Mathematics - 2018-09-26 (page 1 of 3)

12:09 AM

@ted Would you characterize Hartogs' extension theorem as very deep?

I thought I remembered the proof in Griffiths and Harris being relatively straightforward.

12:23 AM

@MikeMiller it changed the face of several complex variables iirc showing how different one variable is from several variable results

So in that sense it's deep I'd say, I think G&H bring it up mainly to show how several variables can be different

@Semiclassical should check out Weyl's QM setup of spherical harmonics, except for this decomposition everything seems so simple and obvious compared to the way e.g. Jackson does it

Hartog's does a simple case also, the general proof of the general theorem is really deep or at least what it's linked to is

Ted Shifrin

@MikeM: No.

Poles occur along divisors, and they're not contained in a compact set.

Semiclassical

12:42 AM

@bolbteppa well tbf

'simple and obvious compared to Jackson'

not exactly a high bar :P

bolbteppa

iirc Jackson just references somewhere else to make things worse :p

Semiclassical

lol, ofc

Mike Miller

@TedShifrin Someone objected to my use on main and give a much simpler argument. Their simpler argument is nice.

I don't know if I immediately understand your argument, unfortunately.

I remembered it being some tricky games with using the Cauchy integral formula on two different factors of $\Bbb C$.

Leaky Nun

wow guys, you're doing several complex variables

that's not good

that's too deep

Mike Miller

vOv

@bolbteppa G&H bring it up because that is a book in complex geometry in multiple variables, in which it is a relevant theorem.

bolbteppa

12:50 AM

@MikeMiller obviously

But they literally say why they bring it up where they bring it up, they could easily have chosen to do the more general theorem there or later, clearly it's historical significance in distinguishing several variables from a single variable is why they gave Hartog's 2 variable proof instead of Osgood's more general proof, kind of ignoring that whole 'they did it because it's relevant' thinking if the general one is more relevant to later arguments :\

Nicolas

@bolbteppa is that decomposition not the eigenfunction expansion for the Laplacian?

bolbteppa

1:09 AM

@Nicolas forgot to include coefficients, - take a homogeneous polynomial of degree $l$ in $u_l = \sum c_{abk} \xi^a \eta^b z^k$ for $a+b+k=l$ and $\xi = x + i y, \eta = x - i y$, that satisfy $\nabla^2 u = (4 \frac{\partial^2}{\partial \xi \partial \eta} + \frac{\partial^2}{\partial z^2})u = 0$, apparently this is all one needs to see that $u_l$ separates into a sum of terms $\sum u^{(m)}$ for $m = -l,\dots,0,\dots,l$ where the $a - b = m$, without any work apparently

Semiclassical

"apparently"

bolbteppa

(At least to a physicist)

Semiclassical

lol

bolbteppa

haha

The implication seems to be that reducing the second order Cartesian partials $\partial_x^2 + \partial_y^2 + \partial_z^2$ to complex variables (and so to first order derivatives in separate variables) somehow makes this obvious

Semiclassical

I mean, the condition that it be an eigenfunction gives $$\sum c_{abk}(4abz^2+k(k-1)\xi\eta)\xi^{a-1}\eta^{b-1}z^{k-2}=0$$

which if you identify term-by-term gives...

Mike Miller

1:13 AM

@bolbteppa That was pretty rude of me, I apologize. I'd remembered Hartogs' being used with some frequency, and no more general theorem. It's been a while, and clearly your memory is fresher than mine.

Semiclassical

$4(a+1)(b+1) c_{a+1,b+1,k}+(k+2)(k+1)c_{a,b,k+2}=0$ I think?

not exactly revelatory

bolbteppa

@MikeMiller no worries, my memory is pretty hazy as well, high likelihood G&H state the general one though lots of sources just do it for a disk in 2-D :p

Mike Miller

I'm not exactly sure what the 'general one' is

Semiclassical

$4(a+1)(b+1)c_{a+1,b+1,k-2}+k(k-1)c_{a,b,k}=0$

is a little nicer since it makes it apparent that both have subscripts adding to $l$

Mike Miller

The statement I know, for context, is that a holomorphic function defined in a domain on the complement of a compact set may be extended to the whole of the domain

but that also might not be how it's phrased anywhere, just what I walked away with in the back of my head

Semiclassical

1:20 AM

one nice feature of the above is that when $k=0,1$ we get $c_{a+1,b+1,-2}=c_{a+1,b+1,-1}=0$ so long as $a,b\geq 0$

so coefficients $c_{abk}$ with $k<0$ should vanish, which seems appropriate

similarly I think we get that $c_{abk}=0$ when $a=-1$ or $b=-1$

so again we naturally are forced to consider polynomials

Nicolas

Is this before or after the spherical harmonics are introduced? because you are defining the space of homogeneous polynomials of degree l that are harmonic. the spherical harmonics are homogeneous polynomials of degree l and harmonic, and a basis of that space, so if you have that, the stated fact follows

