How can one construct surfaces in $\mathbb R^3$ of constant negative Gaussian curvature containing a line in $\mathbb R^3$? (this question is inspired by this MSE post).

1$\begingroup$ Did you mean a constant $<0 \, K$ surfaces containinfg a straight line? $\endgroup$ – Narasimham Oct 26 '17 at 13:10

$\begingroup$ Yes, that's what "line" means. @Narasimham $\endgroup$ – Mikhail Katz Oct 26 '17 at 13:12

$\begingroup$ The question asked above was not "inspired" by that linked MSE post, but rather is a cross post of the linked question. MSE doesn't like simultaneous cross posting (question has since been deleted on MSE, so not all of you will be able to see anything at the other end of the linked post), but it seems MO may not mind crossposted questions. $\endgroup$ – amWhy Oct 29 '17 at 20:36

1$\begingroup$ @amWhy, you are in error. The question at MSE concerned the complete case, whereas this question concerns the incomplete case. $\endgroup$ – Mikhail Katz Oct 30 '17 at 7:48
Given any point $p$ on a surface $S$ of Gauss curvature 1, there exists an open neighborhood $U\subset S$ and $p$centered coordinates $(x,y):U\to\mathbb{R}^2$, whose image is a domain $R = (x,y)(U)\subset\mathbb{R}^2$ and a function $u:R\to (0,\pi/2)$ such that the first and second fundamental forms of the surface are given by $$ \mathrm{I} = \cos^2\!u\,\mathrm{d}x^2 + \sin^2\!u\,\mathrm{d}y^2\tag1 $$ and $$ \mathrm{I\!I} = \cos u\,\sin u\,(\mathrm{d}x^2\mathrm{d}y^2) = \cos u\,\sin u\,(\mathrm{d}x\mathrm{d}y)(\mathrm{d}x+\mathrm{d}y).\tag2 $$ The function $u$ satisfies the sineGordon equation $$ u_{xx}u_{yy} = \cos u\,\sin u.\tag3 $$ Conversely, given a function $u:R\to (0,\tfrac12\pi)$ that satisfies (3) on a simply connected domain $R\subset\mathbb{R}^2$, there exists an immersion $X:R\to\mathbb{E}^3$ whose first and second fundamental forms are given by $\mathrm{I}$ and $\mathrm{I\!I}$ above; this immersion is unique up to rigid motion and has Gauss curvature $1$.
The asymptotic curves of the immersion $X$ are given by holding $x+y$ or $xy$ constant. The angle between the asymptotic curves (appropriately oriented) is $2u$, and the ambient curvatures in $\mathbb{E}^3$ of the asymptotic curves described by holding $x\pm y$ constant are $$ \kappa_{\pm} = 2 (u_x \mp u_y).\tag 4 $$ (Here, $\kappa_{\pm}$, which are the curvatures of the asymptotic curves, are not to be mistaken for the principal curvatures of the surface $X(R)$, which are $\tan u$ and $\cot u$.)
Thus, constructing a surface in $\mathbb{E}^3$ with Gauss curvature $1$ that contains a straight line (which is necessarily asymptotic) is equivalent to finding a (local) solution $u(x,y)$ of the sineGordon equation that satisfies the 'characteristic' initial condition $u_x(x,x)+u_y(x,x) = 0$. (It's called 'characteristic' because the curves $x\pm y = c$ are the characteristic curves of the PDE (3).)
Unfortunately, characteristic initial value problems for nonlinear PDE can be delicate to study; standard techniques are generally not adequate to prove existence or uniqueness. However, this problem can be partially avoided by looking for special solutions that will satisfy the desired initial condition by virtue of other special properties being assumed.
If one seeks a solution to (3) of the form $u(x,y) = \tfrac12\,f(x^2{}y^2)$, one finds that $f$ must satisfy the (singular, nonlinear) ordinary differential equation $$ 4t\,f''(t) + 4f'(t) = \sin\bigl(f(t)\bigr).\tag5 $$ Moreover, since, for such a solution, one has $(u_x \mp u_y) = f'(x^2{}y^2)(x \pm y)$, it will follow that the two lines $x{\pm}y = 0$ in $R$ will be mapped by $X$ into straight lines in $\mathbb{E}^3$.
To see that there are nontrivial solutions to the singular ODE (5),
let $\theta\in(0,\pi)$ be chosen,
and write $f(t) = \theta + \tfrac14\,\sin(\theta)\,t + g(t)$.
Then (5) becomes the equation
$$
4t\,g''(t)+4g'(t)
= \sin\bigl(\theta+\tfrac14\sin\theta\,t+g(t)\bigr) \sin\theta
= G\bigl(\theta,t,g(t)\bigr)
\tag6
$$
The left hand side of (6) is a singular linear differential operator
at $t=0$ with no resonances, while $G(\theta,t,g)$ is an analytic
function on $\mathbb{R}^3$ satisfying $G(\theta,0,0) = 0$.
Thus, there is a unique convergent power series
solution to (6) satisfying $g(0)=0$,
which implies the convergence of the corresponding formal power series for $f$:
$$
\begin{aligned}
f(t) &= \theta + \frac{\sin\theta}{4}\,t
+ \frac{\cos\theta\sin\theta}{64}\,t^2
+ \frac{(3\cos^2\theta2)\sin\theta}{2304}\,t^3\\
&\quad\quad + \frac{\cos\theta\,(18\cos^2\theta17)\sin\theta}{147456}\,t^4 + \cdots,
\end{aligned}
\tag7
$$
yielding a solution to (5) satisfying $f(0) = \theta$ on an open interval around $t=0$. It is easy to see from (5) that this local solution to (5) satisfying
the initial condition $f(0)=\theta$ extends uniquely to a solution to (5)
defined on the entire real line $\mathbb{R}$. The range of $f$
will not be contained in the interval $(0,\pi)$, of course, but, there
will be an interval $\bigl(a(\theta),b(\theta)\bigr)\subset\mathbb{R}$
containing $0$ that $f$ maps into $(0,\pi)$.
Then the corresponding $X$ will immerse the region between the two hyperbolae $x^2y^2 = a(\theta)<0$ and $x^2y^2= b(\theta)>0$ into $\mathbb{E}^3$ as a surface of constant Gauss curvature $1$ and will map the two lines $x\pm y = 0$ into two straight lines that intersect at an angle of $\theta$ in the image surface.
Added remark: (23 October 2017) After doing a literature search on examples of surfaces of constant curvature $1$ in $\mathbb{E}^3$, I have found out that this surface was discovered by M. H. Amsler, Des surfaces à courbure négative constante dans l'espace à trois dimensions et de leurs singularités, Math. Ann. 130 (1955), 234—256.

