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00:01
@BalarkaSen word up
00:40
So, I have a function $\pi(n)$ and I want it to satisfy these constraints:
$\pi(\pi(n))=n$
$n\in\mathbb{N}\iff\pi(n)\in\mathbb{N}$
For a set $p\in\mathbb{N}$, then $\text{gcd}(\pi(n),p)=1$ if $n\neq p$.
Is $\pi(n)=p-n$ the only function that satisfies these constraints?
00:50
Or rather, the family of functions $\pi(n)=rp-n$ for $r\in\mathbb{Z}$
 
4 hours later…
user131753
04:23
@topologicalmagician Paolo Aluffi, Algebra: Chapter 0.
05:02
Sines are pretty, but are painful to listen to
 
2 hours later…
07:16
Good morning all, a question from a "physical" paper but maybe some could know:
I don't understand equation (3). What does "reduced matrix element" mean. By what is it reduced?
These are scalar products in some Hilbert space, and the $\Gamma$ stand for irreducible representations, according to which the Hilber space "vectors" $\Psi$ are labelled, each symmetry species contains some $\gamma$ type vectors. Somehow it seems the construction "factors out" the symmetry part. But how this is done in detail I don't understand.
"each symmetry species contains some γ type vectors." -> "each symmetry species contains some $\mu$, $\nu$ or $\gamma$ vectors.
 
3 hours later…
10:11
How would you call a sum such as $x_1 x_2 + x_2 x_3 + \ldots$?
"a sum in the form $x_1 x_2 + x_2 x_3 + \cdots$"
I think we've long surpassed the medieval practice of using long string of words for what can be written in a few symbols
even Euclid used symbols for the points and edges
 
1 hour later…
11:37
@LeakyNun In text sure, I don't like maths in titles though
Any math geniouses around that can help me out?
$\frac{test}{latex}$
n	f(n)	f'(n)	Estimate
2	0	+2	0
3	2	+2	1.5
4	4	+3	4
5	7	+4	7.5
6	11	+5	12
7	16	+6	17.5
8	22	+7	24
9	29	+8	31.5
10	37	+9	40
f(n) are from observations and I need to find f. After I observed the delta between the values I came to the conclusion that f'(n) = n - 1 after which I then did $\int_{}^{} n -1 dn \rightarrow \frac{(x-2)x}{2}$ as shown in the estimate column. The values are bit off and I can't figure out why. Can someone help? :D
(not sure why my integral equation does not work)
12:33
0
Q: About absolute convergence of complex series

maths student I know I need to incorporate uniqueness theorem of Dirichlet series to get some kind of contradiction, but don't know how to proceed?

13:00
How one can prove that if zeta function has pole at 1 implies there are infinitely many prime; My guess is to show first that summation 1/p diverges which implies that no. of p is infinite; but how to reach 1/p divergent series?
13:13
@topologicalmagician I recommend Basic Algebra I and II by the great Nathan Jacobson.
13:37
Hi guys, anyone got an idea on that one math.stackexchange.com/questions/3113921/… ?
Could any number theory boys lend me a hand with this math.stackexchange.com/questions/3113877/… ?
13:55
Is there a standard mathematical concept / measurement, similar to a standard deviation, for an ordered set?
So, even though the standard deviation is the same for [1, 6, 1, 6, 1, 6] and [1, 1, 1, 6, 6, 6] by this measurement they'd be different
Something like the average of the first derivative would be one way to measure it
But I'm wondering if there's any well-established statistical measurement which does this
All the statistics that I remember learning about never had to do with an ordered list
@IanRiley by which measurement?
@IanRiley: You can look at the $n$th order difference series, thats stuff like $\Delta^n x_i = \Delta^{n-1}x_{i+1}-\Delta^{n-1}x_i$ with $\Delta^0 = 1$. But idk if there is something like a statistical theory on that.
