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20:00
not yet
well, elements of $\Bbb Q(\sqrt{2})$ are all of the form $a + b\sqrt{2}$, a and b both rational.
and automorphism preserve operations.
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how is the morphism defined? from where to where? why is it an isomorphism?
@cc $\Bbb Q(\sqrt{2})$ to itself.
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ok
it preserves operations.
so it's an isomorphism, right?
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20:02
I don't really see one, as an example
well, apply $f$ on $a + b\sqrt{2}$
$f(a + b\sqrt{2}) = f(a) + f(b\sqrt{2})$
(preserving addition)
$f(a) + f(b\sqrt{2}) = f(a) + f(b)f(\sqrt{2})$
(preserving multiplication)
are you with me, @cc?
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hmm, ok, I was just trying to find a concrete example of such $f$
@cc you'll soon get all of them, not just one.
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ok
@BalarkaSen ok by definition
so now $f$ fixes $\Bbb Q$ pointwise, by assumption.
hence $f(a) = a$ and $f(b) = b$.
Bob
Bob
20:05
@BalarkaSen can you take multtipilcation as product ???
@Bob what do you mean?
Bob
Bob
means the product operation you are taking in auto morph as $+$ right ??
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so $f$ is the identity function on $\Bbb Q$, ok
@cc yes, we have assumed that.
7 mins ago, by Balarka Sen
right. so now look at the automorphisms of $\Bbb Q(\sqrt{2})$ which fixes elements of $\Bbb Q$ pointwise.
@cc hence $f(a + b\sqrt{2}) = a + b f(\sqrt{2})$
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ok, what do you do with $f(\sqrt 2)$ now? is it in $\Bbb R \setminus \Bbb Q$?
20:07
@cc note that $\sqrt{2}^2 - 2 = 0$ so $f(\sqrt{2}^2 - 2) = 0$ as auts preserve identity.
$f(\sqrt{2})$ can be either $\sqrt{2}$ or $-\sqrt{2}$.
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indeed
so auts are either identity or conjugation.
@cc Balarka is telling you the modern, field-theoretic version due to Emil Artin. Classically, you start off with a given polynomial, consider the rational relationships between its roots and whether those roots can be swapped around and in what ways while preserving those relationships. You get a permutation group doing this, the Galois group, and if that's solvable (origin of term), then the polynomial is solvable in radicals. This is same as the field extension stuff Balarka is describing.
@AndrewG that's much more group theoretic, as found by Galois.
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ok interesting, does it help finding other roots when you have a polynom of degree n?
20:09
you directly work with pre-Cayley aged groups by doing so.
@cc i am not finished yet -- the automorphisms form a group.
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if there are more than 1 automorphism
the auts are either identity or conjugation.
$\{1, f\}$
$f^2 = 1$
So your group is $\cong \Bbb Z_2$
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that's a simple group, yes
that's precisely your galois group.
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oh ok
so you can do that for all algebraic equations? $x^3-2=0$
20:12
@cc yes.
sure.
But you may have to adjoin more than one elt to get a nontrivial group now.
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how does it look like?
@cc $\Bbb Z_3$
Now $S_3$
=P
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hehe ok
@cc Yeah, and in particular, it means that some degree +5 polies can be solved in radicals -- namely, those with solvable Galois groups. (But there are many, many non-solvable groups of order $\ge 5$, so no "general solution" in radicals.)
@AndrewG well at least you can adjoin roots without knowing it.
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20:14
yes degree 4 equations are already hard to solve in general
@cc and the radical part of the roots of 5th degrees are in general much complicated.
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Is this theory useful to find roots and relations between them, or it's useful for other stuff?
@cc both.
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ok
@AndrewG in a fancy style, some degree +5 polys have solvable deck transformation group of the covering of $\Bbb C$ through the corresponding riemann surface.
=P geometer trapping =P
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20:18
@BalarkaSen Can you explain Riemann surface too :) I've seen this term a lot in 'machine learning'?
