7:00 PM
Any people with number theory/arithmetic geometry background around?

Hey Ted

Otherwise typing out my question would take too long

Right. This isn't bad, since when you repeat the ideal sums of products of elements of the ideal is just the same as ...

Hey Krijn, Felix

Hi DogAteMy
Hi @Tobias

7:01 PM
@TedShifrin Hi
Got the student evaluations today

Oh yeah?

hmm if someone can suggest a number theory book?

@AkivaWeinberger Hey

What kind of number theory, @Petar?
Elementary stuff?

7:01 PM
@TedShifrin Not sure how to finish the sentence...Would it just be an element in $(r^3)$? I'm somewhat confused.

@TedShifrin No, the evaluation is done before the course is fully complete, so that there is time to tell the students about the changes I plan to make based on the evaluations

@user193319: I do believe so. Check it out.
Ohhh ... very different system, @Tobias.
So, what did they say?
I think students appreciate responsiveness to feedback, if their feedback is reasonable ;)

@TedShifrin Generally decent scores (I think). The comments were mostly constructive, which was nice

How can I define a trivial connection on a trivial bundle?

They gave me feedback on my socks today: cool

7:03 PM
Just differentiate as usual, using the trivialization, @quallenjäger.
You were wearing socks, Krijn? I'm shocked!

@TedShifrin i am new to that field

Do you know abstract algebra, @Petar?

@TedShifrin Not just any socks, cool socks

The TAs got amazing scores and great comments, which was good to see, given that the structure of the exercise sessions was not very well liked

Well, they didn't blame the TAs for the department's structure. That's enlightened of them.

7:04 PM
@TedShifrin Yeah. A bit surprising, but a good kind of surprise

@TedShifrin I have read that one can define $\nabla_X e_i=0$ for all $X\inT_pM$ and $e_i$ from smooth section of the vector bundle
@TedShifrin Why can't I do that if the bundle is non trivial?

No, that's wrong, @quallenjäger.

Well, not really the department's structure. The structure of the session was planned by myself and the TAs

Oh. Then they should blame you, @Tobias :P

What did you do to the exercise class

7:06 PM
@quallenjäger: Consider the trivial bundle $\Bbb R^n\times\Bbb R$ on $\Bbb R^n$. You're saying that you can say that any real-valued function has zero derivative.

@TedShifrin hmm i am not sure since we use other terms here but lets say i dont

@Krijn We changed it so that fewer exercises would be done at the board

Do you know about groups and rings, @Petar?

In fact, we cut it all the way down to one exercise per session

Hmm, @Tobias. And what else happened?

7:07 PM
Ah, yes, I see that they would dislike that

The problem with working homework problems is that the students who haven't made a good effort don't learn much, but they think that by copying solutions they've mastered the course.

@TedShifrin This was me, in my undergraduate studies

I used to criticize students a bit who'd come to my office hours without even having read or attempted the exercises.

@TedShifrin Well, the way the sessions used to go was that the students would get some amount of time to do the exercises (because they would invariably not have done them beforehand), then the problems would be done at the board, either by one of the students who had done it or more commonly by the TA

@TedShifrin i have some knowledge in it

7:09 PM
@TedShifrin Right, that deficiency of the board exercises was why we changed it. So now they would be split into smaller groups depending on how well prepared they were and be told to discuss the exercises in those groups with guidance from the TA

OK, @Petar. A classic book is Hardy and Wright. Beautiful stuff in it, but a bit old-fashioned. A more modern book is by Joe Silverman.

Okay, just read some complex analysis for the first time. It says something, perhaps not-rigorously at this point; 'if $f$ and $g$ are holomorphic functions in $\Omega$ which are equal in an arbitrarily small disc in $\Omega$, then $f=g$ everywhere in $\Omega$.

This sounds something like, take a sheaf $\mathcal{O}_X$ of holomorphic functions. Then if $f,g\in \mathcal{O}_X(U)$ have germ $f_x=g_x\in O_{X,x}$, then $f=g$. Is this correct?

Probably an intelligent change, Tobias, despite their whining.

