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1:08 AM
Lately I've been thinking of asking on here every single question of Magnus et al.'s "Combinatorial Group Theory [. . .]", no matter how easy I find it - I could always use the proof verification tag - so that I'll get a better understanding of the area I'm doing my PhD in, instead of self-studying it and wasting time.
 
 
1 hour later…
2:24 AM
@Daminark, I am trying to understand this answer. What I have got is this: if $F$ is a finite dimensional field, then
@Daminark, I am trying to understand this answer. What I have got is this: if $F$ is a finite dimensional field, then there is a bijection which is linear, between row space and column space. Otherwise, there is a bijection between row space and column space, but there may not be any linear bijction.
Have I got that correct?
 
That's not quite the idea
Keep in mind the question, it's whether one can show bijectivity without the linearity bit. The point is that there is a bijection between two finite vector spaces over a given field iff their dimensions are the same
So the fact that they are isomorphic as vector spaces isn't saying anything more than the fact that they have the same size
In the case where you're field is R or C, all non-zero finite dimensional vector spaces have the same size, so the fact that there is a linear isomorphism between the row and column space says a lot more than the fact that they have the same size
 
2:42 AM
@Daminark here?
@Daminark here ?
 
Yeah
 
can we talk about something called Q-spread
 
Go for it
 
can you check page 2 do you understand how $\mathcal{X}$ is constructed ?
 
2:58 AM
Let me get home in a couple minutes and I'll read
 
okay
 
Okay I'm back, let's see how
 
okay
 
Okay so the $p_i$ here seem to be the ones we quotiented out by last time we talked in order to get all these algebraic numbers
 
yeah
I would like to understand how is $\mathcal{X}$ constructed
 
3:12 AM
These polynomials in principle may not be homogeneous so I'm guessing here the point is that we want to homogenize them in order to define a projective variety
 
sure
you will get S
by collecting coeffients
according to the algebra we talked about last time.
 
Well I'm looking at this paragraph
It's later on in the page
 
yeah so I don't understand this
what is exactly the map $\pi : \mathcal{X} \rightarrow S$
 
So $\mathcal{X}$ is gonna be a variety over $\mathbb{Q}^{T+B+N}$ defined by having the $f_i$ (which homogeneous polynomials in the $z_i$) being $0$ as well as the $p_i$ (polynomials in the $y_i$?) being $0$, and I'm guessing this projection is just choosing out the latter points
Because those guys are gonna somehow "come from" $S$
(This is just me feeling it through, double check with someone who knows AG)
 
oh yeah
@loch can you confirm when you come
but I think your right.
this is the only thing that would make sense.
because S is defined over $\mathbb{Q}^{T + N}$
so we just project from $\mathbb{Q}^{T + B + N}$ into $\mathbb{Q}^{T + N}$ agree?
@Daminark ?
 
3:20 AM
That's what it seems to me, yeah
And then "varietize that projection"
 
for example let us take the elliptic curve example they defined earlier
as they showed $S = \mathbb{P}^1$ then $\mathcal{X}$ would be $\mathbb{P}^2$ right ?
and the map would be $[x,y,z] \mapsto [x,y]$ right ?
 
Let me talk this out loud just to be sure
So okay, what's the analog for $X$ here?
When they say $X$ is defined over $k$ by $f_1,\ldots,f_r$
 
for example elliptic curve example
 
Oh oh
I just noticed the elliptic curve lmao
 
yeah
so they computed $S = \mathbb{P}^1$
and $\mathcal{X}$ would be $\mathbb{P}^2$ ?
 
3:30 AM
Hmm, so we define $X$ as the solution to the homogenization of $y^2 = x(x-1)(x-\alpha)$
So it should be a subset of $\mathbb{P}^2(k)$
 
yeah
 
This kinda bothers me
 
why ?
I think in this case it would be $\mathbb{P}^2$
@loch please join discussion when you come
 
So, they define $\mathcal{X}$ as a variety over $\mathbb{Q}$ but there are no $p_i$ to work with here, so I'm not seeing how it's a variety over $\mathbb{Q}$
 
@Daminark k is defined in that way
so X is variety over k
projective variety over k
and I guess we do the construction in order to make k to be some kinda of geometric guy
 
