Mathematics - 2017-12-27 (page 3 of 5)

Balarka Sen

15:00

Sounds like diffing wrt z does give you the third thing

So that's it?

Akiva Weinberger

Yeah I got it that's where the $\frac12e^{z^2}$ bit came from

Now to actually integrate it on that path, I need to identify its endpoints

Balarka Sen

Ya

And plug it in

subtract

boom

:exploding noises:

Akiva Weinberger

It starts at $\begin{bmatrix}1\\-1\\0\end{bmatrix}$ and ends at $\begin{bmatrix}e\\2\\1\end{bmatrix}$ I think

And, tell you what, I'm not actually gonna finish this problem

'cause I'm essentially done

and the last step is boring

Balarka Sen

:p

I never did it because I knew how to do it

Akiva Weinberger

I'm doing it because I didn't know how to do it :P

For some reason I thought it wasn't closed (and thus not exact)

But it turns out everything's fine

Balarka Sen

15:04

If it wasn't closed you'd be flubberducked

Maybe try to break up into an exact and a smaller non-exact part

Akiva Weinberger

Yeah

But then again, the path was horrible

Balarka Sen

Truly

Akiva Weinberger

So, hold on, was it $ydx-xdy$ that gives you the area?

Balarka Sen

Yeah $\iint 1 dydx$ and then use Green's theorem

Akiva Weinberger

This corresponds to the vector field $(y,-x)=R_{90^\circ}(x,y)$

The thing that points counterclockwise everywhere

So whenever I move a particle in that field, the area of its path is the work done to it by that field? Weird

Wait no, it's $\frac12{ydx-xdy}$

Balarka Sen

15:06

Wait, $xdy$ gives you area, right?

Yeah that works too

Akiva Weinberger

Right, so $ydx$ and $-xdy$ work on their one

I guess I see how $(y,0)$ gives you the area (in terms of the work analogy)

Still frickin' weird

Balarka Sen

Kinda.

Akiva Weinberger

But like, moving it horizontally gives you the area under that horizontal line

and moving it vertically gives you the area under the vertical line (i.e. 0)

So if you add it up for a rectangle it works

And most things are almost lots of rectangles

Balarka Sen

Take a point $(a, b)$ and feed $ydx - xdy$ the vector $(v_1, v_2)$. That spits out $bv_1 - av_2$. That's a determinant, right?

Uhh

Akiva Weinberger

?

Balarka Sen

15:10

I want to say it's locally volume

It's det[b, a; v_2, v_1]

Hm

Akiva Weinberger

So 1-forms are vector fields. I'm guessing 2-forms are "little plane thing" fields, but I didn't get that from his introduction of them

Balarka Sen

Well, no, 1-forms are dual to vector fields

They eat vector fields

Akiva Weinberger

But there's a natural correspondence between them

If I try to draw a 1-form I only know how to draw its corresponding vector field

Balarka Sen

Sure, the zero set of the 1-form is your vector field (on the plane, at least. On R^n zero set of the 1-form is a codimension 1 plane field)

And that's where things get icky. Not all plane fields come from a 1-form

Akiva Weinberger

Oh

Balarka Sen

15:13

Maybe I'm wrong. Wait

Akiva Weinberger

Wait, when he said integrating 1-forms is like work from vector fields, that worked in any dimension

Balarka Sen

Ok, I'm wrong. You need to choose a Riemannian metric to construct the 1-form from the plane field, though, I think.

Yeah, that's how it works. If you have a codimension 1 plane field $E$ on a manifold $M$, give it a Riemannian metric to have it's orthogonal line field $E^\perp$

Consider the orthogonal projection $T_x M \to E^\perp_x$ for each point $x \in M$

Akiva Weinberger

OK hold on

Balarka Sen

That is a 1-form at $x$ which vanishes on $E_x$

Akiva Weinberger

Are we not doing subsets of $\Bbb R^n$ anymore

Balarka Sen

15:17

So that gives a 1-form with zero set $E$ on $M$.