Semiclassical

actually, what's the analogue of $L_z Y_{lm}=m Y_{lm}$ in this setup @bolbteppa

bolbteppa

@Mike connected too (which shows up in the d-bar proof of it in the analytic continuation), but yeah that's it, lots of sources do it for a disk or some other set and his original proof was specific too, extending this to meromorphic functions (and beyond) is such a rabbit hole

Semiclassical

One thing we do have in this is Euler's formula: $(\xi\partial_{\xi}+\eta\partial_{\eta}+z\partial_z)u_l = l u_l$

since $u_l$ is a degree-$l$ homogeneous polynomial

no idea if that's pertinent tho

Mike Miller

@bolbteppa Oh I see, as opposed to some specific domain. Makes sense.

I can believe meromorphic functions are a shitshow

bolbteppa

1:35 AM

This is an old book so words like 'basis' are a bit loose, what he does is take the Laplace equation $\nabla^2 u = 0$ in Cartesian coordinates and note that homogeneous polynomial solutions exist, based on this alone we can re-express $u$ in spherical coordinates to get 'solid harmonics' $u = r^l Y_l$ for some $Y$ and then use the Green identities to show the spherical harmonics $Y_l$ are orthogonal, though we don't know what they look like yet, which is the goal

(without doing any hard work like solving an ODE :p ) He then re-expresses the Laplacian in complex $\xi = x + i y, \eta = x - i y$ coordinates as above, and based on this alone he proclaims it is obvious that a homogeneous solution $u_l$ of degree $l$ in $\xi, \eta, z$ (not $x,y,z$ now) decomposes as above

that $u_l$ separates into a sum of terms $u^{(m)}$, $u_l = \sum u^{(m)}$ for $m = -l,\dots,0,\dots,l$ where the $a - b = m$ by argument-of-obviousness

Nicolas

@bolbteppa just had a look in Weyl's book, isn't he just saying that the polynomial $u_l$ contains terms "in which the exponents of $\xi$ and $\eta$ have fixed difference $m$"

Semiclassical

I guess I don't get how it's being separated in to those terms

unless it's literally just as simple as the above

Nicolas

which is pretty clear since you just pick out the parts where the exponent of $z$ is $l - m$

Semiclassical

those would satisfy $(\xi\partial_\xi-\eta\partial_\eta)u^{(m)}=m u^{(m)}$

But besides obvious Euler's theorem stuff, I don't really get how they're supposed to relate to $\partial^2 u=0$

I guess another weird thing is that one usually has $L^2 Y_{lm}=l(l+1)Y_{lm}$

Something seems funky.

($\nabla^2u=0$, not $\partial^2 u=0$)

bolbteppa

It's just a very weird thing to do

For $l = 2$ we have
\begin{align}
u_2 &= c_{abk} \xi^a \eta^b z^k \\
&= c_{200} \xi^2 \eta^0 z^0 + c_{020} \xi^0 \eta^2 z^0 + c_{002} \xi^0 \eta^0 z^2 + c_{110} \xi^1 \eta^1 z^0 + c_{101} \xi^1 \eta^0 z^1 + c_{011} \xi^0 \eta^1 z^1 \\
&= \sum u^{(m)}, \ \ \ a - b = m \\
&= u^{(-2)} + u^{(-1)} + u^{(0)} + u^{(1)} + u^{(2)} \\
&= (c_{020} \xi^0 \eta^2 z^0) + ( c_{011} \xi^0 \eta^1 z^1 ) + (c_{002} \xi^0 \eta^0 z^2 + c_{110} \xi^1 \eta^1 z^0) + ( c_{101} \xi^1 \eta^0 z^1 ) + ( c_{200} \xi^2 \eta^0 z^0 )

Semiclassical

1:47 AM

yeah

i mean, one can do that decomposition

the issue is really why it's particularly insightful to do so

bolbteppa

Like it's so important in spherical harmonic theory that I was hoping he was implying this was obvious by going to complex coordinates

