Mathematics - 2018-03-11 (page 1 of 3)

Xander Henderson

00:41

Q: What do you call someone who reads a paper on category theory?

A: A coauthor.

Emolga

:)

Secret

Q: How to become an author from a coauthor
A: By going past a pair of dual doors

Q: What about becoming a first author?
A: Do a bubble sort on the authors

Secret

01:02

Q: Why is the math chat empty?

A: Because they are being cartesian producted with an empty set

[The Cult of Infinity]

DarkRunner

01:38

Hi

Can someone help me with the latex for prime factorization?

I have $Let\quad n={ { { p^{ { e }_{ 1 } } }_{ 1 } } }*{ { { p^{ { e }_{ 2 } } }_{ 2 } } }*...*{ p^{ { e }_{ n } } }_{ n }$

so far, but that looks really weird

Wow

hello?

Secret

02:05

[The Cult of Infinity]

0. We start with some logic, which give us the symbols and the inference rules

1. The first object to be constructed is the empty object $\varnothing$

user338510

Anyone explain why 1+1=1?

4

user338510

I need help.

Secret

@idk in what algebraic structure you encounter the identity 1+1=1?

user338510

Haven't you heard of 1+1=2?

user338510

But that's incorrect.

Secret

02:15

1+1 can be equal to anything depending on which algebraic structure youare in

Akiva Weinberger

@DarkRunner Let $n$ equal ${p_1}^{e_1}\cdot{p_2}^{e_2}\dotsb{p_n}^{e_n}$.

user338510

I want to know from the true mathematicians, why 1+1=1

Akiva Weinberger

(I guess it depends on if you want $p_1^{e_1}$ or ${p_1}^{e_2}$, they look slightly different) @DarkRunner

@idk …It doesn't

Secret

@idk As I have said, we need to know the details of the algebraic structure to understand how this is derived from the axioms

what exactly are the axioms in your algebraic structure?

user338510

What? You cannot defy the fundamental concept that 1+1 = 1.

Akiva Weinberger

02:16

@idk Assuming "+" means what it normally means and "1" means what it normally means

and "=" as well I guess

user338510

I want to have a good grasp of this fundamental concept.

Secret

@idk well, is your 1 the multiplicative identity, the + addition and = equality as we knew in first order logic?

Akiva Weinberger

Secret, he's trolling

Secret

Akiva: O ok, I see

Akiva Weinberger

@BalarkaSen Yeah, Gauss's lemma is the only time that the symmetry of the connection is used

so it's entirely 'cause of that

user338510

02:19

What? In which aspects am I trolling? (I am... shhh) I am simply asking for help.

Secret

Akiva: Btw, an algebraic structure where 1+1=1 and is distributive is one of the most trivial structure in existence (as we have discussed some years ago), because then every element will be idemponent to itself due to x(1+1)=x1 => x+x=x

so it will form an idempotent semigroup

Akiva Weinberger

If you ask Terrence Howard, though, he'll tell you that 1x1=2

Secret

Terrence Howard

Terrence Dashon Howard (born March 11, 1969) is an American actor and singer. Having his first major roles in the 1995 films Dead Presidents and Mr. Holland's Opus, Howard broke into the mainstream with a succession of television and cinema roles between 2004 and 2006. He was nominated for the Academy Award for Best Actor for his role in Hustle & Flow. Howard has had prominent roles in many other movies including Winnie, Ray, Lackawanna Blues, Crash, Four Brothers, Get Rich or Die Tryin', Idlewild, August Rush, The Brave One and Prisoners. Howard played James Rhodes in the first Iron Man and its...

This guy?

user338510

How would you go about solving for $x$ in $(x^2+3x+2)^2+3(x^2+3x+2)+2=x$?

Akiva Weinberger

He's wrong, and hilariously so, but he won't admit it

@idk I would simplify it, for a start

user338510

02:23

Here, I admit that $1+1=2$, how would you solve it?

