Mathematics - 2016-05-26 (page 1 of 2)

OneRaynyDay

12:15 AM

Hey guys! I was just wondering whether an answer I came up with was correct for the given problem(linear algebra):

Find all 2x2 matrix such that E_7 = R^2.

I thought since the matrix could be composed of any coefficients [[a,b],[c,d]] such that ax+by = 7x and cx + dy = 7y.

arctic tern

@OneRaynyDay what do you mean by E_7?

OneRaynyDay

@arctictern The eigenspace where the eigenvalue is 7.

arctic tern

so matrices A for which det(A-7I)=0?

OneRaynyDay

Yep! But specifically that the span of the matrix is all reals in 2d.

arctic tern

ah, E_7= all of R^2

OneRaynyDay

12:20 AM

I know how to do it if the span is a specific vector, but my intuition fails for all reals in 2D

arctic tern

hint: if e1 and e2 are the two basis coordinate vectors, then Ae1 and Ae2 are the columns of A

actually you don't need a basis

if v is in R^2, then Av=7v, so A acts as multiplication-by-7...

OneRaynyDay

Oh - I see. If e1 and e2 compose the two basis vectors then it spans all reals in 2D right?

Right - So I gave the answer [[7,0],[0,7]], effectively 7I.

But there's no answer for this(even questions darn you), and I don't know if its a unique solution :o

arctic tern

you know the first column of the matrix. you know the second column of the matrix. so you know the whole matrix.

alternatively, distinct matrices correspond to distinct linear transformations and vice-versa. since Av=7v for all v, A must be 7I.

OneRaynyDay

Ah, I see what you mean. So the "for all matrices that..." was just a mind-trick

Since the distinct matrices thing is true/obvious but I wasn't able to pick up on that

arctic tern

I don't see the phrase "for all matrices" anywhere. I see you typed "find all 2x2 matrix" ...

OneRaynyDay

12:26 AM

@arctictern Right - I mean for 2x2's.

Axoren

I bet there's a bunch.

At least one.

usukidoll

hiii

arctic tern

hi

OneRaynyDay

@Axoren a bunch? After arctic convinced me, I'm pretty sure there's only one :)

usukidoll

1:05 AM

anyone here

Surdz

1:16 AM

hi there,

if f:R→R and $f(x)=x^{2}$ is it injective? i think it is not since $f(−1)^{2}$=1 and $f(1)^{2}$=1

arctic tern

you mean since f(-1)=1 and f(1)=1

Surdz

yeap

f(-1)^2=1 and f(1)^2=1

OneRaynyDay

Hey arctic! I'm trying to get the MathJax thing to work, but for some reason when I click on rendering on it doesn't seem to change anything.

Ironically I should know this since the professor who made this is from the same uni I'm in LOL

arctic tern

1:33 AM

@Surdz no, f(-1)=1 and f(1)=1.

@OneRaynyDay you made the mathjax code a thing in your bookmarks bar right? and you click it while you're on this tab?

OneRaynyDay

Yup - I clicked on it and there's literally no difference. Should I refresh or something?

arctic tern

@OneRaynyDay did you not try that already?

OneRaynyDay

@arctictern Tried - and no difference.

arctic tern

does it work on the mathjax page where you got the bookmark thing?

OneRaynyDay

yup - when I clicked it the mathjax page works.

Maybe it doesn't like opera?

arctic tern

1:37 AM

it's possible. I don't know if we've experimented much with opera.

OneRaynyDay

I see. By the way - does RobJohn frequent this chat?

arctic tern

depending on what you consider frequent.

you can always ping him and he'll get back to you. better than silently waiting to talk to him live.

OneRaynyDay

Perhaps once per week? I've never seen him on campus actually. Didn't even know there was a math professor called Rob John.

@robjohn Hello there! Are you still teaching at UCLA?

And ah, @arctictern it works with chrome :)

Axoren

2:05 AM

@OneRaynyDay That's right, there's at least one. There's also at most one. It was a joke about your question stating "all".

