« first day (2611 days earlier)      last day (2706 days later) » 

00:10
sup chat
heya Eric
hi @Ted
I saw you got my note about the geometric analysis lectures.
yup im gonna start watching after i finish homework for the night
You can let me know what you think. I haven't looked yet.
I saw you like your PDE class. What else did you settle on?
00:14
im taking grad algebraic top and rep theory
Ah, that's a good balance ...
all of my profs are fantastic lecturers so it's probably gonna be a fun quarter
Well, you lucked out.
yup yup
although the PDE class looks like it's gonna be tough
You need tough :)
00:17
true enough, i am very hyped for it
pdes r tough
00:33
Hello all, is there a thematic entry on SE for "correct result obtained by applying wrong method" ? I searched a bit, but didn't found, yet I would be surprised that such a topic won't exist yet.
ah, found it !
00:54
Yo @Eric
sup dude
And @Ted sorry I was out during that time. Classes are quite fun!
Eric: Is PDE on MWF or TR?
Sick
Sat in on DCal's class today and I can confirm his accent is quite nice to listen to
is he teaching complex analysis?
01:02
Yup
Today was mostly dull since he went all the way through the very basics of complex numbers and all, though he did talk a bit about field automorphisms so there was that
@AkivaWeinberger Say, do you happen to know how one could define what it means for a function to be recursive as far as $f:\Bbb{Ord}\mapsto \Bbb{Ord}$?
Same with any other function, no? If it's defined based on other arguments
Hm
But one would usually say that, for example, $f:x\mapsto\varepsilon_x$ is recursive
Where $\varepsilon_x$ is the $x$th epsilon number
Hm
No, I don't think that's my problem
*::walks away for a moment to collect himself::*
@AkivaWeinberger I guess my problem is defining "based on other arguments" when dealing with limit ordinal inputs
01:21
I mean, I wouldn't really put a rigorous meaning to what it means for a function to be recursive
Rather, I'd define what it means for a specific definition of a function to be recursive
Like, $F_n=F_{n-1}+F_{n-2}$ is recursive, but $F_n=(\phi^n+\bar\phi{}^n)/\sqrt5$ (or whatever) isn't
Oh, wait
Hehehe
I wanna make a Fibonacci ordinal sequence now
just for funs :-)
$\Bbb N\to{\rm Ord}$ seems possibly interesting, like a generalized Fibonacci function
$f(0)$ and $f(1)$ are random ordinals, the rest are defined by the recurrence relation
Hm, maybe it's not that interesting. Hm
Certainly, $f(0)=\omega$, $f(1)=\omega+1$ is predictable
$f(n)=\omega F_n+1$
I was personally thinking of $\Bbb{Ord} \mapsto\Bbb{Ord}$ :-P
$$F_\alpha = \sup\{F_\gamma+F_\delta:\delta<\gamma<\alpha\}$$
That's the simplest I could cook up
01:28
Hm, interesting
Yeah, that could be interesting
I wonder if it changes if you write $(\delta\ne\gamma)\land(\delta,\gamma<\alpha)$ instead
$F_\omega=\omega \\F_{\omega+1}=\omega2 \\F_{\omega+2}=\omega3\\ F_{\omega+3}=\omega5\\\vdots$
what with noncommutative addition and all
@AkivaWeinberger You mean allow $\delta>\gamma$?
Yeah
(Noncommutative addition, my God. Not even ring theorists go that far)
:P
Well, actually, in a ring with unity it follows from the axioms anyway
Well, if we agree that $F_\alpha$ is strictly increasing, then allowing $\delta>\gamma$ doesn't have much effect
01:31
(Expand $(x+1)(y+1)$ in two ways)
@SimplyBeautifulArt $F_{\omega+1}$ changes, no?
Wait, hold on, the way you wrote it, $F_{\omega+1}$ should just be $\omega$
@AkivaWeinberger But beyond that, there is little change
Hi chat
@AkivaWeinberger $F_{\omega+1}\ge\sup \{\omega+F_\delta: \delta<\omega\}$
01:32
hits drum set
akiva I igot a question about isomorphism theorem
abstract algebra
@SimplyBeautifulArt Oh!
$F_\gamma+F_\delta$, not the other way around
Sorry, I had misread
Yeah, I think that at every fixed point, it kinda goes "bleh" and repeats the Fibonacci numbers
@KasmirKhaan Go ahead
Nothing superbly interesting unfortunately :-(
01:34
we have a homomorphism from f :G-->G' the theorem sais that G/ ker f isomorphic to G'
my question is what is the map from G---> G/H
Is f surjective in that theorem?
I'd expect G/ker f to be isomorphic to im f
well yes i was going to write Im (f)
@KasmirKhaan You mean $G/\ker f\to G/H$ (the isomorphism)?
(Test: $\ker f$ $\im f$)
We started with G--->G'
($\operatorname{im}f$)
01:36
we did G--> G/H then G/H --->G'
what is the map from G---> G/H
The quotient map
x maps to xH
so basicly what the theorem sais
(which equals Hx because H has to be a normal subgroup (because otherwise you couldn't quotient by it))
that if we compose f bar with the quotient map
where f bar is the map from G/H ---> G'
its the same as going from G-->G'
01:39
all right thanks :)
This also shows that all kernels are normal subgroups (otherwise G/ker f doesn't make sense) @KasmirKhaan
Also, all normal subgroups can be kernels (take the quotient map G-->G/H, the kernel is H)
Thus, a normal subgroup could be defined as "something that can be the kernel of something"
I know that the kernels are normal ,but why would 'nt make sense
you mean the multiplication law on cosets would fail right?
02:01
Define a $f\in R_\beta$ if it satisfies one of the following:

