TITLE: "Asymptotic " transformation for principal curvature lines on positive Gauss curvature K surfaces
QUESTION [1 upvotes]: If we have lines of curvature for a  $ \mathbb R^2 $ surface in 3-space $ $(x,y,z)$  as
$$ [ f(u,v), \ g(u,v), \ h(u,v) ] $$ 
for negative Gauss curvature surfaces then  
$$  f(u+v,u-v), g(u+v,u-v) , h(u+v,u-v) $$
represent asymptotic lines whose normal curvature vanishes.
What is characteristic about the corresponding lines for positive K surfaces? What are its properties? I appreciate any remarks.. intuitive, qualitative, or equation-based regarding what curvature vanishes or remains constant, or basically why is it not considered important or interesting.
EDIT1:
How do lines rotate in tangent plane by virtue of such transformation in either case? Secondly are they conceptually linked to (eg Struik's Differential Geometry book) "Imaginaries in surface theory"?

REPLY [0 votes]: If I understand, you're asking:
Let $M$ be a surface in Euclidean $3$-space and $p$ a point of $M$. If the Gaussian curvature at $p$ is negative, then the asymptotic directions at $p$ can be expressed as "linear combinations" of the principal directions. What do the corresponding linear combinations means if the Gaussian curvature is positive at $p$?

Assuming that's correct:
Let $k_{1}$ and $k_{2}$ be the (non-zero) principal curvatures of $M$ at $p$. By a Euclidean motion, we may assume $p$ is the origin, and $T_{p}M$ is the $(x, y)$-plane with the coordinate axes as principal directions. In a sufficiently small neighborhood of $p$, the surface $M$ is a graph
$$
z = \tfrac{1}{2}(k_{1} x^{2} + k_{2}y^{2}) + o(x^{2} + y^{2}).
$$
For purposes of curvatures, we may neglect the higher-order terms and look at the quadratic approximation.
The asymptotic directions at $p$ are the lines of intersection (if any) of $\{z = 0\}$, the tangent plane at $p$, with the graph $z = \tfrac{1}{2}(k_{1} x^{2} + k_{2}y^{2})$. If the Gaussian curvature $k_{1} k_{2}$ is negative, put $k^{2} = -k_{1}/k_{2}$ and write
$$
0 = \tfrac{1}{2}(k_{1} x^{2} + k_{2}y^{2})
  = -\tfrac{1}{2} k_{2} (k^{2}x^{2} - y^{2})
  = -\tfrac{1}{2} k_{2} (kx - y)(kx + y).
$$
That is, the asymptotic directions are the lines $y = \pm kx$.
If the Gaussian curvature is positive, by contrast, the preceding factorization is non-real: Putting $k^{2} = k_{1}/k_{2}$, we have
$$
0 = \tfrac{1}{2}(k_{1} x^{2} + k_{2}y^{2})
  = \tfrac{1}{2} k_{2} (k^{2}x^{2} + y^{2})
  = \tfrac{1}{2} k_{2} (kx - iy)(kx + iy).
$$
Geometrically, the graph of a positive- or negative-definite quadric touches the tangent plane only at the point of tangency, so long as we work over the reals. In that sense, the corresponding directions are non-real, and therefore "uninteresting".
(Incidentally, if $M$ is real-analytic, then $M$ can be viewed locally as the "real points" of a complex surface in $\mathbf{C}^{3}$ by "allowing the power series variables to be complex". In this setting, the preceding factorization has geometric meaning. In retrospect there's no mystery, however, since all quadratic forms are equivalent over $\mathbf{C}$.)
{"set_name": "stack_exchange", "score": 1, "question_id": 1269657}