OneRaynyDay

3:39 AM

@Axoren Gotcha :)

Damon Williams

7:38 AM

/

Hanan N.

8:18 AM

Do you think my solution is right? I'm getting different results on online calculators.

Tobias Kildetoft

Fortunately, checking is just a matter of differentiating the result.

ADG

Hello people, I have no idea about legendre transform but was reading some paper, can you procide some sources to read it.

I only found it in context of thermodynamics and classical mechanics

Tobias Kildetoft

@ADG didn't the paper provide any?

ADG

@TobiasKildetoft they kind of skipped it, if you widh i can tell which line of some paper

i prefer purely mathematical context first

Tobias Kildetoft

@ADG I don't know anything about it myself

ADG

8:24 AM

hmm, no problem

Hanan N.

@TobiasKildetoft where do think my problem is?

Tobias Kildetoft

@HananN. No idea. What did the calculator tell you the answer should be?

ADG

@HananN. I see a simple integration picture you have attached, what du u wish to know

Tobias Kildetoft

And what do you get when you differentiate your answer?

Hanan N.

differentiate the answer i get $ -sin(x)/2cos(x)^2 $

ADG

8:27 AM

you have put the substitution wrongly

Tobias Kildetoft

@HananN. Ahh, you have changed a $1-\cdots$ to a $1 * \cdots$

ADG

try differentiationg 0.5tanx-secx

Hanan N.

so it should be $1+...$ ?

Damon Williams

Hey guys when my lecturer asks us to write about "how Real Analysis/Number theory has influence your thinking about infinity," what do you think he is really looking for? It's the first analysis subject we've ever done

Tobias Kildetoft

8:42 AM

@DamonWilliams Why not ask him? He is the one who set the task

(sounds like a rather weird question to ask though)

Damon Williams

It's a homework question so he wouldn't tell us. But I have no idea what to write to be honest.

Tobias Kildetoft

@DamonWilliams What do you mean he wouldn't tell you? He has set a vague assignment so he needs to clarify it.

Damon Williams

All he said is: "It's meant to be a reflective piece. Write how it's changed your perspective about infinity."

Tobias Kildetoft

What sort of course was this?

Damon Williams

8:58 AM

It's called Real Analysis (also known as Number theory).

Tobias Kildetoft

Those are two completely different topics that do not overlap until a quite advanced level. What sort of place are you studying?

Damon Williams

I'm a 2nd year university student, studying at the university of Melbourne

Very basic level. Most of the subject was an introduction to analysis

Tobias Kildetoft

Right, so how is that number theory?

Damon Williams

My bad, I thought they were the same thing. Other universities teach number theory rather than real analysis in 2nd year

Tobias Kildetoft

@DamonWilliams Most universities teach more than one thing per year

Damon Williams

9:02 AM

Yea I've always thought they were the same. Thanks for clearing that up.

Tobias Kildetoft

I am not sure how you could get that impression. Does this mean you have not had any number theory at all? Like, not even some basic stuff like unique factorization of integers and stuff like that?

Anyway, I have no idea what your lecturer expects you to write. It seems like a weird assignment to me (I mean, what if it has not changed your view at all, should you just write that?)

Damon Williams

To be completely honest, it really hasn't. And to answer your first question, I don't know what number theory refers to in general

Gotta BS my way through it I guess

Tobias Kildetoft

In math, it is the general study of integers and certain generalizations (such as number fields)

Damon Williams

ohhh

Leaky Nun

9:45 AM

@HananN. I believe that the second step is wrong, namely you converted $1-2\sin(x)\ \mathrm dx$ to $1\cdot\mathrm du$

(where $u=2\cos(x)$)

Leaky Nun

10:00 AM

Here:
$\displaystyle\quad\int\frac{0.5-\sin(x)}{\cos^2(x)}\mathrm dx$
$\displaystyle=\int\left[0.5\sec^2(x)-\tan(x)\sec(x)\right]\ \mathrm dx$
$\displaystyle=0.5\tan(x)-\sec(x)+C$

Hanan N.

excellent.
Thanks.