Akiva Weinberger

@idk Oh, wait, no

user338510

Yes akiva I simplified it

Akiva Weinberger

That would be a quartic and those are hard

user338510

And i got solutions

user338510

by factoring

Akiva Weinberger

02:24

Hm

user338510

and trial and error

user338510

It factors to $(x^2 + 2 x + 2) (x^2 + 4 x + 6)=0$, BUT that took completing the square and some trial and error.

Akiva Weinberger

Arright

user338510

Surely, there must be a way to notice that the quartic is in the form $f(f(x))$, right?

Secret

Akiva: Well tbh, if 1 is the multiplicative identity, then 1x1=2 will give you the trivial group, since 2 will then become a multiplicative identity and hence 1=2

but anyway...

Akiva Weinberger

02:25

I noticed it, I don't know what to do with it @idk

user338510

Anyone offer help?

Secret

if it is $(x^2+3x+2)^2+3(x^2+3x+2)+2=0$, then it is easy to factor since that is basically a quadratic equation in disguise, but this is $(x^2+3x+2)^2+3(x^2+3x+2)+2=x$, and factoring becomes nontrivial

user338510

Yes, that's why I am stumped in finding a non-trial and error way to do it.

Akiva Weinberger

Oh, here's a good way to see it has no real solutions

First note that $x^2+2x+2$ is $(x+1)^2+1$ and thus always positive

So $x^2+2x+2>0$

Add $x$, that's $x^2+3x+2=f(x)>x$

So we have $f(x)>x$

Subsitute $f(x)$ into that, we get $f(f(x))>f(x)$

So we have $f(f(x))>f(x)>x$

Thus $f(f(x))$ is never equal to $x$

user338510

$x^2+3x+2=(x+3/2)^2-9/4+2=(x+3/2)^2-1/4$ might not always be greater than 0

Akiva Weinberger

02:35

@idk Right, but the point is that it's always greater than $x$.

user338510

That can't really help in finding the roots.

Semiclassical

Sure it can. It tells you something very important about the roots of $f(f(x))=x.$

Akiva Weinberger

You were asking when $f(f(x))$ was equal to $x$, not $0$.

So I guess you wanted the roots of $f(f(x))-x$.

Secret

Meanwhile, I am thinking about what does $(x^2+3x+2)^2+3(x^2+3x+2)+2=x$ mean in terms of remainders...

Akiva Weinberger

Doesn't help much in finding the complex roots, though, I'll grant you

@Secret What, like, modular arithmetic?

user338510

02:38

There is probably a more elegant solution

Secret

I am trying to see whether there is a way to systemise the lone $x$ on the RHS so we can use the fact that the LHS is a pseudoqudratic on $(x^2+3x+2)$

because it is much easier to solve a pseudoquadratic than a quartic

Akiva Weinberger

Is $f(x)-x$ always a factor of $f(f(x))-x$?

Hm, I guess it is. Weird.

For any polynomial

user338510

No, there must be some reason for your discoveries.

Semiclassical

Fixed points are also points of period 2.

user338510

Which can be explained with a great way.

Akiva Weinberger

02:40

I have a proof

Secret

Something that always bugs me about polynomials is that while they form a vector space, predicting where the roots of even a sum is nontrivial in general

Akiva Weinberger

Doesn't mean it's not weird

Semiclassical

I think it just comes down to the fact that if $f(y)=y$, then $f(f(y))=f(y)=y$.

Akiva Weinberger

Proof: If $f(x)=x$, then $f(f(x))=f(x)=x$. Thus, any root of $f(x)-x$ (real or complex) is a root of $f(f(x))-x$.

Secret

akiva got sniped

Semiclassical

02:42

Which is to say, fixed points are also points of period 2 (and period 3, period 4, etc.)

Akiva Weinberger

Hm, this doesn't necessarily work if $f(x)-x$ has double roots

but there's probably something Zariski-y you can do

In addition, $f(x)$ is alway a factor of $f(f(x)+x)$.

Secret

There's something strange about the behaviour of $f^n(x)$, it seemed to be converging to something...

Akiva Weinberger

Where $f'(x)=0$, $~(f^{\circ n})'(x)=0$.

