Mathematics - 2017-11-17 (page 2 of 8)

@Daminark "free" action vs "faithful" action differ merely on the quantifiers

@io_cantor then such an element has to do nothing right?

just consider what Jacksoja suggests

@Leaky Sure, I'm not talking on the level of being technically imprecise, but mentally we can all fill in the quantifiers and infer the only thing it'd be reasonable for him to ask

@Daminark I don't know which one he is asking about

He's trying to find a geometric example of a non-faithful action

I kind of know matrix groups?

4:03 AM

@Daminark I thought he's asking for non-free action

Okay so think of the action of GL_2(R) on the lines

Okay, lines in R^2?

set of all lines?

Through the origin

got it

identity matrix is obviosly in the stabalizer

@io_cantor can you find all of the elements in the stabilizer?

4:05 AM

and the 180 degree rotation matrix

there are more

ah

all matrices of the form

First do prove that this is an action, good linear algebra execute, but yeah do find the kernel of this action

where you have just stretching or squishing

no rotation

@io_cantor how to express this in the language of linear algebra?

4:06 AM

I don't know how to write a matrix in LaTex lol

you don't need to

and that's a hint

det(A) = product of its eigenvalues

yes?

product of its diagonal entries *

no that isn't right

4:08 AM

My eigenvalue stuff is weak

42 secs ago, by io_cantor

det(A) = product of its eigenvalues

ah, you're correct

this works for any matrix lol

$\[\begin{bmatrix}
a&0\\
0&b\\
\end{bmatrix}\]$

if you use '[', don't use $ as well. I'm not sure if '[' works here.

4:10 AM

@TedShifrin thanks

@io_cantor that isn't right

ugh, that didn't work, either. :(

Hmm

Could you give me a hint as to why it's incorrect?

Characteristic polynomial is $\det(A-tI) = (-1)^n t^n \pm \dots +\det A = (-1)^n(t-\lambda_1)\dots (t-\lambda_n)$.

this matrix just stretches or squishs R^2

so it will give u the same line

for all lines

so it will be in the stabalizer

4:12 AM

No, lines get changed, @io_cantor, unless $a=b$.

But any linear map does map lines through the origin to lines through the origin.

Argh, you're totally correct

That happens sometimes :P

It needs to be the same scalar

value*

Okay, so now I have non-identity matrices in the stabalizer

That was helpful, thanks.

Is there any trivial geometric shape for which this can happen?

This is why there's $PGL(2,\Bbb R)$ to act freely on the space of lines through the origin.

I don't know what that is

What does the P mean?

4:16 AM

projective

$PGL(n) = GL(n)/\{\lambda I: \lambda\ne 0\}$.

yup, projective

ohhh

just as projective space is what you get when you mod out by non-zero scalar multiples.

very interesting!

DogAteMy (again!).

Akiva Weinberger

4:17 AM

Hi

Well, that was a quick visit.

Conjecture: There exists a bijection between the states of a geometric object and the set of actions on that object.
where the actions have some restrictions
they must turn one state into another
So, I cannot find a non-trivial stabalizer.

Right: Look at the orbit/stabilizer theorem.

Argh I keep timing out

Ah you guys spoiled the exercise. But yeah the linear transformations that preserve lines are stretching/squeezing, meaning scalar multiples of the identity

4:23 AM

So while @Jacksoja gave me a permutation that was in the stabalizer, that was not a valid action that turned one state of the square into another
like switching 2 and 3

that was nonsense, Demonark

I understand now

This has been such an enlightening evening

Sarcasm duly noted :D

I can actually explain to people when they ask me what group theory was about, rather then give them the axioms and say this defines a group

group theory is about*

rather than*

it's all about symmetries of objects :)

4:25 AM

yes!

@Ted projective geo is pretty dope

and a group "should" be thought of as a set of actions where we have a way of composing those actions

That's why I wrote a chapter about it in my algebra book, Eric :P

but projective diff geo is a whole other kettle of fish :)

one place where things get particularly interesting is when you have an object with a continuous symmetry

like a circle?

