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7:00 PM
I'll be back in 30 minutes or so. I have to finish a problem for a student.
 
@BalarkaSen Maybe. But it's of fundamental importance to my field
It would be nice to have an actual proof
 
Let $p$ be a prime number. Prove that $$\prod_{k=1}^{p-1} k^{2k-p-1}$$ is an integer number.
 
If one has a continuous vector field fitting the bill, can one smooth it without picking up zeros?
 
yeah you can approximate C^0 maps by C^infty maps arbitrarily close to it
a vector field is a map M --> TM after all
well there's some work to keep it a vector field after smoothing
some pullback shit
 
@LeakyNun hi
 
7:10 PM
@BalarkaSen arbitrarily close works on a compact guy...
 
@KasmirKhaan hi
 
:D
I want to see if I got the isomorphism theorem or not
can I write what I got and you tell me if its right or wrong ?
 
ok
 
All righty
we have a surjective homomorphism from f: G-->G'
we first the a map pi : G---> G/H by sending a---> aH
where H is the normal subgroup of G
then from we have another map , f -bar : G/H ---> G'
f bar ( aH) = f(a)
what the isomorphism theorem says now is that composing fbar with pi
is the same as doing f
But G/H --> G' is an isomorphism
ie f bar is an iso
@LeakyNun well ? :D
 
2 mins ago, by Kasmir Khaan
where H is the normal subgroup of G
you finally define your variables :D
that's a good change
 
7:15 PM
haha :D
well I kinda get what this says
and I know how to prove that this is well defined
why we define fbar (aH) as f(a) and all the details
 
good
 
still I dont get the feel of this theoem
I need to see how it can be used
Can you give me an example and i try to solve it?
 
consider f: S3 -> Z2 defined by the parity
 
hmm let me see
do you mean parity of like even permutation get sets to 0
and odd to 1 ?
 
yes
 
7:18 PM
okay first i have to list the elements of S_3
1, (12) ,(23) , (13) , (123) ,(132)
to find a normal subgroup i can take any of the two of order 3
okay let me see H = {1, (123), (132) } this would work right?
oups
does not work
accually it does
123 and 132 are inverses of each other
and we have the identity
 
@KasmirKhaan
 
@LeakyNun the question is i need to prove that S3 / H is isomorphic to Z/2 right?
 
H is the kernel of the map
you're doing it backward
 
what do you mean ?
 
37 secs ago, by Leaky Nun
H is the kernel of the map
 
7:26 PM
Yes i know that
 
a homomorphism f gives rise to a normal subgroup ker(f)
and a quotient G/ker(f)
 
ah you mean I could get that info from the ker
all right so even permutations will be in the kernel
that is what i got but after a bit of work ><
not nessasry work
Ehm this map is a surjective home f : S3 -->Z/2
well in this case its easy
[G:H] has index 2
and any group with 2 elements are isomorphic
@LeakyNun .
 
what is isomorphic to what?
 
G/H and Z/2
 
do you know the names of G and H?
 
7:31 PM
S3 / kerf isomorphic to Z/2
 
Hi everybody
 
do you know the name of ker f?
how do you call the group of even permutations?
 
correct
 
S3/ A3 isomorphic to Z/2
neat :D
Can you ask me some questions on the theorem of iso
to see if I got it right or not
 
7:32 PM
Are you still here? @Eric
 
How are measures of "nice" subsets of $\Bbb R^n$ defined? For example in physics one often integrates of the energy levels $H^{-1}(E)$ of some smooth function $H$. Specifically right now I am interested in getting a measure on the boundary of a (bounded) convex set.
(lebesgue measure is the only thing that exists as an inducing measure here)
 
Hausdorff measure of the appropriate dimension I suppose?
 
@KasmirKhaan why does the homomorphism need to be surjective?
 
hmm
it does not need to be surjective we can have
 
@Alessandro does this measure agree with the induced measure you get when your subset is a manifold and you look at the measure you get from the induced riemannian metric?
 
7:36 PM
G/ker f isomorphic to imf
 
nice
 
here f :is a hom
:DDDDDD
well i think all I need now is to do some exercice on this
to see how usefull it is =p
anyways
 
consider G=Z and f(n)=((12)(345))^n
 
What is R/Z
 
@KasmirKhaan a circle
 
7:37 PM
but why is that?
I mean arent the cosets of the form a+Z
where a in R
 
runs and hides
 
@TedShifrin HiTed :D
 
@s.harp I don't know what a Riemannian metric is so I'm afraid I can't help with that
 
Hi @Ted
 
but @Ted can :P @s.harp
 
7:38 PM
Well, you should know, @Alessandro :P
 
@TedShifrin BTW I understood the theorem from your book :D
 
hi @Balarka
 
should I?
 
