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Ted Shifrin
Ted Shifrin
01:09
@Danu: Moreover, any locally free sheaf of rank $r$ is the sheaf of sections of some (holomorphic) vector bundle of rank $r$.
Steve
Steve
Hi people, I have a relatively simple question that I can't wrap my head around today for some reason. How do you go from $-\pi$ to $\pi$ on the unit circle? I'm refreshing my math and I don't see it explained anywhere
 
3 hours later…
arctic tern
arctic tern
03:47
@Steve what do you mean?
 
2 hours later…
Danu
Danu
06:07
@TedShifrin that is the preposition I was reading.
Do you have any comment on the Kodaira dimension?
Secret
Secret
06:31
@Steve You add $2\pi$ which basically means you rotate one revolution around the unit circle and back to where you started
 
1 hour later…
user228700
07:40
Hello everyone :-) I was wondering if any of you know the formula for the sum of the squares of the terms of an A.P as in a²+(a+d)²+(a+2d)²+...+(a+(n-1)d)² where a is the first term of the A.P and d is the common difference..?
user228700
I have derived an expression, but I dunno if it's correct. Can anybody please help?
Mike Miller
Mike Miller
07:51
@Danu As with most things in life, understand it for curves and surfaces and you will understand it.
user228700
08:22
Actually, never mind the other doubt. @Secret cleared it for me. I have another one. In inequalities involving real numbers, how should I go about trying to find the min/max value of something?
Secret
Secret
i think you need more context, what kind of inequalities you are having in mind?
user228700
Let me explain; if I have an inequality such as E>a where E is some expression and the only condition on a is given as a is a real non-zero, positive number, how can I find the min. value of E?
user228700
If $a$ were a natural number, then the min. value would be 1 but in this case, $a$ is a real number so..?
user228700
My textbook has given it to be 1 regardless. How is this correct?
Secret
Secret
Given a is a posiive real number
If E is a function of a, then the starting point is to find points where E(a) is stationary (although if E is not differentiable at some points, such cusps can be minima, those you just have to use the graph of E(a) to find out)
If E is in addition a monotonically increasing function of a, then there is no minimum but only an infimum at a=0 (which because of the inequality, E(a) can only approach arbitrarily close to E(0) as $a\rightarrow 0$
If E is not a function of a, then the smallest possible E can only be given by the expression $\lim_{E\rightarrow a^+}…
So if the book said the min is 1 regardless of whether a is natural or real number, then there must be a physical grounds that prevent a from going below 1
user228700
08:34
Wow, okay.
user228700
I guess my book doesn't care about such details sometimes, which is confusing, but alright(They haven't mentioned any such constraint). Thank you!
Secret
Secret
Some guidelines on how to find extrema for functions
Note that in general one cannot write a number that is closest to a, because there a an infinite number of real numbers arbitrarily close to a
user228700
@Secret Noted :-)
Mary Star
Mary Star
09:10
Hello!!
We are given $\mathbb{R}^2$ with the metric of the euclidean norm $\| \cdot \|_2$.
How can we check if the following sets are open?
$\mathbb{R}^2$, $\mathbb{R}^2\setminus \{(0,\frac{1}{n})\mid n\in \mathbb{N}\}$

I think that $\mathbb{R}^2$ is open, or not? But how could we justify it?
s.harp
s.harp
09:39
@MaryStar what definition are you using for open sets in $\mathbb R^2$?
 
4 hours later…
shai horowitz
shai horowitz
13:59
Game Theory speaking is it ever wise to listen to threats? threats are only made because of the thought that you will react to them, if you out right discard any threats does that protect you from certain iterative strategies? If so what are the use of threats?
Semiclassical
Semiclassical
14:48
hi @balarka
 
2 hours later…
Balarka Sen
Balarka Sen
16:24
Hi @Semiclassical
Semiclassical
Semiclassical
hey
Balarka Sen
Balarka Sen
What's up
Semiclassical
Semiclassical
not much.
being pretty lazy today.
Balarka Sen
Balarka Sen
me too
Semiclassical
Semiclassical
you? ah.
anything new math-wise?
Danu
Danu
16:28
@BalarkaSen Do you know stuff about divisors, hypersurfaces and line bundles?
Balarka Sen
Balarka Sen
A bit.
What's the problem?
