so I am guessing that there usually are stronger definitions for a chart, where $\varphi$ has to be a diffeomorphism, or smooth, or something similar.

But you just pointed out that it does not make much sense for a chart alone to be smooth or such

@lattice I don't understand your chart. $f$ takes $|x|$ homeomorphically to $\Bbb R$. How can you guarantee $f$ will take $\Bbb R^2$ homeomorphically to some open subset in $\Bbb R^2$? I mean, why is $(\Bbb R^2, f)$ is chart in the first place?

I mean, you have only defined $f$ on the graph of $|x|$. You never even defined it on all of $\Bbb R^2$.

ah okay

now I am less confused :D

I will continue later with the problem, now I need to leave sadly. But thank you for your help!

Is the directional-derivative supposed to be a number when solved?

$\Bbb R^2$ already has a smooth atructure. You cannot just pick any homeomorphism now and call it a chart, anyway.

A friend is telling me it needs to be a vector, but my book always multiplies the two vectors making a constant.

Nope, it's not. Whatever.

Wait yes it is...

The gradient is a vector. The directional derivative is a number. The directional derivative is just what you get when you take the derivative in a certain direction; so you're reducing this to 1D, back when a derivative have a number always.

Good.

That's what I needed to hear, I've been confusing myself a bit. Someone else got it wrong and was trying to tell me it needed to be a vector.

They were probably thinking of the gradient.

Yeah.

Quick question that might offend every single person in the room: is math speculation?, in the sense that one builds a theory by imposing axioms which, one could say, are true by definition, but what if they're not?, what if we're in a Cartesian world in which we're tricked to believe as obvious things that are not?

It doesn't make sense to say axioms are "true" or "false".

Well people prove things for "Cartesian worlds" and "non-Cartesian worlds". It doesn't matter which one we are in for these things to be true.

to the extent that math is used to describe the physical world, it can certainly be considered 'speculative.' More typically one uses the word 'theoretical', with the understanding that certain models of the universe have more support than others.

to the extent that math is a subject unto itself, though, it is not speculative.

Most of math is about proving statements "if these assumptions are true, then we conclude this". Though any really science or philosophy are really arguing the same kinds of things.

A given mathematical model or concept may not do what you want it to do, but to the extent that the math is well-defined and internally consistent it can't be called 'speculative.'

@BalarkaSen How come, Balarka?, would you mind expanding?

Or to say it more pithily: Whether math exists is a philosophical question. Whether it's useful in a given situation is a practical question.

I have nothing to expand on. Axiom is a thing you assume to be true. What does it mean to say it's not true once you assume that, then?

Anyway the philosophy of mathematics is probably only more interesting than semantics. Both of which are favorites on this site.

It doesn't make sense to say a statement is true or false out of blue. You need a logical system. What is that system?

Actually the implications of the etymology of mathematical terms is probably up there among maths most uninteresting aspects.

E.g., you say "what if they're not? what if we're in a world in which we're tricked to believe ..." which is not a viable objection as the moment you propose your axiom you're in a world where those axioms are true.

@BalarkaSen that's a great point.

That said, what is more reasonable is to ask whether your world is consistent, i.e., whether the axioms contradict each other.

That is a well-defined question.

It's more reasonable to ask a math question :P

Well, more preferable at the very least. But I do consider logic to be mathematics.

Sure. Though the real question is whether it's a part of mathematics that's worth thinking much about :p

this is more or less equivalent to the obnoxious and ever present xkcd about field purity

I don't have any opinions on that. People study logic: I won't call their work worthless.

I just prefer not to think about it too much, is all.

Can you play baseball on a finite field that's pure enough?

and what kind of field, scalar or vector?

now that's a good question

math.SE community is best community.

Slightly more technical question about the lens spaces: Can anybody help me see that the boundary map $d_3$ is trivial (on cellular chains)?

@Miguelgondu absolutely

(reference: Bottom of page 145 in Hatcher; his argument is a bit brief for me)

Just use Mayer-Vietoris on the two solid tori

That should give you everything you want.

Danu doesn't know the solid torus description yet.

I'm doing a construction in terms of 1 cell in every dimension (up to 3)---no solid tori.

if it wasn't zero then you would have an oriented closed manifold with trivial top jokology

I know how to see $S^3$ as solid tori (is that what you're referencing?), but I'd like to do things this other way.

Think simplicially, @Danu.

@BalarkaSen Didn't learn simplicial homology.

Well the two solid tori is the best description.

wtf does "think simplicially" mean when someone is trying to calculate cellular homology

They are isomorphic, it doesn't matter.

christ

Is anybody willing to look at page 145 of Hatcher and expand on his argument? It's the one that was presented to me, but I don't get it.

what ch?

@Danu start by writing down ezplicitly what the CW decomposition is; not just how many cells of each dimension, but their attaching maps

Well that's the problem.

The attaching map for the 3d cell.

The other ones are no problem.

It's given by a rotation of the sphere composed with a reflection along the meridian, not?

I want to say yes to that, but it's not really clear why, to me.

It's quotienting by a rotation.

and a reflection.

Yeah, I know that.

But I don't really see why this is the boundary of the 3-cell, or something.

Are you asking why this agrees with the quotient of S^3 by Z/n definition or what?

I sticking with my donuts.

Reading Hatcher's text is such an enjoyable experience! I should do this more often!

