \in \Gamma,|\Gamma|<\infty}S_{\alpha \beta}$.

All this mathscr is gonna give me an aneurysm

Well, Demonark, maybe that's doing you a favor in the long run!

:0

I certainly don't like the sentence "Let $X\in\mathscr T$."

1:18 AM

$\bigcap_{\beta \in \Gamma,|\Gamma|<\infty}S_{\alpha \beta}=\bigcup_{x\in \bigcap_{\beta \in \Gamma,|\Gamma|<\infty}S_{\alpha \beta}}S_x$. Hence $X\in \mathscr T$

Am I correct?

Are you trying to take an arbitrary element of $\mathscr T$?

I encourage sentences with far fewer symbols and more words.

Yes. sorry. It should be some $U$

Aha.

I don't think you need arbitrary open sets. Basis elements should suffice for any such proof.

Doesn't that simplify things?

what is a subbasis?

okay

1:21 AM

oh yes.

I didn't know if you were being socratic.

In topology, a subbase (or subbasis) for a topological space X with topology T is a subcollection B of T that generates T, in the sense that T is the smallest topology containing B. A slightly different definition is used by some authors, and there are other useful equivalent formulations of the definition; these are discussed below. == Definition == Let X be a topological space with topology T. A subbase of T is usually defined as a subcollection B of T satisfying one of the two following equivalent conditions: The subcollection B generates the topology T. This means that T is the small...

@TedShifrin me?

me

No, JoeShmo.

i was. i meant to write to write that in the adjacent chat on the next window

1:22 AM

Question: What does something need to have in order to be a geometry?

I literally cannot read stuff like you typed, @N.Maneesh. I need words and as few symbols as possible.

@N.Maneesh I know what a sbbasis is

@Rithaniel: I have no idea what your context is. And, despite being a geometer, I may not know how to answer it.

@TedShifrin okay. I will type again.

No, just think things through to be as efficient as possible. I really don't like this stuff.

1:25 AM

its a funny-ly phrased exercise. i want to answer "by definition"

@TedShifrin I think $\bigcap_{\beta \in \Gamma,|\Gamma|<\infty}S_{\alpha \beta}=\bigcup_{x\in \bigcap_{\beta \in \Gamma,|\Gamma|<\infty}S_{\alpha \beta}}S_x$. Hence $X\in \mathscr T$ is the important step. right?

I have no idea what you're typing.

Seriously.

You're using symbols that haven't even been defined.

i second that. i can't parse all of that either

Okay, then it's a more complicated question than just a given definition? Because I read someone state that a non-Euclidean geometry is just a geometry where the parallel postulate does not hold, and, best I've been able to tell there are five axioms that Euclidean geometry is built on, and I'm unsure if these other four axioms are just prerequisites for a geometry, or if the person who stated that could be wrong.

finite intersection of elements of $\mathscr S$ can be the union of elemnts of $\mathscr S$

this thing I was tryiing to prove in the reverse inclusion.

1:31 AM

@Rithaniel: That is reasonable. But there are other notions (look up Klein's notion of geometry on Wiki). I don't know the context of what you're reading, and I don't live in that world of axiomatics for geometry.

using the property of basis $x \in B_1\cap B_2(B_1,B_2\in \mathscr S$). There exist $x\in B_x \subset B_1\cap B_2,$ where $B_x$ is a basis element. So I can write the intersection of $B_1$ and $B_2$ is the union of basis elemnts. @TedShifrin This thing I was trying to prove.

Were the subbasis and the basis you were given related at all? Or is this a totally abstract question?

It's nothing professional, just a person posting something on a comment thread talking about Lovecraft. What constitutes a "non-Euclidean geometry" came up, which got me wondering.

@Rithaniel: Yeah, to most people that would mean the usual axioms without the parallel postulate. But, honestly, I never think of geometry this way.

To me, a non-Euclidean geometry comes from a constant curvature model with non-zero curvature, so the sphere, projective plane, or hyperbolic plane.

Can you can also have a geometry with the metric requirement being relaxed? Would that still be a geometry?

1:40 AM

@TedShifrin Actually, I wanted to prove $\mathscr S$ is a basis for $\mathscr T$. I got this collection is a subbasis for $\mathscr T$.

Are you doing this in a concrete case, or totally abstractly, @N.Maneesh?