Akiva Weinberger

And do we not get plane fields (if that's the right word) for 2-forms?

Balarka Sen

@Akiva Yeah I'm thinking generally. Sorry for the tangents.

@AkivaWeinberger Say you have the 1-form $dz$ on $\Bbb R^3$

Rickyfox

@MarkLStone damnit, that was actually the cause of the problem. I've feeling silly for overlooking that

Balarka Sen

It's zero set is the "horizontal plane field" always parallel to $xy$.

$\Bbb R^3$ is a very specific Euclidean space, so there is in fact a canonical correspondence between 1-forms and vector fields on $\Bbb R^3$ (take the line field perpendicular to the zero set of it, which is a 2-plane field)

Akiva Weinberger

Sure, and the same for $\Bbb R^n$ in general

Balarka Sen

15:21

Ah, well, true. :P

Mike Miller

i an cold

Balarka Sen

I was thinking of 2-forms on $\Bbb R^4$

In which there is no way to get a vector field into the story

2-forms in $\Bbb R^3$ do correspond to vector fields though

(Zero set is a line field)

Akiva Weinberger

The zero set of 2-forms isn't even something in $\Bbb R^n$, is it? It's something in $(\Bbb R^n)^2$, 'cause 2-forms take in two arguments

Balarka Sen

It's not something in $\Bbb R^n$, it's something on $\Bbb R^n$

It's a $(n - 2)$-plane field.

Just like vector fields but stick a plane at each point

Akiva Weinberger

A plane with a maginitude

A little circle I guess

Balarka Sen

15:25

Well you don't always have a inner product imposed on you, unlike R^n

Akiva Weinberger

So when I integrate a surface over a 2-form (which I assume is what's happening next chapter), I shouldn't think of it like integrating it over a 2-plane field

but rather a vector field (in $\Bbb R^3$), or I guess an $n-2$-plane field

Still, though, how do I go from the 2-form to the vector field?

Balarka Sen

Oh, no, integration of a 2-form is integrating it over a 2-plane field

That doesn't have to do anything with the correspondence between forms and plane fields

Akiva Weinberger

OK so I'm still confused

Balarka Sen

Shall I explain the general perspective?

Akiva Weinberger

Sure

I mean, $dx\wedge dy$ is a function, right? How do I go from a function to a plane field?

Balarka Sen

15:32

Ok, maybe I'll answer that first. $dA = dx \wedge dy$ is a function on the space of 2-plane fields.

It eats a plane at the point $(x, y)$, spanned by vectors $\vec{v}$ and $\vec{w}$

And spits out $\det[v_1, w_1; v_2, w_2]$ I think

Akiva Weinberger

Sure

Balarka Sen

So when you say, integrate $dA$ over a domain $D \subset \Bbb R^2$

Akiva Weinberger

Wait, the choice of $\vec v$ and $\vec w$ should matter, no?

Or only insofar as their magnitudes

If they're unit vectors, does it not matter which ones we choose?

Hm, I guess not

Balarka Sen

I think $\iint_D dA$ means integral of $dA(e_1, e_2)$ over $D$

@AkivaWeinberger Yeah, that determinant is the volume of the parallelogram they span after all

Leaky Nun

the purpose of category theory is to show that what is trivial is trivially trivial

@BalarkaSen what do you think?

Balarka Sen

15:39

I think I want a chocolate.

That's more important than category theory.

Akiva Weinberger

Wait no hold on

We care about the angle between them

We're essentially projecting $v$ and $w$ onto the $xy$ plane and taking the determinant of their projections

Balarka Sen

Well, $v$ and $w$ are already elements of $T_{(x, y)} \Bbb R^2$

Akiva Weinberger

Of what?

Balarka Sen

The tangent space of R^2 at (x, y)

I mean they are tangent vectors after all.

Akiva Weinberger

OK?

In any case, I think it makes sense for me to think of $v$ and $w$ as perpendicular unit vectors

And they define a plane, or I'll think of them as defining a small unit circle in the plane they span

Then, I project them and their unit circle onto the xy plane

Balarka Sen

15:43

Mmkay.

