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00:00 - 21:0021:00 - 00:00

12:00 AM
But sir, I count two variables in the equation, actually?
We're thinking of a family of curves. The parameter tells you which curve in the family we're looking at.
You actually should count three, not two.
I see, thanks.
How goes the physical therapy professor @TedShifrin?
12:17 AM
Let $F\colon M\rightarrow N$ be a diffeomorphism of smooth manifolds, $\omega\in\Omega^k(M)$. I want to define $dF(\omega)\colon N\rightarrow\Lambda^n(T^{\ast}N)$ for each $q\in N$ via $dF(\omega)\vert_q(v_1,...,v_k)=\omega\vert_{F^{-1}(q)}(dF^{-1}\vert_q(v_1),...,dF^{-1}\vert_q(v_k))$; this is an alternating $k$-linear form on $T_qN$.
If $X_1,...,X_k$ are smooth vector fields on $N$, then $dF(\omega)(X_1,...,X_k)=\omega(dF^{-1}\circ X_1\circ F,...,dF^{-1}\circ X_k\circ F)\circ F^{-1}$ is smooth, so $dF(\omega)\in\Omega^k(N)$. I feel like this is what the pushforward of a differential form
1:00 AM
You're pulling back by $F^{-1}$, since $F$ is a diffeomorphism. Otherwise you can't do it.
No one ever writes this sort of thing, imho.
Does anyone know why Wikipedia started showing a "tap to display image" message on its articles?
I haven't seen that ...
oh, I'm indeed just pulling back by $F^{-1}$
for some reason I didn't notice that lol
Perhaps it's my device that isn't loading them?
You didn't notice all those $dF^{-1}_q$'s?
1:06 AM
I was initially writing $dF\vert_{F^{-1}(q)}^{-1}$
but yeah, in hindsight it's clear that pulling back by the inverse is the right thing to do
cause of course that's inverse to pulling back to $F$...
so I was just being silly
But, yeah, you always pull back forms. That's why they're called covariant tensors :P
I only know covariant functors
Well, things that pull back are normally contravariant. This mis-nomenclature is the fault of the physicists who looked at how coordinates transform and named it first.
ah, that makes sense
or rather, it makes sense that that doesn't make sense
I'm glad we're agreed nonsensically.
1:19 AM
Q: Is CURED healthy?

Mr.MathoThis post is a follow up post of Is CRUDE healthy?. CURED is the chatroom for closing, reopening, undeleting, deleting, editing posts on MSE. After observing the chatroom for quite some time, I have unfortunately seen that most of the problems highlighted by user samerivertwice are still contin...

What do you think guys?
I don't follow much of the politics here. However, I am currently annoyed by the number of people who post questions and then delete them after someone (e.g., I) has given significant input. They don't know the etiquette that dictates you don't just delete once you know how to do it.
Hi again @robjohn
@TedShifrin Lissarob?
Rob au jus becomes Lissarob
yeah, I've seen quite a few people just delete the entire body of their question after it has been answered
1:29 AM
Sounds like overly competitive students.
what's the purpose? are they cheating and trying to hide it?
@Thorgott I have undeleted those kinds of things.
sure sucks that there's an edit history in that case
@Thorgott yep
yeah, I agree with that course of action
1:30 AM
Hide the evidence
@Thorgott I hate it because others spend time giving answers and they delete their cheating question
There was a sort of interesting question yesterday where the person just deleted. Here it is. I think all such people should be flagged and censured. (Unless it's a question with multiple answers on MSE.)
yeah, it's disrespectful and completely goes against the spirit of the site
I told one guy, and his excuse was that he had figured it out and he didn't understand what I was telling him. (I.e., he doesn't actually know what the Transversality Theorem says and needed an alternative solution.)
We should have a button for flagging once someone deletes.
Cheating has gotten rampant, so no doubt trying to "delete" the evidence is partial explanation.
Online education will make it worse.
1:35 AM
More incentives
Not "will." We're already in the middle of it.
Come September we'll see :-)
It'll make the AOL eternal September look like spring break.
All else staying under "control," of course
The Clarinetist has the right idea, in that there should be more oral exams.
