« first day (2301 days earlier)   

12:09 AM
was my question in the question pool answered already?
the "what circumstances led you here?" question
 
similar questions were answered throughout the transcript, yes.
 
thanks ill start from the top then
 
@StevenSagona Daniel has bookmarked the AMA as a conversation here in case you don't want to scroll up to get to the top
 
@ACuriousMind thanks but the deed is done
@heather I'm surprised you aren't expressing any boredom for your classes. I was in a similar situation as you when I was your age (but before the internet was as good of a tool for learning) and my math classes drove me insane. The only thing getting ahead really ended up doing for me was giving me an ego that'd I'd annoy the students around me.
 
12:25 AM
@StevenSagona Drove you insane because the maths classes were too easy for you ? Or am I misinterpreting ?
 
@anonymous yeah but mostly because I already learned the material so it was pointless
things are really simple at the beginning - they just make you do algebra over and over again
 
@ACuriousMind You here? I need to bounce ideas off of someone
 
@StevenSagona True. That would be definitely boring but I am not sure it would drive one insane :). For me it would be a good time to sleep in class without getting noticed :P. I've faced such situations before :D
However, they sort of help in revising the concepts....
 
@DanielS: About that drink. In the first place, I must warn you that I made it for three people and it was mostly guess-work. I mixed the following ingredients in a bowl using an electric mixer: 3 cups of milk, 1 spoon of instant coffee powder, 5 tablespoons of cocoa powder, 3 "pastries" 5 spoons of sugar and 10 ice cubes. The mixture became quite frothy and warm so I kept it in the freezer to chill for about 10 minutes before gulping it down.
Pastry: Choco pie--
> A choco pie is a snack cake consisting of two small round layers of cake with marshmallow filling with chocolate covering. (@JohnR: See here <---)
 
Who mixes choco pie in a drink ? :P
The marshmallow would affect the flavour
 
12:32 AM
Me, that's who! :-) It was delicious.
 
LOL :D I will try doing that. Choco-pie+Chocolates+Wafers+Pastries+Fruit Cakes + Coffee + Hot Chocolate... in mixer and make a milk shake out of it :'D
 
that sounds healthy
 
obe
yum
 
I think cola and wine might be good addition to your ingredients :') @Kaumudi.H
I've tried coffee with hot chocolate before
But your recipe is a class apart :P
 
12:56 AM
@Kaumudi.H sounds good
 
@ACuriousMind Suppose I have some group action on a pointed space. Can I decompose the group as $F\rtimes T$, where $T$ acts transitively and $F$ leaves the basepoint fixed?
Er, I mean, where $T$ is any subgroup that acts transitively.
 
@0celo7 What about the standard $\mathbb{Z}_2$ action on the sphere by the antipodal map?
 
@ACuriousMind I want the action of $G$ to be transitive, sorry
 
@0celo7 Then consider any transitively acting $G$ on the sphere and the action of $G\times\mathbb{Z}_2$.
Well, I guess your question trivially has the answer "yes" for $F$ the trivial group and $T$ the original group...
 
$T$ is given
I have some transitive action $T$
$T\subset G$
 
1:03 AM
Then you can't, by my example. $G\times\mathbb{Z}_2$ acts transitively but you can't decompose it into a product with $G$ such that the second factor leaves any point fixed.
 
What is the $G\times\Bbb Z_2$ action?
flip, then apply $G$?
 
Hm, you're right, two separate group actions don't give an action of their product
 
@ACuriousMind The context is that I'm trying to compute isometry groups (rather, follow Wolf's proofs) of Minkowski spaces and the model constant curvature spaces
He writes down $I(\Bbb R_s^n)=L\cdot V^n$, where $V^n$ is the translational group and $L=\{f\in I(\Bbb R_s^n):f(0)=0\}$
he uses $\cdot$ for the semidirect product
So I'm wondering how he got that
A priori, why should the isometry group be of that form?
 
Ah, here's a silly counterexample: Consider the circle, and the action of $\mathrm{U}(1)$ on it, which is clearly transitive. You can also define a transitive $\mathrm{U}(1)\times\mathrm{U}(1)$-action just by composing the two rotations. As everything is commutative, there's no problem here, but clearly you cannot decompose the $\mathrm{U}(1)^2$ in your fashion w.r.t. to the $\mathrm{U}(1)$ because nothing but the identity leaves any point fixed.
@0celo7 Given any isometry $f$, you can compose it with a translation to achieve $f(0) = 0$.
Nothing to do with general group actions, everything to do with your specific situation :P
 
what?
 
1:14 AM
For any $f\in I$, we have that $g(0) = 0$ for $g = t_f \circ f$ where $t_f$ is the translation that shifts $f(0)$ to $0$.
 
Hmm. I guess the semidirect product is needed for if you decompose $f\circ f'$ in that way?
 
Now apply the inverse translation to get $f = t_f^{-1}\circ g$ and therefore the claimed decomposition.
@0celo7 Yep, since the second isometry will act on the translation coming from decomposing the first.
 
Right. So for the dS/AdS type spaces I have a transitive action of $O^s(n)$ by isometries. Pick some basepoint $x$. So given an isometry $f$, I find a $t\in O^s(n)$ that moves $f(x)$ to $x$. Now, the group leaving $x$ invariant is just $O^k(n-1)$, where $k=s$ or $s-1$. But that group is the "maximal group preserving the inner product on $T_x\Sigma$" and I don't know what the significance of that is
I think that means that if $l\in I(\Sigma)$ leaves $x$ invariant, then $dl\in O^k(n-1)$
So if $L$ is the stabilizer of $x$, then...hmm
$L\subset O^k(n-1)$
@ACuriousMind why is algebra so hard
Ah! An isometry is determined by its value and first derivative. But the value is fixed since it's a stablizer
So we're looking at $L\to O^k(n-1)$ and that's injective
So it's a subset in a natural way
Then we get $I(\Sigma)\cong O^k(n-1)\rtimes O^s(n)$
but the action is just by multiplication, so we can absorb that factor
 
 
1 hour later…
2:40 AM
@Danu Getting into the good stuff. If $M^{2n}$ is complete and connected with constant sectional curvature $K>0$, then $M$ is isometric to $S^{2n}$ with radius $1/\sqrt K$ or to $\Bbb RP^{2n}$ with the same radius.
@ACuriousMind hola
@Secret hi
 
Hi, I missed the AMA cause Japan time is 8:00, where we are still asleep. Glad to see it went well
 
@SirCumference hello
 
2:59 AM
Howdy
@0celo7 Quick question, this is complete nonsense, right?
 
i think so
Why?
 
@0celo7 It relates back to that paper I'm reading, about pseudo profound bullshit.
 
[Attempt at localising unicorns] Suppose there exists something call antiferromagnetic force and suppose it act by enforcing antiferromagnetic order to whatever is being interacted. Now suppose there exists something called iohydrogen. We might expect that it is a state of hydrogen that tend to antialign any hydrogen it encounters in the way. Now suppose negatively curved means concave. There are no known electrostatic configuration that can induce the reversal of electric charge so that
,hydrogen becomes antihydrogen and vise versa. Therefore we concluded iohydrogen does not exist (as charge reversal of hydrogen just turn it into antihydrogen, and it does not preferentially antialign spins of surrounding (anti)hydrogens
 
3:15 AM
@SirCumference I bet I could come up with some pseudo profound bullshit
 

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