14:48
$\zeta(s) = \displaystyle \prod_{p \in \Bbb P} \frac1{1-p^{-s}}$
if there are finitely many primes then the product would still be ok for $s=1$
@Rudi_Birnbaum yeah I can think of a lot of ways to measure it, I was just wondering if there was a common statistical theory / named variable for it
@mathsstudent ^
Volatility is the most sensical to me, but it seems like that word has been captured by finance/stock people, and used to mean either standard deviation, or a High - Low
@LeakyNun this proof is completly fine but we have to use zeta function has pole at s=1 and then we have to prove that
D'you think the following is a helpful comment?
It might help to look at the amalgamated free product $$\langle a\mid a^n\rangle\ast_{a^p=b^q}\langle b\mid b^m\rangle.$$ — Shaun 17 mins ago
14:55
@LeakyNun any idea about the proof that use pole idea
If $z_0$ is a pole of a holomorphic function $f(z)$ then $\lim_{z \to z_0} f(z) = \infty$
@mathsstudent ($A\Rightarrow B) \Leftrightarrow (\bar{B}\Rightarrow \bar{A}$)
@BalarkaSen but how to show 1/p diverges
The sum of the reciprocals of all prime numbers diverges; that is: ∑ p prime 1 p = 1 2 + 1 3 + 1 5 + 1 7 + ...
@LeakyNun @BalarkaSen any idea about this?
Using the triangle inequality, $|a+b+c|\ge |a|-|b|-|c|$ etc. This works the same way as the diverse root bounds for polynomials. Show that for $Re(s)>c$ and greater than the given fraction, the function value is decidedly non-zero. — LutzL 19 mins ago
15:02
Apply logarithm to the Euler product formula and compute the first order Taylor expansion. It's pretty direct.
Using the triangle inequality, $|a+b+c|\ge |a|-|b|-|c|$ etc. This works the same way as the diverse root bounds for polynomials. Show that for $Re(s)>c$ and greater than the given fraction, the function value is decidedly non-zero. — LutzL 19 mins ago
I'm not going to respond to random questions, just ask and don't ping everyone
0
Q: About absolute convergence of complex series

maths student I know I need to incorporate uniqueness theorem of Dirichlet series to get some kind of contradiction, but don't know how to proceed?

Tschebyscheff said it and I say it again there is a prime between $n$ and $2n$.
there is comment by lutzl which I don't understand can somebody explain that comment
15:10
@BalarkaSen !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
@anakhro !!
Suppose the number of primes if finite. Then $\zeta(1)=c<\infty$. But $\zeta(1)=\infty$. Contradiction. So the number of primes is infinte.
How are you, @BalarkaSen?
Not bad. How about you
Not horrible.
Writing my introduction for my thesis.
So explaining contact geometry like you've never seen it before.
15:42
Does multiplying a curve by a scalar change its curvature?
@gian you'd have to elaborate on what "multiplying a curve by a scalar" means.
sure, linearly
I am sure that once you elaborate on that, you can figure it out yourself by using the definition of curvature.
@Rudi_Birnbaum are you sure.
The parametrization of the curve. For instance, given $\gamma = \lambda\beta$ where $\gamma$ and $\beta$ are curves and $\lambda$ is a scalar, I'd assume the curvature of $\gamma$ is $\lambda$ times the curvature of $\beta$.
Oh, not like $\gamma(t)$ vs. $\gamma(\lambda t)$.
15:46
@anakhro with definition of it in the generic case, I would say yes.
@gian in any case, you can work out the answer to your question just by using the definition of curvature.
@anakhro $\frac{\partial^2 s f(x)}{\partial x^2} = s f''(x)$
The curvature is given by what formula?
One can also use the kissing circles defintion for curvature, that would result in the same.
by geometric intuition
(reciprocal radius of that)
$\frac{\vert T'(t)\vert }{\vert \gamma'(t)}$
15:54
@gian are you using parameterization by arclength?
Is it true that $Aut(G \times H) \simeq Aut(G) \times Aut(H)$, where $G$ and $H$ are groups?
