Mathematics - 2017-05-13 (page 2 of 4)

jackerysmith

08:13

how to activate Mathjax in this chat-room any help?

Martin Sleziak

08:23

Jun 20 '14 at 0:53, by anon

Chat guidelines | $\LaTeX$ in chat

@jackerysmith The above links are also in the room description.

And here is relevant post on meta: Should chat have TeX support?

Kaj Hansen

08:35

@BalarkaSen, you there?

Balarka Sen

Yup

Kaj Hansen

So I've been considering a question recently

I've been wanting a result along the lines of: open subsets of $\mathbb{R}^2$ are $\cong$ to $\mathbb{R}^2$ itsellf. Then I realized this isn't quite strong enough since we won't have an open annulus $\cong \mathbb{R}^2$. So I strengthened the hypothesis to being simply connected. This seems like it should work, but I'm having some trouble justifying it.

Secret

[Weird challenging question inspired from worldbuilding] Formulate Clarke's 3rd Law in terms of mathematical objects and prove/disprove/show undecided the existence of counterexamples to this law

Kaj Hansen

In $\mathbb{R}$, this can be accomplished by screwing with the arctangent function to adjust its range and give a homeo. between $\mathbb{R}$ and an arbitrary open interval.

Secret

It will be very interesting if violation of it is shown to be a mathematical impossibility like the 3rd Law of Thermodynamics (in the absence of infinite processes), because it will mean we have a mathematical example of something that applies only to writing fiction

Kaj Hansen

08:41

Oops, I guess open subsets have nontrivial interior by default, haha. I was originally thinking about closed subsets (where you need nontrivial interior) and a closed ball, which I think should be true as well

Balarka Sen

@KajHansen Correct, simply connected open subsets of R^2 are homeomorphic to R^2.

Kaj Hansen

Oh, hmm. I might be able to get somewhere by working with the $S^2$ compactification for $\mathbb{R}^2$

Balarka Sen

This is a consequence of classification of surfaces.

Secret

But first, we need to express the concept "science" as a mathematical object. One key property of science is induction principle and reproducibility, thus those can help. Perhaps, we can start with some mathematical object that obeys some kind of induction principle

Balarka Sen

@Kaj In fact, simply connected open subsets of C are biholomorphic to the open unit disk by Riemann mapping theorem.

Kaj Hansen

08:44

Ah, that gives an idea. I know that all surfaces in a particular homeomorphism class will have the same euler characteristic. Would the converse hold?

Balarka Sen

Compact surfaces are classified by Euler characteristic, yup!

Kaj Hansen

I.e. everything with the same euler characteristic is homeomorphic? Maybe if we add the condition of no boundary...

Balarka Sen

When I say surfaces they have no boundary, yep.

Kaj Hansen

Thanks for the references @BalarkaSen, this is def. pointing me in the right direction

What do you call surfaces with boundary, out of curiosity?

Balarka Sen

surface with boundary :)

Kaj Hansen

08:46

haha

Balarka Sen

"Manifold with boundary" in general is the usual terminology

Kaj Hansen

The ambiguity is present in Ted's text IIRC

Balarka Sen

Yeah I think so

Kaj Hansen

Now if we consider surfaces with boundary as well, I'm guessing we can find a surface w/ boundary and a surface without that have the same euler characteristic?

In particular, if we're cutting out holes, each hole removed reduces the characteristic by $1$ since that's one less face in the triangulation (but same # of edges and vertices)

Balarka Sen

Exactly. So what's the Euler characteristic of an annulus?

Kaj Hansen

08:49

Whatever that of a circle is, minus 1

I'd have to triangulate a circle to figure it out though; don't have those memorized off the top of my head

Balarka Sen

Hm, why is it $\chi(S^1) - 1$?

Kaj Hansen

A circular disk I meant

Balarka Sen

Ah. Yes, $\chi(D^2) - 1$.

You can triangulate D^2 very easily: it's just a single triangle :)

Kaj Hansen

haha! You're right :P

too much effort

I should really look at this sort of stuff again. Gauss-Bonnet and all that was really fascinating.

My spatial intelligence is really bad though, lol

Alessandro Codenotti

doesn't an annulus have characteristic 0? Unless I messed up counting the cells of course

Kaj Hansen

08:55

Imagine a triangulation of a disk with tons of triangles @AlessandroCodenotti

Remove an interior triangle, and you have an annulus

Everything is the same, but minus one face

Alessandro Codenotti

oh, right, I constructed an annulus as a CW-complex and did the alternating sum of nnumbers of cells, but is way faster

Balarka Sen

@KajHansen V = 4 in your picture, not 3.

