@AlgebraAnalysisManifolds Look

@AlgebraAnalysisManifolds What about Unity:P?

@savick01 Jasper is a Separatist.

@AlgebraAnalysisManifolds Whut? Nah.

Bye

@robjohn Found it! $\boldsymbol{Yeah}$.

Hi all, why is that $\mathbb{R}^k$ is a submanifold of $\mathbb{R}^n} ($k \leq n$)? I'm struggling in defining charts.

@ADR What charts? Don't you like identity?

@savick01 but $\mathbb{R}^k$ is not open in $\mathbb{R}^n$ right?

right

does it matter?

@savick01 in the definition of submanifold it requires that for every point $p$ exists a chart around $p$, so the function identity should map the open neightbor to an open subset in $\mathbb{R}^n$, or what I'm missing here?

With that definition a sphere is not a submanifold of R^{n+1}?

It is not a very good definition...

@savick01 I replaced chart by identity regarding your answer, the definition is 1.9 page 9 of Brocker's "Introduction to Differential Topology"

books.google.com.mx/…

Can't see it

@ADR Notice that it's a chart of the manifold $M$, rather than the set $N$.

@savick01 A subset $N \subset M^{n+k}$ is called an n-dimensional differentiable submanifold of M if, for every point $p$ in $N$, there exists a chart around $p$ $h: U \to U' \subset \mathbb{R}^{n+k}$ so that $h(N \cap U) = U' \cap \mathbb{R}^n$ where we consider $\mathbb{R}^n$ as $R^n \times 0$

The point is, the identity on $\mathbb R^{n+k}$ works fine here.

@user1 Tomorrow I start Analysis II.

Line integrals, Surface integrals, Green's Theorem, Stoke, Gauss. Conservative fields. Applications. Differential equations. Phase Diagrams.

@PeterTamaroff Cool.

@user1 I'm scared it would be too "physical".

Google Chrome is so dramatic!

@AlgebraAnalysisManifolds Dude.

Help.

Analysis II. Line integrals, Surface integrals, Green's Theorem, Stoke, Gauss. Conservative fields. Applications. Differential equations. Phase Diagrams.

What book do I pick?

@anon Yo!

hello

@AlgebraAnalysisManifolds Come one braw!

Give me something.

I used susan colley's vector calculus, and it was very good

\checks profile

ah, it is jasps

indeed

@AlgebraAnalysisManifolds Hello.

Is this chat a figment of my imagination, etc...

@AlgebraAnalysisManifolds How are you?

Hmn.

@AlgebraAnalysisManifolds Let me see.

@AlgebraAnalysisManifolds What ya doing?

@anon @AlgebraAnalysisManifolds Marsden, J., Tromba, A. "Vector Calcus".

@PeterTamaroff Someone told me that this book sucks.

@AlgebraAnalysisManifolds Rudin?

Just don't remember who.

@GustavoBandeira Support your claim!

@PeterTamaroff Well, Jesus told me that in a dream.

@GustavoBandeira Jesus most certainly didn't know math!

At most he knew what a right angle was.

@PeterTamaroff Of course he knew.

@AlgebraAnalysisManifolds But where does he have vector calc?

Jesus was a carpenter, of course he knew math

@anon That's why I mentioned right angles. He knew geometry, probably.

@AlgebraAnalysisManifolds Right, so do you think I should read about differential forms and such things?

What about line integrals and surface integrals?

@PeterTamaroff He knew Topos Theory and programming - for using automated saw and stuff.

@AlgebraAnalysisManifolds I am reading Apostol. I am reading it thoroughly!

@AlgebraAnalysisManifolds Spivak? You mean Manifolds?

@AlgebraAnalysisManifolds Aha.

@AlgebraAnalysisManifolds Remember! Remember!

@AlgebraAnalysisManifolds I've bought a lot of books.

I've bought them here. They were a lot cheap.

Cheap and good books.

I'm also kinda addicted of collecting books. D:

@AlgebraAnalysisManifolds Give some of them to me! =D

@AlgebraAnalysisManifolds I'll go some day.

@AlgebraAnalysisManifolds What holy books?

@AlgebraAnalysisManifolds So to integrate differential forms I must worry only about integrals of continuous functions on compact $n$-dimensional sets?

@AlgebraAnalysisManifolds Got it.

@AlgebraAnalysisManifolds First Spivak?

@AlgebraAnalysisManifolds But you're saying Baby Rudin, right?

Is it really complicated?

@AlgebraAnalysisManifolds True.

Apostol is soft. But nice.