Doesn't specify... although if I did I guess I could remove the denominator
@user193319 No, take $G = H = \Bbb Z_2$
Will do. Thanks!
@anakhro Oh coolio
I'd like to read that
15:57
@gian good to always be aware of if you are parameterized by arclength. Just to be able to check the right formula. However, from what I remember, you are using the correct one (technically for both cases).
@BalarkaSen you can read it when I am done.
I was planning to pick up some symplectic geometry once again
Maybe this weekend I'll start
@anakhro "with definition ": I dropped the word "my" ... so my definition would be the second derivative, for $\Bbb \to \Bbb$ functions. For other cases the reciprocal radius of the kissing circle is better. But also in that case for curves in 2D I think curvature scales linearly. For higher dimensional surfaces its a different question.
How scales the radius of a kissing ball of a 2D surface in 3D if you scale the lengths? I dunno probably differently than linear?
@gian the definition with the cross product is also effective for this problem.
@Rudi_Birnbaum I am wanting him to work it out for himself with a formal proof.
Surely you can understand why that is important for his mathematical well-being.
2
@BalarkaSen the moment you understand symplectic, you understand contact, and vice versa!
Problem: I don't understand either. ;)
 
2 hours later…
17:51
Hi guys, running this by you to make sure I'm not crazy. I am asked to prove --
$A$ is $n \times n$. Prove $\mathbb{C}^n = R(A) \oplus N(A)$. But in general this is not true. Consider $A(x,y) = (y, 0)$. Then $R(A) = \{(x, 0)| x \in \mathbb{C}\} = N(A);$ where The sum is not even direct, and $R(A) + N(A) = \{(x, 0)| x \in \mathbb{C}\} \ne \mathbb{C}^2$
being careful, $A$ is linear -- $A(x, y) + A(a, b) = (y,0)+(b,0) = (y+b, 0) = A(x+a, y+b)$; and $c \cdot A(x,y) = c \cdot (y,0) = (cy, 0) = A(cx, cy)$
A necessary condition as far as I remember is $R(A^k) = R(A^{k+1})$ for the least such $k$, in which case $\mathbb{C}^n = R(A^k) \oplus N(A^k)$
18:46
@JoeShmo: No, it's always true. And, in fact, they're orthogonal complements (either with respect to real inner product or with respect to Hermitian inner product).
You're messing yourself up by writing vectors as rows instead of columns.
Bad, bad JoeShmo.
So your matrix is $\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$. Now check.
@anakhro sorry to disturb you, I read that you were doing contact geometry. I am currently having this course on thermodynamics and I can't help but notice the things like Legendre transformations that appear in classical mechanics as well as in stat mech. Thinking about how a Legendre transformation links a tangent bundle to the cotangent bundle and the symplectic structure of classical mechanics, I was also trying to find something similar in thermodynamics.
My instructor tells me that it has something to do with contact geometry. Do you have any notion of such a similarity?
@TedShifrin are you still teaching the AoPS at UCSD?
and does the class require any prerequisites
tfw you look down and realize you've been wearing your shirt inside out all day
vzn
vzn
19:05
@Semiclassical lol sometimes not a huge problem unless you meet someone eg job interview... reminds me of having shirt collar messed up too :P
yeah
it makes me feel a little silly but really idfc
vzn
vzn
havent seen you in Physics in awhile what are you up to these days
yeah. haven't felt like it
working on a paper unrelated to my thesis work
not planning to talk about it for the time being tho
vzn
vzn
huh whats the topic? (or do you mean not talk about it in here?)
not talking about it here, yeah
not till there's a preprint
vzn
vzn
19:12
ok. physics?
yeah
my part of it is mostly math tho
vzn
vzn
sounds promising :)
19:48
@Rick Nothing to do with UCSD. AoPS is their own program for kids from 2nd grade through high school. I am teaching a calculus class, which has precalculus knowledge as a prerequisite; all of the students in the class are taking AP calculus in high school simultaneously, so I'm trying to make it a bit deeper.
