@TedShifrin the point is that by Sobolev theorems, $g+(\delta R_g)^*u$ is a $C^1$ metric, so the coefficients in front of the leading terms in the PDE are $C^1$
Hm so I guess I was thinking of $f$ and $g$ as (one-dimensional) vectors, which means when $\Bbb R$ transforms by $\varphi$ they don't become $f\circ\varphi$
@Slereah Nah. Take $M = T^2 \times \Bbb R$ with $T^2 \times 0$ foliated by lines and $T^2 \times 1$ foliated using the Reeb foliation. Pretty sure there's no global foliation on $M$ which restricts to those foliations.
I.e., "Reeb foliation and the foliation by lines are not concordant"
And the simplest proof is just doing the old Cauchy-Schwarz idea: $$0\leq \text{tr}[(U_1-U_2 R)^T (U_1-U_2 R)]=\text{tr}(U_1^T U_1)+\text{tr}(U_2 R)^T U_2 R)-2 \text{tr}(U_1^T U_2 R)$$
$\sum_{k=1}^{\infty} [u'_k(t) + \alpha u_k(t)]J_0(\beta_k r) = C$, with $J_0$ being the Bessel function of first kind, $\beta_k$ being its zeros and $C$ some constant. Any chance of solving this?
@BalarkaSen So I saw online that if it's a vector field on $\Bbb R^n$, then $[X,Y]=(DX)Y-(DY)X$, where $DX$ and $DY$ are the Jacobians ($D$ here is used like in Ted's textbook)
(like the matrix made of all the partial derivatives)
What happens if we try to define something as $(DX)Y$? Does this no longer commute with pushforwards/pullbacks, like the Lie bracket has to?
@TedShifrin bonsoir, s'il vous plait comment trouver la dérivée du déterminant? avez vous un pdf ou il est expliqué comment montrer que W'(t)=tr(A) W(t) ?
Non, @Vrouvrou, je n'en connais pas un. Mais si on suppose les $X_i$ indépendants, on écrit $AX_i = \sum b_{ij} X_j$ et on doit employer le fait que det est multilinéaire pour prendre la dérivée.
Abcd: to be fair, the reason that the question was defaced is probably that the person got what they wanted: you solved their homework for them math.stackexchange.com/a/2671020/630
I suspect that they just want to hide it from the professor now
In my case, it's because the flags were piling up, and a different mod was stepping down, so we elected two new mods.
But with becoming a mod on one site, you also become a mod across chat, so flagging things in chat means I'll end up seeing it, even though I can't help with the vandalism in this case
@Ted okay so this is something I said to Balarka with no warranty and I just want to run it by to see if it makes sense since it just came to mind at 2AM in bed
We proved today in class that the Laplacian has a one-sided inverse (surjective), and our prof said that we'd need some regularity theory to get it fully
In my flagging summary under the statistics, there is 1 flag listed as "disputed".
This is the first time I've ever seen a disputed flag in my summary, so I'm just curious, what is a disputed flag? How would you even dispute a flag? Is this the same as voting a flag as invalid?
Return to FAQ in...
So would the idea be that if $\Delta u = \Delta v$, then $u-v$ is harmonic, but we're working in a space of compactly supported functions, so it has to be 0 by analyticity.
Demonark: I guess I don't know the setting here. It's easy to prove using Green's formulas (based on Green/Stokes) that boundary values uniquely determine a harmonic function.
Yeah that's what i was using to show injectivity. But hmm, do you know of a good source for the proof that harmonic functions are analytic, among other things?
So the real identity is, if $[f,g]:=fg'-gf'$, then:$$\left[\frac{f\circ\varphi}{\varphi'}, \frac{g\circ\varphi}{\varphi'}\right]= \frac{[f,g]\circ\varphi}{\varphi'}$$where $f,g,\varphi:\Bbb R\to\Bbb R$ @BalarkaSen
@Akiva 1) It's a folly to think about vector fields in $\Bbb R^n$. A vector field on a manifold $M$ is in general not a smooth function to $\Bbb R^n$; only if $M$ is parallelizable (i.e., $TM \cong M \times \Bbb R^n$) 2) $D$ there means the affine connection.
That's the symmetry condition I told you, $\nabla_X Y - \nabla_Y X = [X, Y]$.
This is the physicists's way of thinking about global objects on manifolds. Locally write a formula, see how it changes under coordinate transformations.
@BalarkaSen To find it at a point, diffeomorph it so that the open set around that point looks like $\Bbb R^n$ and so that $X$ looks constant. Then take the directional derivative of $Y$ with respect to $X$.
That's essentially the same idea as the flowy liney thing you were telling me about before.
Take a manifold $M \cong \Sigma \times \mathbb{R}$, in which there are two submanifolds (of the same dimensions), $S_1$ and $S_2$, such that $S_i \cong \sigma_i \times \mathbb R$. Each of these submanifolds has a foliation $\sigma_{i}(t)$. Does there exist a foliation of $M$, $\Sigma(t)$, such th...
@AkivaWeinberger Let $\gamma^X_p$ denote the integral curve of $X$ starting at a point $p$. Draw the square $\gamma^X_p[0, t] \cup \gamma^Y_{\gamma^X_p(t)}[0, t] \cup \gamma^X_{\gamma^Y_{\gamma^X_p(t)}(t)}[0, t] \cup \gamma^Y_{\gamma^X_{\gamma^Y_{\gamma^X_p(t)}(t)}(t)}[0, t]$, is what I meant, where $X$ and $Y$ are scaled to be unit vector fields, say.
@AkivaWeinberger It won't be a full square in general as in the picture you gave me, right? So like say you start with $p$ then doing this ends you up at $q$
This means the flowline of $[X, Y]$ goes from $p$ to $q$ in time $t$
Graph $x^2$ and $-x^2$ side by side.
