9:09 PM
hi everyone

any help with this?
0

I wanted to better understand dfa. I wanted to build upon a previous question: Creating a DFA that only accepts number of a's that are multiples of 3 But I wanted to go a bit further. Is there any way we can have a DFA that accepts number of a's that are multiples of 3 but does NOT have the sub...

@Theo so would baaa be accepted?

@MaryStar hint : write $\lambda = x+iy, (x,y)\in \Bbb R^2$

@Astyx I see. And at the property that I showed do we have to substitute z=a+bi, or is it correct as I did it?

@LeakyNun yes

9:14 PM
@Theo so it's actually b*(aaa)*b* that you're trying to match?

@LeakyNun so would aaa or bbb (since 0 is divisible by 3)
@LeakyNun yes

I think you would need five states, with three accept states
can you do that yourself?
or six states

@LeakyNun its possible with 5 states? hmm okay I'll try, i thought i would need 7-8.

that was a rough estimate, it doesn't matter

okay i
i'll try and send it here

9:17 PM
@Daminark lol
@Balarka I had a long talk with Farb and it looks like I'll probably end up doing a reading course with him or his wife on gromov stuff

Damn, you and Benson are best mates, it seems

ooh neat

@MaryStar I don't think so

How can I show that $n\ln(1+(\frac{x}{n})^\alpha)$ diverges for $\alpha \in (0,1)$ and $x>0$?

Benson is a bro
Literally one of my favorite profs
His math philosophy lines up so close to mine

9:26 PM
@brot By using an asymptotical equivalent of $\ln(1+x)$ as $x\to 0$

@LeakyNun I believe this works?

accept states?

@LeakyNun a = 0 and b = 1, it was faster for me to add numbers
@LeakyNun q0 and q8, and q5 is a dead state
@LeakyNun so, with {0, 1}, we want a DFA that only accepts number of 0's that are multiples of 3 and no substring 010

oh, my regex was wrong
1*(0(11+)?0(11+)?0(11+)?)*1*
is this right?

@LeakyNun we could also have 1*(000)*1* can we not?

9:34 PM
@Theo that wouldn't accept 01100

@Astyx And then l'Hopital, right?

@LeakyNun oh right, yes then your regex is correct

L'Hopital ?
How ?

@LeakyNun wait my dfa is incorrect, there is no 1 going from q6

@Eric yeah I get the feel that you're something of a Benson/Neves person, also to a lesser degree Schlag

9:39 PM
If I write $\frac{1}{n} \ln(1+(xn)^\alpha)$ and consider $n \to 0$ I can use l'Hopital?

Neves and I are actually weirdly alike as people

@LeakyNun q7 is incorrect too since we get 010, i'm getting stuck on q6 and q7

@Theo I'd probably just read my regex and implement the DFA state by state

@LeakyNun okay will try

9:51 PM
@Eric I don't have much of a picture of Neves as a person, aside from being a total comedian

he's chill and disorganized and we have similar music tastes
and speak the same language obviously

Can relate to the disorganization bit
I mean I dunno if I've seen any profs who are organized, aside from like, Schlag

also have similar math philosophies i think
Danny is pretty organized in my experience

Ah, well I haven't interacted with him much but I could buy that

his lectures and notes are super structured

9:54 PM

no
but like his lecture notes he uses in class

Ah

i sit in front of the table usually so I've seen them laid out and it's extremely well put together

Well that's at least two people then (I at least think Schlag is?)

his office is sometimes a mess but i dont think it's because he's disorganized he just has tea everywhere probably cause he likes tea a lot

9:58 PM
I see. Oh by the way, after a quarter of the guy's class, would you say you recommend it if he's teaching next year?

absolutely

0

Let $X$ be a measurable space and $Y$ a topological space. I am trying to show that if $f_n : X \to Y$ is measurable for each $n$, and the pointwise limit of $\{f_n\}$ exists, then $f(x) = \lim_{n \to \infty} f_n(x)$ is a measurable function. Let $V$ be some open set in $Y$. I was able to show th...

@Daminark are you gonna take the undergrad course in the spring

Why is this not accepted ??
0

I was wondering If it is easier to factor in a non-ufd then it is to factor in a ufd. I can come up with arguments for that , but I also have arguments in the opposite direction. For instance : It should be easier to factor When there are more possibilities ( multiple factorizations in a non-ufd...

0

Consider a non-UFD that only has 2 units ( $-1,1$ ) and the min difference between 2 elements is $1$. Also there are only a finite amount of elements for any given fixed norm. ( Maybe that follows from the other 2 conditions ? ) I wonder about counting the irreducible elements bounded by a lower...

@EricSilva probably not, between this reading course and just reading through some book, prob Aguilar, I'll say that I'm good
I may audit that but I don't want to use up a slot in the spring

10:16 PM
That's fair
Do you know who's teaching it

Nagpal
And I heard he's pretty good so that's something

The sidebar always confuses and amuses me.

Oh yeah Rohit is great I TAed for him

10:36 PM
So I've heard. I'm possibly auditing his commutative algebra

commutative algebra is great, you have to hear it

Yeah I definitely will at some point, it'll be a question of whether being in commutative algebra and basic ring theory simultaneously is a bad idea

@Balarka so it seems the plan is to give something closer to a sketch of the quasifibration part of Dold-Thom and focus instead on the resulting homology bit

How would you make a regex for this? L = {w $\in$ {0, 1}* : w is 0-alternating}, where 0-alternating is either all the symbols in odd positions within w are
0's, or all the symbols in even positions within w are 0's, or both.
I want to construct a nfa from this, but I'm struggling with the regex part

10:44 PM
@Theo so it matches 010001010 and 101110101?