4:26 AM

someone asked me a basic projective geo problem and i hadnt thought about that stuff in ages

yep

but fun stuff nonetheless

I would imagine the group of symmetries of a circle would be isomorphic to Z with a direct product of something

but that doesn't seem right

@Ted wait the linear transformations that preserve lines are the scalar multiples of the identity, no?

R/Z

4:27 AM

multiplicative?

yes, Demonark

well. R/(2pi*Z)

but same difference

That's why you have the projective group :)

$(R/Z)^{x}$?

additive

4:28 AM

So why was it nonsense?

But what's the group of symmetries of the circle, @io_cantor?

Well, you all the rotations, which is isomorphic to Z

Um, no.

that'd be multiples of 360 degree rotations

No, you're right

4:29 AM

that leaves out quite a few of them :)

There's uncountably many rotations

you have a bijection between points on the circle and R

so the rotations are isomorphic to R

well

Nope.

what's the difference between a 360 degree rotation and a 720 degree rotation?

huh

nothing at all

4:30 AM

right

ah!

you want me to make R cyclic

If you're thinking of $e^{it}$, yes :P

I need to mod by something

Hmm, that's just the unit circle though?

what are you trying to do

Yes, the rotation group in two dimensions is just the unit circle!

4:33 AM

oh so that's what you were doing lol

Find the group that the rotations of a circle is isomorphic to

equivalently, you can think of it it in terms of the angles of rotation, with angles theta1,theta2 equivalent if their difference is a multiple of 2pi

so R / (2pi*Z)

:41200883

@TedShifrin so it's just itself?

the two perspectives are related through $t\mapsto e^{i t}$

ah, R/(2pi*Z) makes more sense to me.

4:34 AM

Yup, @io_cantor ... I mean, I'd call it the group $SO(2)$ ($2\times 2$ rotation matrices), but it's isomorphic to the circle as a group.

even though they're exactly the same thing haha

"exactly the same thing" = "isomorphic"

tfw you write out $SO(2) \cong U(1) \cong S^{1} \cong \mathbf{R}/\mathbf{Z}$ every time

wow the connection between modding out and geometry is 'finally' clicking. Even though we're past learning about sylow p-groups lol

the way i'd put it is that we can indicate a configuration of the circle by marking a point on it

and then the set of possible configurations of the circle is just parametrized by the circle itself

4:36 AM

@io_cantor nice

Eric, wtf is tfw?

that feel when

That feeling when

the feels when

wow

lmao

4:36 AM

3 different answers

rolls 3 3/4 eyes

All 3 of us said different things

Tfw sniped

tfw everyone explains internet speak faster than you

Hey, I'm old.

tfw the meanings are the same

4:37 AM

mfw someone doesnt know tfw

my face when

rolls 4 4/5 eyes

there's some ones which I have a genuinely hard time remembering

:0

afaik is one, though now I remember it better

as far as i know

4:38 AM

toik

what if it converges to an irrational number of eyes

that would be weeeeeeeird

toik?

ha ha: that one I know

lol

tfw you don't know toik

tfw udk toik

4:39 AM

tfw idk toik

precisely!

"the way i'd put it is that we can indicate a configuration of the circle by marking a point on it
and then the set of possible configurations of the circle is just parametrized by the circle itself"

suppose you start with the unit circle with a marked point at (1,0)

and then rotate it through an angle theta

then the marked point will just be at (cos theta,sin theta)

ah, got it

so each point on the circle is in bijection with a possible rotation of the circle

4:41 AM

why doesn't this work for, say a rectangle? why cant it be parametrized by itself?

What are the symmetries of a rectangle?

not a continuous symmetry

just 4

right.

oh, i see now

4:41 AM

@io_cantor: Google "homogeneous space."

A homogeneous space M is a space with a transitive group action by a Lie group

OH, so a circle has one orbit

There you go.

it gets trickier from here on out, as one might imagine

but what is a lie group..

It's a manifold? wtf

It's a group that happens to be a manifold.

