Mathematics - 2016-01-27 (page 1 of 2)

I'm an artist

00:29

It's late but amazing results keep flowing ...

L33ter

00:50

More generally, $\Delta$ complexes can be built from collections of disjoint simplices by
identifying various subsimplices spanned by subsets of the vertices, where the identifications
are performed using the canonical linear homeomorphisms that preserve
the orderings of the vertices.

Here what do they mean by identification ?

putting it in same equivalence class or what ?

@PVAL or @MikeMiller

PVAL

@L33ter They are describing a topological space which is homeomorphic to a quotient on a set of disjoint simplices. The identifications are giving you an equivalence relation to get the desired quotient space.

L33ter

I want to understand the identifications issue because I don't understand it that well

Jeff

Hi folks. Can tell how to use epsilon-delta definition to prove lim as x -> 4 of square root x?

L33ter

so from my reading of the book I understand that we look at a face which is (n - 1) simplex which is [v0,...,vn] where the ith one removed

or ith vertice is removed

so by identifications I understand that [v0,..,v_i^,...,vn] ~ [v0,...,vn]

is that correct?

@PVAL ?

I guess this is the ith face of a graph we identify it with an (n - 1) simplex

and so what we mean here we an equivalence class where the elements that generate the equivalence class is the ith face which is identified as the above.

right ?

L33ter

01:23

hi @TedShifrin

Ted Shifrin

01:34

Hi Karim

PVAL

The data of a delta complex is a set of simplices with some identifications between their ordered faces. The identifications on a pair of faces are given by the unique affine homeomorphism between ordered simplices of the same dimension. If $V_1=[v_0,v_1,v_2], W_2=[w_0,w_1,w_2] $ are our simplices in the data of a delta-complex, an example set of identifications could be $\{[w_o,w_1]=[w_1,w_2] ,[v_2,v_0]=[w_0,w_2]\}$.

For instance, there is a unique affine homeomorphism $ \phi$ which sends the face [w_0,w_1] to [w_1,w_2] and in the delta complex the $\phi(x) ~x$. The delta complex is topologized as a quotient of $V_1 \cup V_2$ with the equivalence class described by the above relation (and the other one).

L33ter

back

1 sec reading your thing @pval just wanted to go have a shower

oh I see I understand

thanks @PVAL

$V_1 \cup W_2$ in the case above I guess

but yeah that makes sense

you know after your explanation everything made sense in that thing I was reading in allen hatcher

I was struggling for sometimes trying to understand what he means

Semiclassical

02:31

hi chat

Michael

hello classical

Sorry since I am new to this. What are some types of sequences other than cauchy that converge?

Balarka Sen

04:10

@L33ter A delta complex is a quotient space $\coprod \Delta^n/\sim$, where $\sim$ glues faces by by linear homeomorphisms preserving ordering.

That's all they are saying.

user174558

@Michael Convergent sequences must be Cauchy.

Balarka Sen

04:49

The modified question now needs a point-set topologist who deals with ugly spaces. I doubt if it's still true.

Chaow Wu

05:17

Hello chat, first time here. If I made a post about a question to be answered, is it okay for me to link it here to get attention?

Balarka Sen

05:28

Hi @iwriteonbananas

user174558

06:10

Hello @BalarkaSen. What are you reading now?

L33ter

I see

thanks @BalarkaSen

@TedShifrin hi

there is one guy in my vector calculus class that I am marking show the answer right away without showing any steps

if the answer is right should I still deduct marks ?

user174558

They let undergrads mark papers now?

L33ter

yeah

user174558

If the answer is right and the question is trivial, maybe he did all the steps in his head.

L33ter

yeah I guess but what is weird there is some non-trivial stuff that he just write the answer without doing steps

user174558

06:20

Maybe you can speak to him about it.

L33ter

but his final answer isn't correct

user174558

If the final answer is wrong and no steps are given, then no marks.

user174558

Asking undergrads to mark papers is TERRIBLE...