@Narcissusjewel yup

@Narcissusjewel: You need $\Omega$ connected, of course.

7:10 PM
@BalarkaSen Spicy.

Hey @Bala

This is basically gluing axiom for the presheaf of holomorphic functions

@TedShifrin Good point.
@BalarkaSen <3.

I see Balarka has made it through one more page of Forster. :)

@TedShifrin Yeah. The detailed feedback from some of the comments have given me some ideas for tweaks, but the overall idea still feels solid (and whining is always to be expected when a change means the students need to do more work)

7:11 PM
@TedShifrin ok thanks, i will check them out

@Tobias: I usually found that even the students who complained about my being demanding appreciated me a year or two down the road ...

Well let me try a different way: If the manifold is parallelizable, then I can find a diffeomorphism $\rho:M x \Bbb R^d\rightarrow TM$, $\rho(p,v)=v^iX_i(p)$

Well, not gluing axiom, the locality axiom, I guess.

@TedShifrin That might have just been survivor bias

Hey @Krijn

7:12 PM
Other than that there were some general comments about me being inexperienced, but improving during the course. And a general consensus that representation theory is hard

Long time

@Krijn: That includes people who got not-so-great grades, but â€” sure â€” there were indeed quitters.

Yeah, I'm finishing up my thesis and hoped that someone here could help me out with a small proof
Alas, no number theory in this room so I keep on googling

@BalarkaSen For upgrading to a sheaf you mean here.

@TedShifrin I have gone through Chapter 1, more or less. I gave Daminark what I consider a quality exposition of analytic continuation a few days ago here

7:13 PM
Is the mapping $\rho(p,v)\rightarrow \rho(q,v)$ a parallel transport?

@quallenjäger: You're declaring certain sections to be "constant." Having fixed that fixed basis, you define covariant derivative subject to the product rule, yes. So the connection will depend, of course, on the trivialization.

So at this point I need information of connection?

You should try your colleagues or professors, @Krijn ... Maybe they know more than we do.

Is the connection set once I have choose the basis $X_i(p)$
?

:(

7:15 PM
Yes, once you choose the trivializing sections, the connection is determined.
Good point, @Krijn.

@TedShifrin Thanks, got It
And I have run through the story with the hyperboloid, did you mean that the hyperboloid at infinity is a cone rather than a cylinder?

What would be the cone point?

Hi chat

I mean just "like" a cone at infinity
A cylinder would go straight upwards and that is what bothers me.

If f is a mapping from S onto T , prove that f'f = identity of S

7:19 PM
@Jacksoja What are S and T?

I don't understand how much is enough to prove this

@Jacksoja What does $f'$ mean?

S and T are sets
f' is f invers

Well, I see your point. The surface is, in fact, asymptotic to the cone formed by the asymptotes. I was sloppy.

I can see how if we take s in S f(s) = t
and f' (f(s)) = s

7:21 PM
@Jacksoja the inverse is only defined for invertible, i.e. bijective functions, and it is defined such that the identity you need to prove holds

@TedShifrin Thanks!

@TobiasKildetoft so what is it exactly I should to show?

What if $S = \{ 1,2,3 \}$ and $T = \{ 1,2 \}$ with $f: 1 \mapsto 1, 2 \mapsto 2, 3\mapsto 2$? @Jack?

@Jacksoja The statement you have written is not true

You need more information on $S$ and $T$ here

7:22 PM
f is a 1-1 mapping sorry did not mention that earliar
f is a 1-1 mapping from S onto T , prove that f'f = identity of S

I have the feeling that one need to be extremely cautions in differential geometry.
Its very easy to make mistakes

I am capable of making very easy mistakes in most of mathematics.

Me too :P

All these minus sign, you wish sometimes they disappears, but they simply don't
best way is to erase them and hope no one will see it

@quallenjäger You should try doing Lie algebra cohomology. So many signs
I had a lecturer write up some of it, then say "there is a formula with signs. I refuse to put signs"

7:26 PM

@TobiasKildetoft f is a 1-1 mapping from S onto T , prove that f'f = identity of S
@TobiasKildetoft what I said before makes sense now?
f(s) = t

In complex geometry you must keep track both of signs and of powers of $\sqrt{-1}$.
Lots of folks get both wrong at times.