Akiva Weinberger

That unit circle is now an ellipse, whose area is probably the cosine of the angle between the planes or something

Balarka Sen

Perhaps let's verify $\int_{\varphi(D)} \omega = \int_D \varphi^* \omega$ using that intuition.

Akiva Weinberger

But the point is, the area of it is the value of the function $dx\wedge dy$ at $(v,w)$

And then the zero set is the set of all planes (er, planes with magnitude) perpendicular to the xy plane

And then I guess to recover the actual xy plane I can look at whatever is perpendicular to everything in the zero set?

Balarka Sen

Wait, $dA$ has no zeroes on $\Bbb R^2$.

You're thinking about forms on R^n?

Akiva Weinberger

I was thinking of R^3

Balarka Sen

15:46

Ahh

Akiva Weinberger

But I guess it'd work anywhere

Balarka Sen

Sure

That works

Akiva Weinberger

So plugging in $(v,w)$ into $dx\wedge dy$ is kinda like "dot producting" the plane spanned by $(v,w)$ by the plane spanned by $(\hat\imath,\hat\jmath)$

but the plane spanned by $(v,w)$ has a magnitude determined by the area of the parallelogram determined by $v$ and $w$

So maybe I should think of it as dotproducting two parallelograms together

Balarka Sen

You could integrate 2-forms on non-horizontal domains in $\Bbb R^3$, though.

Akiva Weinberger

What do you mean by non-horizontal domains

Balarka Sen

15:51

Like, consider a map $D \subset \Bbb R^2 \to \Bbb R^3$ which has injective Jacobian (i.e., rank 2).

You could integrate $\omega$ on $\varphi(D)$

Akiva Weinberger

OK

My guess is,

$\int_{\phi(D)}dx\wedge dy$ is gonna be the area of the projection of $\phi(D)$ onto the $xy$ plane

Balarka Sen

Mhm, sounds right. It's $\int_D \phi^*(dx \wedge dy)$ by change of variables.

Akiva Weinberger

Are those two sentences linked

Balarka Sen

So it's $\int_D \det[d\phi] dx \wedge dy$, I guess

Wait, is that really area of projection of $\phi(D)$ on the xy-plane?

No, I doubt this.

Akiva Weinberger

It's not $\det[d\phi]$ @BalarkaSen

$d\phi$ isn't a square matrix

Balarka Sen

16:01

Oh, I mean the matrix $\partial \phi_i/\partial x, \partial \phi_2/\partial y$, right.

Akiva Weinberger

which is the determinant of the derivative of the projection of $\phi$ onto the $xy$ plane

Balarka Sen

Ah, makes sense.

Ok, glad that worked out

@Akiva You see how bad I am at forms? :p

Akiva Weinberger

Now I'm confused again, though, 'cause I thought I had it worked out why 2-forms are planes, but now I realize I have no idea what plane the form $dx\wedge dy+dz\wedge dw$ should be

Also, my phone's autocorrect freaks out whenever I start typing equations

"Yes, I know 'dxdy' isn't a word, just shut up for a moment"

Balarka Sen

On a tangent space in $\Bbb R^4$, it's zero set is given by $\det[\vec{v_1}, \vec{v_2}] + \det[\vec{v_3}, \vec{v_4}] = 0$

Rickyfox

I'm vectorizing a function in order to avoid a big loop. In the iterative version of the function as I use it in the loop, it's quite straight forward, multiplying a 1x15 vector with 15x15 matrix.
Assuming I have 100 instances and want to compute this without a loop I have 100x15 matrix and for each of these 100 rows, I have a 15x15 matrix by which the respective row has to be multiplied. So for the latter I have a three dimensional matrix - is there an operator to perform such a computation?

Akiva Weinberger

16:06

@BalarkaSen Does it make sense to ask for the angle between two planes in 4-space?

I know you can always ask for the angle between lines in any dimension.