Danu
Danu
You know the construction of assigning a line bundle to a divisor?
Balarka Sen
Balarka Sen
Yes.
Danu
Danu
Okay, and do you know the isomorphism $\operatorname{Div}(X)\cong H^0(X,\mathcal K^*_X/\mathcal O^*_X)$?
Balarka Sen
Balarka Sen
What's $\mathcal{K}_X$?
Danu
Danu
16:30
Sheaf of meromorphic functions
Balarka Sen
Balarka Sen
I guess it's not immediately clear to me how to get a divisor from a section of $\mathcal{K}_X^*/\mathcal{O}_X^*$.
Look at the zero set of a meromorphic function or something?
Danu
Danu
It's not so easy
Can you look at Huybrechts (prop. 2.3.9)?
Balarka Sen
Balarka Sen
I don't actually have Huybrechts with me but let me see if I can get a copy.
ok, looking
@Danu Ah I see
Got it.
Danu
Danu
So I'm trying to understand the proof of 2.3.14
The first direction is fine
The second, the only nontrivial step is to somehow get those $g_j$'s from the fact that $\mathcal O(D)$ is trivial.
I don't understand how to do this
Balarka Sen
Balarka Sen
16:48
@Danu The flurry of terminologies is confusing. What does it mean to say $\mathcal{O}(D)$ is trivial, by the way?
Danu
Danu
@BalarkaSen $X\times \Bbb C$
It's a line bundle
Balarka Sen
Balarka Sen
Isomorphic to $D \times \Bbb C$, you mean?
Danu
Danu
$D$ is a divisor hehe
Not a space
Balarka Sen
Balarka Sen
Sheaf of holomorphic functions isn't a bundle in general, though?
Danu
Danu
$\mathcal O(D)$ is the line bundle associated to the divisor $D$
Balarka Sen
Balarka Sen
16:50
Oh, I see.
That's weird notation.
Can you remind me what a principal divisor is? I do not really deal with this algebraic terminologies in the topological category.
Danu
Danu
Divisor arising from a meromorphic function
Zeros minus poles
Balarka Sen
Balarka Sen
OK.
Danu
Danu
So given that $\mathcal O(D)$ is trivial
We need to produce a mero $f$ such that $(f)=D$
This is simple one we have those $g$'s
But how do I produce those $g$'s, and in particular how do I use triviality?
Balarka Sen
Balarka Sen
Yeah I guess it's not clear to me how one does that. Sorry for not being helpful.
Not really familiar enough with the holomorphic category.
Mike Miller
Mike Miller
17:09
It coincides with the notation $O(n)$.
Balarka Sen
Balarka Sen
Ah, good point.
For some reason I confuzzled it with the notation for the ring of global section of the sheaf of holomorphic functions on a variety.
CulDeVu
CulDeVu
I have a simple probability question
Mike Miller
Mike Miller
Do I want to take a course on algebraic cycles, an introduction to modular forms, or neither?
Balarka Sen
Balarka Sen
You'd probably enjoy the former more than the latter.
although that's based on pure guessing and knowing your mathematical taste a bit.
Samuel Yusim
Samuel Yusim
yo @MikeMiller, how's it going?
Mike Miller
Mike Miller
17:16
It's good.
Danu
Danu
@MikeMiller Modular forms
Samuel Yusim
Samuel Yusim
good, good
Balarka Sen
Balarka Sen
A modular forms course would probably involve endless complicated analytic estimates, not just the geometric content
Mike Miller
Mike Miller
With modular forms I might understand what my friends are talking about.
Danu
Danu
They're super* important in lots of mirror symmetry n such stuff
Freudian slip right there haha
Mike Miller
Mike Miller
17:17
Analytic estimates are good.
CulDeVu
CulDeVu
So I was shown a while back a puzzle that simplifies to something like this: "You have 7 bowling balls: 6 are 16 pounds, and 1 is 15 pounds. You have one of those old gravity scales that you can put things on either side of to see which one is heavier. How do you find the 15 pound ball by using the scale only twice?"
Balarka Sen
Balarka Sen
If you're fine with that, go for it.
CulDeVu
CulDeVu
The answer to the puzzle is pretty simple, but I was wanting to approach it from a probability viewpoint
Samuel Yusim
Samuel Yusim
oh yeah, mike, if you ever meet a math grad student at ucla named kevin who studies logic, he's a good friend of mine. he just started there this year
Mike Miller
Mike Miller
I played poker with him two weeks ago. Nice guy, a bit quiet.
