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00:00
if $Im \psi = <\alpha^2>$ then we have $\psi(K) = \alpha^2$ and this completely determines $\psi$. So we have the way $\alpha^2$ acts on H as $h \mapsto h^{-1}$
I would like to get a presentation now for this case?
how would I get a presentation from the semi-direct product ?
@MikeMiller Maybe one can answer that for self-diffeomorphisms? ($k = l$)
presentation for this group.
@Balarka Huh? They're not diffeomorphic otherwise.
I am not sure what you are replying to. I was asking when such diffeomorphisms extend to diffeomorphisms of the handlebodies they bound, assuming $k = l$ (because otherwise the question doesn't make sense)
Maybe I was unclear on what I wrote after the braces.
That's exactly what the OP was asking.
Oh!
No it wasn't. I'm sorry.
00:06
No worries.
This should be significantlt easier then. What you said was the hard one!
Balarka has totally messed up his biological clock again.
Huh, funny.
@TedShifrin The sleep cycle is not homologous to the original one, once again, so...
maybe you need a Dehn twist or two
@Balarka Assuming you were asking when you can extend them to diffeomorphisms, I mean.
00:08
Ah, ok.
Yes, I guess that's what I was asking.
In the OP's question the class of such maps is nullhomologous, but I don't know off the top of my head how to classify them. Maybe one needs to look at it generator-wise. I agree it can't be that hard.
@Balarka: I expected you to deign to respond to my Dehn twist suggestion.
@TedShifrin :P
I tried to think up something stupid, but they were so stupid I didn't reply
LOL, well, good for you for self-censorship :)
00:26
How would one prove $\Bbb{CP}^2 - 0$ is not homeomorphic to an open subset of an affine variety?
Is that even true?
Nah.
Homeomorphic, or algebraically isomorphic?
Just homeomorphic. It's obviously false in that case.
Why obviously false?
That's the normal bundle of $\Bbb P^1$ in $\Bbb P^2$. Embed $\Bbb P^2$ (topologically) in some $\Bbb C^n$; doesn't that give me an open subset of $\Bbb C^n$?
Oh, with the not you meant false.
Mixing categories sounds suspicious to me.
00:31
Actually I am not too sure about what I just said.
Of course topologically $\Bbb P^1$ embeds in some $\Bbb C^n$.
But why would someone use the language of affine variety?
Maybe they mean in the Zariski topology?
There's no reason that open subset of the embedded $\Bbb P^2$ is open in $\Bbb C^n$, I mean to say.
@TedShifrin I was thinking about this.
Oh, well, yes, that's very right.
I assumed complex topology; if they mean Zariski then meh
evening chat
00:33
I read that question as in the Zariski topology, since it's all in the algebro-geometric language.
The OP should have specified.
Hi @Semiclassic
hi @ted
I'm still pushing on that one Riemann surfaces / differentials question, without much to show for it yet
@Balarka I give up. I can't do it for $k=1$.
not surprised, but still frustrating
Yikes.
I haven't been keeping score, Semiclassic.
00:36
nah, i'm just grumbling
Well, grumbling I'm quite used to.
Especially from @Balarka and @MikeM ... and @Danu.
I have an excuse. I'm bitter.
"Bitter. Party of one!"
2
I forget what movie that comes from.
I'm glad there was no internet when I was in grad school.
Google is life.
Bitching in chat also is.
00:39
i'm still trying to understand stuff like $x^s y^r dx/(\partial F/\partial y)$
@Balarka: I just put a comment asking the OP to specify topology.
Good idea.
You don't need the $y^r$ for the basis guys, right, @Semiclassic? Now I'm confuzled.
I think you do if it's not hyperelliptic
And mine isn't
at least, I don't have that presentation of it
mine is stuff like $F(x,y)=(x^2+y^2-1)^2-s(x^3-3xy^2)=0$ where $s$ is a parameter
I'm still curious to understand whether $\Bbb{CP}^2 - 0$ is homeomorphic to a quasiaffine variety or not.
@TedShifrin Zariski topology. Meh meh meh.
00:42