(Actual planes this time)

Balarka Sen

@Akiva I think if you have two planes spanned by $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ then the angle $\theta$ between the planes is defined to be $\det(\vec{a}, \cdots, \vec{d})$ divide by $\det(\vec{a}, \vec{b})$ and $\det(\vec{c}, \vec{d})$.

This should be independent of the bases chosen

Akiva Weinberger

$\det(a,b)$ isn't defined though

Eh, area of the parallelogram they span, close enough

We'll just lose sign information

Semiclassical

I’m forgetting. Is there in general a correspondence between forms and planes? In 3D there is

Akiva Weinberger

Well that's my question

Balarka Sen

@Akiva Yeah I'm sorry, I meant the Gram matrix.

Akiva Weinberger

16:11

Googles

Balarka Sen

@Semiclassical Yep.

Zero set of the form gives a plane field.

Semiclassical

I’m vaguely remembering that the word Grassmannian has something to do with this

Balarka Sen

Nothing beyond the fact that they parameterize the space of plane fields.

Semiclassical

Hmm.

Akiva Weinberger

> Geometrically, the Gram determinant is the square of the volume of the parallelotope formed by the vectors.

Cool

$\det(V^\top V)$ apparently

Balarka Sen

16:13

Yeah.

Akiva Weinberger

@BalarkaSen Right, so we still need to figure out what plane $dx\wedge dy+dz\wedge dw$ corresponds to

Balarka Sen

I just said what it is on a tangent space!

Akiva Weinberger

But that's not a plane

Balarka Sen

Sure it is!! Determinant is a multilinear function

To be clear let's explicitly write this out

Say (a, b, c, d) are the four vectors

If you feed it in your form

Alessandro Codenotti

Hi chat

what's going on?

Akiva Weinberger

16:16

I want to know what plane the form $dx\wedge dy+dz\wedge dw$ corresponds to

I'm pretty sure this is equivalent to writing it as the wedge of two 1-forms tbh

Balarka Sen

Sorry, I'm bad. It eats two vectors (a, b)

And spits $|a_1, b_1; a_2, b_2| + |a_3, b_3; a_3, b_4|$

Akiva Weinberger

Incidentally, the fact that $(dx+dy)\wedge(dx-dy)$ doesn't obey the difference of squares law feels weird

Balarka Sen

So the plane is just given by setting that equal to 0, I guess

That's a bi-linear function on (a, b)

So it is a plane.

Akiva Weinberger

Wut

A plane is a flat 2D surface

And in this context it has a magnitude and orientation I guess

Balarka Sen

That's a flat 2D surface. You're looking at pairs of vectors $(a, b)$ satisfying $T(a, b) = 0$ for some bi-linear function $T$. If $(a, b)$ and $(c, d)$ are two such vectors, $T(a + c, b + d) = T(a, b) + T(c, d) = 0$.

Daminark

16:21

@BalarkaSen ur a plane

Balarka Sen

So the zero set is closed under addition.

It's a vector subspace.

What the hell are you talking about

Akiva Weinberger

I'm so confused

I mean, like, a 2D subspace of $\Bbb R^4$

This sounds like a subspace of $(\Bbb R^4)^2$, no?

Balarka Sen

That's true, that's true.

Akiva Weinberger

OK look

So I did some Googling, and apparently the word I wanted was "bivector"

Balarka Sen

bi-vector you mean

Akiva Weinberger

16:32

> Not all bivectors can be generated as a single exterior product. More precisely, a bivector that can be expressed as an exterior product is called simple; in up to three dimensions all bivectors are simple, but in higher dimensions this is not the case.

> Geometrically, a simple bivector can be interpreted as an oriented plane segment, much as vectors can be thought of as directed line segments.

Balarka Sen

But bi-vectors are wedge product of vectors, not forms.

Alessandro Codenotti

I finally understand why in a metric space we have compact iff complete and totally bounded! Should have looked up this proof a long time ago

Akiva Weinberger

Yeah, but you can associate 2-forms to bivector fields, no?

