@BalarkaSen

@BalarkaSen: want to help me with basic geometry? I'm trying to solve some exercises in a new textbook and I can't ^^

Shoot!

@BalarkaSen: just planar geometry. fix some point $A$ and a line $g$ that doesn't contain $A$. take a point $B \in g$ and now construct $C$ such that the triangle $ABC$ is equilateral. now, if you decide to move $B \in g$ (while leaving $A$ fixed), you will see that all generated $C$ lie on a single line. why?

(ideally a solution not even using similarity and everything that would be taught after that)

hi chat

@Huy Hmm seems like a straightforward exercise. Let me see

Heya @Huy, @Alessandro, @Balarka — long time no see, Huy!

Hi @Ted

Hi @Ted

hey @Ted! I'm fighting my way through high school geometry textbooks and realizing how difficult geometry is.

It can be more interesting than we realize. I highly recommend Pedoe's book Geometry: A Comprehensive Course, if you don't already know it.

@TedShifrin: I'm starting with a brand-new textbook that some Swiss teachers wrote first, for obvious reasons

Hi @TedShifrin!

but thanks a lot for the recommendation, will look into it when I have more time

Brand new is not always fun ... I usually taught out of my books about 4-5 times before publishing.

heya @Ali

it was used for teaching many times before being published, of course

Ah, then it should be fine.

it has surprisingly many challenging and interesting exercises, so I'm trying to solve all those and choose the ones best suited for my geometry class

@Huy: I have an issue with that problem. There are two choices for $C$. What if you don't choose $C$ in a consistent manner as $B$ varies?

Geometry was always my worst in the few math competitions I did in high school

ah, choose $C$ such that the triangle always has the same orientation

I love Geometry at the moment

Aha ... Does the problem say that? :D

no, but it is implied because writing $\Delta ABC$ for a triangle always implies positive orientation over here

Obviously, we can do an analytic geometry solution, but you want one just with geometry?

Hey @BalarkaSen do you remember me ranting about Quivers a while back?

/rant

It's easy if you allow analytic geometry

yes, it comes up in one of the first chapters which deals with isometries

So we want to think about the geometric construction of $C$.

@Ali I do indeed

(even before similarity, Pythagoras, etc.)

Hmm ...

@BalarkaSen OK so I slightly more know what I am talking about now.

Quivers are of interest in algebraic geometry because their representations are a very convenient setting to trickle interesting geometry out of things

Uh, I see

For example categories of quiver representations (which are basically stick a vector space at a node, linear map at an arrow) are similar enough to some very geometry categories

For example if you want to understand P1

you can actually construct some nonsense about quiver representations

Then because representation theorists know their stuff you can say stuff about P1

for projective varieties its very convenient

but it doesn't stop there

Another one is some geometry objects look like moduli spaces of quiver representations so you can understand a variety just by pretending hard enough that its talking about quivers

and you give it to that poor rep theory guy to sort out

but yeah thats just a small crumb about why people care about quivers

Because I gave a very unsatisfactory explanation last time and its annoyed me since

From a more algebraic point of view, every algebra is basically a path algebra if you try hard enough, so you can argue that quivers arise naturally, if you abuse the word enough.

Okie

Hi @Ted! :)

hi @mikeonly, whoever you are

@Huy Wow this is complicated

I know right?

looks so simple at first

@Huy: I worked it out analytically, but now I'm trying to think how I'd discover that line.

wanna now something fun? it's not even considered one of the "challenging" exercises by the author

What book is this?

but I think it's much more difficult than the other problems, unless I am missing something extremely simple

@AliCaglayan I would say that "every algebra" is stretching it a bit

It's actually a nice problem, conceptually. I can wave hands at it — as $B$ moves along the line, you get a pair of pencils of circles, whose intersection point(s) is what you want to track.

@BalarkaSen: ofv.ch/lernmedien/detail/geometrie-1--inkl-ebook/101802 (German book)

@TedShifrin It's only non-trivial because of the lack of tools, I guess

No, but there should be a nice way to see it, @Balarka.

the problem is in section 2, translated "symmetries with respect to lines and reflections, symmetries with respect to points and point reflections, rotational symmetries and rotations, translations"

@Huy: For example, Start with $B$ the closest point to $A$ on $g$.

ok

hi chat

Aha ... Duh. So the line is what you get when you rotate $g$ $60^\circ$ about $A$.

Duh. Duh. Duh.

Oops.

lol

slinks out of the room

I guess I'm really not geometric, after all. hides

dear lord

this is one of the "how was this not obvious?" moments

Uh huh.

Once you listed the topics, it became obvious.

i'm not sure which I find more tedious, math cranks or physics cranks

Stop complaining about us, Semiclassic.

lolno

lol

physics cranks are more entertaining

i'm certainly not talking about this chat

math cranks are garbage cranks

Hi @MikeMiller

Good night, @MikeM

@Huy: Actually, I'm particularly miffed at myself because there's geometry like this in finding one of my favorite points in a triangle. Do you know about the Fermat point?

mmm, steiner trees

no, what about it?