@Jacksoja Except that once the inverse is defined, the identity is the definition of the inverse

@Tobias: Your use of "the identity" is very confusing there.

Oh dear thats why I never touched complex stuff.

7:27 PM
@TedShifrin ahh, right

Get real

Yes exactly
there is also identity on T

People are so used to $1/2\pi i$ in complex analysis they tend to mess up in complex geometry ... when it should be $i/2\pi$ to get signs right.

@Jacksoja By "the identity" I meant the equality (i.e. the one you are trying to show)

But the proof of it , how should one write that ?

7:28 PM
I won't name some very famous people that have gotten it wrong. ;)

I see why the composition sends any eleemnt of s to itself

@Jacksoja how have you defined the inverse?

@Jacksoja: What is the definition of the inverse of a function?
in chorus

well yeah I know what it is
every 1-1 and onto function has an inverse

I even forgot what $1/i$ is

7:29 PM
And what is the definition?
That's a horrible thing, @quallenjäger.
Surely you jest.

f composed with f inverse = identity

In both directions, @Jacksoja.
So the equation you wanted is part of the definition.

so ff' =i_T , and f'f = i_S
in keeping track on what identity map

Backwards maybe?

we just need to see what is the first product act on right?
f :S-->T

7:32 PM
Better :)

@TedShifrin Right corrected it now

Right. And you asked how to prove one of those earlier. But it's half of the definition. That's what Tobias was saying (several times).

all right thanks, these questions on set theory part are easy to see but hard to give proper proof
like if g and f are onto functions
show that the composition is also onto

That's a good question.

I did this part for injective
but surjective was hard to write

7:34 PM
No. It shouldn't be.
It should be easier than injectivity.

sup chat

Let $E\subseteq \mathbb{R}$ be a bounded set and suppose $f:E \rightarrow \mathbb{R}$ is uniformaly continuous Prove that $f(E)$ is bounded.

heya Eric, orbit.

hello
How are you?

Doing just fine, thanks. You haven't had your exam yet, I take it.
Rehi DogAteMy.

7:35 PM
It's today.

Rehi

well g :S-->T ,f : T--->U , i need to show that for every u in U, there is s in S, such that gf(s) = u @TedShifrin

In a few hours :/

Aha.
Right, @Jacksoja.
So?

f(s) = t for some t in T

7:36 PM
This makes no sense at all.
We started with $u$. That's all.

we know all of T is used up because it was given that f is surjective
yes but what i meant to say

What does $gf(s)$ mean?

g(f(s) =g(t) = u for some u in U

No. $u$ is fixed to start with.

and we also know that g is onto

7:37 PM
You have to get your language right.

i should really learn some complex algebraic geo

all right

Yes, $g$ is onto is the right thing to use first.

let u be in U, u =f(t)
t=g(s)

NO.
Oh, your function composition doesn't even make sense, does it?

7:39 PM
that is not how we start this?

Hang on.
Either your function composition doesn't make sense or you got $f$ and $g$ reversed.

well the composition is S-->T-->U

And which function is first, and which function is second?

g:S-->T , f :T--->U

And what does $gf$ mean?

7:41 PM
we do f first

Then this makes no sense.
Details, man.

right we do fg
u=g(t)

OK. So let's start over. You said that for every $u\in U$, we want to find $s\in S$ so that $fg(s) = u$.

right

NO ... how can $u=g(t)$? Look.

7:42 PM
got it backwards again sorry

yeah im wondering how to compute homology for some smooth algebraic curve of degree $d$ in $\mathbb{C}P^{2}$ using pure hatcher stuff so i can show they're orientable

u=f(t)

Pictures help!

hang on ill draw one now

Right. Since $f$ is onto, there is $t\in T$ so that $f(t)=u$.
And you can even put a $t\in T$ with an arrow going to $u\in U$ :)

7:44 PM
right and t = g(s)

Oh, Eric, that's a good question.

f(g(s) = u

@Jacksoja, the words "for some $s\in S$" are crucial.
Do you see the picture with the mappings?

and the only argument here is that we are guanteed to have such an element mapped to u because of ontoness

@EricSilva Degree-genus theorem tells you what genus that dude has, probably?

