np

hi

Alien is an alright movie

nothing scary tho

and it's really repetitive. but it has a scary octopus near the beginning

would recommend 6.5/10

it's not bad but not great

Bit cocky

@Arjun hes into data sci and ml right now. He makes sure statistics and whatever hes doing is detached from mathematics, and makes sure no one claims it to be math. When I asked why he makes this distinction, he said: "so that I keep my subject pure"

lol

I am in the field yes I can confirm, no math there

@Madder The $e_i$ are not numbers - if you phrase it carefully it's the components of vectors that carry the upper indices and components of covectors that carry the lower index

and yes, since the vector itself is a geometric object that does not change under a change of basis, that means that the basis vectors have to transform in the opposite ways to the respective components

(I try not to use the words "covariant" and "contravariant" since they are in my experience more confusing than helpful)

also note that, $e_i$ and $e'_j$ before and after transform are not the "same vectors". so we can't say that $e_i$ transforms into $e'_j$. It's just that we change our choice of basis from $e_i$ to $e'_j$

in fact, no vector transforms under a co ordinate transform, only their components do

so we should not think of $e_i$ transforming into $e'_j$

I made a skecth of the region $\Omega_{f,g}$ using the suggestions of the author, namely to visualize the two separate closed curves $$ \eta = \omega f'(\alpha), \xi = f(\alpha), \alpha \in [0,2\pi]$$

and $$ \eta = \omega f'(\alpha) + \omega_0 g'(\beta), \xi = f(\alpha) + g(\beta) $$

also I understood the case where $\omega_0/\omega$ is irrational, namely that I can represent the line in the plane $x,y$ where $y = \omega_0 t, x = \omega t, t \in \mathbb{R}$ as a set of segments in the square $[0,2\pi]\times[0,2\pi]$

how would all the segments be represented on a two-dim torus?

I can see it on the square only

A torus is just a square whose sides you glued together

I get it but I can't see how the half-line is on the torus

is it like an infinite coil going around it?

like multiple coils?

yes, it winds around it but never closes, i.e. never hits the same spot twice

oh so I thought correctly, it never closes

well the problem is I still don't get what happens in the rational case hahaha

i found the irrational case to be clear

it's the same just the spiral hits itself at some point

No I can't see what happens on the square yet

not on the torus

the books says it becomes a finite set of segments but why??

you just get finitely many line segments - precisely $n$, where $n$ is the smallest integer so that it times the ratio is an integer

@Claudio since the ratio is rational, an integer $n$ as I just described exists (the definition of irrational is that it doesn't)

and if we've normalize the square to side lengths one, then the ratio (let's call it $r$) is the slope of the line segments and each time we traverse the square, we hit the right hand side $r$ further up

i.e. we start at $(0,0)$, hit $(1,r)$, then start again at $(0,r)$ and hit $(1,2r)$ etc.

now if there is an $n$ such that $nr$ is integer, that means we hit $(1,nr)$ after the $n$th time, which is the upper left corner so we start the next go-around at (0,0), i.e. whole process repeats

@ACuriousMind So the upper\lower Indicies refer to the components.

But would it not be easier to denote the base vectors with superscripts? The Dual base vectors are denoted with superscripts. These are obviously covariant (since they are dual elements) and should carry the subscripts themselves.

I find this extremly confusing.

@Madder why "easier"? You need to pick one convention, and the basis vectors and dual basis vectors need to carry opposite indices since they transform oppositely

which one you pick to upper and which one you pick to be lower is a completely arbitrary choice

It is arbitary no discussion about it. It would fit better with the previously established notation about Contravariant vectors.

I'm not sure what "previously established notation" you mean

yes, if you take the $j$-th component of the $i$-th basis vector, you have $(e_i)^j = \delta^j_i$

The previously established notion, is that contravariant vectors, in other terms, those that belong to a vectorspace, have their components denoted by superscripts.

In the books i was looking at, they casually mention that the vectors are denoted by superscripts. Not emphisize the point about the components.

You see how easily then the confusion can arise. Thank you for helping.

Yes, the whole thing is confusing and often not particularly well-explained :P

physics texts often do not distinguish between the components of a vector and a vector, just calling both "the vector"

But then i am even more confused by the following

that's not a tensor, it's just a number

you take the dual vector $(e^\ast)^\mu$ and apply it to the vector $e_\nu$; by definition of the dual, the result is a number, 1 if $\mu=\nu$ and 0 if not

I understand the functationality, i am commenting on the notational part. Maybe i misunderstood something.

and yes, this is confusing in the physicist insistence to deal with everything in components because you can't distinguish this number $\delta^\mu_\nu$ from the (1,1)-tensor $\delta^\mu_\nu ((e^\ast)^\nu \otimes e_\mu)$

since the components of the latter are $\delta^\mu_\nu$

index notation + only dealing with components sucks for trying to think about the abstract mathematical objects, that's just the way it is

This is exactly what i meant.

then we agree index component notation sucks, welcome to the club ;)

@ACuriousMind sorry I had to go do something, I read your message, but $(1,nr)$ should be the (1,1) corner, so the upper right one?