Semiclassical

my impulse remains that one needs to use the homogeneity in a substantive way

bolbteppa

2:28 AM

From
\begin{align}
0 &= \nabla^2 u_2 \\
&= (4 \frac{\partial^2}{\partial \xi \partial \eta} + \frac{\partial^2}{\partial z^2})[(c_{020} \xi^0 \eta^2 z^0) + ( c_{011} \xi^0 \eta^1 z^1 ) + (c_{002} \xi^0 \eta^0 z^2 + c_{110} \xi^1 \eta^1 z^0) + ( c_{101} \xi^1 \eta^0 z^1 ) + ( c_{200} \xi^2 \eta^0 z^0 )] \\
&= 4 \frac{\partial^2}{\partial \xi \partial \eta} [(c_{020} \xi^0 \eta^2 z^0) + ( c_{011} \xi^0 \eta^1 z^1 ) + (c_{002} \xi^0 \eta^0 z^2 + c_{110} \xi^1 \eta^1 z^0) + ( c_{101} \xi^1 \eta^0 z^1 ) + ( c_{200} \xi^2 \eta^0 z^0 )] + 2 c_{002} \xi^0 \eta^0 z^0 \\

For $l = 3$ we have
\begin{align}
u_3 &= c_{300} u^3 v^0 z^0 + c_{030} u^0 v^3 z^0 + c_{003} u^0 v^0 z^3 + c_{210} u^2 v^1 z^0 + c_{120} u^1 v^2 z^0 + c_{201} u^2 v^0 z^1 \\
&+ c_{102} u^1 v^0 z^2 + c_{021} u^0 v^2 z^1 + c_{012} u^0 v^1 z^2 + c_{111} u^1 v^1 z^1 \\
&= u^{(-3)} + \dots u^{(0)} + \dots u^{(3)} \\
&= (c_{030} u^0 v^3 z^0) + (c_{021} u^0 v^2 z^1 ) + (c_{120} u^1 v^2 z^0 + c_{012} u^0 v^1 z^2) + (c_{003} u^0 v^0 z^3 + c_{111} u^1 v^1 z^1) + (c_{210} u^2 v^1 z^0 + c_{102} u^1 v^0 z^2) + (c_{201} u^2 v^0 z^1 ) + (c_{300} u^3 v^0 z^0)

Nicolas

I think you might be overthinking it. We have a harmonic and homogeneous polynomial $u$ in $\xi, \eta, z$, so we can write $u = \sum_{a+b+c=l} \alpha_{abc} \xi^a \eta^b z^c$ with some constraints on the coefficients (because harmonic).

Find the polynomials $u^{(m)}$ where the exponents of $\xi$ and $\eta$ have fixed difference $m = -l, \ldots, l$, ie. just split up the sum. The work begins in the next step, where you have to show that $u^{(m)}$ is essentially unique (this is where I suspect the $\xi,\eta$ might become useful).

bolbteppa

In $(c_{120} u^1 v^2 z^0 + c_{012} u^0 v^1 z^2)$ the $z$ is two powers higher in the second term and the $\partial_z^2$ acts on it, while in the first term $u^1 v^2$ are acted on by the $u,v$ derivatives (switched to $u = \xi$ and $v = \eta$)

Nicolas

This all is just to explicitly construct the polynomial basis of the spherical harmonics.

bolbteppa

Ahh...

You write it this way so that for $c_{abk} u^a v^b z^k$ the term $c_{a-1,b-1,k+2} u^{a-1} v^{b-1} z^{k+2}$ is such that the Laplacian forces these coefficients to be constrained in terms of each other, does it generalize hmm

Well that's the pattern on that term, other terms have a different pattern, lets see

There's some combinatorial way to see the general pattern I think anyway, it's more primitive than 'completeness of Hilbert spaces' :p

Ultradark

2:45 AM

this is fun

lol

Silent

3:32 AM

@Semiclassical, let $C$ be a group, and $A$ be subgroup of $C$, and $A$ be subgroup of $B$ where $B\subset C$. Can we deduce from this that $B$ is subgroup of $C$?

Leaky Nun

4:47 AM

I think that's actually not well-defined

I've never thought of that

we just don't think about groups as sets that way

I guess the answer is this:

@Silent when you said "$A$ be subgroup of $B$", you have a group structure on $B$ implicitly. what is that structure?

Mike Miller

Well-phrased

user 2646

classic well-phrased example of answering a question with a question

::applause::

Karl Kronenfeld

5:43 AM

lol very neat way to make that point @LeakyNun. We would have to give up "the" symmetric group $S_n$, etc, if we were to allow the question to be interpreted as it was intended.

Ninja hatori

@LeakyNun how every non finitely generated set inductively orderd?

It is set of ideals which are not finitely generated?

