@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
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 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
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.
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.
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
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.”
@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} $?
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.
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
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.
@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
I'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...
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
@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
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
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, 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
