Demonark: So you do need the metric, of course. Choose an oriented orthonormal basis for the tangent space and then dual $1$-forms. Call those $\omega_1,\dots,\omega_n$. You OK so far?
@theDoctor Use the Taylor expansion of $e^{i\theta}$ and split into real and complex part. This is a super basic calculus proof you can find in any good textbook
OK, Demonark, we define $\star$ by sending a $k$-form $\omega_I=\omega_{i_1}\wedge\dots\wedge\omega_{i_k}$ to the complementary $(n-k)$-form $\omega_J$, with the order chosen so that $\omega_I\wedge\star\omega_I = \omega_1\wedge\dots\wedge\omega_n$.
So, for example, in $\Bbb R^2$, $\star dx = dy$ and $\star dy = -dx$. (This has things to do with complex, too.)
The derivation I'm fond of is to note that $\frac{d^2}{dt^2}e^{it}=-e^{i t}$, which is the harmonic oscillator ODE with general solution $A\cos t+B\sin t$
and then check initial conditions to get $A=1,B=i$.
Working on diff top homework; do I have the right idea for this argument? Let $f: M \to N$ be smooth and $S \subset N$ an embedded submanifold and $f \pitchfork S$. I want to show that $T_p(f^{-1}(S)) = (df)^{-1}(T_{f(p)}(S))$. Let $\phi$ be a slice chart for $S$ and $\pi$ the projection that sends the image of $S$ in the slice chart to 0. Then, $\pi \circ \phi \circ f$ maps paths in $f^{-1}(S)$ to the zero path. So it maps every vector $T_p(f^{-1}(S))$ to the 0 vector.
So the kernel of the differential of this map has dimension at least $m - (n - s)$, the dimension of $f^{-1}(S)$. But since $f \pitchfork S$, $d\phi$ an isomorphism at $f(p)$ and $\pi$ rank $n-s$, the image of the same differential must be at least dimension $n-s$. But the sum of the two must be $m$. So, the dimension of the image is exactly $n-s$ and the kernel exactly $m-(n-s)$.
@Ted sorry I was out for a second, but yeah I did verify it on a chalkboard and it works, because $d\star df = (\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2})dx\wedge dy$
Okay, and then what info does the Laplacian give? In the plane I know you've got harmonic functions linking in to holomorphic stuff, but... like what info does it carry in general?
That sounds pretty nifty. So wait is this thing the Hodge star? I remember my physics TA some time back was telling us about forms and mentioned this Hodge star but at the time I had no idea what he was going on about
I know when I had cancer, when it got really bad, my students started to flip out, and then I missed the last few weeks of class. They were more upset than I was.
@Ted oh cool, I was doing an exam and there was a question about finite simple groups and their representations, so immediately i thought up a way to turn it into a question about space forms
@TedShifrin Oh you're right my argument isn't totally air tight as stated. That every matrix in my composition is at least rank $n-s$ doesn't guarantee that the product is rank $n-s$. I need to be more careful and say that $d\pi \vert_{T_{f(p)}(S)} = 0$ AND $d\pi \vert_{T_{f(p)}(S)^{\complement}} = \text{id}_{T_{f(p)}(S)^{\complement}}$.
is it true that each open set in $\Bbb R \ ^ 2$ is the disjoint union of rectangles of the form $(a,b) x (c,d)$ (parallel to the axis) ? i know it is the union of such rectangles, but can we force the union to be disjoint?
@Liad that's because the union of overlapping intervals in $\Bbb R$ is an interval while the same doesn't hold for overlapping rectangles in $\Bbb R^2$
start with a wavefunction in position space, sample it, do an FFT on this data to get to momentum space, multiply by $e^{-i tk^2/2}$ to evolve according to a free particle, and then transform back to get the time-evolved wavefunction
being able to do the FT and inverse FT are obviously the key steps, since if these are slow/wrong then the rest is scuppered
@Semiclassical I was sure there's some slick physical way of explaining why in a LR circuit the current lags by a phase difference of $\pi/2$ with the voltage and in a CR circuit it leads by a phase difference of $\pi/2$ with the voltage, but I seem to be forgetting it.
I guess the point is in LR the induction makes the current lag
physically it presumably comes down to the fact that an inductor acts as a choke, i.e. if you try to change the current through it quickly then you get a strong back-EMF