Leaky Nun

context

Ninja hatori

I want to show that each prime ideal is finitely generated then it is Noetherian

Leaky Nun

more context

show me the whole damn proof

Ninja hatori

math.purdue.edu/~sbasu/teaching/fall08/557/Week2.pdf theorem 14 in this pdf

Secret

5:55 AM

[Random]

$P(x+y)=P(x)+P(y)-P(0)+Q(x,y)$

Let $x, y\in \Bbb{T}$. Then:

Suppose $x+y \in \Bbb{A}$. Then:

Leaky Nun

@Ninjahatori what does "inductively ordered" mean?

Secret

Pick $P$ such that $P(x+y)=0$. Then:

Ninja hatori

if every chain of element has upper bound then we say it is inductively orderd?

Leaky Nun

I see

it's the union as usual

Secret

$P(x)+P(y)=P(0)-Q(x,y)$

Ninja hatori

6:01 AM

yes

Leaky Nun

@Ninjahatori if the union is finitely generated, then the generators live on something on the chain, and by finiteness you can take the maximum, and that ideal would be finitely generated

revise the proof of "every ideal f.g. => a.c.c."

Ninja hatori

ok sir

Secret

Then $P(x), P(y) \in \Bbb{T}$

Suppose $Q(x,y) \in \Bbb{A}$. Then:

Pick $R$ such that $R(P(0)-Q(x,y))=0$

Then:

$R(P(x)+P(y))=0$

Then $P(x)+P(y) \in \Bbb{A}$

Now suppose $Q(x,y) \in \Bbb{T}$. Then:

$P(x)+P(y) \in \Bbb{T}$

Now suppose $x+y\in \Bbb{T}$

then:

$P(x+y)-P(x)-P(y)-Q(x,y)=P(0)$

Pick $S$ such that $S(P(0))=0$

$S(P(x+y)-P(x)-P(y)-Q(x,y))=0

Then $P(x+y)-P(x)-P(y)-Q(x,y) \in \Bbb{A}$

Sir Cumference

6:21 AM

Anyone know a good book as an intro to mathematical logic for undergraduates? Something that focuses pretty deeply on it

Leaky Nun

in Logic, Nov 30 '17 at 5:20, by user21820

Good introductory texts for logic here!

Sir Cumference

@LeakyNun cool thx

Soham Chowdhury

@Secret who are you talking to?

38 mins ago, by Secret

[Random]

this should go in a notebook, on a whiteboard, or in a TeX file

Leaky Nun

@SohamChowdhury see starboard, 4th item

Secret

It would be a lot easier if chat rooms don't just froze under disuse. The SE Chat real time rendering of the mathjax and layout has just the right layout to guide my thinking

Even in LaTex you need to rerender all the time, which is less convenient

Leaky Nun

6:35 AM

$Mathematics$ Mathworks (Not the main chat!)

Maths department of SecretLabs SE Branch (chat.stackexchange.c...

Alessandro Codenotti

@SirCumference van Dalen wrote a very readable one

Secret

6:57 AM

25

A: My chat room has been deleted without informing me and without telling the reason?

Your chatroom had been deleted due to inactivity. From the Chat FAQ: (emphasis mine) Rooms will exist indefinitely, so long as there is at least one person actively talking in the room. A room is considered worth retaining if it has more than 15 messages by at least 2 users. Rooms not w...

I need a notebook version of SE chat. This is like the most perfect framework I have ever seen so far for organising notes

Leaky Nun

$Mathematics$ Mathworks (Not the main chat!)

Maths department of SecretLabs SE Branch (chat.stackexchange.c...

Secret

But that room is supposed to be for "publishable material", which is why I made the Rambles chat room (and then the system froze it again and again)

You can sort of see that as I made some effort on that room to make things more readable

Ninja hatori

7:21 AM

@LeakyNun math.stackexchange.com/questions/375454/… in this problem what is meaning of since B is A module via f and how f(s2),f(s) convert to s2 and s here?

Leaky Nun

@Ninjahatori do you know what $S^{-1} B$ means?

Ninja hatori

element such as b/t where b belongs to b and t belongs to s

Leaky Nun

but S isn't a subset of B

Ninja hatori

right but f(S) is subset of B

I don't get the point that if B is A module then how f(s2) becomes s2

Leaky Nun

@Ninjahatori you need to know what $S^{-1}B$ means first

or rather, how it is defined

also, ping me every time you reply, or else I may not read it

Ninja hatori

7:39 AM

@LeakyNun I AM NOT GETTING IT?

Leaky Nun

right, you're not getting it

you don't know how $S^{-1}B$ is defined

$S$ isn't a subset of $B$, so you can't just localize it using the usual construction for rings and their subsets

Ninja hatori

could you please tell?

