Mathematics

12:54 AM

Yo @Eric

Eric Silva

sup dude

And @Ted sorry I was out during that time. Classes are quite fun!

Eric: Is PDE on MWF or TR?

Eric Silva

TR

Sick

Sat in on DCal's class today and I can confirm his accent is quite nice to listen to

Eric Silva

is he teaching complex analysis?

1:02 AM

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

1:21 AM

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

Okay then

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

1:28 AM

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

1:31 AM

(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

Hm

Hi chat

High hat

@AkivaWeinberger $F_{\omega+1}\ge\sup \{\omega+F_\delta: \delta<\omega\}$

1:32 AM

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 :-(

1:34 AM

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'

($\operatorname{im}f$)

1:36 AM

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'

Yeah

1:39 AM

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?

2:01 AM

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

2:15 AM

@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:

Hmm

Essentially you add real numbers but ignore the integer part

$(\Bbb{R}/\Bbb{Q})/\Bbb{N}$

2:17 AM

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?

2:20 AM

@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

2:24 AM

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

2:27 AM

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

sure.

its allready been taking care off

pi will be in the coset {…, 0.14, 1.14, 2.14, 3.14, 4.14, …}

yes exactly

2:28 AM

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

2:30 AM

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

2:37 AM

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

2:42 AM

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 $1+\lambda+1$

My reaction to that: "Don't open your eyes!"