Mathematics - 2017-11-17 (page 3 of 8)

H acting on G/K

You mean the identity coset in $G/K$?

Yes by orbit stab theorem

G acting on G/K

stablizer of K , is K

5:06 AM

OK

does anyone else think the $/$ in $G/K$ is placed weirdly

I'd say $eK$ for clarity.

oh sorry about that

Now we restrict the action of G to H

OK, so restrict the action to $H$.

H acts on G/K

5:07 AM

"The group of all orientation-preserving symmetries is isomorphic to R/Z
as in, if I marked a sequence of points on my circle, and then reflect it across an axis
then the order of the points would be reversed"

But these are just the reflections right?

the stablizer of K in this case is

You're saying the reflection group and the rotation group are both isomorphic to R/Z?

oof

hK= K , wich means that elements that are both in H and K, ie H intersect K

no, I was addressing two things in sequence there

5:08 AM

OK

$\{(z,w) \in \Bbb C^2 | z=w\}$ is a plane by dimensionality argument, but I fail to visualize it (except by projecting it onto one of the complex "axes")

H/ H int K is in bijective correspandance with the orbit of eK

so it make sense that G acting on G/K

will have potentially more out than H acting

If I consider a reflection, it'll flip the orientation of my unit circle

@Semiclassical right.

so if I want my symmetry operation to preserve the orientation of the circle, then I'd better omit reflections

5:09 AM

i mean for one , there is no reason that H would act transtivly on G/K

in that case I'm back to rotations being the only allowed operations

But G does act transitvly

Oh, got it

@Leaky: It's a complex line. Think with vector spaces, not real numbers.

But there's no a priori to preserve orientation is there?

5:09 AM

@TedShifrin was that good way to prove that?

a priori reason

@TedShifrin sure, I know what it is, I can deal with it algebraically, but I want to see it geometrically

i.e. "with my eyes"

well, there is if we want our set of group operations to form a connected manifold

@Leaky: Most of us who do complex geometry usually draw "real" pictures for complex things.

if we include reflections, then we'll have two kinds of symmetry operations: ones which preserve orientation, and ones which don't

5:10 AM

I don't draw CP1 as RP1

and there's no way to smoothly go between those two

@Kasmir: You can say it more explicitly. The number of elements in the $H$-orbit is less than or equal to the number of elements in the $G$-orbit, since $H\subset G$.

The blind can see. There's a fascinating NPR podcast on a blind man who uses clicks to acutally "see". The visual part of his brain actually lights up when he clicks and uses his echolocation

If I'm drawing $\Bbb CP^1$ inside $\Bbb CP^2$, I draw it as a line.

If I'm drawing the Riemann sphere doing complex analysis, I draw a sphere.

@TedShifrin oh thanks ! :D

5:12 AM

1 hour ago, by Leaky Nun

Let $f$ be the diffeomorphism between the torus $\Bbb T^2 := \Bbb S^1 \times \Bbb S^1$ to itself by swapping the two coordinates. $f$ is not homeotopic to the identity map in the ambient space $\Bbb R^3$. Is there an ambient space $\Bbb R^n$ such that $f$ is null-homotopic?

@TedShifrin any idea?

BTW Ted

@io_cantor reminds me of this video game, though I have no idea how realistic it is: youtube.com/watch?v=V6qR8SCUEME

@Leaky: No. Ambient space is totally irrelephant.

If i take topolgy and real analysis in the same time

would that be good?

I don't know your courses, @Kasmir, but I don't recommend it.

5:13 AM

@KasmirKhaan i did but it was lame

I would do real analysis first, and then topology.

Okay, i just want to get in idea of what topology is

@EricSilva: You UC guys do not count.

@Semiclassical that makes sense what you said about connected manifolds

and the teacher of this semester is really good

5:13 AM

@TedShifrin interesting...

i want to be in class for that

tru enuff @Ted

maybe next year diff prof

I thought with enough dimensions you can have homotopy

but, how do we get an isomorphism to O(2)?

5:14 AM

@Kasmir: Talk to the prof about whether he thinks you can do it.

if you treat the two tori as "loops"

@TedShifrin you are so wise TED :D

ill do that, kasmir will go back to work now =p

@Kasmir: Close to 40 years of advising/teaching makes me "wise."

@io_cantor well, every orthogonal matrix either has determinant 1 or determinant -1

Happy work.