Leaky Nun

maybe you should find out how it is defined first

Mike Miller

You're going aggressive lately

2

I like it

Secret

Ninja hatori

7:44 AM

@LeakyNun ok I get your point

Secret

Current status of the number plotter (rational translates of Liuoville numbers omitted else the whole screen will be flooded with blue at this resolution)

Alex Clark

7:55 AM

@Secret I've honestly thought you were trolling for ages

Secret

You know I am not trolling when the chat room is busy, I don't said much even though it is not my bedtime

I like conversations that flows

Alex Clark

But you post a massive number of messages here, in a conversation with yourself, and other people want you to stop it seems

Secret

I am still trying to find a good notebook that can render latex as fast as the chat does, line by line. While it is true Mathworks is not frozen yet, that room is not really a roughwork sheet. I'll try to cut down my ramblings a bit while searching for a better notebook interface

Soham Chowdhury

@Secret well, you can make your own chat :)

just post in it once a week

Nicolas

8:12 AM

@Secret an iPython notebook takes markdown cells with math line by line. there's also Typora (typora.io), but I'm not sure if that ever reached maturity

Abdullah UYU

@Secret I have the intention of writing a replica of this chat. But not in the near future, unfortunately :)

Secret

checking now...

Nicolas

You can also run Jupyter notebooks standalone in a handsome interface (also: in Atom) with nteract.io (alpha version though)

Secret

8:33 AM

Typora seemed to work pretty well

I might be able to shunt most of my roughwork here

Abdullah UYU

9:16 AM

Let $C_I = \{X|X\subseteq I\}$. How can one show that $B\subseteq A\Rightarrow C_B\subseteq C_A$?

ÍgjøgnumMeg

Do it directly using the assumption

Write down what $C_B$ and $C_A$ are

Abdullah UYU

Like how I defined it, right?

ÍgjøgnumMeg

Right, so you'll need to use that containment is a transitive relation (i.e. if $X \subseteq B$ and $B \subseteq A$ then ... ?)

Abdullah UYU

Then $X\subseteq A$.

So I want to show $\forall x\in\{X\,|\,X\subseteq B\}, x\in\{X\,|\,X\subseteq A\}$

It can be written that $\forall x\in\{X\,|\,X\subseteq A\}, x\in\{X\,|\,X\subseteq A\}$

which is tautology.

ÍgjøgnumMeg

Right, so if $X \subseteq B$ and $B \subseteq A$ then $X \subseteq A$. In other words, using $C_B$ and $C_A$, if $X \in C_B$ then... ?

Abdullah UYU

9:23 AM

Then $X\in C_A$ ok. Got it.

I might have to also show that if $X\subseteq B$ and $B\subseteq A$ then $X\subseteq A$.

Jasper Loy

@Secret I don't mind you writing stuff in this chat, but some other chat users may mind. Why don't you write your notes in your personal notebook instead of this chat? For your consideration, please.

Isana Yashiro

9:40 AM

Sorry for interruption, a soft question about linear algebra book: I'm about to finish the book by Friedberg, I want to find a next suitable advanced book. I did search from this site and MO, and currently I have (1)Lax, (2) LA done right by Axler(I haven't browse it) (3) Curtis (4) FDVS by Halmos, (5)numerical LA by trefethenbau; is there any recommendation reading sequence, i.e. which should be read first before another? Or any other more suitable book?

ÍgjøgnumMeg

@Abdullah that's just by how those sets are defined, you've done that already

@Abdullah I see what you mean, you aren't allowed to use (without proof) the fact that containment is a transitive relation

Leaky Nun

10:00 AM

what is an open set? (intuition)

Secret

An open set controls what points you will eventually reach when you start from any point within it

This intuition can be verified with the definition of a closed set, which is a set containing all the limit points. The limit points are all the points you can end up at when you wander in any direction in an open set

actually wait, I might be talking about closures, let me think of this again...

Hmm...

Let $X$ be a set and $\tau$ be a topology. Pick any subset $A \subseteq X$.

ÍgjøgnumMeg

aN oPeN seT iS A SeT wHiHC aRe OpEN

4

Rithaniel

Are open, yes.

Secret

Each $\tau$ specifies a collection of nets (uncountable generalisation of sequences) which controls what points in $X$ you will eventually end up in given any number of points

Intuitively speaking, we can treat $X$ as a vast countryside and $A$ a small region of it. The nets dictates all the possible paths you can follow to move from point to point

The limit points of $A$ are all the destinations you can end up at when you follow any net that contain points in $A$.