4:43 AM

well, a circle is a manifold too :)

a group that tries its best to trick you

including with how you pronounce Lie @EricSilva

haha leeeeeeeee

Manifolds are generalizations of curves and surfaces — they're nice spaces that locally look like Euclidean space, @io_cantor.

@Semi EXACTLY

4:44 AM

to get a feeling for how this'll work

Right, I'm good with basic stuff about manifolds. hausdorff, second countable, locally homeomorphic to R^n

not that i've taken any topology lol

now i'm just trying to understand how groups can be manifolds, by using the circle example

suppose you look at a torus, in the sense of a unit square with edges identified appropriately

Well, a circle is definitely a $1$-dimensional manifold.

you ask that the group operation is a smooth map too

otherwise they're just two structures floating around that have nothing to do with each other

the two relevant operations will then be to shift the square by a certain distance up/down or left/right

4:45 AM

Yeah, group multiplication and inverse both have to be "smooth."

Err I'm having a bit trouble following three different people. I really appreciate the help though! Which comment should I be looking at?

not me, probably :P

Ted

LOL, nah.

omg

everyone is too humble

i'll pick the first.

what do you mean by a unit square and a torus?

4:47 AM

A: not A
B: C
C: not C

how are they related?

Oh, you're right haha, shoulda picked B

too late now

hey that's the klien 4 group

symmetries of a rectangle

um no

whait

4:48 AM

if you imagine stretching that picture like a piece of taffy, you could join up A and A

that'd be a cylinder

you can then wrap that around to join B and B

that'll be a torus

sorry ive just been looking at cayley graphs and that looks like the caley graph for the klien 4 grouop

where A and B are the generators

well, that's probably true as well :)

$\mathbf{R}^{2}/\mathbf{Z}^{2}$ is what he's drawing

Hmm im trying to visualize it

I get the cylinder

Oh

right

got it

kk

i don't really want this to be the torus embedded in R^3, though

if it's that, then it makes sense to rotate around the axis of the torus

4:52 AM

We just think of it as a manifold, not in any ambient space?

but both 'rotations' should be on the same footing

yeah

which is to say, (R/Z)^2

Sorry, I don't see any rotations

I mean,

I can rotate it in one way, like a circle

but what is the other way?

which torus are we talking about? in R^3?

in R^3 the only obvious rotation is the one around the axis of the torus

but the thing is, I could just have easily have chosen to join up B to B first, and then A to A

why does it make a difference? though lets consider it not embedded in anytin

anything

kk

4:54 AM

right

suppose i've got (R/Z)^2 as my manifold

Yeah

and then I take (x,y)-> (x+a,y+b) for all points

right

have I actually changed anything?

4:55 AM

nope

right.

that's one rotation, sure

oh

(x+a,y) and (x, y+b) are the two

yep

translate in the x direction or translate in the y direction

So that's another homogeneous space :)

with (x+1,y)=(x,y)=(x,y+1)

4:56 AM

visually that's the horizontal cross section circle and verticle cross section circle

if that makes sense

right

so the space of possible operations on the torus is just (R/Z)^2 again, i.e. the torus itself

i cant really think of any spaces i care about that aren't homogenous spaces

Oh, I can.

and it's not terribly hard to convince yourself that this would work for any (R/Z)^n

Yeah... I get it, but I'm not comfortable with it yet. I need som epractice

So going back to the circle, we said the group of rotations was isomoprhic to R/Z

4:58 AM

@Ted like what

yep

and i'm arguing that the same is true for the torus, i.e. the torus parametrizes all possible symmetry operations on the torus

Like hypersurfaces of degree $\ne 1,2$ in $\Bbb R^n$ or $\Bbb CP^n$ ... or most surfaces :P

oh i wasnt thinking about things living in other things

Or the twisted cubic, even. :)

i totally care about those

4:59 AM

LOL

Oh it makes more sense now!

this has been extremely cool

and i forgot about CP^n

now, the one where things actually start getting tricky

Well, $\Bbb CP^n$ is certainly a symmetric space :P