12:03 AM
:P

Yeah, Peter. Mathrollatics FTW.

Well I gotta go jogging now. Dawn is just breaking out and its going to be a cloudy day. Depressing weather.
Bye guys see ya later.

@PeterTamaroff Cant find you on facebook, do you post cool stuff there?

@GustavoBandeira Well, that is subjective

12:18 AM
@PeterTamaroff Found.
@PeterTamaroff You have a trollface.
Have you seen WA facebook report tool?

@GustavoBandeira At least I'm not a murderer!

@PeterTamaroff Look

@GustavoBandeira Woah. How do you get that?

Is anyone here familiar with Hatcher's Algebraic Topology? I have a question about a bit of notation in that book.

12:27 AM
@DavidK. @BenjaLim

12:43 AM
I'm back!

@MeAndMath Yay.

I sent you the e- mail

@MeAndMath I got it. Did you find me?

didn't look it yet :-P
I didn't find it...

@DavidK. yes?

12:51 AM
Hey@PeterTamaroff....

@DavidK.which page of hatcher?
@PeterTamaroff ?

@BenjaLim Help the guys insensitive bastard.

@PeterTamaroff who?
@PeterTamaroff I have a Lie theory exam tomorrow.

@BenjaLim Did you go with the Russians?

12:52 AM
@BenjaLim David K.

@BenjaLim Are you familiar with Hatcher's Algebraic Topology? I have a question about a piece of notation that he uses.

Hello

Do you know some book with simple Kirchoff examples about time-constant -thing with capacitors and inductors?
(i.e. you need to calculate the time-constant etc -- looking some training to do more exercises...)
(simple integration but my skills too slow without some training and seeing examples)

@DavidK. which page?
@hhh If you want to ask physics questions, go to physics.se

@BenjaLim Chapter 0, page 3.

hold on
yes

12:55 AM
I mean this thing here. @JayeshBadwaik

When talking about a deformation retraction of a space $X$ onto a subspace $A$ being a homotopy from the identity map of $X$ to a retraction of $X$ onto $A$, a map $\gamma:X\to X$ such that $\gamma(X)=A$ and $\gamma\mid A=\mathbb{1}$ (the identity map)

@BenjaLim "yes yes yes yes yes"

@hhh For god's sake can you post these on physics.se?
@DavidK. yes

@BenjaLim What does the notation $\gamma\mid A$ refer to?

@DavidK. It means the restriction

Technically he should write $r|_A = 1$

@BenjaLim Yes, that is the notation that I am used to. Thanks!!

In Isaacs Algegra, in an exercise in the group theory part he uses the notation $x^\sigma$ for $x\in G, \sigma\in$Aut(G). Is that supposed to mean application of $\sigma$ to $x$?

@DavidK. It is the same
$f\mid A$ means the restriction of $f$ to $A$.
Just different notations.

@PeterTamaroff @BenjaLim Ok great. Thanks!!

12:57 AM
@DavidK. are you learning algebraic topology ?

@BenjaLim Yes. Just started.

@DavidK. What year of uni are you in?

@BenjaLim I'm an undergraduate in my senior year.

third year?

4th

12:58 AM
@DavidK. ah ok.
@DavidK. Let me tell you:
Algebraic topology is bloody hard
don't expect to get all the details
@DavidK. I have been posting so many questions on AT recently

@BenjaLim Yes, I've actually heard that from a few people. My professor is pretty good, but he goes really fast.

@DavidK. Don't expect to understand all the details
best advice I can give you:
Understand the general ideas
@DavidK. And also:
You will need to get used to the handwaving

@BenjaLim As an algebraist, I have actually learned to accept the handwaving quite easily.
That't not what I meant.

@DavidK. the handwaving in hatcher is on an unprecedented level

@BenjaLim "Whiney whiney woo"

1:01 AM
What I meant was that I have surprisingly been able to accept the handwaving, quite easily.

in Hatcher till now?

@BenjaLim Why handwaving? Because the results used are tedious and easy to prove? Or because they require some higher knowledge?

@JayeshBadwaik Look at Hatcher Proposition 1.26, the proof of it :D :D :D :D :D :D
@JayeshBadwaik I know how to prove the proposition rigorously if you only attach one 2-cell

@BenjaLim I see. So basically, sketches and extrapolations.
I had a book on analysis by protter morrey, that had a similar style. However, there, they had explicitly mentioned as such," to allow the student to fill in the proof." and just guidiing him in the right direction.

1:50 AM
Hello folks.

Hello.

@anon How is university?