3:36 AM
Yeah, $X$ is a variety over $k$ but I'm not sure of the process to build $\mathcal{X}$ here
Like, there aren't any $p_i$ in sight as I see it because $k$ is purely transcendental
 
$k = Q(x_1,\ldots,x_T)[y_1,\ldots,y_A] / (p_1,\ldots,p_b)$
using the construction they mentioned
now X is all of that + the ones defined by the $f_i$
 
Well here's the thing, what are the $p_i$ in this particular example?
I think there are none
 
yeah
no $p_i$
it is actually
 
So that's why I'm nervous, we're defining a variety over $\mathbb{Q}$ using this one curve $f$, however that isn't a rational polynomial
 
$Q(\alpha_1)[x]$
I think that is it
agree?
 
3:44 AM
That's where $f = f_1$ lives for me, yeah
 
no
just $Q(\alpha)$ I guess
because we don't have any algebraic guys
 
I meant in the sense of, $f$ is a polynomial so it's an element of $\mathbb{Q}(\alpha)[x]$ (or $z$ if we wanna be consistent with their notation)
 
so we have that + this curve $f$
yeah
 
@loch save us
 
the curve I think lives in $Q(\alpha)[z]$
so once we projectivize things
we will get $\mathbb{P}^2$
I think that is how it works
@loch want your opinion here as well
 
3:48 AM
Okay so, let's write out the curve
 
the curve is given by $y^2 = x(x - 1)(x - \alpha)$
 
$y^2 - x^3 + \alpha x^2 - x^2 - \alpha x$, so when we projectivize it we should have $y^2z - x^3 + \alpha x^2 - x^2 z - \alpha x z$
(Since $\alpha$ is transcendental here we should be thinking about it as a variable
I think)
So $\mathcal{X}$ should be solutions to that guy I think, as a subset of $\mathbb{P}^3$?
 
$\mathbb{P}^2$ ?
why $\mathbb{P}^3$
 
When you think about that as a polynomial over $\mathbb{Q}$, it has 4 variables
Since $\alpha$ is purely transcendental we should just be treating it as an extra variable
 
oh
 
3:54 AM
So you have 2 from the elliptic curve, 1 from $\alpha$, and one to projectivize it
 
yeah
so it is from $\mathbb{P}^3$ to $\mathbb{P}^2$
obtained by projection ?
 
$S$ is just $\mathbb{P}^1$
So we should have a map from $\mathcal{X} \subset \mathbb{P}^3 \to \mathbb{P}^1$
 
I see
how is this defined ?
$[x,y,z,w] \mapsto [x,y]$ ?
 
Hmm, now I'm not so sure. To be honest with this business I think I'm out of my depth to a degree
 
okay I agree with you with everything else
I am not sure about the projection map
@loch waiting your input
 
4:01 AM
Don't tag too much, that might start to get annoying
 
okay
yeah
 
He'll see even the one ping once he checks the chat
 
oh
I didn't know
sorry about that in advance
 
4:56 AM
Hey @Daminark @Adeek
 
5:28 AM
1) There is no non-constant map from $\mathbb{P}^n \rightarrow \mathbb{P}^m$ when $n>m$! To see this - think about what does it mean to give a map to $\mathbb{P}^m$ --- this is to give $m+1$ homogeneous polynomials $f_0,\ldots,f_m$ on $\mathbb{P}^n$. But you don't want anything to map to $[0:0:\ldots:0]$ (because this is not a point on $\mathbb{P}^m$) - but since $n>m$, $n \ge m+1$, and the intersection of $m+1$ hypersurfaces in $\mathbb{P}^n$ is non-empty!

(To parse this I guess you should probably think over $\mathbb{C}$, but the same claim is true generally (i.e. scheme-theoretically))
 
6:05 AM
On this note - I found the claim "$X_{s_1}$ and $X_{s_2}$ are indistinguishable when $s_1,s_2 \not \in \overline{\mathbb{Q}}$ a bit confusing at first. At first glance I found the claim confusing because, over $\mathbb{C}$, elliptic curves are classified by the j-invariant, and certainly giving two different trascendental elements give you different j-invariants in general.