$A$ is closed if $A$ contains all the possible destinations you can reach given a starting point somewhere in $A$

$A$ is neither open or close if some of the nets we carry you to destinations outside of $A$, and others inside of $A$

$A$ is open if any net you follow carry you outside of $A$

The closure of $A$, $\text{cl}(A)$ thus collects all the points in $A$ and all the possible destinations together

Therefore, intuitively speaking a closed set is closed intuitively speaking is because no matter what net you follow, you will always end up somewhere in the set

And an open set is the polar opposite, where any net you follow you will "fall outside of the set"

Now another useful intuition (which I don't know how to prove) is that for any topology $\tau$, let the collection of all neighbourhoods of any point $x \in X$ be its neighbourhood system. Then the more open sets in that neighbourhood system, the less nets it will lead to point $x$. This is why in the indiscrete topology, every point can be reached starting from any point and any net

Likewise, in the discrete topology, only eventually constant nets can reach any point $x$

Alex Clark

10:23 AM

You're doing it again :P

Secret

@AlexClark No, I am answering Leaky's question, I should have tagged him

@LeakyNun

But tbh, I rarely tag people, because I know how annoying that ping noise is, so I often rely on my recipients to get the conversation from the chat flow

Alex Clark

ok

Secret

actually.. the above intuition has a problem: What on earth is a clopen set. I might need to think about this more...

Jasper Loy

@IsanaYashiro After reading Friedberg, if you want something more advanced, see Advanced Linear Algebra by Loehr and forget about all the other books you mentioned which are pretty much on the same level as Friedberg.

Secret

Relevant stuff: proofwiki.org/wiki/Convergence_of_Sequence_in_Discrete_Space

user2236

11:02 AM

@MikeMiller so true, my friend

btw

Why is Mr. Hardy carrying an umbrella on a sunny day?

In his mind, if he holds up an umbrella, it’s definitely not going to rain because God is going to spite him. It’s a funny thing, but it’s difficult as a filmmaker when you have one hour to shoot the scene and you need it to be sunny in rainy England.

Source: NY Times

ÍgjøgnumMeg

@Secret this is a good answer, definitely helpful for me!

user193319

@Faust Do you like Johann von Goethe? I'm actually reading von Goethe's Faust at the moment: it is rather interesting, and very beautifully written (at least the English translation I have is).

ÍgjøgnumMeg

11:28 AM

@user193319 learn German and read the original! lol

although that would definitely be jumping in at the deep end

Jasper Loy

Read The Idiot in the original Russian. =)

Shobhit

12:03 PM

Is anyone familiar with type 2 fuzzy sets?

Rithaniel

So, do you guys have any favorite recreational math puzzles? Ie, things like mathematically solving the lights out puzzle or a particular style of Rubix cube.

user193319

@ÍgjøgnumMeg Haha, that's definitely a long term goal of mine.

@JasperLoy Yes, I need to! Dostoyevsky is a genius! I've read (in English) The Brothers Karamazov, Notes From the Underground, and Crime and Punishment and thoroughly enjoyed all three.

3

Silent

12:21 PM

Let $G$ be a group and $A$ be a two element subset of $G$, where one of those elements of $A$ is identity element $e$ of $G$. So, is it true that $C_G(A)=N_G(A)$? (centralizer equals normalizer?)

Let $A=\{e,x\}$. I think it does, since $g\in G$ lies in normalizer iff $gAg^{-1}=A$, ie, $\{geg^{-1}, gxg^{-1}\}=\{e, x\}$. But since $geg^{-1}=e$, this means $gxg^{-1}=x$ . This implies $g$ lieas in centralizer iff lies in normalizer.

Am I right?

@LeakyNun

Isana Yashiro

@JasperLoy Ok, I will consider the book, thanks for your advise. But the reason I mention those books is that some of them(at latter chapters, I only browse it) are not that easy for me.

Oskar Tegby

12:45 PM

@Akiva: Looking at how to prove $\overline{X+Y}=E$ as seen yesterday. I don't really understand what the closure of $X+Y$ is. Do you have any input here?

Alessandro Codenotti

12:56 PM

Pick any element of $E$ and try to construct an element of $X+Y$ close to it in the $\|\cdot\|_1$ norm. Start with the even indexed digits

Julius

1:14 PM

Hi. Is it true that $E[|Cov(X,Y \mid \mathcal{F})|] \leq E[|Cov(X,Y \mid \mathcal{G})|]$ when $\mathcal{F}\subset \mathcal{G}$?

Leaky Nun

@Silent yes

Shobhit

2:03 PM

hello

need help in latex

$$\bar{A} = \int_{x \in X} \mu_{\bar{A}}(x) /x = \int_{x \in X} \left [\int_{u \in J_{x}} f_{x}(u)/u \right]/ x \hspace{10mm} J_{x} \subseteq [0,1]$$

i want the division sign to be big as well

equal to the size of the square bracket

Julius

Only the one inside the brackets or also the one after them?

