I have been working with metric tensors for a while - and suddenly I have a major confusion (for some unknown reason): the rule for differential forms is $(dx)^{2}=0 because dx\wedge dy=-dy\wedge dx$. So come this doesn't apply for metric tensors? Let's say for the 2-dim Minkowski metric?

@TedShifrin Hello Ted , still here?

i think i made some progress on that question

@eigenvalue tensors are not differential forms

a differential 2-form is a skew-symmetric (2,0)-tensor, or (0,2), I forget the notation.

that means it takes as input two vector fields and spits out s function, with the rule $\omega(X,Y) = -\omega(Y,X)$

@MikeMiller Thanks! Yes... I just realized that. And also that a metric tensor is actually symmetric. For a moment I had a major confusion.

It's ok, confusion is normal

@MikeMiller it all started with the notion of infinitesimal distances dx which are used in deriving the metric tensor. :-/

Yup, it makes sense. dx is still the right name for the operator which sends the tangent vector d/dx to 1.

But one has to take care what kind of multiplication we talk about. People often write $dx dx$ or $(dx)^2$ for $dx \otimes dx$, the operation with $$(dx \otimes dx)(d/dx, d/dx) = 1$.

But the operator $dx \wedge dy$ is defined to be $dx \otimes dy - dy \otimes dx$ --- you have that subtraction there

So $dx \wedge dx = 0$ but $dx \otimes dx$ is a nonzero symmetric (2,0)-tensor :)

@MikeMiller Glad people like yourself are available in this chat to help out when ones brain is about to shut down and weird questions start to emerge

It happens man. If you don't use it you lose it, and frankly we just don't use most of our knowledge!

Partly why good, searchable notes are so useful

$(x \mapsto x^2) \in O(x \mapsto x^3)$

Hi everyone. This message is for graph theory people. Isn't it high time that graph theory theorists agree on notation? D. B. West makes some efforts here [1] [2](faculty.math.illinois.edu/~west/openp/gloss.html). I guess there people out there who wish for some source they could check so that they could stick to standard notation instead of inventing yet another one. How about a cw qn in math.se ?

or is this the wrong platform because graph theory community seems to be using such platforms less?

link [2](faculty.math.illinois.edu/~west/openp/gloss.html) (D.B. West)

So, let $X_1,\ldots,X_n$ be iid continuous random variables with support on $[0,1]$ such that $\text{Max}(X_1,\ldots,X_n)-\text{Min}(X_1,\ldots,X_n)\sim\frac{1}{n}\sum\limits_{i=1}^nX_i$. What is the distribution of these $X_i$?

I am confused about "the set of all natural numbers which are greater than all the natural numbers. Naturally, this set is null set. "

If $A_1=\{1,2,\dots,\}$, and $A_2=\{2,3,\dots\}$, etc, I am having trouble to imagine the intersection of these sets and having trouble to understand that the intersection is an empty set

@Simple Suppose that the intersection is not empty, so there must be $k\in\bigcap A_n$. But this cannot be, since $k\not\in A_{k+1}$

@AlessandroCodenotti Ah I see, and what's the cost of rent like where you're staying?

A little over 300€/month all included iirc, but I'm living in a village close to Bonn, the prices in the city centre are quite crazy

Okay where you're staying the prices seem to be pretty reasonable

Also, can we chat privately via email or on Discord?

Sure, I'm alessandro#1233 on discord

Thanks I'll DM you there soon

Sent you a friend request :) @AlessandroCodenotti

@AkivaWeinberger @BalarkaSen @Semiclassical Hello

I have been struggling with this problem for a while now

I might need some hint or two

let G be a finite abelian group

let m be the maximal order of all elements of G

I took a and b elements on of G st, order of b = m

and order of a = n

gcd(m,n) = 1

so the order of ab is mn right?

how does this not contradict maximality of m ?

I guess that means that there does not exist any element of order n

but how does one conclude that?

In fact, I think this logic shows that the order of an element must be a factor of the maximum

@Jacksoja You've basically just written down a proof by contradiction

but i should assume that a is not the unital element

right?

Suppose a exists. Then ab has order greater than b, contradiction

What do you mean by unital element

I see

sorry i meant identity element

it might be that mn = m

in that case my element a is the idenity for order 1

Oh

Fair

Yeah we need n>1 to have mn>m

I am still not very convinced about this , i want to show that the order of any element a, divides the order of b = m

so far I have this (ab)^mn = 1

so the order of ab divides mn

or should I take (ab)^m ?

Prove that the order of ab is lcm(m,n)

I did that proof

Somehow am not very convinced about this proof

Also, lcm(m,n)gcf(m,n)=mn in general

Hm

I see, but how does that help in getting

Actually I'm less convinced that the order of ab is lcm(m,n)

order of any element has to divide the macimal order?