7:45 PM
I drew the picture like it was bijection and that is bad

Right, ontoness of both.

I don't know algebraic geometry
But this is definitely nontrivial

It's just Euler characteristic for a covering map with branch points, Balarka, Eric.
But is that "in" Hatcher?

Ahh OK
Nope, Riemann-Hurwitz is not there
Or whatever that's called
But it's a good exercise to prove it

How 'bout implicit function theorem to get that it's a complex $1$-manifold, therefore orientable?
@EricSilva: Where did this question come from?

7:47 PM
ya that doesn't take much argument

my prof posed it

Or give a nowhere-vanishing $2$-form on it :P

im trying to use stuff from just this quarter of alg top is the thing

Good luck with that.

cause i know how to do it if i go beyond that stuff
ya im puzzlin

7:48 PM
Ted's $\chi$ hint is pretty good

Lefschetz hyperplane theorem :P

Once you know genus you are all set

just wanted to throw it out there as a cool question :P

You should be able to define a branched covering map to $\Bbb CP^1\subset\Bbb CP^2$.
Draw pictures :)

@TedShifrin Yeah the projection map to a hyperplane should be the branched covering in this case

7:49 PM
But draw the "pseudo-real" pictures that complex geometers draw.
Yup, Balarka.

lol
is rick miranda's book any good
if you're familiar with it

I don't know the book, but I've known Rick for decades. So it's no doubt good. I also like Griffiths's little book on algebraic curves (lectures from China).

@TedShifrin I still cannot get my head around about trivial connection. Basically my frame $X_i(p)$ would still move from point to point. How can I see that the covariant derivative of the frame would vanish?

oh i own a copy of that one
miranda's book is big

Miranda does more than curves, right?
@quallenjäger: In the trivialization, each $X_i$ appears to be constant (like the standard basis in $\Bbb R^k$).

7:53 PM
unsure

LOL, I figure it's time for you to read one of my most popular answers on MSE, @quallenjäger.

Yup. This @quallenjäger.

@TedShifrin So I need to define $\nabla_Y X_i$ is equal to zero? Or this can be shown?

That's defined. You're setting up what your "constant sections" are ...

7:55 PM
@TedShifrin Lol I thought this can be shown, thats why I am struggling.
you mean this one?

@TedShifrin Can I steal Figure 2.5 from Chapter 7 of your book with acknowledgements for an answer I am writing on MSE?

WTH is that, Balarka?

109

To put a slightly different spin on this (before @Zev votes to close :) ). When folks ask me what I teach, I ask them to name their least favorite subject. Occasionally I do get English or history or science. But 90% of the time I get math ... to which I reply "bingo." :) I then engage them in a ...

Fubini's theorem on a triangle :)

Oh good grief.

7:57 PM
The \int \int sin(x)/x dx dy thing

No, @quallenjäger. The one I just linked you to above.

It's a different question
But I need the picture :P

Sure. You may steal.

@TedShifrin Thanks! This is very very helpful to get my head around

Thanks. I am a thief with conscience however, so I shall acknowledge whom I stole

7:59 PM
Hey everyone!

Hi @Daminark

hi Demonark
You found the right link, @quallenjäger?

Yes, thurston 37 th way to think about derivative
?

Right.

I am lack of intuition, thats why I still struggling at different point. I have been looking for many sources for the intuition of the derivative. Thanks for the link

8:01 PM
That post actually took me more than a few minutes to write :P
Back later ...

See you Ted!
@Balarka okay so it seems I don't completely understand the path lifting property so, here's something

$f$ cts, $X$ not compact, but $f(X)$ compact. What's an example of this?

Let $E$ be the union of the two axes in the plane, and let $B$ be the x-axis, let $p$ be the projection. I'm trying to show that this map is not a Serre fibration.
@orbit constant map

No... a singleton is closed

@Daminark Okay

8:09 PM
sorry, that's what you wanted
thanks!