My current potential research topic is on how to use Witt vectors to characterize degree p^2 extensions of p-adic fields. Other topics may come up in the future.
Pretty good. I don't like having to get up in the morning for a lecture I don't even listen to, but...

@peoplepower Yes. Indeed.
@anon Hmmm, pretty interesting stuff I can guess.

@anon Why not witty vectors? Those can surely help more!

1:56 AM
@anon I am supposed to come up with an algorithm to count the number of matrices with prescribed row sum and column sum vectors.

With what kind of entries?

oh, of course non-negative entries.

Was just about to mention that :)

So, we are in $\mathbb Z_+$.
But, I am left to all by myself. I haven't come up with a good algorithm yet.

@KannappanSampath You have been given a row sum and a column sum? and you have to find all the matrices satisfying that?

2:01 AM
@MarianoSuárez-Alvarez Hola.

@JayeshBadwaik I have been given a vector $\underline{r}$ and $\underline{c}$.

Okay, and ?
Okay, got it. So you have a vector of sum for rows and another vector of sum for columns

Now, I need to count the matrices with entries in $\mathbb{Z}_+$ such that the rows add to the vector $\underline{r}$ and the columns add to the vector $\underline{c}^t$. All my vectors are row vectors, unless specified otherwise.

Okay.
In the most basic format, the solution is the intersection of the partition problem me thinks

Bye peoples.

2:04 AM

It might be too much to ask for a nice formula, as he said the problem was to devise an algorithm.
(Presumably an efficient one, otherwise brute force is easy to code for.)
Brute force with a lot of redundancy...

@KannappanSampath See, consider a single row. Then the set of solutions for your main problem is a subset of the set of solutions of the partition problem for that single row.

@JayeshBadwaik Yes. But, the problem is, the partitions could fit in several ways no?
For instance, the partition of $4$--$(2,2)$--in $(2,2,2,1)$ fits in three distinct ways.
So, we'll get three distinct sets of solution...

His problem is one of counting, not enumeration (the latter of which costs much more computationally). Obviously if you wanted to enumerate all $\Bbb Z_{\ge0}$ solutions you could enumerate integer partitions of each row and column component individually and then form two families of matrices by putting together permutations of the row and column partitions respectively and find the intersection of these two families, but that's way infeasible.

@anon ohh okay. I did not notice counting. I read it as enumeration. My bad.

2:10 AM
So, perhaps, I should write some recursion, but I just cannot figure where to look. :(

Essentially it's a linear system of 2n equations in n^2 variables, if you want to invoke linear-algebraic methods.
Maybe it could be a problem of counting lattice points within a simplex in the first $2^n$-tant.

Hmm: an open source says: Except for trivial cases, the counting is done by brute force generation.

2:28 AM
Suppose the vectors and columns of sums are $\underline{r}$ and $\underline{c}$ respectively. And suppose the number of solutions is $S(\underline{r},\underline{c})$

Suppose the sum in the first row is $r_{1}$ and the first column is $c_{1}$. Let $m_{1,1} = \operatorname{min}(r_{1},c_{1})$. Then the element $a_{1,1}$ has the range ${0,1,...,m_{1,1}}$

S(\underline{r},\underline{c}) = \sum\limits_{k=0}^{m_{1,1}} S(\underline{r'},\underline{c'})

Where $r'$ and $c'$ are the appropriately modified vectors. After having derived this formula. We can go backwards,

@KannappanSampath If you want a recursive algorithm you might consider thinking about filling the top row and leftmost column in by generating things $\le$ the given sums, and then calling the algorithm for the left-over $(n-1)\times(n-1)$ minor (via excising said row and column) in each instance, but that seems like a lot of recursion.

@anon I tried this; but the problem, there are several ways a column and a row can be filled. So we'll have to multiply at each stage by that number and so on... But to find that factor itself seems very hard.

I'm unsure of what multiplication you're referring to. You iteratively generate the left corner of the matrix (corner = leftmost column + topmost row) (which corners are possible given the row/column sum vectors is an easy task), and given each corner of the matrix you decrease the row/column sums by the relevant amount and the apply the algorithm to the $(n-1)\times(n-1)$ minor left behind (obviously creates a lot of recursion).