Wait, that might not be true

Oh, it is.

DarkVampiric AbstractArtist

1. Let n = 81^2018 × 12^2121 × 30^3121
.
(a) How many natural numbers are divisors of n? my attempt: 3^(2 x 2 x 2 x 1009) * 3^1 * 2^(2 x 3 x 7 x 101) * 30^3121

Secret

02:57

$(f^{\circ n})'(x)=\frac{df^{\circ n}(x)}{df^{\circ (n-1)}(x)} \frac{df^{\circ (n-1)}(x)}{df^{\circ (n-2)}(x)}\cdots f'(x)$

so if $f'(x)=0$, the derivative of the nested function has to be zero

Akiva Weinberger

@DarkVampiricAbstractArtist Factor it into primes

DarkVampiric AbstractArtist

81 x 12 x 30 = 3^4 x 3^1 x 2^2 x 2^1 x 3^1 x 5^1 = 2^3 x 3^6 x 5^1

Akiva Weinberger

But that ignores the exponents

So 81 is 3^4, so 81^2018 is (3^4)^2018 = 3^(4x2018)

DarkVampiric AbstractArtist

3^4(2 x 1009) * 3^1 * 2^(2 x 3 x 7 x 101) * 30^3121

I think I mentioned that earlier, but how do I do 12^2121 and 30^3121?

Akiva Weinberger

Your factorization of 12^2121 is wrong

DarkVampiric AbstractArtist

03:04

12^(3 x 7 x 101)?

Akiva Weinberger

We don't care about the exponent

Don't factor it

12 is 2^2 x 3

so 12^2121 is (2^2 x 3)^2121, which is

(2^2)^2121 x 3^2121

which is 2^(2x2121) x 3^2121

which is 2^4242 x 3^2121.

So 81^2018 is 3^8072, and 12^2121 is 2^4242 x 3^2121.

Secret

https://www.desmos.com/calculator/csc3qwamfi
Need to figure out how those extra stationary points emerge...

Akiva Weinberger

To factor 30^3121, we factor 30 into 2x3x5

DarkVampiric AbstractArtist

Thanks. I lack a lot of common sense. Does this apply to 30^3121? When (5 x 2 x 3)^3121 = (5^3121) x (2^3121) x (3^3121)?

Akiva Weinberger

~~Yeah, but 6 isn't prime~~

DarkVampiric AbstractArtist

03:08

woops 6 = 2 x 3

Akiva Weinberger

So that's, in the end,

3^8072 times 2^4242 x 3^2121 times 2^3121 x 3^3121 x 5^3121

Collect the things with the same base

(2^4242 x 2^3121) x (3^8072 x 3^2121 x 3^3121) x (5^3121)

2^(4242 + 3121) x 3^(8072 + 2121 + 3121) x 5^3121

2^7363 x 3^13314 x 5^3121

and that's the prime factorization

Now, we need to find all the divisors.

Any divisor is gonna be of the form 2^a x 3^b x 5^c.

(It can't have any other primes in it because then it wouldn't be a divisor of n)

Similarly, a can only be between 0 and 7363, because n only has 7363 twos in it

so there are 7363+1=7364 options for a

b and c, similarly, have 13315 and 3122 options

Secret

There seemed to be another class of fixed points in a quadratic nesting map $f^{\circ n}(x)$: Their rules is to be determined

Akiva Weinberger

So the final answer is 7364x13315x3122

which is 306117282520 @DarkVampiricAbstractArtist

DarkVampiric AbstractArtist

And those are the natural divisors of n? Thanks :)

Akiva Weinberger

306,117,282,520

Why were you asking about such a large number?

DarkVampiric AbstractArtist

03:15

Hmmm. The question asks How many natural numbers are divisors of n? But it seems that you've computed it.

Akiva Weinberger

Yeah, there's lots

'cause n is pretty frickin' huge

DarkVampiric AbstractArtist

Yeah, it's pretty big, but thank you so much for the help :).

