Mathematics - 2017-12-08 (page 3 of 3)

I fucked up the name

Dragneel

@PVAL-inactive Here's how I thought about it. For any integer $n$, $n*16$ seems to result in an integer than ends with either $0,2,4,6,8$, that is, even digits. And since we're looking for $n\equiv 1 (mod 16)$, only odd $n$ integers will satisfy us because of a remainder of $1$.

Tobias Kildetoft

@PVAL-inactive No, $\mathbb{Z}/n\mathbb{Z}$ can have cyclic unit group without $p$ being a prime. It holds precisely for $n = p^r$ or $n = 2p^r$ for odd prime $p$ (well, plus $n=4$).

8:36 PM

I knew it was something like that.

@Dragneel All odd integers though don't have remainder 1 when divided by 16, but it turns out the 4th powers do.

Dragneel

True, good observation.

Anyway, I have to go write an exam, see you later, and thanks for the help guys :)

9:15 PM

A cool generalization to all the "exponentio-polynomial" divisibility problems: if $m$ divides $a + d$, $(b - 1)c$ and $ab - a + c$ then $m$ divides $ab^n + cn + d$ for all natural numbers $n$.

There's a short induction proof. Anyone knows a more natural way to think about this?

Induction is basically bashing all the algebra

Akiva Weinberger

3Blue1Brown has a new video out on the following problem: Choose four points on the sphere uniformly at random. What are the odds that the tetrahedron determined by those points contains the center?

Also: Bye, see you all in ~25 hours

Hi

I have a question. The points of $RP^2$ are the lines of $R^3$ which go through $0$. What are the lines of $RP^2$ then? According to my resources it's the planes containing $0$ of $R^3$ but how can I see that? How is the term "line" even defined for an arbitrary space?

@brot Cool question. So for projective spaces, lines inside $\Bbb{RP}^n$ basically means 1-dimensional projective subspaces. So, the "linearly embedded $\Bbb{RP}^1$'s".

Since $\Bbb{RP}^1$ is the space of 1-dimensional vector subspaces of $\Bbb R^2$, the lines in $\Bbb{RP}^2$ are precisely the subspace of 1-dimensional vector subspaces of a 2-dimensional subspace of $\Bbb R^3$

That's a bit of a world-clutter there, but I hope that makes sense

The term "line" cannot really be defined for an arbitrary space.

The point is if you take a plane $S$ in $\Bbb R^3$ going through the origin, and think of lines lying on $S$ that passes through the origin, those give rise to points on $\Bbb{RP}^3$ which lie on a line.

@PVAL Well we can tell him a bit about geodesics.

9:26 PM

If you have the structure of a Riemannian metric (i,e, lengths of paths) you can talk about geodesics.

In this case there is a natural Riemannian structure on S^2

and the covering map to $\Bbb RP^2$ gives a "spherical" Riemannian structure on $\Bbb RP^2$.

Geodesics are locally length minimizing paths.

The point being that $\Bbb{RP}^n$ has a natural Riemannian metric on it, and under that the geodesics are exactly the $\Bbb{RP}^1$'s coming from the way I described, or rather, the "projective lines", as PVAL says

The geodesics of S^2 are just great circles and the geodesics in RP^2 are just the images of great circles under this projection.

@BalarkaSen I meant $\Bbb{RP}^2$ here, by the way. Unfortunate typo.

I think I get the idea

But what is a projective subspace? Does $\mathbb R P^n$ have vector space structure?

No, it has a projective space structure :)

It's really nothing complicated. A projective subspace of $\Bbb{RP}^n$ is a subspace that comes from a vector subspace in $\Bbb R^{n+1} \setminus \{0\}$ under the quotient map

You can take that to be the definition

9:43 PM

Ok, I see :) and then I can directly observe that in $\mathbb R P^2$ projective lines always intersect at one projective point? that's cool

Quite.

There's another picture I like to think about. Do you know homogeneous coordinates?

You mean like $(x_1:x_2:x_3)$?

yeah

So that's how you denote a point in $\Bbb{RP}^2$, right?

yup

So now consider the set of all points $(x_1 : x_2 : x_3)$ in $\Bbb{RP}^2$ such that $x_3 \neq 0$.

You can scale the coordinates to get $(x_1/x_3 : x_2/x_3 : 1)$

So such points all look like $(x : y : 1)$ in homogeneous coordinates

But there's a bijection between the set of such points and $\Bbb R^2$, right? $(x : y : 1) \mapsto (x, y)$.

On the other hand, consider the complement of the set, which are set of all points $(x_1 : x_2 : x_3)$ such that $x_3 = 0$. Or, well, set of points of the form $(x : y : 0)$. We can ignore the last zero coordinate and just say that this set is bijective to $\Bbb{RP}^1$, bijection being $(x : y : 0) \mapsto (x : y)$

So we have decomposed our projective space as follows: $\Bbb{RP}^2 = \Bbb R^2 \cup \Bbb{RP}^1$

Is this okay?

9:55 PM

hi there, can someone help me with getting the gcd of two polynomials over the field of rationals?

what is the gcd of $x^3 + 2x^2 + 2x + 1$ and $x^2 + x + 2$

@vanaghka euclidean algorithm

Yes I can follow

over $\mathbb{Q}[x]$

@vanaghka doesn't change

am i right with 2?

