Mathematics - 2016-02-14 (page 3 of 3)

Agawa001

7:12 PM

oh, what happened to the server ?

Tobias Kildetoft

@Agawa001 Hardware upgrade

Agawa001

@TobiasKildetoft good, hope questionability wont be payable henceforth

Tobias Kildetoft

@Agawa001 Were those actually words that can be put together like that?

Agawa001

hope that*

plz stop concentrating on my grammar, as mathematicians we are known to priorize content over than appearance

I'm an artist

@TobiasKildetoft John Nash had a phd thesis of 26 pages, so it's not bad if there is much quality.

Tobias Kildetoft

7:20 PM

@Agawa001 I was referring to the actual words, not the grammar. I have no idea what you were trying to say.

@I'manartist True.

Agawa001

@I'manartist ?

I'm an artist

openculture.com/2015/06/…

Agawa001

phd of 26 pages ? lol, my licence thesis is even 4 times width

Balarka Sen

Milnor's exotic spheres paper was 8 pages, if I recall correctly.

Agawa001

thesis =/= paper

mr eyeglasses

7:28 PM

yay i've started getting my rejection letters from schools

Agawa001

@mreyeglasses rejection is my best friend

I'm an artist

Just for the record: despite the discussions we had here (during the time), I have absolutely nothing with no one, and my questions simply arise from curiosity.

@TobiasKildetoft ^^^

Actually, I'm so busy with my research that I lost the meaning of being upset.

hmmm, I'm upset though on the too resistent integrals and series, but at the moment I slayed everything.

user276387

Is this the correct way to set up a Cayley table for dihedral group $D_6$? Where the $r$'s denote rotations and $f$'s denote reflections.

$$\begin{aligned} \begin{array}{cc|c|c|c|}
* && e & r & r^2& r^3& r^4& r^5& r^6& r^7& r^8& r^9& r^{10}& r^{11}\\
&&&&\\
\hline
e && & & \\
\hline
f && & & \\
\hline
rf && & & \\
\hline
r^2f && & & \\
\hline
r^3f && & & \\
\hline
r^4f && & & \\
\hline
r^5f && & & \\
\hline
r^6f && & & \\
\hline
r^7f && & & \\
\hline
r^8f && & & \\
\hline
r^9f && & & \\
\hline
r^{10}f && & & \\
\hline
r^{11}f && & & \\
\hline
\end{array} \end{aligned}$$

Balarka Sen

O_O

Tobias Kildetoft

@user276387 No. It needs to have the same things for rows and columns

And those needs to be all the elements. You have several elements repeated in those lists.

Are you sure you need to set this up? Cayley tables are fairly useless for groups of this size.

user276387

7:38 PM

What do you mean when you say the same things for rows and columns?

Also, which elements have I repeated? Back to the drawing board I guess.

Tobias Kildetoft

I mean that you need to label the rows and columns the same way, namely by picking an ordering of the group elements and using that for both

Semiclassical

now I find myself trying to recall what the Cayley graph (not table) would look like

Balarka Sen

vertices are the elements of the group, edges correspond to multiplication by different generators.

(it's a directed colored graph)

Tobias Kildetoft

@Semiclassical it would depends on your choice of generators

Semiclassical

true.

Balarka Sen

7:43 PM

Oh, and that ^.

Tobias Kildetoft

The right choice is of course as a Coxeter group

Balarka Sen

@TobiasKildetoft What do you mean by the right choice? Curious.

Semiclassical

if i take Wikipedia's first example for $D_4$

Tobias Kildetoft

@BalarkaSen It was meant as manly a joke (though the Coxeter presentation does have many advantages)

Balarka Sen

Oh, OK. I get it.

Semiclassical

7:45 PM

there we go

Balarka Sen

Not every group is a Coxeter group.

Tobias Kildetoft

@BalarkaSen Right (that would have made finite group theory a very different looking topic)

Semiclassical

i'd have to check, but I'd think that the analog for $D_6$ would be a pair of hexagons

Tobias Kildetoft

@Semiclassical Yeah, that is clearly the "wrong" choice :)

Semiclassical

pfffff

Tobias Kildetoft

7:46 PM

Not that I ever actually used a Cayley graph for anything

Balarka Sen

You can avoid the choice of a generator set by looking at the graph coarsely though. This is what one does in geometric group theory.

@Semiclassical That's right. The inner circle corresponds to $\langle a \rangle$, and the outer circle is multiplication by $b$.

Tobias Kildetoft

@BalarkaSen Sure, but usually for infinite groups, right?

Balarka Sen

There is no such thing as a finite group :D

Semiclassical

the other example from Wikipedia for D_4 is this

Tobias Kildetoft

@Semiclassical Yeah, that is the Coxeter presentation

No idea what those F's are though

Semiclassical

7:49 PM

that's just to show the different orientations

Balarka Sen

foo

Semiclassical

start with $e$ as the identity

going to $c$ reflects $F$ across a diagonal, as does going from $b$ to $cb$

Tobias Kildetoft

ahh

Semiclassical

etc. etc. i forget what the Coxeter presentation is about, though I recall a linkage with reflections

which would fit with that diagram

Tobias Kildetoft

Of course, the reason to pick the Coxeter presentation is that then we can easily switch to the KL basis for the group algebra, which has some very nice properties.