@cc oh noes. here we go again. =P
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They use Riemanian manifolds
I don't know any manifolds.
I know what Riemann surface are though.
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so surface $\ne$ manifold
@cc $T^2 - x$, for example, has two roots.
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20:19
yes
@cc So place one sheet of paper after another one sheet of paper corresponding to $\sqrt{x}$ and $-\sqrt{x}$.
@TedShifrin got in a mess. now i have to explain to someone what a riemann surface is.
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here's $f(z) = \sqrt z$
@cc meh. uncomfortable picture.
20:22
Just say graph of multivalued function.
i have got a better one, wait.
@cc: Yes, surface = manifold. As long as it has no bad points (like a cone).
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All bad points I know are in $\Bbb R^2$ like cusp points
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an inflection point is a bad point? or saddle point in 3d
20:25
@cc
No, a bad point is one where there is no tangent plane.
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@BalarkaSen larger view, better
@cc the point $z=0$ is where two planes meet
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yep
since $\sqrt{0} = -\sqrt{0}$
@cc And if you let a point in the neighborhood of $x = 0$ traverse a loop around the origin you'll see the meaning of the picture.
The intersection will never really be made in $\Bbb CP^2$.
I forgot to say hello to you @TedShifrin
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20:28
@BalarkaSen what is $P$ here?
@cc heh. search projective plane.
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ok
@Balarka: The projective plane is no Riemann surface!
@TedShifrin I never said that.
Just checking.
20:29
I said that the apparent intersection will not happen in CP^2
one will just pull a sheet a dimension up.
mr eyeglasses is back!
Hi @TedShifrin
You been on vacation?
I got kicked out of home so I am staying at a church
oh crap ... I'm so sorry. What did your crazy mom do this time?
20:30
yikes, @nabla
It's kind of far from school and there's barely internet here so I haven't been able to go on
@TedShifrin She moved out
She moved out of her own home?
Well it's an apartment but yes, she up and left somewhere
And the lease was up?
I'm not sure
20:32
Wow ... Did she tell you she was gonna do this?
Nice ... looney person, but we knew that. I'm so sorry, mr eyeglasses. Do you have some friends or other family you can turn to?
i'm outta here. i am gonna go do some math with pen and paper in front of me.
Bubye, @Balarka.
I've consulted a school counselor who is trying to help me out
20:34
just a word : @cc never. ever. shower me with so much questions again.
Or a county social worker, maybe.
@Balarka: You have the right to not answer them.
As I do to you.
@TedShifrin i am not as rude as you.
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@BalarkaSen hehe, I see
That is debatable, @Balarka.
@TedShifrin shrugs
byes.
Bob
Bob
20:36
$Z_2$ what it means ?? @TedShifrin
Integers mod 2 ... Like clock arithmetic with only hours 0 and 1.
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@Bob http://www.wolframalpha.com/input/?i=Z_2&a=*C.Z%21_2-_*FiniteGroup- or http://www.wolframalpha.com/input/?i=Z_2&a=*C.Z%21_2-_*ModernAlgebraToken-
Bob
Bob
@TedShifrin @cc got it
Gee, I'm surprised no one's starred @Balarka's rudeness statement.
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$\Bbb Z_n$ for n prime is a field, I think
20:39
@AndrewG: You hiding?
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are all algebraic numbers (numbers solution of a polynom of degree n with integer coefficients) covering all $\Bbb Q$?
what do you mean by covering $\Bbb Q$?
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@TedShifrin "equal"
Are you asking if the set of all algebraic numbers is a countable set (like $\Bbb Q$)?
Certainly not equal. Look at $x^2-2$.
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@TedShifrin but can you find a given polynom with integer coefficients for which each rational number is a root?
20:46
Of course you can. $qx-p$ has $p/q$ as a root. That's not very interesting.