9:57 PM

Nice. So the way you can think about it is, $\Bbb R^2$ is not compact. But you add a "line at infinity" to it to get a compact object $\Bbb {RP}^2 = \Bbb R^2 \cup \Bbb{RP}^1$. So it's a compactification in a sense, like one point compactification, but with adding something more than a point at infinity @brot

If you think like this, then the lines $\ell$ of $\Bbb R^2$ (not necessarily passing through the origin; just a random straight line) become precisely the lines $\ell'$ of $\Bbb{RP}^2$ after the compactification

So it's just your usual concept of a line, souped up somewhat

@vanaghka $\gcd(x^3+2x^2+2x+1,x^2+x+2) = \gcd(x^2+1,x^2+x+2) = \gcd(x^2+1,x+1) = \gcd (2,x+1) = ???$

is $\Bbb Q[x]$ a euclidean domain?

How do you know that $\mathbb R P^2$ is compact?

Good question.

Do you know how $\Bbb{RP}^2$ can be defined as a quotient of $S^2$?

@LeakyNun yes. can i show you what ive worked out so far?

Yes, relating opposing points

10:03 PM

@vanaghka sure

Ok, and then it's compact. I see :)

right

@brot Precisely

it feels kind of weird

because 2 doesn't exactly divide $x+1$

10:11 PM

I don't quite understand the lines part though. So if I have a line in $R^2$ and then compactify it, what happens with the line on $\mathbb RP^1$?

The $\Bbb{RP}^1_\infty$ (line at infinity) is the extra bit you add to compactify $\Bbb R^2$. If $\ell$ was the line in $\Bbb R^2$, in the compactification $\Bbb R^2 \cup \Bbb{RP}^1_\infty$ $\ell$ intersects $\Bbb{RP}^1_\infty$ at one point.

So the line $\ell$ itself gets "compactified" to an $\Bbb{RP}^1$

It's a confusing picture but gets clearer if you think about it

the line at infinity has a point at infinity

double infinity

but not

it's turtles all the way down

worse is if you have RPinfty lmao

i guess $\mathbf{R}P^{\infty}$ is turtles all the way up

these are the kind of bullshit cell decompositions you get when your objects are algebraic though so

10:17 PM

i think the cell decomp of the projective spaces is p good

it's like a 7/10 for general constructions, but like a 9/10 cell complex

eh, it's kinda bad if you want to think about the projective space as an algebraic geometric object

like projective varieties get these kind of structures

i wouldnt know about that

where you have a massive zariski open set as your cell

top cell

and you throw it out

you get a variety again

and take a massive zariski open in it

rinse lather repeat

I should learn some algebraic geo

i agree

10:20 PM

maybe this winter break

i guess you'd immediately go to the complex geometric story

given your background

yeah

I have a copy of griffiths book on complex algebraic curves

coolio

so i was gonna bring that when i go home for break and give it a read

i never progressed much on the Forster thing i had

10:22 PM

@LeakyNun so it doesn't. 2 is the remainder yes. whats the next step after what i've done (if im right??)

rip

I tried to read this book at some point but didnt get very far cause I was doing way too much analysis

i want to restart it again but nobody wants to read it with me

and now im kind of burnt out on analysis

If I think about $\mathbb R P^1$ as the line at infinity, it's intuitive that $\ell$ intersects it at one point. But why can I do so?

well, it's the thing you add to $\Bbb R^2$ to make it compact

like you add a point to $\Bbb R^2$ to make $S^2$

and call that point the "point at infinity"

in this case you add a full projective line, so you call it the line at infinity

10:26 PM

And then parallel lines in $\mathbb R^2$ would intersect $\mathbb R P^1$ at the same point?

Yep

That's why we say the parallel lines "intersect each other at infinity"

the point of intersection lies on the line at infinity

How could I show that?

Phew, the shortest substitution I could write up from $(ab^{-1})(cd^{-1})$ to $(ac)(bd)^{-1}$ with field axioms uses a lot of associative transformations:

$$\begin{split}
&(ab^{-1})(cd^{-1})\\
=&((ab^{-1})c)d^{-1}\\
=&(c(ab^{-1}))d^{-1}\\
=&((ca)b^{-1})d^{-1}\\
=&((ac)b^{-1})d^{-1}\\
=&(ac)(b^{-1}d^{-1})\\
=&((ac)1)(b^{-1}d^{-1})\\
=&((ac)((bd)(bd)^{-1}))(b^{-1}d^{-1})\\
=&((ac)((db)(bd)^{-1}))(b^{-1}d^{-1})\\
=&((ac)((bd)^{-1}(db)))(b^{-1}d^{-1})\\
=&(((ac)(bd)^{-1})(db))(b^{-1}d^{-1})\\
=&((ac)(bd)^{-1})((db)(b^{-1}d^{-1}))\\
=&((ac)(bd)^{-1})(d(b(b^{-1}d^{-1})))\\
=&((ac)(bd)^{-1})(d((bb^{-1})d^{-1}))\\
=&((ac)(bd)^{-1})(d(1d^{-1}))\\
=&((ac)(bd)^{-1})(d(d^{-1}1))\\

@vanaghka oh wait, we're in $\Bbb Q[X]$

$\gcd(2,x+1) = \gcd(2,x+1-2(\frac12x)) = \gcd(2,1) = 1$

11:00 PM

@LeakyNun okay how was that step allowed

user8663905

Question: given $y = e^{x + y} - e^{-x-y}$, can you get all the y's to one side?

11:21 PM

@user8663905 I don't believe so, no

Actually, WA gives the real solution $x=\ln\left(\frac12\left(\sqrt{y^2+4}+y\right)\right)-y$

user8663905

what'd you enter into Wolfram to get that? I'm always struggling to get WA to do what I want it to do.

"solve y=e^(x+y)-e^(-x-y) in reals"

user8663905

Alright, I wonder what the algebra is to get there. Danke for the info