me using $y^r$ is motivated by my seeing this answer and me pretending I know what it means
Oh, the OP answered, @Balarka?
in the comments, yes.
OK, just as I suspected, @Balarka.
What I'd really like to know, though, is how one uses that basis
If I'm doing linear algebra, for instance, I know how to find the components of a vector relative to a given basis---just take inner products with the basis vectors
Yeah, I think you're right, as long as $s+r$ is bounded above by something appropriate.
Well, what are you trying to do?
00:46
Suppose I consider the 1-form $y\,dx$. that may or may not be holomorphic on $F(x,y)=0$---it could have a pole, for instance.
but suppose I subtract off any such poles and am left with something that is holomorphic.
So multiply and divide by $\partial F/\partial y$?
...hm.
that's...not a bad idea at all.
LOL
And then try to get rid of terms whose degrees are too high?
and then $y(\partial F/\partial y)$ will definitely be a polynomial
Yuppers.
But the degree is too high.
00:48
yeah.
But use $dF=0$ on the curve?
Let me know what happens when you play.
Sounds right.
I might go back to a simpler example first and see if I can make it work for that
Sounds smart.
My question sounds difficult, unless I sound stupid.
questions sounds ... go to sleep :P
00:55
or just edit the offending s.
:)
Nah, go to sleeps.
Or, as the modern leet-speaking society would say, sleepz.
But that wouldn't be in line with our previous line.
Balarka, why are you still awake?
Because he's destroyed the homotopy class of his sleep cycle.
00:58
@Krijn celebrating
You're gonna end up sick again, @Balarka. Seriously ...
How can a kid be so smart and yet so not smart regarding sleep schedules
Book-smart kids are often life-smart dummies. Not just kids.
No one asked me what I was celebrating!
And nor shall we.
01:00
Boo.
30 hours without sleep?
Oh well, Krijn didn't get my memo.
You crushed the joke.
I was gonna say "being awake"
Krijn was funnier.
Yay!
Back to maths for me
01:02
Speaking of late hours, isn't it something like 3 AM, @Krijn?
Shhhhhhh.
puts Krijn and Balarka in sleep detention
OK, I really should sleep. Have to do another bunch of homework tomorrow.
puts Balarka on ignore again
01:05
And then off to connections.
@TedShifrin Hi! How are you? It has been quite a while... I needed a break from SE for a bit, haha--too much politics. :P
Not the kind of politics that's been upsetting me, though ...
You still a computer science major?
I am, but I actually just declared a double-major in math
Yippee :)
We're glad to have you back.
Haha I'm glad to be back
01:12
Even if you go to grad school in computer science, it'll be a plus for you.
True! The only remaining class I have for the double-major is "Basic Real Analysis" (I managed to take a bunch of classes that didn't depend on analysis for the major.)
That's not so surprising.
You surely did more algebra/crypto stuff.
Maybe even numerical analysis.
I mainly focused on algebra things. I actually took an Algebraic Combinatorics course, which was really cool
Yup. And even probability is super important and cool.
But it feels like just yesterday that you started college.
Indeed. :) I'm focusing my CS-related studies on Machine Learning, so some probability/stats stuff is helpful.
And it really does feel like yesterday!
It's gone by way too fast.
01:15
Two years?
I actually ended up taking 3, but the plan was originally 2
Good grief. Don't be in such a hurry.
(Adding the math major took an extra year, but it was well worth it)
:P
4 is just fine.
You going to grad school in CS ?
How does 2 years even work?
01:17
I've actually decided to not go to grad school immediately after graduation
A lot of AP and overloads.
@0celo7 I took 60+ credits of work in high school at a community college
I had 60 credits and I'm going to struggle graduating in 4
That's fine, @apnorton.
@0celo: Engineering is very credit-heavy, and you're doing very serious math. So stop bitching.