Leaky Nun

anytime

Ilan Aizelman WS

Getting some difficulties here: math.stackexchange.com/questions/1800622/…

any help is appreciated

and here: math.stackexchange.com/questions/1800627/…

P.Andrews

11:04 AM

I think most would agree that 2*pii=ln(1). Because e^(2*pii) = 1, however ln(1) equals zero doesn't it?

Tobias Kildetoft

@P.Andrews Yes to the last, no to the first

exp is not injective as a complex function

P.Andrews

So is this true for i*pi=log(-1) as well? Anywhere I could read up on this?

Leaky Nun

@P.Andrews proofwiki.org/wiki/Definition:Logarithm

complex logarithm is multi-valued

in particular, $\ln(re^{i\theta})=\ln(r)+i(\theta+2n\pi)$

Tobias Kildetoft

@LeakyNun or alternatively you need to pick a cut to define it

Leaky Nun

where $r,\theta\in\mathbb R$

and $n\in\mathbb Z$

P.Andrews

11:14 AM

Thanks

Axoren

11:30 AM

Is there a better way to end the "Why was 6 afraid of 7?" joke?

SteamyRoot

Because 7 always one-upped 6?

I dunno.

Mats Granvik

Why are these numbers so close to each other?

$$\log (6)=\lim_{s\to 1} \, \left(1-\frac{1}{6^{s-1}}\right) \zeta (s)$$
$$\Im(\rho _1) \approx \pi \Re\left(\left(1-\frac{1}{6^{\frac{1}{\rho _2}-1}}\right) \zeta \left(\frac{1}{\rho _2}\right)\right)+2 \pi$$
$$14.1347251417... \approx 14.1347256418...$$

SteamyRoot

Google tells me a common way to end it is "Because 7 was a registered 6 offender"...

Leaky Nun

12:04 PM

@MatsGranvik Coincidence?

Mats Granvik

@LeakyNun maybe

Slereah

Hey is it possible to write a surreal number as $\{ \mathrm{No} \vert \}$

A number bigger than every surreal number

Since it is a proper class I'm not sure exactly

Tobias Kildetoft

@Slereah Where would that number "live"?

Slereah

That I do not know

But I know that in proper classes you do get odd results like $V = \mathcal P (V)$

(for the universal class)

So I thought I would check

Tobias Kildetoft

@Slereah How do you even define the powerset (powerclass?) of a proper class? That does not seem like it should be a proper class

Slereah

12:13 PM

Hm, lemme check

us.metamath.org/mpegif/df-pw.html

us.metamath.org/mpegif/ncanth.html

Tobias Kildetoft

@Slereah that is pretty much gibberish

So powerclass does not really behave like powerset, since for example the class itself is not an element. So I don't see how this is so strange.

Slereah

Alright

Tobias Kildetoft

So that would be like defining powerset to only consist of the finite subsets of the set.

Slereah

Do you also happen to know if there's some link between Colombeau's generalized numbers and hyperreals, by the way

I've been wondering

Tobias Kildetoft

Anyway, I have never really thought much about the properties of the surreals related to them being a proper class (instead, I ignore all of that)

no idea about what those are

Slereah

12:18 PM

Oh well

Thanks

Axoren

@SteamyRoot That's a good one.

Jester Tran

My lecturer claims probability and stochastic processes is common sense. Thoughts?

Axoren

12:44 PM

That's some horse duke. Unfortunately, humans react to vivid stimuli more than others. We're more likely to fear scary dangers than common dangers (plane crash vs. car crash). We more easily fall into the gambler's ruin, chasing the big payout of a gamble than identifying the constant risk associated with attempting it.

Leaky Nun

how to prove that $\frac12$ is not an integer?

(Sorry if it is too trivial)

I don't have enough sleep

SteamyRoot

I wonder if your lecturer ever had to explain Monty Hall to a person with little to no mathematical knowledge.

Axoren

@LeakyNun Can it be written as a sum of ones?

Leaky Nun

Prove that it cannot?

@JesterTran Then what is he doing there? Improving our common sense?

