Ruled surface of constant K gauss curvature

Question

Ruled surfaces have negative /zero K ; so what are some examples, with parametrization, of a ruled surface with constant negative K ?

EDIT1:

For standard ruled surface we need to integrate linked Reference Equn (14.11):

$$ K= \dfrac{-M ^2}{EG-F^2}=-1$$

K of Ruled surface

Sorry, I misread your question. Up to local isometry all surfaces of constant negative curvature are pseudospheres (which are not ruled). This is by Hilbert's Theorem. — ArtW, Jun 16 '16 at 09:43
Is there any reason to believe ruled surfaces of constant negative curvature exist in $\mathbf{E}^{3}$? (I haven't checked carefully, but would be surprised if they do.) — Andrew D. Hwang, Jun 17 '16 at 11:39
Can it be proved that ruled surfaces cannot have constant $K<0$ ? — Narasimham, Jun 17 '16 at 16:26

score 1 · Accepted Answer · answered Jun 17 '16 at 19:41

1

Every ruled surface may be parametrized in the form $$ x(s, t) = c(t) + su(t), $$ with $u$ a unit-vector-valued function and $c$ a unit-speed curve with $c' \cdot u = 0$. (That is, parametrize the rulings at unit speed, and take $c$ to be a unit-speed curve orthogonal to the rulings.) A short calculation (e.g., B. O'Neill, Elementary Differential Geometry, revised second edition, Exercise 5.4.12, p. 233, compare wikipedia) shows the Gaussian curvature is a function of $t$ divided by a function of $s$ and $t$.

Particularly, the Gaussian curvature is not constant (depends on $s$) unless the numerator vanishes identically. That is, there exists no ruled surface of constant negative Gaussian curvature in three-dimensional Euclidean space, even locally.

answered Jun 17 '16 at 19:41

Andrew D. Hwang

78,195

Can we not solve for standard ruled parametrization, the above edit equation of the asymptote /ruling? – Narasimham Jun 18 '16 at 12:15
As it turns out, you can't: In the standard parametrization,$$-\frac{M^{2}}{EG - F^{2}} = -\frac{\bigl[c'(t) \cdot \bigl(u(t) \times u'(t)\bigr)\bigr]^{2}}{|c'(t) \times u(t) + s u'(t) \times u(t)|^{4}}.$$Consequently, if the numerator is non-zero, then the fraction (i.e., the Gaussian curvature) depends on $s$, and so is non-constant. – Andrew D. Hwang Jun 18 '16 at 12:26
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Thanks, it is not possible with tangential developable, may be possible with binomial developable, shall ask a separate question. – Narasimham Jun 18 '16 at 16:51

score 0 · Answer 2 · answered Jun 16 '16 at 09:51

Since every complete, simply connected surface of constant negative curvature is diffeomorphic to the plane, and in fact isometric to the hyperbolic plane, your first example of such as thing is two-dimensional hyperbolic space. There are lots of different models and associated charts (i.e. parametrizations) readily available online.

If you drop the simply connected condition, then the ruled surface must be a quotient of hyperbolic space by a subgroup of the isometry group. If you want the rulings to be of infinite length, then this leaves you with the pseudosphere, which is homeomorphic to a cylinder. Otherwise you can look at other quotients of hyperbolic space.

As far as incomplete surfaces go, I'm not so sure.

Ruled surface of constant K gauss curvature

2 Answers2

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