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12:36 AM
@BalarkaSen Earlier you mentioned that $\pi_6S^3$ has a $\mathbb{Z}_3$ component. Just wanted to point out that I was reading through Lecture Notes in Algebraic Topology by Kirk and Davis just now and in Corollary 10.13 it's stated that "If $p$ is a prime, the $p$-primary component of $\pi_iS^3$ is zero if $3 < i < 2p$ and is $\mathbb{Z}_p$ if $i = 2p$" which seems to be a stronger result
 
 
2 hours later…
2:14 AM
@Perturbative Yeah a little more work will lead you to that I think.
 
@BalarkaSen sup
 
Let's see. Take the generator $S^3 \to K(\Bbb Z, 3)$ of $\pi_3 K(\Bbb Z, 3)$ and let $F$ be the homotopy fiber, then running the homotopy LES on $F \to S^3 \to K(\Bbb Z, 3)$ gives $\pi_i F \cong \pi_i S^3$ for $i > 3$ and $\pi_i F = 0$ for $i \leq 3$. Take the homotopy fiber of $F \to S^3$ in turn, let's call it $F'$. Then running the homotopy LES on $F' \to F \to S^3$ we see that $\pi_i F' = 0$ for $i > 3$, $\pi_3 F' = 0$, $\pi_2 F' = \Bbb Z$ and $\pi_1 F' = 0$. So $F' = K(\Bbb Z, 2)$.
So run the Serre spectral sequence on $K(\Bbb Z, 2) \to F \to S^3$.
$E^2_{p, q} = H^p(S^3; H^q(K(\Bbb Z, 2))$, which is nonzero only for $p = 0$ and $p = 3$. Just two of those columns full of $H^* K(\Bbb Z, 2)$.
No nonzero differential at the $E^2$ page but there's a good differential at the $E^3$ page, namely $d^3 : E^3_{0, 2} \to E^3_{3, 0}$ and after that no nonzero differential comes out/goes in from those positions, so this guy has to be an isomorphism otherwise $E^\infty$ page will have some nonzero term in $(0, 2)$ or $(3, 0)$ but that'd force $H^2 F$ or $H^3 F$ to be nonzero, which is not possible by Hurewicz + $\pi_2 F = \pi_3 F = 0$.
This is a differential $H^0(S^3; H^2(K(\Bbb Z, 2)) \to H^3(S^3; H^0(K(\Bbb Z, 2))$. Explicitly write the cohomology rings $H^* S^3 = \Bbb Z[x]/(x^2)$ where $|x| = 3$ and $H^* K(\Bbb Z, 2) = \Bbb Z[y]$ where $|y| = 2$. So $d^3(1 \otimes y) = x \otimes 1$. The strategy is to use the product structure in the spectral sequence now
$d^3 : H^0(S^3; H^{2k}(K(\Bbb Z, 2)) \to H^3(S^3; H^{2(k-1)}(K(\Bbb Z, 2))$ is given by $d^3(1 \otimes y^k) = k x \otimes y^{k-1}$ by Leibniz rule. So at the group level $d^3 : E^3_{0, 2k} \to E^3_{3, 2(k-1)}$ is multiplication by $k$.
We know the kernel and cokernel of that! So the spectral sequence degenerates at $E^4$ and in the $E^\infty$ page we have $\Bbb Z/k\Bbb Z$ arranged along the even terms in $3$rd row and nothing else. So $H^{2k+1} F = \Bbb Z/k\Bbb Z$ and $H^{2k} F = 0$.
Go back to homology so that it reads $H_{2k} F = \Bbb Z/k\Bbb Z$ and $H_n F = 0$ if $n$ is odd. Now to compute homotopy groups of $S^3$ out of this...
By induction assume the $p$-primary component of $\pi_i S^3$ is zero for $3 < i \leq k < 2p$ (this is easily checked with $i = 4$ since $\pi_4 S^3 = \Bbb Z_2$ and by hypothesis $p$ is at least $3$). Now $\pi_i S^3 = \pi_i F$, so by Hurewicz theorem mod C (where C is the class of p-groups), $\pi_{i+1} F \cong H_{i+1} F$ mod C.
If $i < 2p$, then $H_{i+1} F$ is either $0$ or $\Bbb Z/n \Bbb Z$ for some $n < p$, so $H_{i+1} F$ has $p$-primary component zero as well. That means $\pi_{i+1} F$ has $p$-primary component zero!
Finally, $\pi_{2p} S^3 = \pi_{2p} F = H_{2p} F = \Bbb Z/p\Bbb Z$ (mod C), i.e., $\pi_{2p} F$ has $p$-primary component $\Bbb Z/p\Bbb Z$. So this proves your assertion :)
Hi @Ryan
 