@TedShifrin often times a lot of questions are deleted by CURED just because they are too simple in their eyes
1:50 AM
@TedShifrin I agree that must be quite annoying
@robjohn Can $9a^3$ ever a perfect cube?
@TedShifrin here is a vector plot of a conservative vector field orthogonal to $y=2+x^3$
@Knight nope
@robjohn How did you conclude something from that image? What is it showing ?
@Knight See my comment to Ted about it
Which comment sir?
2:05 AM
@robjohn So, $9a^3$ can never be a perfect cube ? But sir pardon my ignorance, I’m unable to understand how does that image proves it.
Just give me a little hint and I’ll go all the way through it
@Knight That image has nothing to do with your question
it has to do with the comment that Ted made that I referenced
Okay, doesn’t for $a =3 b^6$ we have $9a^3$ as a perfect cube?
But we really want $a$ to be an integer
@Knight I assumed you were talking about integers, or at least rational numbers.
For $9a^3$ to be a perfect cube, we would need $9^{1/3}a$ to be an integer, which would require $9^{1/3}$ to be rational.
@robjohn But if $a$ is of the form $3b^2$ then $9a^3$ will be $27b^6$ and hence a perfect cube.
2:16 AM
$9^{1/3}$ is an algebraic integer since it is the root of $x^3-9$ and the only way an algebraic integer is rational is if it is an integer. We know that $9^{1/3}$ is not an integer.
@Knight check your math. that is wrong
@robjohn That's easy. The issue is that he wanted it orthogonal to both curves simultaneously. (P.S. I know how to do it, whence my hint at the end.)
Oh yeah!
And now in this case it can never be a perfect cube
Sir, when some writes “ $n \in \mathbb{N_0}$” what does that 0 subscript denote ?
@TedShifrin hey Ted!
2:22 AM
@TedShifrin here is one perpendicular to both $\left(6x^5,4-2y\right)$
@robjohn: OK. I did it analytically, of course.
hi @Stan
There I added the analytic solution
@robjohn: Your graphic solution seems to be very zero at the origin, whereas that isn't.
@TedShifrin That's not the origin... that's $y=2$ and $x=0$
@math4ever Well, there is an appalling number of repetitive "do my homework" going on here. But I'm talking about self-deletions.
2:26 AM
Is Functional Analysis more a tool to use for research in PDEs, or is there active research going on in the subject itself?
@TedShifrin the people in that robotics group are really good. I think i'll learn a lot this summer :)
@robjohn: Sorry. I forgot the stupid $2$. I thought that because the two curves were tangent at the point $(0,2)$ we could make the vector field non-vanishing.
I guess I should check my formula other than in my head.
I'm sure, @Stan.
@Joanna: There is active research, but asking in here is not the way to investigate it sensibly.
@TedShifrin I see -- yeah, true ...
@robjohn Sir what does $\mathbb N_0$ mean? Does it mean set of natural numbers along with zero?
@TedShifrin Should I cold-email professors who work in Functional Analysis?
2:28 AM
@Knight Might be the naturals including $0$
You should do literature searches yourself.
@Knight I'm guessing
@robjohn Thanks.
You should only bug professors, in general, if you're going to try to work with them.
I'm self-studying FA, and it seems like a lot of fun -- and a very natural extension to my previous studies @TedShifrin.
@TedShifrin I see ...
2:30 AM
@robjohn: I can give one with nonzero $y$-coordinate everywhere, can't I? :D
@TedShifrin Must my literature search be only on papers from recent years? And only from the top journals?
I'm not here to give requirements. I'm just suggesting that you do some looking. You can do so much on the internet these days.
@TedShifrin Well, the product of the two potentials dies to order 2, so the gradient vanishes. I haven't looked at your hint. Is there a way around that?
I think I have it
My hint isn't stoooopid.
Of course, the OP (who deleted) never specified nowhere-vanishing.
Interesting that his exercise (presumably) wanted conservative.
no... I don't
2:39 AM
My solution is $C^1$, but not smooth.