Samuel Yusim
Samuel Yusim
17:19
neat
CulDeVu
CulDeVu
My attempt goes like this: The scale can differentiate between 3 states, so the probability that ball $b_n$ is the 15 pound one is $P(b_n) = 1/7$
Balarka Sen
Balarka Sen
@MikeMiller Hi, by the way. Good to see your around.
CulDeVu
CulDeVu
And the probability you want to hit after the first sampling of the scale is $P(b_n | A)$, where $A$ is some additional knowledge that can be gained by sampling with the scale
So $P(b_n | A) = \frac{P(A | b_n)}{7 P(A)} = 1/3$, which implies $P(A) < \frac{3}{7}$
Mike Miller
Mike Miller
@SamuelY I'll mention you next time we talk.
CulDeVu
CulDeVu
So my question is: is there a way to enumerate all the A's that satisfy that requirement?
Samuel Yusim
Samuel Yusim
17:29
@Mike cool!
JeSuis
JeSuis
Hi, I am trying to prove (as an exercise) that for an infinite set $A$ we have $A^2\sim A$.
The exercise define $\mathscr{F}:=\{(X,f): X\subset A \quad\mbox{and}\quad f:X\to \{0,1\}\times X \quad\mbox{and}\quad \vert\Bbb{N}\vert\le \vert X\vert\}$ and an ordered relation as $(X,f)\le(Y,g)$ if $X\subset Y$ and $g$ restricted to $X$ is $f$.

I used Zorn's lemma to deduce that there exists a maximal element noted $(X_0,f_0)$ where $X_0=\bigcup_{(X,f)\in\mathcal{C}}X$ and $\mathcal{C}$ is a the totally ordered set included in $\mathscr{F}.$
Balarka Sen
Balarka Sen
I haven't gotten round to thinking about that exercise of yours on codimension 2 submanifolds of S^n, been pressed with school. (slightly embarrassed to admit I have also been wasting time watching movies and reading) Hopefully will get more time from next week as holiday starts.
Danu
Danu
"wasting time watching movies and reading" WRONG
Being/becoming cultured is not a waste of time
Pedro Tamaroff
Pedro Tamaroff
Exactly what I was going to say.
JeSuis
JeSuis
watching movies= being cultured ??
Balarka Sen
Balarka Sen
17:34
That's true. Was just saying that's cutting out my math-time; not preferable.
Danu
Danu
@JeSuis =becoming cultured, if you watch non-crappy ones.
Pedro Tamaroff
Pedro Tamaroff
@BalarkaSen It is very useful to clean one's mind with entertainment. =)
JeSuis
JeSuis
@Danu :-)
Pedro Tamaroff
Pedro Tamaroff
It usually helps with your mathematical output.
Balarka Sen
Balarka Sen
@PedroTamaroff What have you been doing?
Mike Miller
Mike Miller
17:43
@Balarka As usual, I am ashamed of you.
Or not.
Balarka Sen
Balarka Sen
:(
Pedro Tamaroff
Pedro Tamaroff
@BalarkaSen Taking my last two courses, which are PDEs and an introduction to numerical calculus.
Balarka Sen
Balarka Sen
@MikeMiller Have you seen any Ray films, btw? That's what I have been watching all these weeks.
Pedro Tamaroff
Pedro Tamaroff
@BalarkaSen What are those?
Balarka Sen
Balarka Sen
@PedroTamaroff Nice.
@PedroTamaroff directed by Satyajit Ray.
I like some of his stuff
Mike Miller
Mike Miller
17:47
No. I read a few books last week.
Balarka Sen
Balarka Sen
ah, what book?
Mike Miller
Mike Miller
Einstein manifolds, High-Rise, Tate cohomology, Film as art.
Pedro Tamaroff
Pedro Tamaroff
I've been reading a collection of stories by Hemingway.
Balarka Sen
Balarka Sen
interesting ordering :)
@Pedro for me it's been Borges.
should read more of Hemingway
Mike Miller
Mike Miller
You're right. Generalized Tate cohomology was first, Einstein manifolds third.
