@Semiclassical: about to go offline for a while ... but back later.

What I want to say (and here's where I get into trouble) is that there should be some way to write $\partial_u(dx/y)=A(dx/y)+B (x\,dx/y)$ (up to an exact form)

mmkay, @ted

Plus of course discussing the details of a paper with my former advisor for quite a while did not help (with the voice that is. It helped us both understand the paper a lot better and we both feel like we get the arguments, at least with some additions of our own)

I guess I'm confusing myself as to in what sense $\{dx/y,x\,dx/y\}$ is a basis compared with $dx/y$ alone

I think you should give your seminar with a megaphone

@abenthy Yes.

@BalarkaSen hii

have water on hand

what's up

feeling sleepy...

me too. but i'll probably watch a movie after dinner

What movie?

Now a days I gen go to bed around 12 n wake up at 6

Planning on Apocalypse Now but decisions change pretty quickly.

X-men?

no, that's a different movie you're confusing with

ohhh...I see

any way, your movie taste is very different from mine :P

Harrison Ford is in Apocalypse Now

@Anubhav I figured.

"Apocalypse Now", not Apocalypse last year

lol

did you learn anything interesting these days?

Which, speaking of, I found a great list of films here recently; the ones I watched about exactly are of my taste.

@Anubhav Not sure what you'd find interesting. I learnt the Gauss-Bonnet.

for surface or higher dimension?

Surfaces.

I'm pretty sure if I try to watch any of those movies, I'll fall into sleep in next 5 min :P

@Anubhav Then that is a sign that you ought to be sleeping

That's a pity. But I respect your taste in movies :)

stalker number one D:

Any way, ishould go off line now... I'm actually feeling very sleepy...talk to you tomorrow ... GN

Night.

If I where to write an examen and the assignment is to prove that for all sets M following holds: $\emptyset\subset M$. Would i get all points if i simply write: This doesnt hold for $M=\emptyset$, since the emptyset cannot contain proper subsets. (The hole point is: how literal can i be)

@saturatedexpo I would assume that if the exercise was like that then in the course the notation $\subset$ had referred to inclusions that were not necessarily proper

yes, and so do I. the question is, do we have to assume? Math assignments shouldn't be ambiguous.

If it's ambiguous, I'd ask the person who wrote it.

If they've declared a convention, though, it's fine.

Regardless, the simplest form would seem to be: Prove that the empty set is a subset of any non-empty set.

in that case, the distinction be subset and proper subset is irrelevant

i like $\subset$, because it has a quite different meaning then $\subseteq$. And the assignment itself could all be around that difference.

@saturatedexpo Except it obvious is not about that distinction when phrased like that

@BalarkaSen Thank you :)

popular place this morning

Does someone find the error? "The injectivity radius of every compact Riemannian manifold $(M,g)$ is positive"

@TobiasKildetoft what would be an assignment revolving all around that difference?

Without boundary, I guess.

Well, I could be wrong, but I think the point has injectivity radius zero. Thats what I thought about.

that or infinite injectivity radius :)

but sure, 0-dimensional stuff is silly

Hi @MikeMiller

Apparently I am giving a 5 minute beamer talk.

Where?

At GT

I see. I might go to that.

Funding deadline has passed I think.

Wait fuck.

Oh this isn't the one I was thinking of.

There's one at UGA.

Do I even try and give a proof outline in a 5 minute talk?

No.

I think I just put the theorem I'm talking about with the literature and maybe pretend to give background.

Looking at the participants I think the talk should be understandable to JE and DG.

So thats like 2/70

Just state what you've proved and why it's interesting.

I have no idea how I would be able to present anything in 5 minutes

I remember having trouble figuring out what I might have time for in the 20 minutes I would get at the joint meeting (then realized I did not have any funds to go)

Well I'm going to meet and talk with a specific expert, probably for an extended period of time.

I told my adviser about the option of giving a 5 minute talk, and he said I should do it.

Partially because they couldn't expect me to really do anything in 5 minutes anyway.

I suppose if everyone is already familiar with the basic notation of the subject then that helps

They arent!

then you can forget about any sort of detail I think. And you need to think hard on how you can even get the time to state the main result probably

I think I'm just going to assume some background that many people there probably don't have.

They'll only be lost for 5 minutes so it should be okay.

heh

There was a really poor quality knot theory question on MO , where after a moderator (ToddTrimble) said the question needed to be changed to be appropriate for the site and the question was closed, the OP accused ToddTrimble of being all 5 of the close voters. All of which were non-anonymous people with research contributions in completely different areas.

are highschool math and university math two disjoint sets?