I like the guy.

Learned a big deal from the book.

@FrankScience I am not sure I understand the question. First, an elt $\phi\in G$ is not a map, unless you are identifying $G$ with its double dual in which case I suppose it is a map $G\ni\phi:\hom(G,{\Bbb C}^\times)\to{\Bbb C}^\times:f(-)\mapsto f(\phi)$. Second, no map from a finite group into an infinite group (like ${\Bbb C}^\times$) can be surjective. Also you want to fix an isomorphism between $G$ and its dual in order to speak of "the" dual map $\hat{\phi}$ of an elt $\phi\in G$.

@user1

@PeterTamaroff hi

@user1 I have a sillypants question. Brace yourself.

Consider a function $f:A\subseteq \Bbb R^n\to\Bbb R$, bounded on the set $A$.

ok

Now, define for each $x\in A$ $$\omega_f(x)=\lim_{\delta\to 0^+}\sup\{|f(y')-f(y)|:y',y\in B(x,\delta)\}$$

@anon Bad organized, sorry.

This is usually known as the oscilattion of $f$ at $x$.

@anon Suppose $\phi\colon G\to G'$ is a group homomorphism between two finite abelian groups, and $\hat\phi\colon\hat{G'}\to\hat G$ is its dual map.

The point is to prove that if $A$ is closed, for any $\varepsilon >0$ the set $$B=\{x\in A:\omega_f(x)\geqslant \varepsilon\}$$ is closed, @user1

$x$ should be in the interior of $A$, right?

@user1 Sorry.

No.

or, we could intersect $B(x,\delta)$ with $A$, I guess it's not a big deal.

@anon Namely, $(\hat\phi f)(x)=f(\phi(x))$

@user1 Yes, that.

That is what I missed in the definition.

Because I skipped some stuff.

Anyways, if $x$ is not in $B$, then either $x$ is not in $A$; or it is but the oscillation is smaller than $\varepsilon$. If $x$ is not in $A$, then we get an open $x\in O\subseteq A^c\subseteq B^c$.

@anon Then we'd show that $\phi$ is injective iff $\hat\phi$ is surjective. Yesterday I tried to explain what's $\hat G$, but forgot to explain the conditions, lol.

@PeterTamaroff Of course, I would try to see if $\omega_f$ is continuous.

@user1 Ah, that is a nice thing to do.

@user1 It's not, however.

@FrankScience I was fearing that. =)

@FrankScience Yeah, I figured but the problem may be "removable" in some sense. What's a counterexample?

If the oscillation is smaller than $\varepsilon$, there is a ball $B(x,\delta)$ such that $$\sup\{|f(y')-f(y)|:y,y'\in A\cap B(x,\delta)\}<\varepsilon$$

@user1 A function discontinues at a point.

@PeterTamaroff You'd show that the set of points $x$ such that $\omega_f(x)\ge\epsilon$ is closed, right?

@FrankScience I am doing that yes.

@PeterTamaroff It suffices to show that its complement is open, say the set of points $x$ such that $\omega_f(x)<\epsilon$ is open.

@FrankScience "It suffices..."

I would say "We can..."

But yes, that is also what I am doing! Any help is welcome.

@FrankScience Would you like to see what I think is a proof?

Suppose $\omega_f(x_0)<\epsilon$, then there's a neighborhood of $x_0$, whose oscillation is smaller than $\epsilon$.

@FrankScience Aha.

Then for all $x$ in that neighborhood, $\omega_f(x)<\epsilon$.

Sure. One has to prove that.

It's proved.

Oh?

Details!

Of course it is, but why?

By oscillation of a neighborhood or a open interval $I$, I mean $\sup_{x_1,x_2\in I}\lvert f(x_1)-f(x_2)\rvert$.

@FrankScience We're in $\Bbb R^n$.

The rest work is trivial.

Suppose the nbhd is a ball $B(x,\delta)$. Then pick $y\in B(x,\delta)$. Then pick a ball $B(y,\delta')\subseteq B(x,\delta)$. Then the inclusion means $\omega_f(y)\leq\omega_f(x)<\varepsilon$. Right or left?

I think you get continuity of $f$ in a small enough region around a point $x$ satisfying $\omega_f(x)<\epsilon$.

@FrankScience Why?

Well, instead of an open interval, use the terminology open ball instead.

@PeterTamaroff Seems right.

It's the obvious generalization, even to metric spaces.

@FrankScience Trivial, obvious. Why bother then? =D

Let's be humble, Frank!

We'd introduce some notations.