3:01 AM
Thanks for that proof @BalarkaSen :)
 
It's more for my own practice than anything haha but glad it's helpful
I suppose what one can do is instead of playing of Hurewicz mod C games at the end is localize the spaces involved at the prime ideal $(p)$ from the very onset. I don't know if there's an advantage to that though
 
3:30 AM
Orientations are killing me
I can't even keep it straight what I need for this paper.
 
 
1 hour later…
4:36 AM
@anakhro Isn't that what the parade a few weeks ago was for
 
 
2 hours later…
6:24 AM
 
6:37 AM
and why we're still arguing, to this day
53
Q: "Negative" versus "Minus"

skullpatrolAs a math educator, do you think it is appropriate to insist that students say "negative $0.8$" and not "minus $0.8$" to denote $-0.8$? The so called "textbook answer" regarding this question reads: A number and its opposite are called additive inverses of each other because their sum is zero,...

 
 
2 hours later…
8:10 AM
math.stackexchange.com/questions/3267279/… ; if someone is good in finding the number of zeros, I would appreciate any help
 
 
2 hours later…
@ÍgjøgnumMeg thanks.
@ÍgjøgnumMeg Are you sure that helps with my problem ?
Jose claims my problem could not be solved using standard methods of complex analysis
 
@Kenkar well your question is exactly the same as the one linked, except your polynomial
 
Yeah, but I don't think the same answer would apply.
 
idk man adapt it
 
math, like life, is learning to adapt
 
10:19 AM
@ÍgjøgnumMeg hahah, have you tried yourself ? I've tried using that "technique" which is fairly standard and it didn't help
 
@Kenkar no I'm at work
 
@ÍgjøgnumMeg I see. Well, thanks anyway.
 
Just a suggestion :)
 
:D
I saw another post of this exact problem , just different polynomial , and the guy with the answer proposed some translations on imaginary axis and I don't frankly know what
But it seems to be correct
Just lazy to study that unusual method 2 hours before the exam
And honestly I dont think we are meant to use that method in the exam
 
yeah, 2 hours before is not the time to learn anything new...
 
10:28 AM
definitely yup
 
have something to eat while you review the big picture
 
what does that mean "review the big picture"
 
the "big picture" is an "overview" of the entire course
 
10:45 AM
@RyanUnger Why are you still awake
Why am I still awake
 
um, because the internet doesn't sleep :P
 
oh
well thanks for the suggestion
 
How you get $dx dy = r dr d\theta$ ? Do it work if I incorporate for ellipses?
 
Given a Vector space $V$ , if $\Omega$ is a symplectic form on $V$ and $J \in Aut(V)$, I define the bilinear form $g(v , w) = \Omega(v, Jw)$. If I add the condition that $J^2 = -\mathbb{id}$ does that imply $g$ is symmetric?
I don't think it does, but my book says that $g$ is symmetric. The maximum I was able to prove was that $g( Jv , w) = g(v , Jw)$
 
I got that.
 
11:01 AM
@Albas Who says $\Omega$ has to be $J$-invariant? A priori the symplectic form will have nothing to do with the complex structure.
If it is $J$-invariant, $\Omega(Jv, -w) = \Omega(Jv, J^2w) = \Omega(v, Jw)$ which is the desired symmetry of $g$
 
No they have said that $\Omega$ is $J$-invariant. They are trying to define a Kahler manifold.
 
This is what is colloquially known as a linear Kahler structure, if $g$ is also +ve definite.
@Albas Then it's true, by what I said
 
$$x=r\cos\theta, y=r\sin\theta$$
$$dx=\cos\theta dr - r\sin\theta d\theta, dy=\sin\theta dr + r\cos\theta d\theta$$
$$dx\wedge dy = r\cos^2\theta dr\wedge d\theta -r\sin^2\theta d\theta \wedge dr=rdr\wedge d\theta$$
@AjayMishra What does incorporate for ellipses mean?
 