@saturatedexpo close to it

@PVAL-inactive Yeah, it is generally funny when anonymous people seem to not understand that many people indeed do post under their real names here and on MO (and like that guy yesterday who said people needed to be at "the level of EGA" or they were beneath him, when probably they have done no actual research themselves)

is the folloing true? if $R\subseteq A\times B$, then $R^{-1}\subseteq B\times A$ and therefore: if $(a,b)\in R$, then $(b,a)\in R^{-1}$? AND therefore: $R\cup R^{-1}$ is a symmetrical relation. (but not of A and B, but of some other sets)

What are $A,B,R$?

Ignore me, that was not meaningful

@Danu Sorry, R is a generic relation. A,B generic sets.

$R^{-1}$ is a reverse relation of $R$

@BalarkaSen you read section 4 of MS right

So does it actually follow from the axioms that SW classes are additive under disjoint union?

I think not (the only relevant map I can find is the inclusion of the two parts, but that won't give you a lot of info about the SW class of the bundle on the union, will it?)

Old Macdonald had a form; ei /\ ei = 0

@PhysicsGuy I can't get the wedge to fit into the melody :(

An experimental physicist meets a mathematician in a bar and they start talking. The physicict asks, "What kind of math do you do?" to which the mathematician replies, "Knot theory." The physicist says, "Me neither!"

@Danu ya it will.

For the natural inclusion $i:X \to X \cup Y$, you should know exactly what the map $i^*$ does on cohomology.

Can a mantra in math be: if you know something, try to prove it

@saturatedexpo I would say it stronger: If you have not proved it, you do not know it.

(where the "you" might be a bit less strict than it usually is, as it might also refer to the mathematical community)

@TobiasKildetoft i mean like a mantra in Kant's philosphy. So i might reavulate mine as: if you "see" something, try to prove it. (but yours is equivalent to that)

@saturatedexpo that gets way too philosophical for me

your statement has a merit.

@TobiasKildetoft mmmmm

I have not proved that ZFC is consistent

I have not proved that the integers are a set

@0celo7 Then you don't know it

@0celo7 That last one falls under the "community" thing

Do you know how to prove all of foundational mathematics?

I need to prove that $y(m, n) = \frac{1}{2} (m+n) (m+n+1)+m+1$ is surjective

@NaCl on what space?

@0celo7 No, why would I?

@TobiasKildetoft Then you do not know it

@0celo7 Note my note on the "I".

(which was a "you")

I'm confused

@NaCl do you even know what surjective is? (state the definition or say no ;))

@0celo7 $y:\mathbb{N}_0\times\mathbb{N}_0\to\mathbb{N}$

Many of the things I care about were being thought about long before anyone really thought seriously about "foundational mathematics".

@saturatedexpo A function $f:A\to B$ is surjective if $\{a\in A:f(a)\in B\}$ equals $B$

No.

actually no

Should be ok now

Why not $\forall b\in B\exists a\in A:f(a)=b$.

@NaCl no, the current version is nonsense

as it has a subset of $A$ equal $B$

He's trying to write $f(A)=B$.

@0celo7 Ahh, right

But what he wrote is nonsense.

yes, sorry, I'm not well concentrated

and do you understand it semifully?

He wants: $\{b\in B\mid \exists a\in A:f(a)=b\}=B$.

Man, I really have to learn more english.

However, I'm aware of the definition. I'd have given it to you in my language way quicker and correct

I don't understand your shortings like PDE, ZFC, etc.

@0celo7 Wenn das Bild der Funktion gleich der Zielmenge ist

@PhysicsGuy Well, those are not English per se

@PhysicsGuy That has nothing to do with English.

@NaCl Genau, man schreibt das als $f(A)=B$.

Ok

"image of the domain equals the codomain"

eigentlich geil, viel kürzer :D

Jetzt ergibt das alles mehr Sinn.

Stop with the German

$\forall a\in A: f(a)=B$ WRONG

@saturatedexpo No.

$f(a)$ is not a set.

ouch, correct

@PVAL-inactive Restriction, right.

edited it just for carity ;)

du kannst auch $\forall b\in B \exists a\in A:f(a)=b$ schreiben, aber das ist länger und nicht als klar als einfach $f(A)=B$ zu schreiben (my German isn't great but I hope that's understandable)

its good german

But I don't see how that will allow me to conclude that the SW classes should be additive under disjoint union.