$\Omega_f(V)$ means the oscillation of $f$ in $V$.

The proof is following the bulletins:

1. $\omega_f(x_0)<\epsilon$ implies there's a neighborhood $W\ni x_0$ such that $\Omega_f(W)<2\epsilon$

2. For all $x\in W$, $\omega_f(x)\le\Omega_f(W)$

That's all.

few people here

@PeterTamaroff The rest work is to prove them using $\epsilon$-$\delta$ language or something similar. Is it trivial?

@FrankScience The word trivial shouldn't be used lightly. That is my opinion. I'm off now.

@FrankScience Proposition 16 here.

@anon Well, it's not easy to me.

@anon Thanks. I did guess whether lemma 13 was right. It's right but not immediate. Thanks very much. Another question: how can we determine the order of an element from a conjugacy class through the character table?

I don't know that you can in general. for example if the representation is trivial the character table is full of 1s and so you can't figure out orders of elts from that data.

@anon By character table, I mean the table consisting of all irreducible representations.

derp

@anon Michael Artin's book let me determine the order from some character tables. I don't know whether there's a systematical way.

well, in the abelian case it's easy

you're talking about general finite groups though right?

It's not abelian, since there're some characters $\chi(1)\neq1$

@PeterTamaroff I learnt that word from many mathematical books, not from my experience of learning English.

@anon Yes.

Well, another immediate question: Is a subgroup of an abelian group always a quotient group of that abelian group?

yes

If so, part of character extension lemma could be proved through lifting a character from the subgroup as a quotient group.

@FrankScience you can't determine the order of an element from just the character table

and the fact that quotients of abelian groups are also isomorphic to subgroup follows for example from the fact that the characters are isomorphic to the group (but in a dual way)

Of course, this is not true for infinite abelian groups.

The two non-abelian groups of order $8$ have identical character tables, but their element orders are not the same

@TobiasKildetoft So in what condition it could be determined?

@FrankScience hmm, not sure

@TobiasKildetoft Artin gave many exercises on determining the order. I have no idea.

so looking at the linear characters gives us the order mod the derived subgroup

perhaps you could quote the exercise verbatim

Another easy one: How can we show that a finite simple group that is not of prime order has no nontrivial representation of dimension two? I think it's equivalent to the proposition that $U_2$ hasn't any finite simple group of composite number order.

@anon The table is too large to input.

@TobiasKildetoft Ah, well, in addition to the character table, it gives the size of each conjugacy class. I didn't mention that.

@FrankScience what table? you mean there's a specific table for a specific group, and the exercise was not talking about arbitrary finite groups?

@anon Yes. Though I'm self-studying and there's no assignment, I think I should at least do one of them to get what's involved.

Yeah, I thought there might be a general method. However, now it seems wrong.

The knowledge of the conjugacy class sizes will probably be important

@anon Incidentally, how did you refer that message?

just pasted the url of the permalink

it parses such urls into blockquotes automatically

Ah, I see.

@FrankScience what is the order of the group you have been given the table for?

Well, I'll take a very simple table. I hope it would involve general ideas.

For example, the table of $S_3$

\begin{tablular}{m|mmm}

&1&x&y\\

\chi_1&1&1&1\\

But you said you had been given a specific table

@TobiasKildetoft There're many exercises, each of these gives a specific table.

(btw, if you know what group it is, you don't need to put the table here, I can just pull it up myself)

Well, take $S_3$ as an example. There're three conjugacy classes, with order $1,2,3$, which is for $1,x,y$.

How can we determine the order of $x$, which is $3$, from the column $(1,1,-1)^t$?

$\chi_1$ is trivial, $\chi_2$ is the sign representation, and $\chi_3$ is the third.

so we know that the number of conjugates of $x$ is 2, so the centralizer of $x$ has order $3$.

and we are done since $x$ is not the identity

without even using the table

Right. That example is degenerate.

and same goes for finding the order of $y$ btw

so yeah, $S_3$ is a bad example of this

What about the icosahedral group $I$?

What is the order of that one?

Well, the same argument works.

I see.

Except one of these conjugacy classes

$60$

so you actually just mean $A_5$?

Yes, they're isomorphic.

thought it was 120.

Well, no reflection.

@FrankScience (it is easies to find the table for the alternating groups for me)

I don't know.

btw, is Serre's book a good introduction to linear representations?

by linear, what do you mean?

It seems that I cannot understand the whole stuff Artin introduced.

Representations by $GL$

those are just called representations usually

never read Serre

what type of groups are you interested in? Finite or more general?

