@TedShifrin I consider the more quantitative questions in symplectic geometry pretty geometric. For instance when does there exist a symplectic embedding of $\Pi (B^2(\alpha_i)) \to \Pi (B^2(\beta_i))$, or what are the gromov-witten invariants of this specific projective algebraic variety.
@PVAL: I agree that I am being somewhat arbitrary in my notion of "geometry," and I don't mean it to be entirely serious. I just get annoyed that so many people call Warner a course in differential geometry :P
@Balarka: Cool. It's not really that hard, but, as I said, this is the baby case of some interesting mathematics. These are called isoparametric hypersurfaces, and there's a nontrivial theorem about how many can actually be constant in $n$ dimensions. (So we know that with $n=2$, $0$, $1$, or $2$ will work.)
@Ted For me the biggest distinctions I see regularly are between qualitative and quantitative and the one between constructive and obstructive. Most of the ideas I have had (that worked out or didn't) were entirely on the constructive end.
@TedShifrin Well I know what the Gauss map is and I know what a cuspidal singularity is. I think I spent 5 minutes trying to figure out geometrically what that meant and couldn't.
@TedShifrin: I didn't actually prove the bit about curvature, but the planarity more like (but it should be analogous). It came down to $\mathbf x_{uu} = \Gamma_{uu}^u \mathbf x_u + k_1 \mathbf{n}$, aka $\nabla_{\mathbf{x}_u} \mathbf{x}_u$ is a multiple of $\mathbf{x}_u$ :)
@PVAL Well, part of our theorem was using Chern classes cleverly to count them on a generic surface in $\Bbb P^3$. But I can tell you easily what the really geometric thing is. A cusp on the Gauss map is a critical point of its restriction to the parabolic locus (where the Gauss map drops rank to 1). The awesome thing we proved is that this is the set of points where the parabolic curve intersects the curve of inflection points on asymptotic curves. And then we gave a count in terms of degree.
In classical surface theory a point is parabolic if Gaussian curvature is 0 (and the point isn't a planar point). This is where the Gauss map has rank 1.
Anyhow, @Balarka, if you're fed up, on to parallel translation/transport and geodesics ...
@BalarkaSen Not so bad and actually somewhat geometric. Although very parametrization-dependent. Hence my preference for moving frames ;) ... But the Gauss and Codazzi equations are absurd.
I'm pretty sure. It should still be closed since the projection is just the identity locally so it shouldn't ruin closedness. And non-degeneracy shouldn't be a problem neither...
@TedShifrin Ah so generically the subset of parabolic points in a surface should be some codimension one thing, and generically the asymptotic curves will also be generically some one dimension thing. So you counted the cusps of the Gauss map by doing intersection theory with these curves?
Close, @PVAL. In the complex world, away from the parabolic curve, there are two asymptotic curves through each point. You follow asymptotic curves until you get an inflection point. You mark all those inflection points. They themselves form a new curve (which we call the asymptotic flex curve). Where it meets the parabolic curve is the cusps of Gauss. :)
I would be inclined to remark that your calculation shows that $\mathbf N = \pm \mathbf n$ and then think about the meaning of a line of curvature and the meaning of torsion. :P
@Balarka: I don't think what you're saying is germane. Curvature of the curve is inherent to the curve, not to studying what happens as you move the curve orthogonal to itself.
One of the (relatively famous) Indian - in fact Bengali - film directors of the 20th century apparently were influenced by the French new wave, which got me a bit interested.
@Danu Degeneration and disintegration of culture, in my humble opinion, from whatever I have seen of this world. People don't think anymore. They just "go with the flow". The literature nowadays is mostly crap too.
India has become a strange, strange country.
I don't know much about the rest of India, but during the latter 100 years of the 200-year long British-rule, Bengal (from where I am) produced great literature, and great films. It retains none of that anymore, as far as I am concerned.
@MikeMiller This type of movies (biographies of famous scientists/mathematicians) typically get annoying for this type of reason, in my experience. The trailer of this particular movie seemed to suffer from this, extremely badly. Ted's example of A Beautiful Mind is a good example.
@TedShifrin: Actually, I was rereading our previous conjecture. You want to show that if $k_1$ is constant, the curvature of a u-curve is $k_1$ (absolute value of that, whatever). I stand correct upon my earlier statement that I am skeptic you can do this without looking at different $u$-curves nearby. The $u$-curve on a pseudosphere has different curvature than $k_1$ at one of it's points (because the curve normal and surface normal doesn't agree).
It should not, indeed, does not suffice that $k_1$ is constant along the u-curve. It has to be locally constant on a tubular neighborhood of the $u$-curve.
@Balarka: I don't think what you're saying is germane. Curvature of the curve is inherent to the curve, not to studying what happens as you move the curve orthogonal to itself.
(Funny typo, I wanted to say conversation, not conjecture)
@Balarka: I think $u$ and $v$ reverse on surfaces of revolution. The $v$-curves the way I parametrized are the circles. ... But anyhow, I'm still disagreeing with you.
Interesting, how so? My equation was $\mathbf x_{uu} = \Gamma_{uu}^u \mathbf{x}_u + k_1 \mathbf n$, and since $\mathbf x_{uu} = k \mathbf N$, dotting with $\mathbf n$ just says that the normal curvature is $k_1$ which we already knew :(
Suppose my curve is $\mathbf x(u, v_0)$ (suppose the domain of $\mathbf x$ is $[0, 1]^2$ for clarity's sake). I arclength parametrize by precomposing that with a diffeomorphism $f : I \to I$, so it's $\mathbf x(f(u), v_0)$. $\mathbf x(f(u), v)$ is still a fine parametrization of the whole surface, which is an arclength parametrization for the $u$-curve passing through $v_0$, no?
@TedShifrin I have quick question we have $Aut(Z_5) = Z_5^{*}$ right ? So, if $\alpha$ is generator of $Z_5$ why does the generator $\alpha(h) = h^2$ for all $h \in Z_5$ ?
Exercise 19 (classify surface of revolutions of constant curvature 1, 0 or -1) is funny. I guess there are many C^1-immersed such surfaces (so curvature doesn't classify them).
@arctictern I am trying to classify all groups of order 20 and give presentation for them. So I proved that $G = H \rtimes K$ where $H$ is a sylow 5 subgroups(normal), and K is a sylow 4 subgroup