I will stop bitching
But only to ask about semicontinuity
01:18
I will quote you on that.
@0celo7 I 100% agree with Ted--CS was super easy here :P
I still favor staying in college for the 4 years. More social things, more academic things ...
And @TedShifrin I think I might end up at grad school eventually, but I have some loans I wanted to settle before that.
I pontificated on this frequently as an adviser at UGA.
@TedShifrin I really have 0 feel for $\liminf$ and I have to understand a proof using it. The book gives 0 explanation as to what it actually is
I read the wiki article
01:19
No problem, @apnorton.
It's not scary, @0celo. Way easier than most of the stuff you seem to handle.
And yet you know what a Sobolev space is?
As someone who did it in 3, and whose brother is about to do it in 2, I 100% recommend the 4 years if one can afford it.
Think about all possible convergent subsequences.
Take the inf of their limits.
@MikeMiller Would it surprise you to learn I cannot solve a quadratic?
ROFL ... No.
01:20
I don't much care to.
I clearly care about computations more than most people here.
Am I on that list? I'm not sure why, if so.
You just said you didn't much care to :)
Be consistent.
Why? Is maths?
I don't much care to learn that he cannot solve a quadratic.
01:21
Yeah, right.
All I hear in this chat is "Balarka, do this computation!"
And Balarka has been very obedient and learned a ton.
@0celo: I answered your question. Is there a follow-up?
@TedShifrin My question is how do I actually compute limit inferiors.
You look for convergent subsequences with the smallest possible limits.
I have a specific example
$f:X\to\Bbb R\cup\{\infty\}$ ($X$ metric) is lower semicontinuous at $x\in X$.
I'm trying to understand the sequential characterization of lower semicontinuity.
01:24
OK ... I always have trouble remembering which is "lower" and which is "upper," but in the end it makes sense.
So for $c<f(x)$ there is a nbhd $U$ of $x$ such that $y\in U\implies f(y)>c$.
Right, it's continuous from underneath.
Yah
OK. Go on.
So take $x_n\to x$, then clearly $\exists N\in\Bbb N$ such that $\forall n\ge N: f(x_n)>c$.
01:25
Right.
From this, Jost jumps immediately to $\liminf f(x_n)\ge f(x)$.
Think about contradiction, and choose $c$ appropriately.
@TedShifrin Not sure what a contradiction does, but one can see that $\inf_{n\ge m}f(x_m)>c$ for any $c<f(x)$ as $n$ gets large enough, right?
You have it, @0celo?
for any $c<f(x)$.
fixed
01:29
Well, not quite
LOL
wait
Contradiction means that I suppose $\liminf = c'<f(x)$. Now choose $f(x)>c>c'$.
You can then choose a sequence to get an immediate contradiction.
Choose what sequence?
There's a sequence $x_n$ with $\lim f(x_n)=c'+\epsilon$ for $\epsilon<c-c'$.
$x_n$ is already fixed, no?
01:32
Nooo ... Jost means over all sequences.
That's obvious, but first we prove it for one sequence.
At least I thought he did. But it doesn't matter.
If $\{x_n\}$ is fixed, pass to a subsequence for my statement and call it $\{x_n\}$.
Maybe I should write down what $\liminf$ really means.
There are two different definitions.
But you should prove they're the same.
The one I'm using is better :)
$\forall\epsilon>0$ $\exists N\in\Bbb N$ s.t. $\forall n\ge N$ : $|c'-\sup_{n\ge m}f(x_n)|<\epsilon$.
01:35
I'm using a more powerful one.
Which is?
Scroll up.
@TedShifrin I don't know what this means.
I said it before that.
Take the inf of all subsequential limits.
My thing up there is backwards.
I will crawl into a cave and work on this.
01:37
You needed inf.
Oops, that too.
OK. You'll get it.
It's just one big typo, really.
Contradiction helps. Then you can always sort it out and remove the contradiction if you want.
Ah, a picture helps too.
01:39
Always.
Ok, I got it.
Good boy.
...that's a little weird.
LOL.
Night.
Bye.
@TedShifrin Are you still there?
The idea is that by making $c$ close enough to $f(x)$, I can make $\inf_{m\ge n}f(x_m)$ close to $c$ for $n$ large enough, hence close to $f(x)$