Axoren

That's fairly easy, I feel. if you start from $0$ and add $1$, you've already passed $\frac 1 2$. Any further addition of $1$s will get you further away.

Therefore, it's not a positive integer.

Huy

12:48 PM

ok maybe I've gone completely crazy but proving $A \implies B$ is equivalent to proving $\neg B \implies \neg A$, right?

Leaky Nun

@Huy Yes

Axoren

@Huy One of those mornings?

Leaky Nun

Modus tollens

In propositional logic, modus tollens (or modus tollendo tollens and also denying the consequent) (Latin for "the way that denies by denying") is a valid argument form and a rule of inference. It is an application of the general truth that if a statement is true, then so is its contra-positive. The first to explicitly describe the argument form modus tollens were the Stoics. The inference rule modus tollens validates the inference from implies and the contradictory of to the contradictory of . The modus tollens rule can be stated formally as: where stands for the statement "P implies Q". ...

Damon Williams

Is that logic only mathematical. Or does it apply to pretty much everything in life?

Leaky Nun

@Axoren can it be more formal?

Damon Williams

12:49 PM

lol

Jester Tran

@LeakyNun He's implying the course should not be hard in hopes of removing any rumours or fears about the difficulty of the course

Huy

it's just so weird, I'm reading lecture notes and one lemma contains an if and only if statement, and they first prove $A \implies B$ and then they prove $\neg B \implies \neg A$

that makes zero sense

Leaky Nun

@Huy (out of curiosity) what is that lemma?

Tobias Kildetoft

@Huy but then they have shown the same thing twice

Huy

Lemma 3.12, statement (4), page 33 & 34 in this pdf

@TobiasKildetoft: exactly

Axoren

12:50 PM

@Huy $A \to B$ can be rewritten as $\neg A \lor B$

Then, use commutativity.

$B \lor \neg A$

Do you see what to do next?

Tobias Kildetoft

@Axoren your hints don't really go anywhere as it really is an error.

Huy

@Axoren: I just can't believe the authors of those notes have proven the same thing twice and forgotten the backward direction

Axoren

Oh, they stated it backwards?

I was still reading the document.

Oh, the document is unrelated.

Leaky Nun

$\not\exists n\in\mathbb Z:2n=1$

Axoren

@LeakyNun While that is a statement, I think for a proof you need to show that no such $n$ exists.

Leaky Nun

12:54 PM

Yep, and that is what I am asking for

I hope it isn't too trivial

Axoren

Consider a property of multiplication by $2$: the result is always even on the integers.

Leaky Nun

even => divisible by 2

Axoren

$1$ is not even, therefore, the factor is not an integer.

Leaky Nun

1 is not divisible by 2, therefore 1 is not divisible by 2

Axoren

We're not working with the quantity $\frac 1 2$?

Tobias Kildetoft

12:56 PM

I actually think the most clear proof is by induction

Axoren

I guess you're working with the division of 1 by 2.

Leaky Nun

Basically, you're saying that since 1 is not divisible by 2, so 1/2 is not an integer?

@TobiasKildetoft but... we're dealing with $\mathbb Z$ not $\mathbb N$...

Tobias Kildetoft

@LeakyNun but clearly if it is an integer then it is positive

Jester Tran

@DamonWilliams Yeah, can applied to life

Leaky Nun

@TobiasKildetoft well, ok...

Axoren

12:57 PM

You are working with 0.5, which we know is greater than 0 in other fields.

Or at least I hope so.

It's one of those mornings.

Leaky Nun

well, so it is between 0 and 1

then... what's the next step?

because $\frac12>\frac02=0$ and $\frac12<\frac22=1$

or do I need to prove that as well?

Tobias Kildetoft

In some sense there is no way to do this without specifying what you actually mean by the integers

Axoren

@TobiasKildetoft By induction, would you show that $2n \neq 1$, then $2(n+1) \neq 1$?

And then pick a base case?