Secret

They could be arised from $f''(x)=0$ I suspect

Akiva Weinberger

@Secret Incidentally, Desmos lets you write $f(x)=\text{blah}$ on one line and $f(f(x))$ on another

Makes examining $f(f(x))$ easier

Secret

03:34

adding or removing $x$ is basically a sheer map on the polynomial function

nesting, however is more nontrivial since the map is nonlinear

Interestingly, their finite analogue is easy to solve by using matrices

Perhaps, a numerical way to predict how the roots of a function will vary under sheering is to grab a sequence of partitions on the x values, which at the limiting case, will converge to the continuum. Then solve the matrix equations in each case until there is enough to approximate the limiting case

DarkVampiric AbstractArtist

04:23

woops

3. Decide which of the following statements are true. For those that are false, provide a counter-example (with a short explanation).
For those that are true, provide a proof.(a) If A and B are bounded sets such that inf(A) = inf(B) and sup(A) = sup(B), then A= B. My Attempt: I said that it's true and here's my direct proof: Since we know that InfB∈A, then InfB is the largest element in A. We denote this fact by M=InfB. By definition of B, we know that InfA∈B. Because M is the largest element in B, we must have the inequality infA≤M=InfB, as desired.

Akiva Weinberger

@DarkVampiricAbstractArtist Why would inf(B) be the largest element in A??

DarkVampiric AbstractArtist

The infinimum is said to be the greatest lower bound.

Secret

also let A=(0,1)-{0.5} and B = (0,1)-{0.7}. Then A=/=B but they have the same bound

Akiva Weinberger

@DarkVampiricAbstractArtist Right, lower bound.

If the minimum exists, the infimum equals the minimum.

And I don't see how you got A=B in the end.

DarkVampiric AbstractArtist

So then it's false?

Akiva Weinberger

04:31

It is.

Take A={1,2,4} and B={1,3,4}. Or really any pair of sets that have the same minimum and maximum.

Or (0,1) and [0,1].

The infimum of (0,1) (meaning the interval from 0 to 1 not including the endpoints) is 0. The supremum is 1.

The same is true of (0,1], [0,1), and [0,1].

DarkVampiric AbstractArtist

So the supremum of [0,1] is 1 also?

Akiva Weinberger

Yes

DarkVampiric AbstractArtist

If both sets A and B contain the infininum/supremum, that would mean that A=B?

The infinimum of A and B being 0. Also the supremum of A and B being 1.

Akiva Weinberger

No

Take {1,2,4} and {1,3,4} again

and {1,4} as well

These all have 1 as the infimum (and minimum) and 4 as the supremum (and maximum)

and they all contain the infimum and supremum

DarkVampiric AbstractArtist

Set A and B respectively containing 2 and 3 means that A=/=B?

Akiva Weinberger

04:36

The point is that the contain other elements as well, in addition to the infimum and supremum. And these can be different

@DarkVampiricAbstractArtist Yes

Two sets are equal if and only if they have the same elements

If something's in A and not in B, or vise versa, then A=/=B.

DarkVampiric AbstractArtist

Thank you @AkivaWeinberger

Balarka Sen

@AkivaWeinberger Nailed it

Secret

04:53

hmmm.... this is going nowhere....

Trying to slide from f(x) to f(x)-kx, but a path is missing

DarkVampiric AbstractArtist

it translates to -kx, meaning the parabola moves down.

Secret

I know, I am trying to use that to track the movement and the formation/removal of the roots

That is, trying to find a way so that given you know the roots of $f(x)=0$, find where the roots of $f(x)-kx=0$ are

DarkVampiric AbstractArtist

Try to find where the parabolas intersect and try to eliminate them from on there.

Akiva Weinberger

05:15

@BalarkaSen Hey, random question

So, we know how to find the area of a loop $\gamma:S^1\to\Bbb R^2$

Say that $\pi_{xy}$ is the projection of $\Bbb R^3$ onto the $xy$-plane, similarly for $\pi_{xz}$ and $\pi_{yz}$

and $\pi_P$ is the projection onto some random plane $P$

Balarka Sen

mmk

Secret

Actually nvm, roots of $f(x)$ does not necessary have anything to do with the roots of $f(x)-kx$ since $f(x)$ can have no real roots to start with, and it is impossible to predict where the roots are without full knowledge of the values of $f(x)$

Ted Shifrin

Howdy @Balarka and DogAteMy

Balarka Sen

@Ted!