5:15 AM

and there's an easy bijection between those two sets

haha am sure you were wise Before =p and thanks :)

$SL_n(R) \neq O(3)$

whoops

@Leaky: The point is that that map is orientation-reversing, and therefore cannot be homotopic to something orientation-preserving. Nothing to do with where the torus sits.

so it suffices to look at those with determinant 1, i.e. SO(2)

$SL_2(R) \neq O(2)$

5:16 AM

$SL_2$ is very non-compact, for starters.

@TedShifrin so other maps that aren't naturally homotopic might be homotopic in another space, if they are orientation-preserving?

It also is way too big.

so we want to argue that SO(2) is isomorphic to R/Z

No, @Leaky. Ambient space is not relevant. Your homotopy has to stay inside whatever space you're talking about.

@TedShifrin ok thanks

5:17 AM

Unless you specify that you allow the deformation to occur in the big space. But that's not a homotopy of a map $f\colon X\to X$.

Oh, so would the group of symmetries of the circle be isomorphic to $(\mathbb{R}/mathbb{Z})^2$?

I do allow it to occur in the ambient space

how do you call that?

no

Hm. We have that the rotations are isomoprhic to R/Z and the reflectiosn are isomorphic to SO(2) which is isomorphic (we said) to R/Z

if we're only looking at orientation-preserving operations, it's R/Z

noooooo

5:18 AM

oops okay what

I guess ambient homotopy or isotopy, @Leaky.

how about this

I said that the set of 2-by-2 orthogonal matrices can be split into those with det=1 and det=-1, and there's a bijection between these

@TedShifrin I see

I take some time to think about it and get back later when I've mused over it?

I really appreciated this though. Very eye-opening

5:19 AM

so we can describe the set of such matrices as O(2) ~= SO(2) x Z/2

@Leaky: An example of that is the sphere eversion theorem of Smale.

Much more interesting than rudin lmao

in the sense of Z2 isomorphic to {1,-1} under multiplication

What is Z2?

Z/2

5:20 AM

@TedShifrin ah, right

ah

SO(2) is what'll end up being ~= R/Z

(lazy notation for isomorphic)

okay I'll think this through. I really thank you

One last question

main point is that O(2) will be a manifold, but it'll consist of two disconnected pieces

I learned in class that GLn(R)/SLn(R) ~= R^x

5:22 AM

Rigt.

What's the geometric interpretation of this?

And why R^x? why not the additive group?

I don't have a great intuition for GLn(R) geometrically

Yeah, the orientation preserving and the orientation flipping. okay

but wouldn't it just be the fact that any matrix can be written as A = A/det(A)*det(A) ?

A/det(A) is manifestly det=1

You're scaling a matrix with determinant $1$ to make its determinant be whatever you want.

No, @Semiclassic.

You need an $n$th root.

5:23 AM

oh, right

Oh! So it's all scalar multiples of a matrix!

yeah, I forget that det(aI)=a^n

Back with a different question: math.stackexchange.com/questions/2524217/…

and that's obviosly isomorphic to R^x

I don't know how to find the expectation of this mixed distribution.

5:24 AM

well

wait,

You have to be careful

it is as long as the matrices are nonsingular :)

yes, that's true

What if the determinant is negative and $n$ is even?

5:24 AM

oKay, so let's take n = 2

We're not trying to map backwards. We're just trying to map forwards.

hrm, yea

$SL$ is the kernel of the determinant homomorphism.

But you can't quite do it as I glibly said by scaling, unless $n$ is odd.

right.

i can believe that it's just details, though

Well, it's deeper than that. It's a question of when that short exact sequence splits and you get a semi-direct product structure of $GL$.

5:26 AM

ugh

once i start seeing the phrase "semi-direct product" I know I'm out

SL is normal because it's the kernel of a homomorphism. And so GL/SL is a group. It is isomorphic to R^x because we can consider any matrix as an elemnt in R^4 and R is isomoprhic to R4 when looking in terms of sets. But when we get rid of all multiples of the tuples, we get back R? I'm trying to intuit it lol

We've had to do a lot of semi direct stuff.. sadly i skipped all of those questions on the homework

Huh?

Ignore that. That was just be rambling.

me*

Anyway, I'm out. Thanks for all of the help! Good night!

Night.

or good morning, wherever you are

5:29 AM

wait, SL is normal??

definitely night

yeah!

it must be

of course it is.

Trying to come up with a title for that question so that people don't think it's too hard and ignore it.

how do you see that a villarceau circle is a circle?

5:30 AM

Oh, that's far from obvious.

Unless you think up in $S^3$. But I think of it in terms of slicing the torus in $\Bbb R^3$.

That's a challenge problem in my multivariable math book :P

I was gonna say: if it acts like a circle and it quacks like a circle...