Hm

it is

i picked their order to be relativly prime

How do you know that you can do that

I can first prove the special case

then work the general case, that is how i started

Hi everyone I am struggling with a coordinate geometry problem. And problem is related to straight line.

Like I have three straight line forming an triangle, and I have a point p(x, y) to prove that point lie within the triangle.

I use earlier method such as sum of distance of point p from each of the line is the less than side length.

But if I have a coordinate which is not an integer like (a, sinA) how do I prove that it lies inside the train ke or not.

Any help is fully appreciated.

@Jacksoja I outlined a proof earlier

click through to see the part you're interested in

@MikeMiller thanks, looking at it right now

@MikeMiller what you mean by replacing x with gcd(m,n) x ?

I mean work with z = (m,n)x. This is an element whose order is coprime to the order of y.

That reduces us to the case you're interested in

I'm having great difficulty in understanding this $$ \textrm{ $\left(P and Q\right)$ or R} \implies \textrm{P or R} $$ as a first a conclusion of (P and Q) or R.

Please help me in seeing that.

@LukasHeger Hi lucas!

Sanity check: If $\operatorname{SL}_2(\Bbb Z) \cdot M \cdot \operatorname{SL}_2(\Bbb Z) = \bigsqcup_{j} \operatorname{SL}_2(\Bbb Z) \cdot M_j$ and I multiply both sides with $\operatorname{SL}_2(\Bbb Z)$ then I get a representation of $\operatorname{SL}_2(\Bbb Z) \cdot M \cdot \operatorname{SL}_2(\Bbb Z)$ as a (not necessarily disjoint) union of $\operatorname{SL}_2(\Bbb Z)$-double cosets right?

I mean I have an obvious representation like that anyway (just $\operatorname{SL}_2(\Bbb Z) \cdot M \cdot \operatorname{SL}_2(\Bbb Z)$) but my text seems to do this lol

Seems fine

Hi @Balarka @Edward

Hi @Alessandro

@YuvrajSingh... How are your straight lines defined? For instance, if you have your triangle as the intersection of three half-spaces $\mathbf{N}_i]\cdot\mathbf{p}\geq r_i$ for $i=1\ldots 3$ (i.e., essentially its definition as a 'convex body', then testing if a point is in your triangle is just testing that it liers in all three half-spaces.

Hey @Alessandro

und @Balarka

usw.

I have to go

seeya

bye

Tell me some cool facts about the Sorgenfrey plane @Alessandro

It shows that being normal is not preserved under finite products

Pinging again for a hopefully-different crowd: has anyone here ever seen the notion of '3d origami' (or arguably 4d) being studied, from a constructibility perspective? That is, taking all the construction axioms of origami (for instance, 'given two points we can find the line that reflects one onto the other') to some suitable analogs ('given two lines we can find the plane that reflects one onto the other') a dimension higher?

It also shows that being Lindelöf is not preserved by finite products

@StevenStadnicki hi.

The lens are x-2y+2=0,and x+y=0 and x-y-pi=0

These were the equations, if you can elaborate your last comment it will be helpful.

@YuvrajSingh... The line $x+y=0$, for instance, divides the plane into two (closed) half-planes: $x+y\geq 0$ and $x+y\leq 0$. If you find the vertex of your triangle opposite this line, it will fall into one or the other of these half-planes; in this case that vertex is $(2+2\pi, 2+\pi)$ Since this vertex satisfies $x+y\geq 0$, then that half-plane will be one of the ones that defines your triangle.

@StevenStadnicki ok

Okay back

lol

Ty @Balarka, just saw your "seems fine"

Once you have all three half-planes, testing whether a point is in that triangle is just testing whether it satisfies all three inequalities.

Is it possible to follow yourself on facebook? On stackexchange you can star your own question and since facebook is more of a social media than stackexchange you should be able to follow yourself on facebook. Or is it a barbers paradox? A facebook user is a facebook user that does not follow him/herself.

@Lukas Themen der nicht-kommutativen Algebra clashes with algebraische Zahlentheorie lol

damn :/

@Jacksoja hey

yeah sucks

well it says "Voraussichtlich Dienstags 14-16"

so, whether or not that changes will decide it lol

I'll come to the Vorbesprechung anyway lol

So, I have this question: Let $\mathbb{F}$ be a field and $V$ a vector space of countably infinite dimension over $\mathbb{F}$. If we let $R=Hom_{\mathbb{F}}(V,V)$ then $R$ is a ring with identity (with multiplication given by function composition). Show that as a (left) module over itself, $R$ is isomorphic to $\oplus_{i=1}^n R$ for every positive integer $n$.

My thought is to simply show that $R\cong R\oplus R$, but I'm having trouble constructing the isomorphism

My first thought is to try making $\phi(f)=(f,f')$ where $f$ is some homomorphism and $f'$ is a homomorphism whose image is the kernel of $f$. But I don't know if that pans out

Yeah, it wouldn't be well defined because there might be several homomorphisms that map to the kernel of $f$

@Rithaniel Splitting lemma?