Semiclassical

evening @ted

Ted Shifrin

05:19

Hi @Semiclassic

Akiva Weinberger

If $\gamma$ and $\delta$ are loops in $\Bbb R^3$, and the areas of $\pi_{xy}\circ\gamma$, $\pi_{xz}\circ\gamma$, and $\pi_{yz}\circ\gamma$ equal the corresponding areas for $\delta$,

will the area of $\pi_A\circ\gamma$ equal the area of $\pi_A\circ\delta$?

Balarka Sen

Hmm

Semiclassical

$\pi_{xy}$ = projection onto the xy-plane?

Ted Shifrin

and $\pi_A = \pi_P$? :)

Akiva Weinberger

In other words, if two loops in 3-space have the same areas after being projected to the coordinate planes, will they have the same areas after being projected to any plane?

Yes, and whoops yes

Balarka Sen

05:21

Yes, this is true

Akiva Weinberger

Weird

Ted Shifrin

You should have kept studying my multivariable book and learned 2-forms.

Balarka Sen

Area of a planar region $A$ in $\Bbb R^3$ satisfies $A^2 = A_x^2 + A_y^2 + A_z^2$

Akiva Weinberger

Oh, is this explicitly done somewhere in there?

@BalarkaSen Yeah but $\gamma$ isn't planar

Balarka Sen

Where $A_x$, $A_y$, $A_z$ are the projections onto the $yz$, $xz$ and $xy$ planes

$\pi_A \circ \gamma$ is

Akiva Weinberger

05:22

It's just a random loop in 3-space, it doesn't need to lie in any one plane

Ted Shifrin

True, @Balarka, but, even better, the area $2$-form for a random plane $P$ is a (random) linear combination of $dx\wedge dy$, $dy\wedge dz$, and $dz\wedge dx$, with the coefficients the coefficients of the unit normal, DogAteMy.

Balarka Sen

@AkivaWeinberger $\pi_A \circ \gamma$ is planar and has the same projections as $\gamma$, no?

Akiva Weinberger

@BalarkaSen Oh. But then we have to compare $\pi_{xy}\circ\pi_A\circ\gamma$ to $\pi_{xy}\circ\gamma$

Balarka Sen

I think they have the same areas, Akiva

Um

Akiva Weinberger

@BalarkaSen No, take $A$ to be a coordinate plane

Balarka Sen

05:23

Maybe there's some fuckery involved

Akiva Weinberger

@TedShifrin Ah. This makes sense

Balarka Sen

@TedShifrin Ah yes right

Akiva Weinberger

@BalarkaSen Yeah I don't think your thought works

Ted Shifrin

The loops are red herrings. You're talking about surfaces.

Akiva Weinberger

If a surface is a map from $D^2\to\Bbb R^3$, we don't need the entire map to find the areas of the projections

We just need the boundary of the map, $\partial D^2\to\Bbb R^3$.

Ted Shifrin

05:26

But I'm suggesting that's the wrong way to think about it.

Balarka Sen

Maybe you could choose an appropriate surface bounding the loop to help your computation

Akiva Weinberger

In any case, this came about 'cause I was thinking about generalizing the idea of the area of a loop in a plane

Area is a function that takes in a loop in $\Bbb R^2$ and outputs something in $\Bbb R$.

The generalization would take in a loop in $\Bbb R^3$ and output something in $\Bbb R^3$

namely, the triple of areas of the projections

Ted Shifrin

This is very coordinate-dependent.

Akiva Weinberger

which is apparently enough information to get the area of call projections.

Ted Shifrin

I believe what I suggested if there's some sort of convexity. The loop might have crazy projections with different signs on different pieces.