Generally, to prove that a function $f:A\to B$ is surjective you need to show for any arbitrary element $b\in B$ there exists an element $a\in A$ such that $f(a)\in B$

@NaCl show that it doesnt hold for one element of $\mathbb{N}$ (in der Zielmenge)

@Danu So use the naturality under bundle maps.

I thought we were to show that it is surjective

yes

@saturatedexpo It should hold, as I need to prove that $y:\mathbb{N}_0\times\mathbb{N}_0\to\mathbb{N},y = \frac{1}{2} (m+n) (m+n+1)+m+1$ is bijective

@PVAL-inactive Of course I tried that, but all I get is $w(\xi_1)=\iota^*w(\xi)$ where $\xi_1$ is $\xi$ restricted to $M_1$, and similarly for $\iota_2$.

However, proving injectivity for it is quite easy, I'm just not sure about the surjectivity-part

Yes and there is a unique class which does that.

@NaCl ok, then write down 2 equations (do you know which?)

$\mathbb{N}_0=\mathbb{N}\setminus\{0\}$?

$\mathbb{N_0}=\{0,1,2....\}$

with zero, presumably.

no wait, maybe the opposite

$\Bbb N$ does not contain 0

@PVAL-inactive I know I'm supposed to get $w(\xi_1)+w(\xi_2)$, but I just don't quite see how this works.

some people include $0$ and some don't (I do), so I wasn't sure about the notation

@saturatedexpo No

I guess I should think concretely, in terms of cochains with support on the different parts, or something.

@Danu The cohomology of a disjoint union $H^*(X_1 \cup X_2)$ splits as a product $H^*(X_1) \times H^*(X_2)$ where the map $H^*(X_1 \cup X_2) \to H^*(X_1) \times H^*(X_2)$ is the induced map from the inclusion on each factor. This plus the naturality of the SW classes gives you what you want.

@PVAL-inactive So I'm not sure if I knew this. I think I was supposed to, probably. Thanks.

Hmm, it's definitely not in the notes on the courses I took.

Welp, Mayer-Vietoris or something

This is completely obvious say in the context of singular homology, as the chains split like this as any map from a simplex to a space factors through an inclusion of one of the spaces connected components.

Any map from a simplex lives inside a single component of the target.

Yeah

@saturatedexpo I thought I'd have to show that $y$ has an inverse, but $\mathbb{N}\to\mathbb{N}_0\times\mathbb{N}_0$ is not as trivial as, for example, $\mathbb{N}\to\mathbb{Q}$.

@Danu Mayer-Vietoris probably isn't so nice because you need to think about the homology of the empty space.

@NaCl i have a very similar assignment and post my solution to that. i hope it explains yours ;)

but i'm working on it myself^^

@saturatedexpo as I was fast enough to read, I simply can resolve $n=\frac{1}{2} (m+n) (m+n+1)+m+1$ and $m=\frac{1}{2} (m+n)(m+n+1)+m+1$ to find the inverse or what?

@PVAL-inactive It's not trivial?

@NaCl I'm not sure. because i work on Q x Q-> Q x Q. But yours is A x A -> A. So my intuition just tells me, that there exists no surjection.

(or at least no bijection)

Someone has an idea about this

@saturatedexpo Unless $A$ is finite there are clearly plenty of bijections

@TobiasKildetoft thanks for clarification.

@saturatedexpo There should, as there do exist at least two bijective functions $f:\mathbb{N}\times\mathbb{N}\to\mathbb{N}$. Take for example Cantor's pairing function or Szudzik's pairing function

@Danu It's $0$ in non-negative degrees and $\Bbb Z$ in degree -1. Essentially it is defined so it fits into the associated Mayer-Vietoris sequence defined by the disjoint union, so using Mayer Vietoris is essentially circular.

@NaCl yes then what you posted before would be the way to go. Altho that might be the tediousest way to go.

I'll scribble around, thanks for your input!

(thatswhy i just thought of looking for a counterexample, like squareroot 2)

@PVAL-inactive Heh.

@NaCl the way is wrong i think. and im sorry that i mentioned it ;)

okay

It makes sense that it is wrong, as you set elements from $A$ equal to elements from $B$

exactly

@PVAL-inactive Oh that's cute.

@Tobias just wondering, there is some AC needed to say that there are bijections $A^2\to A$ for every set $A$, right?