$\chi$ of disk = 1.

Alessandro is right.

Kaj Hansen

Derp, forgot the middle vertex @BalarkaSen

Sorry bout that

Balarka Sen

Right. No worries.

Kaj Hansen

Kind of interesting looking at homeomorphism classes in this way

Like, poking out a single point and cutting out a closed disk are equivalent

In the one, you lose a vertex, in the other, a face. But same euler characteristic in the end

Balarka Sen

09:00

In any case, annulus has $\chi = 0$, but so does the torus.

So with boundary, $\chi$ does not classify surfaces.

@KajHansen Yeah, exactly.

The fancy result is that "homotopy equivalent spaces have the same Euler characteristic"

Kaj Hansen

It's kinda funny

I have this really bad habit

Where I will think of homeomorphisms and linear transformations and such as not being instantaneous. I.e. not just immediately expanding x4 in the case of the matrix $4I$, but instead doing it in a smooth motion.

And I always catch myself and have this reminder of "ahhh, so that's what homotopies are good for"

I've never actually studied much at all w.r.t. them though

Homotopy equivalence feels like a stronger result. Are there spaces that are homeomorphic but not homotopy equivalent @BalarkaSen ?

I'm not sure that question even makes sense how I've asked it

Balarka Sen

R^2 is homotopy equivalent to a point, but not homeomorphic :)

You should learn algebraic topology at some point.

Kaj Hansen

I really should. I've been curious on looking into homology too. That the jordan curve theorem needs so much machinery intrigues me; I'd love to be able to understand its proof

Balarka Sen

@KajHansen The reason why JCT is really hard is because curves can be really pathological.

My favorite example is the Osgood curve, which is a simple closed curve of positive area.

For smooth curves, you do not truly need homology. Actually, that reminds to that a better idea for you is to study differential topology (which a bunch of people in this chat are into right now).

That's a really good starter, and you can integrate bits of algebraic topology into it with no problem.

Kaj Hansen

Huh. Hopefully I can find a lecture series. I like watching those in conjunction w/ a textbook

Secret

09:13

[random]
Proving something is essentially a zero sum game, you either win, or the opposite win. The only case where it is not zero sum is that it is equivalent to the halting problem and hence undecidable

and this is why mathematics is so powerful, because the set of ambigurities is empty

Balarka Sen

Hm, I don't know of a lecture series, but we can always discuss things. The rest of the manifold topology crew (I think?) are @Akiva, @Astyx, @Daminark, @Alessandro by the way.

If you want a textbook reco, let me know.

Kaj Hansen

Very cool. I know of Guilleman & Pollack

Secret

I think you forgot Danu, he is also into manifolds if I recall

Balarka Sen

Great, so you have my favorite book. GP + Milnor's intro to differential topology is pretty much a self-contained course.

Hatcher's AT for algebraic topology, of course. One should be able to integrate fundamental group, covering spaces, homology theory etc into it with no trouble.

What is truly missing in this is differential forms (although GP talks about it a bit, but I don't like that bit) and vector bundles. Hatcher's VBKT is very good for the latter but I literally don't know anything about forms. I am sure we can pick these up when we are sufficiently into difftop.

I don't know how in-depth all of you want to study this stuff though (I certainly haven't). Starting with GP is probably the first step, with picking up some multicalc (inverse function theorem and pals) from before.

Mike Miller

I just watched the first two episodes of One-Punch Man. pretty good show

Kaj Hansen

09:20

"and pals" haha

Alessandro Codenotti

Ted looks like a Bond villain in the starred message

Balarka Sen

lol

Jason Bourne

09:56

Hey @KajHansen how are you? Jasper here.

How can people who like manifolds not like Lee's trilogy? Hmm.

Jason Bourne

10:21

@MikeMiller I watched the ten episodes of Hindsight. Very good show, since Laura Ramsey is so pretty, but then the stupid company cancelled the second season and now there is nobody who wants to take up the project even though the script is already written.

Alessandro Codenotti

11:46

Nevermind, I read it backward

conchild

i have multiple models, each producing some estimate or inference about a variable, but i want just a single distribution over that variable. i need to average them somehow. what is the correct averaging procedure?

the models are Bayesian networks

Infinity

Hello everyone. Having the fraction ((3^i)/Root(3^i)), one can multiply the nominator and denominator by the conjugate and get Root(3^i). I just wanted to ask if the same can be applied when the expression is in a summation (Sigma) ?

Leaky Nun

12:30

@Infinity why not?

Infinity

I checked it in Mathway.com and it said that it cannot be simplified!!