@TedShifrin I am indeed thinking in S^3

all of those problems are arisen from my exploration of the Hopf map

Oh, well, then it's the fiber of the Hopf map, so of course it's a circle.

exactly

It's the intersection of $S^3$ with a two-plane through the origin, so it's a round circle.

Much more interesting to look at the torus problem in $\Bbb R^3$.

5:32 AM

what is the torus problem?

I had to come back! I thought of something!

Show that if you slice a torus with a bitangent plane you get two interlocking circles.

@io_cantor aka "eureka!"

What's the connection between the symmetric group and the group of symmetries for an object?

@LeakyNun nah i dont think so

symmetric group is on a finite number of discrete objects

5:33 AM

@io_cantor: The symmetric group is the group of symmetries of an abstract set with $n$ elements.

Not as interesting as you thought.

and permutations don't have preserve any kind of structure

Well, if we look at a rectangle. The symmetric group of {1,2,3,4} is much bigger than its symmetry group

sure.

Right.

Same thing with a regular $n$-gon.

Oh, permutations dont preserve structure

But automorphisms do!

OH!

5:34 AM

right.

The symmetries of a (regular) $n$-gon embed as a subgroup of $S_n$.

So the automoprhims of an object is precisly the symmetry group!

@TedShifrin aka $\Bbb D_n \hookrightarrow \Bbb S^n$

Is that correct?

What, @Leaky?!!

5:35 AM

50 secs ago, by Ted Shifrin

The symmetries of a (regular) $n$-gon embed as a subgroup of $S_n$.

am i right?

dihedral group

You typed $S^n$ rather than $S_n$.

oops, $\Bbb S_n$

pebkac

for an irregular $n$-gon, it can be a tiny symmetry group, of course.

oh

5:36 AM

@Semiclassical is that a counterexample?

the orientation-preserving symmetries of an n-gon, though, are just generated by the permutation (123...n)

which is analogous to the orientatoin-preserving symmetries of a circle being generated by infinitesimal rotations

So the symmetric group is the group of symmetries of an object with the most minimal amount of structure

@io_cantor is what a counterexample

dihedreal group

oh

nah

i was just saying that that's what Dn was

5:38 AM

i asked if the automorphisms are the symmetry group

ah

@TedShifrin how would you choose the representatives in $\Bbb S^3$ under group action by $\Bbb S^1$?

I'm thinking $\Re(z)=0$

I wouldn't. The Hopf bundle has no global section.

scratch that

You can do it on a very large open set, but not everywhere.

I can do it via decomposition as two solid tori

I believe

5:40 AM

Not in a continuous way you can't.

2 questions:

1) The symmetric group is the group of symmetries of an abstract object with the minimal amount of structure

well it's continuous in each torus

2) The symmetry group of an object is the automorphism group

No, I think it degenerates.

they're one vertical disc in each torus

5:40 AM

Those do not glue, @Leaky.

I'd say the symmetric group is the symmetry group of a set of n distinct objects.

@io_cantor: It's sort of a vague circular discussion.

It depends what structure you endow your set/space with.

I'm considering concrete examples like the symmetries of a rectangle

then looking at S_4 and finding the automorphims of S_4

@TedShifrin I think they do. The boundary becomes a horizontal circle in the other torus, and if you trace the horizontal annulus, half of it would be identified with the other half, which gives you a mobius loop, but the central circle is also identified to one point, which gives you back another vertical disk

and seeing if they match up with the klien 4 group

@Semiclassical hm okay

5:43 AM

Vertical in one torus should become horizontal in the other, @Leaky. But maybe I don't know what you're talking about. But I guarantee you that there's no global continuous way to do it.

@TedShifrin I acknowledged that

2 mins ago, by Leaky Nun

well it's continuous in each torus

Well, you seemed to argue with me when you said "I think they do."

I thought you were trying to argue these two things patch together continuously to give a global answer.

oh, we interpreted "glue" in different ways

I know the two solid tori glue.

I interpreted it as "glue as long as you consider two points in the same orbit as one point"

whereas you interpreted it as "glue in the larger space before the quotient"

5:45 AM

Aha! Wikipedia has answered it for me:
"In mathematics, an automorphism is an isomorphism from a mathematical object to itself. It is, in some sense, a symmetry of the object, and a way of mapping the object to itself while preserving all of its structure. The set of all automorphisms of an object forms a group, called the automorphism group. It is, loosely speaking, the symmetry group of the object."

Alright, now actually good night

@io_cantor: Moral of the story — you don't need us when you have Wiki.