I realised that this doesn't contradict anything. One can think of this as saying that the two elliptic curves are actually "isomorphic", and that you can give a map from one to the other -- but this is not going to res
 
 
1 hour later…
7:30 AM
it's so nice that whenever you want to talk about something there are people talking with you about it.
 
7:45 AM
except analytic number theory
nobody here talks to me about analytic number theory
 
8:11 AM
@LeakyNun modular forms are part of analytic number theory
 
well i'm mainly referring to PNT
 
 
1 hour later…
9:38 AM
Hello!!
I want to check if any of the elements 4, 5, 7 is prime in $\mathbb{Z}_{12}$.
For that do we have to check the divisors in $\mathbb{Z}_{12}$ ?
 
9:57 AM
Hello @LeakyNun !! Do you have an idea about my question?
 
5 and 7 are units
4 = 2x2
so none of them are prime
 
Why are 5 and 7 not primes because of the fact that they are units? @LeakyNun
 
look up the definition of prime
 
Let x be unit, then there is a y such that x*y=1. How does it follow that the only divisors of x are 1 and x? @LeakyNun
 
no it doesn't
 
10:02 AM
Oh sorry, how do know from the fact that x*y=1 which are the divisors of x? @LeakyNun
 
if x is a unit then every unit is a divisor of x and every divisor of x is a unit
 
Why does this hold? @LeakyNun
 
go prove it yourself
 
10:16 AM
@MaryStar Also, you did not quite the correct definition of being prime
 
Which is the mistake? @TobiasKildetoft
 
@MaryStar Look up the definition of prime element in a ring
 
hi @TobiasKildetoft
 
@LeakyNun Hi
 
11:00 AM
Hi all
 
11:10 AM
Shame how when you're confused about the Weierstrass $\wp$-function you can't just email Weierstrass
(Kauffman, on the other hand…)
 
11:29 AM
o..o
 
11:49 AM
Writing down the $\wp$ is probably one of my biggest struggles atm
Yesterday I gave a quick talk and introduced it, literally just said "I have no idea how to write it, the tex code is \wp, I'll just use $\varphi$ instead"
 
It's like you want to draw an 8 by starting at the top and going left
but on your way up you change you mind and make it a p
 
12:15 PM
@Daminark oh you had your talk on $(a+ib)(c+id)$? how did it go?
 
did you F.O.I.L. them? :P
@MatheinBoulomenos what would be the German acronym for F.O.I.L.?
 
It went pretty well, I think. Managed to get through what I intended to without going too fast, so that was nice
 
did you use the full 15 minutes?
 
12:37 PM
@user1732 we don't have an acronym for that
I was very confused when I heard people "use FOIL" for the first time
@Daminark sounds great!
we don't explicitly teach multiplying two binomials here in Germany and we don't have a mnemonic for that. We still teach people how to use distributivity, of course
 
12:55 PM
0
Q: Motivation behind using complex numbers in combinatorics

alxchenThere's a common technique in couting/tiling problems in Olympiad which is to use complex numbers. I'm giving some examples: (IMO 1995 P6) Let $p > 2$ be a prime number and let $A = \{1, \cdots, 2p \}$. Find the number of subsets of $A$ each havinig $p$ elements and whose sum is divisible by $...

 
@AkivaWeinberger am I the only one who writes 8 by exactly starting from the top and going left?
 
I think I do too, but the first stroke takes me down the right side of the top loop and the left side of the bottom loop
rather than the left side of the top loop and the right side of the bottom loop
 
1:17 PM
@LeakyNun hat tip
Morning
 
1:27 PM
hi
 
@Akiva wait what? Do you write the two loops separately or something?
 
@Daminark no, the first stroke goes like S but reflected
(my first stroke goes like S, no reflected)
 
Ah okay
Yeah I always start from the middle and go left and up
 
most people I know do that also
 
I do the one on the left of that image
 
1:39 PM
Okay yeah I felt something was off because somehow you start from the top, go left, and yet trace out the right side of the top loop first
 
I do the "S" first.
top down
 
@Daminark Wait, sorry, "from the middle and go left and up"?
 
@AkivaWeinberger That's plain crazy :/
 
@rschwieb yay
@Semiclassical @TedShifrin why can we apply the nabla formalism to div and curl even for polar coordinates?
 