2:40 AM
@anon I got it!!!!!
0

I think I know how to show that given any $X \in \mathfrak{g}, Y \in \mathfrak{h}$ that $e^Xe^Ye^{-X} \in H$. Now $$\begin{eqnarray*} e^{X}e^Ye^{-X} &=& 1 + e^XYe^{-X} + e^{X}\frac{Y^2}{2!}e^{-X} + \ldots\\ &=& 1 + e^{\textrm{ad}_X}(Y) + \frac{1}{2!}(e^XYe^{-X})(e^{X}Ye^{X}) + \... I knew I had to use ad somehow @KannappanSampath I guess my method is good. Start with the problem of the sum of all row vectors zero and column vectors zero. For that we have one solution, namely the zero matrix. Now let only one row and one column have the sum 1. Count the number of such matrices. It will be m \times n where m and n are the number of rows. Next, consider with two rows of sum one and two columns of sum 1. Calculate this, we will get mC2 \times nC2. Continue this, till we are left with a matrix with all columns 1 and all rows 1. If rows are not equal to columns, then obviously, one of them will Looks nice, I will have to look at it better later though. I will write a code in the afternoon, have to run now. bye @anon The feeling is so satisfying!! @JayeshBadwaik Sure, later. :-) I think I'll first think of (0,1)-matrices. What do you think @anon? 2:46 AM What's a (0,1)-matrix? oh, duh, 0s and 1s Sounds like a good entry point to check out. @anon, can I ask you a number theory question? sure Ok, it is about the artin symbol. 3:02 AM I don't know that much algebraic number theory.. Given a number field K and a prime ideal \mathfrak{p} \subset O_K. Let L /K be a galois extension in which \mathfrak{p} does not ramify. L/K? For any prime ideal \mathfrak{P} in L  containing \mathfrak{p}, we define \sigma , the unique element of Gal(L/k), for which \sigma(\alpha)=\alpha^{N(\mathfrak{p})} (\pmod {\mathfrak{P}}) \pmod{\mathfrak{P}} Ah, thank you for your patience Now, given any element \tau in the galois group prove that  \sigma for \tau (\mathfrak{P}) is the same as the element \tau \sigma \tau^{-1} There. I can make no headway on proving it. Apparently it is "easy" 3:09 AM @anon Hmm, thank you. Uniqueness, surely I'll be here after some time. Could you expand on that, @ZhenLin? Hey Apply \tau to both sides of \sigma(\alpha)-\alpha^{N(\mathfrak{p})}\in(\mathfrak{P}) and then replace \alpha with \tau^{-1}(\alpha), right? 3:10 AM You defined \sigma by a uniqueness property, so use it! heh, that's right I'm not sure I'm getting it. In fact, I am sure I am not getting it. Show that \tau\sigma_{\mathfrak{P}}\tau^{-1} satisfies the defining property of \sigma_{\tau(\mathfrak{P})}, and by uniqueness of the latter they are one and the same that's supposed to be a subscript, but \mathfrak{P} is a big letter So \tau\sigma_{\mathfrak{P}}\tau^{-1}(\alpha)= \tau(\tau^{-1}(\alpha))^{N(\beta)} \pmod{\mathfrak P} , for any \alpha yeah, now I can use that \tau is an isomorphism In other words, show that for all \alpha, we have$$(\tau\sigma\tau^{-1})(\alpha)\equiv\alpha^{N(\tau(\mathfrak{P}))}\pmod{\mathfrak{\tau(P)}}$$given that you know that$$\sigma(\alpha)\equiv\alpha^{N(\mathfrak{P})}\pmod{\mathfrak{P}}$$for all \alpha. You can do this by noting that N(\tau(\mathfrak{P}))=N(\mathfrak{P}) and \alpha^{N}=\tau\left(\big(\tau^{-1}(\alpha)\big)^{N}\right) and x\in X\iff f(x)\in f(X) if f is a bijection. 3:17 AM Ah, yes. So, it is just the notation that is messy. Thank you so much. @FortuonPaendrag I'm not sure that's true. Do we always have \tau(\mathfrak{P})=\mathfrak{P}? (I'm considering the modulus you're using.) @FortuonPaendrag What's \mathfrak{P} meaning? Good night, Fortuon. =) @GustavoBandeira It's a prime ideal of the ring of integers associated to the field. @anon We have no guarantee of that. \tau can be as weird as possible in the Galois Group. Yes, so you need what you wrote on that line to be modulo t(P), not just mod P. 3:21 AM But, we mistakenly typed N(\mathfrak{P}) instead of N(\mathfrak{p}) though. @GustavoBandeira Good evening indeed! @anon Thanks. =) @anon Ah, you are saying the same thing, Honestly I do not see the point of the base field k or the lower prime ideal \mathfrak{p}, are they even relevant? (Or is it part of the setup to more parts of an overall problem?) Yes! Very much so. If you see the definiing property of the artin symbol, it is an element of Gal(L/k) oh. still, \mathfrak{p} though... 3:24 AM and next thing, we are raising the arbitrary element \alpha to N(\mathfrak{p}). oh, that's supposed to be small \frak p And all this defining depends on the fact that \mathfrak{p} does not ramify in L well, I'll leave the details to you then :) But, thank you @anon and @Zhen . I hope you have a good evening :) Or whatever time it is for Zhen @FortuonPaendrag Do you know about Umbral Calculus? 3:29 AM I've been scribbling the \mathfrak{P} symbol all over my notebooks. I find it incredibly cool. @GustavoBandeira Yes, I have heard about it, but do not really know much about it. I'm going to ask it. No one asked about it before. You mean ask for an explanation? Did you try Wikipedia? It is quite decent. @FortuonPaendrag Yep. The wikipedia article is open here, but it's not so clear. I've also found this: Yeah, but wikipedia has all that and more. But if you need more clarity, go ahead and ask. Although, IMHO, you would benefit more from strengthening your essentials before trying to understand these "higher" concepts. @FortuonPaendrag I'm not trying to fight umbral calculus - I mean, for understanding it. I'm in search of: 1 - A simple explanation on some aspects of it; 2 - To make people know that it exists, it may be useful for someone.$$(y+x)^n=\sum_{k=0}^n{n\choose k}y^{n-k} x^k
What's the meaning of the parentheses with N and K?