Semiclassical

Right answer, no reasoning shown---maybe they knew what they were doing and were just lazy. Wrong answer, no reasoning shown---that problem's dead on arrival

user174558

Even grad students might be bad enough to mark them wrongly...

Semiclassical

06:22

(right answer, wrong reasoning = headache to grade)

peer review stuff like that in classes isn't necessarily terrible, but it's the exception rather than the rule

it can possibly work in small course settings. beyond that, and it just doesn't make sense

user174558

I seriously hate this idea of letting students mark each other's work.

user174558

Both of them might be wrong and not know it, thinking they are both right.

Semiclassical

i did have something like that for one course in undergrad, though i entirely forget which one

user174558

Maybe some lazy teacher invented this "pedagogy" and passed it off.

user174558

By my standards, it seems almost all math teachers I have seen are lousy.

Semiclassical

06:26

and for that it was that it was a small course setting, and each week one student would peer review everyone else's homework

but i don't think it was literally just "grading"

user174558

I was referring to school teachers above.

L33ter

I had high score in my vector calc @Jasper

I got 95 in vector calc

user174558

@L33ter Good for you.

Semiclassical

introductory math education in the US isn't very inspiring or sensible

not sure why. but that's what i blame the attitude of "Oh, I'm bad at math, haha"

user174558

It is very interesting that I am often being judged by people whom I should be judging instead. This statement might come across as arrogant.

user174558

06:30

You have an organisation full of lousy people, and they all promote lousy people to the top ranks.

user174558

They fail to appreciate good people when they see it, because they are all too lousy.

user174558

And then they say stupid things to me, which make me laugh right now.

user174558

I should have told them in the face, do you know how stupid you are?

user174558

Wow, the way they scolded me, I cannot stand it...

Huy

@Jasper =(

user174558

06:34

@Huy I especially like the love scenes in Episode 2 of Star Wars, lol.

Huy

@Jasper: I thought you didn't like the movies and stopped watching?

user174558

@Huy Well, I have finished 1,2,3,4. Will watch 5,6,7 soon.

Huy

@Jasper: You should have sticked to my order.

user174558

The Puppet Master has about 10 movies, and the order is confusing, lol.

user174558

Professor Anthony Knapp wants me to tell others that the digital second edition of his "Basic algebra" and "Advanced algebra" are available from his website now. However, they may not be printed as Springer has the rights to the print material. Please star this.

user174558

06:39

His "Basic Real Analysis" and "Advanced Real Analysis" will be available later this year, the digital second editions, on his website.

user174558

Of course, Springer has already printed all four books.

user174558

Let us thank him for this gift to the mathematical community!

L33ter

06:51

wow

Balarka Sen

07:05

@Jasper I have school exams coming up, unfortunately. But I am studying multivariable calculus for now.

@L33ter So, you understood $\Delta$ complexes now?

L33ter

07:38

yes @BalarkaSen sorry was marking assignments

thanks a lot

I am going to sleep nights

Jake1234

09:05

Guys I've been wondering... do you know of any nice intuitive explanations for symmetry of second derivatives?

Tobias Kildetoft

@BalarkaSen Now my new paper with my mentor is on arXiv

Balarka Sen

@Jake1234 The graph of the function is nice. Moving infinitismally along x axis and then infsmly along y-axis lands you onto the same point you get by moving inftsmly along y and then x axis.

@TobiasKildetoft Cool! Link?

Tobias Kildetoft

@BalarkaSen arxiv.org/abs/1601.06975

Jake1234

@BalarkaSen I'm not sure how you move infinitismally along x and then along y, I don't even know what you mean by moving infinitismally along an axis.

Balarka Sen

@TobiasKildetoft Nice, congratulations. Which journal are you planning to publish that in?

@Jake1234 I'm having dinner right now, but if you can be around for a few seconds I can explain better.

Tobias Kildetoft

09:13

@BalarkaSen Not sure yet. The results are fairly elementary (I think the main deep result being used is Perron-Frobenius, plus a single use of the Schauder fixed point theorem)

Jake1234

Sure.