@TedShifrin If not conservative, then it might be easier
2:57 AM
@TedShifrin $\left(\frac{3x^5}{\sqrt{\left|x^6-(2-y)^2\right|}},\frac{2-y}{\sqrt{\left|x^6-(2-y)^2\right|}}\right)$
Not continuous at $(0,2)$
@robjohn: Maybe I'm being stooopid. But I'm taking the vector field $(\text{sgn}(2-y)3x|x|,1)$.
3:12 AM
@TedShifrin I was considering one with sgn's and things, but I wasn't getting one that looked continuous. Let me try that one.
@TedShifrin is that conservative?
@TedShifrin not continuous across $y=2$
Oh, I was thinking only about at the point $(0,2)$. Hmm.
So bump function games interfere with conservativeness.
So I wonder if there's no nowhere-zero conservative field that works.
I can define it fine (and $C^1$) along the curve, but any extension is likely to be non-conservative.
3:28 AM
hi Ted and robjohn, what are you discussing?
A (nowhere-vanishing?) conservative vector field orthogonal to the two curves $y=2\pm x^3$. (The $2$ is a red herring.)
. o O ( I have a red herring-bone sweater )
Now that is a red herring.
3:50 AM
@robjohn I am having issues with another moderator. May I have a conversation with you?
@RithikKapoor sure, but it might be better to talk to the Community Managers, since any chat room I create can be seen by any other mod.
I also have no authority over another mod.
I would like to try and resolve this issue with another mod before contacting anyone from SE. Can we move to another room?
4:06 AM
@TedShifrin Why is 2 a red herring? Are there easy examples when there's no 2 there?
4:40 AM
@Pig he means that $y=\pm x^3$ is just as good
oh i see, cool
@TedShifrin: the more I look at it, the more I think that if the field is conservative and continuous, it may need to vanish at $(0,2)$.
5:00 AM
@robjohn: The original question specified nothing about non-vanishing. But we could take the 0 vector field, so I don't like that.
What about $y=0$ and $y=x^2$?
i may be missing something, but what's wrong with a bump function kind of argument?
Conservative force field.
Bump functions don't play nice.
conservative means it's df right?
if you look at the $1$-form, yes.
if you take an open neighborhood of y = x^3 U y = -x^3
and find a smooth function that's constantly 1 on this neighborhood
5:04 AM
The issue is local at the origin, so I don't see how this helps.
But if we allow the vector field to vanish at the intersection of the two curves, it's easy.
If the two curves are $g=0$ and $h=0$, we just take $f=gh$.
The question is: If the two curves are tangent at the point of intersection, is there a way to rig a vector field that does not vanish at the intersection point?
5:46 AM
1 hour later…
7:11 AM
@TedShifrin do I need the GRE general test or the GRE math test?
Good morning.
is there a difference between $\frac{d f(x(t),y(t),z(t))}{dt} $ and $\frac{\partial f(x(t),y(t),z(t))}{\partial t}$ ? i know how the full derivative to t of a skalar function looks like, however i am not so sure what is the partial derivitave in this case?
1 hour later…
8:19 AM
Can someone please give me the links where problems of perfect cubes are solved? I mean questions like these “Prove that $1+4m^2$ can never be a perfect cube”. I want some examples of how to go for problems like these where we are given an expression in some variable and are asked to prove that it cannot be a perfect cube.
1 hour later…
9:46 AM
Good morning: if anyone of you knows by chance some simple convex geometry/toric geometry, I posted some days ago this bountied question which I really like to understand


If any of you can help me, I'd really appreciate. Have a nice day and apologize for the spam ;)
10:38 AM
@LeakyNun unless things changed u need both
11:16 AM
@robjohn Can you please give me a hint for this problem $$\text{Given two expressions}\\ t-6, ~~~~~ 4-\frac{t}{3} ~~~~~~~~~~~~~~t\in\mathbb Z \\ \text{Prove that there doesn’t exist any value of $t$ such that both expressions can be perfect cubes}$$
First, I'd replace $t\mapsto3t$
Then you are showing that $3(t-2),4-t$ are not both cubes.
We could also replace $t\mapsto t+2$
Then you are showing that $3t,2-t$ are not both cubes.