@Daminark Ah OK

@LeakyNun the first one yes, not the second one since it has 111

oops
I meant, 1010001010

@LeakyNun yes
@LeakyNun that matches

but then you never know whether you're in the first pattern or the second pattern

10:47 PM
@Daminark Whatever you said to me so far proves that $\pi_n \text{SP}$ is a homology theory in the Eilenberg-Maclane sense though, no?
Well modulo Mayer-Vietoris
That's how you show $\pi_n \text{SP} \cong H_n$, right?
after noting they agree at a pt

Well, so the theorem itself I mentioned should be the key to verifying the axioms, yeah

@LeakyNun which is why im struggling with it

But you'll need that quotient bit to get the exact sequence

Just got back from a math talk: math.umn.edu/seminar/…

SP(A) --> SP(X) --> SP(X/A) thing, yeah
@Semiclassical oh schnoop dawg

10:49 PM
the technical details were beyond me, but I followed it well enough
So that was neat

Sick! Hope it was fun

Yeah
Though attending talks at 3:30 pm are always a bit of a risky endeavor

Why so?

soooo sleeeeepy

It’s about the time of day where it’s easy to fall asleep :p

10:52 PM
Ah

Ding

@LeakyNun 1*(01*01*)* + 1*01*(01*01*)* + 0*
@LeakyNun would this do it?

how do you even +

They also mebtioned quadratic differentials which is something I’d wanted to understand a while back
(I didn’t succeed, alas. But somehow it connects the kind of integrals I was interested in with Teichmuller theory?)

yeah you can think of the Teichmuller space as the space of quadratic differentials
It's a proof-worthy statement

11:00 PM
:D
What's up, guys?!

@LeakyNun yeah sorry, i think the one i gave is incorrect

Demonark: I'm the very model of disorganization :P

@LeakyNun do you have any approaches for the regex?

@BalarkaSen yeah, they had statements like that

@Theo no idea
it's still three things union together

11:03 PM
Something like quadratic differentials are the cotangent bundle for the moduli space of genus-g curves

What does the half-plane modulo the action of $SL_2(\Bbb Z)$ like? does it have some nice description?

A precise statement found via google: “The moduli space of holomorphic quadratic differentials is isomorphic to the cotangent bundle over the moduli space of complex structures.“

@LeakyNun no worries! thanks for the help though! :)

@MatheinBoulomenos It's well uh like a cone

11:05 PM
So there :p

with the base having two non-manifold points

@TedShifrin nice

(which are the "orbifold points" of order 2 and 3)
and the top of the cone going to infinity

Wait why is this not a manifold? is the action of $SL_2(\Bbb Z)$ not properly discontinuous?

It fixes points on the boundary of the upper half plane

11:15 PM
hi
does anyone know baout conditionalvariance?

I meant the upper half plane without the real line

I once did, @AbcD

@MatheinBoulomenos well, sure, then it is properly discontinuous. But that's not the moduli space of the torus...

11:17 PM
Depends on what your questions are.

You don't want to include the real line. The points on the real line don't give you a lattice

@Balarka: But you're looking at $1$ and $\tau\in\Bbb H$ to get the lattice, and $\tau\notin\Bbb R$.

Oh fair enough
Blah
So you get three cusps
I get it

But the fundamental domain will have 3 points "at $\infty$"

and the two cusps at the bottom have that Z/3 and Z/2-symmetry

11:20 PM
Hmm ... Maybe one of 'em isn't at infinity. You get the intersection of the vertical ray and the semicircle at the top of the semicircle, I recall.

Oh
Right
It's on the middle of the semicircle

sup chat

"o" <-- this isn't a circle, not because it's imperfect: even if I manage to arrange the atoms such that it's perfectly continuous and perfectly circular, it still isn't a circle

a circle is the abstract concept in the Platonic universe
it's the equivalence class of all those things that look like a circle
@MatheinBoulomenos hi
what's the Galois group of the general quintic in F_p?

11:26 PM
if you have an irreducible quintic it's cyclic of order 5

no, not that

the general quintic over any field has Galois group $S_5$, if I remember the definition of a general quintic correctly

"the general quintic", a well-defined thing
oops, I forgot the definition

The definition I know of general polynomials involves the extension $F(x_1, \dots, x_n)$ over the field of symmetric rational functions over $F$ in $x_1, \dots x_n$
this will always have Galois group $S_n$, regardless of the field $F$
the general polynomial of degree $n$ is then just $(T-x_1) \dots (T-x_n)$
which has symmetric polynomials as coefficients
because the symmetric rational functions are those rational functions fixed by the action of $S_n$ permuting the variables, this has Galois group $S_n$
By a theorem of Artin

DogAteMy!

11:34 PM
@MatheinBoulomenos thanks

the theorem of Artin I meant is if $G$ is a finite group of field automorphisms acting on a field $F$, then $F/F^G$ is Galois with Galois group $G$
that's an important step in the modern proofs of Galois theory
and also interesting on its own

11:52 PM
@TedShifrin …I don't remember being online at 6:33
But, uh, hello