(I see now you wrote $J \in \text{Aut}(V)$, by which you mean it's a symplectic automorphism)
 
Hello all. If I want to show that a $1$-form $\omega$ defined on $\Bbb R^4$ in Cartesian coordinates is nonvanishing when I pull it back to the $3$-sphere over the natural inclusion map, do I actually need to bother pulling it back with respect to a parametrisation
Say it's just something nice like $\alpha = xdy -ydx + z dt -t dz$, can I just observe that for no $x,y,z,t\in\Bbb R^4$ do all $4$ of $x,-y,z,-t$ vanish
 
11:09 AM
@BalarkaSen In the second bit $\Omega(Jv , J^2w) = \Omega(v, Jw)$ you are using the fact that $\Omega(Jv , Jw) = \Omega(v , w)$ right?
 
Yes, @Albas.
 
Cool cool.
 
@user681391 That won't do. You have to show that the form is nonvanishing restricted to any tangent space of the 3-sphere.
Think of the form $xdx + ydy$ on $\Bbb R^2$. Pulled back to the unit circle it's the zero form (why?).
 
Is there any easy way to prove darboux theorem? The one I have done is quite tedious. It involves taking the derivative with time of this 1-parameter family of differential forms pulled back under a diffeomorphism between two manifolds.
 
That's the Moser trick, I don't think it's tedious at all. That's why it's called a trick!
There are hands-on proofs (eg, Arnold's book has one) which are tedious
 
11:16 AM
But deriving that huge ass formula is tedious.
 
No.
 
Well I did it for a manifold of the form $M \times I$ and then generalized it to any two manifolds $M$ and $M'$
There must be some way to not do all this.
How to use the 1 parameter family in the proof is slick, but deriving that mess is not.
 
I don't understand what's complicated about it. Given any 1-parameter family of $k$-forms $\alpha_t$ you want a flow which takes $\alpha_1$ to $\alpha_0$. That won't be a global vector field in general, it would be a time-dependent one, let's call it $X_t$. So you want to solve for $\phi_t^* \alpha_0 = \alpha_t$ where $\phi_t$ is infinitsimally generated by $X_t$
Differentiate that to get $\mathcal{L}_{X_t} \alpha_0 = \partial_t \alpha_t$
 
@BalarkaSen when you type (why?) in parenthesis I think maybe I'm suppose to see why without pulling back? But when I pull back I get the zero form so it is definitely true
 
This is the starting point. The Moser lemma says that if $\alpha_t$ is a family of symplectic forms in a specific cohomology class, so that $\alpha_t = \alpha_0 + d\beta_t$ for some $\beta_t$, then the solution to $\mathcal{L}_{X_t} \alpha_0 = \partial_t \alpha_t$ is given by uh setting $X_t$ to be the $\alpha_t$-dual of $\partial_t \beta_t$
 
11:22 AM
So good point, I'll try pulling $\alpha$ back
over the parametrisation:
$(u,v,w)\mapsto(\sin^2w\sin^2v\sin^2u,\sin^2w\sin^2v\cos^2u,\sin^2w\cos^2v,\cos^2w)$
 
The way I have seen it done is surely not like that.
 
Typo: The formula should have been $\mathcal{L}_{X_t} \alpha_t = \partial_t \alpha_t$. My differentiation was wrong.
Or at least misleading notations
The last thing I said is a very easy check: Let $X_t$ be such that $\partial_t\beta_t = \alpha_t(X_t, -)$. Then $\mathcal{L}_{X_t} \alpha_t = \iota_{X_t} d\alpha_t + d \iota_{X_t} \alpha_t = d \partial_t \beta_t = \partial_t d\beta_t = \partial_t \alpha_t$
 
This is so much better way to phrase it.
It seems easy now
 
So that indeed is a solution. But why? That was always the problem for me in the Moser trick; conceptually it's not at all clear why it's happening
@Albas I read this version from Eliashberg-Mishachev, they have a crash course in symplectic/contact geometry if you want to check that out
 
Sure. I have seminar to attend right now but I will rant about that trick sometime later.
 
