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6:06 AM
someone already answered on freenode. meaning two oriented knots are equivalent in R^3 iff they are in S^3
 
ah
but the question is weird
there are oriented knots which can be deformed to themselves with the opposite orientation already in R^3
 
yes
 
well-definedness of orientation is not a good way to phrase whatever it is you meant :-)
 
indeed
 
say hi to the folks at #math from me :-)
 
6:10 AM
(-:
<antonfire> tell him to go fuck himself
what did you do? :o
 
What's the problem with that antonfire dude?
 
He's always insulting people for no reason
 
@MarianoSuárez-Alvarez Hey I have a super quick question
 
shoot
@FernandoMartin IRC is a dangerous world
 
6:15 AM
Is it true $\Bbb{P}^2 \cong \Bbb{P}^1 \times_k \Bbb{A}^1$?
 
Are there no mods at #math? Or worse, is he a mod?
 
@MarianoSuárez-Alvarez I am reading this proposition in qing liu saying that $\text{Proj} ( B \otimes_A C)= \text{Proj} B \times_{Spec A} Spec C$ where $B$ is a graded $A$ - algebra
 
so $\Bbb{P}^2 \cong \Bbb{P}^1 \times \Bbb{A}^1$ is true ? @MarianoSuárez-Alvarez
Geometrically I can't get this right in my head
 
that is not true
how did you get that?
 
6:18 AM
well $\Bbb{P}^2 =\operatorname{Proj} k[x,y,z]$
$\Bbb{A}^1 = Spec k[z]$
 
(That would give you a non-constant map from P^2 to A^1, and there are none!)
 
$\Bbb{P}^1 =\operatorname{Proj} k[x,y]$
 
what is A, B and C?
 
$A = k, B=k[x,y]$, $C= k[z]$ @MarianoSuárez-Alvarez
 
Notice that in that statement B is a graded algebra but C isn't
 
6:20 AM
right but what's the problem?
It doesn't matter if $C$ is graded or not
 
well
if you say so!
 
It seems that isomorphism involving fiber products says exactly that the projective plane is the affine line times the projective line @MarianoSuárez-Alvarez
@MarianoSuárez-Alvarez I'm confused
 
No, it doesn't
 
why not?
 
What is the graded algebra $B\otimes_AC$ in your case?
what is its degree-zero component?
 
6:22 AM
$B_0 \otimes_A C$
 
and that is?
 
ah ok that is $k[z]$
which is not the same as the degree $0$ component of $k[x,y,z]$
 
since $B\otimes_AC$ is very different to $k[x,y,z]$, your isomorphism has very little hope!
 
I'm confused because to me they are the same ring
just the grading is messed up
 
The scheme $Proj(B\otimes_AC)$ in your situation is what people write $\mathbf P^1_C$.
Proj is very very very sentitive to grading!
 
6:24 AM
right.
 
for example, if you pick $A=k[x,y]$ and grade it so that it is all in degree zero, what is $Proj(A)$?
 
um
wait it's all prime ideals? @MarianoSuárez-Alvarez
because no the irrelevant ideal is the emptyset
 
all prime ideals of what?
 
wait it's none of them
proj(A) is the emptyset!
Because every prime contains the irrelevant ideal which is the emptyset.
 
6:27 AM
wow this is screwed
 
as we said, Proj is very sensitive to the grading :-)
 
my god this proj is really confusing
 
the irrelevant ideal is not the emptyset
the emptyset is never an ideal of anything
I promise that you'll get used to Proj
 
why do you say so?
 
you just have to keep going
why do I say what?
I say many things!
 
6:28 AM
@MarianoSuárez-Alvarez say I will get used to it
 
well, most people who insist get used to proj and all the paraphernalia
 
insist on?
 
persevere is a more precise verb
 
@MarianoSuárez-Alvarez man so much stuff in AG and so many technicalities
my god sometimes it's very overwhelming
 
you have to get past the technicalities, and then it becomes geometry
 
6:32 AM
right. But right now the stuff on divisors is very confusing
 
divisors are just hypersurfaces :-)
well, "just"
 
that's a very simple way of putting it mariano.
@MarianoSuárez-Alvarez well linear combinations of
 
yup
I would say that the other way round
 
@MarianoSuárez-Alvarez If I look at the conic $xy - z^2$ in $\Bbb{P}^2$
 
divisors are a very complicated way of putting hypersurfaces :-)
 
