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12:14 AM
related niceity: "almost-sure sense" or "almost sure sense"?
 
12:33 AM
@MatheinBoulomenos does 3x^3+4y^3+5z^3 (the famous counter-example of Hasse's local-global principle) have a solution in $\Bbb A_\Bbb Q$?
 
 
1 hour later…
1:48 AM
Hi, anyone think without formal graduate education in math, it will be diffcult to generate a new proof for the general cases of Carley Hamiliton theorem?
 
2:00 AM
Given how well-trod that ground is, I think the main work would be verifying that any particular proof is actually novel
 
 
3 hours later…
4:32 AM
So
Israeli Elections 2: Elective Boogaloo
 
 
6 hours later…
11:00 AM
Hi Math SE I'm looking at a Cauchy-like matrix
The elements have the form $(y_j-y_i)/(x_i-x_j)$
So if I use mathematica to get an educated first guess about the closed form, I'm getting the heuristic
"Given that this matrix is $N\times N$, ...
Pick a set of $N/2$ elements $A$ and consider the set $N/2$ elements $B$ disjoint from $A$, one term in the summation that makes up our determinant will be the product of the differences of each element of $A$ with each element of $B$
Up to the sign that we can extract from multilinearity, so how would I write this expression in simple sigma pi notation :p
 
 
2 hours later…
1:26 PM
@TedShifrin Nah, my university has a habit of buying many copies of a book and missing out crucial ones because they spend their funding on many copies. For some reason they have 8 copies of Bourbaki's algebra but not a single copy of Arnold's ODE, Dusa Mcduff's Symplectic geometry, and not even classics like Dirac's book on QM.
 
1:38 PM
@SayanChattopadhyay diracs book on qm is pretty bad
 
2:20 PM
There’s an acronym for HW questions I’m forgetting
To describe ones where the OP only states the question and just says “I don’t know where to start”
I’ve gotten very tired of that
 
can one make a commutative diagram for an integer lattice, s.t. the objects are integers lol
and if you start anywhere on the lattice the path will lead eventually to (1,1)
maybe this doesn't make sense because the morphisms would be sending integers to integers?
 
@Semiclassical psq , problem statement question
 
2:36 PM
"Sure you can. You just have to decide what the mappings between the integers are, and what composition is." -User: jgon
4
Q: Can we have the category of integers?

digitalis_In beginning to learn category theory, I've tried to come up with my own examples of categories. Could we have a category Int, where the objects are integers, and arrows, all mappings between integers? If you can't (which I suspect might be the case), what can't be an object/arrow?

 
2:51 PM
@s.harp there it is
Which could more vulgarly be put as “piece-of-s*** question”
 
maybe they really don't know where to start?
 
3:44 PM
@s.harp I like it. It's terse and not for a first intro but I prefer it over other books like Sakurai which are physicsy in nature. For mathematics there's always Takhtajan
Though you need books like Sakurai to do problems
 
You haven’t done QM if you haven’t suffered through problems
Samurai has some really hard ones
 
@SayanChattopadhyay who does the book benifit if its too hard for a first intro, too basic for the grad student, is too theoretical for the experimtenal physicists and too unrigorous for the theoretical physicists
 
4:06 PM
Given two vectors $\mathbf{u}$ and $\mathbf{v}$, how would one prove that an equilateral triangle, where the norm of $\mathbf{u}$ equals that of $\mathbf{v}$ and $\mathbf{u-v}$, is also equiangular?
Obviously $\cos{(\theta)}=\frac{ \mathbf{u} \dot \mathbf{v} }{ \lvert \mathbf{u}\rvert \lvert \mathbf{v} \rvert}$, where $\lvert\mathbf{u}\rvert=\lvert\mathbf{v}\rvert$.
 
@RyanUnger Sakurai? Yeah it does
@s.harp I don't believe its too unrigorous for theoretical physicists. Many places its still a mandatory reading as a first course. I liked the Dirac-Sakurai combo. Though I guess you could just do sakurai instead.
 
4:51 PM
@schn: Write out $\|\mathbf u-\mathbf v\|^2$.
 
Evening :)
 
Good morning, @ÍgjøgnumMeg.
 
Hiya @Ted lol
 
@TedShifrin $\|\mathbf u-\mathbf v\|^2=\langle {\textbf{u}-\textbf{v}},{\mathbf{u}-\textbf{v}} \rangle$
 
Proceed.
 
4:56 PM
@Ted got my timetable!
not too packed
 
Timetable?
 
enough time for lunch+library from 1-4 each day
err my schedule
for uni
 
Oh, you mean you know what courses you're taking. Do you have any TA duties?
 