Yeah, I start from the middle of the 8, then trace out the left side of the top loop, and continue from there
 
1:42 PM
So like that
 
Yup
 
I honestly never considered that as a possibility
 
That's the only way I had ever known of
When you guys were doing it differently I was like huh
 
hey
 
OK so Google says Leaky's right
 
1:45 PM
what
is there even a right order
 
Well, mine is objectively the best
'cause it looks the best
 
Leaky's literally cannot be right
Because he goes to the left
 
it seems that loch has answered our question yesterday
I just woke up
 
Actually funny thing is, all of us have our first strokes going toward the left
 
'cept akiva
 
1:47 PM
Also I read "pull lines and push curves" on the calligraphy subreddit at some point which gives me validation
 
I am just gonna have breakfast mediadate and check it out
morning @Daminark
 
Hey Adeek
 
@LeakyNun Hm?
 
Akiva's traces out the right side of the 8 first but his pencil is still moving right to left
 
oh interesting
 
1:52 PM
@loch I still don't understand explicitly what is $\mathcal{X}$
so it is just the one defined by the projectivization
 
looks like a fan blade to me
 
of the equation defining
the elliptic curve
and we think of it living in $\mathbb{P}^2 \times \mathbb{P}^1$ instead of $\mathbb{P}^1$ ?
if so what is the projection explicitly is it $[x,y,z] \times [w,m] \mapsto [w,m]$ ?
good your here loch
 
Hi @Adeek, I have to leave shortly but I'll be back in an hour or so

but let me just say that I'm also slightly confused by what he wrote later about projectivising the equation. I don't think it's meant to be interpreted as "Here you have a bunch of equations in variables $x_i, y_j, z_k$ (say you have $n,m,l$ of them respectively), now consider your variety sitting in affine space $\mathbb{Q}^{n+m+l}$ and take its projective completion", because if you do this I don't think you necessarily have a map to $S$ basically for the same reason I said earlier about 1)
 
oh
ok can we discuss this in an hour. I will go have breakfast and mediadate
I will also be back in an hour @loch
 
I think, in the example for elliptic curves if you think of it sitting in $\mathbb{P}^2 \times \mathbb{P}^1$ it'd make sense, and I suspect in general when he says projectivsation perhaps he's also thinking something along these lines - but maybe I'll have to think/google a bit and see :p
 
1:59 PM
good morning as well.
hm
 
Okay now I feel a little less bad about fucking it up that hard
 
yeah I want to know in general how is it constructed
 
@LeakyNun 90% of people draw an 8 by completing an S
 
interesting
 
but anyway I will go have breakfast and medidate cya in hour @loch
I think I know how to do it
I think it is $X_k \times k$
so we have the equation $y^2 = x(x - 1)(x - \alpha)$ we projectivize that
in $\mathbb{P}^2$ because we only have the variables x and y
then we will have $\mathbb{P}^2 \times k \cong \mathbb{P}^2 \times \mathbb{P}^1$
but anyway I will discuss the rest when I come brb
 
2:06 PM
Challenge: Tie this out of string
(or headphones or charger cable)
Then untie it 'cause it's the unknot apparently
 
apparently
 
Does anyone have any ideas for this question about intuition for Gaussian: math.stackexchange.com/questions/2823278/…
 
2:19 PM
On second thought, don't use headphones
Use your height in yarn
 
Let $S_1, S_2$ be two sets and define $S_1+S_2=\{s_1+s_2, s_1\in S_1, s_2\in S_2\},$ how to prove $S_1+S_2\subseteq S_1\cup S_2$?
 
@Adeek Right yeah you have $y^2=x(x-1)(x-a)$, if you projectivise this equation in $\mathbb{P}^2$ you'd get $Y^2Z = X(X-Z)(X-aZ)$, and you're suggesting that we're varying $a$, so we're looking at the family defined in $\mathbb{P}^2 \times \mathbb{A}^1_a$ (where I'm using $a$ to denote the coordinate in $\mathbb{A}^1$) by the same equation, and actually we want $\mathbb{P}^1$ instead because we want something projective. Is that what you were saying?
 
lol, ok I solve it, that's my teacher's typo, it should be $\supseteq$
 
2:35 PM
Did anybody discuss that freakish way of drawing an eight where you just simply do two circles? Ugh.
 