Semiclassical

7:51 PM

KL basis?

Tobias Kildetoft

Kazhdan-Lusztig basis

Semiclassical

ah. no idea what that is

Noble Mushtak

I am sorry to interrupt this discussion, but for me, something seems to have happened to the Math.SE site. It seems to be down and says that it's not accepting any comments, questions, or answers, yet I can still see others post new questions and answers. What is happening?

Tobias Kildetoft

For the dihedral groups you just take the sum of all elements which are smaller in the Bruhat order

@NobleMushtak Try to refresh. It should be back up as far as I can tell

Noble Mushtak

OK, it's back. Thank you!

user19405892

8:42 PM

hi

i just thought of a really interesting geometry/number theory problem

Balarka Sen

8:54 PM

Wee, I have just found a beautiful counterexample to what I thought should be true.

Hmm, but apparently this raises further questions on how much true the original claim can be.

oops my counterexample apparently doesn't work. sigh.

Mambo

Hello

user276387

Still on the subject drawing the Cayley table for $D_6$. Apologies since this takes up so much space. But I'm reposting this.

$$\begin{aligned} \begin{array}{cc|c|c|c|}
* && e & r & r^2& r^3& r^4& r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f\\
&&&&\\
\hline
e && e & r & r^2& r^3& r^4& r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f \\
\hline
r && r & r^2 & r^3& r^4& r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f & f \\
\hline
r^2 && r^2 & r^3& r^4& r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f & f & rf \\
\hline
r^3 && r^3& r^4& r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f & f & rf & r^2f \\
\hline
r^4 && r^4& r^5& f& rf& r^2f& r^3f& r^{4}f& r^{5}f & f & rf & r^2f & r^3f\\

I'm an artist

YYYYUUUPPPPPPIIIIIIIII, ANOTHER ABSOLUTELY AMAZING RESULT IN PLACE!!!!!!!!

user276387

Is that any better?

Balarka Sen

OK, counterexample fixed. Now it works, I believe.

Agawa001

9:40 PM

@I'manartist when u said john nash, i though u meant jasper :D

I'm an artist

@Agawa001 lol :-)

Agawa001

i saw that movie "beautiful mind", the guy had very serious dating problems

Link

Hi, I have um a matrix $A$ and need to find a matrix $P$ such that $PA = R$, where $R$ is the reduced row echelon form of $A$, I've found $R$, but unsure what to do next, anyone know?

Balarka Sen

@Link Row reduction is the same as multiplying by the elementary matrices.

Link

@BalarkaSen Um, can you elaborate what that implies? I'm not sure how to use that information.

Balarka Sen

9:51 PM

Do you know what elementary matrices are?

Agawa001

be back to my code [switch to CS mode]

Mike Miller

@Link: Every time you do one of your three "moves" (addition of rows, scaling of rows, swapping of rows) you get there by multiplying to the left by an elementary matrix. So write down these as you do them, and what you get is that to reach $R$ you multiplied by a bunch of elementary matrices in order. Then $P$ is the product of those elementary matrices.

Link

@MikeMiller However $A$ is a $2x3$ matrix, so I don't think I can multiply a host of $2x3$ matrixes since the columns $=/=$ rows for subsquent matrices?

Balarka Sen

The elementary matrices won't be $2\times 3$. They will all be $2 \times 2$.

Link

@BalarkaSen Okay.

Balarka Sen

9:58 PM

I mean, row reduction cares about what happens to the rows. In this case you have $2$ rows, so the elementary matrices are $2 \times 2$. The columns are irrelevant.

Link

@BalarkaSen How do you represent adding one row to another with an elementary matrix?

Balarka Sen

If you're adding row $i$ to row $j$, take the identity matrix and replace a $0$ by a $1$ on the $(j, i)$-th place.

Link

So adding row $1$ to row $2$ is simply:

$\begin{bmatrix}1 & 0\\ 1 & 1\end{bmatrix}$ in this case, correct?

Balarka Sen

Yes.

Link

Okay. Thank you.

user19405892

10:25 PM

hi

skull petrol

hi

Balarka Sen

Phew, got loads done today.

user19405892

do there exist numbers, besides 0, where there are multiple combinations (x y) that equal it?

for example for 21 it

s just 7 choose 2

Balarka Sen

@user19405892 $\binom{5}{2} = \binom{5}{3}$. There are a buckload of these. All comes from realizing them as coefficients of the binomial expansion.

Specifically, $\binom{n}{k} = \binom{n}{n-k}$ in general.

I'm off to bed, can't stay awake any more.

skull petrol

Later pal.

Balarka Sen

10:39 PM

See ya, @skull.

