8:00 PM
then plug that $\mu(R)$ into $s$ and $r$---or, more specifically, into the expression $(r/s)^2$

Given sidelengths $a$, $b$, and $c$, is there a simple expression of coordinates of vertices of a triangle with those sidelengths
in any dimension

then differentiate with respect to R and find the R for which it's maximized

I'm guessing no

which I got and this verifies the bound
at which point I noticed that what I got happened to be isosceles
and that let me get it a bit more simply
But uh
this was pretty horrible

Arccos a fifth

8:02 PM
definitely not suitable for a journal problem solution
So while I'm confident enough in the condition for equality, I don't have an actual solution to the inequality
not a respectable one anyways

What's the area of the triangle with vertices $(A,0,0)$, $(0,B,0)$, $(0,0,C)$
…Are all triangles of that form acute?
(Or right, if one of $A,B,C$ happens to be zero)
Conversely, are all acute triangles congruent to something of that form?

Note that the relative vectors between the first vertex and the other two are $(-A,B,0)$ and $(-A,0,C)$
Hence the dot product is $A^2$ which is positive non-negative

Positive dot product

And that means no obtuse angles allowed

hi, i need to find an embedding of $C_{60}$ in $S_{12}$ . anyone can help?

8:06 PM
As for the area, I guess you'd use heron's law?

$C_{60}$ is the cyclic group with 60 elements

Oh but you know what you could do
$(A,0,0)$, $(0,B,0)$, $(0,C,0)$

you've got $(a,b,c) = (\sqrt{B^2+C^2},\sqrt{A^2+C^2},\sqrt{A^2+B^2})$

and uh throw away the third coordinate
for obtuse triangles

in which case you're back to 2d?

8:07 PM
Oh that works for all triangles doesn't it
Eh whatever
I don't want to do out Heron's

me either

What's the distance from $(0,0,0)$ to the plane through those points

@MatheinBoulomenos maybe you have an idea?

well
the plane would be $x/A+y/B+z/C=1$
so the normal vector is $(1/A,1/B,1/C)$
hence the point of closest approach should be $(t/A,t/B,t/C)$ for some $t$
plugging in we get $t=(A^{-2}+B^{-2}+C^{-2})^{-1}$
and therefore the distance is $t\sqrt{A^{-2}+B^{-2}+C^{-2}} = 1/\sqrt{A^{-2}+B^{-2}+C^{-2}}$

And so the distance is $\sqrt t$?

8:11 PM
@user123 $60=3 \cdot 4 \cdot 5$ and $12=3+4+5$

Sniped again
Aight so the volume of the tetrahedron of those four points and the origin is $ABC/6$
which is one-third base times height

slick

Any wave theorists out there? What happens when two first harmonic waves collide?

so we get that the area of the triangle is $\frac12ABC\sqrt{A^{-2}+B^{-2}+C^{-2}}$ I think

Do they cancel eachother out?

8:12 PM
Seems legit

Which is $\frac12\sqrt{B^2C^2+A^2C^2+A^2B^2}$

@user123 so, if you take for example $\sigma=(1,2,3)(4,5,6,7)(8,9,10,11,12) \in S_{12}$, then this has order $60$

The problem is
I forget why I wanted to know that value

yeah, I was wondering

so if you map a generator of $C_{60}$ to $\sigma$, you get an embedding of $C_{60}$ into $S_{12}$

8:14 PM
i only got the last step :P

Oh I think I wanted to find the inradius
'cause area is $sr$

of course it would be an embedding if we found an element of order 60
but how did you come up with that?

@akiva one line of approach to this is to consider equation (4) here: mathworld.wolfram.com/Incenter.html

@user123 if you have a product of disjoint cycles, then the order is the lcm of the cycle lengths

Problem is the semiperimeter is ugly

8:15 PM
right!! ahh i forgot this.

and equate that to the inradius
To get a (admittedly horrible looking) constraint on $a,b,c$

@MatheinBoulomenos thanks a lot!

@user123 np

and then somehow show that $(r/s)^2$ is maximized when $a,b,c$ are as in that triangle

8:17 PM
I can't get myself to care that much tho

@MatheinBoulomenos maybe you could help me with (g) ? i only need the $= \cap Core(H_i)$ equation
the other equation is simple, but this one im not sure
$G\X$ is the set of equivalence classes under $x~y$ iff $y \in O(x)$ where $O(x)$ is the orbit of $x$

if the top three strings are moving down and the bottom three strings are moving up do they cancel each other out?

@AkivaWeinberger yeeeep

@user123 you can reduce it to showing that for $X=G/H$, we have $\cap_{x \in X}\mathrm{stab}_G(x)=\mathrm{Core}(H)$

8:22 PM
One cute thing about that, I guess
Suppose you restrict $a,b,c$ to $[0,1]$ each
For what region of the $(a,b,c)$ cube is the numerator of that expression positive?

btw in (f) the $H_i$ are the stabilizers right ? @MatheinBoulomenos

If you were just looking at that numerator directly, that'd be pretty tough

now just compute that in this case $\mathrm{stab}_G(g)=gHg^{-1}$

but geometrically you know that the distance is well-defined so long as the triangle exists
so the numerator should be positive so long as $a+b>c,b+c>a,c+a>b$
Which I think would be a cone?