@Rithaniel: There are all sorts of finite projective planes, for example, where there's nothing like what I'm talking about. In some settings, a geometry is just a notion of points and lines in a plane. With certain properties.

abstractly

can you give the exact problem statement

I created an excercise like this Let $\mathscr S$ be a subbasis for a topology $\mathscr T'$ on $X$. Let $\mathscr S$ be a basis for a topology $\mathscr T$ on $X$. prove $\mathscr T=\mathscr T'$.

So why can't I just say that I get a basis by taking finite intersections of subbasis elements?

Wait. YOU created it?

Well, as you created it, it's nonsense.

1:42 AM

:(

why?

The subbasis gives you a basis. The two bases have to give you the same topology.

i was so puzzled by the question

It's nonsense because you have no hypotheses relating the subbasis and basis, although you used the same letter (which I assume is a mistake).

I'm getting too old to be in this chatroom.

youre the spirit animal of this chatroom

I resign.

1:44 AM

Hey, there needs to be someone to tell us when we're being stupid.

@Rithaniel I generally like to say that geometry is the study of properties and symmetries of rigid things, while topology is ahout flexible things.

Well, at least you should give Klein's definition :P

Hmmmm, I like that. Kind of a philosophical approach to it.

I'm loosely familiar with topologies at this point. Maybe I'll try to take some abstract geometry courses next semester.

sorry! It was otherway round

Is that even a thing, though? "Abstract geometry"

1:47 AM

There are not such things, typically, @Rithaniel. Typical colleges/universities may have an "axiomatic" geometry course, typically aimed at future secondary teachers (who don't appreciate it), as well as a differential geometry course. But "abstract geometry"?

I recommend you get Pedoe's book, Geometry: A Comprehensive Course.

One finds a flavor of geometry they like and explore that

Oh wow, that's only $20

Let's see if I can find a hardcover, though.

There isn't one.

But it's not going to be horribly axiomatic.

Really? No hardcover? That's unfortunate.

@MikeM: I was proud to include a chapter on affine and projective geometry in my algebra book, to broaden people's horizons. But hardly anyone every reads/teaches it.

@Rithaniel: I don't believe so. Dover reissued it years ago. I have no idea even who the original publisher was.

1:52 AM

I created an excercise like this Let $\mathscr S$ be a basis fa topology $\mathscr T$ and if $\mathscr S$ be a subbasis for a topology $\mathscr T'$. then $\mathscr T=\mathscr T'$.

@TedShifrin

Wait, @N.Maneesh. Are you intentionally using the same letter for the basis and subbasis?

yes

We missed this hours ago.

OK. So you start with a subbasis that is already a basis.

yes

Geez.

1:55 AM

sorry for incorrect preposition earlier.

I wanted to prove both generate the same topology.

So you need to show that any finite intersection of subbasis elements, which is a finite intersection of basis elements, is again a union of basis elements. And we know that's true.

You never need to think about the topologies.

using the property of basis $x \in B_1\cap B_2(B_1,B_2\in \mathscr S$). There exist $x\in B_x \subset B_1\cap B_2,$ where $B_x$ is a basis element. So I can write the intersection of $B_1$ and $B_2$ is the union of basis elemnts. @TedShifrin This thing I was trying to prove.

Right, and that's true. What do you mean "trying to prove"?

You take the union over all $x\in B_1\cap B_2$. (In principle, you need a finite intersection, not just two, but same thing.)

sorry I copied from previous chat

@TedShifrin Nice, but not surprising.

1:58 AM

LOL, no.

Thank you very much @TedShifrin. We met after a long time :)

I always taught several weeks of projective geometry stuff, figuring classic projective geometry (beginnings of algebraic geometry) would be more fun for most people than weeks of Galois theory.

@N.Maneesh We met? :)

@TedShifrin yes, occasionally, 5 months ago, I disturbed everyone in this chat with questions :P. Now, I am busy with my lectures. :)

Sorry, I can't keep track unless people are regulars.

It's okay.

2:03 AM

howdy loch

"More fun than weeks of Galois theory"
:'(

Demonark: Remember that only epsilon of my students even considered going to graduate school in mathematics. ... And I did several weeks of Galois theory.

Did someone say Galois theory?