Akiva Weinberger

05:28

@TedShifrin Fair. I guess the codomain is just "a three-dimensional vector space", whose basis depends on your choice of basis of $\Bbb R^3$ (where the loops live).

Ted Shifrin

So you need to be very careful even defining what you mean if the loop is knotted.

Balarka Sen

I can come up with a loop which projects to a circle in the xy-plane and something with zero areas when projected to the yz and xz-planes

Akiva Weinberger

And this is connected to $\dim(SO(2))=1$, $\dim(SO(3))=3$, I think.

What's $\dim(SO(n))$ in general?

Is it $\binom n2$?

Ted Shifrin

Yes.

Balarka Sen

n(n-1)/2

Ted Shifrin

05:29

Nope.

Akiva Weinberger

I guess that would match the number of planes to project to.

@BalarkaSen -

Balarka Sen

sniped.

Ted Shifrin

It also matches $\dim \bigwedge^2\Bbb R^{n*}$. :)

Akiva Weinberger

@TedShifrin Sure, but that's not a hard problem. $\oint ydx-xdy$, right?

Ted Shifrin

But I'm still a bit troubled by what area means if the region is non-simple.

Um, with a factor of $-1/2$.

Akiva Weinberger

05:31

Regions where the curve winds around them with a winding number of $n$ get counted $n$ times

@TedShifrin Right

Ted Shifrin

It still makes sense with my surface notion, as long as the projections are interpreted as oriented $2$-chains.

Akiva Weinberger

This all connects back with holonomy

That's the word, right? Parallel transport around loops?

Ted Shifrin

Yup.

Akiva Weinberger

Which is why I was thinking about it

Balarka Sen

@AkivaWeinberger How so?

Akiva Weinberger

05:33

If you have a loop in $S^2$, parallel transporting around it gives a rotation (i.e. an element of $SO(2)$) in the tangent plane.

And, in fact, that rotation is proportional to the area of the region that the loop surrounds.

Ted Shifrin

And Stokes's Theorem computes it for you by integrating the curvature $2$-form over a region the curve bounds.

Yup, 'cuz constant curvature for the 2-sphere.

Akiva Weinberger

Well, it's the area times the curvature (or the area divided by the radius squared), I guess.

Balarka Sen

I have never actually done the angle computation with the holonomy

Akiva Weinberger

So as the radius goes to infinity, you approach the plane.

Ted Shifrin

You quit reading my notes before you got to that, Balarka. That was Chapter 3.

Akiva Weinberger

05:35

And the area becomes an element of the tangent space of $SO(2)$.

Balarka Sen

mm I see @Ted

Akiva Weinberger

@BalarkaSen Well, it's additive, you can see that much

Balarka Sen

Mhm I agree

Akiva Weinberger

So the natural question is, what happens when we repeat this in $S^3$?

So I guess you get an element of $SO(3)$.

And, as the radius goes to infinity and whatever, and the space approaches the plane, the result becomes an element of the tangent space of $SO(3)$.

Which is a 3D vector space.

And I think that gives you the same answer as the projecting-onto-coordinate-planes construction

MatheinBoulomenos

@TedShifrin sorry for replying to this so much later, but can you clarify this a bit? Do you mean that we can also go the other direction, i.e. associate to a representation of the fundamental group of $X$ a flat bundle $E \to X$ that induces this representation via holonomy? If yes, is this functorial?

Ted Shifrin

05:39

Yup, @Mathein, you just use the standard associated bundle construction, as I recall.

Akiva Weinberger

$SO(3)$ isn't commutative, though, so I guess this means you get some weirdness if you stick with measuring loops in $S^3$ rather than inflating it into $\Bbb R^3$

But, yeah, this construction isn't dependent on the basis, which meant that the projection-onto-planes thing must not be dependent on the basis either, which seemed odd

Balarka Sen

@MatheinBoulomenos Say you have a representation $\rho : \pi_1 X \to GL_n(\Bbb R)$

Akiva Weinberger

which is why I came to the chat asking if it was true

Balarka Sen

Consider $\widetilde{X} \times \Bbb R^k/(x, v) \sim (gx, \rho(g)(v))$

This should be your flat bundle, connection coming from the flat connection on $\widetilde{X} \times \Bbb R^k$, corresponding to $\rho$

Ted Shifrin

DogAteMy: You can still integrate an $\mathfrak{so}(3)$-valued $2$-form over the $2$-chain. I guess the exponential is still an element of $SO(3)$.