@Alessandro Right (it might actually be equivalent to AC, though I can never remember which ones are)

@NaCl i remember the proper way: i give you (x,y) in N (the Codomain). And you have to give me (w,z) that satisfie that (you can make them dependable from x and y)

@saturatedexpo And for that you need the inverse

wait, what. How is $(x,y)\in\mathbb{N}$?

oh

@MikeMiller I think during an algebraic topology lecture of Tye's I got very confused about some quotient of the empty space he was using.

well, that sucks because you got A x A-> A. So you have to give me z.

@MikeMiller and he was just like "believe me this is what it is".

@PVAL-inactive A quotient of the empty space?

Tye is funny.

Well there was some homology computation where he used that $H^-1(\varphi)=\Bbb Z$. I do not remember the details at all.

@PVAL-inactive Where $\varphi$ means the empty space?

ya

But then again, I like to study the kernels of the iterates of the map on matrices in char $p$ that raises each entry to the $p$'th power.

They have a name and everything

I've got a question: Is there any place where I can find any information regarding what would be good schools in a subfield (in my case I'm interested mostly in geometry/topology) to do my PhD at? I'm particularly interested in non-US schools (@Tobias, any ideas?), the US ones are easy to find.

@Danu No idea about that, Look at where the good researchers in your area are I suppose

@danu Yes. Wherever a good professor in that subfield is at that moment. Ask that good professor.

I never really looked around for where to do my PhD

I just went to what I thought was the best all-around department I got into, because I really didn't know what I wanted to specialize in.

I mean, I only applied for one place.

Generally though everyone I got a letter from had some very strong things to say about where I should apply and where I should go.

@PVAL-inactive @Tobias I understand that that'd be nice, but the reality is that deadlines are soon and I'm not at a level yet where I know who are eminent researchers in my area(s).

@Danu Sorry, no idea then

Let alone what exactly my areas are.

Then the bigger issue is finding appropriate people willing to write you letters.

@Danu Well, at least that makes you open to options

@PVAL-inactive I think I'll be okay in that regard.

@Danu Ask those people for their opinion then

Yeah, I will do that too.

I'm clear enough on what I want to say that I want to study geometry/topology of manifolds, but not exactly what part of that (and I realize that that's still a huge area).

So the letters will come from the corresponding professors in my department

i have a very beginner question about randomness

So no idea about the inversion thing? :(

@AlexanderBollbach what is it

its coming

in a lecture, the guy says something will transform into a random next state..

with a certain probability

i hope that helps you

(x,y are obviously in Q)

so based or that, my question is, can something be random if there's a known probability to transforming into that thing?

@AlexanderBollbach Would not not call dice random?

i guess i'm just looking for a concise formal definition of random

i mean i would call it random.. the die outcome

but that is different from perhaps the layperson usage of random as in "anything could happen"

@NaCl but i simply don't know how to approach that if A x A-> A ( or (a,b)->(c) )

@AlexanderBollbach That would be a weird use of the word

well i'm a weird person

but to summarize, something can have a 15% prob of occurring and still be random

@Semiclassical: Back. I question your partial derivative.

@saturatedexpo The function I try to invert is a very well known function I wasn't aware of. There is a proof on the wikipedia article.

hi @Ted

@NaCl that's nice :)

Hi @Ted

Another standard bijection $\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ is the one sending $(n,m)$ to $2^m(2n+1)-1$

Hi @Alessndro & Danu

I'm stuck on Semiclassic's question.

Which one?

$h:Q^2->Q$ $(x,y)\to x-y$ injective? Let h(x,y)=0. To be injective only one ordered pair can be mapped to 0. but (1,1) and (2,2) get mapped to 0. (both 1 and 2 are element of Q) Is this ok?

Up there ^^^

yep, that's not injective

I'm slowly being driven crazy by the signs and complex conjugations involved in curvature calculations.

I swear, I spent like 6 hours on showing that the curvature of the dual has the opposite sign due to that stuff.

I've written enough papers with them that I'm not crazy.

@TedShifrin I don't see it.

@TedShifrin I think I've finally got it right...

What's the connection form of the dual?

I'd say just $-A^T$

So calculate curvature using $A$ skew-hermitian.

Oh, I have that, don't worry

$F_{\nabla^*}=-F^T_\nabla$

My problem was with the Hermitian structures, haha

The anti-linearity of the isomorphism $v\mapsto h(\cdot,v)=v^*$ tripped me up for the longest time.