:D

And one of my best students ever went to MIT and did a Ph.D. in algebraic geometry, so it didn't hurt him. (Of course, he didn't stay in mathematics. More from the way the economy works and having a family and kids to support.)

heya @Fargle

2:07 AM

How can we replace the additive inverse condition of a v.s

Yeah, probably gonna pull a lot of us out of academia

Why replace it, @Jacksoja?

with 0.v =0 for all v inV ?

Heya @Ted.

@TedShifrin it is a question I have,

2:07 AM

Oh ... Then you can do it, @Jacksoja.

What should be $-v$?

Projective geometry definitely is nifty

I mean by replace what do they mean ?

I really should go back and try to learn Galois theory. I think I'm strong enough on the other algebra to grasp the basics of it now, and that was the portion of algebra that I was skipping back when my life was less together.

They mean that you should assume $0\cdot v = 0$ and deduce existence of additive inverses.

is it if we assume one of them we can derive the other ?@TedShifrin

2:08 AM

@Fargle d o i t, Galois theory is so much fun

Look at a book by Stewart on Galois theory, @Fargle. It's beautiful.

Or Emil Artin's original pamphlet.

Will do. Is it particularly accessible or inaccessible?

@Jacksoja: See above. I already answered that.

0v= (0+0)v = 0v+0v

@Fargle Yes

2:09 AM

then 0v=0

Not helpful, but try something else.

Knowing $0\cdot v = 0$, what is your candidate for $-v$?

(I didn't actually read either, just wanted to take the "Is it X or Y?" "Yes" opportunity)

-1v ?

Demonark's obnoxiousness quotient doesn't diminish.

and i need to use the distributive law

2:10 AM

You mean $(-1)v$? @Jacksoja

Yes

Yes. Do that.

but it does not seem I proved anything new

New?

(-1)v+ (1)v = (-1+1)v =0v =0

2:11 AM

You're assuming a different axiom and deducing you have additive inverses.

Right. That's it.

I mean does not mean I added anything extra to what has been said

They're asking you to show you get the same stuff starting from a different place.

so in short, 0v=0 forces an additive inverse for all elements in V

Right, whereas you usually use existence of additive inverses to prove that property.

So it's, as you said, an equivalent formulation.

okay thanks Ted

2:13 AM

Sure.

so in the other direction

using that we have additive inverse to prove 0v=0 for all v in V

Right

0v =(0+0)v = 0v+0v , and now we add -0v to both sides

here we had to assume the existence of such thing

-0v right?

Sure.

And then from that you deduce that $(-1)v = -v$.

these questions i always had trouble with idonno why

2:15 AM

Yeah, this stuff gets really confusing.

(-1)v =-v is easiar

No it isn't.

Be careful.

(-1)v+(1)v = (-1+1) v = 0

But you need $0\cdot v = 0$ first.

hence (-1) v is the additive identity of v

and by uniqness of inverses no ?

2:17 AM

You need first to show $0\cdot v = 0$, then to deduce $(-1)v=-v$.

okay but i showed that before

But you can't say it's easier.

The first thing is part of it.

i meant easiar for me not in general

LOL ...

shrugs

because I understand that property

what it means to be additive inverse

but the rest seems hard, because i usually dont understand what i need to show

2:18 AM

Most of my students tried to use $(-1)v=-v$ to prove $0\cdot v=0$, and that doesn't work.

(From the usual axioms.)

is there a way to get better at this?

is it important ?

Practice. And this is my least favorite part of math.

or should I not spend too much time on for now

Right, do real proofs. :)

okay thanks ! I may return to these once I have time for it

seems something that i know the answer to, but hard to show axiomaticlly

2:21 AM

I am not a big fan of formality in mathematics, but some people are different, and some teachers are different.

to what extent not a fan?

A larger extent.

but the main question is , does it help understanding the subject better?

or is it just a matter of style?

for example?

If that question had a single answer there would be no debate.

2:22 AM

I think these are baby proofs that are far simpler than most complicated proofs that you have to deal with later.

okay

I take your word on that Ted, if my teacher tells me otherwise , i will give him your number

so await a phone call ^^

LOL, @Jacksoja. Feel free. I don't say that lots of people don't disagree with me. But they're wrong.

@TedShifrin haha joking, doing self study course,but thanks for the help again

I'm not joking. They're wrong.

MikeM might be on my side. But maybe not.

I'm mostly on your side. I think it depends on the student. I also think lots of students think that the formality is helping their understanding when in fact it is hiding their lack of understanding.

