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0celo7
0celo7
10:00 PM
the ratio test where $r=1$
 
Balarka Sen
Balarka Sen
if $P$ proves $Q$ is wrong, it proves "not $Q$" is right.
 
0celo7
0celo7
no, not that one...
erm...
Yeah, no, the ratio test
if you fail the ratio test you diverge, but you can pass it you can still converge or diverge
that's my impression of mod 2 theory
 
Balarka Sen
Balarka Sen
weird philosophy.
but ponder on
i have to flee now
 
0celo7
0celo7
why
 
Balarka Sen
Balarka Sen
it's dead of the night
 
0celo7
0celo7
10:34 PM
@BalarkaSen I think you can get the two components by picking inward and outward pointing sections.
@BalarkaSen I'm trying to imagine what happens if the Mobius band were the normal bundle to the circle.
I need to construct a Mobius band.
this thing is freaky
@BalarkaSen ...what the hell
I can get on the "other side" of the zero section circle without crosssing the circle...
@BalarkaSen I'm so confused
It would SEEM that if I cut out the zero section it's connected
but that makes literally zero sense.
if I cut out this line it will fall apart.
 
Balarka Sen
Balarka Sen
11:00 PM
@0celo7 That's the idea.
@0celo7 Normal bundle inside what? R^2, right?
@0celo7 Hmm? Cutting out the zero section in Mobius bundle does give me a connected thing.
 
0celo7
0celo7
@BalarkaSen well I know I can't embed it in R2 so this is probably wrong
@BalarkaSen HOW
I made one out of a paper I had laying around
 
Balarka Sen
Balarka Sen
What do you mean by how?
 
0celo7
0celo7
If I cut out the zero section it will fall apart
 
Balarka Sen
Balarka Sen
What?
 
0celo7
0celo7
where will it be connected
 
Balarka Sen
Balarka Sen
11:02 PM
Are you sure you made a meobius band?
With a half twist, not a full twist?
 
0celo7
0celo7
tries to make a full twist
er, I made a half twist
 
Balarka Sen
Balarka Sen
Then that's nonsense. How did you cut it?
Along the center circle, or cut a nbhd of the center circle out?
 
0celo7
0celo7
the center circle
I cannot fathom it staying together
 
Balarka Sen
Balarka Sen
You get a disconnected thing, you say?
 
0celo7
0celo7
I have not cut it because I do not have tape or scissors
at this location
 
Balarka Sen
Balarka Sen
11:05 PM
Oh. lol.
 
0celo7
0celo7
I know that it's supposed to be connected
but I would bet my life against it
I cannot fathom it being connected after the cut
wait, does it become a larger Mobius band?
 
Balarka Sen
Balarka Sen
It becomes a larger Moebius band with a full twist, yes.
Larger as in longer
 
0celo7
0celo7
@BalarkaSen Oh, then if $NX-X$ is always connected I see a proof.
But I don't see where I need TNT
and I don't know how to prove that it's connected
 
Balarka Sen
Balarka Sen
You don't need tubular nbhd theorem to prove that.
Um, a proof of what?
 
0celo7
0celo7
@BalarkaSen You said I did.
@BalarkaSen $NX-X$ is connected if it's not trivial.
 
Balarka Sen
Balarka Sen
11:09 PM
@0celo7 In different context. You can prove $NX - X$ is connected if it's nontrivial without TNT, but how do you propose to prove Jordan-Brouwer for compact hypersurfaces in $\Bbb R^n$ from it?
 
0celo7
0celo7
@BalarkaSen I want to prove the $NX-X$ thing first.
 
Balarka Sen
Balarka Sen
Well, do it.
 
0celo7
0celo7
I don't know how...
Assuming $NX\not\approx X\times\Bbb R$...
 
Balarka Sen
Balarka Sen
You're allowed to give $NX$ a Riemannian metric.
 
0celo7
0celo7
Oh god, what?
 
Balarka Sen
Balarka Sen
11:12 PM
You know what a Riemannian metric on a bundle is, differential geometer.
 
0celo7
0celo7
Yes I do
But I do not see how that helps!
Something geodesic?
 
Balarka Sen
Balarka Sen
You won't, not immediately.
No geodesic business, nothing more than a metric is needed.
I feel nauseated and shivery. Grmph.
 
0celo7
0celo7
Hopefully not Nash embedding?
 
Balarka Sen
Balarka Sen
No lol
I don't know differential geometry beyond what a metric is, and I did it (after considerable thinking, of course). Maybe that helps you to understand how much is needed to prove it.
 
