# Looking for the formula for a given graph of a quartic

Quatric with symetry and two nodes

Hi, I found that curve in the net and guess how to find out the formula for that. Can someone support me? Thanks very much.

• if you tell us enough about the source of the diagram, as well as your background in mathematics – Will Jagy Oct 12 '15 at 0:27
• An awesomely weird graph please tell us your level of understanding of mathematics so that we can help you out – Sujith Sizon Oct 12 '15 at 1:09
• Oh and btw welcome to MathSE – Sujith Sizon Oct 12 '15 at 1:10
• "about the source of the diagram"; Universität Bayreuth. But already deleted. – TheNexpert Oct 13 '15 at 10:40

• Let $~\Big(x^2+y^2\Big)^2~-~\displaystyle\sum_{k=0}^3A_k\Big(x^k+y^k\Big)~=~B_4\Big(x^3y+xy^3\Big)~+~B_3\Big(x^2y+xy^2\Big)$.

• Playing around in Desmos or GeoGebra, we get $A_3=-1$, $A_2=5$, $A_1=\dfrac32$ , $A_0=-1.595$,
$B_4=1$, $B_3=3$, which yield a good approximation.

• Have fun, and feel free to adjust the parameters as you please ! :-$)$

$\newcommand{\PSP}{{\mathbb P}}$

If one takes the intersection of the image $X = s(\PSP^1_k \times \PSP^1_k)$ of the Segre--imbedding $s: \PSP^1_k \times \PSP^1_k \to \PSP^3_k$ (given as $X= V(x_0 x_1 - x_2 x_3)$ in $\mathrm{proj}(k[x_0,x_1,x_2,x_3]))$ and a certain quadric surface $F_2$ in $\PSP^3_k$ and calls the result $C_4$, then the projection of $C_4$ to a plane $H = \PSP^2_k \subseteq \PSP^3_k$ from a point $O$ on $X$ but not on $F_2$ and not on $H$ gives a curve of the sought for type. To put it into other terms, $C_4$ is an effective divisor of type $(2,2)$ on $X$ which is smooth for a generic $F_2$. (See Hartshorne, III. Ex. 5.6)

This discussion with some other cases too can be found in Felix Klein, "Vorlesungen über die Entwicklung der Mathematik im 19. Jahrhundert", Springer Reprint 1979, p. 315ff., p. 318, especially p. 319, Fig. 29

I tried to reproduce analytically what is depicted in Fig. 29 of Klein's book. It turned out to be astonishingly difficult to get exactly the topological type of the quartic, that was sought by the OP, but now, I think I have succeeded.

One intersects the hyperboloid

$$u_0^2-u_1^2 - u_2^2 + u_3^2 = 0$$

with the cylinder (our surface $F_2$)

$$c u_0^2 +(u_1 - d u_3 - a u_2)^2 - b u_2^2=0$$

and considers only the affine patch $u_2 = 1$, giving the affine coordinates $U_0=u_0/u_2, U_1 = u_1/u_2, U_3 = u_3/u_2$.

In these coordinates the projecting point O is $(U_0=0,U_1=0,U_3=1)$ and as receiving plane I chose $U_1 = -10$. Putting all this into Maple and doing a lot of eliminations I got for the resulting quartic the following (admittedly not so nice as in the other answer) expression:

$$\begin{multline} \left( -b+{d}^{2}+2\,da+{a}^{2} \right) {x}^{4}+ \left( \left( 2\,{a }^{2}-2\,{d}^{2}+4\,c-2\,b \right) {y}^{2}+ \left( -4\,{a}^{2}+4\,b-40 \,d-40\,a-8\,c+4\,{d}^{2} \right) y+4\,c+40\,d-198\,{a}^{2}+198\,b-202 \,{d}^{2}-400\,da+40\,a \right) {x}^{2}+ \left( {a}^{2}-2\,da-b+{d}^{2 } \right) {y}^{4}+ \left( -40\,a-4\,{a}^{2}+8\,da+40\,d-4\,{d}^{2}+4\, b \right) {y}^{3}+ \left( 206\,{d}^{2}-194\,{a}^{2}+194\,b-12\,da+400+ 120\,a-120\,d \right) {y}^{2}+ \left( -800+3880\,a-396\,b-404\,{d}^{2} +396\,{a}^{2}+8\,da+4120\,d \right) y+400-9801\,b-4040\,d-3960\,a+ 19998\,da+9801\,{a}^{2}+10201\,{d}^{2} \end{multline}$$

The depicted curve results from choosing $a=2.6,b=2.3,c=0.8,d=-1$ in the equation above.

• Great! Thank you very much! – TheNexpert Oct 13 '15 at 10:38