3:41 AM
I understand. But I feel like you would have a better feel for these ideas once you take more classes. So, I am ashamed to ask, as you told me once.. but what is your math background?
It is the number of ways to choose $k$ distinct objects from $n$ objects.

@FortuonPaendrag Relax with that, people have the right to forget.
2

Or alternately if you understand $n!$ means, it is equal to $\frac{n!}{k!(n-k)!}$

Yep, I understand factorial.
I don't knowhow to describe my math background, but it's not too much.

Anyway, I am going to be off to bed, so can we talk another time?

Ok. =)

3:45 AM
Good night :)

Thanks for the information on the N and K in the parentheses.

user19161
4:14 AM
@GustavoBandeira Do you mean the n choose k part? Well, that is just the number of ways to choose k objects out of n objects, where distinct arrangements of the k objects constitute the same way. You may read up more on "combinations" and "permutations".

hi all

user19161
@jshin47 Hello. What's up?

Im looking for a bit of quick help on a lemma I need

user19161
Why not just post on the main site?

I guess I will!
I just thought it was rly small
like a yes or no answer]

user19161
4:18 AM
What is it anyway?

I just want to know if I am going down the right road
I just want it to be true that the map A -> (A_transpose)_inverse defines some isomorphism between
GLn(F) over some field F (n X n matrices) onto itself
well I dont need it to be onto
just a map from itself to itself
And I believe it is
very fuzzily in my mind its somehow connected to the old change of basis stuff ...
but Im not sure if that is the right road to go down
and that is my question
apologies for not knowing tex.

4:40 AM
@jshin47 What doubts do you have?

@PeterTamaroff Hola!
@DylanMoreland, you can be opening Pandora's box with that question! :-)

Hm, true.

4:56 AM
The road to Pandora's box is the most traveled ;-)

5:10 AM
Does that mean anything

@WillHunting Yes, I got it. =)

@DylanMoreland In my opinion, it means most people ask questions that are overly complicated to answer quickly.

6:07 AM
I'm beginning to notice a pattern... it seems my university only has access to "articles" on SpringerLink, but not "books"...

6:49 AM
You can ask your librarian... S/he surely knows the precise access details :-)
People should consider examples before making conjectures...

@MarianoSuárez-Alvarez It is quiet possible though. It was that way at my univ too. We had access to only articles, no books. Confirmed with authorities.

I did not say it was impossible
I just meant that there is generally no need to guess at the pattern: one can ask!
most big deals with springer are in fact only papers

Could someone help me with trying to understand Qiaochu Yuan's answer here: math.stackexchange.com/questions/191296/…

Suppose $F$ is a module with a subset $B\subseteq F$ which has the correct property
let $L$ be the free module on the set $B$

I should mention that I don't know anything about category theory

6:55 AM
the natural set map $B\to L$ has a unique extension $f:F\to L$ which is a map of modules, precisely because of the hypothesis made on $B$

Right

on the other hand, there is a map of modules $g:L\to F$ which maps the free generators of $L$ to "themselves"
show that these two maps $f$ and $g$ are mutually inverse, using the uniqueness property assumed on $B$ and the correspondingn property of $L$
Qiaochu has the interesting habit of assuming everyone learns Yoneda's lemma from early childhood
:-)

OK, I don't know how to show that they are mutually inverse

well, think

I think I can get one direction