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5:00 AM
$\Bbb{R}^0$ is just a point?
 
Yeah
Sniped
 
well, I don't know how one can even define tangent vectors of isolated points
 
360
@AkivaWeinberger thanks for the paper. it's a fun read
 
@Secret it's 0 dimensional stuff so gonna be one of those things where you define it by convention so that your theorems work nicely
(e.g. degree of 0 polynomial being negative infinity)
 
5:04 AM
I guess it's a vector in a 0-dimensional vector space
 
yeah, 0 dimensions are weird..
(The polynomial case kinda make sense though, since there is no well defined powers of x in the zero polynomial and defining its degree that way will ensure its degree stays unchanged when you differentiate it)
 
$\{0\}$
Oh, my convention was always that $\deg(0)=\pi$
 
Lmao
 
Heh, apparently this sort of thing is only possible for 1, 2, 4, or 8 squares
(product of the sum of N squares being the sum of N squares)
 
(Actually might play with this idea later in mathworks: Define an alternate polynomial like algebra such that it has the property that $p \neq 0 : deg (\frac{d p}{dx}) = deg (p)$)
 
5:07 AM
It's moot anyway since every number is the sum of four squares, and thus the sum of N squares for all $N\ge4$ (fill it up with copies of $0^2$)
but I guess they still work in more general settings like rings and whatnot
 
Well, one example use of the negative infinity convention is that it is now true that $deg(fg) = deg(f) + deg(g)$ without having to specify that it's non-zero
etc etc
Also hello @Ted!
 
There's a similar thing with 16 squares but you need division; it expresses it as the sum of 16 squares of rationals
Hey, Ted! We were talking about how $\sin$ was analytic (someone was unsure), and then discussed the radius of convergence of $\dfrac1{e^x+1}$ which then managed to get segued into a discussion about the quaternions and normed division algebras in general
 
OEIS don't have finite sequence, thought I can just lookup how many times 1,2,4,8 appeared in mathematics
 
There's that one which goes 1,2,4,8,16,31
about cutting up a circle
 
Alerting the drama here
What are we doing?
 
5:12 AM
Found it
 
Balarka: Akiva give a nice summary on what happened in the past hour
 
And then Secret summed up for Balarka how Akiva summed up the last hour to Ted
(I won't continue)
 
And then repeat the above $\omega$ times :P
 
Re the 1,2,4,8 thing, that's a nice puzzle. I can't remember why it does that, though
 
@Akiva @Daminark Small subtlety: $TS^2$ is not isomorphic to $S^2 \times \Bbb R^2$ as a vector bundle over $S^2$ because of hairy ball. There's a little bit of trickery (We Are Number One tune starts) involved in proving that they are not homeomorphic as manifolds.
 
5:15 AM
There's a proof for the normed finite division algebra case (which I never get to look in detail), but for the sum of squares identity, I have no idea
 
@BalarkaSen I mentioned that
'round here
Or do you mean to say that that's not the only reason?
 
Well that's the reasoning to prove they are not isomorphic as vector bundles.
 
@Secret Oh, no, I meant the image
with the circle divisions
 
@Balarka is it easier to prove that they're not diffeomorphic?
 
You have $n$ points on a circle, join them together with lines such that no three are concurrent, as in the image, and ask how many regions you get
 
5:17 AM
Yes, mildly.
It's hairy ball theorem, truly, but a little fuckery is required. $TS^2$ has a submanifold with self-intersection number 2
$S^2 \times \Bbb R^2$ has no such thing
 
@BalarkaSen Hm. I guess it's not too hard to triangulate $TS^2$
Do the homology groups agree?
 
They do :)
I mean both are homotopy equivalent to S^2
Contract the plane fibers
 
Oh, right
 
What does not agree is the cup product structure
 
Doesn't that only depend on homotopy type?
 
5:19 AM
And not even that. It's the compactly supported cup product structure
@AkivaWeinberger Not if you think compactly supported. Consider the Moebius strip and the actual strip
If $\alpha$ is the center circle,
 
Minus boundary (so they're not compact)?
 
Right
$\alpha \cup \alpha$ (compactly supported cup product) is nonzero in the first and zero in the latter
Because perturbing the center circle a little bit makes it intersect back with itself
 
Makes sense
Would this be different for the compact versions (with the boundaries)?
 
I gotta go
Sorry
 
Give me a yes or no, at least!
Well then.
But yeah I guess in $S^2\times\Bbb R^2$, you can shift your canonical copy of $S^2$ in there "up" a bit so it doesn't intersect itself anymore
but for the canonical copy of $S^2$ in $TS^2$, doing so would solve the hairy ball problem.
@BalarkaSen I'm gon' ping you so you can answer my question when you get back and/or wake up
 
5:33 AM
Howdy, DogAteMy
Thanks for the hello, Demonark.
I just figured out I can make a video with my doc cam or even live stream. Amazing, technology. Still, pretty boring to just write on a piece of paper and drone on — I preferred having live students to interact with.
Ah, I see the difficult question came up about whether non-isomorphic vector bundles can be homeomorphic (diffeomorphic) as manifolds.
And on cue MikeM shows up.
 