Balarka Sen

I guess k-forms eat k-vectors

Akiva Weinberger

In any case, the thing I wanted is not a simple bivector, so

I mean, I assume it isn't

So it doesn't correspond to a plane

Balarka Sen

16:34

@AkivaWeinberger 2-forms are functions on the space of bivector fields.

Look, this is a difference between V^* and V

Akiva Weinberger

But there's a one-to-one correspondence between 2-forms and bivector fields, no?

In subsets of $\Bbb R^n$ at least

Balarka Sen

Yes, just like there is an isomorphism between V^* and V (for f.d. V), just upto a metric.

Eric Silva

Y'all talking bout multivectors?

Balarka Sen

Yeah we're confusing ourselves because we're doing it in R^n.

There's waaaay too many dualities on R^n

Akiva Weinberger

I like the fact that dot products exist

Eric Silva

16:39

difference between 2-forms and 2-vectors is just $\wedge^{2}\mathbb{R}^{n\ast}$ vs $\wedge^{2}\mathbb{R}^{n}$

Akiva Weinberger

You have a linear function from $\Bbb R^n$ to $\Bbb R$, I'm like, "That's just dotting with this vector $\bf a$ over here"

Easy to think about

Balarka Sen

@Eric Akiva wanted a correspondence between 2-forms and 2-plane fields

(Which doesn't exist because, as he points out, bivectors do not correspond to planes always)

Akiva Weinberger

The real word is "simple bivector", as I've since learned

Eric Silva

i see

Akiva Weinberger

I guess this is kinda like how double rotations exist in 4D

Eric Silva

16:40

well in $\mathbb{R}^{2}$ and $\mathbb{R}^{3}$ all 2-vectors are simple

but yeah in 4d ur f u c k e d

Balarka Sen

tru

Akiva Weinberger

In 3D it's fine to represent rotations as planes but in 4D you can't

And you can add planes in 3D and you always get more planes

Hold on, wait, so

I can think of bivectors as rotations

and then 2-forms are bivector fields ('cause $\Bbb R^n=\Bbb R^{n*}$), so I guess 2-forms are rotations at every point?

But now I'm dizzy

Eric Silva

that equality stresses me out

Balarka Sen

I hate these noncanonical ways you're identifying everything with everything

Akiva Weinberger

It's canonical in $(\Bbb R^n,\cdot)$, isn't it?

Eric Silva

16:43

yeah

Akiva Weinberger

I don't need to specify a coordinate system, I just need angles and shit

Eric Silva

for the things I've had to use $k$-vectors for it's important to have them distinguished from $k$-forms for sanity reasons

Balarka Sen

Sure but

That ^

it's breaking my mind

Eric Silva

$k$-vector fields don't really show up if you're not gonna integrate forms in my experience

Akiva Weinberger

I guess I think visually and I need some sort of image in my mind for k-forms

Balarka Sen

16:46

Why not as a volume operator on k-vector fields

Akiva Weinberger

"Volume operator"?

Balarka Sen

Eats a k many vectors, spits bars about their volume

A$AP k-form

Eric Silva

lol

Akiva Weinberger

"Spits bars" lol

Mike Miller

@Akiva I think visually but I like to think of k-fields as just associating volumes to k-manifolds... where we take the weighted average over the manifold weighted by the k-form

Akiva Weinberger

16:47

("Uh, summa lumma dumma lumma")

Balarka Sen

don't quote Eminem

Mike Miller

I think there's a visual enough picture here, draw a submanifold, color the tangent planes, asdign numbers to those

Balarka Sen

he got Revived

he's an outcast now

Mike Miller

Then that gives you a function on the manifold to integrate, more or less

Akiva Weinberger

I basically know one rap and it's that

Mike Miller

16:48

moms spaghetti

Akiva Weinberger

Well, actually only the fast verse tbh

Mike Miller

Artist: Apologetix Song: The Real Sin Savior

►

Akiva Weinberger

I can't give m'f'ing audiences a feeling like they're levitating :(

@MikeMiller "Color the tangent planes"?

Balarka Sen

Aesop Rock - Hot Dogs (Official Video)

►

this is good

Akiva Weinberger

Is $f(x)dx$ always closed?