7:38 PM
no hi for poor kasmir?
 
I ran and hid from poor Kasmir :D
 
@Ted y u use a lowercase h for me
 
Oh, hell, @Balarka, it's more than you deserved.
 
i am hurt
 
7:39 PM
Yeah, I know.
 
@TedShifrin you had an example with R/Z i could not get
 
@KasmirKhaan consider a set of representatives [0,1)
 
direct after the proof of the iso theorem
 
where you wrap back to 0 if you pass 1
just like a circle
 
@s.harp: If you're working with something other than an open set of $\Bbb R^n$, Lebesgue measure won't be good. You either have to use its manifold structure or use Hausdorff measure of the appropriate dimension.
 
7:41 PM
"It was on analytic and algebraic topology of locally Euclidean metrization of infinitely differentiable Riemannian manifold--Боже мой!"
3
 
@LeakyNun oh thats a neat trick :D
 
hi there self deprecating memelord
 
i was thinking of it like this, each irrational number in its own coset (ofc up to addintion by Z)
 
@Ted thanks, the Hausdorff mesaure is looking like the thing I am looking for
 
@Leaky: That's the topological picture. But the argument is just basic complex numbers, Kasmir.
 
7:41 PM
like sqrt(2) + Z ect
hmm
 
right
 
@Fargle: 3 stars for Tom Lehrer.
 
Hi @Balarka
 
but many people know nothing about complex numbers, I swear
 
@Ted: out of?
(Also hi)
 
7:42 PM
Haven't seen you in ages, Fargle!
 
@KasmirKhaan how much do you know about complex numbers?
 
@Kasmir: I gave the homomorphism explicitly using $\phi(t) = e^{2\pi it}$!!
 
I have a friend in Minsk, who has a friend in Pinsk, whose friend in Omsk has friend in Tomsk has friend in Akmolinsk
 
He's had a complex variables course, Leaky.
 
@LeakyNun id say normal knowledge =p
 
7:43 PM
lol Kasmir hasn't been listening
 
Whose friend in Alexandrovsk, has friend in Petropavlovsk, whose friend is now working somehow on the problem in Dnepopetrov
 
Ted gave you the homomorphism
 
I hope I have that right
 
@TedShifrin Been busy. How goes it?
 
Yeah, Kasmir needs to do more work on his own. :)
 
7:43 PM
@TedShifrin I know you gave the map, but i wanted to understand what R/Z is
grrrr
 
Other than oral surgery and toothaches on the other side of my mouth, fine, Fargle.
 
@TedShifrin i hope you heal fast :D
 
Well, it's topologically a circle, as Leaky just showed, and as a group it is the circle group, as well. Addition in $\Bbb R/\Bbb Z$ corresponds to multiplication in $S$.
 
@KasmirKhaan I thought you have the isomorphism theorem!
 
Thanks, Kasmir. The surgery is healing well. It's all the other pain that isn't fun.
 
7:45 PM
@BalarkaSen "And when his work is done, haha! begins the fun--from Dnepropetrovsk to Petropavlovsk by way of Iliysk and Novorossiysk to Alexandrovsk to Akmolinsk to Tomsk to Omsk to Pinsk to Minsk to meeee the news will run!"
 
f:R->circle is a homorphism with kernel Z, so R/Z is isomorphic to a circle! @KasmirKhaan
 
That's exactly the example he's referring to.
 
YES TO MEEEE (epic dank music here) THE NEWS WILL RUN
 
@TedShifrin I hope that improves.
 
Well I asked that Q only to prepare for another Q
what is R/Q
 
7:46 PM
Me too, Fargle, thanks. Any fun math for you these days?
 
@KasmirKhaan topologically, R/Q is a mess
you should know what C/R is!
 
Don't get Leaky started on $\Bbb R/\Bbb Q$.
 
@TedShifrin oh I already finished
 
haha
 
7:46 PM
Good :)
 
Okay then Ill keep doing exercices
 
R/Q just gets ze indiscrete topology
 
@KasmirKhaan describe the elements of C/R
 
but the idea of [0,1) i find to be awesome
 
@BalarkaSen what the hell
of course not
 
7:47 PM
However, topologically, $\Bbb R/\sim$ with $x\sim y\iff x-y\in\Bbb Q$ is super nice
 
@AlessandroCodenotti how?
 
are we talking of group quotient or subspace quotient
 
@TedShifrin My research is delving into numerically solving stochastic ODEs, which is...to put it lightly, not to my tastes, but hopefully it can get me a paper. Outside of that, I have a class in mathematical models (...) and a class on the Lebesgue integral (!!!).
 
ti+R where t is in R
 
@KasmirKhaan have you come across the definition of quotient in terms of equivalence relation?
 