Tobias Kildetoft

But they do have some nice applications in explaining some things that previously required big machinery

Jake1234

I mean I get the proof, and it all seems nice... but I feel like I'm missing the big picture.

Balarka Sen

09:28

Ok, back.

Jake1234

I think I understand what you mean now ... math.stackexchange.com/questions/942538/…

?

answer by rahul

Balarka Sen

@Jake1234 Oh. I thought you were asking about the proof. The proof is a corollary of the fact that, in the graph of $z = f(x,y)$, moving from $f(a, b)$ to $f(a,b+h)$ by moving through the y-coordinate, and then to $f(a + k, b + h)$ by moving through the x-coordinate has the same final effect (landing on to $f(a+k, b+h)$) as going to $f(a + k, b)$ and then to $f(a + k, b + h)$.

@Jake1234 Yes, Ted's comment/Rahul's answer is what I was hinting at. But it's still a (geometric) proof, not a geometric visualization of the symmetry.

Jake1234

hm

Balarka Sen

@Jake1234 Ok, here's a possible visualization. If mixed 2nd partials of a function $f : \Bbb R^2 \to \Bbb R$ are symmetric, then the Hessian matrix $Hf = [\partial^2f/\partial x_i \partial x_j]$ is symmetric.

Thus, you get a quadratic form corresponding to $Hf$.

Fact: This form is the best quadratic approximation to $f$ near any point.

So a reformulation of that is that the graph of $f$ admits a best quadratic approximation.

I don't know if you can get a more direct visualization that this.

Jake1234

09:44

hm

Alright, thanks, I'll look into it.

Balarka Sen

No problem. I'd just recommend carrying on with a vague picture in mind - anything beyond the first derivative is very hard to visualize.

I'm an artist

10:34

Evinda

10:51

Hello @DanielFischer !!!

I want to show that if $\rho(x,y)$ is a metric on $X$ then $\sigma(x,y)= \min \{ 1, \rho(x,y) \}$ is a mteric.

I have a question about this property: $\sigma (x,z) \leq \sigma (x,y)+ \sigma (y,z)$.

We have $\sigma (x,z)= \min \{ 1, \rho(x,z)\}$.
How can we show that this is smaller than $\min \{ 1, \rho (x,y)\}+ \min \{ 1, \rho (y,z) \}$ ? @DanielFischer

Tobias Kildetoft

11:04

@Evinda Split into cases depending on which entry gives the min

Evinda

@TobiasKildetoft Do we have to take the cases $\rho(x,z)>1$ and $\rho(x,z)<1$ or $\rho(x,y)+ \rho(y,z)>1$ and $\rho(x,y)+ \rho(y,z)<1$ ?

Tobias Kildetoft

@Evinda Why would the last one play any role?

Lembik

hi all.

what tags should math.stackexchange.com/questions/1626887/… have?

Evinda

@TobiasKildetoft I thought this:

Suppose that $1< \rho(x,y), \forall x,y \in X$. Then $\sigma(x,z)=1 \leq 1+1= \sigma(x,y)+ \sigma(y,z)$.
Suppose that $1> \rho(x,y), \forall x,y \in X$. Then $\sigma(x,z)= \rho(x,z) \leq \rho(x,y)+ \rho(x,z)= \sigma(x,y)+ \sigma(x,z)$

But what if for example $\rho(x,y)<1$ and $\rho(y,z)>1$ ?

Tobias Kildetoft

@Evinda Well, write up what you need to show in that case

Evinda

11:13

@TobiasKildetoft We want to show that $\min \{ 1, \rho(x,z)\} \leq \min \{ 1, \rho(x,y) \} + \min \{ 1, \rho(y,z)\}$

Tobias Kildetoft

@Evinda I meant in the specific case you now asked about

Evinda

@TobiasKildetoft Then we want to show that $\min \{ 1, \rho(x,z) \} \leq \min \rho(x,y)+1$, right?