Oh, it's exact isn't it 'cause it's just $dF$

Mike Miller

16:53

@AkivaWeinberger The picture I have had the tangent plane for each point "shine" a little brighter than the rest of the planes in the Grassmannian at that point. To each we associate a number in any case

This is almost certainly unparseable

mb

Akiva Weinberger

They shine brightly if the form gives them a high rating?

Mike Miller

it's a weighted average, that's all i got for you

Yeah

I started half-assing the description after I decided it wasn't helpful :P

Akiva Weinberger

Is $2xydx+x^2dy$ exact

Ah, it's $d(x^2y)$

Mike Miller

yes closed forms on R^k are exact

Gabriel Romon

@BalarkaSen since when did you become a memester ?

Balarka Sen

16:56

hahah some time ago

Akiva Weinberger

@GabrielRomon Have you met him

Gabriel Romon

@AkivaWeinberger he was very different a few years ago

Mike Miller

yes

Akiva Weinberger

OK so this exercise, after throwing away exact bits, is asking for $\oint_C2xydy$ on a certain contour

Mike Miller

he's degraded but improved

Eric Silva

16:58

@MikeMiller this is basically how i picture rectifiable varifolds

Akiva Weinberger

@GabrielRomon Hm, whom do I blame then

Dami probably

:P

Mike Miller

I started it but I gave epsilon to Dami's 1-2epsilon

Akiva Weinberger

Re: This exercise, it's not exact so I guess I'm meant to just parametrize the thingy and actually plug it in

and actually do work

So I'm not gonna finish it

(It's 8.3.16 if you're curious)

Balarka Sen

8.3. has some good ex's i haven't thought about

I like the planimeter though

Akiva Weinberger

Yeah I haven't done that one yet

I've also never understood planimeters

Balarka Sen

17:04

It's pretty much what you were talking about before

Daminark

@MikeMiller that is an interesting way of phrasing it. Not inaccurate for sure

Leaky Nun

so $PSL(7^2)$ is generated by $\overline{\begin{bmatrix}1&1\\0&1\end{bmatrix}}$, $\overline{\begin{bmatrix}4&0\\0&1\end{bmatrix}}$, and $\overline{\begin{bmatrix}0&-1\\1&0\end{bmatrix}}$, which acts on the 1-dimensional subspaces of $V(7^2)$ labelled $\{0,1,2,3,4,5,6,\infty\}$ with cyclic notation respectively $(0123456)(\infty)$, $(0)(142)(635)(\infty)$, and $(0\infty)(16)(23)(45)$. So, this gives us an injective representation on $V(2^8)$.

I have checked that the orbit of $0+1+2+4$ is a $4$-dimensional subspace. However, how can I confirm whether the resulting representation is still injective?

Akiva Weinberger

PSL = Pakistan Sign Language

incidentally

Leaky Nun

who should I ask?

Balarka Sen

@Daminark u want to know the pf of structural stability

if not go to gulag

Daminark

17:11

Structural stability of what?

Balarka Sen

@Daminark Oh I pinged you something here

I guess the ping didn't go through

Akiva Weinberger

@BalarkaSen Is that Tintin?

Balarka Sen

yeah the first book

in the land of soviets

nobody shall speak of the second book, however

Daminark

Ah, maybe tonight?

Balarka Sen

For sure!

We also encountered your master theorem from Brin-Stuck in K-H

Daminark

17:18

I see

Balarka Sen

It's easy to get Stuck in Brin-Stuck but reading Katok-Hasselblatt is a Hassel

Every frigging dynamics textbook you see a bad pun coming in

Daminark

I don't remember puns in mine. What did yours have?

Balarka Sen

I just wrote them down!

not intentional puns, just puns on the names of the authors lol

Hirsch-Pugh is just Hirsch with the readers

Akiva Weinberger

@BalarkaSen There's an exercise that says, if a loop intersects itself finitely many times and doesn't hit the origin, then [the integral that determines the winding number] is always an integer

($\frac1{2\pi}\int_C\frac{-ydx+xdy}{x^2+y^2}$)

Why do we care that it intersects itself finitely many times? Is it just to make the exercise easier?

