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02:00 - 12:0012:00 - 00:00

2:35 AM
@Ottavio The Weyl tensor becomes $R - s/(2n(n-1)) (g \mathsf{KN} g)$ on an Einstein manifold, because the middle term vanishes. Vanishing of this object is a necessary and sufficient condition for the manifold to be conformally flat.
You can thus prove $\Bbb{CP}^2$ is an Einstein manifold which is not conformally flat very easily from here.
Because $R$ being proportional to $g \mathsf{KN} g$ would mean it has constant sectional curvature, which is only true in the complex directions on $\Bbb{CP}^2$. The sectional curvature swings between $1$ and $4$ if you vary over all planes (NOT strictly quarter-pinched, that forces your manifold to be a sphere by a famous theorem)
I am going to $\mathsf{KN}$ from now because $\bigcirc \!\!\!\! \wedge$ is a pain in the ass
 
3:09 AM
I don't like KN how about like
lame MSE doesn't have any good approximations either
 
what are you doing man
 
i want a better symbol
 
$㋬㋩㋫㋑⧁⧀ⓥⓋ⎉㉠㉦$
 
i thought a lot about it but couldnt find it
$\huge{㉦}$
Nah
 
$g \mathbin㉦ g$
 
3:13 AM
Just need to rotate $\huge{Ⓥ}$
Oh wow it's as bad as $\huge{\bigcirc \!\!\!\!\!\!\!\!\!\vee}$
 
i tried rotating with a style in chat, doesnt work
 
Hey
This will be pretty if it rotates
 
which, the $\vee$?
hmm what is it lol if not vee
 
$$\require{HTML} \style{display: inline-block; transform: rotate(180deg)}{\large{\bigcirc \!\!\vee}}$$
 
lol i could just turn off mathjax to see it
 
3:15 AM
WTf
 
LOL
 
It looks awful
$\require{HTML} \style{display: inline-block; transform: rotate(180deg)}{\large{\bigcirc \!\!\!\!\!\vee}}$
 
$\large{\bigcirc\hspace{-1.05em}\wedge}$ $$\large{\bigcirc\hspace{-1.05em}\wedge}$$
 
Yes look at that beauty
Horrible Calvin
$$\require{HTML} \style{display: inline-block; transform: rotate(180deg)}{\large{\bigcirc \!\!\!\!\!\vee}}$$
Absolute beauty
$\newcommand{\KL}{\require{HTML} \style{display: inline-block; transform: rotate(180deg)}{\, \large{\bigcirc \!\!\!\!\!\vee}}}$
 
you should use yours for an noncommutative operator
 
3:19 AM
$g \KL g$
Um
Why is the spacing bad
come on man i give up
 
cuz you started with some weirdly spaced thing to begin with
 
$\newcommand{\KL}{\require{HTML} \style{display: inline-block; transform: rotate(180deg)}{\large{\, \bigcirc \!\!\!\!\!\vee \,}}}$
$g \KL g$
 
lol
 
Terrible
 
no dont do that
use DeclareMathOperator
wait
no, you want binary operator
i suppose \newcommand{...}{\mathbin{blah}}?
 
3:22 AM
Ah
 
$\newcommand{\calvinstester}{\mathbin{X}}$ $a\calvinstester b$
yup works
 
$\newcommand{\KL}{\mathbin{\require{HTML} \style{display: inline-block; transform: rotate(180deg)}{\large{\bigcirc \!\!\!\!\!\vee}}}}$
$g \KL g$
Yes, still sort of weird but can be dealt with
 
but your wedge is not centered on my screen and also why the heck are you rotating a vee
 
the wedge is perfect on my screen! it touches the perimeter of the circle
 
you want it to be asymmetrical?
 