11:30 AM
@user681391 You can sort of see in two ways, one is that if $p = (a, b)$ is a point on the unit circle with (unit) tangent vector $v = (-b, a)$ then if $\omega = xdx + ydy$ then $\omega(p, v) = a(-b) + b(a) = 0$.
This means $\omega$ vanishes on every tangent space to the unit circle, which is the same as saying the pullback is zero
The other is to note that $x^2 + y^2 = 1$, so just "differentiating that", $2xdx + 2ydy = 0$ :P
This is the implicit function theorem at work, if you want
This is the less tedious, without pullbacks, way to do it
 
Oh, right, that's true
So I can define a $3$-frame at every point for the tangent space at each point, given by $T_xS^3=\{v\in\Bbb R^4\mid x\perp v\}$ and check it for that for example
 
Yup
Too many differential forms people in chat today. Scary
 
@user681391 don't mind, I got that.
 
@BalarkaSen Is this not normal?
pun non-intended
 
Not exactly :)
(Pun intended)
 
11:43 AM
Haha that was pretty smooth
Some people feel that puns polarize chat rooms though, half the people like them, and the other half have closed minds
The half with closed minds can just vanish though
 
lmao well our resident expert on differential forms (@TedShifrin) is a local punmeister
you should come by when he's around
 
Is there any reason why you should expect something like darboux theorem for symplectic two forms? Some kind of geometry picture associated with it? I kind of have a feel of why symplectic geometry must be governed by global invariants rather than local because using the symplectic form you can define the volume form which gives you a definition of orientability and in a neighbourhood around a point its always orientable.
I don't think this has got to do anything with locally having the same structure but this was just my thought process behind looking at darboux theorem.
 
Having an $n$-th root of the volume form is much stronger than having a volume form, of course.
@Albas What do you know about nondegenerate skew-symmetric billinear forms on a vector space?
 
Umm you can show that the vector space has to be even dimensional
It will generate an isomorphism between the vector space and its dual
But that's nothing special
 
12:01 PM
Good morning
 
Hullo
 
@Albas What's a normal form for the matrix of such a thing?
Hi @ÍgjøgnumMeg @Rithaniel
 
What've you guys got going today?
 
A whole lotta nothing
 
Heya @Balarka @Rithaniel
Sitting at my desk pretending to work, as usual
looking at commute distances from Schwetzingen to Heidelberg lol
 
12:06 PM
Hey, having nothing going on can be relaxing on occasion.
I'm going to be sorting files and unstapling/stapling things.
 
If I want to write it in a 2 by 2 form then the diagonal will be 0 and the identity matrix on the top right row and negative of the identity matrix on the bottom left, if I am not wrong
That gives you the standard symplectic structure on $\Bbb{R}^{2n}$
 
What's going on in Heidelberg?
 
@Rithaniel I am doing my master's there
 
Oh, excellent
I'm still trying to figure out where I want to persue my Ph.D. How did you select Heidelberg?
 
It has a good reputation, has a wide variety of courses in my area of interest, nice city, and I hate my own country
hahaha
 
12:14 PM
(I suppose I should set my eyes on a masters first, but people at my current college are trying to get me to go for it there.)
 
oh and it has people working in the area I potentially would like to pursue for research
 
@Albas Right, so there's always a basis $(e_1, \cdots, e_n, f_1, \cdots, f_n)$ on the vector space such that the billinear form is $e^1 \wedge f^1 + \cdots + e^n \wedge f^n$
This is the "linear Darboux theorem". A not so hard linear algebra exercise
 
Fair enough. My biggest hurdle right now is figuring out what colleges are around which would be in my area of interest.
 
Just scour google and call places up and stuff
What are you interested in?
And where are you from? lol
 
All I know are the big names, like Berkley.
I'm in America. South Carolina more specifically.
I'm generally interested in algebra, topology, and geometry.
 
12:17 PM
@BalarkaSen Ahh makes sense. Neat.
 
Alright that's not so specific lol
 
Leaning heavily towards algebra, though.
 
If $\omega$ is an almost symplectic form on $M$, then for every point $p \in M$, you can find basis $(e_1, \cdots, e_n, f_1, \cdots, f_n)$ such that $\omega = \sum_i e^i \wedge f^i$. In fancy language $(e_1, \cdots, f_n)$ constitute an abstract section of the frame bundle $\text{Fr}(TM) \to M$ along which $\omega$ is the standard symplectic form.
The condition $d\omega = 0$ is an integrability condition, which says this abstract section of the frame bundle can be chosen to be locally integrable, i.e., comes as derivative of a local parametrization $\phi : U \subset \Bbb R^{2n} \to M$.
 