We don't discuss heresies in this chat
 
yeah @loch
I think that makes the most sense
and actually it works out
 
I always start at the top and go left...
 
that is the fiber over the point $x \in \mathbb{P}^1$ is precisely the original variety
@loch
I will go and mediadate I will be back in 30 min
I will be back to discuss things with you loch
 
Norman J Wildberger made me do it.
 
2:40 PM
@Adeek Yeah I agree. I'm pretty sure this is what he meant when he talks about projectivisation later - for the general construction - also because that way you do always get a map to $S$, and if you interpret the statement as take my polynomials in $\mathbb{Q}^N$ for some big $N$ and then take its projective closure I don't think you always get a map to $S$, contrary to what he suggests...
 
I write numbers like this
but I know some people write one like a line and seven without the line in the middle
How do you write one and seven?
 
2:56 PM
I don't use the line for 7's, but i did start using it for lower case z's while teaching
 
I use it for neither
I do the 1 as a line and 7 without the line in the middle
 
I use the line for seven and z actually
 
back
@loch here ?
 
hey @Adeek
 
hey :D
yeah I completely agree with you
good this is understood now
This spread business is very elegan
elegant
I think it is used all over research papers this method because you can put hodge structures on the fibres
do you know hodge theory @loch ?
 
3:03 PM
@Semiclassical This is a nice read. And yeah, I don't know these diagrams exists until now. It does shows how orderly the chaotic region actually is, with the saddle node bifurcation being the hard to predict things
 
i sat in a hodge theory course last year but i didn't really put enough effort to it to understand much - basically i think nowadays if someone talks about hodge theory i can nod my head and kind of follow :p but i absolutely have to review it at some point
 
yeah me 2 I have kinda shallow understanding of hodge theory.
but I need it for research later.
I will start working on it in the fall.
 
cool! I guess since you're interested in hodge conjecture stuff there's no way you can avoid hodge theory haha
 
yeah
 
3:12 PM
So, the reason why it looks like random mush is because most of these period cycles are repelling
 
huh
in the logistic map, is every a smaller than 4 just making a cycle (but with a crazy high period) or is there actual chaos for a < 4?
 
@loch I decided after my thesis defence is done to dedicate 2 hours per day to studying physics
so hopefully 2 years from now I will have big background in physics for mathematical physics
 
@G.Ãœnther Well according to this reading material Semi found, the chaos is actually a dense collection of period cycles appearing and they are all highly repelling
 
(if only i could dedicate 2 hours per day to actually working)

but that sounds good! when is your thesis defense?
 
and most of these arises from what is called a saddle point bifrucation, and you can see that phenomenon gets more common the further you go down
 
3:16 PM
Hello everyone.
 
hmm, that's interesting.
 
I am not ready yet I will be ready though by next Tuesday
 
what exactly are those saddle point bifurcations though?
 
I am just covering now certain background material and from tomorrow I will start reviewing my tehsis @loch
 
they form things like period 3, 5 , 7 cycles
 
3:18 PM
I have a question regarding a proof that I have found. I don't understand how the base case is supposed to be equal in this one.
 
I will be ready with everything by Next tuesday and start practicing my talk Wed and Thursday.
 
It's said that $$f^1(x) = f(x) = \frac{x}{(\sqrt{1+x^2})} = \frac{x}{(\sqrt{1+1x^2})}$$ is the same but how so?
 
@Adeek i see. Good luck with that!
 
thanks
my thesis is like 124 pages
hehe so I will probably miss something while questioned haha
@loch I could send it to you if you want
 
I don't mind! but I can't guarantee I'll spend much time looking at it
 
3:27 PM
okay just skim through it you will like it
email ?
 
@Secret On that site you've linked there's a lot more interseting stuff I didn't know. Thanks for sharing. Now I'm kinda interested how coloring periodic points would look with King's Dream fractal
But I won't dive into that now
 
@loch ?
got it
did you get it ?
 
cool! anyway i got to go - i look forward to looking at your thesis
yeah i did
 
awesome
me 2 I will go to work
cya l8ers
 
2k+: Please approve this review: math.stackexchange.com/review/suggested-edits/1051093 , it is non-trivial but useful.
 