So if you set $f(x)-c=\epsilon$, then the Wiki definition works.
And let $c\to f(x)$ (from below)
01:58
does this proof look good?
Almost, just add $n \neq 0$ at the start and you're good to go.
alright thanks :}
im self-learning, so i have nobody to help me
which sucks
guys does anyone know anything about graphical convolution?
Not me.
like taking two arbitrarily defined functions and using graphical inspection to convolve them?
02:13
I wish I knew
Every time I see a convolution it blows my mind
02:42
Hey @arctictern here ?
I would like to ask if H,K are finite groups with relatively prime order.
I would like to prove that $Aut(H \times K) = Aut(H) \times Aut(K)$
02:53
okay
suppose $\alpha$ is an automorphism
show $\alpha(H\times 1)\subseteq H\times 1$ and $\alpha(1\times K)\subseteq 1\times K$
get equality because of size
so @arctictern are you a grad student?
so the way I am doing this is as follows $(\theta_1,\theta_2) \mapsto \psi$ where the way $\psi$ works on H is as follows $(h,k) \mapsto (\theta_1(h),\theta_2(k))$
@0celo7 undergrad technically. behind on gen eds but I've worked with some other faculty on math stuff without credit, am angling to get bachelors and masters at the same time with retroactive credit
Now this map is homomorphism and injective
just have to get surjectivity
@arctictern Worked without credit?
03:03
@0celo7 like, we talk weekly about things (lie theory, algebraic number theory, random stuff). I do about a 75 min lecture every week, so it's about the equivalent of taking a course.
@arctictern have you heard of Allufi chapter 0 ?
I am currently reading it in my spare time it is very nice right ?
I think so
You give a lecture?
03:04
@0celo7 presentation, whatever. like, crash course on rep theory, tensor products, string diagrams from qiaochu's blog, octonions, classical groups, etc. etc.
lots of random things
to a handful of other faculty
that is awesome @arctictern
interesting
which university ?
the one I go to
I mean which university is it
if you don't mind me asking.
03:11
he prefers to be anonymous
though maybe I can trick him into coming to UCLA to work with raphael rouquier
cool
Tern, unlike social science and humanities, the MA doesn't much matter for going for a PhD. But you probably know this.
Anonymity is good.
03:46
@arctictern is $Aut(Z_p \times Z_p ... \times Z_p) = Gl_n(Z_p)$ ?
oh ok is there an obvious way to see that ?
I mean yeah I could create a map as a matrix
preserving group structure
<-> preserving addition
<-> preserving multiplication by integers
<-> preserving multiplication by integers mod p
hm yeah so I think I got a good way of proving it
So $Z_p$ is a vector space over itself right ?
04:00
the basis of the vector space $Z_p \times ... \times Z_p$ is just $(1,0,0,....,0),(0,1,0,..,0),etc$ So we can define a map $\phi : Aut(Z_p \times Z_p ... \times Z_p) \rightarrow GL_n(Z_p)$ as follows
so elements of $Aut(Z_p \times ... \times Z_p)$ are just maps $\psi$ so we can define the first coloumn of the matrix by where it sends the first basis vector etc.
and I think this should work right ?
04:26
@MikeMiller @arctictern I think I have much better solution. If I show that any endomorphism is a vector space of dimension $n^2$ then I am automatically done.
> endomorphism is a vector space
Because we know any finite dimensional vector space of same dimension are isomorphic right ?
because then that would Show that $Aut(Z_p \times ... \times Z_p)$ is a vector space of dimension $n^2$
Aut(Zp x ... x Zp) is not a vector space
an endomorphism is not a vector space
@arctictern kek
oh ok...
is true that if $e_1,...,e_n$ is basis for $Z_p \times ... \times Z_p$ then if $\phi \in Aut(Z_p \times ... \times Z_p)$ then $\phi(e_i)$ are linearly indepedent.
yeah I guess this should be true
yeah this is true
04:45
@TedShifrin here ?
@arctictern I solved it would you like to discuss it ?
 