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:)) n=1 is enough, I'm stupid
What's more interesting is that the set of algebraic numbers is in one-to-one correspondence with $\Bbb Z$ or $\Bbb Q$.
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it equivalently means that it's countable?
Yas
You get the existence of transcendental numbers for free.
20:49
mr @Pedro !
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transcendental numbers are numbers not algebraic :)
Well, mr @Pedro ... Federer played a great match, but he didn't quite make it.
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is there a notation for transcendental numbers for a given degree for example "transcendal numbers up to degree 1" = $\Bbb R \setminus \Bbb Q$
20:58
They all sit inside $\Bbb C$, @cc.
@TedShifrin He seems a bit out of tune.
He is.
Getting old.
That was the best match I'd seen him play in about 2 years, @Pedro. But, yeah, he too cannot escape age entirely.
@TedShifrin I'm getting old, too.
You poor baby ...
Dang it Ted, look at my profile.
21:02
Oh right, happy belated birthday!!
Now you're an adult. Scary, isn't it?
@TedShifrin It's not belated, it's today. You'd know if we were Facebook palsies.
@TedShifrin Hehe, I'm an USA-adult.
Right :) You are always welcome to make me a friend ... I just don't initiate such things ...
I didn't know that.
Well, happy birthday, kiddo. YOu're right — it is all downhill now :)
Bob
Bob
@PedroTamaroff happy birth day
21:09
@TedShifrin I took a crash course on homological algebra yesterday night. I had to nap just some time ago to recover.
LOL ... who'd you take it from?
@PedroTamaroff birthday??? I have integrat gifts then to offer :D Compute in an easy way $$\int_0^{\infty} \frac{\log(x)}{(1+e^x)^2} \ dx$$ :-)
Are you doing anything to celebrate your birthday, @Pedro, other than the trip?
@TedShifrin I am hoping to do something with some friends after I'm done with exams.
@TedShifrin Jacobson's BAII.
@Chris'ssis inb4 leibniz differentiation
21:12
Ah @Pedro ... well, Jacobson's a good teacher ... :) Your parents aren't making you a nice dinner, @Pedro?
i've computed those things before @Chris'ssis
@TedShifrin Yes, we had a great brunch.
That's very nice, @Pedro.
@TedShifrin One of my sisters bought some pain au chocolat, a cinnamon roll and a second french thing... I cannot recall the name.
give me a hint?
It was covered in chocolate and had a yellow pastry inside.
21:17
oh, pastry cream inside ... I know exactly what it is
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paris-brest :s
@Chris'ssis how do you open that gift?
no, individual servings with chocolate on top @cc
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just a hint
it's an éclair, @Pedro @cc
@cc It's already opened by Pedro. :-)
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21:18
Leibniz differentiation, ok
@TedShifrin Yes, that. =)
I didn't just fall off the turnip truck, @Pedro :)
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I'd just want fruits as a birthday cake :)
Yum ... pêches blanches...
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or persimmons, or figs
21:33
Yum figues ...
21:49
@PedroTamaroff this integral requires some work (many students would fail to get the right answer) $$\int_0^{\infty} \frac{\log(x)}{(1+e^x)^2} \ dx$$ :-) Give it to your class and see if any get the right answer (without software help)
@Chris'ssis Nah, consider $\int_0^{\infty}x^s/(1+e^x)^2dx$; diff w.r.t $s$, profit.
(Provided one knows a formula for that, which involves $\zeta$ and $\Gamma$.)
Still nice.
@PedroTamaroff From that point there is still some work to do.
@Chris'ssis Why? =/
@PedroTamaroff At a certain point you need to know how to compute a certain limit in terms of the derivative of the zeta function.
@Chris'ssis Well, still I guess we can agree the idea is clear.
21:53
@PedroTamaroff Sure, but my point is that I feel that many would be in trouble with it in a real test.