I just wanted to make sure

Simply Beautiful Art

I'd simplify it to 3^(i/2)

Assuming root = √

@Infinity

Infinity

@SimplyBeautifulArt In that case, is there a summation formula for that?

Simply Beautiful Art

You mean sum(√(3^i)) from i=0 to n?

satyatech

Guys please help me out how to find the probability in this trial thing..

Infinity

12:44

No, sum(3^(i/2)) from i=0 to n

Simply Beautiful Art

@Infinity That's basically what I said

It's a geometric series

sum r^n, where r=√3

satyatech

@SimplyBeautifulArt can u help mah bruh

Infinity

You mean (1 - x^n) / (1 - x) can also be applied to sum(3^(i/2))?

Leaky Nun

@Infinity yes

Simply Beautiful Art

@satyatech I do not do probabilities

Leaky Nun

12:48

@satyatech do you know binormial distribution?

satyatech

Yah

@LeakyNun

Fawad

Hi @satyatech what's up

Leaky Nun

@satyatech Do you understand that the number of event A is following a binormial distribution?

@satyatech Can you see that the number is following B(3, 0.4)?

satyatech

Yah I got it thanks

Waiting

@Hippalectryon how are you doing? :-)

Hippalectryon

13:03

@Waiting Great :-) and you ?

Waiting

@Hippalectryon You're always great. :D Not that bad, doing some kind of research.

Leaky Nun

@Hippalectryon :o you are here

Hippalectryon

@LeakyNun :D

Leaky Nun

the mighty creator of chemobot

Hippalectryon

Chemobot is dead :( SE changed its authentfication API

I have to fix it

Leaky Nun

13:06

oh

Astyx

@Hippa o/

Hippalectryon

@Astyx \o

Astyx

@Balarka Are you there ?

Leaky Nun

13:23

@Fawad why do you suddenly ask question like that

Fawad

13:53

@LeakyNun it was my brother .___.

Balarka Sen

@Astyx I am now.

Leaky Nun

@Fawad why would you let your brother use your account?

Astyx

I had a (small) question on differential forms if you have time ? @Balarka

Balarka Sen

Sure, go ahead (you're already into forms?)

Fawad

@LeakyNun me using SE my brother:I will also ask one question, me:ok no arguments

Balarka Sen

13:55

@Alessandro I see your ping in my inbox. No, that's exactly the point: annulus and torus have the same $\chi$, but are not homotopy equivalent.

Leaky Nun

@Fawad ...

Alessandro Codenotti

@BalarkaSen I know, homotopy equivalent implies same characteristic, not the other way around, I read it backward earlier

Balarka Sen

Ah, ok.

satyatech

Is " Ni-Y-G-Y-G-Y-A " a bad word , intelligent people will understand what I mean and I did n't say anything inappropriate this time cause it's your thinking problem this tym.

@Everybody

Leaky Nun

@satyatech it can be considered offensive, regardless of intelligence

Astyx

14:05

In his lectures, Ted defines the differential forms $dx_I$ where $I$ is a family of indices $i_k$, and then $dx_I$ is the function that take $|I|$ vectors and returns the determinant of the matrix where the $k, l$ entry is the $k$-th coefficient of the $l$-th vector.
My guess is you can define those without relying on the actual coordinates by taking a familly of vectors $F = (v_1, \dots, v_k)$ and give $dx_F$ (I don't know how to note that) some sense.
Unless I'm mistaken you could do that by setting $dx_F (u_1, \dots, u_k)$ to be the determinant of the matrix where the $i,j$-th entry is $…

@Balarka ^

Balarka Sen

@Astyx That is correct, if $|I| = k$, $dx_I(v_1, v_2, \cdots, v_k)$ is the determinant of the matrices whose entries are $dx_i(v_j)$ for $i \in I$ and $j \in \{1, 2, \cdots, k\}$. You don't need a basis to do this.

Does Ted really use a basis?

Fawad

Hey @LeakyNun talk with my brother. He have some doubts.

Astyx

In the lecture I'm watching yes, he uses $\Bbb R^n$ and the canonical basis. Maybe I'm not watching the right ones ?

Leaky Nun

@Fawad alright

Fawad

Why we can't say pi equal to half circle? @LeakyNun

Balarka Sen

14:11

@Astyx Actually, sorry :P You need basis to define $dx_i$.

Astyx

But can you not equivalently define it on any familly of vectors ?

Balarka Sen

No! Whenever you say "component", you are saying that you can write the vector as a linear combination of a given set of independent vectors. That's the most natural notion.

Leaky Nun

@Fawad why would you say so?