$$f(\alpha)=\alpha+1\\f(0)=\gamma,f(\alpha)=\sup\{g(f(\delta)):\delta<\alpha\}\\f(0)=\gamma,f(\alpha)=\sup\{\varphi_\eta(g,f(\delta)):\delta<\alpha,\eta<\min\{\alpha,\omega^{\mathrm{CK}}_\phi\},\phi<\beta\}$$

where

$$\gamma\in l_\beta\\g\in R_\beta\\\varphi_\eta(g,\delta)=\begin{cases}\sup\{g(\varphi_\psi(g,\delta)):\psi<\eta\},&\eta>0\\\delta,&\eta=0\end{cases}\\l_0=\{0,1,2,3,\dots\}\\m_\beta=\bigcup_{\zeta<\beta}l_\zeta\\l_\beta(0)=m_\beta\cup\sup(m_\beta)\\l_\beta(n+1)=l_\beta(n)\cup\{f(\pi):\pi\in l_\beta(n),f\in R_\sigma,
I'm heading to bed, but I wonder if this can define recursive functions @AkivaWeinberger
Hopefully its also well-defined
02:15
@AkivaWeinberger how to make sense of R/Z
I know Z is a normal subgroup of R
but what do the cosets look like ?
@KasmirKhaan It's also called the "circle group"
[Random]
Breaking reality:
Essentially you add real numbers but ignore the integer part
$(\Bbb{R}/\Bbb{Q})/\Bbb{N}$
02:17
Imagine adding angles but ignoring multiples of 360 degrees
but with the idea of G/H
we have a cosets in the form
aH where a in G
I dont see the analogy here
@KasmirKhaan Remember these are additive groups
Actually no, maybe do it this way:
$(S^1 / \Bbb{Q}) /\Bbb{Z}$
So you would have the elements be of the form a+Z
okay but where does it say to ignore integer part?
pi+Z
for example is 1 coset right?
02:20
@AkivaWeinberger why even call it addition then :S
Hi semi
Actually, what is a nontrivial normal subgroup of $\Bbb{R}/\Bbb{Q}$?
these stuff is all comfusing to me
the index is inifinity on these cases
a+Z , a in R
each irratianal number is in its own coset
@KasmirKhaan not quite. for instance, $\sqrt{2}$ and $\sqrt{2}+1$ are both irrational but differ by an integer.
so they're in the same coset.
yeah thats what i said =p
irrational + Z
02:24
well, i was objecting to "each irrational number is in its own coset"
both sqrt(2) and sqrt(2)+1 are irrational, but they define the same coset.
Yeah but i meant if we ignore the addition by Z
each irrational number will have its own coset
well sqrt2 +Z
sqrt 3 +Z
How do you even define a coset in that case?
hmm
let me think if i got this right
02:27
The point is this. Given distinct irrational numbers a,b, it need not be the case that a+Z and b+Z represent distinct cosets.
yes i know but i meant
soemthing like sqrt 3
will be the representative of sqrt 3 +Z
so we dont need to worry about sqrt 3 +17
its allready been taking care off
pi will be in the coset {…, 0.14, 1.14, 2.14, 3.14, 4.14, …}
yes exactly
02:28
Well, 0.14…, not 0.14, but you know what I mean
I was about to say :P
yes yes =p
Right sure so like (0.7+Z)+(0.4+Z)=(0.1+Z)
'cause 1.1+Z and 0.1+Z are the same cosets
those are rationals
yeah. the setting here is R/Q not R/Z
02:30
ah =p
I'm not sure how one gets a handle on R/Q tbh
Some discussion on that here: math.stackexchange.com/q/182247/137524
i like Asaf's answer.
(and comparing his answer to the bottom one is an object lesson in why using paragraphs is essential)
> Any set of representatives for ℝ/ℚ cannot be measured.
wait, a nonmeasurable set?
Yep.
the comments to that answer reference the Vitali set
which is really weird
once again, infinitary set theory is cray cray
Well, Leaky, simpleart and I were currently somewhere in the wilderness of choiceless universes, and we are so far doing fine
well, that's certainly your choice :P
02:37
So I suspect when we return to the choiceful universe, we will be fine also
because the wilderness has a reputation of being even weirder than constructs from AC
@HernanEscobarSánchez are you sure this is your full name ? did not miss something ?
what is the axiom of choice?
Also need to check later: Whether the cantor set can be found inside the set of irrationals
considering 1/4 is in the cantor set
probably not
@Semiclassical Because it's the most logical extension of addition to the ordinals
Note that $\Bbb N\subset{\rm Ord}$
and ordinal addition on the finite ordinals (i.e. $\Bbb N$) is just regular addition
You could also think of it as a special case of order type addition
If $\alpha$ is the order type of an ordered set $A$ and $\beta$ is the order type of B, then $\alpha+\beta$ is the order type of the disjoint union of $A$ and $B$, with every element of $A$ placed before those of $B$
Will try to find the normal subgroups of $\Bbb{R}/\Bbb{Q}$ later, need to get back to the chemistry symposium
02:42
For example, if we call the order type of $\Bbb R$ "$\lambda$", then (since $\lambda$ is also the order type of $(0,1)$) we have that the order type of $[0,1]$ is $1+\lambda+1$
(The order type of a finite set with $n$ elements is just called $n$, so $4+5=9$ and such hold)
('cause if you smash a four-element ordered set and a five-element set together you get a nine-element set)
[double-checks that four plus five is, indeed, nine]
@Semiclassical Incidentally, that's why Indiana Jones shouldn't have bothered
He should have just let them show it to Hitler like they wanted to do
 