@robjohn t-2 ?
11:22 AM
What we did after getting $3(t-2)$ and $(4-t)$?
We replaced $t\mapsto t+2$
How $4-t$ got transformed?
Can we please use different letters? Like $t \mapsto u+2$
Okay, we finally got to prove $$\text{$3u$ and $2-u$ both cannot be a perfect cube}$$
Suppose that $3u$ is a cube. we must have $u=9a^3$. Then $2-u=2-9a^3=b^3$ means that $9a^3+b^3=2$. Now look at the cubic residues mod $9$.
11:41 AM
@robjohn “cube residues mod 9” means? You mean I should look for remainders ?
Find the contradiction
Remainder can never be 2, is that a contradiction ?
@Knight Yes, because $9a^3+b^3=2$ says that $b^3\equiv2\pmod9$
11:51 AM
@robjohn Thank you so much sir.
3 hours later…
3:11 PM
Hello chat
1 hour later…
4:23 PM
is the boundary of a compact manifold with boundary itself compact?
that's the same as asking whether it's a closed subspace. take a convergent sequence, it eventually lies in some chart and then it can be checked on that chart where it's true since the boundary of the half-space is closed, so should be
show that the complement is open
oh yeah, that sounds good too
same idea essentially, test on a chart, which is a diffeo and sends boundary to points and interior points to interior points
3 hours later…
7:41 PM
Thanks, @robjohn!
@amWhy That was not a favor. That was to stop everyone from commenting there. As the room name said, it was for two people to talk. The rest were like unwanted people showing up at your house and complaining about your hospitality.
@robjohn I understand that. But you need to understand that the user you opened the room with had a lot of baggage, and that when opening a room with someone, that baggage is likely to come with them. Perhaps be more cautious with whom you open a chat,
@amWhy But that user left (perhaps not wholly of their own volition) and was not talking any more. It was the conversation after that the needed deleting.
@amWhy I will open another room to keep issues from being discussed here.
@robjohn okay.
I did not expect a peanut gallery to come there and go wild.
7:51 PM
Are you specifically referring to me?
@Knight no one person
Q: Paracompactness in second countable spaces

topologicalorientablesurfaceIf $X$ is second countable, locally compact hausdorff space then $X$ is paracompact. The first part of the proof is something like: Let $(K_j)$ be an exhaustion of compact sets. Put $A_j=K_{j+1}\backslash intK_j$ and $W_j = intK_{j+2}\backslash K_{j-1}$. Since $A_j$ is is closed in a compact sub...

@robjohn Okay
@Knight No, not at all. Just the publicity the chat robjohn intended for a conversation with one other user. I'm just as much if not more, someone robjohn considers a person from the peanut gallery.
7:54 PM
@robjohn Just be more careful, and more apprised of math.se issues before you act so carelessly in the future. Where's the chatroom you were going to open. Or was that a threat just to get me off your case?
@amWhy It was not careless as far as the original conversation went. Next time, it will be a private room so no one else can participate.
@robjohn You still haven't answered my question. Where the chatroom you were going to open, or was that just a bluff?
@amWhy what did you find threatening? I did not intend any threats.
@amWhy You were in the chatroom I opened.
@robjohn Like you, I'd rather not discuss that here. I do not see the chatroom you opened. Where is this? Don't make me out to be a fool.
@amWhy you were IN the chatroom. I froze it and deleted the comments.
8:01 PM
@robjohn that will define hurt peanut sales.
@amWhy Oh, you're asking what happened to the chat room? It was frozen. Maybe that hides it.
@robjohn No I am asking about this promise.
@skullpatrol Do you have nothing better to do?
@amWhy 'I said that next time [I create a room for a conversation with someone] it will be a private room. That was not a threat.
@robjohn Was this a threat or a promise?
Hi @TedShifrin
8:05 PM
@amWhy It was a statement. It does not apply to conversations about math, just conversations about personal matters.
@robjohn I assumed you meant that you did not want me discussing this matter here, and that you would, (by now have) create another chatroom.
@amWhy No. Sorry you read it that way, but that is not what I meant.