Doc cam?
 
Document camera ... your teachers probably use those in their classrooms occasionally.
 
hi @TedShifrin
 
They don't
 
hi Karim
 
5:37 AM
but suddenly I want one
 
ROFL
 
I think my master thesis will be like 150 pages @TedShifrin
:D
I am happy
 
How interactive are your classes?, most math lectures I been to often have professors using doc cams alot and wrote stuff on, and some even just decided to play youtube videos of their prerecorded lectures
 
Even I used one occasionally in college teaching, DogAteMy. Projecting a book up on the screen or something from someone's piece of paper.
 
Because, you know, how else will I post videos to YouTube of me drawing on a piece of paper?
Other than making something out of cardboard to prop my phone on, I guess
 
5:39 AM
@TedShifrin if everyone had your teaching abilities then I would like attending classes
 
Secret: I had a lot of interaction with students. And in certain classes I tried to get them to tell me how to do proofs, make up examples, etc. Particularly did that in algebra and point-set topology.
 
With advanced math I noticed that people can't really teach it
unless they are talented in teaching like you @TedShifrin :)
 
I imagine it's a practice-makes perfect sort of thing
 
I have known a surprising number of excellent teachers, Karim, but also some horrid ones.
 
but if you don't get a lot of feedback it's hard to improve
 
5:40 AM
The geometry kids in the AoPS class I substituted for were extremely interactive (a bunch of 12-13 yr olds, I think).
 
Maybe it's easier to, like, "train" yourself by tutoring one person about something, before trying to do it in front of a class
 
yeah
Also, with advanced math you need to introduce all ideas
 
DogAteMy, our IT fellow back at UGA tried to talk me into using a regular webcam with a whiteboard on an easel or something, but I think I made the right decision.
 
because missing 1 key idea can ruin people focus in class.
 
Ted: That's good, cause often the students have to just do it in order to understand some of the proofs, kinda like in fine arts where one just have to do it before they will get what will happen and why it is the case.

Abstract agebraic proofs are often quite abstract for most students thus guiding them should help them to learn better on what kind of thinking process to be used and proof strategies, I think...
 
5:41 AM
especially with like advanced things
 
One-on-one tutoring is surprisingly different from teaching, DogAteMy. But it gives you some insights, if you do enough, so that you know what students will find difficult or misunderstand.
 
What you want to do is have everything relevant on the board so that they need to use their memory as little as possible
 
Our set-up in the UGA classrooms was terrible. When I used Mathematica, displaying it on the screen, the screen covered almost all the blackboard. Terrible design. I bitched.
 
Same reason I think there's no point in trying to do hard multiplication problems in your head if you have paper available
 
@TedShifrin did you know that there is connections between algebraic curves and differential equations ?
that is kinda very weird !!
 
5:43 AM
So, DogAteMy, we gonna have our own exploring differential topology YouTube channel? :P
 
At one point I wanted to make a thing on Brouwer's fixed point theorem, which never happened
 
Semiclassic can give you a lecture on that, Karim. Also the whole notion of monodromy ties together both algebraic varieties and differential equations with singularities.
 
I mean, it still can, it just hasn't happened yet
 
Akiva: Some problems are simply way too hard to do mentally, for me, if a problem have more than three inequalities, then I have to do it on paper to even get my bearings
 
@TedShifrin very cool
 
5:44 AM
DogAteMy: Here's a problem I'm going to give my AoPS kids tomorrow. You might like it.
 
@Semiclassical maybe you can tell me something about this ?
 
I think part of the reason 3Blue1Brown so good is that he generally leaves important stuff on screen
 
Take a regular $n$-gon inscribed in the unit circle. What is the product of the lengths of the segments from one vertex to the $n-1$ remaining vertices?
 
and draws your attention to the parts of the screen that's relevant at a particular moment
 
I miss blackboard teaching.
 
5:45 AM
@TedShifrin That's a classic, isn't it?
 
Oh, is it?
I put it in my algebra book, but I don't remember where I found it (or if I found it at all).
 
It's $\prod|1-\omega|$ over the roots of unity.
 
Well, of course. :)
We're doing roots of unity tomorrow.
 
And $\prod(x-\omega)$ is $x^{n-1}+\dotsb+1$.
 
And no, I'm not teaching them Galois theory.
 
5:46 AM
So then you get $n$?
 
Aha. That's the easiest way to do it, yes. There are other ways :)
 
And you can check for an equilateral triangle ($\sqrt3\times\sqrt3$) and for a square ($\sqrt2\times2\times\sqrt2$) to be sure
 
Another way is to let $u=1-\omega$, so $u$ satisfies a polynomial and you want the product of its roots.
 
What's that end up being, $(1-x)^n-1$?
 
Nope.
 
Oh, that divided by $x$?
 
OK.
 
@orbit-stabilizer What about it?
 
@orbit: Crazy cuz he's been through medical school? Interesting he went to UCSD and majored in math. I wonder if @PVAL knew him.
 
seal team 3, math undergrad, MD from harvard, now he's becoming an astronaut
he's only 33
 
5:50 AM
Honestly I thought there was only one SEAL team
and we just called it 6 so that the enemy would think there's more of them
 
Dunno, not american.
 