6:35 AM
I guess the variety being isomorphic to $\Bbb{P}^1$ means its divisor class group is just $\Bbb{Z}$
 
P^1, you mean
 
yea sorry using the veronese embedding
 
Indeed, the divisor class group is invariant under isomorphism
 
right. I was getting worried because the coordinate ring is not (one is a UFD the other not)
I'm guessing the class group is generated by $[(x,y)]$
just by looking at where the point $(0,1)$ gets sent under the veronese embedding @MarianoSuárez-Alvarez
@MarianoSuárez-Alvarez I've got to go
@MarianoSuárez-Alvarez Thanks for the discussion. See ya
 
The group is generated by the class of a point.
 
6:40 AM
right. and $[(x,y)]$ is the point $(0,0,1)$ :D @MarianoSuárez-Alvarez
 
Right. Thanks @MarianoSuárez-Alvarez
@MarianoSuárez-Alvarez I gave a counter example here, I think it's right: math.stackexchange.com/questions/533951/…
 
gtg @MarianoSuárez-Alvarez. Bye!
Thanks.
 
 
2 hours later…
8:21 AM
Hi @amWhy how are you?
 
 
1 hour later…
9:34 AM
Hello! The integral $\int\limits_{S^1} \frac{1}{\vartheta \cdot x} \, d\vartheta$, where $x \in \mathbb R^2 \setminus 0$ doesn't converge, right?
 
9:48 AM
In $(ℤ/5ℤ)[X]/(X^2+2X+2)$. We have that $0=X^2+2X+2=X^2+2X-3=(X+3)(X-1)$. Therefore $(ℤ/5ℤ)[X]/(X^2+2X+2)$ is not a domain. Is this correct ?
 
 
1 hour later…
10:56 AM
@robjohn I recently thought about whether there is a compact non-finite hausdorff field.
 
11:18 AM
subtractive tetration, explain this equation:

a = 1;
b = 1;
c = 1;
Do[
a = N[Exp[-a - b - c], 30];
b = N[Exp[-a - b - c], 30];
c = N[Exp[-a - b - c], 30];
, {n, 1, 175}]
a
N[LambertW[3], 30]/3

0.3499696316546799867
0.349969631654679986662899023518
 
11:29 AM
I appear to have attracted a downvoting foe recently. Someone has been through my (somewhat recent, but off-frontpage answers). math.stackexchange.com/users/1269/kahen?tab=reputation
 
11:55 AM
hello :)
 
 
1 hour later…
1:21 PM
1
Q: Problems related to symmetry

user43418I am currently studying the chapter entitled "Symmetry" from Michael Artin's book "Algebra" and am having some difficulties understanding the material. It is dealing with isometries, dihedral groups, ... Would anybody recommend any other book for the lesson and somewhere were I can find problems...

 
hi
 
 
2 hours later…
3:10 PM
@skull hi, Skullzinho
@mats I always read your name as GraviNk >.<
 
3:40 PM
it's a bit irritating that editing bumps the question up to the top of the front page
 
4:01 PM
Yes, it is
 
4:37 PM
I think the set of all binary operations $(a,b) (c,d) \mapsto (p(a,b,c,d), q(a,b,c,d))$, that preserve non-coprimeness, forms an ideal of $(\Bbb{Z}[x_1, \dots, x_4])^2$.
 
@PeterSheldrick I am glad, there's nothing worse than a question not being answered simply because it's been pushed to the background.
Guys bbc.co.uk/news/technology-24608277 a microsoft "glitch" that affects only 17 people.
 
In recent news, I took a clean poop this morning
o_o
 
I'm happy for you, truly.
 
I'm a regular mathematician now, get it?
a real log-a-rhythm
 
Where are you getting these jokes from? The big bang theory?
 
4:47 PM
I don't get it
All i know is that the topology of my poop was metrizeable
 
5:23 PM
Hi. There are two water containers A & B. A has radius r and B has radius r/2. They are both put out in rain. Assuming both are getting equal amount of rain, what would be the ratio of amount of water in both containers?
can someone please tell what would be the answer?
 