Balarka Sen
Balarka Sen
17:55
hah
oh, I wanted to ask something but I forgot. I wanted to revise my differential forms but wanted a book which actually does something with them than just introduce them and say "this is the gateway to de Rham theory", because I am getting demotivated by rereading and doing exercises.
can you recommend me a book like that? preferably not Bott-Tu and the like, because that seems to do algebraic topology with forms, which I am not really keen to learn
(I should mention I have studied what's in chapter 1-3 in G-P)
Mike Miller
Mike Miller
I don't know what you're asking. What do you mean review differential forms but not do de Rham theory? What do you want to do...?
Balarka Sen
Balarka Sen
I am good with learning de Rham theory. But not homotopy theory, which I think is a major part of what Bott-Tu does?
Mike Miller
Mike Miller
Yeah, I mean, in the second half of the book.
Balarka Sen
Balarka Sen
Would prefer more geometric aspects of it than algebraic.
@MikeMiller Ah.
Mike Miller
Mike Miller
Geometry of differential forms, Morita.
Balarka Sen
Balarka Sen
18:04
Thanks.
Mike Miller
Mike Miller
There's also a workbook in differential geometry by some authors I don't remember that I always recommend my students and they never use.
Balarka Sen
Balarka Sen
I'd look at it if you tell me what the book is.
Krijn
Krijn
@Balarka, you should really do some Algebraic Number Theory
It's superb
Balarka Sen
Balarka Sen
I'd like to. But that's a long way from where I am.
Did you learn something new?
Mike Miller
Mike Miller
18:20
books.google.com/…
Krijn
Krijn
Today we calculated the Dirichlet density of the number of primes $p$ such that $2$ is a square mod $p$, or a cube mod $p$ and even a fourth power mod $p$.
And it was surprisingly easy
Balarka Sen
Balarka Sen
Thanks, @MikeM.
@Krijn interesting
what's the Dirichlet density again?
0celo7
0celo7
Probably too much to hope for, but when is a function space second countable in the compact open topology?
Are there any general results?
Krijn
Krijn
@BalarkaSen Ehhh, for a number field $K$ with $P$ a subset of prime ideals of its ring of integers $\mathcal{O}_K$ it is defined as $$\lim_{s \to 1} \frac{\sum_{\mathfrak{p} \in P} \frac{1}{N(\mathfrak{p})^s}}{\sum_{\mathfrak{p} \in \text{Max}(\mathcal{O}_K)} \frac{1}{N(\mathfrak{p})^s}}$$
Pedro Tamaroff
Pedro Tamaroff
@Krijn Yay for Dirichlet series.
Balarka Sen
Balarka Sen
18:29
@Krijn I see
Krijn
Krijn
@PedroTamaroff I still need to develop intuition for this though
I keep thinking of it as a chance too much
Pedro Tamaroff
Pedro Tamaroff
@Krijn Ah?
Krijn
Krijn
Like, the density of prime ideals with $f_\mathbb{Q} = 1$ (residue class degree or something?) is 1, but that doesn't mean that almost every prime ideal has $f = 1$, right?
Pedro Tamaroff
Pedro Tamaroff
@Krijn Well, you haven't given meaning to "almost every".
Mike Miller
Mike Miller
@0celo7 jstor.org/stable/2320076?seq=1#page_scan_tab_contents
Krijn
Krijn
18:39
@Pedro I can't :(
Pedro Tamaroff
Pedro Tamaroff
@Krijn The point is that your definition of density gives meaning to that.
Krijn
Krijn
The number of primes which are $\pm 1 \mod{8}$ have Dirichlet density $\frac{1}{2}$. Does that imply that for big $n$, there will be almost the same amount of primes $\pm 1 \mod 8$ and $\pm 3 \mod 8$?
Pedro Tamaroff
Pedro Tamaroff
I have no idea how Dirichlet density and usual density relate.
Krijn
Krijn
I probably think of it too much as usual density
0celo7
0celo7
19:19
@MikeMiller Many thanks.
SAWblade
SAWblade
If $f:A \to A$ is surjective, is it also injective?
Krijn
Krijn
No, take $A = \mathbb{N}$ and send all positive numbers $n \mapsto n-1$ and send $0 \mapsto 0$.
SAWblade
SAWblade
darn
Thanks! :)
SteamyRoot
SteamyRoot
It holds for finite $A$ though.