Good night :P

haha, hardly true

@Semiclassic @EricSilva: Agh — someone else is saying "vectorization." :(

that is a phrase, for better or worse, isn't it

no, it's a single word :P

5:47 AM

@TedShifrin what I find most amazing is that you can decompose $\Bbb S^3$ in two different ways such that the "equator" is $\Bbb S^2$ in one way and $\Bbb T^2$ in the other way. hardly any analogue exists for $\Bbb S^2$

roll

so $\Bbb S^2$ and $\Bbb T^2$ in some sense are quite similar

(as long as you take it to the fourth dimension)

@Leaky: There are lots of things that work differently in different dimensions. That's not surprising. In even dimension, you have nonzero Euler characteristic which stops you from foliating the sphere. In odd dimension, Euler characteristic of the sphere is zero and you have foliations ... in this case, by the Hopf circles.

interesting

wait, $\Bbb S^2$ and $\Bbb T^2$ have different genus though...

right, $\Bbb T^2$ can be foliated by circles. You were talking spheres.

But Euler characteristic is the relevant thing.

Euler characteristic of the torus is likewise 0, as it is for $S^3$.

5:49 AM

I wasn't replying to the foliation

hell, $\Bbb T^2$ can be foliated by trivial circles or villarceau circles

have fun doing the latter

villakdsfjaksjceau

Actually, the Villarceau circles overlap in pairs.

I'm not sure you get a foliation.

that's weeeeeeiiird

I'm only taking one of them

and rotating it around

it's the preimage of a circle under the Hopf map

@TedShifrin there are two families of villarceau circles, each family gives a foliation

OK, I yield. I hadn't thought about that recently.

I just assign the homework question :P

5:52 AM

@TedShifrin I was exploring the Hopf map for the whole day

quite literally

although strictly speaking, the phrase "villarceau circles" means a pair of them, that do intersect (one from each aforementioned family)

@Leaky: You should definitely look at Berger's two-volume book Geometry. All sorts of beautiful and deep stuff in there. And Pedoe's book, too.

Right, @anon.

hell, you can foliate it by trefoil knots

?

Where did that come from?

5:53 AM

Pedoe is good

trefoil knots are just a villarceau but faster right

Huh?

I imagine villarceau as going around the horizontal coordinate as fast as the vertical coordinate

for one thing, trefoil knots are ... knots. ever tried to embed a knot into 2d?

handsome folks, hi again, I need help understanding the deeper meaning of permutation representation

5:53 AM

Oh, you're talking about (2,3) curves on the tori.

trefoil knot is rotating three times as fast right

ah

You seem to be busy ill come back later ._.

> deeper meaning

does everything have deeper meaning?

I feel that I did not get the idea properly

5:54 AM

I don't know a deeper meaning.

from what i know now

I only know a shallower meaning.

yeah, just using nice representatives of an element from pi_1(T^2) and then applying an S^1 to get its orbit

But anon knows all :)

If we have G acting of a set X

then there is a homomorphism from G to S_X

5:55 AM

@anon it's literally $\Bbb S^1 \times \Bbb S^1$

villarceau is travelling along the second coordinate as fast as the first

permutation Group of order of X elements

trefoil is three times as fast

what is comfusing to me , or what i dont see clearly here

so in a sense villarceau is $ab$ and trefoil is $a^3b$ in your $\pi_1(\Bbb T^2)$ notation

the Wording of a homomorpshim here, i mean i know what it is ofc

5:56 AM

we have $\pi_1(\Bbb T^2) = \pi_1(\Bbb S^1) \times \pi_1(\Bbb S^1)$ right

but i seem to fail getting it in this context

@KasmirKhaan we've already talked about this

I know anon and you explained that part to me

@KasmirKhaan think of $g$ as an operation on $X$

g(hx)=(gh)x means (applying g)o(applying h)=(applying g*h), so the group operation * turns into permutation composition o

5:57 AM

i.e. it transforms elements of $X$

just like $\rho$ is a transformation of the square

Yes but when using that idea

it maps the vertices to its neighbour

so it is a permutation

like in proving A_5 is simple

I got lost

@KasmirKhaan what is the set in question?

anyone have any suggestions for what TeX command i should use for a cup product?

A: Who uses the "\smile" symbol, and what for?

5:59 AM

7

It is very common in Mathematics, more specific in Algebraic Topology. It is used to denote the cup product.

If H is a subgroup of G of index n, then G acts on G/H , the permutation represntation f: G--> S_G/H iso to S_n

cool

G is simple ==> ker f = {e}

elements of $G$ definitely permutes the cosets of $H$