@skullpatrol is 8 my answer? can i get it in percentages? like:50:50 or maybe a little explanation plz. my math is not good
now that i have thought about it, its simple. 50%(A):100%(B). Hope im right.
 
The first thing I thought of was the formula for the volume of a sphere V = (4/3)pi *r^3.
and V' = (4/3)pi *(r/2)^3
 
5:53 PM
@skullpatrol hi
 
@Charlie hi
@Charlie how are you?
 
@skullpatrol I'm fine, and you?
 
@Charlie sleepy ;O
 
@skullpatrol oh...
 
@Hussain @Skullpatrol If they both get equal amounts of rain, the ratio of the water in both containers will be 1
 
5:56 PM
Hi @kevin
 
Howdy @Charlier
 
@KevinDriscoll i'm good, though it's a bit too hot
and how are you doing, @kevin ?
 
Hot, really? Its getting cold here. I'm doing well, just need to get some more work done.
 
@KevinDriscoll 30°C
 
@Charlie I'm American, but I know what you mean
 
5:59 PM
XD
I've never seen an author saying I'd have fun reading his work. I saw yesterday, and he was right
lots of fun
 
I wouldve also accepted 303°K
 
:D
 
@KevinDriscoll Kelvin does not have a degree symbol.
iirc
 
oh yes, it's true!
 
@Skullpatrol That's correct, but I think its BS. So I use one anyway
 
6:04 PM
:D
 
:)
 
:)
;O
 
;O
:)
@KevinDriscoll What have you been doing, Kev ?
 
@Charlie just trying to compute this complicated finite part integral
 
@KevinDriscoll ohes
 
6:10 PM
I've been procrastinating on it because I dont wnana try my idea and see that it doesnt work
 
@KevinDriscoll i do that quite frequently :/
 
6:25 PM
hi @ian
 
Hej, can somebody help me to write a code for calculation of coefficients in cubic terms of simple characteristic polynomials ?
 
Bye @Charlie and et al
 
@skullpatrol Bye, Skull :)
@Irina lets try
 
what's the idea Charlie ?
 
6:32 PM
@Irina what do you know about what you wanna do?
if you have an idea, then it's easier for us
and what language?
 
well, let me see how to explain!
The best way might be to try in Mathematica;
 
ok
i don't work much with mathematica
 
then what do you think about the case? In which language you suggest ?
 
I only know C :)
 
any idea of coefficients computation with C ?
 
6:36 PM
maybe
@Irina you said it's characteristic polynomials, right?
 
Right!
it's easier to see the case with this polynomials'
I am solving physics problem, but got stuck in coefficients calculation in front of cubic terms;
 
@Irina could you share the problem you are stuck?
 
I dont know what cubic terms of simple characteristic polynomials means
characteristic polynomials of what?
 
here is the type of equation: one might get an exact code for coefficients:
 
but characteristic ppolynomials supposed to come from a matrix...no?
 
6:43 PM
@Charlie Or a differential equation or some other things
 
I have hug matrix
which solved;
 
@KevinDriscoll ah, ok :)
 
I know some C, C++, & D, can help
 
Now, this looks simple, you have to take the coefficients, which are numbers;
 
6:45 PM
coefficients in front of cubic terms, get it ?
 
yes, i think i do
 
Calculate the char polynomial of a $3\times 3$ matrix or $n \times n$?
 
@EnjoysMath Its a "huge" matrix
 
$3\times 3$ seems huge enough
 
you are free to choose n
 
6:46 PM
for me
:D
 
take N=6 for example
 
What exactly are you trying to calculate?
 
physical charges
 
No I mean mathematically
 
coefficients of cubic terms
 
6:48 PM
@Irina can you calculate the eigenvalues of the matrix?
 
it's done Kevin;
Mathematica calculates coefficients of linear terms
 
well then this should be pretty simple because the char poly is just $\prod_{i} (x-\lambda_i)$
where the $\lambda_i$ are the eigenvalues
 
oo, come on man
doesn't work; :)
 
Why not?
 