Astyx

What I'm saying is that you can replace "i-th component" by "$\langle x_i, v_i \rangle$". then in more generality you wouldn't need the $x_i$'s to form an orthogonal basis, and it would coincide with the definition when they do

Fawad

I had one problem in that

Leaky Nun

14:15

@Fawad show me

Balarka Sen

@Astyx That does not seem like a very useful notion. Think of all your $v_i$'s lying on 1-dimensional subspace. I am pretty sure $dx_I$ becomes 0 then.

Astyx

You'd also get the nice property that it becomes 0 when the family is not linearily independant, which generalises the fact that it's 0 whenever one $i$ appears twice

Ixion

Hi, I would like to ask you if there exists a pdf about the evaluation of the period of a periodic function. Could you suggest a good one? Thanks.

Astyx

It's not really doing much more in fact, it's just taking a more abstract approach maybe ?

Balarka Sen

@Astyx So it's 0 when the family is not linear independent. That means the only true case you should look at when it is linearly independent, in which case you get back the usual notion :P

Fawad

14:17

pi radians equals to 180 degrees ,180 equals to half circle,,why we can't say pi equal to half circle @LeakyNun

Astyx

@BalarkaSen Yes that's my question :)

Balarka Sen

@Astyx Well, it's not always useful to abstractize everything in sight. $dx_I$ is not an arcane algebraic object; it's a measure of volume.

Leaky Nun

@Fawad because 180 degrees is an angle, and "half a circle" is a geometrical object

you can say "angle subtended by half a circle"

Balarka Sen

$dx_I(v_1, \cdots, v_k)$ measures the volume of the parallelpiped spanned by $v_1, \cdots, v_k$ when projected to the $I$-subspace.

Leaky Nun

@Fawad 180 is not equal to half circle either

Balarka Sen

14:19

"$I$-subspace" means the $x_{i_1} x_{i_2} \cdots x_{i_k}$-plane, where $I = \{i_1, \cdots, i_k\}$.

Fawad

Ok right bro thank u

Astyx

The reason I thought of this is because I found the definition of the exterior product relying on distribution sloppy (how can we be sure nothing fails ?)

(even though I know it is not)

Balarka Sen

Hm, what do you mean by relying on distribution?

Astyx

Well he defined the $dx_I$'s, then stated $dx_I\land dx_J = dx_{I,J}$ and "then extend[ed] by imposing the distributive property" $(dx_I+dx_J)\land dx_K = dx_I\land dx_K + dx_J \land dx_K$

(with coefficients in front actually)

rudresh dwivedi

Please tell me a non-invertible function which maps two poisitve integers into one and from the output it is impossible to invert and get the two inputs.

Leaky Nun

14:27

@rudreshdwivedi a+b

rudresh dwivedi

@Leaky Nun It could be guessed...

Leaky Nun

@rudreshdwivedi it is impossible to invert and uniquely get two inputs

@rudreshdwivedi if a+b=9, what are a and b?

rudresh dwivedi

@Leaky Nun If someone knows that you performed addition, he can apply brutee-force

Balarka Sen

@Astyx Right. I can tell you another, coordinate-free way to do this if you want. I don't find it a lot inspiring, but you might like it.

Astyx

Please do @Balarka

Balarka Sen

14:31

let me finish my snacks first

Astyx

Sure :p

rudresh dwivedi

@Leaky Nun I need a non-invertible function which should also be revocable

Leaky Nun

@rudreshdwivedi but you can't exactly know what a and b are. There are many possibilities. That is what we call non-invertible.

Astyx

$(a,b)\mapsto a$ @rudreshdwivedi

rudresh dwivedi

@Leaky Nun suppose I have two numbers [1 2] , I map them to a row vector [2 0 1] with a key [3 1] i.e.to the third and frst position. Now if someone compromises [2 0 1], we chnage the key and derives something new

Balarka Sen

14:41

@Astyx Done. OK, let's see.

Astyx

I'm all ears

Or eyes

Balarka Sen

Say $V$ is a vector space. An alternative $k$-multilinear form is a map $T : V \times V \times \cdots \times V = V^k \to \Bbb R$ such that $T$ is linear in each component, and $T(\cdots, v_i, \cdots, v_j, \cdots) = -T(\cdots, v_j, \cdots, v_i, \cdots)$ (this is the "alternating" condition)

Modulo a choice of basis, $dx_I$ for $|I| = k$ is an alternating $k$-multilinear form.

Now, call the space of all alternating k-multlinear forms as $\bigwedge^k V^*$. This is a vector space, because scalar multiplying or adding alternating multilinear forms you get alternating multilinear forms.

