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12:00 AM
hm, I woke up
 
mee too
 
I present to you the following!
 
I have a question
 
$|G|<\infty$ and $G$ has a unique group of order $d$ then I am trying to prove $H\unlhd G$ and here is my proof, is this correct: (writing)
take $g\in G$ then $gH=H$ that means $\forall g\in G \implies g\in H$ so $G=H$
also if $\forall g \in G, gG=G$, that mean $G$ is abelian (though I am unsure) so $G=H\unlhd G$
 
12:10 AM
$g\in G$ does not imply $gH=H$, and of course $g\in G\implies g\in G$ that's obvious, and none of that implies $G=H$
you should also say $H$ is your subgroup of order $d$
and $gG=G$ does not imply $G$ is abelian
so pretty much nothing you're saying makes any sense
 
$|gH|=|H|=d$ and this is unique
@anon it was a typo
 
$gH$ is not a subgroup
you need to say $gHg^{-1}$
 
was this trivial
 
$gHg^{-1}=H$ for all $g\in G$ is the definition of normal, so fairly trivial
 
no, I mean coming up with this normal subgroup for the proof, was it trivial?
 
12:13 AM
what?
 
When you say "Hg", I think of mercury.
 
well that completes the proof, thanks @anon
 
gG
 
the good elements of the group of Games act on themselves by multiplication
 
A few rare visitors to this chat tonight.
 
12:17 AM
Yeah hey guys it's my first time on chat
3 - 6x3 + 2
 
Welcome @Daminark
 
Only for genius
Someone help ples
 
the answer is 28
 
@Daminark you have a post from 8 hours ago , damn his head looks like a tic tac
 
You are a genius.
 
12:18 AM
lets keep this chat professional
 
Whoops I got -1/12...
And lol aight
 
i am a professional
let's talk about 1 + 2 + ... = -1/12
 
what is a cycle type ?
do they mean even / odd
or n-cycle
 
I think it means that you write it as a product of disjoint cycles
And then just list the sizes of the cycles
 
Math has cycles?
 
12:23 AM
Is index of a group defined even if it is not normal because the quotient group is then not defined?
 
Yup, context is cycle notation for permutations in algebra
@RE60K I think you can just define it as the number of cosets
 
ah they meant for example (123) (45) ( 89) is ( 3,2,2)
but thanks :)
 
Well yeah that's what I said :P
 
I think that is you said but did not get it untill now =p
:D
 
Sniped
 
12:26 AM
if they cone out to be prime for a finite group (G) 's subgroup (H) does that mean $\{gH|g\in G\}$ is cyclic?
 
Also note that disjoint cycles commute
So this is unordered
 
Good point
 
therefore $|gH|=p$? and $H$ is cyclic?
rather $g_1,g_2\in G$ $g_1Hg_2H=g_2Hg_1H$ because $\{gH|g\in G\}$ is cyclic hence abelian
So $G$ is abelian?
 
Hi mathematicians. I admire you've taken the math field for your studies. I would like to know how you mathematicians think out of the box when it comes to solve math problems? Any books? Any tricks?
 
<--- never professional
@Kasmir: Of course it's not that hard. So what did you find?
 
12:37 AM
lol
hi @Ted
 
Oy vey, Balarka, are you still awake?
 
ah no i just woke up
 
Oh, maybe that's better :D
 
much better, yep. i got sleep for about 4 hours
 
Well, that's better than 0.
 
12:38 AM
true
 
Oh, I missed @anon. Hi :)
And Pedro is happily a graduate student in Ireland now.
I may have to stop hanging out here. The whole room is studying algebra. I miss analysis and geometry.
@KasmirKhaan So how many of each sign?
 
@TedShifrin sticks out hand furiously in a crowd of algebraists
 
@Balarka is still doing geometry I think
Oh lol sniped but by my internet
 
Balarka, you were the algebraist ne plus ultra a few years ago :P
 
right now Riemann surfaces
 
12:43 AM
Yes, RS will be acceptable. :P
 
but I am also doing a reading course on ODE's and dynamics which i started this week
 
I don't know if I still have my problem sets to send.
That's good, Balarka. What the hell will you do when you get to university?
I took a grad course in dynamical systems my first year in grad school and would have gone into that had I not had the opportunity to work with Chern.
 
Ah, hah.
 
Okay so I saw a Riemann surface once, defined as "gluing copies of the complex plane together along branch cuts so that you can make a smooth choice of branch" (yeah it was in the old physics stuff)
 
The professor I'm doing the reading course under is actually one of the people who gave a series of talks on dynamics on the workshop I visited
probably my favorite series of talks
 
12:45 AM
That's a totally naïve description.
 