Oh, I missed that he enlisted right out of high school before he went to college 6 years later.
 
But, damn, he is incredibly driven. And he has a wife and kid.
 
> He will report for duty in August 2017.
I see that updating this website was not anyone's priority
 
and he's a time traveler!?!
 
5:53 AM
Lol
 
stop the simulation.
now it's just too obvious
 
@TedShifrin Drat
 
what is the best way of putting an arrow with on the top of an arrow name of a function
like f denoting function from X to Y ?
 
So, it came across my Facebook feed (and I've shared it) that if you call the White House feedback # to speak with someone, they give a biased diatribe about how the Democrats shut down the government and so no one can take your call.
\vec, Karim
although for things that are bigger you need \overrightarrow ... can always write a macro.
 
5:55 AM
$A\xrightarrow fB$
\xrightarrow
 
nice thanks
 
@TedShifrin Little letter on an arrow, not the other way around
 
Oh, I misread.
No, you misread.
He wants something like $\vec f\colon X\to Y$.
Or did he mean $X\overset{f}\to Y$?
 
ha
$\xrightarrow{\vec\to}$ONWARD
 
5:58 AM
His sentence was garbled. I quit.
 
FedE$\to$x
 
Karim: You should probably learn about using some CD (commutative diagram) stuff, too, although it doesn't do non-horizontal/vertical lines.
OK, I'm going back to Australian Open tennis.
Bye.
 
Watching? Participating? Insulting?
 
Yes.
 
Have fun.
 
6:00 AM
Night :)
 
@Ted I don't think he was on the research track.
He looks vaguely familiar.
 
oh yeah @TedShifrin
I use XY pic
nights btw @TedShifrin
 
Morning] @TedShifrin
 
Let's all ping him as many times in a row as we can
 
Diagram of an ordinal sardine:
 
6:09 AM
i think thats the wierdest thing ive seen so far today...
congrats @Secret
 
perfect fusion of biology and set theory
 
6:23 AM
Does power set exist for an uncountable set?
Is the power set of an uncountable set a sigma algebra?
 
power sets is defined for any set
and the resulting set is strictly of larger cardinality than the set itself
 
@Yashas Sigma algebra means that it's closed under countable unions and intersections, as well as complements, right?
And the empty set and full set are in there
The power set of any set is a sigma algebra, yes.
It's closed under arbitrary unions and intersections, which means it's definitely closed under countable unions and intersections.
 
I am back
 
YAY
1 hour ago, by Akiva Weinberger
Would this be different for the compact versions (with the boundaries)?
 
Well, you need to use an awkward notion of cup product again. The self-intersection number of the center circle does not change (it's 2 and 0 respectively)
 
6:34 AM
I suppose at some point I should sit down with a Hatcher, a pencil, and a piece of paper and just work it out by hand
A Hatcher, a pencil, a piece of paper, and a hand
 
I need to work out a calculation explicitly
:(
Differential forms are too hard
 
6:49 AM
@Akiva Btw, I think just one-point compactifying and thinking about cup product on that compact space is enough
 
The one-point compactifiction of $X\times\Bbb R$ for any $X$ sounds annoying
 
Nah, it's $X \times [0, 1]/X \times \{0, 1\}$
 
It's like $SX$ but you glue the two cone points together
 
For compact $X$
 
the smash product of the one-point compactification of X with the one-point compactification of R?
 
6:50 AM
I am pretty sure
 
Yeah, sure, I just don't like the idea of a double cone point.
Although, actually, this is $S^2\times\Bbb R^2$, not $S^2\times\Bbb R$.
Random side note: Apparently $\rm\Bbb OP^2$ exists but $\rm\Bbb OP^3$ doesn't?
 
@AkivaWeinberger Should just be homotopy equivalent to $\Sigma X \vee S^1$ by "thinning away" at the cone point
 
And $\rm\Bbb OP^1$ as well, though I think that's just $S^8$
@BalarkaSen Hm, yeah
 
Why? $\rm\Bbb CP^1$ is $S^2$, it seems to be the same dimension as the field it's on
 
6:55 AM
@anon Would that be "smash product of X with the one point compactification of R" if X is compact? That cannot be right; consider X = S^1
 
wait nvm
 
What's the one-point compactification of $S^2\times\Bbb R^2$? It sounds like you get a higher-dimensional cone point
 
Do actual topology please.
 
$S^2\times S^2/S^2\times\{0\}$?
 
That's it.
 
6:57 AM
Which I can't visualize well enough to homotope it to something nicer
 
Whereas the one point compactification of $TS^2$ is $S^2 \times S^2$ mod the diagonal copy of $S^2$
I think
 
(like you can for the other thing being the wedge of a thing and a circle)
And I guess the diagonal of $(S^2)^2$ sits differently in the space than a horizontal section does?
 
Very differently.
 
In a way that's not true for $(S^1)^2$ (torus) or $(S^3)^2$
 
The diagonal has self-intersection number 2
@Akiva Right
 

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