@Astyx Note that $dx_I$ are nothing but a choice of basis for this space (let's call it $A^k$ for notational convenience).

Whoops :P

Astyx

Right, I agree

Balarka Sen

Now we want to define wedge. This is going to be a map $A^i \times A^j \to A^{i+j}$.

One way to think about it is, for a moment, forget about the alternation, and just think about $k$-multilinear forms. $T, S$ be $k$- and $\ell$-multilinear forms respectively.

$(T\otimes S)(v_1, \cdots, v_k, w_1, \cdots, w_\ell) = T(v_1, \dots, v_k)S(w_1, \cdots, w_\ell)$ should, I think, be a $k+\ell$-multilinear form.

Astyx

The RHS operation being wedge right ?

No, usual product ?

Balarka Sen

14:56

@Astyx Well, this isn't wedge, because even if $T, S$ alternate, $T \otimes S$ may not be alternating.

Astyx

Yup, fair enough

Balarka Sen

"$\otimes$" is known as tensor product, by the way.

Astyx

Oh cool, I had heard of it without knowing what it is

SoumyoB

okay a bit unrelated to math but I'm filling my address in a webpage, and if it says 'Street Address (Line 1)' and 'Street Address (Line 2)' in two different lines, can I write part of my address in the first one and the rest of my address that doesn't fit in there in the next one?

Balarka Sen

@Astyx I am trying to remember how to get an alternating multilinear $k$-form $T_1$ out of a multilinear $k$-form $T_0$. This should be an averaging construction: look at various expressions of the for $(-1)^{\sign(\sigma)}T_0(v_{\sigma(1)}, \cdots, v_{\sigma(k)})$ where $\sigma$ is a permutation of $\{1, \cdots, k\}$ (an element of $S_k)$ and $\sign(\sigma)$ is $+1$ or $-1$ depending on if it is even or odd number of transpositions.

Astyx

15:06

yes, the signature of $\sigma$ I know of those

Fawad

What is $\operatorname*{cis}$ ?

Balarka Sen

I think you take average of those. $1/k! \sum (-1)^{\text{sgn}(\sigma)} T_0(v_{\sigma(1)}, \cdots, v_{\sigma(k)})$

Astyx

And substract ?

Hippalectryon

@LeakyNun Chemobot is back :D

Leaky Nun

@Hippalectryon nice

Fawad

15:12

@Hippalectryon and it's stronger than ever

Astyx

Oh no, I'm being silly

Balarka Sen

@Astyx I think that is an alternating $k$-multilinear form.

Astyx

Yup, I agree

Hippalectryon

@Fawad :D

Balarka Sen

So call this the "alternatization". You can define $T \wedge S$ to be the alternatization of $T \otimes S$.

Leaky Nun

15:14

@Fawad $\operatorname{cis}(x) := \cos(x) + i\sin(x)$

Fawad

@LeakyNun so it is short representation?

Leaky Nun

yes

Astyx

@BalarkaSen And this coincides with $\wedge$ as we defined it earlier ?

Sha Vuklia

Guys, we know that $V(x_1)=-\int_{x_0}^{x_1}F(x)\,dx$ and $V(x_2)=-\int_{x_0}^{x_2}F(x)\, dx$. So we have
$$
V(x_1)-V(x_2)=\int_{x_0}^{x_2}F(x)\,dx-\int_{x_0}^{x_1}F(x)\,dx=\int_{x_0}^{x_2}F(x)\,dx+\int_{x_1}^{x_0}F(x)\,dx
$$
Now how do I know for sure that I can add these two integrals? Because I've learned that we need $c\in[a,b]$, such that
$$
\int_a^bf=\int_a^cf+\int_c^bf.
$$
For all we know we have $x_0<x_2<x_1$. How can I do the calculation in that case?
$$
\int_{x_0}^{x_2}F(x)\,dx-\int_{x_0}^{x_1}F(x)\,dx=\dots=\int_{x_1}^{x_2}F(x)\,dx.

oh maybe I could change the order here: $\int_a^bf=\int_a^cf+\int_c^bf$ so show it. Oh yea, I think that's it. I can show that your "$c$" can be anywhere, by just changing the order.

Balarka Sen

@Astyx Modulo constants, yeah. I think you miss a factor of $k!\ell!$

Which gives unrealistic results like the volume of the $k$-cube is $1/k!$ :P

Astyx

15:24

Also a question : if $S\in \bigwedge^k V^*$, does there always exists a matrix $B\in M_{n,k}(\Bbb R)$ such that $S(X) = \det(B^TX)$ (where $X\in M_{n,k}$ is the matrix of the $v_i$s in the basis $x_i$) ?