Tobias Kildetoft

@Axoren Right, though probably it is even easier to observe that the only solution to $x+y = 1$ in the natural numbers is with either $x$ or $y$ equal to $1$

Leaky Nun

@TobiasKildetoft You know, $0\in\mathbb Z\land n\in\mathbb Z\iff S(n)\in\mathbb Z$

Tobias Kildetoft

1:01 PM

but all these things come back to what the integers are

Leaky Nun

(I don't know if this is the formal definition)

Axoren

The successor is the best way to talk about naturals.

However, integers have a negative side.

Leaky Nun

What's the formal definition of that?

Axoren

Depends.

Many places, I've seen integers defined as a 2-tuple of naturals.

Leaky Nun

I see

Axoren

1:03 PM

It's not explicitly helpful to get that deep into the construction, I feel.

What are you trying to prove this for?

Leaky Nun

for fun

Axoren

That's a shame, it won't be.

Leaky Nun

well...

Axoren

Many of the proofs of $2+2$ are ridiculous, but at least those operate strictly in the naturals.

Leaky Nun

Then let's only deal with the naturals?

Well, $2+2=2+S(1)=S(2+1)=S(S(2))=S(3)=4$

Axoren

1:05 PM

Super easy: use the successor function and induction.

Leaky Nun

how would I use the inductive step?

Axoren

5 mins ago, by Axoren

@TobiasKildetoft By induction, would you show that $2n \neq 1$, then $2(n+1) \neq 1$?

Algebra on integers is well-defined.

Leaky Nun

Yes, and how do I prove that?

Axoren

Start with two base cases: $n = 0, 1$

$2(0) = 0$, $2(1) = 2$

For all $1 \le k < n$, $2k > 1$,

This is slightly easier.

Leaky Nun

well, okay...

Axoren

1:09 PM

If $2n > 1$, $2n +2 > 1$

As a result, the natural $n + 1$ is not half of 1.

By induction, no naturals are half of 1.

I made a bad change of variables I shouldn't have done.

Long story short, induction. (Which I guess is exactly what induction is.)

Leaky Nun

ok, thanks

Slereah

1/2 an integer means 2n = 1, which is 2n = s(1) *n = n * 1 + n = n+n, we have $n \geq 0$ and $n+n \geq n$ or $1 \geq n$

So either $n$ is 0 or 1 but that's pretty easy to prove false

Guess I didn't need to decompose the product tho

Oh well

Agawa001

this room is the only one where moderators are never absent

Slereah

Math never rests

Agawa001

1:25 PM

atleast one of them i mean *

Axoren

The moderators are a sleep, post contradictions.

Huy

maybe you're a sleep

Axoren

I can't deny that.

Slereah

$A \neq A$

$p \wedge \neg p$

$\neg (p \lor \neg p)$

Agawa001

and this is the only room where moderators are more rushed to help

Axoren

1:28 PM

$\perp \not \vdash \psi$

Slereah

$\exists A. A = \{ x \vert x \notin x\}$

Axoren

$\forall (x, A), x \not \in A$

Slereah

Well, that's not a contradiction if you consider the empty universe :p

Or a universe with just one set

Mats Granvik

All opposites have something in common. Black and white are both colors. True and False are both values. 1 and 0 are both numbers. 0 and Infinity are numbers.

Jester Tran

@Axoren what does this mean?

@MatsGranvik "0 and Infinity are numbers"???

Slereah

1:33 PM

Depends how you define number, I guess

Mats Granvik

@JesterTran ok that last one was a bit of a stretch. But they are something, everything is something.

Slereah

Is nothing something

Jester Tran

@Slereah lol

Mats Granvik

@Slereah Outside the set of everything is nothing. So in a Venn diagram everything is nothing.

Slereah

I don't think I am high enough to carry on this topic

Damon Williams

1:40 PM

give me contact eye

Slereah

By the way, anyone got an idea on this

2

Q: Coordinates of a plane with a handle

I am trying to find the appropriate coordinates for a plane with a handle (of topology $\mathbb{R}^2 \# \mathbb{T}^2$), without having to use several coordinate patches. My current intuition is to use two cyclical coordinates, $\theta$ corresponding to some angle around the half torus embedded ...

