Or possibly the issue is how to conclude from here? There is a standard argument one uses now, but of course "standard" depends on the reader and where they come from. :)
I have here an (exemplary) definition for chain $C_i$ and co-chain $C^i$ with a finite simplicial complex $X$, and have difficulties to see how both can be of a "similar kind", isnt a set of functions something quite different from a vector space?
@MikeMiller Ok, I guess I'll start pinging:) So for me an Hamiltonian is just a map h: M \to R that we can use to define an hamiltonian v.f using the symplectic form. Is the map into the intersection of the two tangents defined as h \mapsto X_h where X_h is the vector field obtained using dh and the symplectic form?
@Riccardo Yes, I am thinking of the associated Hamiltonian vector field. Good point.
The end is what I call the "standard argument". If shows up all the time when studying transversality, and is worth internalizing. I am on my phone so it will take me some time to write down.
I'm going to use TeX to make it more readable - see TeX in chat to the right
Suppose we have chosen a finite dimensional vector space H of Hamiltonian vector fields. Consider the map $F: H x L_1 -> M$, given by $F(V, x) = $ time-1 flow of $x$ along $V$.
F stands for Flow.
The assumption on H about its image in each intersection tells us that $F$ is transverse to $L_2$ along $\{0\} \times L_1$.
By compactness of $L_1$, there is a neighborhood $U \subset H$ so that $F: U \times L_1 \to M$ is transverse to $L_2$. (First you get a neighborhood inside $H \times L_1$ then apply the 'tube lemma' to see existence of $U$.)
There is a projection map $I_U \to U$. I will call the fiber above a vector field $V$ "$I_V$".
1) $I_V$ is the set of intersection points of $F_V L_1$ with $L_2$. (The first term means we have flowed $L_1$ along $V$.)
2) Exercise, but a very important one: $F_V L_1$ intersects $L_2$ transversely iff $V$ is a regular value of the projection $I_U \to U$.
(The proof should amount to symbol-pushing)
So consider that projection. By Sard's theorem, critical values are measure $0$ in $U$, and therefore regular values in $U$ exist - in fact arbitrarily small regular values.
Let's say I have a smooth manifold $M$, with a chart $(U, (x^i))$ on $M$. In this chart can I think of $\frac{\partial}{\partial x^i}$ as a function from $U \to TM$ with $$\frac{\partial}{\partial x^i} (p) = \frac{\partial}{\partial x^i} \bigg|_{p}$$
Ohh it turns out these are actually the coordinate vector fields
Let $\rho_i, i\in\Bbb N$ be the imaginary parts of the non-trivial zeros of the Riemann $\zeta$ function: $\zeta(\frac{1}{2}\pm\imath \rho_i)=0$, $(\forall i)$. Does anonye know if anything (in case what) is known on the Fourier-Transform of a "zeta-zero-Dirac-comb": $$ \mathcal{F}\left \{ \sum_{i=1}^{\infty} \delta(t - \rho_i ) + \delta(t + \rho_i)\right \}[t] $$
Hello. Let $I\vartriangleleft A$ be an ideal of a commutative ring and let $S\subset A$ be a multiplicative system. It seems $S+I\subset A$ is also a multiplicative system. Is the saturation of $1+I$ given by $A^\times +I$? (For the zero ideal this holds by definition.)
I've noticed some similarities between the story of completing a metric space and taking algebraic closure of a field. My question is whether these two stories can be generalized.
Metric space
Fix a metric space $(X, d_X)$. Consider isometries from it to other metric spaces (i.e. the under cate...
@LeakyNun I am thinking about the case of $R[x_1,\dots ,x_n]$ and $I= \left\langle x_1,\dots ,x_n \right\rangle $. Sorry if I'm blind but I think the saturation of $1+I$ is $A^\times +I$, no? (Divisors of polynomials whose value at zero is one are the polynomials whose value at zero is invertible.)
@LeakyNun I don't know. I can express the saturation in terms of primes (since a prime intersects $1+I$ iff it and $I$ are not comaximal). So the saturation would be the complement of the union of prime ideals of $A$ not comaximal with $I$. This is probably not what you're trying to tell me
Then $tu\in 1+I$ says $t$ is invertible in the quotient ring. For this to imply invertibility in $A$ itself it's enough that $I$ be contained in the Jacobson radical.
@LeakyNun @Arrow easy generalization: the saturation of $S+I$ is the preimage of the saturation of the image of $S$ in $R/I$ for $S \subset (R,\cdot)$ a submonoid and $I$ an ideal
The MIT course was indubitably 18.03 Differential Equations. The Laplace transform is used (as is the Fourier transform) to turn differential equations into polynomial equations.
@TedShifrin what's a good intuition for Frobenius's theorem (integrable iff involutive)? I don't really have a good feel for it other than the proof. So basically the only thing I have is that involutive --> we can make a "foliated" chart --> integrable. But the first arrow is kind of a "surprise" to me.
It's a surprise in that I don't see any reason other than the proof to say this is possible.
The intuition comes, I suppose, from the converse. If you have a foliation by submanifolds, then the distribution given by the tangent bundle is of course involutive (brackets of tangent vector fields are again vector fields).
I like thinking in terms of differential forms, of course, so understanding when $\omega=0$ has an integrating factor is a good starting place (for a $1$-form $\omega$).
@TedShifrin I am not quite sure what an integrating factor would be with respect to $\omega=0$? Is it a smooth $f$ such that $f\omega$... I am not sure.
@TedShifrin what do you mean by the intuition coming from the converse? Why might it give intuition for why involutive --> integrable? Purely because you have this link between the two in the other direction?
@anakhro: I mean, it's not unusual that you get intuition for a $P\iff Q$ by understanding one direction and seeing why the reverse direction follows. Not always, but sometimes.