@user123 yes, they are stabilizers for a system of representatives for $G \backslash X$

8:24 PM
So the region over which that numerator is positive is actually pretty simple

@MatheinBoulomenos and the union is disjoint because $X$ is a disjoint union of the orbits?

@user123 yes

alright, just wanted to make sure :P

um
I said that
but once I plot it in mathematica I get something substantially weirder...

$\mathrm{stab}_G(g)=gHg^{-1}$ for $X=G/H$ is just a special case of $\mathrm{stab}_G(gx)=g\mathrm{stab}_G(x)g^{-1}$

8:26 PM
namely, I find that the numerator is positive on a bigger region than the cone I stated
so $d$ having a real value is evidently a necessary but not sufficient criterion to have a triangle with those sides :3
(that the area be real, by contrast, is a necessary and sufficient condition)

who is $H$ here ?@MatheinBoulomenos
i mean when you write $X = G/H$

@user123 I just considered the case that $I$ has only has one element and $H_1=:H$, the general case can be deduced from that

Oh.
alright

the main thing to do is to prove how the stabilizer changes for elements in teh same orbit (the formula that I wrote above). And of course the stabilizer for $\overline{1}$ in $G/H$ is $H$

@MatheinBoulomenos cool . thanks. are you familiar with $Q_{2^n}$ ?

8:39 PM
@user123 I know the definition, but not much else

i think for you it is enough :P am i right that this group has only one element of order 2?
$Q_{2^n} = <x,y| x^{2^{n-2}} = y^2 , y^4 =1 , yxy^{-1} =x^{-1}>$
i think im wrong now.

@user123 yes, this is correct, $y^2$ is the unique element of order $2$

but then i dont get the question im solving: we are asked to show that $k=2^n$ is the smallest $k$ s.t. there exists an embedding of $Q_{2^n}$ into $S_k$
the hint is to identify the elements of order 2 in $Q_{2^n}$
in $S_m$ there are a lot of elements of order $2$ for every given $m$ (in general)

8:54 PM
@user123 so let $z$ be the unique element of order $2$ in $Q_{2^n}$. By Cauchy's theorem, every non-trivial subgroup of $Q_{2^n}$ contains $z$. Suppose that we have an embedding of $Q_{2^n}$ into $S_k$ for $k < 2^n$, i.e. a faithful action on a set of $k$ elements for $k < 2^n$
let $\alpha:Q_{2^n} \to S_k$ be that embedding
let $X$ be the $k$-element set on which $S_k$ (and hence $Q_{2^n}$) acts. You know that $\mathrm{ker}(\alpha)= \cap_{x \in X} \mathrm{stab}_G(x)$
do you see how this helps together with the fact that $z$ is contained in every nontrivial subgroup of $Q_{2^n}$?

$z\in ker(\alpha)$
so its in $\cap_x stab_G(x)$

but that means that $\alpha$ is not an embedding

rightt!!!
but i couldn't figure where $k<2^n$ was crucial

I think you don't understand the proof
you show it the other way around: $z$ is in $\mathrm{stab}_G(x)$ for each $x$ (by orbit-stabilizer, since the orbit has less than $2^n$ elements)

me too :P
wait , $ker(\alpha)$ is a subgroup of $Q_{2^n}$
this was my thought

9:00 PM
yeah, okay, but we don't know that it is not trivial at this point

so by what you said it must contain z

Hi @Mathei

no it might still be trivial
Hi @Alessandro

ohh, any non-trivial..

Is the semester going to begin next week at your uni too?

9:02 PM
so orbit-stabilzer says that $|\mathrm{stab}_G(x)| |\mathrm{Orb}_G(x)| =|G|$
@AlessandroCodenotti we still have a bit left

I see

G here is $S_k$ right?

no, $G=Q_{2^n}$

right right we have an homomorphism from $Q_{2^n}$

Now here's the part where $k < 2^n$ is crucial: $|\mathrm{Orb}_G(x)| \leq k < |G|$, since $G$ acts on a $k$ element set, so $|\mathrm{stab}_G(x)|=1$ is impossible
thus $\mathrm{stab}_G(x)$ is a non-trivial subgroup of $G$ and thus it contains $z$
and so $z$ must also be contained in $\cap_{x \in X} \mathrm{stab}_G(x) = \mathrm{ker}(\alpha)$

9:06 PM
very nice ! thanks

Hi all :)

Hi @ÍgjøgnumMeg!

Hi @ÍgjøgnumMeg how was your interview?

Hey :) It went well I think

great!

9:11 PM
There were no mathematicians in the room so I had to give a layperson's explanation of FLT and Iwasawa theory
lol

Sounds challenging

It was fine, they asked me what my research interests were and what kinda of impact they would have and stuff
and I basically had to say
"Some number theorists might be interested"

Well, in 3 weeks' time I will know the outcome