Never got any answer, not quite sure if it's been done before

Jim Wilson

math.stackexchange.com/questions/1800524/… Anyone here know how to do part (d)? I've confirmed that (a), (b) and (c) are correct

Graham did help. But I still can't figure out the MGF function using the definition

Damon Williams

Uhhh degenerative function

Sorry though, I've forgotten all about this stuff

Jim Wilson

1:58 PM

I don't think it's that hard. But I can solve it and don't know why.

Andy Miles

@MikeMiller I see. Thanks anyway. Do you have a grasp of the correspondence between Lie-algebra-valued forms on principal bundles ("connection forms, Ehresmann connections") and Lie-algebra-valued forms on the base manifold ("gauge potentials")? After all, that's what Nakahara is trying to explain.

Jim Wilson

I keep getting undefined or infinity for the limiting mgf

Andy Miles

An Ehresmann connection $\omega$ on a principal bundle $P$ can always be pulled back to the base manifold $M$ via a section $\sigma:M\to P$. The pullback $\mathcal A=\sigma^*\omega$ is then a gauge potential, i.e. a Lie-algebra-valued form on the base manifold (or on an open subset $M$).

Via theorem 10.1 in Nakahara's book, this correspondence can be bijective, i.e. one can construct an Ehresmann connection from the gauge potential.

Mike Miller

It's the other way around that's easy. Principal bundles do not usually have sections.

The construction you do in both directions can be carried out pointwise. It's not much more than linear algebra. Another place this is explained is Kobayashi-Nomizu's first book.

Andy Miles

2:14 PM

Nakahara does the "lifting construction" basically by pulling back the gauge potential via the projection $\pi:P\to M$, so he defines $\omega:=\pi^*\mathcal A$.

Do you know an exact construction in this direction?

Constructing an Ehresmann connection from a gauge potential?

Mike Miller

Didn't you just do it?

Andy Miles

Well, Nakahara's definition has more elements in it. Basically it's the pullback of the projection, yes, but there's some sort of adjoint action $g^{-1}\pi^*\mathcal A g$, plus there's the addition of a term $g^{-1}\mathrm{d}_P g$ where $\mathrm_P$ is the exterior derivative on $P$ and $g$ is the canonical local trivialization.

Mike Miller

Well, it seems like with your extra data you constructed a lift up above.

Andy Miles

I'd like to understand why these things are necessary, and how it's possible to "add the exterior derivative" to a Lie-algebra-valued form. After all, the exterior derivative is a differential operator. Adding it to a Lie-algebra-valued form seems to me like adding apples to oranges.

Mike Miller

Fair enough.

The reason you need to add such a thing is that there's a list of axioms the form on the total space of the bundle must satisfy. One is that, on each fiber, it's the tautological form of the Lie group. Your pullback is not that tautological form.

Andy Miles

2:33 PM

I see. Gonna check that. And the addition thing? How can $g^{-1}\mathrm{d}_P g$ be added to a Lie-algebra-valued one-form?

Mike Miller

You're not adding $d_P$. You're adding the exterior derivative of the multiplication-by-g map.

Axoren

@JesterTran It means that given a contradiction, you can't conclude anything. Paired with the principle of explosion, it itself is a contradiction. Further, by the principle of explosion, if you accept $\perp \not \vdash \psi$ as an axiom, you can conclude anything.

It's read as "$\text{contradiction } \not \to \text{ any logical statement}$"

Eric Stucky

@TheGreatDuck: I had a flash of insight last night and I have discovered something you might be interested in.

Briefly, it is a very strong connection between the implied integral and the ordinary integral.

Ping me and we can talk :)

[I'm going to try to write an answer to your question, but it might be a while; there are quite a few details.]

Andy Miles

2:51 PM

@MikeMiller Getting closer. Why can the multiplication-by-g map be considered a differential form?

Mike Miller

It's not.

Andy Miles

How can the exterior derivative (which usually acts on differential forms) act on the multiplication-by-g map?

Mary Star

3:34 PM

Hello!!