ATM I am a bit lazy to compute $\int_0^\infty x^s/(1+e^x)^2dx$ but IIRC it was $\Gamma(s+1)\eta(s)$ or something like that?
In mathematics, in the area of analytic number theory, the Dirichlet eta function is defined by the following Dirichlet series, which converges for any complex number having real part > 0: :\eta(s) = \sum_{n=1}^{\infty}{(-1)^{n-1} \over n^s} = \frac{1}{1^s} - \frac{1}{2^s} + \frac{1}{3^s} - \frac{1}{4^s} + \cdots This Dirichlet series is the alternating sum corresponding to the Dirichlet series expansion of the Riemann zeta function, ζ(s) — and for this reason the Dirichlet eta function is also known as the alternating zeta function, also denoted ζ*(s). The following simple rel...
@Chris'ssis That's unnecessary.
@PedroTamaroff There is the integral you asked about.
Ah, use some [alt text](link) there.
$\Gamma(s)\eta(s+1)$ then?
Heh.
21:58
@PedroTamaroff This is just a tiny bit of the work to the series $$\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} (-1)^{k+n} \frac{\log(k+n)}{k+n}$$ that can be computed in lots of ways.
I didn't decide yet which one is the fastest way.
@MikeMiller skyone
heya @Mike
@Mike: Here's a cool question I haven't had time to figure out yet.
Busy with my myriad own problems, @Ted
LOL
Call your psychiatrist?
22:05
I don't think they'll be able to help :)
@TedShifrin Harder than I thought - obvious idea is to use LES of a pair, with the identification $H_\ast(M, K) \cong H_\ast(M/K)$, but naturality is garbage here, since we have a map $f: M/K \rightarrow M/K$ rather than a map $M \rightarrow M$...
Nah, cell structure makes it easy if you're talking about the problem I linked to. But I haven't thought about Lefschetz.
How does cell structure make it easy? He ruled out the trivial cases.
no, the homology is easy ... but the cup product structure ... I dunno.
It has cell structure of $D^0 \cup D^{2k}\cup D^{2k+2}\cup\dots\cup D^{2n}$.
Oh, no, calculating the homology is easy - I wanted to use naturality of the sequence to learn some stuff about the induced map $f_\ast$ by the way it acts on $H_\ast(\Bbb{CP}^n)$ etc
Yes, I know
I think the maps are the same on levels $2k$ through $2n$, but I haven't thought too hard.
22:14
pick $k$ large enough and the cup product structure is useless
It's the cup product structure that gives the fixed point property for $\Bbb CP^n$, so I'm guessing we don't have anything in this setting.
So we need an example of a specific map for $k$ and $n$, right?
I'm trying to think of a nice case like $\Bbb{CP}^3/\Bbb{CP}^1$. I'm a bit worried that this fella is the only nontrivial CW-complex with those cells!
Bah, ignore that
"Up to homotopy" is garbage for FPP unless you're proving FPP with some homotopy invariant structure anyway
Anyhow, didn't mean to distract you :P
22:18
I'd rather be distracted, to be honest
Sick of chasing my error
LOL
Welcome to math.
yup
@TedShifrin Now I'm more interested in whether or not $\Bbb{CP}^3/\Bbb{CP}^1$ is homotopy equivalent to $S^4 \vee S^6$.
Well, be careful. Why isn't $\Bbb CP^2$ then $S^0\vee S^2\vee S^4$?
@TedShifrin Hopf map.
We're killing off part of the attaching map, I'm thinking about that. :)
Right ... of course. So where did the map $S^5\to S^4$ go?
oops, I mean $S^5\to\Bbb CP^2$.
But we don't have $\Bbb CP^2$, so the map degenerates.
22:25
Degenerates in some way.
It's one of two maps, up to homotopy.
@TedShifrin I have an only slightly discontinuous function $f$. $\int_0^1 f(x)\sin(n\pi x)dx$ should be getting smaller as $n$ grows large, yes?