Tobias Kildetoft

@Evinda Well, without the min on the right side

Evinda

@TobiasKildetoft Oh yes, right

@TobiasKildetoft How can we show this? How could we use the fact that $\rho(x,z) \leq \rho(x,y)+ \rho(y,z)$ ?

Tobias Kildetoft

@Evinda You don't even need that.

Just use the definition of min

Evinda

11:51

@TobiasKildetoft Is it right like that?
If $1< \rho(x,z)$ then $\min \{ 1, \rho(x,z) \} =1 \leq \rho(x,y)+ 1 \Rightarrow \sigma (x,z) \leq \sigma (x,y)+ \sigma (y,z)$.
If $1> \rho (x,z) $ then $\min \{ 1, \rho(x,z)\}= <1 \leq 1+ \rho(x,y) \Rightarrow \sigma(x,z) \leq \sigma(x,y)+ \sigma(y,z)$

Tobias Kildetoft

@Evinda Yeah

Evinda

12:24

So we have to distinguish these three cases, right? @TobiasKildetoft

Tobias Kildetoft

@Evinda Not sure about "have to", but it certainly helps

Evinda

@TobiasKildetoft I mean that then the answer is complete, we haven't left any case... Right?

Tobias Kildetoft

@Evinda I haven't really thought about whether we have

Evinda

We could have $\rho(x,y)>1$ for any $x,y \in X$, $\rho(x,y)<1$ for any $x,y \in X$ or $\rho(x,z)>1$ and $\rho(x,y)<1$ for $x,y,z \in X$. So I think that we have checked all the cases... @TobiasKildetoft

Partly Putrid Pile of Pus

12:59

Is there any meaning in saying that a topological space is over a field?

(just by itself, no mention of metrics, norms, inner products)

user153330

@NaCl actually i felt off very tired and i sleeped so i didn't even start hahahaha but anyway it was fun

Tobias Kildetoft

@PartlyPutridPileofPus Not immediately, no. One might speak of the category of topological spaces over some field, but that would be with some specified topology on the field, and the rest of the field structure would not matter anyway

Partly Putrid Pile of Pus

@TobiasKildetoft Thanks

Tobias Kildetoft

@PartlyPutridPileofPus Where did you encounter this?

Partly Putrid Pile of Pus

In what I assumed was a misheard spoken answer, and I was trying to reconcile this with the fact that a vectorspace contains its underlying field when the vectorspace has nonzero dimension, i.e. $k$-vectorspace of dimension $1$ with basis $\{1\}$

I.e. trying to work out, if the concept did exist, would it imply that the underlying field is contained in the topological space

Evinda

13:09

@TobiasKildetoft Could I also ask you something else?
We have a function of two variables and find the maximum as for the one variable and the maximum as for the second one.
How can we find the maximum of the function as for both variables using the above?
There are also some restrictions.
Can we write somehow a recurrence relation?

Tobias Kildetoft

@Evinda What do you mean by finding the maximum for one variable?

Evinda

We set the second variable as a constant. @TobiasKildetoft

Tobias Kildetoft

@Evinda but where the max is will depend on what that constant is

Do you mean you take the derivative with respect to one of the variables and set it equal to $0$?

Evinda

Yes, in order to find where the maximum is achieved... That's why I thought to solve $g'(x_1)=f'(x_1,c)=0$ @TobiasKildetoft

Tobias Kildetoft

@Evinda Well, look up the result that allows you to do this. There should be one that says something along the lines of the maximum being achieved at a point where the gradient is $0$ (or on the edge of the domain)

Partly Putrid Pile of Pus

13:15

@Evinda You mean taking a partial derivative then, surely?

user305938

0

Q: Find the characteristic polynomial of this matrix

Hi fellow mathematicians, I've tried to find the characteristic polynomial of the following matrix $A=\begin{pmatrix} 0 & 0 & \cdots & 0 & -a_n \\ 1 & \ddots & \ddots & \ddots & \vdots \\ 0 & \ddots & \ddots & \ddots & \vdots \\ \vdots & \ddots& \ddots & 0 & -a_2 \\ 0 &...