Shobhit

the double integral divided by x, i want the division operator "/" to be as big as the last square bracket

Julius

Then try
\left.\int_{x \in X} \left [\int_{u \in J_{x}} f_{x}(u)/u \right]\middle/ x \right.

abenthy

3:04 PM

Any number theorist wants to take a look at my question (8 days without an answer) ? I would highly appreciate it. math.stackexchange.com/questions/2921518/…

user525966

Is there a name for the type of proof in a Hilbert system where you just write things out line by line

Jasper Loy

3:54 PM

@IsanaYashiro If I recall correctly, Friedberg actually covers a lot of material as well. However, the proofs are very detailed, which is why you might find it easier to read. The books by Halmos usually use less symbols and more words than other authors, truly explaining the essence of the subject. Every word is well used and there is no verbosity.

Abcd

Can someone teach me Cayley Hamilton theorem? Or provide any good link to it? (just interested in the application, not in the proof)

Jasper Loy

4:10 PM

@Abcd Read a good book on linear algebra, for example, Linear Algebra by Peter Petersen.

Math geek

math.stackexchange.com/users/585505/math-geek?tab=questionsm,.\

Jasper Loy

@Abcd Sorry, I did not see the application part, ignore me.

Abcd

@Mathgeek you have just posted the link to your profile

@JasperLoy Okay ...

Math geek

4:29 PM

0

Q: Why is the equality $|L\backslash M_H|=|L|-|M_H|$true? Why $|L\backslash M_H|\ge|U|$?

Doubt on underline 1 What do you mean by the edge subgraph of $G$ defined by $L$?IsIt a subgraph with all edges from $L$? Doubt on underline 2 Why is the equality $|L \backslash M_H|=|L|-|M_H|$true? Why $|L\backslash M_H|\ge|U|$? Can you please explain?

@Abcd sorry

Ted Shifrin

@Abcd: The only application I can think of immediately is to use $p_A(A)=0$ to give a polynomial formula for $A^{-1}$ when $A$ is invertible.

Abcd

@TedShifrin What does $p_a(A)$ mean? Matrix polynomial of A?

Leaky Nun

@TedShifrin and that has deep implications to algebraic number theory!

Ted Shifrin

$p_A$ is the characteristic polynomial of $A$. Cayley-Hamilton is the equation I wrote — that when you plug $A$ into that polynomial, you get $0$.

Jasper Loy

By the way, there is a third edition of Peter Petersen's Riemannian Geometry.

Leaky Nun

4:31 PM

btw it's $0$ as in the matrix $0$.

Abcd

What's the characteristic polynomial?

Ted Shifrin

It has implications lots of places, @Leaky, but I don't think that's @Abcd's concern.

The polynomial you write down to solve for eigenvalues, @Abcd. $p_A(t) = \det(A-tI)$.

Abcd

@TedShifrin Can you tell me about it without the usage of eigen-stuff?

Ted Shifrin

@Jasper: I don't know that particular book of Petersen. But I will say that his differential geometry book (in first edition) was FULL of errors and very hard to read. What's more, when I emailed him with comments about some of the slips/errors, he was very dismissive and arrogant about it.

I just gave you the definition, @Abcd.

Abcd

@TedShifrin Yeah, but I want to learn it from scratch. Coz I dont know aught about it.

Jasper Loy

4:34 PM

@TedShifrin Oh dear, are you like good friends with him? =)

Ted Shifrin

Start by writing down a $2\times 2$ matrix, @Abcd, and work out what I just wrote. ... Alternatively, feel free to watch a few of my lectures on this to learn about it.

Abcd

@TedShifrin where's your lecture available

Ted Shifrin

Nope, @Jasper. But I met him when he was just starting out.

Mike Miller

Tired

Jasper Loy

@TedShifrin I emailed him once about his books. He was very nice to me. =)

Ted Shifrin

4:35 PM

@Abcd: Start with 3510 lecture 47 here‌.

@MikeM: Are you in transit?

Mike Miller

I didn't like Peter's book/class, but that wasn't his fault; it was just a matter of my taste.

Home

Ted Shifrin

Ah, welcome home.

Mike Miller

Just bad sleep last night

Ted Shifrin

I was very unimpressed by his irresponsible way of handling things with our mutual friend, Mike.

Travel is always sleep-stressful.