@MikeMiller I think this is much easier to see simplicially. after taking the boundary the triangles one identifies explicitly cancels out, and in the end nothing remains. dunno what can be more simpler than that. not sure why you yelled at me :)

@PVAL Eh... 2?

@MaryStar @MaryStar Is $K[x]$ artinian when K is a field?

@Danu I found it.

@BalarkaSen So I just went through this construction in terms of some cells.

I think this is the construction Hatcher gives, too.

@MaryStar There's a sort of obvious infinite decreasing chain.

I think you can probably find it on your own.

Maybe I can try to give you my way of trying to think about it

I'll soon have to go to sleep though. Maybe others here can help.

I start with some unattached 3-cell (i.e. closed disk). Its boundary is the same as the boundary of the lens before the quotient.

@Balarka Because that's not what was asked for.

My TA did something like this:

And "I would like to undersdand how to calculate cellular homology" is a completely reasonable request.

So this identification of the boundary of a separate 3-cell with the boundary of the lens-before-quotient induces an ident. on the same spaces, but with 1-skeleton (circle) collapsed

This is an identification between wedge sums of two 2-spheres

@Danu I have a meeting in a few minutes but if nobody has helped when I'm done in a couple hours I'll take a look.

If I wanted a CW description for a lens space I'd start with the description as a union of solid tori and then do something analogous to Heegard.

I don't really like Hatcher's way of doing it. It seems somewhat convuluted to me.

Recalling that the cellular chains are homology classes in $H_2(X^{(2)}/X^{(1)})$

So, in the case when $K$ is a field the decreasing sequence of ideals of $K[x]$ is infinite? @PVAL

That's exactly what Danu's CW decomposition is though. It's the standard handlebody decomposition but collapsing the handles to appropriate dinensional cels.

@MikeMiller All right. The thing I did and proving $d_3 = 0$ in cellular homology are more or less equivalent to me, but I suppose, yes.

I now look at a generator of $H_2(X^{(2)}/X^{(1)})$

You apparently can't show that $d_3$ is zero in cellular homology because you made great efforts to not just say how to do so.

@MaryStar You need to find a chain of decreasing ideals which is infinite.

@MaryStar There should be a somewhat obvious such chain to try in $K[x]$.

First of all, does the wedge sum induce direct sums on homology?

And you need to see that it indeed doesn't terminate.

Oh, nvm.

Yes, @Danu.

I did learn that, I think.

You mayer-vietoris thickening of the two wedged pieces.

Ok, I need to head to bed. G'night, all.

Bye!

OK, so I somehow should be doing the following. I start with a generator of $a\in H_2(\partial D^3\cong S^2)$ and see where it maps to after quotienting.

@MaryStar What ideals does $\mathbb R$ have?

First of all, it has the obvious ideals, $\mathbb{R}$ and $0$. How can we find the other ideals? @Krijn

@Mary If an ideal contains 1, what can we say about the ideal?

@Mary And if an ideal is not $(0)$ it must contain some $a \in \mathbb{R}$ nonzero. So can you find an element of $\mathbb R$ such that the ideal must contain 1?

If the ideal contains 1, it is the whole ring, or not? @Krijn

True!

@Mary Your first answer is correct

Your second isn't

I got stuck right now... Does $1$ have to be contained in each ideal of $\mathbb{R}$ and so we conclude that the only non-zero ideal is the whole ring? @Krijn

No

If 1 is contained in an ideal then that ideal is the whole ring

So if our ring is a field, then every nonzero element has an inverse

So 1 must be in every nonzero ideal

If our ring is not a field, for example take $\mathbb Z$ things go very different

Here $(2) = \{ z \in \mathbb{Z} \ | \ 2 \text{ divides } z \}$ is an ideal

Which does not contain 1

Guys, if I have a convex, closed, plane curve, could it be self-intersecting? I'd guess yes ( do a loop inside at some point) - but I'm not sure.

what do you mean by a convex curve?

If I take a tangent at any point of the curve, all the points of the curve are on one half.

That example I wrote is wrong, and I'm changing my opinion to no, it's not possible.

lol

That's also my opinion. Convex curve should be equivalent to "is a Jordan curve, and its interior + itself is a convex closed subset of the plane".

Unfortunately I don't see how to do it yet, so I'm going to abandon shit.

nevermind

Certianly one can prove that if it's a Jordan curve its interior is convex. But showing it's Jordan seems a little tricky. It can't be done entirely locally near the self-intersection points, since you might intersect tangentially to yourself with both curves above the tangent line. Eg take the graphs of $x^2$ and $x^4$ near zero and then a little bit further away close that up into a single curve.

This is of course not convex.

Hi everyone

just by chance.. someone have this book in pdf version?

google download free textbooks reddit

I've searched on libgen but not successed

@PVAL of course I had search it :) that's because I ask..

Eh, I think you can get a proof by working in two cases (transverse intersection and tangential intersection) and in the tangential case reducing it to a picture like I have above and considering the tangent line of the top curve at $x = -\varepsilon$.

I mean if you take a circle and then cut down the middle, this seems to be convex and self intersecting

That doesn't sound like a curve?

Will do a circle and then do a loop inside the circle

if you take a tangent inside the loop, it will not have the curve on one half

oh nvm

@MikeMiller Thanks for the replies. In the tangential case - what do you mean by $x = -\varepsilon$ - going a bit further along the loop?