But I've also seen it as a 1-D complex manifold
 
You do better with reading courses than most US grad students, Balarka.
 
meh, it's a description that works if you want to do contour integration in the presence of branch cuts and such.
 
Yup, that's the correct definition, Demonark, but understanding the former is worthwhile. However, it is thinking of explicit branched coverings of $\Bbb CP^1$.
 
Bob
recently I worked on this problem: math.stackexchange.com/questions/2425919/… I am wondering what the level of the problem is.
 
12:46 AM
Right, the point is every compact Riemann surface branches over the Riemann sphere
So you can think about what happens near the branch points if you loop around
(Branched cover means covering space away from finitely many points btw)
 
@Bob: You keep asking questions like that. What's the point?
 
Bob
To see how I am doing. I think I have asked a question like this only once before. Maybe twice.
 
It's probability at the college level. It's typically an upper-division course that requires multivariable calculus. But there's a lot more stuff in such a course.
You're doing the same stuff as before, best I can recall. It's still probability.
 
Bob
I thank you Ted S.
 
Well because it was from some time ago the only example I recall is the log
 
12:50 AM
That's out of the purview of algebraic geometry. That's noncompact and analytic, Demonark.
Log is cool because it's infinitely multivalued.
 
stuff keeps winding
The deck transformation group is Z under the usual covering projection
 
oh oh ... Balarka's gonna move into country western songs.
 
haha
 
Eh, Log is fairly tame
 
The idea is that you want to cut off the negative real numbers so you can integrate around a loop, but you lose all the info about it being multivalued. Now, I can see why that gives you a helicoid
 
12:52 AM
Lambert-W is...worse.
 
@Semiclassical I knew you'd say Lambert-W
 
I guess I am enough an algebraic geometer that this sort of stuff has never captivated me.
 
Understanding the branch structure of the Lambert-W function gave me a headache.
 
Yes, Semiclassic has become utterly predictable.
I won't tell you how many headaches students have given me :D
 
just because I'm predictable doesn't mean I'm wrong :P
 
12:53 AM
And why that does it all. But I think something was said about this being the graph of the function. That's what I'm not sure about
 
Demonark, start with a baby problem. Think about $w=\sqrt z$ ... or $z=w^2$.
 
Also @Ted I didn't actually know this had anything to do with algebraic geometry (though I didn't really know what that is so...)
 
Algebraic curves.
 
it's 1 dimensional complex algebraic geometry
 
By GAGA, for compact, analytic = algebraic.
 
12:55 AM
right. pretty surprising theorem though
 
I take it for granted :)
The proof I read in Griffiths/Adams was not that bad.
 
I forget the statement. Every compact analytic (aka can be locally cut out by analytic functions) submanifold of CP^n is also algebraic (aka can be locally cut out by algebraic functions)?
 
But don't ask me to berember.
Not locally. Globally.
 
What does it mean to be globally cut out by functions? CP^n doesn't have global functions on it
I thought it means that on affine charts
 
Oh, I meant globally in the conclusion.
Homogenous polynomials.
 
12:58 AM
Ah, I see
Ok, yes, aka it's an algebraic subvariety of CP^n
 
nods
 
If something is on every affine chart cut out by polynomials, then surely it's globally an algebraic subvariety though?
 
Yeah, that should follow from vanishing cohomology that tells you things patch.
 
Hm, I see.
 
Okay so for the square root, we don't want to circle around 0, right?
 
1:03 AM
Well, if you do, you go from one branch to the other.
Notice from the diff top point of view that the projection map $(z,w)\mapsto z$ is a covering map away from $z=w=0$.
 
Is infinity also a problem? A circle around infinity is one around 0 so a good branch cut is probably the negative reals
 
yup
 
Good question, Demonark. Are there two points over infinity or just one?
 
I think two, because if you go to it along positive numbers (principal root) you're approaching infinity, if you do it along negative numbers you're picking up i
 
By that logic, why don't you pick up a circle's worth of points?
 
1:11 AM
Good question, and the square root has just two values... Oh Riemann sphere, right?
 
Small hint: Think $z \mapsto 1/z$
I won't explain that.
 
The base space is the Riemann sphere. We don't know what's upstairs.
Balarka has learned to be obtuse (learning from Ted, I presume).
 
1:29 AM
Okay, uh, I'm sorry if you go blind reading this but I'm gonna say one point over infinity, because sqrt(1/z) when you go to infinity goes to 0, and at that point you're single-valued. 1/z is also single-valued so...
(Also sorry I'm taking forever to respond, I'm still grappling with this)
 
Grappling is fine. You don't quite have it right, but you're close.
 
