Hello,


I am interested in proofs for why the only irreducible doubly ruled surfaces in ${\mathbb R}^3$ are the one sheeted hyperboloid and the hyperbolic paraboloid. While many books and papers state that this is "well known", I could hardly find any sources that give more details. I only found the following two:


1. In the book "Mathematical Omnibus: Thirty Lectures on Classic Mathematics" by Fuchs and Tabachnikov there is a proof relying on rather unusual tools. The proof heavily relies on the property that the neighborhood of every (non-singular) point behaves similarly to a plane.
2. Various places state that we can take three lines from one generating family, and these should intersect every line of the second family. I am not sure how to prove such a claim, and couldn't find a reference with more details (it does seem much simpler in the complex projective space, where one could rely on plucker coordinates).


Could anyone provide references to proofs of this property? Or describe a proof different from the one I mentioned in item 1?


Many thanks!
Adam 



Proofs for doubly ruled surfaces
The classical proof via differential geometry goes like this: 


Suppose that the surface in $\mathbb{R}^3$ is smooth and parametrize it locally in the form $X(s,t)$ where the two rulings are defined by holding either $s$ or $t$ constant. This is a local argument, so, for simplicity, I'll assume that the domain of $X$ is a rectangle in the $st$-plane. Of course, one can reparametrize in $s$ and/or $t$ separately, and this will turn out to be useful at some point in the calculation. 


The two tangent vector fields $X\_s$ and $X\_t$ are linearly independent and are the tangents to the two rulings. Since $X\_{ss}$ is the acceleration of the $t$-ruling, it follows that $X\_{ss} = f X\_s$ for some function $f$. Similarly $X\_{tt} = g X\_t$ for some function $g$.
Note that, if one reparametrized, using $\bar s$ and $\bar t$ instead of $s$ and $t$, then one would have
$$
X\_{\bar s} = \frac{ds}{d\bar s}\ X\_s\quad\text{and}\quad
X\_{\bar s\bar s} = \left(f + \frac{d^2s}{d\bar s^2}\left(\frac{ds}{d\bar s}\right)^{-1}\right)\ X\_{\bar s},
$$
with similar formulae for $X\_{\bar t}$ and $X\_{\bar t\bar t}$. This will be useful below.


Since the surface does not lie in a plane, $X\_{st}$ cannot be a linear combination of $X\_s$ and $X\_t$ (otherwise, the plane spanned by $X\_s$ and $X\_t$ would be fixed, and the surface would lie in plane). This means that $X\_s$, $X\_t$, $X\_{st}$ is a basis of $\mathbb{R}^3$, and, as such, there are equations of the form
$$
\begin{pmatrix} dX\_s& dX\_t & dX\_{st}
\end{pmatrix}
= \begin{pmatrix} X\_s& X\_t & X\_{st}
\end{pmatrix}
\begin{pmatrix} f\ ds& 0 & f\_t\ ds\\\\
0 & g\ dt & g\_s\ dt\\\\
dt & ds & f\ ds + g\ dt
\end{pmatrix}
$$
(The equations for $dX\_{st}$ follow since $(X\_{st})\_s = (f X\_s)\_t = f\_t\ X\_s + f\ X\_{st}$, etc.) By comparing partials, or by using the structure equations above (i.e., expanding out the consequences of $d(d(X\_{st}))=0$, etc.), one sees that $d(f\ ds + g\ dt) = 0$, so that there must exist a function $h$ such that $f = 2h\_s$ and $g = 2h\_t$. (The coefficient of $2$ avoids some fractions later.)
The equation now becomes
$$
\begin{pmatrix} dX\_s& dX\_t & dX\_{st}
\end{pmatrix}
= \begin{pmatrix} X\_s& X\_t & X\_{st}
\end{pmatrix}
\begin{pmatrix} 2h\_s\ ds& 0 & 2h\_{st}\ ds\\\\
0 & 2h\_t\ dt & 2h\_{st}\ dt\\\\
dt & ds & 2h\_s\ ds + 2h\_t\ dt
\end{pmatrix}
$$
Moreover, the structure equations now imply that $d(e^{-2h}h\_{st})=0$, so $h\_{st} = C\ e^{2h}$ for some constant $C$. By adding a constant to $h$, one can reduce to the case that $C$ is one of $0$, $1$, or $-1$.