1 hour later…
04:12
Baka
04:51
@Semiclassical Just a reminder that your graph theory room is close to 14 days interval after which it can be frozen.
@Secret "the" cantor set: not really; "a" cantor set: maybe
@Secret interesting question
@Secret yes, you can't have all sets measurable with countable additivity and translativity
@KasmirKhaan every number in [0,1) defines a coset; nothing to do with irrationals
yeah, R/Z is much nicer than R/Q
R/Z: oh, that's just a circle
R/Q: ...yikes
05:11
2 hours ago, by Secret
Will try to find the normal subgroups of $\Bbb{R}/\Bbb{Q}$ later, need to get back to the chemistry symposium
Don't think of it as the set of representatives: think of it as set of cosets
That way, it is easily seen that $\langle \pi+\Bbb Q \rangle$ is a subgroup
(obligatory remark: since it is abelian, every subgroup is normal)
horrible question: What can be said about $(\mathbb{R}/\mathbb{Q})/\langle \pi+\mathbb{Q}\rangle$?
@Semiclassical the same about $\Bbb R/\Bbb Q$
Why the hell latex doesn't shows up in windows Edge ? That's one of the reason windows edge sux.
@AlexKChen you need to enable latex using the link in the room description
I know, but it isn't working in Edge, but the exact thing worked in Chrome
05:18
@LeakyNun so, not much? :)
Should have listened to the guy before installing windows
@Semiclassical well it's uncountable and non-measurable :)
@AlexKChen can't you still use chrome?
@Semiclassical are you going to prevent your room from freezing?
05:21
i dunno, tbh. it was there for discussing a specific problem and I'm not really looking at it now
so if it's served it's purpose then there's little sense having it outlive that
@Semiclassical I've raised a question there
hopefully you can read it
sure, I can read it. i just have no hope of answering it :P
6hrs of testing my brain fried pancake
hope jasper got some help today
hes been pretty down lately
05:28
@AlexKChen to find a finite field of order $p^k$, firstly find a polynomial of degree $k$ which is irreducible modulo $p$. Then, pretend $\omega$ is a root of that polynomial, and your field elements would be polynomials of $\omega$ modulo $p$ of degree less than $k$.
$i$ failed for $p=5$ because it represented the root of $x^2+1$ which is reducible modulo $5$: $x^2+1=(x+2)(x-2)$
So you can't find the order of a prime which is 1 modulo 4, because it has -1 as a quadratic residue, right ?
@AlexKChen it doesn't work for order of a prime which is 1 modulo 4 because it has -1 as a quadratic residue, but you can always find other polynomials
Wrong ?
you don't have to stick to $x^2+1$
Yup I once found $4x^2+1$ works too
05:30
@AlexKChen $4x^2+1=4(x+1)(x-1)$ shouldn't work
on the other hand you can try $x^2-2$
No I mean it works perfectly for primes not of the form 4k+1
@AlexKChen oh ok
OK, so this is impossible to progress further without using python, so should I use class when writing code to manipulate stuff with fields ?
BTW, thanks, this looks way too much interesting :thumbup:
@AlexKChen I refuse to give you any opinion on whether you should use class :P
No well why ? You don't use class ?
05:35