So, I have a group generated by $a,b$ where $a^4=b^4=1$ and the subgroup $\langle a^2,b\rangle$ is the dihedral group on the square. As a guess, is this, by chance, the group of rotations on the octahedron?
20:36
@Rithaniel, do you know what the group of symmetries of the octahedron is? Or do you mean octagon, maybe?
I'm specifically referring to rotations, not general symmetries. Though, I imagine that there could be obvious things I'm missing and that the "group of rotations on an octahedron" isn't applicable here. (The reason I mentioned it is that when I noticed I had the dihedral group as a subgroup, I immediately interpreted $a$ as a "half reflection")
The group of proper symmetries (all rotations) of the octahedron is a group of order 24. I don't follow you.
So you have relations involving $a$ and $b$. What are they? Just $a^2b = b^{-1}a^2$?
Well, a few I have are that $a^4=b^4=1,\: a^2b=b^{-1}a^2,\: b^2a=ab^2$
(Switched to the symbols I was using in my notes there at the end)
Oh, so there's another relation.
At any rate, isn't this a group of order $16$?
I don't know for sure. How'd you arrive at 16?
20:46
Every element is a word $a^jb^k$ for $0\le j,k\le 3$.
(I'm looking at a subgroup of $S_8$. Specifically $\langle (1234),(1537)(2648)\rangle$. I got all these relations from scratch work.)
I don't remember how many isomorphism classes of nonabelian groups of order $16$ there are.
Ohh ...
What is your goal here?
Well, to work with a concrete example to try to develop intuition for more general cases.
You're trying to understand an arbitrary subgroup of $S_8$?
@TedShifrin today I realized that $F^\ast \mathrm d = \mathrm d F^\ast$ is harder to prove than I thought
20:52
You should talk to Mathein and Tobias and others ... probably representation theory is going to be the way to go.
@Leaky: To do it brute force is very tricky (it took my roommate and me about a week when we were taking graduate differential geometry). However, it's easy to prove.
very interesting
Yes, and I'm trying to predict what the behavior of the subgroup will be as I go, and checking to see if it lines up with what the subgroup actually ends up being.
Can you prove it for $0$-forms? :) @Leaky
@Rithaniel, $S_8$ is extremely complicated. Even doing this for $A_5$ and $S_5$ would be challenging, except that there there's some geometry going on.
But you should chat with the algebra guys, not me.
@TedShifrin it still needs a large amount of unfolding stuffs
Alright, well thank you for the engagement. It always helps.
20:57
@Leaky: You should have a 2- or 3-word answer to me for the case of $0$-forms.
"(the) chain rule"?
Right.
yes that's the essence of it, but proving that it actually works still needs quite a bit of unfolding
well, depending on definitions, perhaps
20:58
What is $d(F^*g)$?
$\mathrm d(F^\ast g) = d(g \circ F) = \partial_i (g \circ F) \mathrm dx^i$
OK, now it shouldn't be hard to see that that is $F^*(dg) = F^*(g_\alpha dy^\alpha)$.
$F^\ast \mathrm dg_pv = (\mathrm dg)_{F(p)} DF_p v$
Don't evaluate on vectors.
Work with forms.
Beginners always want to evaluate too much.
well I am a beginner
21:02
Well, I said you have bad tendencies.
What is $F^*(g_\alpha dy^\alpha)$?
$F^\ast \mathrm dg = F^\ast (g_j \mathrm dy^j) = F^\ast(g_j) F^\ast(\mathrm dy^j) = (g_j \circ F) F^\ast(\mathrm dy^j)$
(pullback commutes with wedge, 0, 1)
Do NOT use the same indices for different spaces.
I used $i$ before
That will totally mess you up. $i$ and $j$ are adjacent and should be in the same range of numbers.
Use a different enough letter to help keep track ... often you have multiple indices and you need a good way to keep track.