I don't :P
 
Well, good — no excuse not to learn lots and work hard. :)
 
4:58 PM
there#s nothing on fridays so I guess I'll be in the library all the time
yis
 
I was never a library inhabitant.
 
I spent most of my undergrad in the library, though here it was mostly empty
 
But I guess you don't have a (shared) office.
 
depends where you are I guess
nah I'm just a lowly master student
 
I mean, I went to the library to look things up or take out books/journals, but I didn't hang out there.
Now, we look everything up on our computers ...
 
5:00 PM
Right lol
lol I only have 2 hrs a week of not-number theory
how delightful
 
Yuck.
 
:D
at least I'll have some formal background in topology by the end of the semester lol
 
Hey @ÍgjøgnumMeg how’s analytic nt?
 
@TedShifrin $\langle {\textbf{u}-\textbf{v}},{\mathbf{u}-\textbf{v}} \rangle=\langle \textbf{u},\mathbf{u} \rangle-2\langle \textbf{u},\textbf{v} \rangle+\langle \textbf{v},\mathbf{v} \rangle=2(\|\mathbf{u}\|^2-\langle \textbf{u},\textbf{v} \rangle)$
 
Does anyone here have any idea what the OP is talking about here?
And what does that equal, @schn?
 
5:04 PM
@Ultradark idk yet.. lol
looks loggy
 
How does one analyze the stability of a vector field
 
@Ultra: Please go read books and learn rather than constantly posting questions in here.
 
Okay I will go do that
but aren’t questions good?
 
@TedShifrin Since $\cos{(\theta)}=\dfrac{ \langle {\mathbf{u}},{\mathbf{v}} \rangle }{ \|\mathbf{u}\|\|\mathbf{v}\|}=\dfrac{ \langle {\mathbf{u}},{\mathbf{v}} \rangle }{ \|\mathbf{u}\|^2}$, but $\|\mathbf u-\mathbf v\|^2=\|\mathbf{u}\|^2=2(\|\mathbf{u}\|^2-\langle \textbf{u},\textbf{v} \rangle)$, it follows that $\dfrac{ \langle {\mathbf{u}},{\mathbf{v}} \rangle }{ \|\mathbf{u}\|^2}=\dfrac{\|\mathbf{u}\|^2}{2\|\mathbf{u}\|^2}=\dfrac{1}{2}$. Thanks!
 
There you go.
 
@Ultra: IMHO, you ask vague questions that you don't even have any understanding of. Occasional focused questions, where you have some specific thing you don't understand or are stuck on, are fine.
 
@Akiva very cool
 
5:35 PM
@Almo Oh I am really sorry about that. I was trying to access the link on my mobile device and I accidentally clicked the flag option. I clicked it back again and it said that it was no longer flagged.
 
@Sayan: You're flagging me???!!! You're dead to me.
 
Nooooooooooooooooooo :p
 
Hello @people
 
hi @Tobias
 
5:38 PM
Hey @Tobias
 
(if that did not ping you, it means the system does not consider you "people")
 
I'm not @people.
 
Nor am I @people.
 
I am already dead
 
5:39 PM
only to @Ted
to me you're just not @people
 
nani?!?
 
omae wa mu..
 
shindeiru
 
meeemes
 
5:39 PM
lol
 
me-mes
 
@TedShifrin I think achille’s comment there is as plausible an explanation as I could come up with
 
I can't wait to pepper a professor with questions in 15 minutes
 
I really wanted the OP to answer my question in a semi-cogent fashion, @Semiclassic :)
 
Lol
If they were that incoherent to start with, I don’t have much hope for improvement
 
5:43 PM
Achille's comment is what I naturally assumed, too. It's not like I was clueless :P
 
@SayanChattopadhyay HAHA ok! :)
 
@Ultradark :)
 
(:)
 
@ted I forget if I pinged you, but I have convinced myself that the problem with that de Finetti quote I gave is on the part of the translation and not de Finetti himself
 
5:46 PM
c(:>)-|-<
@Ultradark how is AA course going?
*or school in general
 
And that includes the bit about almost-sure equality. He does have that earlier in the paper, and his wording in fact indicates to me that he meant almost-sure equality with a fixed value in that sentence
 
It's probably an error in the paper
 
it's not bad @ShineOnYouCrazyDiamond
 
The translation misses that entirely
 
but my professor apparently did not know that polynomials form a ring
 
5:49 PM
@Ultradark that's good
@Ultradark were they a TA?
 
and a student had to remind him
 
The one bit which de Finetti doesn’t consider is whether his inner product space is complete
 
no, the prof has a large age
 
Translations are often suspect, @Semiclassic.
 