Take a look at it guys ^

Partly Putrid Pile of Pus

@user305938 What type of matrix is that? I have forgotten

Companion matrix was it...?

Companion matrix

In linear algebra, the Frobenius companion matrix of the monic polynomial is the square matrix defined as With this convention, and on the basis v1, ... , vn, one has (for i < n), and v1 generates V as a K[C]-module: C cycles basis vectors. Some authors use the transpose of this matrix, which (dually) cycles coordinates, and is more convenient for some purposes, like linear recurrence relations. == CharacterizationEdit == The characteristic polynomial as well as the minimal polynomial of C(p) are equal to p. In this sense, the matrix C(p) is the "companion" of the polynomial p. If A is an n...

hhh

Can someone recommend here some books on Algebraic Graph Theory? I am trying to understand what the polynomials are: some cuts?

Partly Putrid Pile of Pus

The answer is in that page @user305938

Evinda

I want to find the maximum of the following:

$\max (x_1^3-11 x_1^2+40 x_1+x_2^3-8x_2^2+21 x_2) \\ x_1+x_2 \leq 5.4 \\ x_1, x_2 \leq 0$

We set $f(x_1, x_2)=x_1^3-11 x_1^2+40 x_1+x_2^3-8x_2^2+21 x_2$

Then if we consider $x_2$ as a constant, let $s$ then $g(x_1)=f(x_1,s)=x_1^3-11 x_1^2+40 x_1+s^3-8s^2+21 s$ and $g'(x_1)=0 \Rightarrow x_1=\frac{10}{3}$ or $4$

Similarly, if we consider $x_1$ as a constant we get $x_2=\frac{7}{3}$ or $x_2=3$.

How do we continue? @TobiasKildetoft @PartlyPutridPileofPus

user305938

13:20

Thx a lot!!

Sometimes you only need sth to search for

Partly Putrid Pile of Pus

@user305938 No problem, I am super surprised the name jumped into my head when I saw it

I saw it around a year ago

user305938

You have a beautiful mind then

Partly Putrid Pile of Pus

I can't immediately see why that is the characteristic polynomial, so I am going to try to work it out after I finish what I'm working on

user305938

Thank you :)

Partly Putrid Pile of Pus

@Evinda They are both negative and sum to less than $5.4$, what?

@Evinda That is meant to be $x_1,x_2\geq 0$?

Evinda

13:24

@PartlyPutridPileofPus Yes, I am sorry... It was a typo

Partly Putrid Pile of Pus

Np

Give me a min to get my lagrangian hat on

Evinda

If my $x_1, x_2$ are right for no combination of $x_1, x_2$, $x_1+ x_2 \leq 5.4$.
Or am I wrong? @PartlyPutridPileofPus

Partly Putrid Pile of Pus

I believe you are correct

We take the partial derivatives with respect to each variable, use quadratic equation to find viable $x_1,x_2$

And then adding the three possible combinations of $x_1,x_2$ all yield values greater than $5.4$, with minimal $x_1+x_2=5.6\overline{6}$

Evinda

13:41

So there is no maximum right?

But could we also right a recurrence relation to deduce this? @PartlyPutridPileofPus

Partly Putrid Pile of Pus

How so?

Chaow Wu

Hi chat, anyone willing to guide me through a proof of the replacement theorem (linear algebra) for infinite dimensional vector spaces?

Partly Putrid Pile of Pus

Do you morally understand why the theorem is true?

Chaow Wu

yes, i understand the induction part on finite dimensions

Partly Putrid Pile of Pus

Actually, I am not sure how to think of this for infinite dimensions

Chaow Wu

13:53

:/ teacher left this exercise in our advanced lin algebra class and no one knows how to do it haha

Partly Putrid Pile of Pus

Yes, I am not sure how to do it for infinite dimensions

Hopefully someone else can help

Need some indexing set, yada yada

Tobias Kildetoft

@ChaowWu What is the replacement theorem?

