I know we can bound the triple point on quintics in cp^3 by 5. But how to write down quintics with 5 ordinary triple point (here are simple elliptic singularity)explicitly?


1 Answer 1


Choose five points $P_i$ in general linear position (all choices are equivalent under ${\rm PGL}_4$, so you might as well put four of them at the coordinate vectors and the fifth at $(1:1:1:1)$); then at each $P_i$ the condition of a triple point imposes $1+3+6 = 10$ linear conditions on the space of quintics, which has dimension ${8 \choose 5} = 56$, so you get a linear system of quintics with a triple point at each $P_i$, of projective dimension $56 - 5\cdot 10 - 1 = 5$.

A quick Google search turns up a paper

Stephan Endrass, Ulf Persson, and Jan Stevens: Surfaces with triple points, J. Algebraic Geom. 12 (2003), 307--320. arXiv: math/0010163

which gives on pages 5-6 an alternative description. Start with a cubic surface that has a triple point (so is a cone over a cubic plane curve), and apply a "reciprocal transformation": choose four points $p_i$ on the cubic (other than the cubic singularity), use projective coordinates $(x_1:x_2:x_3:x_4)$ that make $p_i$ the coordinate points, and apply the birational involution $$ (x_1:x_2:x_3:x_4) \leftarrow - \ - \rightarrow (1/x_1:1/x_2:1/x_3:1/x_4). $$ As remarked at the end of this section, "The above analysis is very elementary, and parts of it have not too surprisingly already appeared in the literature", citing a 1952 paper in Italian:

D. Gallarati, Sulle superficie del quinto ordine dotate di punti tripli, Rend. Accad. Naz. Lincei, serie VIII, XII (1952), 70--75.

  • $\begingroup$ Many thanks, Noam,could you explain why triple point put 10 restrictions? $\endgroup$
    – xin fu
    Sep 3, 2017 at 14:02
  • 2
    $\begingroup$ Look at the Taylor expansion of a function in three variables about a point. There at 10 coefficients that must vanish for the zero-set of the function to have (at least) a triple point there: the value, three derivatives, and six second derivatives. $\endgroup$ Sep 3, 2017 at 14:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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