If I take the point set $\{ w: |w|=1 \}$ and map it under $f^{-1}$ then I get the imaginary axis back?

sure

Okay good

imagine this as the prototypical example: on the real axis, cycle (-1 0 1 inf), fix i and -i, and bend the imaginary axis to the unit circle and the unit circle to the imaginary axis. (draw this for aid.) this prototypical example can be postcomposed with a rotation around 0, and that rotates the axes around.

What do you mean "cycle"?

permutations

-1 scoots right to 0, 0 scoots to 1, 1 scoots to inf, and inf scoots to -1

(simultaneously)

$(-1,0,1,\infty) \to (\infty, -1,0,1)$

uh, that's scooting everything left, I said scoot right

>.<

meaning the points themselves on the real axis

That's scooting to the right inside the parenthesis

I'm talking about the points. on the real axis.

you're the one who decided to write them in a list

so if 0 is scooched to the right, it's sent to 1 not -1

How can you say inf is scooted to -1?

Is it like in snake, it goes through the wall on one side and come back at the opposite?

visualize it. -1 scoots right to 0. 0 scoots right to 1. 1 scoots right to inf. then inf has to wrap around from the LHS.

yes.

Would $f(i) = -i$?

no

i and -i are fixed points in the prototypical example I'm talking about

they're treated like antipodal points on a sphere, and the real axis is like the equator

the left side of the unit circle buckles inwards for -1 to go to 0, while the right side of the unit sphere explodes outwards for 1 to go to inf

meanwhile, the imaginary axis inside of the circle bends to the right to become the right half of the unit circle, while the outside parts of the imaginary axis fold inwards from the left to become the left half of the unit circle

Is the left side $e^{i \theta}, \pi /2 \leq \theta \leq 3 \pi /2$?

yes

buckles inwards to form a line?

or what do you mean buckles inwards

yes

to form the line segment between -i and i

right

and the right side forms what, the real axis?

no

then in what direction?

well, the left side of the unit circle becomes the line segment between -i and i. the point 1 goes to infinity. so what do you think happens with the rest of the right side of the unit circle?

It forms the halfplane $Re z > 0$?

a one-dimensional arc does not get mapped to the whole half-plane

Oh, right it's a circle... not a disc

visualize it. as you're pushing the left side of the unit circle inwards, what is happening to the right side of the unit circle?

keep in mind 1 will eventually be sent all the way to infinity

Maybe, just maybe, it forms the positive (or non-negative) real axis

:D

no. I'm not currently asking you where it ends up. I'm asking what happens along the way

suppose you ever so slightly push the left side inwards. what's the effect on the right side?

It gets slightly pushed slightly outwards?

yes

now keep pushing it outwards more and more. keep in mind, the arc will always be anchored to +i and -i. so, let's say we keep pushing the arc more and more outwards. so much so, 1 ends up at infinity. where does the rest of the arc go?

the rest of the arc must be connected to +i, infinity, and -i...

A triangle with vertices at $\pm i, \infty$

which is?

the rest of the imaginary axis

the arc between 1 and +i blows up to the top part of the imaginary axis, and the arc between 1 and -i blows up to the bottom part of the imaginary axis

altogether, the unit circle is mapped to the imaginary axis

that's a good visualization of it, but pretend the fixed point (dark blotch) in the upper left is +i

I have no idea what I'm looking at lol

imagine it as "hyperbolic" rotation of the plane (well, half-plane, but the same thing happens to the bottom half) around i

I learned about Möbius transformations (or conformal mappings in general) just today, so this is all very new to me

the pattern in which it colors the points of the plane isn't really important (that has to do with crazier stuff), what's important is to notice how the points are flowing under the "hyperbolic" rotation

so, start the video, and imagine a vertical line through the fixed point. what happens to that line as time passes?

It curves?

right

More to the point, it curves inwards

well, the top part curve inwards from the left, while the base part curves outwards to the right

I would agree with that I guess, yes

pretty pictures

@arctictern Is it true that $f(z_1) = \frac{1}{f(z_2)}$ if $z_1, z_2$ are symmetric with respect to a line or circle $C_z$ that gets mapped under a Möbius transformation $f$?

is it true when f(z)=z?

No :(

@arctictern Can you say $0$ is symmetric to $\infty$ with respect to the unit circle?

And if yes, how?

still not entirely sure what you mean by symmetric

explain again?

is that like inversion across a circle / reflection across a line?

Two points $z_1$ and $z_2$ are said to be symmetric with respect to a circle $C$ if every straight line or circle passing through $z_1$ and $z_2$ intersects $C$ orthogonally

Hm okay then it makes sense

Can you construct a circle that goes through $0$ and $\infty$?

Or is that nonsense?

the generalized circles through 0 and inf would be the lines through the origin

Right

So orthogonal to the unit circle

At intersection

Hi, excuse me, it may looks absurd. But four month later, nobody can solve this. math.stackexchange.com/questions/1798556/…

tfw you work for two days on a problem and then learn something in class that lets you solve it in minutes

I was given a problem in analysis we didn't know how to solve because the prof wanted us to "struggle" (his word)

i feel that

the struggling, i mean

i struggle constantly

@0celo7 would you like to discuss a analysis problem

depends what it is

you know about $L_p$ spaces?

not yet

oh ok

How can I get the common area here please? ctrlv.in/868116 Should I first subtract the circle from the petal to get the outside petal, and then subtract the outside petal from the whole petal to get the inside petal?

my problem was "How many injections $f: \{1,2,...,n\} \to \{1,2,...,z\}$ exist where $\nexists i \in \{1,2,...,n\}$ such that $f(i) =i$ and $n \leq z$?"

@0celo7 struggling is good

O thats cool it highlights replies now

turns out the answer is $\text{exd}(n,z) = n!\sum_{i=0}^{n} (-1)^{i}\binom{z-i}{n-i}\frac{1}{i!}$

which is neato

much easier to do inclusion/exclusion than it is to solve $a(n,z) = (n-z)a(n-1,z-1) + (n-z+1)(n-1)a(n-2,z-1)$

oops

should be $z-n$

Anyone into that ?

Hi ocelo

Can you integrate paths with values in a Lie group?

there's a product integral where you can integrate values in a lie algebra to get values in a lie group

makes sense - that seems like the natural thing

can you do the same thing for maps out of, say, anything you can integrate over

I imagine the outputs would need to interact somehow

for accumulation to occur

can't quite parse what you mean

you're thinking the order is a necessity?

hello :3

guess eithwrwise the noncommutativity of the codomain could cause trouble

with normal integrals you can add the outputs, with elements of the lie algebra you can scale them by \Delta h, exponentiate and multiply in some order. if the outputs are just elements of a barren set, say...

oh, the codomain has an operation on it

$\nabla f(x_0,y_0)$ is said to be orthogonal to $f(x_0,y_0) = k$ right?