@BalarkaSen Really cool !

@Sha Take $a =x_1, b= x_0, c=x_2$ ?

Sha Vuklia

yea I see what's going on @Astyx

you can still write: $\int_a^bf=\int_a^cf+\int_c^bf.$

even if $c<a<b$ for example

int hat case, your $\int_a^c f$ will go in the "opposite" direction

so you subtract, technically

Astyx

Oh right I misread you. Yes this still holds, because you can write $\int_{x_0}^{x_2} = \int_{x_0}^{x_1} + \int_{x_1}^{x_2}$

Sha Vuklia

ohh okay, yea that's also a way to explain it

Astyx

So decompose the first integral and the first term of the RHS vanishes with the one you substract

Sha Vuklia

but you can also show it generally with the $a,b,c$ thing :P and then you don't have to think about it ever again

Astyx

15:31

Well yeah, but this is how you show it :p

Sha Vuklia

oh right haha okay

Leaky Nun

@ShaVuklia sorry for my late reply; not sure if you still need help; but you could just use the fundamental theorem of calculus if you're uncertain.

Sha Vuklia

uh yea I think it's solved :P

Leaky Nun

alright

Astyx

Thanks for the help @Balarka I think I'll stop here from now, I'll probably continue later this evening

I think I wanted to make things more abstract to make sense of them in infinite dimension, when we don't have an orthonormal basis

Sha Vuklia

15:48

hey @Astyx still about those integrals: you've showed a technique to show it for one specific case. However, there are 6 cases in total to consider (6 combinations of $a,b,c$). Is there a way to show it at once? Or should we treat the remaining 5 cases separately? (because the theorem holds for $a<c<b$)

Astyx

What six cases do you mean ?

Leaky Nun

@ShaVuklia just use the Fundamental Theorem of calculus. No need to care about orders.

Sha Vuklia

So we're talking about this: $\int_a^bf=\int_a^cf+\int_c^bf$. We could have $a<b<c$, or $c<b<a$, or $b<a<c$, etc.

@Leaky I have no idea how

ohh wait

maybe.. I do

Astyx

Once you shown it for $a\le c\le b$ and $a\le b\le c$ you're done

Sha Vuklia

The case $a\leq c\leq b$ is considered in the theorem, so that means all I need to show is $a\leq b\leq c$.

but let me think why that holds

Astyx

15:52

(Hint: take the opposite of the integrals)

Sha Vuklia

You mean to show it holds for $a\leq b\leq c$? Or you mean to show that all cases are covered then?

I think the latter

Astyx

The latter

Sha Vuklia

ok.. but if I consider the opposites, then I'm technically showing it one for one after all? @Astyx

hey @Ted

Ted Shifrin

hey @Sha, @Astyx

Sha Vuklia

but maybe it's easier to "see" it, if you realise you're just changing signs anyways?

Astyx

15:58

@Sha Mmm yeah. In any case the important thing is that you know why it's true. Writting out all six cases is tedious and useless imo

Sha Vuklia

so you don't really have to write it out, but you just see it works

Astyx

@Ted Salut

I've been watching your lecture on differential forms

May I ask a few questions ?

Ted Shifrin

Fall asleep yet?

Sha Vuklia

lol:P

Astyx

No, I was really excited by them actually :p

Thus my questions

Ted Shifrin

15:59

OK, questions.

Astyx

So you define the differential forms $dx_I$ relying on the canonical orthogonal basis of $\Bbb R$ right ? So I was wondering wether one could have a more abstract approach with any familly of vectors $(x_1, \dots, x_k)$ instead of an orthonormal basis by taking $dx_I(v_1, dots, v_k) = \det((\langle x_iv_j\rangle)_{i,j})$ and thus generalise in infinite dimension

(for $k$-diff forms where $k$ is finite)

Ted Shifrin

There are definitely more abstract approaches — I was trying to be as concrete as possible. You just need a basis for $V^*$. I guess I've never thought about existence of dual bases in infinite dimensions.

No, $dx_i$ is based on differentiating a coordinate function, not vectors.

You really need a basis for $L(V,\Bbb R)$.

Do we always have dual bases in infinite dimensions? I know we can run into problems when $V=V^{**}$ fails.

Jing Weng

In this i.imgur.com/HMhlb5tl.png , how can I write e(h) in terms of $\epsilon_1 (h)$ and $\epsilon_2(h)$?

Astyx

What are you saying "No" to ?

Ted Shifrin

Just write out all the terms, @JingWeng.

I'm saying "no" to the family of vectors $x_i$, @Astyx. Those are actually dual vectors.