3:24 AM
the \wedge gives you a bad result
@CalvinKhor I want the legs of the wedge to touch the circle
Goddamnit
 
I thought \vee$\vee$ was literally a rotated \wedge$\wedge$
$\newcommand{\notKL}{\mathbin{\require{HTML} \style{display: inline-block; transform: rotate(0deg)}{\large{\bigcirc \!\!\!\!\!\wedge}}}}$
$g \notKL g$
 
That's fucked, see? It's totally not centered
And the legs don't simultaneously touch the circle
Mine is obviously better
 
LMAO
no I parse it as visually different
 
i thought you were trolling hahahahaha
 
3:29 AM
ill show you a screenshot but im too lazy
 
thats k i trust lol
 
ill make do with yours
$$R = \frac{s}{2n(n-1)} g \KL g$$
 
$g \nparallel g$
that's my KN product
 
$\mathsf{K\!\!\!N}$
 
lmao
 
3:33 AM
Thats my KN product
 
Is anyone able to provide me with a sanity check on some vector calculus. I have a gradient type derivative problem which goes as follows: I have an expression $$ \left( \left[\mathbf{A} \times \mathbf{B}\right] \cdot \mathbf{C} \right)^2 $$ and I want to calculate the derivative with respect to $\mathbf{A}$. I get $$ -2 \left( \left[\mathbf{A} \times \mathbf{B}\right] \cdot \mathbf{C} \right) \left( \mathbf{B}\times \mathbf{C}\right) $$.
I am not sure if the order of the second vector product is correct
 
@MikeMiller is an expert in vector calculus
he can help you for sure
 
lol
 
write it in coordinates
 
The second vector product is a number scaling a vector. It doesn't make sense to write $\vec{v}2$, you've written it in the right order
not an endorsement of your answer which I haven't thought about
 
3:38 AM
dude you're onto something
what if
we think of the the vector space
as acting on the base field
 
@MikeMiller thanks for that. So if it was the same problem except without the square it would be $$ - \left( mathbf{B} \times \mathbf{C} \right) $$, is that correct?
 
I mean I don't really know what you mean by derivative with respect to $A$ here. I would say that your thing is a function $f(\bf A) = [\bf A \times \bf B] \cdot \bf C$. But the derivative of a function $f: \Bbb R^3 \to \Bbb R$ isn't a vector
 
is it possible to upload image in chat room ?
 
Instead it's a linear map
Something that you plug a vector into to get an output
@ronakjain Didn't you ask this earlier today and get an answer
 
please you tell me i have just completed 100+ reputations
 
3:43 AM
@MikeMiller the representative from riesz rep?
 
Ah sorry. I have a scalar potential and I'm calculating the gradient (in the same sense that we would calculate the gradient with respect to Cartesian coordinates) of the potential with resect to $mathbf{A}$. I have a number of these type of expressions popping up and just making sure that I haven't lost my mind.
 
@CalvinKhor I just wanted that said
@ronakjain Nah I think you should look at the last time you asked this and check the answer you got then
 
fair
 
i could not find that . The answer is lost in chats.
 
all these answers, lost in chats... like tears in rain
4
 
3:46 AM
Please tell me are you able to upload image in chat room. is there any upload image option in your chat option. I am not having such option instead i have completed the 100+ reputation criterion.
 
@Rumplestillskin OK, fair enough. Yeah the gradient w/r/t A is B x C.
 
If i get 100+ reputation in mathematics than will i be able to upload image in other sites also.
 
I would write this out carefully but I feel like we'd have a language barrier
@ronakjain Yes I am able to upload images in the chatroom
 
@MikeMiller What is requirement for that option.
 
Who's to say
 
3:48 AM
@MikeMiller Thanks!
 
Love this panel
 
I'm skeptical of your minus sign though
Here's how I would explain it in my language
The derivative of F(A) = [A x B] * C at A is DF_A(D) = [D x B] * C; I want the vector v so that [D x B] * C = v * D. (By definition this is the gradient of F at A) This follows from the vector triple product identity [D x B] * C = [B x C] * D, so
 
Hey Mike Miller. Please tell me what will i have to do for upload image option.
 
@Rumplestillskin $A\times B\cdot C = \epsilon_{ijk} a_j b_k c_i$, so $$\partial_{a_l}(A\times B\cdot C) = \partial_{a_l}\epsilon_{ijk} a_j b_k c_i = \epsilon_{ijk}\delta_{lj} b_k c_i = \epsilon_{ilk} b_k c_i = \epsilon_{lki} b_k c_i = (B\times C)_l $$
 
Jesus christ Calvin
Get a room
 
3:50 AM
lol
 
@ronakjain "Do you really want to hear? It gets complicated!"
 
