So you've got a vector field $X$ on a manifold. Say at every point $p$, $X$ gives you a vector $v:=X(p)$ tangent to the manifold at $p$. And if $f$ is a differentiable function from the manifold to $\Bbb R$, you can consider the directional derivative of $f$ at a point $p$ in the direction of $v=X(p)$.
I wonder if someone wrote some humongous indices notation like $A_{\beta}$ and then let $\beta = \begin{pmatrix}a& b & c \\ d & e & f \\ g & h & i\end{pmatrix}$ where each entry is an index?
If you have two vector fields, you can look at $X(Yf)$ and $Y(Xf)$, which are essentially second derivatives of $f$ (first in the direction of $X(p)$ and then in the direction of $Y(p)$, or the other way around)
However, if you look at $XYf-YXf$, apparently the second derivatives cancel
@Alessandro There should be an indoor game where a bunch of people throws each other random math as much as they possibly can and everyone participating has to :thonkfast: about every thing
Yeah I mean you don't really have a multiplication operator to write it as aba^-1b^-1 fully rigorously, which is why I was suggesting to pass to flows.
@Abcd ah I forgot you are dealing with generic n, cause for specific values you can just plug in the midpoint and show it is greater, but plugging in n/2 will require evaluating (n/2)! which needs the gamma function, thus too complicated
@Semiclassical I think I finally found the bible which, assuming you read a 120 year old 800 page old complex analysis book before beginning it, would make those Dubrovin notes people.sissa.it/~dubrovin/rsnleq_web.pdf readable without any magic :p
If you go at this rate 12 years form now on you'll find yourself reading old Mesopotamian and Egyptian earliest mathematical texts to understand complex algebraic geometry or whatever the shit you want to understand.
I applaud your endeavor and bid you a good luck in your journey.
I recently found out that that 800-page complex book (by Forsyth) was literally the first English translation of European analysis of the 19'th century, in the 1890's, and changed English-speaking math, but was considered not rigorous enough once people read it and thought about it, but then Whittaker came along as the first rigorous analysis book in English, so you can't really go much earlier apart from those Euler texts, Gauss Disquisitiones, Euclid, Newton, and that's it maybe :p
and that Forsyth book is quite literally insane in the second half
It holds that the set {a} is an affine subspace in $\mathbb{R}^2$, right? Can it be that a is an affine subspace? Or only the set that contains one point and not just one point?
Suppose we have the set $C=\{x\in \mathbb{R}^3 \mid \langle x, \begin{pmatrix}-1 \\ 1\end{pmatrix}\rangle=1\}$. To check if this is an affine subspace do we have to check if we can write that set in the form a+U ? @AkivaWeinberger
at $\langle x, \begin{pmatrix}-1 \\ 1\end{pmatrix}\rangle=1$ the left side is the dot product, or not? But $x$ has 3 components, how is it defined? I got stuck right noe.
I'm trying to show that $\Bbb{R}^\omega$ with the uniform metric is not 2nd countable or separable (either one will work, since they are equivalent for metric spaces). I could use a hint.
My idea was to transfer the proof that $\Bbb{R}_\ell$ isn't countable into the $\Bbb{R}^\omega$ setting. If $\mathcal{B}$ is a basis for $\Bbb{R}_\ell$, then $x \mapsto B_x \subseteq [x,x+1)$ is an injection between $\Bbb{R}$ and $\mathcal{B}$, which shows that $\mathcal{B}$ is uncountable. How do I do something similar in $\Bbb{R}^\omega$?
The standard way to get that is to relate this sequence to the other way of getting $e = \sum 1/n!$. You can easily see that series sums to less than $3$.
What are we allowed to use?
With some calculus, of course, it's quite easy.
@user193319: Did you figure out your $\Bbb R^\omega$ yet? You're right that the easiest thing to do is give an uncountable number of disjoint open sets.
@Abcd perhaps this is unsolicited advice, but you shouldn't let your reach exceed your grasp. It's a bit safer to be sure you can at least understand the proofs of what you're doing. You don't need to memorize them, but you should definitely be able to understand them if they're in front of you.
@AkivaWeinberger there is a distinction to be drawn, though, between people who know the statement of the Riemann hypothesis and people who understand why proving it is so hard.
@TedShifrin That's what I was trying. I thought that mapping $x = (x_1,x_2,...) \in (0,1) \times \{0\} \times ...$ to $B(x,x_1)$ would work, but it doesn't...
@Abcd: OK, in order for an infinite series to converge, the individual terms must in fact approach 0. Or, you look at the formula for the finite sums and look at what's left over and want to know what makes that go to 0.
No, @user193319. Here's a huge hint. Allow only integer coordinates for the centers of your balls.
You're not using anything at all about the structure of $\Bbb R^\omega$ the way you're doing it.
@AkivaWeinberger The former evaluates $f(\varphi(x))g'(\varphi(x))\varphi'(x) - f'(\varphi(x))g(\varphi(x))\varphi'(x)$, which is $([f, g] \circ \varphi)(x)\cdot \varphi'(x)$ isn't it.
@BalarkaSen The only detail to check is what happens if you have a $C^1$ metric whose scalar curvature equals a smooth function. It's probably smooth but who knows
Let $g$ be a smooth metric and $\delta R_g$ the linearization of the scalar curvature at $g$. Then my PDE is really $\mathrm{scal}(g+(\delta R_g)^*u)=f,$ where $u\in W^{4,p}(M)$ and $f\in C^\infty(M)$.
This should imply $u\in C^\infty$ by some theorems in Morrey but I want an independent proof
@TedShifrin Okay. I claim that $\{B(x,1/3) \mid x \in \Bbb{Z}^\omega \}$ is a uncountable collection of open disjoint. Suppose that $z \in B(x,1/3) \cap B(y,1/3)$, where $x \neq y$ which means there exists a $k$ such that $x_k < y_k$. Then $\rho(x,z) < 1/3$ and $\rho(y,z) < 1/3$ implies $|x_i - z_i| < 1/3$ and $|y_i-z_i| < 1/3$, resp, for every $i$. Hence $|x_k - y_k| \le |x_k - z_k| + |y_k - z_k| \le 1/3 + 1/3 = 2/3$, implying that $|x_k - y_k| = 0$, because they're integers, or $x_k = y_k$...
![[They do not move.]](https://imgs.xkcd.com/comics/interaction.png "[They do not move.]")