2$\begingroup$ Thanks, Robert. This seems like a Math Monthly article. $\endgroup$ – Mikhail Katz Oct 22 '17 at 10:50

$\begingroup$ Can one say anything about the behavior of the surface as one goes to infinity along the line? The representation by the sineGordon equation doesn't seem to shed immediate light on the global geometric behavior of the surface. For example, does the surface tend to "twist" around embedded line and if so can one say anything about the rate of such twisting? $\endgroup$ – Mikhail Katz Oct 22 '17 at 11:35

2$\begingroup$ @MikhailKatz: Yes, what one can read off from the structure equations is that the surface normal spins around each line at unit speed as one moves at unit speed along the line. Thus, the surface twists around the embedded line, with the surface normal coming back to be the normal to the pair of lines every $2\pi$ units. The surface contains both lines, of course, but the two twisting 'ribbons' get progressively thinner as you move away from the cross. $\endgroup$ – Robert Bryant Oct 22 '17 at 13:03

$\begingroup$ Robert Bryant : Before (in the MSE since closed thread) you said the tractricoid does not contain any straight lines and now you mention here asymptotic curved (straight) lines of Chebychev net satisfying Sine Gordon equation as straight lines of hyperbolic geometry with vanishing $\kappa_g$ etc., and that to me sounds contradictory. Is it the difference between $E^3,R^3$ that I miss or is there something bigger I am still missing? $\endgroup$ – Narasimham Oct 29 '17 at 21:41

$\begingroup$ @Narasimham: Perhaps there is some confusion about what is meant by the term 'straight line'. The tractricoid is a specific surface in $\mathbb{E}^3$ that has Gauss curvature $1$. It has geodesics (none of which are straight lines in $\mathbb{E}^3$), as do all sufficiently smooth surfaces in $\mathbb{E}^3$. However, there are surfaces with Gauss curvature $1$ have some geodesics that are straight lines, i.e., geodesics in $\mathbb{E}^3$. Of course, any straight line in $\mathbb{E}^3$ that lies in a smooth surface $S\subset\mathbb{E}^3$ is both a geodesic and an asymptotic curve in $S$. $\endgroup$ – Robert Bryant Oct 29 '17 at 23:28