Why ? You just tell me the requirement for the option.
 
why do it geometrically when it comes out so easily by direct computation
 
dIrEkt cOmpUtaTioN
 
also apologies for all subscripts...
 
3:52 AM
So it sounds like the answer is no
I'll accept that
 
Please tell me i really require that option. I am new to stackexchange chat.
 
You must simply continue diligently, without ever truly knowing if you have the option
“Devotion is diligence without assurance. Faith is a way of saying "Yes, I pre-accept the terms of the universe and I embrace in advance what I am presently incapable of understanding.”

― Elizabeth Gilbert, *Eat, Pray, Love*
 
@CalvinKhor hence @Rumplestillskin you have an extra minus sign
 
@CalvinKhor @MikeMiller many thanks !
 
@Rumplestillskin yw!
 
3:59 AM
Hallo goyuz today we're going to do a prank where I pretend to cheat on my gorlfriend. Seek that upload button and smash sobscroibe
 
Thanks
 
@BalarkaSen Can you present a (sufficiently nice, which you may determine; certainly includes metrizable) topological space with a computable countable amount of data in the following sense?
 
@CalvinKhor then just for sanity if we had the following problem $$ [(A \times B) \cdot C]^2$$ then the derivative with respect to $B$ ( all quantities assumed to be vectors ) would result in the following ( using index notation ) $ 2 (\epsilon_{ijk} A_j B_k C_i ) (\epsilon_{ijk} A_j \delta_{lk} C_i) = - 2 (\epsilon_{ijk} A_j B_k C_i ) \epsilon_{lji A_j C_i} $?
 
@MikeMiller In what sense?
I assume you were writing something
 
Howdy. Sorry to disrupt. I haven't slept in a while so my train of thought might be missing a few wheels. I don't even have ChatJax so here goes nothing.
Ordinals vs Cardinals (infinite sets)
If I have, say $\{a,b,c\}\cup\mathbb{N}$, is the ordinal number of this set $3+\omega_0$ or is it still $\omega_0$? I think we can agree its cardinality is $|\mathbb{N}|$. I just want to make sure I'm understanding this topic as much as I'm pretending to.
 
4:15 AM
$3 + \omega_0 = \omega_0$
I think I nice way to think of ordinals when doing arithmetic is thinking of them as queues
$\omega_0$ is a queue of infinite length
$3+ \omega_0$ is the exact same shape, a queue of infinite length
$\omega_0 + 3$ on the other hand is different. It's a queue of infinite length, but then there's "somehow" a cut and an extra little line of $3$ behind the infinite length queue
Essentially, we can relabel the $3 + \omega_0$ case so that it looks identical to the $\omega_0$ case, whereas we can't do the same with the $\omega_0 + 3$ case
 
Ok... so it's not a commutative f...field.... In other words, it matters where the union is placed?
 
Yeah, ordinal arithmetic isn't commutative
The wikipedia article on ordinal arithmetic is actually very nice for understanding this in my opinion
 
Wow interesting, thank you
I actually read it but I have some mistrust for wikipedia in general
 
@BalarkaSen That message didn't send on my end so I gave up
Anyway the answer is probably no
 
4:47 AM
@Rumplestillskin It might be too late, but I think your minus sign is wrong. After all, $(A\times B)\cdot C = (B\times C)\cdot A$.
Then the derivative with respect to $A$ is obviously $B\times C$. Perhaps @MikeM already did this.
 
yeah Mike and Calvin pointed that out later
 
I'm always late to the party.
 
I like being late to the party
Cut to the food
That's what I care about
 
Well, some parties if you're late you miss all the food.
 
The parties I have been to either consisted of haranguing people in the initial segment or a bunch of stoners and drunkards getting smashed. I always got the food.
Just in lieu of coming late
 
4:49 AM
In lieu of?
Or in view of?
 
in view of, thanks
 
Count on me to be obnoxious.
 
I thought that was my job.
 
It's your secondary job. Your primary one is worst humor.
 