I listened to a talk by Witten yesterday for a general audience, and it was nearly indistibguishable from a crank talk. Minus some indignation.

What point $P$ has the property that it minimizes $AP+BP+CP$?

You can interpret/find it physically, too.

Was it a resurgence theory talk?

I like that execise

I've given about 5 lectures on it to various audiences, @Balarka.

@TedShifrin my advisor gave me the four point version of that as an exercise in spontaneous symmetry breaking

@KevinDriscoll cranks don’t have such soothing voices

Right, @Semiclassic — we've discussed this before.

lmao @Mike

this is r/nocontext material right there

@KevinDriscoll on that note, have you listened to any more of the resurgence talks?

I heard Witten give the plenary lecture at an AMS meeting. I found it pretty much incomprehensible.

lol

This was probably 20 years ago.

@TedShifrin: should this be solved non-analytically? because analytically I don't think it is too hard

Yeah it can be done analytically

What do you mean by analytically?

You can solve it by multivariable calculus, for example.

But actually construct the point!

I mean it's a minimization problem

hm. I'll think about it.

Haha no, does he work on that? It was a talk for a general audience that was sort of a summary of what work he was doing from 2010-2011. Mostly things about basics of knots, jones polynomials, khovanov cohomology and their relation to Chern-Simmons and some more complicated 3+1D theories

how much knowledge of geometry do I need to construct it?=

tbf, computing steiner trees analytically for more than a few points is reeeally hard

but for the three point case it shouldn't be terrible

It's a fascinating question, @Huy. The other amazing thing (which I only discovered just in time to put it in my multivariable math book) is that if you take the circumcenter, incenter, and centroid, they're all collinear and one of 'em divides the other segment into a 2:1 ratio. (I may have the points wrong. I'll check.)

@Huy: Nothing besides the kind of thing we were just talking about.

"The other amazing thing is [weird triangle center stuff]."

But you have to know which point you want to construct first :P

Triangle centers are weird.

Cool, but weird

@TedShifrin: yes, I think that's Euler's line (in German)

Right, I think the centroid divides it into 2:1

I never studied any of that stuff or the 9-point circle. Too bad. Beautiful stuff.

That's not hard to prove

No, not incenter. Intersection of altitudes.

Using vector algebra, it's all beautiful ... I'm agonna do this with my 9th graders precalculus class :P

@KevinDriscoll But those things you mentioned were generally way before 2010

Barycenter?

we have a book of super hard geometry exercises that old crazy maths teachers used to throw at young and inexperienced maths teachers to intimidate them

No, centroid is barycenter, isn't it?

@KevinDriscoll I know he's done some stuff related to resurgence theory

Oh yes you're right

in particular, I know he's talked about lefschetz thimbles

@Huy: You're gonna be lots of fun this next year :P

and resurgence people like that as a way to handle stokes phenomenon

I guess it is the same as the orthocenter

@TedShifrin: ?

No, orthocenter is intersection of perpendicular bisectors, @Balarka.

With geometry questions, @Huy.

@Semiclassical I did watch a few of the resurgence talks at KITP. Your integral came up. But nothing truly comprehensible.

Let $A \subseteq X$ be an uncountable set, $\mathcal{B}$ a countable basis for for $X$. Then each point $x$ has a countable basis $\mathcal{B}_x$. But $\bigcup_{x \in A} \mathcal{B}_x \subseteq \mathcal{B}$ is a uncountable union of countable sets which is a contradiction unless....What? What can I conclude from this? That some of the sets in the union are equal; that they overlap?

oh, right

yeah, I figured

the rush to field theory is what I always find frustrating

@TedShifrin Those two points agree

@user193319: You use at least one set (probably more) uncountably many times.

9th grade has "algebra" and geometry classes. the first time I taught them, I hated geometry because it was so much to prepare. now I'm doing it for the second time and it is so much more fun than algebra and my students love it much more too.

No, @Balarka.

@TobiasKildetoft Ya Khovanov was in like 2000 I think. The thign Witten was doing specifically was trying to find a more direct or natural 3+1D theory related to Khovanov cohomology than the one that had already been constructed a few years earlier

even though we both agree that it is very difficult.

It's much more interesting and not just about writing formulas, @Huy. I'm trying to put more geometric stuff into the precalculus course I'm teaching than the people originally wanted. They're fond of tricky algebraic manipulations of trig formulas. That's fun, but ultimately less useful if you're going to actually use trig in physics and math.

@KevinDriscoll I never really learned much about Khovanov cohomology (and especially not how it related to physics). It just keeps being mentioned as motivation for categorification.

Ya I know literally nothing about it except the name and that it associates vector spaces to .... something. I don't have any idea what cohomology is.

@Balarka: For example, draw an obtuse isosceles triangle to make it clear.

@TedShifrin Wait. Perpendicular bisectors intersect at the circumcenter, does it not?