I'm confus, then
 
6:52 PM
yes Charlie, I am also confused actually
 
From Michael Artin's Algebra: " The char polynomial of an $n \times n$ matrix $A$ has the form $p(t) = t^n - (trace A) t^{n-1} + \text{ (intermediate terms) } + (-1)^n(det A)$. There is a formula for the cubic term coefficient. You have to find that out.
 
it seems simple, but general strategy doesn't work; or, maybe i am not accurate with changes in physics problem
 
charges or changes?
What's the physics problem?
 
generally, calculating charges in higher dimensional CFT's
 
I thought you were gonna say something like that
 
6:55 PM
but the issue comes in cubic term
 
CFT?
 
Conformal Field theory
 
Yes
 
@EnjoysMath Its a formalism used in both high-energy and some condensed matter applications
 
great, let's solve her problem and blow up the world :D
 
6:56 PM
Sadly, my physics knowledge is useless here
 
well, friends; it seems a bit hard to handle now; my brain blows up
 
@Irina what about that determinant formula I posted, does that help?
*char poly formula
 
Greetings
 
maybe tomorrow post-doc can push me
Greetings :)
@Enjoys Math; problem seems more serious
 
@Irina as i said, post the original problem, maybe from it we can understand better
:)
@Chris'ssis helloes
 
6:59 PM
@Charlie hello! :-)
 
@Charlie Sir :) should I open my project here ?
 
@Irina yep
 
@Irina howya doing?
 
@Charlie Hi! Just lurking around.
 
7:00 PM
@IanMateus hehe studying a lot, recently?
 
@IanMateusI have something very cute to show you.
@IanMateus $$\sum_{n=1}^{\infty} \log\left(1+\frac{1}{n }\right) \log\left(1+\frac{1}{2n }\right) \log\left(1+\frac{1}{2n+1 }\right)$$
Isn't it lovely? :-)
 
@Charlie Nope, waiting another week to come.
 
@IanMateus haha que ano cê tá?
 
@Chris'ssis I don't think I've ever dealt with a product of logarithms before.
@Charlie terceiro ano
 
@IanMateus ah, tá na cara na faculdade já ;)
 
7:02 PM
@IanMateus there is a "first time" for all things. :D
 
@Chris'ssis Nice, I'll try to think something about it
 
@Enjoys Math: Thanks for Michael Artin 's book;
 
@Charlie aham, não sei o que eu vou fazer ainda :(
@Chris'ssis how did you find it?
 
@IanMateus não sabe qual curso, ou qual caminho tomar para chegar lá?
 
7:04 PM
@IanMateus I received it from a kid by e-mail, he challenged me.
 
@Charlie curso, acho que vou fazer bacharelado em matemática
 
@Enjoys Math: yea, right; you seems to be really enjoying math ;)
 
@IanMateus pense bem
 
@Chris'ssis does it have a nice elementary closed form?
 
@IanMateus It should have.
 
7:06 PM
I do enjoy math, I wonder how everyone knows that
 
:)
you wonder?
 
@Irina is your research on the string theory/quantum gravity side or the condensed matter side?
 
@Kevin, string theory
I am struggling in CFT's :)
 
@Irina Okay, that was my guess when u mentioned high-dimensional stuff
@Irina Well, that's no problem! Just go into the bulk and do some calculations there!!
 
are you working in this stuff too? Kevin ?
hahahaha, that was fun
 
7:11 PM
@Irina No, I do many-body quantum theory
 
I just smoked some weed, hit me with problems :)
^-^
 
bulk-boundary conditions sometimes disappointing me intensely
@Enjoy, are you math student ?
 
But I spent the past 3 or 4 days watching the fuzz or fire talks at UCSB about the firewall problem @Irina
 
@Kevin, that the part of fun
 
I'm a student of math, but not currently a college student
 
7:13 PM
@Kevin, What horrifies me is the idea that someone might use noncommutative field theory to resolve the naked singularity in the 2d dilaton gravity black holes. I can feel it is coming.
@Enjoy, keep it up;
 
You must be psychic :)
 
@EnjoysMath when I read your location, I thought of this
 
I have no idea what you're saying btw, but it sounds incredibly advanced. Are you from the future or just really into mathematics?
 
@EnjoysMath You have to be into mathematics to study string theory
@Irina I was unaware that there was a naked singularity problem
 
@Kevin, there is :)
you are researching Kevin ?
 