Partly Putrid Pile of Pus

Yes, that would be a good idea, can you state it how you are given it. Normally the theorem seems to force a finite condition on $V$s dimension

Chaow Wu

let B be a basis for a vector space V, and let S be alinearly independent subset of V. There exists a subset S1 of B such that S U S1 is a basis for V

Partly Putrid Pile of Pus

Oh, that's different somewhat

Tobias Kildetoft

13:56

@ChaowWu Just repeat the proof of the existence of a basis, but using linearly independent subsets containing the given one instead of all of them

Ohh, missed that the new basis had to be taken from the old one

Chaow Wu

I do think you need to use the maximal principle im just not quite sure how

Tobias Kildetoft

But still, you can probably consider all possible linearly independent subsets consisting of the given vectors along with ones from the chosen basis

Chaow Wu

so how do i write that out? let F denote the family of all linearly independent subsets of V containing S and elements of B?

Tobias Kildetoft

@ChaowWu yeah. Then show that this collection satisfies the requirements to apply Zorn

Chaow Wu

aah, i wasnt paying attention last semester when we saw zorn -.- time to go learn haha, but thank you very much for the help! really appreciated

oh wait, Zorn is just the axiom of choice, which is basically the maximal principle right?

Tobias Kildetoft

14:05

@ChaowWu Not sure what the maximum principle is

Joshua A

yesterday, by Balarka Sen

For example, sheaves and ringed spaces come after regular maps and algebraic varieties, not before :D

Is this true?

Let $(V,O_V)$ and $(W,O_W)$ be algebraic prevarieties. A map $\phi :V\to W$ is said to be regular, if it is a morphism of $k$-ringed spaces

Should not ringed spaces come prior to algebraic varieties?

An algebraic variety is a ringed space that is blah blah

@BalarkaSen

Tobias Kildetoft

@JoshuaA That is an unusual way to define algebraic varieties as an introduction

Joshua A

Could you recommend me a text that gives a standard introduction please @TobiasKildetoft

Tobias Kildetoft

@JoshuaA Mumford is pretty good as I recall

Joshua A

14:21

"Algebraic Geometry I
Complex Projective Varieties" @TobiasKildetoft ?

Thanks @TobiasKildetoft

Mambo

Problem with the proof of Peter Weyl Theorem for compact groups

Any help will be appreciated

user305938

14:38

I'm trying to find 2 complex matrices with the same spectrum but with different traces and determinants

any ideas?

3x3 matrices

Partly Putrid Pile of Pus

Spectrum is the set of eigenvalues right?

user305938

yes

Mambo

Take any two complex numbers $a$ and $b$

Build the matrices with $a$ as two eigenvalues and another with $b$ as two eigenvalues

Partly Putrid Pile of Pus

Both just diagonal matrices?

Mambo

Why not

user305938

14:43

mambo I'm not sure if I understood what you said

"with $a$ as two eigenvalues" what do you mean?

Partly Putrid Pile of Pus

eigenvalues are $a,a,b$ and $a,b,b$

Huy

diag(1,1,-1) and diag(1,-1,-1) have same spectrum but different determinant and trace

Mambo

yes

Exactly

Partly Putrid Pile of Pus

Because the spectrum is still $\{a,b\}$

user305938

ohh ok

Partly Putrid Pile of Pus

14:45

But trace is $2a+b$ and $a+2b$

user305938

is it ok to use diag(1,1,-1) and diag(1,-1,-1) when it says complex matrices?

Partly Putrid Pile of Pus

det is $a^2b$ and $b^2a$

They are complex

user305938

of course R is also a part of C

but im not sure if this term complex implies that I have to use i

Partly Putrid Pile of Pus

I don't think it would, that would be strange

user305938

alright ok, thx a lot!

Huy

14:46

otherwise replace 1 by i...

Partly Putrid Pile of Pus

^

user305938

diag(i,i,-i) and diag(i,-i,-i) works?

Partly Putrid Pile of Pus

Or leave $a,b\in \Bbb C$ arbitrary ($a\ne b$)

Well just make sure $a^2b\ne b^2a$

user305938

ok I got it now, thanks for your support stackmath, you really save my bachelors degree

Mambo

@Huy May I ask what are studying now?