I've been studying a particular kind of non-associative magma these last couple of months, though, in my free time.
I gather it would be wise to be more specific, then?
 
@BalarkaSen I had thought of that condition as coming from the Hamiltonian vector field(written using the Poisson bracket) and taking the lie derivative of the two form along it using the magic formula, which did give that condition.
 
12:20 PM
Sure, that's one way to intuit it.
 
@BalarkaSen But this is something new, thanks.
 
Or, "it's just the right condition for Darboux to hold"
 
Yea I was going to say that exactly
What are some interesting things people do with symplectic geometry. I am interested in understanding Geometric Quantization but besides that? You can always try to find global invariants but besides that?
Finding global invariants is not a small deal but just asking.
 
@Albas OK, so you might ask, a Riemannian metric on a tangent space is just a positive definite symmetric form, so you can cook up an abstract section of the frame bundle along which the Riemannian metric is just the usual dot product. What's the local integrability condition there? That's the vanishing of the Riemann curvature tensor
 
@Rithaniel I suppose it's good to be a bit more specific when going into a masters because you're really aiming at a PhD at that point (I assume)
 
12:25 PM
But studying flat metrics is just... not that interesting. One way to think about that would be that vanishing of the Riemann curvature tensor is a second order differential restriction on the metric; on a normal coordinates the second order term in the Taylor expansion of the metric is $R$
So what's the first order restriction? Vanishing of the torsion.
The analogue of Darboux for a metric then is existence of normal coordinates
 
Ohhhh
That's nice.
 
There are other examples scattered throughout geometry. If $J$ is an "almost complex structure" on $M$, i.e., $J : TM \to TM$ is a bundle automorphism such that $J^2 = -I$ fiberwise, then the integrabiliy condition for $J$ is the vanishing of the Nijenhuis tensor $J[X, Y] - [JX, Y] - [X, JY] - J[JX, JY] = 0$. If this happens, then $J$ integrates to an actual complex structure on $M$. The analogue of Darboux is existence of complex charts
That's the Newlander-Nirenberg theorem
Just so happens that for symplectic structures $d\omega = 0$ turns out to be a pretty simple integrability condition
 
Yeah. Hmm this does make me wonder. Having this two form imposes this cohomology condition on the manifold that $H^2(M)$ (sorry if this is the wrong notation) is not trivial. Are there similar things in cases such as the complex structure?
 
Oh plenty admitting a complex structure is a strong topological restriction
 
@BalarkaSen Why so?
 
12:34 PM
Well, some people get a masters without then pursuing a Ph.D. afterwards, but that won't be the case for me. So specificity is probably what I need.
Which is difficult because there is so much that I want to know. I don't want to hedge my potential topics.
 
are you interested in applied math?
 
More abstract, to be honest.
 
This person once told me that pure math is something that hasn't been applied till now
 
mathematical physics can be very abstract
 
@Albas Even admitting an almost complex structure is a strong restriction; only spheres which admit almost complex structures are $S^2$ and $S^6$ (I think a proof involves characteristic classes). I also think there are nontrivial cohomological obstructions, since there are natural bigradings on the de Rham complex arising out of holomorphic and anti-holomorphic form which I think relates to Hodge theory?
 
12:40 PM
sup
 
@BalarkaSen I have a strange message from you
 
@BalarkaSen Oh another thing to learn then, the list just keeps on increasing.
 
Heya Ryan
 
It's pretty believable that complex structures are very rigid if you ask me, but I'd be hard pushed to give concrete examples of the topological obstructions, not knowing much about them myself
@RyanUnger Strange message from me?
 
12:41 PM
@Rithaniel hi
@BalarkaSen "why are you still up"
I recieved it while I was not up
but it was not a crazy time in any case
 
oh but ur icon popped up on chat
 
my phone does strange things if I odn't close chat in chrome
I think it refreshes in the background
 
I tend to leave my computer on over the night, though I disconnect the internet
 
I just leave it
@ÉricoMeloSilva
@BalarkaSen someone is talking about Fourier-Mukai
 
whats that
 
12:46 PM
Something the Langlands people are on about
idk what they're doing at a geometry conference
 
o shit lmao
 
I mean real geometry
like geometry with real numbers
f'(x)
lo yeah I need that talk
someone asked me for advice on getting into Princeton
 
lmao
 
I sent him my CV like what more can I do
it's all luck
 
Luck never made a man wise.
 
