Lines on degree 2n-3 Fermat hypersufaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T05:46:08Z http://mathoverflow.net/feeds/question/118029 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118029/lines-on-degree-2n-3-fermat-hypersufaces Lines on degree 2n-3 Fermat hypersufaces aglearner 2013-01-04T08:02:47Z 2013-01-04T13:05:51Z <p>It is well known that a generic hypersurface of degree $2n-3$ in $\mathbb CP^n$ has finite number of lines. I would like to ask a couple of questions about lines on Fermat hypersurfaces and their symmetries: </p> <p>$$\sum_{i=1}^{n+1}x_i^{2n-3}=0.$$</p> <p>Fermat hypersurfaces have a group of automorphisms of order $(2n-3)^n(n+1)!$. In the case $n=3$ (the case of cubic) this group is acting transitively on the collection of $27$ lines and this rases some questions.</p> <p>The first question is pedagogical, I plan to use it for teaching and really want to know the answer. </p> <p><em>Question 1.</em> Is there some slick way to give a high-school proof of the fact that there are exactly $27$ lines on Femat cubic in $\mathbb CP^3$ using (or not) the symmetries of the cubic but without using any theory at all?</p> <p>Further questions are not for teaching, I am just curious about them.</p> <p><em>Question 2.</em> Is it known that a Fermat hypersurface of degree $2n-3$ has finite number of lines for any $n$? Is it known that these lines are never multiple?</p> <p><em>Question 3.</em> Can one say something about the number of orbits of the action of symmetries on lines on a Fermat hypersurface of degree $2n-3$? For example, what happen in the case of quintic, $n=5$? According to wiki a generic quintic has $2875=125\cdot 23$ lines, so if Fermat quintic is generic, there should be more than one orbit in the action on lines on it. What is the number of orbits?</p> <p>I would be happy to know the answer on any of these questions.</p> http://mathoverflow.net/questions/118029/lines-on-degree-2n-3-fermat-hypersufaces/118032#118032 Answer by Sasha for Lines on degree 2n-3 Fermat hypersufaces Sasha 2013-01-04T08:24:08Z 2013-01-04T08:24:08Z <p>Assume for example that $n = 2k + 1$ is odd. Let $\xi^{2n-3} = -1$. Then for any $(y_0,y_1,\dots,y_k) \in \mathbb{CP}^k$ the point $(y_0,\xi y_0,y_1,\xi y_1,\dots,y_k, \xi y_k)$ is on the Fermat hypersurface. So, it contains $\mathbb{CP}^k$. In particular, if $k \ge 2$ (and so $n \ge 5$) the number of lines is infinite. A similar argument works for even $n \ge 6$.</p> http://mathoverflow.net/questions/118029/lines-on-degree-2n-3-fermat-hypersufaces/118038#118038 Answer by Johannes NordstrÃ¶m for Lines on degree 2n-3 Fermat hypersufaces Johannes NordstrÃ¶m 2013-01-04T10:46:26Z 2013-01-04T13:05:51Z <p>Regarding question 1, any line in the Fermat cubic <code>$C = \{X_0^3 + X_1^3 + X_2^3 + X_3^3 = 0\}$</code> must meet the coordinate hyperplane <code>$H_0 = \{X_0 = 0\}$</code>. So which points <code>$x \in (C \cap H_0)$</code> can lie on lines? If <code>$Y, Z$</code> are homogenous coordinates on <code>$T_x(C \cap H_0) \cong \mathbb{P}^1$</code>, then the restriction of <code>$X_0^3 + X_1^3 + X_2^3 + X_3^3$</code> to <code>$T_x C$</code> is of the form <code>$X_0^3 + F(Y,Z)$</code> for a homogeneous cubic <code>$F$</code>. For <code>$x$</code> to lie on a line, <code>$X_0^3 + F$</code> must factorise, so <code>$F$</code> is a cube. This means that <code>$x$</code> is an inflection point of the plane cubic curve <code>$C \cap H_0 = \{X_1^3 + X_2^3 + X_3^3 = 0\}$</code>. The inflection points are given by intersection with the zero set of the Hessian determinant <code>$216X_1X_2X_3$</code>. Hence the intersection of any line in <code>$C$</code> with any coordinate hyperplane must actually have two corrdinates equal to 0, and it follows that the lines consist of <code>$\{X_0^3 + X_1^3 = X_2^3 + X_3^3 = 0\}$</code> and its two images under permutating the coordinates (9 lines in each).</p> <p>P.S. Here is a related exercise I like. Once one has identified the 27 lines in the Fermat cubic <code>$C$</code>, one can use the symmetries of <code>$C$</code> to guess how to arrange 6 points in <code>$\mathbb{P}^2$</code> so that the blow-up is isomorphic to <code>$C$</code>, and then write down an explicit rational map <code>$\mathbb{P}^2 \dashrightarrow \mathbb{P}^3$</code> that maps birationally onto <code>$C$</code>.</p>