*you

Huy

14:52

@Mambo: the proof of the bigon criterion

Mambo

@Huy I meant in your research

Huy

I'm just a student, not a researcher

Mambo

Masters?

Huy

yes

Mambo

Oh Good

Huy

14:54

why?

Partly Putrid Pile of Pus

Where does one learn Lie groups? What textbook is a good introduction?

Mambo

I am also doing masters

@PartlyPutridPileofPus What are you doing actually?

Huy

@PartlyPutridPileofPus: what's your background?

Partly Putrid Pile of Pus

Second year student, know field theory

Strong algebra background

Huy

do you know (differential) topology?

Partly Putrid Pile of Pus

14:56

I don't, is this a good way to reach Lie groups?

Huy

you might want to study Lie algebras and representations first if you have a strong algebra background, but I'm no expert

Lie groups are smooth manifolds

Partly Putrid Pile of Pus

Okay. So Differential topology will unlock Lie groups. Could you recommend a textbook for DT then?

Mambo

You should concentrate on multivariable calculus and Differential geometry first @PartlyPutridPileofPus

Huy

Lee's books are often recommended for first courses, but I don't know

why do you want to study Lie theory anyways?

Anonymous - a group

Hello! If a prime p divides a number n, it is called a prime factor. Is there a name for the largest power of p that divides n?

Mambo

14:59

You give it a name

for your convenience

Anonymous - a group

So there is no standard name for this?

Semiclassical

@I'manartist that's the kind of problem which looks worse than it perhaps is, since one can replace $Ei$ with an appropriate integral representation

Partly Putrid Pile of Pus

@Huy Lie groups are mentioned periodically, and I always am left out of the loop

Mambo

@Huy Do you have any background in Harmonic analysis?

Partly Putrid Pile of Pus

@Mambo Sorry to continually bother, do you have a recommendation for multivariable calculus and DG texts?

Semiclassical

15:01

which is to say, it's better to think of it as a triple integral in elementary functions rather an double integral in special functions

Mambo

DG - Thorpe

Partly Putrid Pile of Pus

I have done Calc I through III, but haven't done any DG

Huy

@PartlyPutridPileofPus do you think other people in the courses where they are mentioned have already studied them? a book or course on Lie theory requires quite a bit of a background.

@Mambo hardly, I know some very basics of Fourier analysis but that's it.

Partly Putrid Pile of Pus

@Huy I am not so worried about the other students, it's just something I would like to explore for now

Huy

@Mambo: Unless you consider studying the Laplacian on some manifolds harmonic analysis.

Mambo

15:03

Multivariable calculus-Function of several real variables Martin Moskowitz

Partly Putrid Pile of Pus

Thanks @Mambo

Mambo

@Huy I have issues with Peter Weyl Theorem

Huy

@PartlyPutridPileofPus Sitwell's book on springer is probably the one with least prerequisites, but I don't know if it's good. I just needed it for an elementary proof. Maybe it would be useful for you, but as I said, I'm no expert.

@Mambo: Sorry, I won't be able to help.

Mambo

@PartlyPutridPileofPus There is very classic book for Lie Groups : Claude Chevalley

You can read it.

Huy

@PartlyPutridPileofPus: so you know topology already?

Mambo

15:06

I think it's probably the first concise book written on lie groups

Claude Chevalley was algebraic topologist

Partly Putrid Pile of Pus

@Huy I have picked up a decent amount of point set topology I would say

(For a 2nd year student)

I'll check all of these texts out, thanks for the recommendations

@Huy Sorry, could I have another of Sitwell's names? I can't find anything

Huy

sorry, misremembered the name

springer.com/br/book/9780387782140

Partly Putrid Pile of Pus

Oh, thank you

Chaow Wu

is a zero dimensional vector space just {0}? cant seem to find the exact phrase in my textbook

Huy

yes

Partly Putrid Pile of Pus

15:11

Yep

Mambo

Yes it is

Chaow Wu

thank you :) was slightly unsure

Mambo

@Huy Are you applying places?