@Rumplestillskin that notation is kinda a mess now, you can't repeat an index more than twice. But $[A\times B\cdot C]^2 =[(-B\times A)\cdot C]^2 = [B\times A\cdot C]^2$ i.e. its symmetric in $A,B$ so you should get the same formula in the end. i.e. $\nabla_B [A\times B\cdot C]^2 = - 2 (A\times B\cdot C)A\times C = 2(B\times A \cdot C)A\times C$ (this last one is just exchanging $B\leftrightarrow A$ from the previous formula)
where the subscript in B denotes the variable that is being differentiated in.....sorry for choice of notation lol
 
5:05 AM
@CalvinKhor Oooops, I knew that but was being a bit hasty! Perfect! Thanks for that!
 
@Rumplestillskin yw, sorry I was busy and left the room
 
@TedShifrin Of course! Excellent thank you also
@CalvinKhor No worries that was very helpful! All the best
 
 
2 hours later…
7:13 AM
rehi @Ted, @Edward
 
yo @Balarka
 
sup
 
0
Q: Equality of fields being deduced from isomorphism of Galois groups

Edward EvansI'm reading and attempting to understand a proof of the local Kronecker-Weber theorem in https://arxiv.org/abs/math/0606108 (page 17). Let $K$ be a local field and $\sigma \in W(K^{LT}/K)$ with $\sigma|_{\hat{K}} = \operatorname{Frob}_K^{n}$ (i.e. $v(\sigma) = n > 0$). Let $L = K(\mu_n)$ and exte...

 
@EdwardEvans maths is a big lie
 
7:26 AM
Hi all! Is U(N) a subgroup of the compact version of the symplectic group, USp(N)?
 
USp(n) is all nxn matrices A with quaternionic entries such that $A^* A = I$ where $A^*$ is quaternionic conjugate transpose. Complex numbers are in particular quaternionic and complex conjugation of a complex number is it's quaternionic conjugate, so yeah
U(n) embeds in USp(n)
 
@BalarkaSen ok thanks!
 
Here's a fact I learnt a few days ago. USp(2)/USp(1) is an exotic 7-sphere
It's homeomorphic but not diffeomorphic to S^7. I didn't know exotic spheres could be homogeneous spaces
 
@EdwardEvans do you know anything about completion?
in a first countable space do sequences and nets capture the same information?
 
7:57 AM
Sequences already capture closures in first-countable spaces and those capture the entire topology
 
8:35 AM
what if we're in a group with a countable neighbourhood basis at 0?
and "information" includes the uniform structure?
 
9:03 AM
Is anyone amount you a jee aspirant ?
 
 
1 hour later…
10:05 AM
Hey @feynhat
 
Hi. 'sup?
 
What are quaternionic analytic functions $f : \Bbb H \to \Bbb H$? Would we just want that $df$ commutes with $I, J, K$?
Polynomials with quaternionic coefficients are examples, yes?
Just because $df$ is obviously multiplication by a quaternion, so a quaternionic linear map
 
Why can't we just use the limit of the difference quotient?
 
Should be equivalent I think if there is any truth to the world
Because for complex analytic maps the difference quotient thing is the same as saying $df$ commutes with $J = (0, 1|-1, 0)$
(Cauchy-Riemann equations)
 
@BalarkaSen I think even "differentiable" for quaternions is not very well understood and people are trying to get analysis to work there. Or maybe I'm completely misremembering a talk from a few years ago
 
10:12 AM
What's wrong with saying $df$ commutes with $I, J, K$ though
Really, just $I, J$ suffices because $IJ = -K$
Don't see an immediate problem
 
I doesn't even commute with J
 
Maybe that's a fair point. So even multiplication by $i$ is not analytic
 
rip non-commutative analysis
 
There should be some simple workable definition. What if I just demand $df$, as a matrix in $M_4(\Bbb R)$, to be in $\Bbb H$?
$df$ is a quaternionic linear map. That's all.
All quaternionic polynomials are then by definition analytic
 
sounds reasonable
 
10:17 AM
OK question
 
that's analogous to the complex definition
 
Do these have an open mapping theorem
 
Hi can I ask a combinatorics question?
 
interesting question, but I dunno
I'd start by figuring out when $df$ is in $\mathbb{H}$
 
Yeah it's not clear at all
 
10:18 AM
you should get some PDE
 
right
 
why are you even thinking about analytic quaternionic maps
differentiable*
no clue how close these terms are for those maps
 