Yes, @Balarka. But you said the altitudes did.

At least, I thought you said that.

I know cohomology in only the most baby form

How do you insert emojis in here? :smile:

@KevinDriscoll It associates chain complexes to knots such that the graded euler characteristic is the Jones polynomial (or maybe it is just graded cohomology)

@ChinmaySarupria: We type :D.

@ChinmaySarupria: you get an overview of all available emojis by pressing ALT + F4

Hmm, what on a Mac?

e.g. being able to construct a scalar potential on a subset of 3D space tells you about the topology of that subset

@Huy I know what happens when you press Alt + F4

@TedShifrin Yeah I'm getting confused. So am I right to say orthocenter is the intersection of the altitudes?

or something like that. i'm not really thinking properly right now

@TedShifrin That didn't work, did it?

Ugh, @Balarka. Now I'm getting confuzled.

Haha. There's too many points flying around

Probably that's right. I forget what the orthocenter is.

Yeah, that's the definition.

It's just not the center of an interesting circle :P

Yeah

I was thinking of the inscribed triangle

Anyhow, I love that 2:1 thingy that shows up yet again.

The center of which is a different point

That's the incenter. Intersection of angle bisectors.

Right; there's a small geometry argument there.

By drawing the six subtriangles and fiddling around

@Semiclassical About right.

the better description is in terms of closed and exact forms yadda yadda

You ask if a given vector field is the gradient field of something

right.

@Semiclassic: Yeah, we can discuss electrostatic versus magnetic ($H^1$ versus $H^2$) sometime if you want.

i suspect I know quite a bit already from the physics

Of course.

hmm

It's all just language.

are H^1 and H^2 related in the R^3 case? I want to say poincare but I'm not sure if that's just me being tired

Yes, Poincare.

kk

Compact, please ... or compact supports at least.

Right, of course, with compactly supported cohomology.

heh, not all the physics cases respect that insistence

e.g. potential nonzero at all finite x but vanishing at infinity

Hai

Heeeeere comes potato! @Daminark

howdy Demonark

though if the point is that one can do without such cases then I can buy that

Daminark, of course, would define $H^n(X; G)$ as $[X, B^n G]$.

Peter May skitz

And let's see him compute for us.

Lol, I mean all I want is to build a few definitions, show they're equivalent, and then use whatever is best in a given context

Actually you can prove the Eilenberg-Steenrod axioms from $[X, B^nG]$

that's...nice

It's a good exercise

I worked it out once upon a long long time ago

I'd like to see one understand Poincaré duality in terms of transverse intersection of submanifolds using Peter May's way.

Oh, hah, rekt. Hm, I think there is a way.

Um, no, not that I know of.

Algebraists don't care about it.

Cup product is really complicated to understand in that definition. It corresponds to taking smash products

Even Peter May believes quite strongly in being eclectic, surprisingly enough. He says he prefers classifying space business to cochain complexes as the starting definition because apparently a cohomology theory only has this cochain business if it's ordinary cohomology

main place I've seen poincare duality is in the case of graph duality

i think that's what I'm dimly remembering, anyway

I'm not sure I stumbled over that, Semiclassic.

you probably have, I'm just saying it weirdly

So he wants to first present the definition that generalizes nicely, and then derive cochain in the case of ordinary stuff. I'm not deep enough into this stuff to say if I think that's the best method of presentation or not, both sides have something going for them

i have in mind: take a planar graph, put a vertex inside each face, and draw edges between vertices of adjacent faces

Right, dual cell structures

Oh, that's the way Poincaré understood it.

As Balarka just said. It's rather beautiful.

obviously that's the very simplest case

I thought you meant just plain graph theory.

but still a good one

nah

right now the duality i'm struggling with is why this FFT doesn't make sense to me

to which the answer is probably "you're doing it wrong"

I have a physics/geometry question, when writing the Laplacian in $\Bbb R^n$ in polar coordinates we get $\frac{\partial^2}{\partial r^2}+\frac{n-1}r\frac{\partial}{\partial r}+\frac1{r^2}\Delta_{S^{n-1}}$, where the professor said that to properly make sense of $\Delta_{S^{n-1}}$ we'd need to do some geometry and to avoid that we can use the trick of setting $\Delta_{S^{n-1}}u=\Delta U\big|_{S^{n-1}}$ where $U$ is defined as $U(x)=u\left(\frac x{||x||}\right)$ for $x\neq 0$.

What's the proper way to treat this laplacian without resorting to this trick?

Is (-1)^pi a nonsensical question, or merely indeterminate?

depends on how you define it

It is not a question, it's an expression, and is sensible, yes.

@Alessandro: You can define it for any Riemannian manifold.

You need to use the metric.

but the typical interpretation would be $(-1)^\pi =(e^{i\pi})^{\pi}=e^{i \pi^2}=\cos(\pi^2)+i\sin(\pi^2)$

Have you learned diff forms yet?