Right. I had forgotten, but just looked up $\pi_{n+1}(S^n) \cong \Bbb Z_2$. I should have a framed cobordism argument.
Ah, that's beyond me, unfortunately.
Riemann-Lebesgue Lemma ... As long as $f$ is in $L^1$ you should be ok.
Absolutely.
It's the indicator function of the union of three intervals... that's pretty damn nice.
nevermind, of course it's growing small
Basically, you get a framed $1$-manifold, and the framing of its normal bundle is given by an element of $\pi_1(SO(n-1))\cong \Bbb Z_2$ (for $n>3$).
OK, dindin time for me.
22:41
The only frames I know are around paintings
Snooker, @Mike?
Hi @Ted.
@DanielFischer Herro Daniel-san.
Hola Pedro, happy anniversary.
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23:15
@Chris'ssis I failed :) I tried to work with $\frac{\log x}{1+e^{\alpha x}}$ which leads to $f''(\alpha) =\int_0^{\infty} \frac{\log x \log(1+\alpha e^{ x})}{x^2}dx $ but nothing to do then, or with $\frac{\log {\alpha x}}{1+e^{ x}}$ , with $\frac{\log x}{1+\alpha e^{ x}}$, no better things
@DanielFischer I spent the past few hours trying to find a certain error in a computation. I found it, so now my results aren't completely off. Instead, they're off from what I'm expecting by exactly one.
@Chris'ssis Didn't you see my comments?
Work with $x^s/(1+e^x)^2$.
@MikeMiller Have you checked your expectations?
Yes, actually.
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@PedroTamaroff ok :)
23:17
What sort of computation, @Mike?
@DanielFischer Harmonic measure, but mostly it was solving a PDE with Fourier stuff.
@cc First, you should determine what $$\int_0^{\infty}\frac{x^s}{(1+e^x)^2}dx$$ is.
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seeing what you said it's $\Gamma(s+1)\eta(s)$, I'll try to prove it and link it to the other integral
@MikeMiller I'm sorry to hear that, I won't offer my help checking the stuff then.
I am not sure it equals that. Check it just in case.
23:22
@DanielFischer Not even remotely worth your time :)
I was just whinging.
@DanielFischer Actually, I think I know what it is, and you might be able to help me on it (it's pretty simple).
I have a continuous function $f$ on $[0,1]$ with $f(0)=1$. I wrote that it has a Fourier sine series, convergent everywhere but the boundary, $$f(x) = 1+\sum_{n=1}^\infty a_n \sin(n\pi x).$$
That $1$ shouldn't be there, should it?
@Mike, whining, thou meanst?
Hi @DanielF
@MikeMiller Not in a sine series.
@TedShifrin Nope.
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23:29
@PedroTamaroff $$\Gamma(s+1)\eta(s) = \int_0^{\infty}\frac{x^s}{1+e^x}dx$$ without the squared denominator
whingeing?
It has an e?
@DanielFischer Is my statement still true as I said it, then, if I nuke the 1?
Ah. For the squared denominator, use the infinite series of $x/(1+x)^2$.
Or something like that.
@MikeMiller I think so. $\{\sin (n\pi x)\}$ is, iirc, an ONB of $L^2([0,1])$.
Not quite, @DanielFischer, if you want orthonormal there needs to be a $\sqrt{2}$ out front.
But that's what I thought too, roughly. Thanks. I was being silly.
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23:33
wait sorry
$\Gamma(s+1)\eta(s+1) = \int_0^{\infty}\frac{x^s}{1+e^x}dx$
@MikeMiller Heh, I didn't say what measure ;)
You got me there!
Well, I suppose having an out of place 1 would explain why everything I got was exactly one off, @DanielFischer
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@PedroTamaroff maybe decomposing the other part of the square $1/(1+e^x) = \sum_0^{\infty} (-1)^n e^{nx}$
@MikeMiller I think it would.
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