Suppose that we have a matrix with characteristic polynomial $(x+2)^2(x-5)^3$.
How can we find the elementary divisors?
Are the elementary divisors all the elements of the form $(x+2)^i(x-5)^3$, with $1\leq i \leq 2, 1\leq j\leq 3$ ?

Eric Stucky

I don't think there's enough information to answer that, Mary.

See this question, for instance.

Obliv

4:10 PM

does it take $n = m$ exponents for $a^{n} \equiv 0 ~(\mod m)$ if $(a,m) = 1$?

if $a^n = a + a + a...$ n times of course.

ah nvm yes that's true

Is this a true statement: Only permutations of $S_n$ of order $n$ are of length $n$?

Puzzled417

4:44 PM

hi

Obliv

hello

SteamyRoot

@Obliv what do you mean with "a permutation of length $n$"?

Obliv

cycle length

SteamyRoot

A single cycle of length $n$, or would $(1 2)(3 4 5)$ also be length $5$?

Obliv

a single cycle, sorry.

if it were not true, there would be disjoint cycles with $lcm = n$ which I don't think is possible.. not sure

SteamyRoot

4:55 PM

Well, in $S_6$, $(1 2)(3 4 5)$ and $(1 2 3 4 5 6)$ both have order $6$?

Obliv

damn it

well that throws my proof out the window haha

Mary

5:36 PM

hi, can I show you my forum question on this chat?

Axoren

Oh jeez, someone starred my horse duke message.

I'm glad I used less colorful language than I could have.

Mary

my question is about representation of lie algebras, I am not able to find a basis of weight vectors for a linear lie algebra, here it ishttp://math.stackexchange.com/questions/1800883/how-to-find-a-basis-of-weight-‌vectors

0

Q: How to find a basis of weight vectors

I have to following Lie Algebra $L=\{x\in End(\mathbb{C}^6)\colon x^tS+Sx=0\}$, where $S=[\begin{smallmatrix} 0&I_3 \\ I_3&0 \end{smallmatrix}]$, and the subalgebra $H$ given by the diagonal elements in $L$. I have found a basis of $L$ and $H$, which are, respectively: $B_L=\{e_{ij}-e_{i+3j+3},...

representation-theory vectors lie-algebras

Axoren

6:13 PM

I asked this before at some point, but is there any notion for a graph whose vertex set is continuous?

Tobias Kildetoft

@Axoren what does it mean for a set to be continuous?

Axoren

My mistake. Dense in the reals*

Slereah

If I have the parametric equation $$x(t) = r\cos_e(t),\ y(t) = \sin_e (t)$$

Axoren

And the edges to be continuous functions from $[0, 1] \to V$

I dozed away while reasking the question.

Slereah

Such that $\cos_e(t)$ is a function such that over the range of $e$, it spans every possible conic section

Is $\cos_e$ continuous at $e=1$?

I don't think you can continuously deform an ellipse into a parabola, no?

Axoren

6:24 PM

A parabola can be considered the limit of an ellipsoid as its major radius tends towards infinity.

Slereah

Would that function be continuous, though

Axoren

At some point, I suspect it would break.

Slereah

I don't think so since I think it would be like... $\cos t \rightarrow t^2$?

Axoren

Given that $\infty$ isn't in the domain of that function.

Slereah

Unfortunate

Welp, I think my attempts to make a coordinate system for a plane with a handle have been met with too many obstacles

Probably time to drop that one

Axoren

6:28 PM

Is the handle hollow?

Slereah

It's the manifold $\Bbb R^2 \# T^2$

Kari

Are there any nice layman term summaries of the isomorphism theorems' statements?

Rajesh Dachiraju

Anyone familiar with Linear Algebra

Tobias Kildetoft

@Kari I don't see how that would make much sense. The theorems require you to know the terms involved which makes you not a layman

Kari

Rather, a simplified version, @TobiasKildetoft?

Tobias Kildetoft

6:31 PM

@Kari simpler than what? And which particular isomorphism theorems?

Kari

The first, second and third ones.

