The set of 5-tuples of lines in $\mathbf{P}^3$ is parametrized by the 20-dimensional product of Grassmannians $G(2,4)^{\times 5}$. The set of cubic surfaces is parametrized by a 19-dimensional projective space.

I can present a line in $\mathbf{P}^3$ as a $2 \times 4$ matrix (the projectivization of the row span) and five lines in $\mathbf{P}^3$ as a $10 \times 4$ matrix. Which rational function of this matrix vanishes on the set of tuples of lines that are contained in a cubic surface?

  • 1
    $\begingroup$ I don't know about functions (this should be an exercise), but, assuming the five lines skew, there must exist six other skew lines, each intersecting all five. The 27 lines in cubic surfaces is a very classical theory! $\endgroup$ – Alex Degtyarev Nov 6 '14 at 20:53

It's a $20 \times 20$ determinant. Take each $2 \times 4$ matrix $$L = \begin{pmatrix} s & t & u & v \\ w & x & y & z \\ \end{pmatrix}$$ and turn it into the $4 \times 20$ matrix $$M := \begin{pmatrix} s^3 & s^2 t & s^2 u & \cdots & v^3\\ 3 s^2 w & 2 swt+s^2x & 2swu+s^2 y & \cdots & 3 v^2 z \\ 3 s w^2 & 2 swx+w^2 t & 2swy + w^2 u & \cdots & 3 v z^2 \\ w^3 & w^2 x & w^2 y & \cdots & z^3 \\ \end{pmatrix}$$ If the pattern isn't so clear, the columns are indexed by the $20$ cubic monomials in $4$ variables. For each monomial, plugin $( a \ b ) \begin{pmatrix} s & t & u & v \\ w & x & y & z \\ \end{pmatrix}$ for the $4$ entries and collect coefficients of $a$ and $b$. For example, the monomial $s^2 t$ becomes $(as+bw)^2 (at+bu)$; expanding and collecting coefficients gives the column $(s^2 t,\ 2swt+s^2 x, \ 2swx + w^2 t, \ w^2 x)^T$. A cubic vanishes on the row span of $L$ if and only if that cubic is in the kernel of $M$.

Stack up the $5$ $M$ matrices and take the determinant to get your answer.


I would write it differently. First, consider the universal space $M$ consisting of tuples $(S,L_1,\dots,L_5)$, where $S$ is a cubic surface and $L_i$ are lines on $S$. This space lies inside the product $P^{19}\times Gr(2,4)^5$ and can be described as the zero locus of a canonical global section of the vector bundle $$ E := O(1) \boxtimes (S^3U_1^* \oplus S^3U_2^* \oplus S^3U_3^* \oplus S^3U_4^* \oplus S^3U_5^*) $$ on it, where $U_i$ is the tautological bundle on the $i$-th copy of the Grassmannian. Consequently, the structure sheaf of $M$ has the following Koszul resolution $$ 0 \to \Lambda^{20}E^* \to \dots \to \Lambda^2E^* \to E^* \to O \to O_M \to 0. $$ We are interested in the image of $M$ in $Gr(2,4)^5$, so let us push forward the above resolution to $Gr(2,4)^5$. Note that $E^*$ restricts as a sum of $O(-1)$ to any fiber $P^{19}$ of the projection, hence the $p$-th term restricts as a sum of $O(-p)$. Thus restriction of almost all terms are acyclic, and they do not contribute to the pushforward. The only terms which do are the first and the last. The first gives $O$, and the last gives $$ \det(S^3U_1 \oplus \dots \oplus S^3U_5) = \otimes_{i=1}^5\det(S^3U_i) = O(-6,-6,-6,-6,-6). $$ This means that the equation of the image of $M$ is a hypersurface of polydegree $(6,6,6,6,6)$ on the product of the Grassmannians. Of course its equation is the one given by David Speyer --- the determinant is a homogeneous polynomial of polydegree $(12,12,12,12,12)$ in coefficients of matrices, but if you write it in terms of the Plucker coordinates (which are quadratic in coefficients of matirces), it will have polydegree $(6,6,6,6,6)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.