7:18 PM
@Irina this is in a 2D bulk theory with a 1D CFT at the boundary?
 
you are genius Kevin;
 
hehe, if my guess is correct itjust means I've been spending too many hours listening to Susskind and not enough hours on my own research
 
:) you are also in PhD ?
 
I am. Right now I'm just trying to compute some nasty integrals
 
I am also doing lots of technical work
it terrifies me!
 
7:23 PM
@EnjoysMath so you study on your own?
 
@Irina My advisor and I have alreayd generated an accurate numerical solution to this problem, but I have spent the last 4 months trying to develop an approximate analytical solution with basically no progress. My advisor seems totally happy with a numerical solution, but it annoys me. I have 1 or 2 ideas left to try and if that doesn't work I'll be moving on ot something else
 
being physics student is hard to enjoy math stuff reall Kevin
technical work is part of the work, i ma say most of the work :)
 
@Irina It can be. My advisor has a very good intuitive grasp of mathematics so he can kind of 'cheat' and get the right answer. I'm not so experienced so I have to do things the rigorous(ish) way. Which means I spend a lot more time thinging about some technical math questions
 
I'm in the same track; It distracts me from the main project dealing with codes and proving formulas..
 
@Irina Have you kept up with this Firewall problem? Do you have an opinion?
 
7:30 PM
@Kevin, I heard a nice review of attempts to find theories where the entire black hole entropy can be understood in terms of entanglement of matter fields in the background (related to "induced gravity"), and of the disasters that befell information in 2d dilaton gravity.
I am studying things, but idea comes in time, I am not much experienced with strings, but trying my best Kevin :)
 
@Irina Of course. These things take years of study and experience.
 
and I am just a 1st year graduate student Kevin;
people seems to be silent Kevin :)
 
@Irina It seems that it'd be nice if the entropy can be explined in terms of entanglement, but there do seem to be some serious problems with that idea. To salvage it, it seems that we'll either have to totally redefine what we think happens at the event horizon, or admit that the equivalence principle fails, or admit that there is information loss in the universe. None of which seems particularly nice.
I think people are silent because no one knows what's going on (as Polchinski admitted during his talk on the issue)
UNless you mean in the chat. In that case, its just cuz its the middle of the day and they've been overrun by physics students!
 
@Charlie, yep, not by choice, I just don't know any mathematicians around here. OTOH I don't because I interact with other math people on MSE!
 
Here is almost 10 pm :)
 
7:39 PM
@EnjoysMath i see
 
Does weed help anyone here think creatively about math problems?
 
@Kevin, holography is growing intensely nowadays;
Particularly, the talks by CM theorists about topological entanglement entropy, classification of topological phases, and open questions in their field more generally were inspiring me.
 
@Irina Yes. It is something of a blessing for people working on this, becuase even as we exclude more and more of the supersymmetric theories at the LHC, a whole new field has been opened up that uses holography. I just heard a talk about this the other week by Brian Swingle (who is at Harvard)
He has been at least somewhat trying to incorporate tensor networks and holography (Juan Maldacena is doing something similar I think)
 
yes, Maldacena is building exact models @Kevin;
@Kevin, are you familiar to Moonshine stuff ?
 
7:48 PM
@Irina No
I dont really know any gorup theory
 
@Kevin, No technical prerequisities, but a serious interest in modern physics is desirable! ;)
 
Haha, I see. At some point in the next few years I'll sit down and really hammer out this groups stuff. BUt its gonna be hard and I dont have any use for it yet
 
@KevinDriscoll it's a fascinating subject, kevin
 
@Kevin, what about wormholes ?
 
@Irina Everything I know about wormholes I learned just 2 or 3 days ago from a talk Susskind gave on Einstein-Rosen bridges
 
7:55 PM
@Kevin, history has shown that down the path of wormholes lies nothing but madness, despair, and bad science fiction movies. People are resisting bravely, for now... :-)
 
@Irina Yes, the only thing I see people suggesting now is that there may be non-travesable wormholes. You can tgo through them or sue them to send information, but perhaps htey entangle the inside of the BH to some other region of space (perhaps the boundary even)
 
@Kevin, Sounds interesting;
 

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