Proofs for doubly ruled surfaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T16:12:43Z http://mathoverflow.net/feeds/question/109389 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109389/proofs-for-doubly-ruled-surfaces Proofs for doubly ruled surfaces Adam Sheffer 2012-10-11T16:08:52Z 2012-10-13T15:51:41Z <p>Hello,</p> <p>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:</p> <ol> <li><p>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.</p></li> <li><p>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).</p></li> </ol> <p>Could anyone provide references to proofs of this property? Or describe a proof different from the one I mentioned in item 1?</p> <p>Many thanks! Adam </p> http://mathoverflow.net/questions/109389/proofs-for-doubly-ruled-surfaces/109399#109399 Answer by Dick Palais for Proofs for doubly ruled surfaces Dick Palais 2012-10-11T17:46:47Z 2012-10-11T17:46:47Z <p>Sorry, I had a bad reference</p> http://mathoverflow.net/questions/109389/proofs-for-doubly-ruled-surfaces/109408#109408 Answer by Sue for Proofs for doubly ruled surfaces Sue 2012-10-11T19:39:22Z 2012-10-11T19:39:22Z <p>I don't have it in front of me right now, but I believe that there is an old-fashioned proof in the book</p> <p>G. Salmon, A Treatise on the Analytic Geometry of Three Dimensions, Vol. 2, 5th edition Hodges, Figgis And Co. Ltd. (1915).</p> <p>I learned about this from the beautiful paper "On the Erd ̀‹os distinct distance problem in the plane" by Guth and Katz.</p> http://mathoverflow.net/questions/109389/proofs-for-doubly-ruled-surfaces/109493#109493 Answer by Robert Bryant for Proofs for doubly ruled surfaces Robert Bryant 2012-10-12T21:13:19Z 2012-10-13T14:56:29Z <p>The classical proof via differential geometry goes like this: </p> <p>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. </p> <p>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.</p> <p>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 $$\begin{pmatrix} dX_s&amp; dX_t &amp; dX_{st} \end{pmatrix} = \begin{pmatrix} X_s&amp; X_t &amp; X_{st} \end{pmatrix} \begin{pmatrix} f\ ds&amp; 0 &amp; f_t\ ds\\ 0 &amp; g\ dt &amp; g_s\ dt\\ dt &amp; ds &amp; f\ ds + g\ dt \end{pmatrix}$$ (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 $$\begin{pmatrix} dX_s&amp; dX_t &amp; dX_{st} \end{pmatrix} = \begin{pmatrix} X_s&amp; X_t &amp; X_{st} \end{pmatrix} \begin{pmatrix} 2h_s\ ds&amp; 0 &amp; 2h_{st}\ ds\\ 0 &amp; 2h_t\ dt &amp; 2h_{st}\ dt\\ dt &amp; ds &amp; 2h_s\ ds + 2h_t\ dt \end{pmatrix}$$ 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 $-1$.</p> <p>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.</p> <p>Now, consider the case $C\not=0$. The structure equations have become</p> <p>$$\begin{pmatrix} dX_s&amp; dX_t &amp; dX_{st} \end{pmatrix} = \begin{pmatrix} X_s&amp; X_t &amp; X_{st} \end{pmatrix} \begin{pmatrix} 2h_s\ ds&amp; 0 &amp; 2C\ e^{2h}\ ds\\ 0 &amp; 2h_t\ dt &amp; 2C\ e^{2h}\ dt\\ dt &amp; ds &amp; 2\ dh \end{pmatrix}$$ 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. </p> <p>To this end, consider the new frame field $$\begin{pmatrix}E_1 &amp; E_2 &amp; E_3\end{pmatrix} = \begin{pmatrix}e^{-h}X_{s} &amp; e^{-h}X_{t} &amp; e^{-2h}X_{st}\end{pmatrix}.$$ Calculation shows that it satisfies the structure equation $$\begin{pmatrix} dE_1&amp; dE_2 &amp; dE_3 \end{pmatrix} = \begin{pmatrix} E_1&amp; E_2 &amp; E_3 \end{pmatrix} \begin{pmatrix} h_s\ ds-h_t\ dt&amp; 0 &amp; 2C\ e^h\ ds\\ 0 &amp; h_t\ dt - h_s\ ds &amp; 2C\ e^h\ dt\\ e^h\ dt &amp; e^h\ ds &amp; 0 \end{pmatrix}.$$ 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 $$\begin{pmatrix} x_1&amp; 0 &amp; 2C\ x_2\\ 0 &amp; -x_1 &amp; 2C\ x_3\\ x_3 &amp; x_2 &amp; 0 \end{pmatrix}.$$ 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 $$Q = \begin{pmatrix} 0&amp; 1 &amp; 0\\ 1 &amp; 0 &amp; 0\\ 0 &amp; 0 &amp; -2C \end{pmatrix}.$$ 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. </p> http://mathoverflow.net/questions/109389/proofs-for-doubly-ruled-surfaces/109538#109538 Answer by Agol for Proofs for doubly ruled surfaces Agol 2012-10-13T15:51:41Z 2012-10-13T15:51:41Z <p>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.</p> <p>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 $\Sigma$.</p> <p>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 $l_2,l_2',l_2''$. </p> <p>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. </p> <p>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 <a href="http://mathworld.wolfram.com/SkewLines.html" rel="nofollow">Hilbert-Cohn Vossen</a>, 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 $l_2''$. </p> <p>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. </p> <p>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 <code>$P_1=\{(1,0,0,0),(0,1,0,0)\}, P_2=\{(0,0,1,0),(0,0,0,1)\}$</code>. 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 <code>$\{(1,0,1,0),(0,1,0,1)\}$</code>, 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. </p> <p>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. </p> <p>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. </p> <p>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 $\mathbb{R}^3$. </p>