The simplest conceptual proof of fundamental theorem of algebra is as follows: A polynomial $f : \Bbb C \to \Bbb C$ is a complex analytic map sending $\infty$ to $\infty$, so extends to a complex map $f : \Bbb{CP}^1 \to \Bbb{CP}^1$. By open mapping theorem it is open, domain is compact so the domain is closed and open hence surjective, so it must have a zero.
I know there's a version of FTA for quaternionic polynomials, and $\Bbb{HP}^1$ is a hypercomplex manifold
It has two obvious quaternionic charts
So there should be a conceptual analogue of this proof
 
I'd rather develop non-commutative Galois theory for division algebras
what's the FTA analogue for quaternions?
 
The usual topological proof of FTA pushes through essentially
I want an analytic proof
 
10:27 AM
I see
guess you'll have to develop quaternionic analysis
but actually the matrix definition might be problematic
 
Need to figure out what PDE is $df \in \Bbb H$ man
 
does that definition actually avoid the "i not commuting with j" conundrum?
 
It does, right? $f(x) = i x$ is now differentiable, $df = I$
which is in $\Bbb H$
I guess next step is to check $ixj$ lol
what even is that
Come on it has to be linear or something
Just evaluate on $x = a + bi + cj + dk$
Clearly
 
hmm, so being quaternionic-linear is actually not the same as commuting with the basis vectors
 
yeah
 
10:36 AM
but shouldn't that be what linearity means
 
seems not
 
nah, it has to be what linearity means
what breaks down is that the matrix being in H is not the same thing as being quaternionic-linear
both of these options sound wrong
am I tripping
 
yeah it's funky what $\Bbb H$ as a module over itself means. Left module? Right module? What the fuck?
$\Bbb H$ is a $(\Bbb H, \Bbb H)$-bimodule
I mean, so, a quaternionic structure on a vector space $V$ should definitely be three automorphisms $I, J, K$ of $V$ such that $I^2 = J^2 = K^2 = IJK = -1$
Three complex structures compatible in this sense
That automatically makes $V$ a $(\Bbb H, \Bbb H)$-bimodule
 
Let $f\colon\mathbb{H}\rightarrow\mathbb{H}$. If we consider $\mathbb{H}\cong\mathbb{R}^4$ (not canonical, but we fix the most reasonable choice), then we think of f as a map $f\colon\mathbb{R}^4\rightarrow\mathbb{R}^4$, which possesses a Jacobian $df\in M_4(\mathbb{R})$. If $df\in\mathbb{H}$, we get a map $df\colon\mathbb{H}\rightarrow\mathbb{H}$ via left multiplication. This ought to be the derivative of $f$.
 
Yeah just fix one side, work left-modules.
 
10:44 AM
But $df$ is not $\mathbb{H}$-linear in general
 
What is $\Bbb H$-linear mean
You mean $(\Bbb H, \Bbb H)$-linear?
 
no matter which side
 
You have to compromise one sidedness I think that's the only issue
@Thorgott No I mean what is the definition of $\Bbb H$-linearity according to you
 
$df(x+y)=df(x)+df(y)$ and $df(xy)=xdf(y)$ for $x,y\in\mathbb{H}$, but that doesn't work anymore
 
Forget the derivative
A map $T : V \to W$ of left $\Bbb H$-modules is left $\Bbb H$-linear if $T(qv) = q T(v)$?
Then yeah take $V = W = \Bbb H$, $T(v) = \alpha v$ multiplication by a quaternion is not $\Bbb H$-linear
 
10:49 AM
yeah, what else should it mean?
the issue is that $\operatorname{End}_R(R)=Z(R)$, I guess
 
yeah the issue is algebra is useless here
 
that's why non-commutative things suck
 
It is OK to say $df \in \Bbb H$ though. I do wonder what is the restriction it poses
 
it's still multiplication with a scalar, so it's not a nonsensical generalization
 
yeah
 
10:51 AM
it's just that multiplication with a scalar and endomorphism as module over itself are the same notion in the commutative case
but the latter would be a stupid requirement here, cause it would mean to ask the Jacobian to sit inside the reals inside the quaternions
 
correct
i agree
 
ok, so we agree this is a good definition
 
What does a typical quaternion look like as a 4x4 matrix man
 
now write down the PDE
$\begin{pmatrix}z&w\\-\overline{w}&\overline{z}\end{pmatrix}$
now represent each $z,w$ as $2\times2$ matrix
 