Abcd

in JEE/High School Maths Problems, Sep 20 at 15:13, by Abcd

> Let $a,b,c \in \mathbb R$ such that no two of them are equal and
> satisfy $\det\begin{bmatrix}2a&b&c\\b&c&2a\\c&2a&b\end{bmatrix} = 0$,
> then the equation $24ax^2 + 4bx +c=0$ has:
>
> a) atleast one root in $[0,\frac 12]$
>
> b) at least one root in $[-\frac 12, 0)$
>
> c) at least one root in $[1,2]$
>
> d) none of these

Jasper Loy

4:36 PM

John Lee's Introduction to Riemannian Manifolds second edition is already selling now.

Abcd

is Cayley Hamilton needed for this^^^?

Ted Shifrin

@Abcd: That has nothing to do with Cayley-Hamilton.

Mike Miller

@Ted Yes, of course.

Jasper Loy

There are good prices on Book Depository. Grab it while it is hot!

Abcd

in JEE/High School Maths Problems, Sep 20 at 15:33, by Avnish Kabaj

@Abcd Cayley Hamilton se nahi ho Raha?

Mike Miller

4:37 PM

I do not feel comfortable talking too loudly about things like that, though.

Abcd

@TedShifrin translation to English: Abcd you are not able to solve it using Cayley Hamilton?

Ted Shifrin

I think that person is wrong, but let me think. Did you compute the determinant? What did you get?

Mike Miller

Wow! This is a great question and an even better answer.

10

Q: Can $C^1$ mappings with derivative of low rank be approximated by smooth maps?

Asked once on SE-mathematics. Let $U$ be an open subset in $\mathbb{R}^n$, $m\in\mathbb{N}$, $1\leq m<n$ and let $$\mathcal{C}^k_{\leq m}(U,\mathbb{R}^n):=\lbrace g\in\mathcal{C}^k(U,\mathbb{R}^n)\mid\dim \operatorname{im} Df(x)\leq m\:\forall x\in U\rbrace,$$ where $\mathcal{C}^k(U,\mathbb{R}^n...

fa.functional-analysis gn.general-topology real-analysis differential-topology geometric-measure-theory

Abcd

@TedShifrin I asked that question on main let me search

Jasper Loy

Hello @MatsGranvik hope you are well.

Ted Shifrin

4:39 PM

I think understanding eigenvalues and eigenvectors is more relevant, @Abcd. The matrix has the property that every row sums to the same thing, so $(1,1,1)$ is an eigenvector and $2a+b+c$ is an eigenvalue.

Mats Granvik

Abcd

6

Q: Location of roots of quadratic equation

Let $a,b,c \in \mathbb R$ such that no two of them are equal and satisfy $$\det\begin{bmatrix}2a&b&c\\b&c&2a\\c&2a&b\end{bmatrix} = 0 ,$$ then the equation $24ax^2 + 4bx +c=0$ has: a) atleast one root in $[0,\frac 12]$ b) at least one root in $[-\frac 12, 0)$ c) at least on...

determinant

Mats Granvik

@JasperLoy

Jasper Loy

@MatsGranvik Don't forget me when you solve Riemann! =)

Abcd

@TedShifrin Oh, what will you do after that?

Ted Shifrin

4:42 PM

The matrix also has the property that you're cyclically permuting the entries from one row to the next.

Mats Granvik

I probably should add the Im[Log[Zeta[1/2+I*t]]] that the animated gif converges to.

Abcd

its a symmetric matrix

Mike Miller

@Ted I was just messing around on math genealogy on the plane trip and found out that someone whose paper I use had a PhD student that did some serious calculations that are useful to me.

Ted Shifrin

But it's more than symmetric.

Mike Miller

These don't seem to have been published outside the thesis.

Ted Shifrin

4:43 PM

Very cool, @MikeM. Have you contacted said person?

Mike Miller

So I only knew about this from messing around on the genealogy

No - not in math anymore

But I probably will eventually; I'm sure they will be amused

Ted Shifrin

Ah.

Mike Miller

cse.msu.edu/~ander501

Ted Shifrin

So $a$, $b$, $c$ are chosen so as to make that matrix singular (noninvertible). That's what $\det = 0$ tells us.

The determinant is the product of the eigenvalues, and we already said $2a+b+c$ is one eigenvalue. We're told now that $0$ is another eigenvalue. I still haven't figured out what the polynomial $24ax^2+4bc+c$ has to do with things.

Secret

4:58 PM

Behold the prettiness:

Semiclassical

@Abcd this problem showed up on the main site recently

$Mathematics$ Mathworks (Not the main chat!)

$Mathematics$ Mathworks (Not the main chat!)

Transcript for

$Mathematics$ Mathematics