2:26 AM

Learning formal rules for negating quantifiers makes it "easier," but doesn't help in the long run.

Mathein is a good example of someone for whom the abstraction helps him understand better, and puts that understanding to good use.

One has to develop intuition.

Mathein is a very sophisticated breed.

And, ultimately, he does have intuitive understanding to communicate most of the time.

Yes, but I think sometimes built originally from formalities. But note that above I said that most students use it to hide misunderstanding.

Like me!

I think it is rarer for someone to do this effectively.

2:28 AM

We can't all aspire to be Quillens.

Yeah, Fargle doesn't know anything.

<3

I know Fargle doesn't know anything. He's years behind on diff geo and diff top. :P

I don't even know prealgebra don't make fun of me guys I'm only 8

ROFL

:)

@MikeMiller <3

2:29 AM

Which brings me to one of my pedagogical concerns. I bring in extra — typically harder or more theoretical — problems to my AoPS kids for the best few to think about. One of my weakest students insists on working on those, gets nowhere, and doesn't work on the more mundane stuff — to his detriment. I've told him twice he needs to do that, but "pride" stops him every time. Sigh.

Should I just not hand it to him? That would be rude. But it might be necessary.

Give him warning you're thinking of doing that.

he has to train the pride out of himself. sooner or later it will dawn on him

We have an exam coming up in two weeks. Originally I was going to include a few of my own more conceptual questions, but now I'm thinking I won't do it.

@JoeShmo: It's tough when you're in 10th or 11th grade.

well, you have to hold people up to high standards. and communicate, in the most non-condescending way, that pride will only hold your back

This is a very non-traditional teaching situation. I miss "traditional" college teaching.

2:35 AM

if you don't, you're doing him a disservice.

when I was in college, I remember a conversation I had with a marketing professor

he gave three exams and the students grade for the course was the highest of the three exams

as a result, a lot of students talk more marketing courses and he liked that

I would never grade like that, but that's irrelephant.

Hmm, has that kid given much indication as to why he dislikes the mundane problems?

If it's one of those, he's just not interested much by the material in general but once it gets to the tricky stuff he's excited? Desire to not be behind? Etc?

He doesn't look at them, Demonark. It's like he's doing mine to "please" me, even though he gets nowhere, even with my help. And then he struggles on on-line homework. Not good.

i guess he marketed his own class

2:38 AM

his class was very popular

No, I write the problems for the two best students who are more bored with the standard stuff, Demonark.

the math department was not

I'm not surprised, @Bob. But so what.

maybe the math department should lower its standards

whats the material?

2:38 AM

Oh I meant bored in the sense of, maybe he just doesn't even like the subject much but the extra problems are more fun by virtue of being puzzles

But yeah hmm

@Bob, that's destined for failure. The physics and engineering departments will quickly start teaching their own math classes and there won't be any math students. That's just crap. ... Of course, there's no excuse for bad teaching. But incompetent certification on top of it is destined for failure.

Demonark: I don't think he's good enough for that.

okay

Students who actually need to use the math in their work (e.g., physics and engineering) can't afford to get high grades for incompetence. That's what makes America great. Ideas like that.

On the other hand, I don't excuse bad teaching and bad grading. That does occur too.

i was thinking the same thing @Daminark

i don't know how math departments i've seen can lower their standards any more, given the students i see taking real analysis

2:41 AM

We're talking just about calculus grades here, Mike, but, yeah, you're right. I had students in diff geo who shouldn't have passed Calc I. Seriously.

LOL

I'll stick with the students (many not so great) who took many courses from me because they respected the challenge and learned a ton, even if they only got B's and C's.

I'm still not sure why the standards got as low as they are. Like, what's stopping universities from just failing a few people out and then everyone else gets the idea and gets their stuff together

I was on university teaching award committees and saw evidence of plenty of teachers like Bob describes. Loved because they made it fun, they were fair, and they made it fun. Nowhere was the word "challenge" or "learning" mentioned.

Demonark: Money. Students are commodities.

Even without MAGA.

I appreciate when a professor grades work mercilessly. It let's me know what I need to improve upon.

2:44 AM

I have a friend who was teaching high school math and he was up for tenure

and he was told that he did not use enough colors in his presentations and therefore he was not getting tenure

@Rithaniel: Most professors don't grade the work themselves.