$F^\ast \mathrm dg = F^\ast (g_a \mathrm dy^a) = F^\ast(g_a) F^\ast(\mathrm dy^a) = (g_a\circ F) F^\ast(\mathrm dy^a)$
21:04
OK, and what is $F^*dy^\alpha$?
a headache (at least for me)
Answer: It's $dF^\alpha$.
You can write it out in terms of $dx^i$, of course.
am I supposed to prove that without evaluating on vectors?
I would, yes.
Of course, it's a baby case of what we're trying to prove. $F^*y^\alpha = F^\alpha$.
you would prove that without evaluating or you would evaluate on vectors?
21:06
Without.
interesting
Hi @Ted
hi @Balarka
@BalarkaSen! :P
21:07
It's the basic premise that if $f(x)=y$, then $f^*dy = df$ for scalar-valued functions.
It's not hard to check $df^* = f^*d$ basis-wise either, but Ted's trick (prove it for 0-forms, then upgrade) is easier.
As I commented to Leaky, a friend and I spent a solid week of grad school trying to do it bare-hands. There were some complicated lemmas involved.
Of course, I have no recollection of details and threw everything out when I retired.
Gives the rest of us mortals some hope. I got badly stuck in some integration-over-fiber computations a few weeks ago
Kept miscomputing
I remember that Spivak (who taught the course) made a big deal about equality of mixed partials and called it "IT" (because it was the key to all of diff geo, hence "That's IT"). My recollection is that we had something we called Super-IT that we needed for this proof.
Preparing the de Rham complex is very complicated
All the manipulation with forms are
21:12
Well, @Balarka, when McCrory and I wrote our beautiful little paper on cusps of Gauss maps on surfaces, there was an intricate Chern class computation that we each did something like 12 times, and we had 23 different answers. I would have given a kingdom for Mathematica or something back then ... It would have been so much easier than messing up high school algebra with zillions of variables.
And I'm pretty good at computing :P
Or I was.
That's when you write "the reader can easily verify that this computation gives the desired result"
Just so many places to make mistakes.
@Alessandro: Well, we had to get the desired result twice first :P
Ultimately, it was fine.
If you're confident that you will get it if the calculation is done properly you don't need to do it!
21:14
Hey everyone!
We were trying to prove a theorem, @Alessandro. We didn't know the answer a priori.
hi Demonark
The nerd arrives
tfw exposed
Now I can't remember which computation it was in that paper ...
I had forgotten how much neat stuff is in that paper. Including an application of the Lefschetz FPT.
21:17
Hi chat
this place is full of nerds
Bye chat
Salut @Astyx
Au revoir @Astyx
hi Eric.
Ça va bien ?
21:17
Oui, plus ou moins, merci. Je suis déjà encore vieux :P
J'ai mes examens la semaine prochaine
Ah, bonne chance!
Bonne chance
Sniped
a @Balarka: There are bunches of things in that paper you'd like :P
Merci !
21:19
$F^\ast \mathrm dy^a_p [v]_x = \mathrm dy^a_{F(p)} DF_p[v]_x = \mathrm dy^a_{F(p)} [D(y \circ F \circ x^{-1})(v)]_y = e^a D(y \circ F \circ x^{-1})(v) = D(y^a \circ F \circ x^{-1})(v)$
$\mathrm dF^\ast y^a_p [v]_x = \mathrm d(y^a \circ F)_p [v]_x = (D(- \circ x^{-1})(v)) (y^a \circ F) = D(y^a \circ F \circ x^{-1})(v)$
(sorry Ted)
@Eric btw next week we're starting Morse theory
I'm not looking at it, @Leaky. You should be able to remove all $v$'s.
@Daminark - .... .- - / .. ... / -. .. -.-. .
too bad ill barely be around for the next 4 weeks
id have loved to sit in on those
Visits?
21:20
Visitation trips, @Eric?
Cool.
Your name will be MUD if you don't go meet Rafe and tell him who you are.