Balarka Sen

05:41

$\widetilde{X}$ is the universal cover, $g$ runs over all element of $\pi_1 X$

MatheinBoulomenos

Thanks @BalarkaSen @TedShifrin

Ted Shifrin

@Balarka: Probably need a $g^{-1}$ in there somewhere.

Akiva Weinberger

@TedShifrin Is that another way of getting the 3-vector of areas?

Ted Shifrin

I hadn't thought of it, but, yeah, if you think about what $\mathfrak{so}(3)$ is.

MatheinBoulomenos

I need to look into bundles more. We did some stuff in diff geo class, but not that much

Balarka Sen

05:43

@TedShifrin Hmm, maybe $g^{-1}$ acts on the first factor

Ted Shifrin

@Mathein: Complex geometers (and topologists) make a much bigger deal out of bundles than most Riemannian geometers do.

@Balarka: One or the other, yeah.

Daminark

Hey there everyone!

MatheinBoulomenos

bundles feel like an algebraic aspect of geometry, so I guess it's not surprising that I find them interesting

Ted Shifrin

I suppose there's always the left/right confusion, too.

Akiva Weinberger

@Daminark Hi!

Balarka Sen

05:44

@TedShifrin Yeah, ugh

Ted Shifrin

@Mathein: I always taught way more with bundles in my diff geo courses.

MatheinBoulomenos

@Daminark Hi!

Ted Shifrin

Hi Demonark

Balarka Sen

Hi @Daminark

Akiva Weinberger

I wonder, then - is there a way to modify the same construction to give us the volume of regions in $\Bbb R^3$?

Balarka Sen

05:45

I learnt this construction during my fiddling with foliations, from Candel and Conlon

Akiva Weinberger

I guess we'd need some notion of "parallel transport" across a surface…

MatheinBoulomenos

@TedShifrin I think I would've liked your diff geo course. Doing more bundles and forms (iirc) than usual sounds like an approach to diff geo I like

Balarka Sen

@AkivaWeinberger Jesus god

Ted Shifrin

I know it pains you to say that, @Mathein. :)

Akiva Weinberger

@Daminark I had a weird thought, which is that, while the area of a loop in $\Bbb R^2$ is a number (i.e. a 1-dimensional vector), the area of a loop in $\Bbb R^3$ should be a three-dimensional vector.

MatheinBoulomenos

05:46

@TedShifrin why would it pain me?

Balarka Sen

Maybe parallel transporting a 2-vector along a surface? I'm going to fuck outta there

Ted Shifrin

@Balarka, @Mathein: I may not have mentioned this exercise before, but it's given short shrift in diff geo books. Take a symmetric space $G/H$ (like the sphere, $SO(n+1)/SO(n)$). Give its tangent bundle in terms of representation theory.

DogAteMy: Of course, the area of a loop is far from well-defined.

MatheinBoulomenos

Hmm, symmetric spaces are topics for the second diff geo course here (which I didn't take)

Akiva Weinberger

Right, but this three-vector is.

Daminark

The area of a loop? I've never heard of this before

Ted Shifrin

05:48

That's cuz it doesn't make sense, Demonark :P

Akiva Weinberger

@Daminark In the plane the idea makes perfect sense

Balarka Sen

@TedShifrin Can you minimize over the area of all surfaces which bounds the loop? (Assume loop is embedded)

Akiva Weinberger

It's the area of the region the loop surrounds (modulo some weirdness with winding numbers if the loop intersects itself).

Ted Shifrin

@Mathein: So you get a splitting $\mathfrak g = \mathfrak h \oplus \mathfrak m$, and $\mathfrak m$ is $ad(H)$-invariant.