12:51 PM
@BalarkaSen your friend and I were sitting together trying to dechpyer Sasha Logunov's lectures
little success
 
what was it on
 
Yau's conjecture on the area of Laplace nodal sets
 
urk
 
he gave the proof for analytic metrics
it was incomprehensible but somehow inspiring
I asked Sasha to email me the secret notes he got the proof from
secret as in they used to be on Fanhua Lin's webpage but have disappeared
 
Is it true @ryan that Langland's office is suppose to be Einstein's old office?
 
12:54 PM
I haven't been to the IAS yet
it's not on campus
 
let me know, if you find out
 
1:15 PM
 
1:36 PM
Woo Woooo
 
I'm trying to see why $$dx^1\wedge dx^2+dx^3\wedge dx^4+\dots + dx^{2n-1}\wedge dx^{2n}$$ is indecomposable, so I took arbitrary $1$-forms $\alpha = \sum_{i=1}^{2n} a^idx^i$ and $\beta = \sum_{i=1}^{2n}b^idx^i$ and then took $$\alpha\wedge \beta =\sum_{1\leq i<j\leq 2n} (a^ib^j-a^jb^i)dx^i\wedge dx^j$$ which gives me lots of equalities, but I can't see how to deduce that these are all impossible to satisfy
 
2:22 PM
@user681391 If a 2-form $\omega = \alpha \wedge \beta$ is decomposable then $\omega \wedge \omega = 0$ (playing off of the fact that square of a 1-form is zero)
(In fact verify that if $\omega$ is your form then not only $\omega \wedge \omega \neq 0$ but $\omega^{\wedge n} \neq 0$)
 
2:47 PM
$\omega^n = n! dx^1\wedge \dots\wedge dx^n$ I think, since I only need to move even position things to even position things and odd position things to odd position things, which uses an even number of transpositions in each case
Oh right, $\alpha\wedge \beta = (-1)^{deg(\alpha)deg(\beta)}\beta\wedge \alpha$ which means for odd degree $\alpha\wedge\alpha = 0$ but for even degree they commute
Was meant to say $n!dx^1\wedge\dots\wedge dx^{2n}$
Wait no $n!$ was right
 
3:08 PM
How do you deal with differential $1$-forms in the coordinate-free setting?
An arbitrary $1$-form is of the form $\sum_{i=1}^n g_id(f_i)$?
and how do I apply $d$?
$$d(\sum_{i=1}^n g_id(f_i)=\sum_{i=1}^n d(g_id(f_i))=\sum_{i=1}^n (d(g_i)d(f_i)+ g_i d^2(f_i))$$
$$=\sum_{i=1}^n d(g_i)d(f_i)?$$
Using leibniz and linearity
 
 
1 hour later…
4:17 PM
@user681391 There's a horrible coordinate-free definition of $d\omega$ you can find online, but I don't know how it is derived
There's a user Ted that comes in here who can probably help later
 
4:36 PM
Why does my name keep getting tossed around? :D
@user681391 This is correct, except i would insist that you put the wedge product in there. People make sloppy mistakes when they don't.
 
Apparently this room is filled with differential topology/geometry students, and you're very famous :)
 
Ha ha ... There is a lot more geometry than there was when I first started chatting 4 or 5 years ago.
BTW, the coordinate-free expression is in textbooks, not just on-line. And it is highly useful (e.g., for understanding/proving the Frobenius theorem).
 
like this?
$$d(\sum_{i=1}^n g_id(f_i)=\sum_{i=1}^n d(g_id(f_i))=\sum_{i=1}^n (d(g_i)\wedge d(f_i)+ g_i d^2(f_i))$$
$$=\sum_{i=1}^n d(g_i)\wedge d(f_i)?$$
 
Yup :)
You don't need the parentheses, though. Everyone writes $dg_i\wedge df_i$.
 