1 hour later…
05:54
0
Q: Commutative group-likeness formed by compressibly repeated substrings of a formal string?

Enjoys MathLet $S \in \Sigma^*$ be a string over a finite alphabet. A compressibly repeated substring is a substring $t \leqslant S$ (let that mean substring), such that $t\gamma t \leqslant S$ and $|t| \geq 3$, or $t\gamma t \gamma' t \leqslant S$ and $|t| = 2$, for some strings $\gamma, \gamma'\in \Sigma^...

What up.
:P
Therfore P=NP =P
lol
Probably not quite there yet, but I rarely see this many conditions met without finding a full on interesting algebraic structure.
06:47
@TedShifrin What conical pictures?
 
2 hours later…
user228700
08:32
Hey everyone :-)
user228700
No line can be divided externally in the ratio 1:1, correct?
what does that even mean?
user228700
External division..?
yes, what is that?
user228700
Yeah, no, it definitely can't. That would give us 1-1 in the denominator. In fact, we can't have any point dividing a line in any ratio $m:n$ in which $m=n$.
I have no idea what they mean by dividing externally there. The line segment is not divided into pieces.
user228700
No, not exactly...
user228700
Anyway.
user228700
Perhaps this will help:
user228700
(It's what helped me!)
10:05
Hi everyone, any hint/explanation, please?

Prove that $(\sqrt{3}+1)^n+(\sqrt{3}-i)^n=2^{n+1}\cos(n\pi/6)$
10:17
is there a good approximation for ${n^2 \choose n}$ ?
10:38
@Lembik use stirling
@Lozansky hi
10:59
$(\sqrt{3}+i)^n+(\sqrt{3}-i)^n$
$=(2\exp(i\pi/6))^n+(2\exp(-i\pi/6))^n$
$=2^n\exp(ni\pi/6)+2^n\exp(-ni\pi/6)$
$=2^n(\cos(n\pi/6)+i\sin(n\pi/6))+2^n(\cos(n\pi/6)-i\sin(n\pi/6))$
$=2^n\cos(n\pi/6)+2^n\cos(n\pi/6)$
$=2^{n+1}\cos(n\pi/6)$
notes: $\exp(iz)\equiv\cos(z)+i\sin(z)\\\exp(z)\equiv e^z$ (Euler's Formula)
11:18
0
A: Is that right asymptotic formula?

Claude LeiboviciHint $$\log \left(\binom{n^2}{n}\right)=\log\left(\frac{(n^2)!}{n!(n^2-n)!}\right)=\log((n^2)!)-\log(n!)-\log((n^2-n)!)$$ Now, use Stirling approximation $$\log(m!)=m (\log (m)-1)+\frac{1}{2} \left(\log (2 \pi )+\log (m)\right)+O\left(\frac{1}{m}\right)$$ Apply it to each term and simplify.

11:30
How to prove that $\dfrac{\sin{A}}{\sin{A'}} = \dfrac{\sin{B}}{\sin{B'}}$ if and only if $A = A' & B = B'$?
@deostroll what is A' and B'?
Angles...
that is obviously false
Okay, so why does it turn out to be false then?
for example, A=B and A'=B'
11:39
Okay, I missed it...you have to account of that condition too in the question...
what about A=180°-B and A'=B'?
what if A, B, A' and B' are just 4 random numbers which satisfy that equation?
I should also mention that $A, B, A', B'$ are values in the interval of $[0, \pi]$ only...
it does not matter
So random values in that interval...doesn't simply satisfy that equality...right?
sin(0.1)/sin(0.2) = sin(0.3)/sin(0.628692394) = 0.5025104592
the point is that you can just generate 3 random numbers for A, B, and A', and B' would be $\arcsin(\sin(A')\times\sin(B)\div\sin(A))$
11:46
@DHMO , Thanks for explanation.
you are welcome
 
1 hour later…
12:57
@DHMO Hi there
@Lozansky hi

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