I don't know enough Python to tell you whether you should use class
Oh OK.
Great, now you're talking about Python in the Math room :P
pythons math whats the diffrence?\
@LeakyNun That's a lot of cosets (continuumly many)
@Secret what is?
05:41
Pick any irrational number $r$ and form $\langle r+ \Bbb{Q}\rangle$, that's already a coset and a normal subgroup
and no two of them overlap
$\sqrt{2}$ and $\sqrt{2}+1$ are both irrational and generate the same coset.
or do you mean, no two of the cosets overlap?
uh, I think I should be saying: Any two irrational numbers whose difference is not a rational
e.g. $e, \sqrt{2}$
@Semiclassical come on you are being hard
but still, yeah
pick any element in $\Bbb R/\Bbb Q$ and have it generate a subgroup
thus there are continuumly many subgroups
pick $a$ and generate $a\Bbb Q + \langle \Bbb Q \rangle$
this gives a subgroup isomorphic to $\Bbb Q$
in a sense you are dividing by $\Bbb Q$ twice
this amazes me
@Semiclassical can you fact check
05:51
i'm not nearly awake enough to fact check that
ok thx
@Secret
I came with a very interesting problem recently: the set of primes is $P$ and let $S \in P$ such that $S$ and $P-S$ is both infinite. Prove there exists a integral irreducible polynomial $Q(x)$ such that all terms of $S$ doesn't divides $Q(a)$ for all $a$, and another polynomial $R(x)$ such that the set of prime divisors of $R(\mathbb{Z})$ is subset of $S$.
@AlexKChen $S \subseteq P$
Morning @TedShifrin
Morning, Faust.
05:54
How was your day?
Other than toothaches, OK ... didn't play so well at bridge, but ... meh.
@TedShifrin care to fact check my rant above?
hmm i never was that fond of bridge
regarding dividing by $\Bbb Q$ twice
i prefer strategy games
05:56
I like bridge a lot. There's lots of strategy, actually.
hmm have to explain it to me sometime
I don't understand your notation at all, Leaky.
my friends say im not allowed to play games of chance
well they wont let me play with them =(
I mean, $\Bbb Q^2$ is isomorphic to $\Bbb Q[\sqrt2]$ so you can just do $\Bbb R/\Bbb Q[\sqrt2]$ right @Secret @TedShifrin
Isomorphic as vector space, yes.
05:57
as groups
Yeah.
Aren't you just saying $\Bbb R$ is uncountably generated as a $\Bbb Q$-vector space? I don't see the point of modding out.
It must be morning for Alessandro :)
Well, exciting as this is, I guess I'm calling it a night.
06:08
@TedShifrin well i'm stupid
Hardly. But you and I think very differently.
because I was thinking about subgroups of $\Bbb R/\Bbb Q$
But those are in one-to-one correspondence with $\Bbb Q$-vector subspaces of $\Bbb R$.
@TedShifrin hmm, really?
For example, $\langle \pi+\Bbb Q \rangle$ would be a subgroup of $\Bbb R/\Bbb Q$
I don't know what that notation means.
06:14
the subgroup generated by $\pi+\Bbb Q$ as a coset
i.e. $\{\cdots,-2\pi+\Bbb Q,-\pi+\Bbb Q,0+\Bbb Q,\pi+\Bbb Q,2\pi+\Bbb Q,\cdots\}$
Oh, blah. Yeah. It's too late.
I'm going to bed.
good night
06:26
@Waiting hahah I'm doing great, how about you?
07:09
BTW, Can you guess what I'm writing: BaDaAhSeJaYeGaBaSeJaHeYeJa.... KyMrUsMaSaKiJaHyJiKa... VaSrKaBr.... KyAmKiNaRiSeYeRaWaAyJiPaChDh...
Hint: It's the first two letters of each word in a song lyric.
 