Ah ok and then put $z = (a, b|-b, a)$
This is gonna b a dope PDE
lmao
 
10:54 AM
PDE with 16 equations
you can probably cut some down via symmetry
then check whether it's elliptic
 
Must be elliptic
The decoupled equation is just that each component is harmonic
Elliptic too electric boogaloo
So everything goes through
 
so quaternionic-differentiable implies quaternionic-smooth already
does it also imply quaternionic-analytic
 
I mean as a function $f : \Bbb R^4 \to \Bbb R^4$ it's analytic
That forces open mapping right
 
but why is it analytic?
 
I mean $\Delta f_i = 0$, $1 \leq i \leq 4$ is an elliptic PDE
 
11:04 AM
do we have a Cauchy formula for quaternionic maps
 
if $f$ solves that we're analytic
 
I thought elliptic regularity only implies smooth and not yet analytic?
 
I always thought analyticity goes through
I mean components are harmonic so analytic
 
yeah, I guess you can reduce it to the complex case regardless
but then we should also be able to get a Cauchy formula for quaternionic maps, right?
 
Quaternionic Taylor series should be harder I guess
 
11:06 AM
hmm, actually
 
yeah because you're dividing by $1/(z - a)$ quaternion
thats annoying af
 
for quaternionic polynomials, we should use a noncommuting variable
or at least, it makes a difference whether we do or don't
 
Yeah but doesn't for example $f(x) = ixj$ also have $df \in \Bbb H$
you mean like $ixjx$ could be annoying?
Don't see why
 
so quaternionic-analytic is a weaker notion than analytic as map R^4->R^4
but we already got the latter by being quaternionic-differentiable
 
Ah i see ur point
 
11:08 AM
something's afoot
 
the game?
 
$ixjx$ is a polynomial in the broader sense, what does it's Taylor series in the narrower sense look like?
surely something isn't right when quaternionic-analytic is suddenly a weaker notion than quaternionic-differentiable
 
Yeah because R^4 -> R^4 projection to R^3 is quaternionic analytic by our defn?
That doesn't sound right
No I mean
that satisfies coordinatewise Laplacian = 0
but not our initial $df \in \Bbb H$ thing ofc
we need more than coordinatwise analytic for open mapping theorem for sure
I mean $f : \Bbb R^2 \to \Bbb R^2$, $f(x, y) = (xy, y)$ is coordinatewise harmonic but not an open map
 
how is the projection q-analytic
 
It's not, it has coordinatwise Laplacian = 0 is what I meant
we need to work much more for open mapping theorem
very annoying
I give up man this is too annoying
 
11:23 AM
but the Laplacian thing works for C, why doesnt it work for H
 
The real and imaginary parts of a complex map are not just harmonic they are related, one is the harmonic conjugate of the other
You need something like this here
 
and this isn't the case here anymore?
 
I dunno
 
it should be
I mean, you still have the C-R equations
 
How do you prove open mapping theorem map? $f(z) = z^n g(z)$ where $g$ is nonzero, that's the crucial thing, right?
and $z^n$ is an open map, that's what one needs
We need a quaternionic Taylor theorem like you were saying
This is bullshit man
 
11:27 AM
yeah, so a) figure out Cauchy integral formula for quaternionic maps (hopefully exists), b) figure out whether 1/(z-a) is quaternionic-analytic (hopefully; I assume you actually need to use noncommuting variables for this), c) do the usual trick
 
I $\Bbb H$ate it man
 
is there a quaternionic geometric series
 
What is the quaternionic $1$-form $f(q) dq$?
$(f_1 + i f_2 + j f_3 + k f_4)(dx + i dy + j dz + k dw)$
What the actual fuck
Maybe that's closed
 
this looks interesting
our definition was wrong it seems
 
11:47 AM
"Thus
a theory of quaternionic power series will be the same as a theory of real-analytic
functions on R
4
."
huh
 
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