Hi chat
 
Hi @Eric
 
Yo @Eric
 
What's going down
 
1:41 AM
listening to finnish death metal
lol
 
ur ew
well i guess i can't call it death metal; black metal perhaps
 
One of the few genres I don't like
Any interesting math lately?
 
Complex analysis sorta?
 
@Eric i like the non-mediocre lyric. i'm not super into extreme death growlish stuff
 
1:45 AM
I scroll up and see Riemann surfaces stuff
@Balarka I used to enjoy metal but I just don't anymore idk why
 
maybe you have turned into a normie as you grew up
i kind of hated it before i got into the good stuff
now i see the appeal
 
I still wouldn't quite call myself a normie
 
eheheh
i was trolling you there
 
<- is aware
 
@Ted okay so 1/sqrt blows up as we go to 0 to a unique value
Because sqrt at 0 is single-valued, as is 1/z
But the pre-image of 0 under the whole thing is the pre-image of infinity.
 
1:59 AM
Hi Eric.
Demonark: So you need to think of parametrizations at $\infinity$ in domain and range. You need a conjugation. But you understand what's going on, methinks.
 
HAYLO
 
@TedShifrin Would you mind telling me if an argument makes sense if i email it to you?
w8 i can prolyl type it
 
not based on that message
 
@Semiclassical so tutorial was literal BS, we didn't do the quiz but spent an hour doing the "practice problems" I ended up just reading the text book and then the prof from the first section of the class showed up. I sent him an email asking about required readings and he never answered.
 
lol
 
2:04 AM
Very confused about what's happening in that class.
 
Assume $ Sup(S) + Sup(T) > Sup(S+T) $ where S and T are bounded subsets of the reals does my assumption that $ \exists \alpha \in \mathbb{R} $ s.t $ Sup(S) + Sup(T)- \alpha \geq Sup(S+T) $
make sense?
why can make this assumption
 
Okay, so 0 and infinity are precisely the two points for which sqrt is single-valued
 
OK, Demonark.
 
@TedShifrin can you recommend a text to learn to graph functions intuitively?
 
You need $\alpha>0$, Faust.
 
2:08 AM
w/o any type of graphing software.
 
Yikes, no, Dodsy. That's what precalc classes are for.
 
:(
 
can assume that it exists and is greater than 0?
 
Look at a precalc book.
Yes.
 
why
what magical property makes it so?
 
2:08 AM
What does it mean for $a>b$? It means $a=b+\delta$ for some $\delta>0$.
 
ok ok ! ;)
just wanted to make sure i wasnt assuming somthing outlandish
whats if
$ Sup(S) + Sup(T) $ is negative?
thats why i didnt assume $\alpha >0 $
nvm
 
so I know that a cubic function has 2 "changes"
 
Not always, Dodsy.
 
:o
 
3, actually. But sometimes 0.
 
2:11 AM
@Dodsy $x^3$
 
what counts as 3
 
Up-down-up
 
yes that's what I meant
 
Oh, 2 changes. Never mind.
But sometimes 0.
 
no problem, I'm probably using the wrong terminology :D
 
2:11 AM
No, I'm being dumb.
 
it would have 3 intervals of increasing and decreasing though?
 
But I'm about to go eat dinner, so I'm disappearing.
 
:o
 
Yes.
 
but you are my only hope Ted!
 
2:12 AM
Invisibru Tes
 
No, I'm not. You don't need me.
 
we all need you, mein freund.
du bist wunderschon
my german seems to get worse after a couple drinks!
but my math seems better.
 
LOL. Bye for now.
 
bye Ted.
 
See you @Ted!
 
2:15 AM
maybe you guys could help me
 
I CAN TRY
sorry caps
 
Okay so I'm not really supposed to discuss this with pplz
:o
but I'll give an example because I really want to learn this stuff
so this isn't the real question
the real question can not be solved by factoring
but basically it's under a square root sign
I'm supposed to express the function as a union of dijoint intervals
so I know that $f(x) \in \mathbb{R} : f(x) > 0$
but the function under the root is a cubic
so I'm looking at the functions graph for areas of the graph where $f(x) > 0$ but am having a hard time getting a coherent view of the graph.
feel like I'm going to be instantly expelled now
>_>
anyways I'm not supposed to use any electronic resources ( no desmos)
or talk to anyone about it.
 
@Dodsy I repeat: ew
(I'll refrain from commentary on the above.)
 
thanks I guess haha
any good resources?
 
