Proofs for doubly ruled surfaces - MathOverflow

Proofs for doubly ruled surfaces

2012-10-11T16:08:52Z

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:

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.
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

Answer by Dick Palais for Proofs for doubly ruled surfaces

2012-10-11T17:46:47Z

Sorry, I had a bad reference

Answer by Sue for Proofs for doubly ruled surfaces

2012-10-11T19:39:22Z

I don't have it in front of me right now, but I believe that there is an old-fashioned proof in the book

G. Salmon, A Treatise on the Analytic Geometry of Three Dimensions, Vol. 2, 5th edition Hodges, Figgis And Co. Ltd. (1915).

I learned about this from the beautiful paper "On the Erd ̋os distinct distance problem in the plane" by Guth and Katz.

Answer by Robert Bryant for Proofs for doubly ruled surfaces

2012-10-12T21:13:19Z

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.

Answer by Agol for Proofs for doubly ruled surfaces

2012-10-13T15:51:41Z

Here's a sketch of an argument for approach 2, under the mild hypothesis that the lines in the family of doubly ruled lines varies continuously, which I think is intuitively clear, but requires some justification.

Take a point $p$ on a doubly ruled surface $\Sigma$, and take two lines $l_1, l_2$ going through $p$. For any nearby points $p_1 \in l_1, p_2 \in l_2$, there are lines $l_2'$ intersecting $l_1$ in $p_1$, and $l_1'$ intersecting $l_2$ in $p_2$. But since $l_1$ and $l_2$ intersect on the surface $\Sigma$, and since the family varies continuously by hypothesis, $l_1'$ and $l_2'$ must intersect when $p_1,p_2$ are close enough to $p$ by general position on $\Sigma$. Similarly, for any point $q$ near $p$, there are lines $l_1'$ and $l_2'$ going through $q$ meeting $l_2$ and $l_1$ respectively by continuity of the family and general position on $\Sigma$.

Now, take points $p_1, p_1'$ on $l_1$ near $p$, together with lines $l_2'$ and $l_2''$ meeting $l_1$ in $p_1$ and $p_1'$ respectively. Then $l_1'$ must meet both $l_2'$ and $l_2''$ for points $p_2$ near $p$ on $l_2$. Thus, one sees a neighborhood $p\in U\subset \Sigma$ such that any point in $U$ lies on a line meeting these three lines $l_2,l_2',l_2''$.

If any pair of these lines is coplanar (i.e. intersect or are parallel), then the portion of surface $U$ near $p$ must be planar. Otherwise, one has 3 skew lines.

Now, I claim that for 3 skew lines, there is a unique surface of lines meeting all three lines, which is either a hyperbolic paraboloid or hyperboloid. This is proved by Hilbert-Cohn Vossen, but I'll give a sketch of the proof. The uniqueness follows because for any point $p$ in $l_2$, the plane spanned by $p$ and $l_2'$ meets $l_2''$ uniquely in a point $q$, and thus every point on $l_2$ goes through a unique line $\overline{pq}$ meeting $l_2'$ and $l_2''$.

Three skew lines in $\mathbb{R}^3$ give three projective lines in $\mathbb{RP}^3$ which do not intersect. I claim that this configuration is unique up to projective transformation.

Three skew projective lines in $\mathbb{RP}^3$ correspond to three planes $P_1,P_2,P_3$ in $\mathbb{R}^4$ meeting pairwise only in the origin. I claim $GL_4(\mathbb{R})$ acts transitively on such configurations. Take basis vectors for $P_1, P_2$, then these form a basis for $\mathbb{R}^4$. Thus up to linear transformation, we may assume $P_1=\{(1,0,0,0),(0,1,0,0)\}, P_2=\{(0,0,1,0),(0,0,0,1)\}$ . Now, the subspace $P_3$ has a basis of two vectors, such that the first two coordinates are linearly independent in $P_1$, and the second two coordinates are linearly independent in $P_2$ since we assume that the planes intersect only in the origin. We may take linear transformations in $GL_2(\mathbb{R})\times GL_2(\mathbb{R})$ stabilizing $P_1, P_2$ and sending these two vectors to $\{(1,0,1,0),(0,1,0,1)\}$ , and thus we have normalized the three planes, so that the action is transitive. The corresponding action of $PGL_4(\mathbb{R})$ is thus also transitive on skew projective lines.

Take a piece of a hyperbolic paraboloid, and take 3 skew lines lying on it. The three skew lines uniquely determine the paraboloid, since it is the surface of lines meeting the three skew lines. Then there is a projective transformation taking $l_2,l_2',l_2''$ to these three skew lines, and therefore sending a portion of $\Sigma$ near $p$ to a hyperbolic paraboloid by uniqueness.

In fact, when we compactify a hyperbolic paraboloid or a hyperboloid in $\mathbb{RP}^3$, we get a 2-torus with two foliations by projective lines. So up to projective transformation, there is only one such surface.

Now the surface $\Sigma$ has patches which are hyperbolic paraboloids or hyperboloids. But one can see that each such surface lies in a unique doubly ruled torus in $\mathbb{RP}^3$, so $\Sigma$ must be identically such a surface intersected with $\mathbb{R}^3$.