Well, I'll send you the psets then and whatever source he's using
i have to fly like every week
@TedShifrin lmao ill go meet him
He mentioned Milnor at the beginning of the quarter but during the midterm he was reading a different one so idk
21:21
You might remind him that you know me ...
@MatheinBoulomenos hi
hi @LeakyNun
The computation I was messing up was when I was trying to recollect the proof that if $f_t : M \to N$ are homotopic then $f_0^* - f_1^* = dP + Pd$ because my batchmate, who's reading forms, asked me what the intuition for $P$ is. There is a slick proof using Cartan's magic formula, by looking at the 1-parameter family of forms $\omega_t = f_t^* \eta$, which is obtained from taking Lie derivative along $\partial/\partial t$ in $M \times I$ of $F^* \omega$ (where $F$ is the homotopy map).
one of the problems in our mod rep pset involves computing the number and dimension of the simple modules of $K[G]$ where $K = \overline{\Bbb F_2}$ and $G = S_5$
rip me @MatheinBoulomenos
21:24
In which case $P$ becomes, like, $\int \iota_{\partial/\partial t}$
Right. But the chain homotopy is just integration over the fiber.
As you just said.
Right, that's what I was trying to explain
I usually comment (especially in the grad course) that this is adjoint to $\times [0,1]$ on singular/simplicial homology.
You can then test it out on $\sigma\times [0,1]$ by Stokes's Theorem.
I don't remember if you and I have discussed that.
is there any value to the theorem $H^k_c(M) = H^{k+1}_c(M \times \Bbb R)$?
Mike and I get annoyed when people ask for "values" of theorems like that.
21:26
@LeakyNun Yes. It's a special case of Thom isomorphism.
Balarka's answer is the fancy answer, yes.
@TedShifrin Vaguely, I think. I get what you are saying though.
is it easy to show that every compact manifold has a (finite) good cover?
I should draw the diagram and check all that someday.
Yes, you should, a @Balarka :P
With a Riemannian metric it's easy.
Convexity of geodesic balls.
21:28
@LeakyNun Compact smooth manifold, but no. You need geodesic charts
Ted sniped me but so it is
You can also use Morse theory :)
I don't know why my lecturer seems to have skipped that part
It's a technical detail you can believe for now.
I kept wondering if I'm stupid
and also how difficult is partition of unity?
21:29
Everyone should go through partitions of unity once (or more times).
Not difficult.
I taught it in my YouTube lectures.
It's not that hard.
I saw your youtube lectures, you constructed some by hand on R^n
You need an explicit construction using normality of manifolds.
But I did the general argument, I think, by covering with half-balls of parametrizations ...
You just need to construct bump functions on R^n, then set it 0 outside the compact support in the chart
21:30
balls of half radius, I mean
It's worth working out carefully, starting with $e^{-1/x}$.
Agreed.
but to ensure that they add up to 1?
Chapter 1.1. exercise in Guillemin Pollack
@LeakyNun Scale
that's the easiest part lol
oh...
21:31
@Balarka: Do you believe this? Because the torus in $\Bbb R^3$ is very non-flat, does it feel to you like he's got the right notion of constant torsion? I need to work that out.
He'd be right on a (flat) cylinder.
i.e. given any positive $\rho_i \in C^\infty(M)$ I can basically... $\rho_i / \sum \rho_i$?
Right, because the sum is locally finite.
...
I want to hide in a hole
Again, you can't reuse the same index.
Different reason here.
@TedShifrin That's a good point
21:33
throws dirt on top of Leaky in the hole
Which one is a good point? :P
Non-flatness of the torus making existence of such a helix slightly unbelievable
I don't think that can happen
Maybe the component of the torsion along the surface?
I'm not sure, as I just posted there, what the OP means by a helix.
Presumably he means a curve making a constant angle with the meridian circles? (Not the parallel circles ... you and I always have problems here.)
God, which ones are parallel and meridian again? Meridians are the short circles to me, but long circles to you?
21:38
No, we agree on short circles.