Hi, skull.
Hi, tos.
8:07 PM
@robjohn Then say what you mean and mean what your say. And don't mislead, like you did.
@amWhy I did
@TedShifrin: how goes?
I'm trying to show that every locally compact, second countable, hausdorff space is paracompact
Hi there. Sorry to interrupt. I was wondering how one can determine the induced map on the second homology, for the cover $f:T^2\to T^2$ given by $(z_1,z_2)\mapsto(z_1^{d_1},z_2^{d_2})$ where $d_1,d_2$ are positive integers. One can probably just 'guess' what it is, but I'm not sure how to show it. Perhaps this can be done using cellular homology in an easy way? Or otherwise, which homology should I be thinking about?
Thanks for agreeing with me, @robjohn.
Let $(K_j)$ be an exhaustion of compact sets.
Put $A_j=K_{j+1}\backslash intK_j$ and $W_j = intK_{j+2}\backslash K_{j-1}$. Since $A_j$ is is closed in a compact subspace, it follows that $A_j$ is compact in $X$. Since $W_j$ is the intersection of two open subsets, it is therefore open. Note that $K_j=\varnothing \iff j<1$. Moreover, by construction, $A_j\subseteq W_j$ for each $j$.
But I can't see
8:08 PM
Hi, @robjohn and AmWhy.
why: $X=\bigcup_{j=1}^{\infty}\bigcup_{k=1}^{n_j}V_{x_i}$.
@amWhy I am not being drawn into "where did I mislead?" Because I don't think I did.
@TedShifrin Hi! How are you!
What does that map tell you, user574847?
@amWhy Only because you edited your comment.
8:09 PM
Well, on homotopy, it sends the generating loops to loops that go around $d_1$-times as fast, and $d_2$-times as fast.
@robjohn No, I think you agreed with me, but you are afraid of admitting it! ;D
What does this mean? @amWhy
Like on $\pi_1$ I mean
On $H_2$ though, err
We're talking second homology?
8:10 PM
@skullpatrol Do I know you?
I was commenting on peanut sales @amWhy
To lighten the mood.
@skullpatrol Do I know you?
@TedShifrin I guess I should think of it as wrapping the open $2$-cell around the torus $d_1$-times in one direction, and $d_2$-times in the other direction
But that's not exactly rigorous
Use the ignore button if you don't @amWhy
End of story
Yes, you're counting how many times the fundamental class covers the fundamental class in the image. This is called the degree of the map.
8:12 PM
@skullpatrol Just curious, whether I'm supposed to know you. And copy your response, to you.
That's your choice
I guess I know that $H_2(T^2)\cong \Bbb Z$, so it's orientable, and I can pick a generator $1$ or $-1$, but it's not clear to me what that generator is geometrically
Thinking simplicially? Cellularly?
Which is easiest, for induced maps?
Lets say cellularly, since it's the one I least understand
you want to understand it geometrically ... I don't know what you know in terms of smooth stuff, cohomology, whatever.
8:16 PM
I only know stuff from chapter 1 and 2 of hatcher
I was hoping to take a chain map from $0\to \Bbb Z\to \Bbb Z^2\to \Bbb Z\to 0$ to itself, induced by my covering map above
Hatcher talks about degree, doesn't he?
He does, yeah
What's the degree of this map?
The number of sheets, which isn't immediately clear to me (but as a guess it's $d_1d_2$)
How many times does any point in the image get hit?
8:20 PM
Ahh true, it is $d_1d_2$ by that logic
Okay so you can immediately deduce it's $d_1d_2$ if you know the degree is the number of sheets
Yes. And thinking simplicially you can prove it.
Triangulate so that every simplex is evenly covered by simplices.
I'll have to think about that triangulation argument more
How do you see the fundamental class, i.e., the generator of $H_2$, simplicislly?
Well, cheating, I'd deform $\Delta^2$ to a square, and fit it onto that standard square that defines the torus
Probably a really bad answer though, so I'll try to think of something better :P
Try to answer this so it works in general.