7:48 PM
Yes
 
@KasmirKhaan therefore C/R is isomorphic to?
 
Well, that's cool, Fargle. Are you settling down to a stable existence (as opposed to your previous chaos)?
 
Oooh, Lobachevsky :3
 
@KasmirKhaan then you should know that R/Z corresponds to x~y = x-y in Z
 
@LeakyNun my guess would be R^2
 
7:48 PM
i.e. the fractional part
@KasmirKhaan no!
 
@TedShifrin Starting to. Having to wake up at 7 every day helps.
 
let me think
 
where does the square come from?
 
I'm glad, @Fargle. And I'm glad you sent Ken a note a while back.
 
7:49 PM
@LeakyNun Why? Q is dense in R.
 
Is the inclusion $\ell^p\subseteq\ell^q$ strict for $p<q$?
 
@TedShifrin Me too! I don't do it often--I have a strange reputation as a student under my various professors, as I think I've harped on before.
 
@BalarkaSen oh, you mean that topology
 
@Alessandro: Can you find something in $\ell^2$ that's not in $\ell^1$?
 
In the quotient topology something is open in R/Q if it's preimage is open in R; so it's an interval
 
7:50 PM
@BalarkaSen you're right, sorry
 
@LeakyNun I meant the indiscrete topology :P Not sure what "that" means
 
@TedShifrin The harmonic series
 
@BalarkaSen I thought you were talking about the topology on a set of representatives
 
Oh, that reminds me, @Fargle. I have more than a dozen recs to do for a former student/advisee who's applying to grad schools. I think my days of doing this are coming to an end soon.
 
blah that's the nonmeasurable bullshit
 
7:51 PM
OK, @Alessandro. Can you generalize?
 
@TedShifrin I'll have to angle in while I can! :)
 
i dont care (grapefruit is sour, yes)
 
<---- loves grapefruits
@Fargle: Sure, if you do more from my book or from diff geo :P
 
@TedShifrin Nothing in this world is free. I accept that.
It may be quite a time before I can work on it--once the paper is finished, I'll get on it.
 
@TedShifrin $n^{-1/p}$ is not in $\ell^q$ for $q\le p$
 
7:53 PM
@BalarkaSen I've never heard of quotient topology though
 
@Fargle You need to teach me ODEs
 
what is the condition on Q?
 
@BalarkaSen Just basic ODE stuff? I can do that at some point I think
 
@Fargle Yeah I'm learning about them on a reading course thingy
 
I'll need to refresh myself on stuff like variation of parameters and such techniques but the rest of a first course should be doable
 
7:53 PM
@LeakyNun is it a line ?
 
Cool!
@LeakyNun ze condition?
quotient space is defined whenever you have an equivalence relation on your space
 
@BalarkaSen for a quotient topology to exist
@KasmirKhaan what is it isomorphic to?
@BalarkaSen oh thanks
 
@Fargle Do you want to think about a Lebesgue integral doubt I had yesterday and half solved in the meantime?
 
@AlessandroCodenotti I'm not very far, but I can give it a think
 
I dont know the map
f : C---> ?
 
7:56 PM
you already listed the cosets man
recall your isomorphism theorems
 
I've been on a loose mode for too long, I should use my holidays constructively
 
f: G --> G' is a homomorphism
ker(f) = H is a normal subgroup of G
G/H is isomorphic to im(f) with f-bar(aH) = f(a)
 
Why am I a chronic passive-active procrastinator?
 
@Fargle I have a decreasing sequence of measurable sets $E_n$, such that $\bigcap E_n=E$. I'm trying to work out conditions under which $\lim\limits_{n\to\infty}\int_{E_n}f(x)\mathrm{d}x=\int_E f(x)\mathrm{d}x$
 
okay here H = R normal subroup of C
the image of C is whole C right?
 
7:57 PM
Surely I need $L(E_1)<\infty$ otherwise it's not even true that $L(E_1)\to L(E)$ (Or $L(E_i)<\infty$ for some $i$ which is the same)
 
@KasmirKhaan we don't even have a function ><
whence the image
 
@AlessandroCodenotti Hmmmmmmmm
 
i really dont get it
 
@Balarka: Are you a passive-aggressive procrastinator (not to mention prognosticator)?
 
let me think
 
7:59 PM
I think I can prove that this holds if $f$ is nonnegative on $E_1$ and $\int_{E_1}f\mathrm{d}x<\infty$, because in this case the function $A\mapsto\int_Af\mathrm{d}x$ is an outer measure and this becomes the continuity theorem for outer measures
 

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