@Ted: One thing that's nice to know, but I feel like nobody ever writes down (it's probably in Hirsch as an exercise, of course) is that every bundle has a section transverse to the zero section (or more generally you can achieve transversality of sections of a fiber bundle).

I remember figuring that out in graduate school. It's a great exercise. How to wiggle a section and turn it back into a section.

hi .are all subgroups of $Z_4xZ_6$ normal subgroup?

I would have to check notes, but I think I showed the G&P class that in the context I was mentioning to @Balarka.

@TedShifrin I'll probably assign it to our friend.

LOL

I truly should retire completely now. Interesting conversation going on with Ivo Terek in answers/comments on one of his questions.

He pointed out that if a submanifold has nontrivial cup product with itself, it can't have trivial normal bundle. This leads IMO obviously to a couple things that I don't want to give him the names of, lest he dive into a rabbit hole.

But I do want him to be able to define, say, $e(E)$, whatever that is. ;) Whence the exercise.

Well, self-intersection, or cup product of P-dual. :P

@Albas entire functions have no zeros in the whole complex plane, e.g. exp(z). however modulo entire functions (and some fudge factors to secure convergence) every meromorphic has a weierstrass factorization (keyword to search)

@memonto what's your guess? it's abelian...

heya anon :)

yo

Another year down? :)

guess so. teaching algebra more over summer.

advisor had a week-long hopf-galois module theory seminar again, got to see a bunch of big names at his pool party at the end.

well, that's cool. They should just have you teach the majors algebra course :P

@arctictern thank you. gn :)

@MikeM: G&P do define the Euler characteristic as the self-intersection of the diagonal in $X\times X$, and then there's the exercise to see it's the self-intersection of the zero-section of $TX$ via the tubular nbhd theorem.

To define that, you needn't intersect with a section, of course, but if you have one ...

@TedShifrin Right.

set theory question: what set is equal to its complement? Is it even possible?

What do you think, @paulpaul?

I think not

@MikeM: He will get to essentially that in G&P, so be patient.

For the tangent bundle, but it translates more generally.

I forget, they actually do the Whitney idea?

Good, @paulpaul.

No, @MikeM. They move all the zeroes of a section into a little ball and then use the Hopf Degree Theorem.

@paulpaul: I can think of only one possible scenario, where your universe is empty.

The same proof works in an arbitrary setting but I find it a little less tasty.

I guess I don't follow that remark. That idea works for any vector bundle, clearly.

(with the right rank, before you correct me)

I agree. I meant I like the Whitney trick more.

@TedShifrin then probability theory question: among the students who take probability theory class we pick one, let event A denote all students that are teenagers, B that they live in a dorm, C that they don't smoke, what is "A intersect B intersect complement_of_C" when B=complement_of_B?

Ohhh ... I think Hirsch does the same thing I said.

Huh? I bet that's a typo, @paulpaul, based on what we just discussed.

I believe it

@TedShifrin that's weird then. there are other questions, describe ABnotC when a)ABnotC=A -- my answer: all students that are teens live in dorm and smoke b) notC is a subset of B -- my answer: all students that smoke also live in a dorm, so we are picking from a set of students who are teens, live in dorm and smoke. c) notA=B, my answer--all students who live in a dorm are not teens, and we pick from the set BnotC, is it correct?

i actually screwed it up a little, because in the book it says B means "students don't smoke", lol, and C means "students live in a dorm", but anyways

for a) shouldn't you say all teenage students?

I'm not reading all of this, sorry.

yea, but according to ABnotC=A all students who are teenagers live in a dorm and smoke. But yea, we are picking a student from a set of all teenage students

Right, so you need to say that.

well, when notB=B i think that we don't have any students who smoke or don't smoke, that that's only possible when the set is empty?

the universal set

I remember a task "prove that all cars on mars are blue and red", the solution is:"the set of cars is empty, that's why they're red and blue"

Quick question: Does summation of lg(k) = Θ(n lgn) ?

whats k

is k is a natural number and you sum from 1 to n, then you get logN! which is approximately nlogn

yes, thanks alot

@Square-root you may be interested in stirling's approximation for n!

yea, that's what it's called

Hello

@DanielFischer Do you maybe have an idea?

Hi @iwriteonbananas

@BalarkaSen Hello

How's things?