Profile curves is what I use to be non-ambiguous.
Got it.
The ones that are actually parallel are parallels. :P
Varying radiii ...
The meridians are geodesics which do not hit each other! They are parallel!
Joking, I know what you mean. Not constant distance in the metric
@TedShifrin That'd make sense
I wonder if I can even write down a parametrization for a helix on the torus ....
Too hard for my soul
21:43
I can do it on the flat torus in $\Bbb R^4$ easily enough.
Hmm, I have a few spare hours. I'll see ....
Maybe you can write down coordinates tangential to the meridian and the parallel curves, and use Euler's formula since it makes constant angle with the meridians
Well, I have the obvious parametrization with $\theta,\phi$ as usual. How do you want me to use Euler?
Oh, for normal curvature?
But how does that help with curvature and torsion?
I suppose it doesn't.
21:48
So I have to go around the meridian circle at constant rate and move from one to the "next" in such a way as to keep constant angle.
Does the Villarceau circle satisfy this?
Whoa. Certainly doesn't look helical :P
It's a (1, 1)-knot so that was my first guess
Probably not though
Almost surely not (knot).
Apparently the angle with the parallel curves is constant. They are loxodromes in the torus.
That's obvious because the plane cuts them all at the same angle.
21:56
Well, I guess since the two sets of circles on the torus are orthogonal, that's not surprising: constant angle with one set means constant angle with the other set.
I'm calculating, though, how quickly you have to turn horizontally if you move at constant speed around the meridian circles.
Let me know if you have a general formula
@TedShifrin what do you mean by $F^a$?
I shall take leave now, it's 3:30 AM in the morning
As I surmised, the turning rate depends on where you are along the meridian circle. You need $(2+\cos t)\theta'(t) = \text{constant}$.
Yikes, gnight, @Balarka. I'll update you. Won't be around for 4 days or so.
Aha
Oh, where are you going?
21:59
Going on a holiday with my bf for a few days.
Enjoy!
@Leaky: $F=(F^1,\dots,F^k)$. I mean components.
Thanks, a @Balarka!
Actually I'll smoke some more before sleeping. But cya, all!
$F^\ast \mathrm dg = F^\ast ((\partial_a g) \mathrm dy^a) = ((\partial_a g) \circ F) (F^\ast \mathrm dy^a) = ((\partial_a g) \circ F) \mathrm dF^a = ((\partial_a g) \circ F) (\partial_i F^a) \mathrm dx^i$
$\mathrm d(g \circ F) = \partial_i (g \circ F) \mathrm dx^i = D(g \circ F)(\partial_i) \mathrm dx^i = (Dg)_{F(p)} (DF(\partial_i)) \mathrm dx^i$
idk they look the same
I don't really know what to do without eating a vector
You didn't write out $D(g\circ F)$ explicitly to match up with the first line.
$D(g\circ F) = (g\circ F)_i dx^i = (g_\alpha\circ F) F^\alpha_i dx^i$.
22:15
conclusion: I don't know enough about smooth functions (maybe because they're not linear)
your last step should be the chain rule
how would you write $F=(F^1, \cdots, F^k)$ in einstein notation?
22:54
Oh, $F = F^\alpha e_\alpha$ (where the $e_\alpha$ are the standard basis vectors).
Ok so $(Dg)_{F(p)}(DF(\partial_i)) \mathrm dx^i = (Dg)_{F(p)}(DF^a(\partial_i) e_a) \mathrm dx^i = (Dg)_{F(p)}((\partial_i F^a) e_a) \mathrm dx^i = (\partial_i F^a) (\partial_a g)_{F(p)} \mathrm dx^i$
bingo
LOL, so you've done the proof for $0$-forms. Now figure out why you're done. :)
yeah I've seen that part
Oh, cool.
but now I still need to show that $F^\ast \mathrm dy^a = \mathrm dF^a$ without evaluating at vectors
00:00 - 23:0023:00 - 00:00

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