1 hour later…
08:29
Now to try writing a popular scientific description of my project. And in Danish no less, which means I may have to make up some words as I go.
Hello!! Is someone familiar with the exterior product?
@MaryStar Sure
@TobiasKildetoft I have posted my question also in the main... I will send the link:
0
Q: For the exterior product it holds that $\omega\land \sigma=(-1)^{k\ell}\sigma\land \omega$

Mary StarTwo of the properties of the exterior product are the following: Let $\psi_1, \ldots , \psi_k, n_1, \ldots , n_{\ell}\in V^{\star}$ then it holds that $$\left (\psi_1\land \ldots \land \psi_k\right )\land \left (n_1\land \ldots \land n_{\ell}\right )=\psi_1\land \ldots \land \psi_k\land n_1\la...

Do you maybe have an idea?
[Random] $\pi +e=\frac{a}{b}$ $b(\pi+e)=a$
@MaryStar How could the second property follow from the first one?
the first property is basically just associativity
08:38
Maybe they mean that we can prove the second property using (besides others) the first property? @TobiasKildetoft
@MaryStar They mean you can prove the second property from knowing the general properties of exterior powers (possibly)
$b(\pi+e)=a,b(i\pi+ie)=ia,e^{b(i\pi+ie)}=e^{ia}$
Ah ok. But how could we continue from $\omega\land \sigma=\left (\psi_1\land \ldots \land \psi_k\right )\land \left (n_1\land \ldots \land n_{\ell}\right ) \\ = \psi_1\land \ldots \land \psi_k\land n_1\land \ldots \land n_{\ell}$ ? Do have have to apply a permutation on the indices? @TobiasKildetoft
@MaryStar You obviously have to use one of the other properties of the exterior product (more or less the defining property)
$e^{bi\pi}e^{bie}=e^{ia},(\cos(b\pi)+i\sin (b\pi))(\cos (be) +i\sin (be))=\cos(a)+i\sin (a)$
Now we are getting somewhere...
$((-1)^b+0)(\cos (be)+i\sin (be)) =\cos (a) +i\sin (a)$
$(-1)^b\cos (be) + (-1)^b i\sin (be) =\cos (a)+i\sin (a)$ and now...:
$(-1)^b\cos (be) =\cos (a),(-1)^b\sin (be) = \sin(a)$
If $b$ even:
09:02
@TobiasKildetoft I have changed a little my post. Could you take a look at it if it is correct so far?
$\cos (be) = \cos (a),\sin (be)=\sin (a) \implies be = a+2n\pi,be = a+2m\pi$
@MaryStar Actually, I just realized that what I told you were also commented on that post 5 hours ago
@TobiasKildetoft Oh yes... I didn't see that I have a comment.
If $b$ odd:
$-\cos (be) = \cos (a), -\sin (be) =\sin (a) \implies \cos (\pi - be) = \cos (a), \sin (-be)$
$= \sin (a) \implies \pi - be = a+2n\pi, -be = a + 2m\pi\implies - be = a+(2n-1)\pi, -be = a + 2m\pi \implies a + (2n-1)\pi = a+ 2m\pi \implies (2n-1) = 2m$ contradiction
If $b$ even (cont.):
$2n\pi = 2m\pi \implies 2n=2m$.
Therefore if rational: $\pi + e =\frac{a}{b}$ and $b$ even

« first day (2611 days earlier)      last day (2706 days later) »