4:25 PM
conjecture: if $f : \Bbb C \to \Bbb C$ has an essential singularity at $a$ then $\operatorname{Res}(f'/f,a) = 0$
@AkivaWeinberger hi
 
@LeakyNun Suppose $f = e^g$ where $g$ has an essential singularity at $a$. Then $f'/f = g'$. But I can cook up functions with essential singularities and whatever residue I want there.
 
then how can I find the "winding number"?
 
Not sure I understand the question
 
How do I find $\operatorname{Res}(f'/f,a)$?
 
write down a power series running over integer powers of $z-a$ centered at $a$ and calculate the coefficient of $\frac{1}{z-a}$
 
4:33 PM
I can't read it out from the Laurent series of $f$?
 
What does a general element of $\Bbb{Z}[t_1,t_2,t_3,...]$ look like?
 
not clear to me how to calculate the power series of $\frac{1}{f}$ when the power series of $f$ is bi-infinite
 
@user193319 it looks like an element of $\Bbb Z[t_1, t_2, \cdots, t_m]$ for some $m$
(elements of polynomial ring are "locally finite")
@MikeMiller how about if $f$ is just infinite to the left?
 
Ah, thanks!
 
@LeakyNun Seems fine then, so there are no positive powers?
 
4:35 PM
right
like $\exp(1/z)$
(for this one I know I can use the $e^g$ trick, but can I read it out from the Laurent series?)
 
$f = a_0 + a_1/z + a_2/z^2 + \cdots$, then $f' = -a_1/z^2 - 2a_2/z^3 - ...$
 
then the residue is 0
 
the residue of $f'$ is
oh I see
@LeakyNun I made a mistake above, as you might have noticed: the residue of $g'$ is zero
 
@loch here?
 
for any function $g$...
rip me
 
4:40 PM
lol @MikeMiller
 
how do we even multiply two laurent series together
 
@LeakyNun So it seems the answer is yes for functions with power series in $\frac 1 z$, but it is not clear to me for arbitrary functions with essential singularity
@LeakyNun $$(a_0 + a_1 z + a_2 z^2 + \cdots)(b_0 + b_1 z + b_2 z^2 + \cdots)$$ $$= a_0 b_0 + (b_0 a_1 + a_0 b_1) z + (b_0 a_2 + a_1 b_1 + a_0 b_2) z^2 + \cdots$$
this uses the fact that there are only finitely many terms that could contribute to one coefficients
if your series are bi-infinite there are problems
 
@MikeMiller $f(z) = z^{-1} \exp(z^{-1})$, $f'(z) = -z^{-3}(z+1)\exp(z^{-1})$, $f'/f = -z^{-2}(z+1) = -z^{-1} - z^{-2}$
you need $a_0 \ne 0$ for that to work
 
Ah nice
looks like my part here is done
[zooms out]
 
so we know the right-infinite case. for the left-infinite case, let $f(z) = z^m g(1/z)$, then $f'(z) = mz^{m-1} g(1/z) - z^{m-2} g'(1/z)$, and the required residue is $m$ again
and we still don't know the bi-infinite case
hi @MatheinBoulomenos
uw0tm8 the new math.SE layout
ok it isn't just math.SE
 
4:54 PM
its terrible
i do not like it sam i am
 
@MikeMiller can Res(f'/f,a) ever be a non-integer?
 
interesting. i guess not.
it's clearly scale-invariant
(cf)' = c f'
 
WA refuses to expand the Laurent series of f'/f where f=e^z+e^(1/z)
and it refuses to find its residue
I'm forced to see what the integral gives
it gave 0
ok now how does WA know the integral but not the residue
 
5:16 PM
0
Q: Argument principle for essential singularities

Kenny LauLet $f : D \to \Bbb C$ be meromorphic where $D$ is an open connected subset of $\Bbb C$ that contains $0$. Assume that $f$ has an essential singularity at $0$. Is there any way to find $\operatorname{Res}\left(\frac {f'} f, 0\right)$ from the Laurent series of $f$ at $0$?

 
5:50 PM
@loch ping me when you come
 
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