Well, that's total crap, @Bob.

@TedShifrin That is what he told me and I believe him

Although I do use colors, and I had an ancient colleague, who was to boot a horrid teacher, tell me that boardwork and presentation shouldn't matter in teaching.

@Bob: I'm not not believing him. I'm saying it's crap. And lots of crap occurs at all levels of teaching.

@Rithaniel in principle yes, but when you pay to get a degree, do you appreciate it?

@Bob: I'd like to have watched him myself and come to my own conclusions.

2:45 AM

I feel like any one student still has more to lose by failing than the uni does by losing them. And I do think that once some failings start, students begin to realize that their future is at risk, they'll start to work

Well he has a new job today

Nah, I don't want that at all, @JoeShmo

Demonark: Have you not heard of helicopter parents? (I presume yours are not that.)

I was actually advocating for the opposite

I'm totally lost now, @Rithaniel.

2:47 AM

I have but I feel like they don't really need to be listened to, just hang up if they start to complain

LOL ... then they complain to the dean and to the president. Your view is a bit simplistic.

I'd appreciate if my professors were to fail me if I kept getting everything wrong.

speaking of getting things wrong

If I just pass without learning anything, then what is the purpose of college?

Some of the students I've had who're fondest of me, @Rithaniel, are ones who got bad grades. They never thought of being angry at me for them. But others ....

2:48 AM

Why do the dean and president not just ignore them?

@Rithaniel: A credential that says "You may make money now."

Demonark, they can't do that. Money speaks.

if on Monday I post to Math stack exchange a differential equation of a certain form with an incorrect solution and then somebody points on my error.

and I do the same thing on Wed with a different differential equation of the same form, is that okay?

WTF is your point, Bob?

There's nothing professional going on at MSE, for starters. It's all voluntary, with no professional responsibility or pay.

I did that and somebody complained

Yeah, that's a foolish reason to seek a college education, in my view.

2:50 AM

or at least pointed it out

Well, good that they complained.

But there's no obligation to hold anyone to any standard other than personal integrity.

I complain about lots of incompetence and ultimately just shrug and walk away many times.

But what in the world does this have to do with what we're talking about?

@TedShifrin thank you for listening

good night

@TedShifrin am trying to see how the covariant derivative acts on tensors , I write a tensor in its basis representation then apply it , I get the usual big sum of tensor and then the covariant derivative runs through each term just like the lebibniz rule , am I on the right track?

That's too vague for me to answer.

So I have that the covariant derivative acts on s tensor t by the Leibniz rule

2:55 AM

Yes.

Shouldn't you finally learn some ChatJax to type math?

Yaaa

I would rather learn math though...

Well, sometimes you have to do both.

I need to capitalize on the little concentration that I have

And if you're in graduate school, you're going to have to type lots of things — including your thesis work — in LaTeX.

But your right

2:56 AM

I don't care about your excuses.

Anyhow, yes, you use the Leibniz rule to take the covariant derivative of $s\otimes t$. So?

I write the tensor in its basis rep

So I get this huge sum of basis vectors and basis covectors with scalers

well, scalar functions

Ya

so you have to differentiate those scalar functions and the various basis vectors/covectors

So it's a giant product rule.

Damn , ya

2:59 AM

OK. So what's the question?

So how does this thing act on covectors?

The covariant derivative

Is it apparent from the def ?

Think about the contraction rule. If $\sigma(s) = \text{constant}$, then $0 = d(\sigma(s)) = \nabla\sigma(s) + \sigma(\nabla s)$.

And we have functions rather than scalers couse we are dealing with tensor field rather than locally ?

Huh?

No, it's still all local.

I though when you fix a point on the manifold the functions of the tensor become scalers

3:02 AM

The dual basis has covariant derivative the (transpose of?) the negative of the covariant derivative of the basis.

If we're differentiating, we're working LOCALLY, not AT A POINT.

In a chart?

Yes.

Ok I see

I'm going to cook dinner. Night for now, y'all.

And this thing acting on covectors spits out the negative of whatever it spits out for the dual

?

With a dx instead of a basis vector

Take care

1 hour later…

4:06 AM

$$ \text{distance2banana} = b^3 / \theta^2 $$

i see you've been to xkcd today

hello

4:46 AM

Which one do you prefer: $\tan^2 x+C$ or $\sec^2 x+C$?

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