Remember that $dx_i$ is a notation for the dual basis vectors $\phi_i$, dual to the basis $e_j$.

Astyx

16:09

Yes, it coincides with what I said I think

Ted Shifrin

No, @Astyx. You're taking $x_i$ as elements of the vector space, not elements of the dual space.

If you're interested in the infinite-dimensional case (which I'm not, particularly), look at Lang's book on Differentiable Manifolds. He does infinite-dimensional manifolds throughout, including forms. But integration only makes sense on finite-dimensional manifolds :)

Astyx

If you take $x_i = e_i$ you get (with what I said) $dx_i(x) = \det(\langle e_i, x\rangle) = \langle e_i, x\rangle = \Phi_i(x)$, no ?

Ted Shifrin

But I'm telling you $x_i$ are NOT vectors. They're linear functions.

Astyx

Oh right, it's just that my notations are terrible

Ted Shifrin

Heya Demonark

Jing Weng

16:12

I get $\f(x) \epsilon_1(h)+ f^{'}(x)*h g^{'}(x)+ f^{'}*h\epsilon_1(h)+\epsilon_2(h)g^{'}(x)*h+\epsilon_2(h)\epsilon_1(h)+g(x)\eps‌ilon_2(h)+g(x)f^{'}(x)*h$ = $\epsilon(h)$

Ted Shifrin

Don't use * when writing multiplication, @JingWeng. We are not computers.

Astyx

Take a familly of vectors $(u_1, \dots u_k)$ in $V$ and set $dx_I(v_1\dots v_k) = \det \langle u_i, v_j\rangle$

When $(u_1 \dots u_k) = (e_1, \dots e_k)$, we find what we had before

Anyway I'll check that reference, thanks

Daminark

How's it going Ted? And hey everyone!

Astyx

Hi @Dami

Ted Shifrin

@Astyx: I insist that you start with linear functionals. You actually don't need a dot product. Then you can build $\phi_I (v_1,\dots,v_k) = \det\big[\phi_i(v_j)\big]$.

The Jumble isn't so good today, Demonark, but I'll give it to you if you want.

Jing Weng

16:16

$f(x) \epsilon_1(h)+ f^{'}(x)h g^{'}(x)+ f^{'}h\epsilon_1(h)+\epsilon_2(h)g^{'}(x)h+\epsilon_2(h)\epsilon_1(h)+g(x) \epsilon_2 (h)+g(x)f^{'}(x)h$, but I don't know how to handle the $f^{'}(x)h g^{'}(x)$ when showing $\frac{\epsilon (h)}{\|\h|}$ approaches 0.

Ted Shifrin

@JingWeng: Shouldn't that term have $\|h\|^2$? Look carefully.

Daminark

Let's see it nonetheless

Ted Shifrin

OK, Demonark. "He asked her to marry him, and she said 'yes.' Then he ..." (8/1/5) AOSDROPPAOSTTE

Jing Weng

@TedShifrin Yes, it should be $\|h\| ^2$ since $h\circ h$

Ted Shifrin

Right.

Astyx

16:18

@Ted But $\Phi_i$ is intrisinqually the dot product with $e_i$ (where the dot product is defined with $e_1, \dots e_n$ being orthonormal) right ?

Ted Shifrin

Well, I'm using dot product to give the isomorphism $V^*\cong V$ in this case, @Astyx, but you don't need to. Of course, if you want to think of oriented volumes of parallelepipeds, an inner product structure will be nice.

But still, the point is that $dx_i\in L(\Bbb R^n,\Bbb R)$. $x_i$ is NOT a vector, it's a (coordinate) function.

Anyhow, @Astyx, other questions?

Daminark

Proposed a toast?

Astyx

Yes I agree about the $x_i$, I just had bad notations

Ted Shifrin

Yup @Demonark.

Jing Weng

I know the rest of the terms go to zero , but not the constant term $\frac{f^{'}(x)g^{'}(x)\|h\|^2}{\|h\|}$

as h approaches 0.

never mind

Ted Shifrin

16:22

Huh? @JingWeng Simplify!

Jing Weng

yeah nevermind

Ted Shifrin

OK :)

Astyx

No other question for now ! :) (although I'm still 100% convinced by what you said, nevermind I'll have it figured someday)

Jason Bourne

@MatsGranvik LOL, no need to delete that. I value everything you say.

Astyx

Thanks

I can delate what he said :p @JasonBourne

Ted Shifrin

16:23

Speaking of multilinear algebra, @Astyx, this question might interest you.

Jason Bourne

@Astyx Oh I think there is no harm in that, so what was it?