SAWblade
SAWblade
Related question. If $f: A \to A$ is surjective, then does $\mathbb{Im}f = A?$
Balarka Sen
Balarka Sen
19:33
@Krijn I never got back anything from you about that movie :) But I understand if you're distracted. Just curious how people other than me or people of this bit of the world thinks of it.
Pedro Tamaroff
Pedro Tamaroff
@SAWblade That's the definition of surjectivity.
Krijn
Krijn
@BalarkaSen Ya, some friends showed up and then I ended up in a bar and stuff
Pedro Tamaroff
Pedro Tamaroff
Namely, a function is surjective if every point of its codomain is in the image of some point in the domain, @SAWblade.
Balarka Sen
Balarka Sen
Fair distraction.
SAWblade
SAWblade
Okay, so my friend and I are struggling to understand something then. If $f:A \to A$ is surjective, then $\text{Im} f = A$, and then by the first isomorphism theorem $G/\ker f \cong \text{Im}f = A$, which implies the kernel is always trivial. Is that true? xD
Krijn
Krijn
19:36
@SAWblade It depends, you apparently are talking about objects with some structure on them
SAWblade
SAWblade
Groups!
Krijn
Krijn
I took $A$ to be some set, but you seem to talk about $A$ to be a vector space, and $f$ a linear map, or a group and a homomorphism
Balarka Sen
Balarka Sen
$A/G \cong A$ for some subgroup $G$ of $A$ does not mean $G$ is trivial in general.
Krijn
Krijn
Well yeah, then the situation is different
SAWblade
SAWblade
It doesn't! Okay, awesome, that's what was confusing us. xD
Can you give me an example? :0
SteamyRoot
SteamyRoot
19:40
I suppose that you could construct such group by making infinite direct products or so
Balarka Sen
Balarka Sen
Take $A = \Bbb Z/2 \oplus \Bbb Z/2 \oplus \cdots$. Look at the endomorphism of $A$ which sends $(a_1, a_2, a_3, \cdots)$ to $(a_2, a_3, \cdots)$. This is surjective with kernel isomorphic to the first copy of $\Bbb Z/2$.
SteamyRoot
SteamyRoot
mathoverflow.net/questions/17221/… has a nice example
par
par
Hi friends. Quick question for you (I apologize if this is the wrong venue for it). How does one go about requesting for an existing question to be migrated to Math Overflow?
Balarka Sen
Balarka Sen
The endomorphism there is known as the left shift operator.
@SteamyRoot It's hard to come up with non-Hopfian but finitely presented group.
Pedro Tamaroff
Pedro Tamaroff
Wrap the circle about itself a few times, @SAWblade.
Balarka can tell you about it.
Quotient, you get the circle.
Balarka Sen
Balarka Sen
19:46
Right, also a good and geometric example.
Danu
Danu
So I realized Huybrechts only calls a result a "theorem" if he's unable to prove it in the book.
I don't agree with this choice, but it's interesting :P
SAWblade
SAWblade
.
Oops. xD
Thanks for the quick answers, all. :)
Ted Shifrin
Ted Shifrin
20:02
hi mr @Pedro, no-longer-starving @Danu
and @Balarka and @Krijn ...
@Balarka: Re differential forms beyond where you are, look at a wonderful little book by Darling (RDR initials or something like that).
Danu
Danu
@TedShifrin Figs and yoghurt :3
Also, did you have any comment on the Kodaira dimension?
Any intuition for what the heck it means? And why is it $-\infty$ if the canonical ring equals $\Bbb C$?
Ted Shifrin
Ted Shifrin
There isn't much to comment besides the definition and the fact that it's important in classification of higher-dimensional complex varieties.
Krijn
Krijn
Hey @Ted
Danu
Danu
@TedShifrin So why is it defined like that?
Ted Shifrin
Ted Shifrin
Because of things like Kodaira embedding ... which you'll get to eventually.
Danu
Danu
20:06
Okay
Ted Shifrin
Ted Shifrin
If a line bundle is positive, it may still not have enough sections to embed you in $\Bbb P^N$ for some $N$, but if you take a high enough tensor power, it will work.
And if it's negative, fuggetaboudit.
@Danu: BTW, I bet Huybrechts proves Kodaira vanishing/embedding and calls them theorems.
Danu
Danu
@TedShifrin Whatcha wanna bet?