[Chemistry] Example of stuff that is not working:
 
2:30 AM
:)
 
so far I'm just throwing points into the equation
 
Well, there is always this place called MSE... :P
 
oh god
 
Anything that does not look continuous is screwed
 
2:31 AM
That looks like the scatter plot equivalent of a Rorschach test.
 
it does, some of my collegues also comment it looks like a butterfly
As for the context, this is a potential energy surface profile as I rotate the base of some boron containing rhodium complex, I strongly suspecting there's a minima at around 175 degrees (the horizontal axis) but the SCF energies is so screwed that nothing can be determined from it
What is certain, is that there are at least two minima located somewhere at 75 and 300
Currently figuring how to fix it. But at least, it is a chemistry problem, and no longer a computer one
 
@Semiclassical I solved it
I just sketched a quick graph
and plotted 5 pointes
 
Huzzah
 
and figured it out
hopefully it's right.
 
yeah. being able to get a graph sketch is essential, whether via technology or by hand
 
2:38 AM
now I've gotta write it up in latex and make another graph of a diferent function.
 
In the context of chemistry, having a potential energy surface profile tells you alot of how different isomers are related energetically, which is why when nearly 90% of the potential energy surface is blank is not very illuminating
and later on when moving on to transition states, it tells you the reaction profile such as how fast it is and how thermodynamically favourable it is
I think it is safe to say that regardless of what discipline, being able to plot a graph is always useful
 
ah, activation energy
 
zhk
2:56 AM
Hello Every one
How to find this inverse Laplace? ANy suggestion?

Secondly, What is E_a,_a(---) in 12.17.54?
 
$x \sqrt 6$
sorry
testing out latex :3
 
re
 
3:33 AM
@Semiclassical let's say I'm trying to say that a graph only exists when a function is greater than a certain number, how would I word that?
 
Probably you'd say something about the domain.
 
yeah but I'm looking for a better term than "exists"
 
e.g. $\sqrt{-x^2-1}$ won't have a graph because the domain of sqrt is nonnegative reals and the range of -x^2-1 is negative reals.
 
like what is the technical term for a graph not existing at a point
like let's say my domain is $[6,\infty)$
what would 5 be?
if x isn't part of the domain
 
Not in the domain of your function.
That is the technical word.
 
3:37 AM
would it be weird to say that if $x \geq 6$ then $x$ is a point in $\mathbb{R}^2$
or is that convoluting the situation.
 
well, yes. $x$ is just a real number
it's not a pair of real numbers.
 
right
 
for that matter, y(x) would just be a single real number as well.
so if you want to have a point in the plane, neither of them is enough.
 
sigh.
so I was thinking today
if I say that my interval is $(3,\infty)$
it's the same as saying
$[2,\infty)$
right?
or do I have a poor understanding of open/closed intervals
@Semiclassical
 
Well, both have infinity. So that part is the same. What does the other endpoint signify?
 
3:45 AM
I thought it meant does it include the number or not.
so the square bracket would include 2
 
That's the bracket ( or [, sure. But, how would you represent those intervals with inequalities?
 
$\{x\in \mathbb{R}: x > 3 \}$
$\{x \in \mathbb{R}: x \geq 2\}$
now I seem to have made a mistkae somewhere
damage control
sorry
I went the wrong way
I should have said that $(3,\infty) = [4,\infty)$ , right?
 
well, here's the question: is $x>3$ the same as $x\geq 2$?
 
no I made a mistake
^^
 
or, simlarly, is $x>3$ the same as $x\geq 4$?
 
3:49 AM
I think so.
 
To put it another way: Is every real number that's bigger than three also at least as big as four?
 
ah I am focusing on the naturals.
 
very foolish.
perhaps I shouldn't drink on wednesdays anymore.
 
what would be correct is that [3,infty) differs from (3,infty) only in that 3 is there or not there
so $[3,\infty)=\{3\}\cup (3,\infty)$
 
3:51 AM
but if $\{x \in \mathbb{N}\}$ we have a different story?
for my above example
of (3 vs [4
 
right.
 
so do mathematicians make that distinction when discussing naturals often?
 
do they prefer using open or closed notation
 
tbh I think one usually only uses interval notation when doing real numbers? but don't quote me on that.
 
3:53 AM
no worries.
 
and using open vs. closed really depends on the problem of interest.
 
yeah true.
 
as an example, if you were doing y=1/x, then you definitely wouldn't want to include x=0.
so in that case you'd want to talk about either R-{0} (i.e. every real number but 0) or equivalently as $(-\infty,0)\cup(0,\infty)$
 
that's an interesting union.
 
the former is more convenient when you've only got a handful of cases to exclude
the latter is more convenient when you've only got a handful of cases to include
 
3:55 AM
I like that one.
 
it really does come down to the problem and, if that's not decisive, to taste
(and sometimes they're both annoying, in which case you throw up your hands and do the best you can)
 
How do you know how many words to put into your answers?
like I'm kind of skirting the whole thing
just saying "well the square root of any real number must be > 0 so ...."
is that sufficient?
 

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