No problem!
So we need to find a map $f:D^n \to E$ and $g:D^n\times [0,1] \to B$ such that $g(x,0) = p(f(x))$ but for which you can't lift $g$ to $E$

Yup

@TedShifrin bonsoir, vous allez bien ?

And given that $E$ is a bit of a tight space, chances are we can do it for $n = 1$
Wait hold on I'm no longer convinced that this isn't true

Try $n = 0$

8:16 PM
@orbit-stabilizer $f=\sin$
$X=(0,100)$
$f[X]=[-1,1]$

Because you could just define $h(x,t) = g(x,t)$ and that'll be in $E$
Oh hmm

@LeakyNun yeah, that works too. Sighhhh

why sigh?

good evening, all

failure is imminent

8:17 PM
@orbit-stabilizer hence compact

indeed for me @orbit-stabilizer
I have my real analysis exam today and I haven't been able to sleep for the last two nights because of pseudo insomnia

I have mine today as well!

@Balarka wait $D^0$? What?

And it's not looking good

neither is mine
I'm so sleepy right now I can barely keep my eyes open
yet can't sleep
anyway best of luck @orbit-stabilizer

8:18 PM
@Daminark Yes? Consider $f(\{pt\}) = (0, 1)$ mapping to a point on the $y$-axis in $E$ with nonzero height.

What does yours cover?
you too

several topics, the last ones being extreme value theorem, generalized mean value theorem etc

Rudin?

Oh... I've never heard of people phrasing it as $D^0$ before, like that just didn't register as a thing which existed

Ah
Yeah D^0 is a point

8:22 PM
Okay so you can map $g(*,t) = t$ and $f(x) = (0,7)$
Sorry that was a mess

Yup

@orbit-stabilizer kinda
I'll catch you guys later, the exam starts in 12 minutes!

wow, Good luck!

Oh so the reason we can't lift $g$ is because it won't make the $f$ triangle commute?

@Daminark Do you have to parse it in vile language like this?

8:24 PM
Yeah because if $h$ lifts $g$

Like, is that absolutely necessary?
:P

Kek
That's the first thing that came to mind

What an obscene person

But yeah so if $h$ lifts $g$, you need $h( * ,0) = f( * ) = (0,7)$ to make one side work

@AkivaWeinberger hello, please do you know something about Differential equation of Clairaut ?

8:28 PM
But then $h( * , \epsilon) = (\epsilon, 0)$
So that fails continuity
Okay sick

yup

So that's a quasifibration which isn't a Serre fibration

Yup
Wait
That's a qifib?
Oh
Of course

I've encountered an interesting metric on $\Bbb R$ (on math.SE): $d(x,y) = \begin{cases} |x|+|y| & x \ne y \\ 0 & x = y \end{cases}$
cc @Secret
$0$ is the closest point to any point other than itself
I wonder what sort of topology this induces
$\{a\}$ is open for each $a \ne 0$
and so is $(-x,x)$ for all $x > 0$
in particular $\{0\}$ is not open
any set containing $0$ is closed

@Daminark did you prove serre fibrations are qifibrations

8:45 PM
Yeah

@Daminark any idea?

I proved that $p:E\to B$ is a quasifibration iff the inclusion of the fibers into the homotopy fibers is a weak equivalence

yeah

And you know that if $p$ is a Serre vibration, that inclusion is a homotopy equivalence
You could just do the argument directly though

by lifting, yeah?

8:48 PM
What I had in mind was that you take the homotopy sequence of the pair $(E,p^{-1}(b))$ and compare (compair it kek) to the homotopy sequence of a Serre fibration

Pah, I'm a cat person anyway

i posted two show off comments lmao

Fucking hell $\mathbb{R}^n$ is absolute garbage
15

lool
-Amin, 2017

8:54 PM
Lord, those two quotes in succession though
"All open sets in $\mathbb{R}^n$ are homeomorphic"
"Fucking hell $\mathbb{R}^n$ is absolute garbage"

That is beautiful

You can guess what realization happened in the middle
Yo @Eric

sup