0celo7
0celo7
I don't know what one does with a metric and without a connection.
Morse theory?
Gradient flow?
 
Balarka Sen
Balarka Sen
11:16 PM
Do I look like I know any of that?
 
0celo7
0celo7
You're a sentient green square, what kind of question is that
 
Balarka Sen
Balarka Sen
My laptop's charging out, I may be kicked outta my computer soon.
 
0celo7
0celo7
@BalarkaSen You didn't use the metric to put the metric topology on $NX$ either?
 
Balarka Sen
Balarka Sen
Nah
 
0celo7
0celo7
(which is the normal one, but maybe something falls out if you do that)
 
Balarka Sen
Balarka Sen
11:18 PM
Don't worry about the metric. Think about the picture, the metric will show its use in due course
 
0celo7
0celo7
the only picture I have is the Mobius band
and it's clear that it happens because of the "twist"
 
Balarka Sen
Balarka Sen
Good picture. Work with it.
 
0celo7
0celo7
I'm hung up on why you need a metric
Lengths of vectors I guess
 
Balarka Sen
Balarka Sen
All you have to do is to prove $NX - X$ is disconnected implies $NX$ is trivial.
@0celo7 Good guess.
OK, any time now my laptop will kick me out
 
0celo7
0celo7
HMMMM
You need the metric to get the length of vectors in $\Bbb R$
along each fiber
that's how you get you diffeomorphism
got it
wait, no
@BalarkaSen Ok, if it is disconnected...
then you can diffeomorph one part to $X\times\Bbb R^+$
and the other to $X\times\Bbb R^-$
 
Balarka Sen
Balarka Sen
11:23 PM
Go on.
 
0celo7
0celo7
and the technical details of that diffeomorphism involve a metric
and then $X$ should fit right in the middle, so you get $NX=X\times\Bbb R$
basically, you use the metric to map the fiber on each half to either $\Bbb R^+$ or $\Bbb R^-$.
 
Balarka Sen
Balarka Sen
Correct
On the right track. Can be simplified a bit perhaps, but right so far
 
0celo7
0celo7
How do you simplify it?
 
Balarka Sen
Balarka Sen
$S^0$ sits inside $\Bbb R$.
 
0celo7
0celo7
...hmm...sphere bundle, eh?
 
Balarka Sen
Balarka Sen
11:26 PM
nod.
 
0celo7
0celo7
and you need a metric to define that
ah
and then you take your curve between points on the sphere bundle
 
Balarka Sen
Balarka Sen
nods.
 
0celo7
0celo7
and homotope it through the zero section
and change the intersection number
I don't see how the sphere bundle thing is any easier to prove...oh no wait I do
Ok
Is that basically it?
 
Balarka Sen
Balarka Sen
I don't understand what you just said
 
0celo7
0celo7
I don't either, I don't see how the sphere bundle thing is easier to prove than triviality
I'm not even sure what you want to prove with the sphere bundle
 
Balarka Sen
Balarka Sen
11:28 PM
That's why you think.
 
0celo7
0celo7
@BalarkaSen I have my solution
 
Balarka Sen
Balarka Sen
OK?
 
0celo7
0celo7
although I'm not sure I even need a metric.
at least given GP's definition of the normal bundle.
 
Balarka Sen
Balarka Sen
(I may shut off any second now)
 
0celo7
0celo7
@BalarkaSen ok then explain what your sphere bundle thing is
quickly
 
Balarka Sen
Balarka Sen
11:30 PM
no, figure it out
 
0celo7
0celo7
figure what out
I'm not sure what you're trying to do
 
Balarka Sen
Balarka Sen
Figure out what I was trying to do.
:)
 
0celo7
0celo7
do you not need to prove that $NX-X$ is connected?
 
Balarka Sen
Balarka Sen
I will prove $NX - X$ is disconnected implies $NX$ is trivial, yes.
 
0celo7
0celo7
Using the sphere bundle...
wat
 
Balarka Sen
Balarka Sen
11:32 PM
Think about your only existing example of a normal bundle.
Well, line bundle.
 
0celo7
0celo7
Is the $S^0$ bundle connected as well?
 
Balarka Sen
Balarka Sen
Hypothesis is that $NX - X$ is disconnected, not connected.
"$NX - X$ disconnected iff $NX$ is trivial"
One direction is trivial, so don't worry about that
 
0celo7
0celo7
I agree.
Well...find the S0 bundle, then take the span of the unit vectors
Nope, no clue what you're triyng to do
My way seems simpler
@BalarkaSen please give a hint
 
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