I haven't checked yet
Ted Shifrin
Ted Shifrin
If I'm right, you owe me a dinner :)
Danu
Danu
When you come visit?
Ted Shifrin
Ted Shifrin
Yup.
Danu
Danu
20:10
It'll be a student-level dinner.
Same for you?
Ted Shifrin
Ted Shifrin
OK. :)
Danu
Danu
Also, if he doesn't fully prove them, but calls them theorems, then nobody wins.
Ted Shifrin
Ted Shifrin
LOL ... he wouldn't dare.
Danu
Danu
PROPOSITION
Proposition 5.2.2 (Kodaira vanishing)
lololol
Proposition 5.3.1 (Kodaira embedding theorem)
I win!
Ted Shifrin
Ted Shifrin
But the index calls it theorem and he refers to them as theorems. He's a dope.
Danu
Danu
20:13
No weaseling out of it :P
Ted Shifrin
Ted Shifrin
"Let us now come to the proof of the Kodaira vanishing theorem."
Danu
Danu
Proposition!
Ted Shifrin
Ted Shifrin
If some of the deepest results can't be called theorems, then the reader can't distinguish between proposition and theorem. I fire him.
Danu
Danu
Obviously, the header of the relevant tex environment determines it
It's ridiculous though
Ted Shifrin
Ted Shifrin
Having written 4 books in LaTeX, and having revised my own style documents, I do not accept that excuse.
Danu
Danu
20:15
I'm expecting a dinner :P
Ted Shifrin
Ted Shifrin
I suppose you are.
Danu
Danu
:D :D :D
Ted Shifrin
Ted Shifrin
From the preface: "In general, all results are proved except for assertions presented as 'theorems', indicating that they are beyond the scope of this book." What crap.
Danu
Danu
Oh!
I was just sure because he did the same with "Siegel's theorem"
(called it a proposition in the end, while referring to it as a theorem throughout the text)
Ted Shifrin
Ted Shifrin
I would instead put Theorem*, with a notation in the preface that such things do not have proofs.
Danu
Danu
20:17
Yeah, it's a strange choice
Ted Shifrin
Ted Shifrin
I mean, Kodaira's deep theorems and Ted's proposition should not get equal weight.
Danu
Danu
Lol
Ted Shifrin
Ted Shifrin
He also misspelled Griffiths in his bibliography. Sigh.
Danu
Danu
But what if Ted was a really nice guy!
@TedShifrin Precise reference? I'm compiling typos (have like 20+ already!)
Ted Shifrin
Ted Shifrin
Irrelephant.
Reference 59.
Danu
Danu
20:19
The best one I found so far is
"moduli spaces of shaves"
Ted Shifrin
Ted Shifrin
Well, I have pages of notes on mistakes/typos in Griffiths/Harris, to be fair. And lots of mistakes in my own books which the publisher is too lazy to fix.
Danu
Danu
Which is, surprisingly, not a nonsensical sentence :D
@TedShifrin I just hope Huybrechts doesn't already know about them.
Ted Shifrin
Ted Shifrin
well, there are closer shaves and less closer shaves. Sounds fine to me, of coarse.
Danu
Danu
Do you at least have the errata online?
Ted Shifrin
Ted Shifrin
Have you checked his website for errata?
Danu
Danu
20:20
@TedShifrin I see what you did there
@TedShifrin I tried to find some, didn't find any.
Ted Shifrin
Ted Shifrin
Yes, I do for my books. And for Guillemin & Pollack. Never typed up everything in Griffiths/Harris.
Danu
Danu
nice
Hahahaha
From Huybrechts' website:
Under books:
> (with M. Lehn) The geometry of moduli spaces of shaves. Aspects of Mathematics E 31, Vieweg (1997), 260 pages.
Same error :D
Ted Shifrin
Ted Shifrin
BTW, the stuff you and I were discussing about $\mathcal O(D)$ you can find discussed in my complex geometry notes pp. 19-20-...
Danu
Danu
Cool
Ted Shifrin
Ted Shifrin
Hmm, today I look green again.
Obliv
Obliv
20:31
"A single vector parallel to a plane is not enough to convey the
“direction” of the plane, but a vector perpendicular to the plane does completely specify
its direction" - I'm skeptical about this. If a vector is parallel to a 3D plane then surely that specifies the plane's "direction". If it is orthogonal, then the vector specifies the plane's direction once more. how does the former not completely specify its direction?