So he doesn’t quite have the full modern concept of a Hilbert space of random variables.
 
5:51 PM
but I think he actually knew, it's just maybe was distracted
 
But he’s got the geometric content just fine
@TedShifrin yeah
It’s frustrating but not surprising.
 
@Ultradark probably, or some people lose their minds as they age
Even if they study math so that doesn't exclude us
 
Professors who are losing their minds due to old age should not keep teaching. I retired way early partly because I saw people who should have retired still teaching.
 
Therefore we should do math that supports cure for neurological diseases
 
5:54 PM
@TedShifrin perhaps they weren't paid enough to retire early enough
 
A lot of professors hang on because they have nothing else to do, too. But it really isn't good, and there are lots of young people going jobless because of it.
 
It doesn’t help, no
 
You can always find something more to do. Never stop growing.
 
Of course, the unis are probably happy to be able to keep the young people in indefinite full-time-but-untenured state
 
Write a book, take up programming and make a video game, or even just hike across one continent or another.
 
5:58 PM
Well, it's true that a retirement needn't necessarily convert to a tenure-track job. But still.
@Rithaniel: My body won't allow much hiking at all, sorry.
 
Heh, yeah, that's fair.
Get a helicopter and helicopter across one continent or another.
 
Making the carbon footprint worse. No thanks.
 
(Disclaimer: I make joke)
 
OK, I'm going shopping. Do some good math(s) whilst I'm gone.
 
Slightly more serious suggestion: Horseback riding across one continent or another.
 
6:09 PM
While Ted's gone, let's do some math
 
I am looking for a book recommendation on functional analysis. I am currently following Reed and Simon's functional analysis. Are there other books that I can follow. Reed and Simon is a bit wordy and confusing sometimes
 
Let $G$ act on a connected graph $\Gamma$, and let $F$ be the associated fundamental domain. Is the orbit equal or homeomorphic to $F$?
 
7:02 PM
What's the fundamental domain associated to a graph?
 
@AkivaWeinberger presumably the one associated to the group action
 
is it unique?
 
Oh I think I see?
So like if the graph is the infinite grid $\Bbb Z^2$ with edges between adjacent vertices
and $G$ is the set of translations
then the fundamental domain would be a vertex plus two edges coming out of it?
Not sure I understand
 
Someone mentioned hackertyper.net to me today. It is pretty awesome
 
7:24 PM
Fundamental domains are not unique.
I suspect they are the same up to translation by the group; but this is a mere suspicion
 
I put this question up for bounty but didn't receive an answer. Any hints are welcome. math.stackexchange.com/questions/3316658/…
 
So, thinking of quaternions as a complex vector space, every quaternion can be written as $z_1 + z_2j$, right?
For $z_1,z_2 \in \Bbb{C}$?
And $j^2 = -1$?
I'm trying to show that $SU(2)$ is isomorphic to $S^3$ as groups, but I've been thinking of $S^3$ as living in the quaternions as a real vector space. I defined a which I believe is a isomorphism, but having four real parameters is a pain in the___especially with the matrix multiplication.
And I keep screwing up the calculations...I have negatives where there shouldn't be negatives.
 
7:44 PM
@user193319 Yes: just note that $jz=\bar zj$
 
The map I claim is an isomorphism is $\begin{pmatrix} a + bi & c + d i \\ -a +b i & c - di \\ \end{pmatrix} \mapsto a + bi + cj +dk$.
But verifying that this is a group homomorphism is literally hell.
 
Not $\begin{pmatrix}a+bi&c+di\\-c+di&a-bi\end{pmatrix}$?
which is what Wikipedia has
 
@AkivaWeinberger Oh, so the map I should define is $\begin{pmatrix} z & w \\ - \overline{z} & \overline{w} \end{pmatrix} \mapsto z + wj$?
 
You can define a representation $Q_8 \to \text{GL}_2(\Bbb C)$ which sends $i, j, k$ to various 2x2 complex matrices of determinant 1. I believe that will extend to an isomorphism $S^3 \to SU(2)$ though I have not checked.
 
$\begin{pmatrix}z&w\\-\overline w&\overline z\end{pmatrix}$
Each variable is found on the diagonals
 
7:47 PM
@AkivaWeinberger Oh, and map that to $z + wj$?
 
...that sounds so much easier...I suspected there was an easier way. So, if $z + wj \in S^3$, then $|z|^2 + |w|^2 = 1$, right?
 