Apparently $d\omega(X,Y)=X\omega(Y)-Y\omega(X)-\omega([X,Y])$
 
4:42 PM
Yes, that is the highly useful formula to which I was referring.
 
ok
I'm trying to prove it, and I see that everything everywhere is linear, so I think I can just check it for an arbitrary $gdf$
So $\omega = gdf$ and $d(gdf)=d(g)\wedge d(f)$
So we let $X,Y$ be vector fields on our manifold $M$
 
First, check that the right-hand side defines a tensor — i.e., is linear over $C^\infty$ functions. Then it's easy to check by choosing $X=\partial/\partial x^i$ and $Y=\partial/\partial x^j$.
Be careful what you mean by "is linear."
 
Oh, so you prove the coordinate-free even still by fixing coordinates for the time being?
 
Yes.
 
Ohhhhhhhhhhh
No wonder I had no idea how to approach this lol
 
4:45 PM
:)
 
4:59 PM
@BalarkaSen I just saw $\log\log$ appear in a geometric analysis talk
the number theorists have invaded
 
Time to implement security measures
 
5:16 PM
@RyanUnger marco is a v nice cool guy
 
@ÉricoMeloSilva I know him
 
cool
i graded for him a couple years ago
 
@ÉricoMeloSilva he was actually here for a talk the day before my 21st
we did shots at midnight
uh, here = back in TN
 
Can someone tell me how a parametrisation of a torus in $\Bbb R^4$ gives an orientation to the torus?
Can you just pullback the standard orientation form from $\Bbb R^4$?
Right, of course
 
5:32 PM
 
For some dumb reason I ruled it out thinking that would always vanish since the dimension of the torus is 2 and $\Bbb R^4$ is $4$-dimensional lol
 
That's correct. You can't do that.
Except in the case of a closed hypersurface, you don't get a canonical orientation (indeed, it might not even be orientable — take the Klein bottle in $\Bbb R^4$) on a submanifold.
 
Yeah I just checked and it did vanish sadly
So it wasn't dumb oops
 
Well, of course. If you have a parametrization by an open subset of $\Bbb R^2$, then the orientation on $\Bbb R^2$ passes to the submanifold.
 
So how does $\Phi(u,v)=(\cos u,\sin u,\cos v, \sin v)$ give an orientation to $T$
 
5:38 PM
You orient the submanifold so that your parametrization is orientation-preserving.
 
Oh
Well I guess I have $\Phi:[0,2\pi)\times [0,2\pi)$ which isn't open
Or I can fail to completely parametrise by excluding $0$ from both?
I'll be honest, I always found the parametrisation coming from $(0,2\pi)$ confusing
@TedShifrin So I can pass the orientation of $\Bbb R^2$ to all but an arc along the torus?
 
Why? We all know that you can parametrize a circle by $[0,2\pi)$ unofficially, and by two charts $(-\pi,\pi)$, $(\pi/2,3\pi/2)$ or whatever, officially.
Yes, but that extends continuously with no problem (unlike what happens if you remove a circle from the Klein bottle and orient what's left).
 
Okay! That's excellent
Although I'm not entirely sure how to actually pass it from $\Bbb R^2$ to $T$
I take the standard orientation form $dx^1\wedge dx^2$ on $\Bbb R^2$
which restricts to $(0,2\pi)\times (0,2\pi)$
 
You can pull back the $2$-form by the inverse of the parametrization, or you can designate what will be the oriented basis of the tangent plane at each point.
 
I wish my professor was as enlightening!
 
5:50 PM
LOL, thanks for the compliment. I guess 40 years of teaching experience helps :P
 
I imagine it does :D
 
Teaching style also changes from person to person. Like, I've had professors who just throw numbers out and professors who describe the closure axiom for groups by saying "If I add two and two, I had better not get a chicken."
 
Yes, lots of us have our pet bits of humo(u)r.
I even had my high-school AoPS kids trained to ask "Self ..." when I said, "You might ask yourself ..." :)
heya @Tobias
 
Hi @TedShifrin
 
@Rithaniel: Generally, you can tell which teachers enjoy that part of their job and which of them think it's something they just have to do.
But sometimes even the ones who enjoy it aren't so effective. :(
How's life in the real world treating you, @Tobias?
 
