Discovery is the privilege of the child: the child who has no fear of being once again wrong, of looking like an idiot, of not being serious, of not doing things like everyone else.
We're making $C_1$ the reals from $\epsilon$ to $N$, $C_2$ a big semicircle from $N$ to $-N$, $C_3$ the reals from $-N$ to $-\epsilon$, and $C_4$ the semicircle from $-\epsilon$ to $\epsilon$. And then we divide it by $1+e^{\pi i/3}$ @TedShifrin
I got chewed out for not being nice to a gentle flower on main yesterday (despite hours of having helped over the months) and now this room is nuts. I think I need a long sabbatical.
@TedShifrin idk of any books that focus on dealing with a general manifold as ambient but I'm sure it's talked about somewhere in the standard books I haven't read thoroughly enough
@TedShifrin Ah gotcha. Those would be nice too. I think I will have to go through some of the harder computations of your diffgeom text and multcalc text soon.
Oh, right, it's diffgeo in English (we shorten it to geodiff in Italian because adjectives and nouns are swapped)
So we defined today the area of a subset of a surface in $\Bbb R^3$ via the first fundamental form (it was done in a hurry tbh). How do I see that this agrees with the Hausdorff $2 dimensional measure?
So that's the real reason for the formula you were given using the first fundamental form, @Alessandro. Now the Hausdorff measure question is tangential to that, really.
As a differential $k$-form, it assigns to a basis for the tangent space at $x$ the induced $k$-dimensional volume of the $k$-dimensional parallelepiped spanned.
"given a singular foliation without isochore singularities on a surface, we say that a focus $x_0$ and a saddle $x_1$ are in simple elimination position if when one positively orients the foliation near $x_1$, one and only one stable separatrix connects to $x_0$"
Well, do it for a subset of a general plane and see why it agrees with the formula you have. That should convince you, since locally a surface is well-approximated by a piece of its tangent plane.
ohhh, you're talking about a zero of a vector field @anakhronizein.
Because the integral isn't over the whole real line, so we reasoned that the integrand is ALMOST even, in that for the left half plane it's $e^{i \pi /6}$ times the integral over the righht half plane
Sure it's pretty and when you sit down and work through it your like yeah wow elegance wow but when you like need to do concrete things with it you just go back to multi at the end of the day
So say I wanna prove that the cone, which I'll call $C$ not a smooth manifold (as defined in G&P). I could prove part of it the following way: Assume that a nbhd of the vertex $v \in C$ is diffeomorphic to a nbhd $U^*$ in $\mathbb{R}^n$ for some $n > 1$. Let $\phi$ denote the diffeomorphism. Observe that $U \setminus \{v\}$ is not path connected whereas $U^* \setminus \{\phi(V)\}$ is path connected a contradiction.
Also what about the pushforward measures of the Lebesgue measure through the parametrizations? Those are just a mess I think, they depend on the parametrization and I'm not even sure they can be glued together on the whole surface
@TedShifrin yeah I thought so. Are there "area-preserving" parametrizations in the sense that the push-forward measure agrees with the one coming from the ambient space?
@Perturbative: No, that isn't. Then you're right that it's just a topological argument. Pulling out the vertex disconnects it, whereas that doesn't happen in a disk (dimension > 1, of course).
Can anyone provide complex analysis video lectures link, other that nptel once? (I am asking this, because i just got to know that this awesome lecture series on multivariable calculus exists, but neither youtube nor google was recommending that).
@TedShifrin I was in dark! Just curious, do you take rigorous approach in your lectures or 'engineering' approach as taken in MIT OCW multivariable calculus?
@AlessandroCodenotti The key is to use local graph parametrizations. Then the graph measure agrees with the Hausdorff measure by the area formula, and the graph measure agrees with the surface measure more or less by definition, or some basic Riemannian geometry if you define it intrinsically.
@Akiva @GFauxPas: So your sign was originally correct. You pick up the $e^{\pi i/3}$ from the branch of cube root, but there is no minus sign. You go backwards, but you change the limits, so it's $1+e^{\pi i/3}$, as you originally said. The residue computation then gives $$\pi (2\cos \pi/6)/|1+e^{\pi i/3}|^2.$$
For instance, somehow these polynomials in the form of x^2+/-y^2=c, aka conic sections, are all somehow parameterized curiously by exponential functions, cosh and sinh instead of f(x,y) = t, P(t)
Conics are very special, @JohnJoe. Ellipses are all done with $\sin$ and $\cos$ and hyperbolas are done with $\sinh$ and $\cosh$. You can also parametrize conics by rational functions. But other algebraic curves, in general, can't be parametrized explicitly at all.
I never claim to know everything after 40+ years as a published mathematician, but somehow I end up being called rude and destructive to all the little flowers.
I mean, is there a philosophical reason to prefer one coordinate system than another, if they're compatible in the sense of differentiable manifolds for example?
It's interesting, though. All the conics are projectively equivalent, and you can put the ellipses and hyperbolas in a family using exponentials, but there's a discontinuity as you pass through parabolas/lines, and only the rational parametrization of conics seems to extend reasonably across those parabolas/lines, I guess.
a lecturer of mine was telling me that his lecturer in his first year would write $(x) f \circ g$ because he felt that the convention of writing $g \circ f (x)$ but first applying $f$ and then $g$ was too confusing for his students
Does anybody have some link for good video lessons of a complete course in Complex Analysis?
Grateful.