Huy

applying places?

Mambo

PhD?

Huy

15:14

no

Partly Putrid Pile of Pus

So Naive Lie theory doesn't define Smooth manifolds, it just works with the matrix Lie groups, so this is probably good for me for now

Huy

@PartlyPutridPileofPus: exactly

Mike Miller

morning

Huy

morning Mike

Mambo

It's 9 PM here

Ron Ford

15:17

@Anonymous-agroup The largest prime $p$ dividing $n$ is often denoted ord$_p(n)$.

Mambo

Does anyone here have background in harmonic analysis?

iwriteonbananas

15:32

Morning, Mike

Mike Miller

how's it

iwriteonbananas

Quite good, I was working on some alg top problems today. This one was fun:

Find a closed 3-manifold $M$ with $H_1(M)=\Bbb Z/17 \oplus \Bbb Z^2$

Now I'm working on diffgeo problems

Specifically, I'm trying to prove that parallel transport between two points is curve-independet iff the manifold is flat

Mike Miller

ah cool you're getting into holonomy

holonomy is the shit

iwriteonbananas

Glad to hear that, though I've never heard that word :P

Mike Miller

Given a loop based at $x$ parallel transport gives you an isometry $T_x M \to T_x M$, right?

iwriteonbananas

15:46

Indeed, I think I proved last week that parallel transport for the LC-conn is an isometry

Mike Miller

So consider the set of Isometries $T_x \to T_x$ arising from parallel transport along a piecewise smooth loop. That's closed under multiplication (do one loop then the other) and under inversion (follow the loop backwards). So it's a group.

iwriteonbananas

Interesting.

Mike Miller

It's called the holonomy group at $x$; harder to see but still true is that it is a Lie group.

Chaow Wu

hello again chat, can anyone help me with the following proof? Show that dim(W1) + dim(W2) = dim(W1+W2) + dim(W1 n W2).

So what i did was set x1...xk as the basis of W1 intersect W2. Then I said i could extend this basis for w1 and w2, so B1= x1...xk,y1..ym and B2 = x1... xk, z1... zn. I take the union of the two and claim that it is a basis for B1+B2. the generating part is trivial, but i cant seem to prove linear independence. Ive tried contradiction, but that gives me that z1 is a linear combination of xs ys and zs, which doesnt look like much of a contradiction.

iwriteonbananas

So the group is trivial when $M$ is flat

Mike Miller

15:50

This is essentially independent of basepoint. So we just say the holonomy group of $M$ is a Lie subgroup of $O(n)$, defined up to conjugation.

iwriteonbananas

Ok

Mike Miller

Not necessarily. Flatness just says that parallel transport only depends on the homotopy class of the loop, so you have a surjective map $\pi_1(M,x) \to \text{Hol}(M,x)$.

A lot of the time it's desirable to work in simply connected manifolds because this stuff is better behaved there, eg the holonomy group is a closed subgroup.

iwriteonbananas

Ah, okay

Mike Miller

henceforth for convenience everything is simply connected.

iwriteonbananas

Nod

Mike Miller

15:55

holonomy is a pretty powerful invariant. for a "generic" Riemannian manifold you have holonomy $O(n)$; for a generic oriented one, you have holonomy $SO(n)$.

First fact: Note that because it's a subgroup of $SO(n)$, holonomy is really a group $G$ equipped with a representation (up to conjugacy). Suppose this representation was reducible, like the action of $O(k) \times O(n-k)$ on $\Bbb R^n$.

iwriteonbananas

It's a subgroup of $SO(n)$? Was that meant to be $O(n)$?

Mike Miller

yeah

iwriteonbananas

Ok, sure

Mike Miller

Then de Rham's theorem says your manifold is actually locally isometric to a product of Riemannian manifolds.

So holonomy can actually capture completely whether or not you locally have the structure of a product.

if you're complete, that's actually a GLOBAL isometric decomposition.

iwriteonbananas

That's definitely neat.

Wow

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