Artin mostly only introduced finite.

for finite groups and for self-study, James and Liebeck is probably a good choice

Come back to the representation of $A_5$. How can we deduce that the order of $(12)(34)$ is $2$?

Its centralizer is of order $4$.

@FrankScience by knowing how to multiply the elements

@TobiasKildetoft What?

the problem is that we don't know that a specific column corresponds to that cycle type

@TobiasKildetoft The collection of irreducible representations is unique, right?

@FrankScience when doing this problem, you should forget that you know that conjugacy classes correspond to cycle types

you should try to only use what is given in the table (ignoring the labels of the classes)

Greetings noble souls

@TobiasKildetoft You said that you can easily obtain a table for the known group.

@FrankScience sure, I just ask GAP

Its column is $1,-1,-1,0,1$

right, so we know that it has order either $2$ or $4$

hmm, so how can we rule out the order being 4

so if that was the case, the elements that commute with it are precisely the powers of the element itself

I don't know how to make use of non-1D characters.

so if we consider $x^2$ then the centralizer of that will also contain the centralizer of $x$, so the order of its centralizer will be divisible by $4$, but it will be a different conjugacy class than $x$. This is not possible

since the only other centralizer whose order is divisible by $4$ is the one for the identity element

once again, we really only used the conjugacy class sizes

Generally, if there's a 1D character $\chi$ such that $\chi(g)$ is a primitive root of order $n$, then $n$ must divides the order of $g$, right?

yes

But I cannot get any information from $\chi$ such that $\chi(1)\neq1$

@FrankScience I am also not seeing a nice way to do that (at least in these questions)

for non-simple groups it might be a bit easier, since you get some normal subgroups to play with from looking at the table

but even then, I am not sure you get much in terms of finding element orders

So we can only first try to obtain information from those $\chi(1)=1$, and the information from the sizes of conjugacy classes.

Well, another question: can you see any obvious reason that $U_2$ hasn't nontrivial non-prime-order simple groups?

by $U_2$, you mean what group?

The unitary group.

hmm

Its origin is to prove that a finite simple group of non-prime order has no nontrivial representation of dimension $2$. If so, that group is embedded in $GL_2$, and since it's finite, there's a conjugate group in $U_2$.

right

so what does the center of $U_2$ look like?

$e^{i\theta}I$

ok, and any such subgroup will intersect trivially with the center

and we can actually assume it is contained in $SU_2$

since otherwise, the kernel of the determinant map will be a non-trivial normal subgroup

A simple subgroup of $e^{i\theta}I$ should be of prime order, since it's abelian.

$SU_2$ is just the 3D sphere $S^3$.

@TobiasKildetoft Why is it nontrivial?

@FrankScience why is what non-trivial?

@TobiasKildetoft If it's not contained in $SU_2$, the $\ker\det=\{I\}$

well, by definition, $SU_2$ is the subgroup where the determinant is $1$, so if out group is not contained in that, it will have elements with determina t $1$ and element with determinant not equal to $1$

but the set of those with determinant $1$ is a normal subgroup

I mean, what about the case that the subgroup only contains one element with determinant $1$, viz. $I$? The other case is that all of elements are with determinant $1$, which is you said, viz. a subgroup of $SU_2$.

@FrankScience since those with determinant $1$ are the kernel of the determinant map, if only the identoty satisfies that, then the group is isomorphic to a subgroup of an abelian group, which we assumed it not to be

Well, I was stupid. Then, is it easy now?

@FrankScience I am still not seeing precisely how to do the final piece, but at least we have some reductions now

I see.

but I need to go now. I'll be back later

Is $PSL_n(\mathbb C)$ isomorphic to $GL_n(\mathbb C)/Z$ where $Z$ is the center?

Goodbye!

yes

@anon what do you know about the minor of a matrix corresponding to an entry if the matrix is not square?

like, what information does the minor tell us?

no I mean what is meant by say the minor of an $m \times n$ matrix $A = (a_{ij})$ corresponding to element $A_{2,1}$ say

@anon with $m\neq n$

ah, dunno

ok

@anon what does this mean here

look at "other applications" en.wikipedia.org/wiki/Minor_%28linear_algebra%29

@anon

@anon Hey

busy atm

do you any relationship between the minors of a matrix and its kernel

sorry, I was on the phone with two people plus I'm trying to cook bacon and eggs

the wikipedia section clearly defines what the minors of nonsquares are

dunno immediately what relation they have to kernel image and whatnot

@MarianoSuárez-Alvarez Hey

@TobiasKildetoft

Hey slimlim =)

@BenjaLim hi

Hi @BabakS. how are you pal :-)

@TobiasKildetoft hey

do you know why the plucker ideal is a prime ideal?