@Balarka: Sure, there's a least-area surface bounding it (by old hard stuff). Not sure this has anything to do with what DogAteMy is doing, though.

Akiva Weinberger

@Daminark Say $\gamma$ is a loop in $\Bbb R^3$, and say $A_{xy}(\gamma)$ is area of the projection of $\gamma$ onto the $xy$ plane.

Balarka Sen

05:50

@TedShifrin Yeah that was an unrelated question. Very cool

How does one prove that?

Ted Shifrin

That's old Douglas-Rado stuff at the beginning of minimal surface theory in the early 20th century.

Akiva Weinberger

Then the vector I'm considering is the vector those three values ('cause of the three coordinate planes)

Ted Shifrin

Presumably Eric could lecture you on all this stuff when he's done with finals.

Balarka Sen

That would be dope

Akiva Weinberger

I guess $(A_{yz}(\gamma),A_{xz}(\gamma),A_{xy}(\gamma))$ @Daminark

Ted Shifrin

05:51

DogAteMy: Indeed, if the projection of the loop intersects itself.

Akiva Weinberger

The strange thing is that this doesn't depend on the choice of basis

MatheinBoulomenos

@TedShifrin very nice! I like that

Ted Shifrin

I tolerate a certain amount of algebra, @Mathein :P

Akiva Weinberger

which is why I feel like it's the natural definition of "area" for loops in 3-space

(despite being a 3-vector and not a number)

Balarka Sen

@MatheinBoulomenos More topology apologist intuition: If you have a flat vector bundle $E$ over $M$ then it's horizontal subbundle (complementary to kernel of $d\pi : dE \to dM$) as defined by the connection (parallel transport lifts a path on $M$ to a path on the horizontal subbundle) is integrable

If you integrate it, you'll get a foliation on $E$ transverse to the fibers of $\pi : E \to M$

These are called foliated bundles

Ted Shifrin

05:54

@Mathein: So you should be able to construct the tangent bundle of $G/H$ as an associated bundle built out of $G\times\mathfrak m$.

MatheinBoulomenos

I don't even know what a foliation is and I forgot what an integrable submanifold is

Ted Shifrin

@Mathein: I can paste in the actual exercise I wrote for my grad class if you are actually interested.

MatheinBoulomenos

@TedShifrin yes, I asked the question so I'm interested. But I don't think I can solve it, though

Ted Shifrin

Well, this is on my question, not yours. :)

MatheinBoulomenos

But it's still on associated bundles, so it's close enough

Balarka Sen

05:57

@TedShifrin I think the tangent bundle should be a twisted $\mathfrak{g}/\mathfrak{h}$-bundle on $G/H$.

Well, I'm saying nothing new. But I mean it should come from $G \times \mathfrak{g}$ using some sort of similar identification I wrote above

Hm

Ted Shifrin

Suppose $H$ is a closed Lie subgroup of $G$, and write $\mathfrak g \cong \mathfrak h\oplus\mathfrak m$ with ($[\mathfrak h,\mathfrak h]\subset \mathfrak h$ and) $[\mathfrak h,\mathfrak m]\subset \mathfrak m$. Consider the homogeneous space $M = H\backslash G$. Prove that $TM \cong G\times_{\text{Ad}(H)} \mathfrak m$.

Hints: "Recall" that if $\rho: H\to GL(V)$ is a (right) representation, we can consider
$$G\times_\rho V = \{(g,v)\in G\times V\}\big/ \big(g,v\big)\sim \big(hg,v\cdot\rho(h^{-1})\big)\,.$$
If $\pi\: G\to H\backslash G$ is the canonical map, define $\Phi\:G\times\mathfrak m \to T(H\backslash G)$ by $\Phi(g,v) = (\pi(g),\pi_*(R_{g*}v))$, and check this induces a well-defined isomorphism. (Remark: For those of you acquainted with the term, here we're thinking of $G$ as a principal bundle with fiber $H$ over $H\backslash G$.)

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