Consider the case $C=0$ (which needs to be treated separately in any case). Then $h\_{st}=0$, so that, in particular, $f=2h\_s$ is a function of $s$ alone and $g=2h\_t$ is a function of $t$ alone. Using the change of variables formulae mentioned above, one can then change variables in $s$ so as to arrange that $f = 0$ and change variables in $t$ to arrange that $g=0$. Thus, the equations have reduced to
$$
d(X\_s) = X\_{st}\ dt,\qquad d(X\_t) = X\_{st}\ ds,\qquad d(X\_{st})=0.
$$
Thus $X\_{st} = v\_3$ where $v\_3$ is a constant vector. Then $d(X\_s-tv\_3) = 0$ and $d(X\_t - s v\_3) = 0$, so there exist constant vectors $v\_1$ and $v\_2$ so that $X\_s = v\_1 + t v\_3$ and so that $X\_t = v\_2 + s v\_3$. Finally, this implies that 
$$
dX = X\_s\ ds + X\_t\ dt = (v\_1+tv\_3)\ ds + (v\_2 + s v\_3)\ dt 
= d\bigl(sv\_1+t v\_2 + st v\_3 \bigr), 
$$
so that
$$
X = v\_0 + sv\_1+t v\_2 + st v\_3
$$
for some constant vector $v\_0$. Thus, $X(s,t)$ parametrizes a hyperbolic paraboloid.


Now, consider the case $C\not=0$. The structure equations have become


$$
\begin{pmatrix} dX\_s& dX\_t & dX\_{st}
\end{pmatrix}
= \begin{pmatrix} X\_s& X\_t & X\_{st}
\end{pmatrix}
\begin{pmatrix} 2h\_s\ ds& 0 & 2C\ e^{2h}\ ds\\\\
0 & 2h\_t\ dt & 2C\ e^{2h}\ dt\\\\
dt & ds & 2\ dh
\end{pmatrix}
$$
where $h\_{st} = C\ e^{2h}$. Moreover, one calculates
$$
d(e^{-2h} X\_{st}) = 2C\ (X\_s\ ds + X\_t\ dt) = 2C\ dX,
$$
so $e^{-2h}X\_{st} = 2C\ X + v\_0$ for some constant vector $v\_0$. In particular, showing that the vector-valued function $E\_3 = e^{-2h}X\_{st}$
takes values in a hyperboloid of $1$-sheet will finish the proof. 


To this end, consider the new frame field
$$
\begin{pmatrix}E\_1 & E\_2 & E\_3\end{pmatrix}
= \begin{pmatrix}e^{-h}X\_{s} & e^{-h}X\_{t} & e^{-2h}X\_{st}\end{pmatrix}.
$$
Calculation shows that it satisfies the structure equation
$$
\begin{pmatrix} dE\_1& dE\_2 & dE\_3
\end{pmatrix}
= \begin{pmatrix} E\_1& E\_2 & E\_3
\end{pmatrix}
\begin{pmatrix} h\_s\ ds-h\_t\ dt& 0 & 2C\ e^h\ ds\\\\
0 & h\_t\ dt - h\_s\ ds & 2C\ e^h\ dt\\\\
e^h\ dt & e^h\ ds & 0
\end{pmatrix}.
$$
Note that the $3$-by-$3$ matrix of $1$-forms on the right takes values in the vector space
$\frak{g}$ consisting of matrices of the form
$$
\begin{pmatrix} x\_1& 0 & 2C\ x\_2\\\\
0 & -x\_1 & 2C\ x\_3\\\\
x\_3 & x\_2 & 0
\end{pmatrix}.
$$ 
This is, of course, the Lie algebra of the subgroup $O(Q)\subset GL(3,\mathbb{R})$ consisting of the matrices that satisfy $A^TQA = Q$, where
$$
Q = \begin{pmatrix} 0& 1 & 0\\\\
1 & 0 & 0\\\\
0 & 0 & -2C
\end{pmatrix}.
$$
It follows that there is an invertible linear transformation $L$ of $\mathbb{R}^3$ such that the matrix $LE$ takes values in $O(Q)$, where $E = (E\_1\ E\_2\ E\_3)$. In particular $L$ carries the image of $E\_3$ into the hyperboloid of $1$-sheet $2x\_1x\_2 - 2C x\_3^2 = -2C$. By the remark above, it follows that $X(s,t)$ must be the image of this quadric under an affine transformation of $\mathbb{R}^3$, as was to be shown. 