Balarka Sen

I think so, yes.

Well

I'm worried about things like the Hawaiian earring

Yeah that causes problems

@Akiva

It's not immediately clear to me how to prove the integral definition of winding number can deal with computing the winding number of the loop traversing the Hawaiian earring.

Akiva Weinberger

17:40

@BalarkaSen That hits the origin

Also, sorry, my internet died

Balarka Sen

I know. Just translate it so that it doesn't.

Akiva Weinberger

But yeah, I see how he wants us to do it - use the self-intersections to break it into finitely many pieces, each of which is homotopic/homologous to a circle

Balarka Sen

Yes, exactly.

Well, after making it transverse to itself if you want to be super-rigorous

But yeah bottom line every intersection looks like an "X" after making it self-transverse

Akiva Weinberger

If, instead, he said it alternates between the positive $x$-axis and the negative $x$-axis finitely many times, that would make it easier as well

And that's always true anyway

Balarka Sen

Ya

Akiva Weinberger

17:43

('Cause it's exact on the plane minus the positive y-axis, and its exact on the plane minus the negative y-axis)

Alessandro Codenotti

With the finite intersections thing it breaks the plane in an unbounded part and a finite number of bounded parts

Akiva Weinberger

('cause it's $d(\arctan(y/x))$, as he shows in an earlier example, so you can mess with the output range of the arctangent)

Balarka Sen

It's exact on the complement on the negative reals on an appropriate branch, sure.

So I guess an alternative way to do it is to count the number of oriented intersections with the negative real axis

Akiva Weinberger

Sorry, my internet is still bad

I'm back

Balarka Sen

If it hits the negative real axis $n$ times, order it using the order on the reals by $p_1 < p_2 < \cdots < p_n$. If $p_i$ and $p_{i+1}$ are two consecutive points with opposite oriented intersection w/ the negative real axis, cut the loop off at those points and cap them off by lines to cancel them

Akiva Weinberger

17:50

Nah there should be a way to do this without caring about orientations of intersections

I knew this at one point

Trying to remember

Balarka Sen

Do it for all points so that finally every point has +ve oriented intersection

Or all -ve

you don't need to do this to prove it's an integer, anyway. it's a sum of $\pm 1$'s

Akiva Weinberger

We don't know that it hits the negative real axis finitely many times, only that it alternates between the two rays finitely many times

to do it in full generality

Oh, OK. So you break it into finitely many pieces, on each of which it either only hits the +ve $x$-axis or only hits the -ve $x$-axis

i.e. on each piece it doesn't hit one of the rays

And then on each piece it's contained in either $\Bbb R^2\setminus[0,\infty)$ or $\Bbb R^2\setminus(-\infty,0]$

and it's exact on both of those things (mess with the range of the arctangent in the $d(\arctan(y/x))$ thing)

Balarka Sen

@AkivaWeinberger That's easily fixed, I think. Make it transverse to the x-axis.

Then it hits the x-axis at a discrete set

If it's not finite, you'd have an infinite discrete subset of [0, 1]

Which is bull

Akiva Weinberger

I don't need your transversiness here

(Infinite compact discrete…)

philmcole

Hello. Can I consider $\mathbb R$ as a closed interval?

Akiva Weinberger

17:58

Sure, it's an interval because it's $(-\infty,\infty)$ and it's closed because that's how topologies work

Balarka Sen

Sure, I'm not trying to find the most elementary proof

Akiva Weinberger

Yeah but I am

philmcole

I want to give an example for an unbounded, continuous function on a closed interval. And the only thing that comes to my mind is some function on the whole of $\mathbb R$.

Ok thanks!

Akiva Weinberger

In any case now you just have this telescoping sum if you do things mod 1 ('cause the two potential functions are equal mod 1), so mod 1 it's zero, or whatever

so it's an integer @BalarkaSen

Balarka Sen

ok

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$Mathematics$ Mathematics