Ted Shifrin

Ah, @skull has returned to his original moniker?

skull petrol

Yes Professor @TedShifrin

Semiclassical

hi @ted

user777

Hello

I have a question. I want to prove the general case of XOR operation by mathematical induction. But I have a problem to do this. At first thought, I think it is recursive mathematical induction;

Mike Miller

10:56 PM

Morning @Ted.

user777

Suppose that the basis step is for k=2. That means we have $2^2$ possible values. We can show that using 'addition modular 2' that the result would be as expected. Now, how to show the general case for any non-negative integer?

Ted Shifrin

11:15 PM

Goodnight, @MikeM. Hi, @Semiclassic.

L33ter

hey @TedShifrin

Ted Shifrin

Hi Karim

L33ter

I was starting to solve you know some questions in munkres book that I didn't solve last semester. I would like to finish it's excerises before going to grad school and the whole book. It is amazing how fast I see the answer now that I have done the course.

I used to before you know in my free 10 min or 1 hour something to watch something funny, but I decided to use my time effectively and solve problems in my free time.

btw @TedShifrin I am starting to improve in my English. I got 70% in my first essay in English 110, and next time I will get a higher mark.

Balarka Sen

Hi @TedShifrin.

L33ter

11:30 PM

@BalarkaSen

how are you

Balarka Sen

fantastic.

L33ter

me 2

Can you answer me this question when is this specific topology a topology on X and why ?$\tau_{\infty} = \{ U : X - U \ is \ infinite \ or \ empty \ or \ all \ of \ X \}$

Balarka Sen

have done anymore topology, @Karim? you were stuck on delta complexes all the time I talked to you :(

L33ter

I just came with a nice argument

yes

I completely understand it now

and I am just reading on singular homology

Balarka Sen

@L33ter I can't parse the question. Can you rephrase?

L33ter

11:34 PM

Can you give me a condition on a set X that would turn this collection into a topology ?

Also why if you have any other condition on X, then it fails to be a topology.

I modified this question from munkres to actually answer it in full generality. I am gonna do the same with all problems from munkres.

My claim is it will be a topology if X is finite, and if X is infinite;whether, it is countably infinite or uncountable infinite, then it will not be a topology.

Balarka Sen

Well, obviously that can't be a topology for general $X$. It might as well be possible that $U, V$ both have infinite complement but $U \cap V$ has finite nonempty complement, e.g., $[0, 1]$ and $\Bbb R - [0, 1)$ in $\Bbb R$.

@L33ter If $X$ is finite, this would be the same as the trivial topology.

L33ter

yeah

Here is one nice argument that I came up with. First suppose that X is countably infinite, so it has a bijection between itself and N, so one way to construct a two open sets that have their union not being element of the topology is as follows.

Mike Miller

@L33ter Any infinite set $X$ is in bijection with $X \sqcup \Bbb N$. It might be more convenient to think about your question for infinite sets in terms of the latter set.

L33ter

yeahhhhh

yeahhhh exactly @MikeMiller

that was the idea I used in my general proof

Because in the case when it is countably infinite, then you can seperate the set into even and odd after taking the bijection between itself and N. Then you can construct two open sets that have finite set as follows : First you take the set $U_1$ to be the even minus one particular even number, and $U_2$ to be the odd numbers. So, once you take the union it will give you that particular even number, which is finite set.

Balarka Sen

I don't see the necessity of doing this. $A$ be a finite subset of $X$. Consider $X - A$. Partition that into two infinite sets $U$ and $V$ and take complement of $U$ and $V$ in $X$. $U^c$ and $V^c$ both have infinite complement, but $U^c \cap V^c = (U \cup V)^c$ has finite complement, namely, $A$.

That immediately proves that $\tau_\infty$ cannot be a topology for infinite sets $X$, right?

L33ter

11:47 PM

yeah

that is easier good one @BalarkaSen

Akiva Weinberger

Hello @TedShifrin

And Balarka

L33ter

hi @TedShifrin

Balarka Sen

Hi mr. pathologi(cal topologi)st and meridianal confusionist.

Akiva Weinberger

I must have missed the "pathologist" conversation.

I can be pathological, too! Especially with respect to lying.

Balarka Sen

Which lie is that?

Way too many ways to lie in English.

skull petrol

11:57 PM

More than other languages?

Balarka Sen

Haha, what I meant is that "lie" has several meanings in English.

Akiva Weinberger

Telling untruths.

Ted Shifrin

Balarka, you're supposed to be asleep. Hi, DogAteMy.

Akiva Weinberger

Hi!

Balarka Sen

A pathological liar and lying down pathologically can mean subtly different things.

Akiva Weinberger

11:59 PM

@BalarkaSen I think that would be called "hypersomnia."

L33ter

@BalarkaSen we are doing cryptography now in my algebra class.

It is very interesting

Balarka Sen

@TedShifrin Well, I was sneaking out to see if you had arrived. I have done quite a few bragable problems from your book, and have also thought about the theory. Going to send you a progress report e-mail soon.

L33ter

We were actually talking about the engima or whatever it is called and how it was done

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