Astyx

He sais he also didn't believe in exercise IIRC, and something else I forgot afterwards

Ted Shifrin

I think I'll just ignore all the people who claim to be mathematicians and don't believe in exercises.

Astyx

Bonne résolution :)

Jason Bourne

@Astyx Something like one has to do a few just to see how to compute some things?

Ted Shifrin

16:27

Puisque nous ne faisons que des exercices ici ...

Astyx

@Jason Not quite. I'll tell you if I remember it

Jason Bourne

@Astyx Yeah, of course he may have deleted it because he no longer agrees with it. I have been trying to delve into the intricacies of the French R, which can be the voiced uvular fricative or the voiced uvular trill if emphasized, it seems.

Ted Shifrin

Gesundheit @Mr. Bond :)

Jason Bourne

Hello! I prefer Theodore to Ted. =)

Leaky Nun

2

A: Proving $|A \cup B|=|A|$

Pick a well-ordering for $A$ and label the elements of $A$ as $a_x$ where $x$ is an ordinal. Label the elements of $B$ as $b_k$ where $k$ is a natural number, and let $n$ be the number of elements of $B$. Define a bijection $A\cup B \to A$ as follows: $$f(v) = \begin{cases} a_{x+n}& v = a_x \la...

Does this require AoC?

Ted Shifrin

16:31

My mother used to call me Theodore when she was angry, Mr. Bond. :)

Well-ordering is equivalent to AoC, @Leaky.

Leaky Nun

@TedShifrin I mean, does this require well-ordering?

I mean, this problem, not my solution.

Daminark

So I've started looking at Lang's manifolds book and whoa, he is so sassy in his preface

Jason Bourne

Axiom of choice has at least 4 other equivalents I have seen, sometimes even treated in analysis books.

Ted Shifrin

Lang was sassy in general.

Jason Bourne

Why treat Banach manifolds when you might as well do the Frechet ones?

So I don't really like that Lang manifolds book either.

Ted Shifrin

16:32

Ultimate generality usually makes for terrible pedagogy.

You're just ridiculous.

Jason Bourne

@Daminark The book you should look at is Fundamentals of Differential Geometry, which supersedes his previous 2 manifolds books.

Ted Shifrin

Not sure that I'd ever tell someone to learn geometry from Lang, anyhow.

Jason Bourne

I really looked to see that everything in the first book is in the second, and everything in the second is in the third, so the third is the best.

Ted Shifrin

@Leaky: You should ask someone like @Alessandro or @Secret, who think about set theory and ordinal/cardinal arithmetic.

Jason Bourne

However, many people still reference the first two books.

Ted Shifrin

16:35

You have no idea how to judge quality of a book, Mr. Bond, seriously.

Jason Bourne

OK. =)

I haven't studied much set theory, but I have Enderton's trilogy on logic.

Daminark

For specifically set theory, I've heard good things about Halmos

Jason Bourne

Halmos covers too little. I would not buy that book.

Essentially does axiomatic set theory informally, using words, but that makes it beautiful writing.

His book on finite dimensional vector spaces is written quite similarly.

But his explanations are very clear, and he always takes care of the empty set properly.

Secret

Let me just quickly check whether uncountable sets needs AoC... If your $A$ is countable that is $x \in \{\omega,...,\omega_1\}$ then you will be fine to use countable choice as the comments said

Meow Mix

Hi

Daminark

16:46

Hey Zach!

Meow Mix

How are you

Astyx

Yo Zach, long time no see

Meow Mix

Hi @Astyx

Daminark

Everything's doing alright, how about you?

Secret

ok... while I seemed to find something, it appears my AoC background is not good enough to give a confident answer to that question...

Meow Mix

16:49

Tired, still working on the animation.

And also maybe a little bit of Zelda.

Leaky Nun

@Secret alright. no problem.

Secret

So yeah, as long your set $A$ is countable, then countable choice is sufficient, otherwise we might need to ask @alessandro or @PaulPlummer

Alessandro Codenotti

What's the question?

Leaky Nun

0

Q: Proving $|A \cup B|=|A|$

i have to prove the following: If $A$ is infinite and $B$ is finite then: $$|A\cup B|=|A|$$ I've tried to use the fact that there's a bijection from $N$ to $A$, and use it to send objects from$B$ to $A$, and send $A$'s elements to themselves, but what do i do with the elements from $B$ which get ...

set-theory

Does this require AoC?

Alessandro Codenotti

Hmmm I'm not very familiar with choiceless cardinalities but I don't think it's needed after we agree on what "infinite" and "$|A|$" mean without AC

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