Hmmm, I don't know it

Wikipedia to the rescue

Ooooooh, I get what you're saying. Find a SES that splits

But, now I'm down the rabbit hole of trying to find the appropriate homomorphisms (again)

@Rithaniel Use that as a vector space, $V \cong V^{\oplus n}$

Does there exist a global vector field which is nowhere zero (besides two points) on a lamination and tangent to the lamination?

If a lamination of a 2 manifold has leaves that converge at two points at infinity does this say anything about the curvature of the manifold?

I guessI should look at the foliated exchange theorem

@LukasHeger How are you ?

Don't see much of you lately , busy with studies?

I'm doing fine, thanks. How are you?

yes, I'm quite busy

not bad thanks for asking

Oh then I wont ask my question then haha

Good luck mate!

@TedShifrin Hello Ted!

Hey Jacksoja. Did you come to a place where you were satisfied about your group theory question?

@Rithaniel Haha not yet

But I have couple more questions about other topics too

I wish one of the Ted's were here to help me out :D

I could give a pointer on the group theory question, if you need it

Yes that would be welcome!

I have made some progress tho wanted to discuss it

But not sure if it it is good or not

Well, you asked if the max order of an element is $m$, then you could find an element of order $n$ such that $\text{gcd}(m,n)=1$, then the order of the product of those two elements would be $nm$. This is correct. However, take a concrete example. Look at elements $\mathbb{Z}_5$ (the cyclic group of order 5).

Can you find an element of order 4 in that group?

Yes 2 is a generator

But what is the order of 2?

Add 2 to itself repeatedly

the group am working in is F*_p

multiplicative not additive

That's not the cyclic group of order 5, though

I do not assume cyclic in the proof

we have to prove that in fact haha

only abelian and finite

the cyclic group of order 5 is an example of an abelian, finite group. Again, we're just working with a concrete example

yes in that case

order of elements is just 1 and 5

Try another example, such as $\mathbb{Z}_7\oplus\mathbb{Z}_7$. See if you can find a pattern.

I cant use direct sum of groups

it has to be done with elementary topics

it is from heirstein book , challenge question

from what i have written now

I took an element of order m

and another of n

b has order m ( maximal ) and a has order n

and first case, gcd (m,n) = 1

need to somehow arrive at contradiction

my attempt is , (ab)^mn = 1

so order of ab has to divide mn

order of ab is either m or n

it can also be 1 and mn

but going to argue why it has to be m

@Rithaniel Still here? :D

@Jacksoja: You asked me about this a few days ago, and then Lucas pursued it and you'll find an actual proof in the chat transcript if you go back there. In fact, Thorgott was a bit trickier even than the proof I suggested.

You have to reduce to the relatively prime case in a somewhat tricky manner.

@TedShifrin I saw that proof Ted!

am working on my own proof

can I argue like this as a continuation

the order of ab cannot be n nor mn if a is not the identity

since if (ab)^n = 1

then a^n = b^n = 1

which cannot be true , so it has to be of order m

so (ab) ^m = 1 , using commutativity we get a^m = 1

and the special case is proven correct?

Huh?

what part did not make sense

I have to go back and read your preceding notation.

okay but i used order of b = m

m is maximal order

order of a is n

i picked a such that , a is not the idenity

and gcd (m,n) =1

Which of course can't actually happen.

yes that is how i excluded mn

as a possible order of ab

@TedS I'm afraid my proof was so tricky that it contained an error! Lucas discovered it a short while after. On the other hand, Mike posted an elegant direct argument a while after.

Why are $m$ and $n$ the only divisors of $mn$?

since mn would be larger than m the maxiaml order

Oh, @Thorgott. Well, I'm very happy with my own proof :)

it has to be either m or n because m and n are coprime

but it cannot be n either

No, @Jacksoja. Doesn't $4\cdot 9$ have lots of factors other than $4$ and $9$?

yes but those would be less than

n and m right?

No, not necessarily.

oh

you mean one factor of each

that is possible yeah

For example.

okay thanks!

Relatively prime integers can have lots of factors :)

that argument is trash now haha

this is why I need Ted in my life :D

I never thought of one problem as much as this one

looks easy but it is not

@Thorgott Can you walk me through your proof?

Thorgott just confessed his proof had a flaw.

just for curiosity , might get some idea from it

Indeed, it was fallacious

but Ted, how did you prove it?

I don't know what Mike's proof is. Mine was to assume that $d=\gcd(m,n)<n$.

I mean ,without using fancy methods

How does one quote messages again?

You have to select permalink.

Ah, it just does that automatically. Neat.

wait is it possible that this works?