@TobiasKildetoft @anon @MarianoSuárez-Alvarez math.stackexchange.com/questions/465800/…

Is the maximal ideal factorization always unique?

I want to put dy/dx in fraction form. How to do that. I am writing it as $\frac(dy,dx)$, it is giving me weird result. can anyone pls help

@Ramit you need to put {} around both numerator and denominator

thanks Tobias

$\frac{\mathrm{d}y}{\mathrm{d}x}$ becomes $\frac{\mathrm{d}y}{\mathrm{d}x}$

I'm trying to find a cute solution to $$\lim_{m\to\infty} \sum_{k=0}^{2013} (-1)^k \binom{2013}{k}H_{m+k}$$

(working on it right now)

@Chris'ssis Hi =)

@robjohn Hey

@N3buchadnezzar Hi :-)

Maple sucks

humpf

Can not even evaluate simple integrals

@LoganM Hey

hi

hi, can anyone pls tell me how to write e to the power xy. i am writing $e^(xy)$. It's appearing weird.

@Ramit $$\mathrm{e}^{xy}$$

'\mathrm{e}^{xy}'

thanks!

@N3buchadnezzar what's up?

Just about to travel to the city where I study.

@robjohn You?

@Ramit you need to use {} to group TeX elements

@N3buchadnezzar You have a particular place in which to study? Sounds restrictive :-)

@N3buchadnezzar Not much, just being amazed that an answer I gave this morning has gotten 43 votes so far. Excuse me, 44.

Well I live in Bergen, but study at Trondheim.

@robjohn impressive, does it deserves the votes?

Ah I saw that, well it follows the curve I posted a few days ago

@N3buchadnezzar I have given up on making any such judgements. Let's just say that I've put a lot more effort into answers that have gotten far fewer votes.

Seems simple answers + if it is linked to at another website recieves far more votes.

I just posted an interesting integral, and as usual nobody gave it a look.

@N3buchadnezzar answer or question?

Oh, question. I also recently posted a nice answer, I think it even will be my second accepted answer!

But seems the OP wants to wait a few days before accepting.

@N3buchadnezzar some people accept early and sometimes change their minds. Often the people who take time to accept, stick with their original decision.

@robjohn Ahoy.

@PeterTamaroff Hey there... I finally got a runaway answer that I am kind of baffled by. It must just be referenced on some other site.

@robjohn Come again?

@PeterTamaroff An answer that has currently gotten 49 votes this morning.

@robjohn Aha!

@robjohn It's on the top of the multicollider (currently 231, so it'll be there for a while). Low level MSE posts often end up there and then get a ridiculous number of upvotes from the rest of the network (mostly SO).

Hi, i've a few doubts. I wish somebody could answer. Do all non-singular matrices have eigenvalues? If no, then why?

all matrices have eigenvalues over an algebraically closed field. without this property there can exist matrices without eigenvalues in the scalar field (e.g. rotations in 2D real plane have nonreal eigenvalues).

@robjohn I found it via the "hot questions" feature on the SE network

I actually have a question about the answer

the answer to why is in the characteristic polynomial

I'd say that twice as likely as 50% is 75%, but applying the math in your answer I get that "twice as likely as 50%" is 66.67%. I guess I'm just wrong and twice as likely as 50% is in fact 2/3, but why?

one must ponder the interpretation of the phrase "twice as likely as [blah]"

Well it occurs twice as often as [blah]. Instead of every other time (1/2) it would occur 3/4 times.. i guess :P Guess I'm a math noob

the interpretation in robjohn's answer is the odds are twice as large

so, if originally the chances are 50% (i.e. 1:1 odds) then "twice as likely" here means 2:1 odds, i.e. approx 67%

that is, "twice as likely" is to mean that the success-to-failure rate is twice as large

hmm, okay I see the difference

I'll have to think about why that matches bayesian reasoning later

Hi guys.

@HenningMakholm Hi

@HenningMakholm Hey there... haven't seen you for a while.

@robjohn Um, no ... came down with a bad case of My Little Pony and have been mostly mucking around on pony forums for several months. Am getting better now, though. :-)

@HenningMakholm I'm not sure I should continue this line of questioning...

@robjohn You do what you need to do ...

@AlgebraAnalysisManifolds Why am I not surprised?

Too bad.