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
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), 307320. arXiv: math/0010163
which gives on pages 56 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), 7075.

$\begingroup$ Many thanks, Noam,could you explain why triple point put 10 restrictions? $\endgroup$– xin fuSep 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 zeroset 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