Danu
Danu
@TedShifrin Still orange to me!
Ted Shifrin
Ted Shifrin
Tarnation!
@Obliv, the vector (1,0,0) is parallel to the xy-plane, but also to infinitely many other planes. However, only the scalar multiples of (0,0,1) are orthogonal to that plane.
(@Obliv: Think of a revolving door rotating about the x-axis. All those planes are parallel to the x-axis.)
Well, I guess I'm leaving.
Obliv
Obliv
@TedS I see. Thanks man
Ted Shifrin
Ted Shifrin
You understand, @Obliv?
0celo7
0celo7
@TedShifrin Could you hold one one sec please?
What should I read before Chapter 2 of Hirsch to get prepared for function spaces?
Obliv
Obliv
20:43
Yes. Many planes may share a common parallel vector that are not necessarily parallel to each other. Non-parallel planes may not share a common orthogonal vector, correct? @tedS
Ted Shifrin
Ted Shifrin
Stuff in Munkres on compact-open topology would help, @0celo.
Correct, @Obliv :)
0celo7
0celo7
@TedShifrin Yeah, I'm doing that.
But he (Hirsch) makes a claim that the weak topology is second countable -- neither my advisor nor analysis prof know how to approach it.
*topology prof
Danu
Danu
@TedShifrin I've just spent half an hour searching my computer for the files you sent me---I somehow can't find them!!!
Do you maybe have a (partial) file name of one or more of the files, so I can search my hard drive with those? I'm sure they must be on there, since I remember having looked at them!
Ted Shifrin
Ted Shifrin
I'll think about it @0celo. It should follow if you can do it for maps on Euclidean spaces.
@Danu: Let me look.
Danu
Danu
Ah, I found some of it!!
I found your difftop stuff
Ted Shifrin
Ted Shifrin
20:49
Same email. It's labeled 18.969
Danu
Danu
Ah, never mind. I found the rest, too :D
I put it in a really bad spot!
0celo7
0celo7
@TedShifrin I'm trying to read the proof of the Meyers-Steenrod theorem. It's not going well because I'm not at all familiar with the compact-open topology. Munkres' two pages on it aren't very helpful...
Ted Shifrin
Ted Shifrin
Well, @0celo, I'll leave it to you to work on it.
@Danu: I didn't realize computers had good spots and bad spots.
Huy
Huy
@TedShifrin: one of my former students just added me on facebook. that means they didn't all hate me !!!
Ted Shifrin
Ted Shifrin
Don't be so sure, @Huy. I've had former students add me too.
Huy
Huy
20:53
haha
Mike Miller
Mike Miller
Which Meyers-Steenrod?
0celo7
0celo7
@MikeMiller Isometry group of a Riemannian manifold is a Lie group.
Danu
Danu
@TedShifrin I really order everything very "well" in my folders.
So if I put it in a path that I didn't expect (months later) then it's very hard to find.
I put your notes in my "thesis material" folder
Mike Miller
Mike Miller
Create a theory of Frechet manifolds and prove that it's a finite dimensional closed subgroup of a Frechet Lie group.
Danu
Danu
While I should've put it in the "Mathematics" folder
Ted Shifrin
Ted Shifrin
20:55
Ohhhh @Danu.
0celo7
0celo7
@MikeMiller ...is that something someone's done?
Danu
Danu
Frechet manifolds is a thing
0celo7
0celo7
@Danu Of course.
But has someone shown what Mike said?
Mike Miller
Mike Miller
Oh, once you have Frechet Lie groups the result is a triviality.
Of course Isom(M) is closed in Diff(M), once you know that isometries are diffeomorphisms.
Then the Lie algebra is the space of Killing fields.
Danu
Danu
have fuuuun 0celo7 lel
Mike Miller
Mike Miller
20:57
The "once you have" is the tricky part.
0celo7
0celo7
@MikeMiller Of course?
With what topology?
Compact-open?
Mike Miller
Mike Miller
That's not the correct topology on Diff(M).
0celo7
0celo7
That's the topology everyone's working with!
(The original paper, Kob-No, Petersen)
Mike Miller
Mike Miller
I don't think that coincides with the C^infty topology even if M is compact, but I forget these things.
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