@Claudio Yes, since $nr \mod 1 = 0$, so it's the upper right/lower left corner

in the grid you drew there it means you exactly hit a grid point at $x = n$

yeah I see, so when I bring it back to the single square I go back to the origini

so this is a closed coil

i keep thinking of a single coil but it's just a long coil that closes on itself at some point

@ACuriousMind thanks for the help :P

no worries

great book btw, it's called elements of mechanics by Gallavotti

it's for math students tho

but this little part about quasiperiodic motions is very readable hahah

like for the irrational case it uses the geometric interpretation of rotations by an angle $2\pi \omega_0/\omega$

and you must show that u can find an integer $n$ s.t. n successive rotations take you in an angular neighborhood of an angle you fixed on the unit circle

the rest of the book is much scarier though, I'd say a 10/10 difficulty boss

People ask me, why i am named Mad?.

Well gentlemen, would you let me know, how is it possible to lay eyes upon such work and not be mad?

Is there any reason on gods green earth, why they need to immediatly flip the order of the defintion?

@Madder they didn't

A multilinear map $V^\ast \times V \to \mathbb{R}$ is an element of $V\otimes V^\ast$, it's right

because a map $V\to \mathbb{R}$ is an element of $V^\ast$ and vice versa

I understand your second statement. A map from V to the field is an element of the dual. But i do not get the first one. Is the dual of $ V^* \times V = V \otimes V^*$? i am unfamiliar with tensors.

but here you have multilinear maps, and more or less by definition of the tensor product the space of such multilinear maps $V^\ast \times V\to \mathbb{R}$ is the tensor product of their duals, $V\otimes V^\ast$

Oh i see.

in fact, since you're unfamiliar with tensors, just take this as the definition of the $\otimes$ sign and trust that the "flipping" of the order is meaningful in the greater picture

@ACuriousMind I mean would it be such a hassle if they flipped the above order (first box) such that the duals appear in the second place?

@Madder then you would be complaining that they flipped the order of $k$ and $l$, no? :P

@ACuriousMind Yes

so they can't win either way, you just don't like that the definition exchanges the spaces with their duals :P

but that's what the definition does, the maps $V^\ast \times V^\ast \times V\to \mathbb{R}$ are $V\otimes V\otimes V^\ast$, you can't write that without flipping the order of something

Is the reasoning hard to understand? because i can not see it from what you wrote so far.

it's so it is consistent with other more abstract definitions of the tensor product

Alright. I will suffice with that for now. Thanks

but also because this way you get the indices right - you will go on to describe these maps by collections of components like ${T^{ij}}_k$, and the index position then corresponds to this being components of an element $T \in V\otimes V\otimes V^\ast$, which can eat (by contracting the indices) two elements from $V^\ast$ and one from $V$

There's a MathJax upgrade in the works. You can opt in to the beta test.

@ACuriousMind on a tangent, there's a pure mathematician I know who I'm told refuses to use Einstein notation and writes out all summations explicitly. It baffles me.

@qwerty My judgement depends on whether this is a differential geometer or not :P

@ACuriousMind I believe they do analysis of some kind so my suspicion is that they rarely need it anyway

in that case it's not that surprising

outside of differential geometry math doesn't have a strong tradition of using the summation convention

to have such strong opinions on the conventions of another field when you don't... wait, ok, point taken :p

strong opinions on things that are outside your scope is possibly normal

but on what you were saying with Madder... I will need to think if there's a "nice" way to please both mathematicians and physicists with vectors and index notation. I may be mistaken but I think the reason we conflate something like j^mu and j is because we've already picked a basis a lot of the time and talking directly about the components is more convenient?

@qwerty oh, I fully agree that plenty of computations are more convenient in index notation (and the explicit final computation may require plugging in the actual components), but that doesn't mean we can't spend a bit of time discussing the difference between an abstract vector $v = v^\mu \partial_\mu$ and its components $v^\mu$ before we introduce "the vector $v^\mu$" as a shorthand. It's a fine shorthand, but many physics texts do not seem to really understand that it is a shorthand

and if you don't discuss this, you get confusions such as the one above

sure, but if I weren't writing a GR or DG textbook, say something like a review paper instead where you'd still want to spend a bit of time being precise and clear...

Wald attempts to bridge the gap with his "abstract index notation" in his books, but that also made no one happy, really :P

a prof once told me "if you need to invent new notation that is surely a red flag things are going wrong"

and I took her very seriously and the advice was very useful ;p

as a rule of thumb that's good - most newly invented notations suck, and if you really think you need a new one, you should know exactly why beyond your own aesthetics

I think the other reason I personally use index notation is that it makes it neat and clear if you're talking about a three vector or four vector

in the abstract version you have to start putting arrows and tildes on things instead of just boldface

@ACuriousMind It made me Happy :P