Yes
You'd want to check $\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}0&1\\-1&0\end{pmatrix}=-I$
 
I think $i$ to $(0, -1; 1, 0)$ and $j$ to $(i, 0; 0, -i)$ works?
 
and $\begin{pmatrix}0&i\\i&0\end{pmatrix}\begin{pmatrix}0&i\\i&0\end{pmatrix}=-I$
and $\begin{pmatrix}i&0\\0&-i\end{pmatrix}\begin{pmatrix}0&i\\0&-i\end{pmatrix}=-I$
 
7:49 PM
It's not clear to me why I need to do that...Will it become clear when I try to prove that the above map is an isomorphism?
 
@AkivaWeinberger Cool, same idea.
 
I'm just writing out $i^2=j^2=k^2=ijk=-1$
which is the defining property of the quaternions
 
Oh, for my Lie algebras class we aren't defining $i$,$j$, and $k$ in terms of matrices.
Those symbols are defined to satisfy those relations, and then $\Bbb{H}$ is just real linear combinations of those with multiplication extended linearly.
 
But you're trying to find an isomorphism from the quaternions to matrices
 
But I know that $\Bbb{H}$ can be regarded as a $\Bbb{C}$-vector space also with basis $1$ and $j$, so I can use that to simplify the calculations.
Quaternions in $S^3$ to matrices in $SU(2)$.
 
7:52 PM
so I guess I really was writing out $f(i)^2=f(j)^2=f(k)^2=f(i)f(j)f(k)=f(-1)$
 
Where $f$ is the alleged isomorphism?
 
Yes
$f:S^3\subset\Bbb H\to SU(2)$
 
Okay. I see. Let me give this a shot. Thanks!
 
I made some typos @user193319
but I'm pretty sure $\begin{pmatrix}z&w\\-\overline w&\overline z\end{pmatrix}$ is it
 
Yeah, I just checked my book and that is right.
 
7:54 PM
In mathematics, the quaternions are a number system that extends the complex numbers. They were first described by Irish mathematician William Rowan Hamilton in 1843 and applied to mechanics in three-dimensional space. A feature of quaternions is that multiplication of two quaternions is noncommutative. Hamilton defined a quaternion as the quotient of two directed lines in a three-dimensional space or equivalently as the quotient of two vectors.Quaternions are generally represented in the form: a + b i +...
 
So, I have this question that I'm having a lot of trouble understanding.
Let $f:\mathbb{R}^4\rightarrow\mathbb{R}^4$ be a linear map defined by $f(e_i)=ie_i$ for $1\leq i\leq 4$ where $(e_1 \ldots,e_4)=I_4$ and let $$ H=\begin{pmatrix}1 & 1 & 1 & 1 \cr 1 & -1 & 1 & -1 \cr 1 & 1 & -1 & -1 \cr 1 & -1 & -1 & 1 \cr\end{pmatrix}.$$ Find the matrix of $f$ under the basis $H$ of $\mathbb{R}^4$.
What is being asked for?
 
So $f(e_1)=e_1$, $~f(e_2)=2e_2$, etc
which means its matrix is $A:=\begin{pmatrix}1&&&\\&2&&\\&&3&\\&&&4\end{pmatrix}$
but
you now have a new basis
which I think is $e'_1=(1,1,1,1)$, $~e'_2=(1,-1,1,-1)$, etc
the columns of $H$
which means $e'_i=He_i$
so
correct me if I'm wrong
but I think the change-of-basis is gonna be $H^{-1}AH$
'cause you want to figure out $f(e'_i)$ in terms of $e'_i$
@Rithaniel
 
8:10 PM
Hmmm, so conjugation of $A$ by $H$?
This particular professor does not make topics entirely clear, I'm afraid.
I'll research change-of-basis for a moment online.
 
8:27 PM
I think I get it. So, the $H$ takes a vector from our $H$ basis into the standard basis, $f$ is the linear transformation, and $H^{-1}$ takes it back to our $H$ basis. Is that right?
Thank you for the help @AkivaWeinberger
 
ABC
8:43 PM
Guys goodevening,
I have a question:
Residue of Complex conjugate is the Complex conjugate of Residue?
 
9:05 PM
you know the complex conjugate of a holomorphic function is anti-holomorphic?
 
ABC
@s.harp Yes I know
 
9:25 PM
how do you define the residue for a anti-holomorphic function?
note for exampple that the integral of $\overline z$ over the circle around $0$ or radius $R$ is equal to $2\pi R^2$
 
 
2 hours later…
11:43 PM
@Rithaniel You can also see my YouTube video on this (and probably others would be helpful).
 
11:55 PM
@s.harp which is great if you want your integral to be the area inside your curve. Less great if you want (local) path-independence...
 
@Rithaniel Yes I believe so
 

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