5:55 PM
Good
 
Indeed. It's always great to find a professor who really enjoys teaching.
 
though my steak lunch with the higher-ups (and the other new employees) got postponed to a later date. I was looking forward to the company paying me both for the lunch and for the time spent eating it.
 
I'm glad to hear it, @Tobias :)
LOL ... Provided good red wine comes with lunch?
 
Though, it also sucks when that professor doesn't get the attention of the class.
 
Not sure about the wine
 
5:57 PM
I'll send you a bottle, @Tobias :)
 
Heh, if they find it appropriate to drink wine during working hour as part of the lunch then I have no doubt it will be included
 
The hell with appropriate :D
 
it is the "appropriate" part I am less certain of :)
 
I seem to be writing down complete nonsense.

So for $(u,v)\in (0,2\pi)\times (0,2\pi)$ we take $$\Phi(u,v)=(\cos u,\sin u, \cos v,\sin v)$$ and then where it's defined $\Phi^{-1}(x,y,z,w)=(\arccos(x),\arcsin(y))$

in which case $(\Phi^{-1})^*(dx^1\wedge dx^2)=d(\arccos(x))\wedge d(\arcsin(y))=-\frac{1}{\sqrt{1-x^2}}dx \wedge \frac{1}{\sqrt{1-y^2}} dy$

where I feel like I'm messing everything up somehow lol
Unless that's right
My main concern is there should be u's and v's lol
Maybe it's better to say $\Phi^{-1}(\cos(u),\sin(u),\cos(v),\sin(v))=(u,v)$ since it's bijective and then proceed as I did, in which case I end up actually having some nice $u,v$ terms and:
$$-\frac{1}{\sqrt{1-u^2}\sqrt{1-v^2}} du\wedge dv$$
Nah that's bad, then $u,v=1$ destroys me
Maybe I want to choose $(u,v)\in (1,2\pi+1)^2$ or something...
 
6:15 PM
I am currently looking through the codebase we are taking over from another company. One thing it needs to do get criminal records for a bunch of people to see if they are eligible for a hunting permit. When it sorts the people into those with and without a criminal record, the people who wrote the original code decided to call the lists badBoys and goodBoys.
 
ha that's cool
 
Haha that's pretty cool
The company's Santa simulator
Wait is it just $du\wedge dv$
If it is, I might just go to have a nap
since my brain has deformation retracted
 
7:02 PM
How can I prove that in field theory, if E/K and F/E are regular extensions, then so is F/K?
 
What's a regular extension?
 
7:18 PM
$E/K$ is regular if $K$ is algebraically closed in $E$, @F.White
 
7:40 PM
@JaakkoSeppälä an extension is regular iff it has no non-trivial finite subextension
 
@F.White This is messed up a bit. You want $(\arccos x,\arccos z)$, and then you just note that $dx\wedge dz$ gives a positive $2$-form. I actually think the right coordinates to use on the torus are $(u,v)$ in the first place. And, of course, $du\wedge dv$ gives the $2$-form. Why do we need ambient coordinates? That's not so natural, but it's fine.
 
Let $B$ be a Banach algebra and let $B^*$ denote its associated continuous dual, which can be normed in a natural way (sup. norm over unit ball in $B$). Is it true that $\phi_n \to \phi$ in $B^*$ if and only if $\phi_n \to \phi$ uniformly on the unit ball in $B$?
 
@user193319 express both conditions in epsilon-delta
 
howdy @Leaky
 
7:55 PM
@TedShifrin this is related to what I was doing yesterday
But he wedges by something that is 0 on the sphere
So I have no idea how he is concluding what he does.
 
Wait, @anakhro. We're trying to find the divergence of $X$ on the sphere?
 
Yes.
Well, just trying to show it is positive on the north hemisphere, and negative on the south.
 
Well, you want to compare $\iota_X(\Omega)$ with $\Omega$ (up to sign).
 
$(\text{div}_\Omega(X))\Omega = d(\iota_X\Omega) = 2\,dx\wedge dy$ as we calculated yesterday
 
I don't like what he's doing with no explanation, but he's comparing the wedge product of normal $1$-form (which annihilates the sphere) and the area $2$-form. (Forget about $\iota_X(\Omega)$ for now.) Think about $(x\,dx + \dots)\wedge\Omega$. Is this $+$ the standard volume form or $-$ the standard volume form?
 
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