Tobias Kildetoft

@Kari the numbering is not universal (nor is the algebraic structure involved)

Kari

Just like how Burnside's lemma can be reduced to "The number of orbits is equal to the average number of points fixed by an element of G".

Axoren

Personally, I like talking to gardeners and bakers about isomorphisms. They have such wonderful insights.

Kari

Ah, I didn't know it wasn't a universal numbering.

math.hmc.edu/~omar/math171F14/m171.notes.10.13.pdf

This page has the three I am referring to, @TobiasKildetoft.

Tobias Kildetoft

6:32 PM

the first one is sometimes numbered 0 instead, and I have seen the other two switched up many places

Slereah

@Axoren A simplistic wisdom

Much more practical than all the science of the scholars

Tobias Kildetoft

@Kari Anyway, I am not sure if I can give you a simpler description. They all have fairly intuitive explanations

Slereah

I guess I'll have to do with the gross coordinates that you get :p

Just stitching coordinate patches together

Kari

I'll try a different way of phrasing then! How can we think of the isomorphism $G/\ker(\varphi) \cong \text{image}(\varphi)$, @TobiasKildetoft? So far, I just see it as the cosets of the kernel being isomorphic to the image but that doesn't help me much.

Tobias Kildetoft

@Kari Hmm, actually I don't think I have a very compelling intuition for that one (it just sort of "clicks" at some point). The other are easier in that sense

Kari

6:38 PM

Ah, I see. I'll try and study examples a bit more until I get it!

Tobias Kildetoft

@Kari For the others, when you take $HK/K$ you would expect to get left with just $H$ (seeing the quotient as in some sense opposite to the multiplication). But you have to take into account that when you multiply there will be some overlap if the subgroups have nontrivial intersection, so you end up also killing off this overlap

Eric Stucky

Karl, the way I usually think about that one is that transporting the structure out of a group is rather difficult-- specifically, all possible ways of doing it are already somehow contained in the structure of the group itself.

Kari

I'm not sure what you mean by 'some sense opposite to the multiplication', @TobiasKildetoft.

Tobias Kildetoft

@Kari Well, it is sort of a division (it is not really that, but it behaves a bit like it)

Eric Stucky

Well, $H\times K/(1\times K)=H$; it's sort of a one-way inverse, for instance.

Kari

6:42 PM

What do you mean by transport, @EricStucky?

Ah, I see @TobiasKildetoft.

Eric Stucky

Nothing rigorous, but that is what a morphism is for, yes?

You want to take the structure of the group and find it somewhere else.

Kari

I understand what you mean now, yes.

$(H \times K)/(1 \times K)$ is a bit confusing.

Eric Stucky

So that theorem is something like: you don't actually have to look anywhere else. If you can just understand the group well (e.g. you know quotients) then you know all the ways of seeing its structure in any context.

Kari

How does knowing the quotients to help us understand the group, @EricStucky?

I don't really see the power in quotient groups yet.

Eric Stucky

Ah, sorry, I meant the other way

Although, certainly understanding quotients does help us understand the group, in the sense that we then understand morphisms out of it :P

But

I mean, "knowing the group well" in that informal description is supposed to mean "knowing all of the group's quotients."

(afk)

Tobias Kildetoft

6:49 PM

Probably one of the easiest way to see how useful the theorem can be is to use it to show that $GL_n(k)/SL_n(k)\cong k^{\times}$ for any field $k$.

Which might get somewhat annoying to do directly

Mary Star

Are the elementary divisors of the characteristic polynomial $(x+2)^2(x-5)^3$ the following?
- $(x+2)^2, (x-5)^3$
- $(x+2), (x-5)^3, (x+2)$
- $(x+2), (x-5)^2, (x+2), (x-5)$
- $(x+2), (x+2), (x-5), (x-5), (x-5)$
- $(x+2)^2, (x-5)^2, (x-5)$
- $(x+2)^2, (x-5), (x-5), (x-5)$

Eric Stucky

Mary, you still don't have enough information to answer that question.

Here's the link again