8:28 PM
Oh wait, I'm thinking singularly, rather than simplicially I guess
The generator is given by a formal sum of two $2$-simplicies
We triangulate the space like you suggest I guess, and take a formal sum of the top-dim simplicies to get the fundamental class (assuming it's orientable)
Right, compatibly orienting those $2$-simplices.
OK, now you should have your proof.
In the case of your problem, the covering map is orientation-preserving ...
Right, this looks more clear I think
Since I can in particular pay attention to the edges of the square that I triangulated (while I apply my map)
Am I correct, that if I apply $(2,3)$ for example, that I end up getting a triangulation of the square that defines $T^2$, by $12$ little triangles, that are $6$ copies of my original triangulation?
Well, you're not talking about a triangulation, first of all. Second of all, do you get open covering? I imagine much smaller triangles than this.
But you don't need to do it concretely.
Oh right, that's not an open cover
Oops. I swear I typed even covering.
8:41 PM
Do this for a circle first --- see if you can construct a CW structure on $S^1$ (I'll call it $\Delta_d$) with exactly $d$ vertices and $d$ edges so that the map $z^d$ is a cellular map $(S^1, \Delta_d) \to (S^1, \Delta_1)$. Then you can calculate explicitly what it does in homology
Then do it for a torus by taking a product of models like this
Okay, doing so now
whats cookin
Howdy, A.
i cant fucken sleep
hi ted
8:45 PM
@MikeMiller You just take $d$ $0$-cells and join them pairwise to make a nice chain. For that $z^d$ to make sense, you think of them as being placed at the $d$-th roots of unity, and you think of the cellular map as just 'going along $\Delta_d$ at speed d'
Which gives a d-fold cover of $\Delta_1$
in Calculus and analysis, 2 days ago, by Simple
Suppose $f(z)$ is holomorphic in a punctured disc $D_r(z_0) − \{z_0\}$. Suppose also that $|f(z)| \leq A|z − z_0| ^{−1+\epsilon}$ for some $\epsilon> 0$, and all $z$ near $z_0$. Show that the singularity of $f$ at $z_0$ is removable.
You get $0\to \Bbb Z^d\stackrel{0}\to \Bbb Z^d\to 0$ as the cellular complex and $0\to \Bbb Z\stackrel{0}\to\Bbb Z\to 0$ for the other one
stucked on this problem
noone has any idea on the problem I mentioned above?
8:50 PM
Hello chat
Apparently Chern classes of a rank $n$ bundle $E/M$ can be read off from the cohomology ring $H^*(\Bbb P(E))$. By Leray-Hirsch, as a module over $H^*(M)$ it's $H^*(M)/(x^n)$ where $|x| = 2$ is the fiberwise hyperplane class ($O(1)$), so $x^n = \sum_{i = 0}^{n-1} c_i(E) x^i$ for some coefficients $c_i(E)$. These are Chern classes.
Err, wrong map for the first complex above, but can't edit. I mean $0\to \Bbb Z^d\to \Bbb Z^d\to 0$, where in the image I have every pair $v_i-v_{i-1}$, so that it has image isomorphic to $\Bbb Z^{d-1}$ as one would expect
I'm messing up some index but OK
I meant $x^n + \sum_{i = 0}^{n-1} c_{n-i}(E) x^i = 0$
Ah, okay, I think I get it, thanks Mike. So for $T^2$ I have CW-structure whose open $i$-cells are the products of the open $i$-cells in $S^1$ and $S^1$. Also this way sort of can be thought of as given a triangulation in the proper sense i guess
Yeah you have to turn those squares into two triangles but it works. I suggested the cellular approach because squares are more natural for products like this
8:57 PM
@Simple Why are you stuck? What are you trying?
@BalarkaSen I think you just need to check this for the universal example
It's simple enough to prove
seems random tho
Says Chern classes measure how far the cohomology ring is from being a product, yeah?
I mean OK I can compute $H^*(\Bbb P(E))$ now but seems backward to define Chern classes using this
Yeah, which is a strange way of understanding it
Apparently Grothendieck came up with this
It's useful for computations in Chow rings I think. Strange as a definition
8:59 PM
Thats what my AG roommate told me
also nutcase btw
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