Got a headache and hence having trouble focussing. Oh well

Have you ever done that barycentric subdivision exercise in Hatcher?

yikes

@iwriteonbananas which one?

two subdivisions turn a delta complex into a simplicial complex

there are a lot of them

@iwriteonbananas yep!

it's a good fact to keep in mind

Sweet, feel like telling me about it?

Sure.

There's two parts to it, right?

(1) one barycentric subdivision gives a delta complex where each simplex has distinct veritces

(2) if we barycentrically subdivide a delta complex where each simplex has distinct vertices, we get a simplicial complex

@iwriteonbananas That's right.

I think the first part was easy, unless I did something wrong

Did it by induction on the dimension of the simplices.

Anyways, I'd like to hear what you did.

Also, I wasn't able to prove (2)

I mean, you throw lines from the barycenter. Barycenter is distinct from anything else because the interior of simplices is left alone in delta complexes. And the ends of each of the two lines (two more vertices there) also cannot be identified because those live in a face of a codimension less, with one of them in the interior.

So those can not be the same either.

That proves (1). Does that make sense?

I am thinking triangles here but the same logic inductively works for higher dimensional simplices.

Yeah, sure, makes sense

So we want to prove (2) now. Remember that simplicial complexes are delta complexes where each face is determined by their vertices.

Yea

How does this work out for 1-simplices? Suppose you have two lines pasted along the two endpoints. You just mark the barycenters, i.e., midpoints.

That makes all the simplices distinct.

yep

Suppose you have two 2-simplices then with same vertices. They can just be made distinct similarly by just marking the barycenters in each and joining, and each triangle is distinct because of the existence of the barycenter which is not identified with anything else.

Each triangle will have vertex set {some vertex of the original triangles, some midpoint of some edge, some barycenter}. By construction no two things can have the same vertex sets, because the barycenter is not identified with any other points in the picture.

Are the two 2-simplices in from the once-subdivided delta complex?

Yeah, two 2-simplices with same endpoints, but each having distinct vertices.

So it can be e.g., two 2-simplices glued along the boundary, or just at the three vertices, or two vertices and an edge or whatever.

ok

So

Let's take two simplices in the twice subdivided complex

Uh-huh.

So they arise from the once subdivided complex by taking some simplex, marking its barycenter and drawing new edges. We take two of those new simplices.

What does it mean to say two simplices being the same if they have different vertices?

whoops i typed the wrong thing

If the two simplices have the same set of vertices, why are they equal?

I still don't see it

@iwriteonbananas Any 2-simplex in the resulting twice divided thing must have one vertex the barycenter of the simplex, one the barycenter/midpoint of a edge, and one an existing vertex.

Sure

@iwriteonbananas If there are two triangles with same set of vertices, then each of them must have the barycenter of some single triangle in common, because both of them has the barycenter as a vertex. That means both of those triangles are in the subdivision of some single triangle.

But then they can never be equal!

It all comes down to the fact that in delta complexes you leaave the interior of simplices alone, so barycenter cannot get identified with anything else. A barycenter of a simplex just lives alone.

Ok, I'm still a bit confused. What exactly are we trying to show? How did we show that?

We're trying to show after we subdivide twice there are no two triangles with the same set of vertices.

no two different triangles*

okay

Right. Distinct triangles.

So we start with two distinct triangles in the twice subdivided complex. They are of the form $[b,v_0,v_1,]$ and $[b,w_0,w_1]$ with $b$ the barycenter of a triangle in the once subdivided thing

and $v_0\neq v_1$, $w_0\neq w_1$

Yes.

Because $v_0, w_0$ are barycenters of the edges of each of the triangles.

Yes

Now what's the next step?

Also, no, you mean $[b, v_0, v_1]$ and $[b', w_0, w_1]$.

$b$ and $b'$ might be two different barycenters, who knows?

We started with two random triangles, I mean.

Sure, but that's easy because then the vertices certainly are distinct

No, don't be so quick, we'll get to that. This notational conflict will confuse you.

ok

Anyway, we're claiming the vertices $b, v_0, v_1$ are the same as $b', w_0, w_1$. That means $b = b'$ (you can easily convince yourself that $b = v_0$ or $b = w_1$ can't happen).

So $b = b'$ is the barycenter of the same triangle. That means these two triangles we began with are triangles coming from subdivision of the same triangle in the complex, yeah?

The logic here is faulty

We started with distinct triangles and we're trying to show that the sets of vertices are not the same

No.

We're doing proof by contradiction. We started with two random triangles, and we're trying to show that having same vertices implies same triangles.

Ok, sure we can do that

Sorry, got distracted.

So, do we agree that the triangles come from the same subdivision?

no prob

Yeah, agree

When can two triangles coming from subdivision of the same triangle have the same set of vertices?

Well, they must be equal, but why?

We have a big triangle $T$ and two subtriangle $T_1$ and $T_2$ in the subdivision. We assumed the vertices of $T$ are distinct. You should prove if $T_1, T_2$ has the same vertices then $T_1 = T_2$.

The assumption that the vertices of $T$ are distinct is essential here. Otherwise I can take, like, a cone obtained from pasting two sides of a triangle.

Yes

Do you see why this is true?

I see it in the pictures I draw but how do I prove it explicitly? Why's $T_1=T_2$?

The vertices of any subdivisional triangle of $T$ is {barycenter, midpoint of an edge, vertex}.

sure

@iwriteonbananas "Barycenter" vertex of both $T_1$ and $T_2$ are same by construction.

Actually...

Mhm?

since the vertices are distinct

Yep, go on.

and the two triangles have the same vertices

of course they coincide

both are the convex hull of the same set of vertices

If both the "midpoint of an edge" and "vertex" are the same then $T_1$ and $T_2$ must have an identical edge, because edges are identified by a linear homeomorphism.

@iwriteonbananas yes, but there's some trouble involved in putting that "of course" into words :P I am sure you can do it.

It's not hard.

Ehh

A triangle with distinct vertices in a delta complex is just the convex hull of its vertices, no?

so two triangles with the same set of distinct vertices are the convex hull of the same set of vertices, so they coincide

There's more words involved, because there's the thing about quotienting vertices involved.

In particular you haven't made clear how you used the assumption about $T$.

It's visually quite clear, it's just troubling to put it into words, is all.

I mean, you're not just looking at a triangle you're looking at the image of the triangle under the characteristic map. So appropriate justification is needed to use the thing about convex hulls, 'cause that only works in R^n, where everything is unidentified.

True I guess

Here's a proof with words. $T_1$ and $T_2$ must share the same vertex of $T$, by the assumption about distinct vertices of $T$. Thus for $T_1$ and $T_2$ to be distinct, the midpoint vertex of each of $T_1$ and $T_2$ must be the same. They can't be image of two different vertices under the char map, because otherwise two vertices at the other end of the edge containing those midpoints must be identified (because edges are pasted by linear homeomorphisms).

So $T_1$ and $T_2$ share the same midpoints of an edge.

That implies $T_1$ and $T_2$ are the same.

that moment when you check the date of a question after answering it

I really don't know what you just did, @BalarkaSen

"They can't be image of two different vertices under the char map" I meant "... be image of two different midpoints under .."

@iwriteonbananas You should just figure it out. It's hard to explain this without a picture.

I am just using the fact that if image of two vertices and midpoints of two different edges of a triangle under the char map are the same, then the two edges must be identified under the char map (because char map identifies edges by linear homeomorphisms), contradiction to the assumption about $T$.

@BalarkaSen Hm

I mean what we're really trying to show is: If $\sigma_1,\sigma_2$ are any two n-simplices in a delta complex and they have the same set of distinct vertices, then $\sigma_1=\sigma_2$

That's not true, though. Take two triangles and paste by identity along the boundary.

God damnit

You're confused on a rather geometrically obvious point. Maybe take a break and come back to it later.

It happens to me once in a while. Taking a short break, drinking some water and walking a bit helps.

yeah, i feel like shit right now, i'm gonna listen to an audiobook and drink green tea

thanks and see ya later!

don't worry, I have been way sillier.

see ya.

hi guys do you know where i can find a list of at least 100 integers that can be written as the sum of 3 cubes? and what formula is used to get the solution?

@lynob OEIS.org is a good place to look. It almost certainly has such a list, and probably has info on generating them as well.

Hi all

@robjohn are you around?

@user1618033 yes

@robjohn I send you something privately immediately in db.

@user1618033 okay