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I watched a video from Wildberger in the Differential Geometry series ( first, or third lecture, I don't remember ) where he says the following.

The general format of a cubic curve is $$a x^3 + b y^3 + c x^2 y + d y^2 x + e x^2 + f y^2 + g x y + h x + i y + j = 0$$

and that all of these can be reduced ( transformed, in fact ) to the canonical form $$y^2 = a x^3 + b x + c.$$

He then explains that there are basically three types, one of them being the elliptic curve with no singularities and two others with singularities.

Is this correct? And how do these transformations work?

P.S. Allegedly ( says Wildberger ) Newton studied cubics intensively and categorized them in 80 different types. Is there some book where this is described?

EDIT. I am studying this question which might be a duplicate. Explicit Derivation of Weierstrass Normal Form for Cubic Curve - Couldn't find this before because I did not have the notion of Weierstrass normal form.

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nilo de roock
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  • This site uses MathJax, which means you can typset formulas using the same commands you would in TeX. I edited for you. About your questions: 1. Yes, this is correct. 2. See "Weierstrass form" on Wikipedia. 3. No need for "allegedly"; Newton did indeed classify plane cubics into types. I think he described 72 types. I don't know a good source, but googling "newton classification plane cubics" returns lots of relevant things. – Relapsarian Jul 07 '15 at 09:43
  • It's perhaps worth saying that this is canonical form only when the characteristic underlying field does not have characteristic $2$ or $3$ (those characteristics have their own forms, however). (Since this is tagged differential-geometry presumably you're only interested in the real and perhaps complex cases anyway.) – Travis Willse Jul 07 '15 at 09:45
  • Thanks. I am basically interested in the transformations as such. Can you give an example? Are Newton's categories still relevant? – nilo de roock Jul 07 '15 at 09:46
  • Oh, yes, real case only. – nilo de roock Jul 07 '15 at 09:46
  • (And actually, the canonical form isn't quite reduced; by applying a the transformation $(x, y) \mapsto (a^{-1 / 3} x, y)$, we can reduce the above form to $$y^2 = x^3 + p x + q .)$$ – Travis Willse Jul 07 '15 at 09:59
  • Why not use an Answer @Travis? I know that transform from solving the Cubic but in this case it yields ( in your Latex )$$ \frac{c x^2 y}{a^{2/3}}+\frac{e x^2}{a^{2/3}}+\frac{b x^3}{a}+\frac{d x y^2}{\sqrt[3]{a}}+\frac{g x y}{\sqrt[3]{a}}+\frac{h x}{\sqrt[3]{a}}+f y^2+i y+j+x^3$$ – nilo de roock Jul 07 '15 at 10:10
  • @ndroock1 To be clear, I mean applying that particular transformation to the partially reduced form $y^2 = a x^3 + b x + c$. – Travis Willse Jul 07 '15 at 10:29
  • Look at this question. – Jan-Magnus Økland Jul 07 '15 at 10:33
  • @Jan-MagnusØkland - I found a likely better duplicate. See my edit. This takes some time to sink in for me. – nilo de roock Jul 07 '15 at 10:44

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If you'd like to work it out yourself, the outline in Reid's Undergraduate algebraic geometry, exercises 2.9–2.10 is good. I'll give a sketch following another book, Dolgachev's Lectures on invariant theory, specifically §10.3.

Let $C$ be an irreducible cubic curve, and say it's nonsingular. Let $P$ be an inflection point on $C$, which you can find by computing the Hessian and intersecting $C$ with the Hessian curve, and choose a system of coordinates such that $P = [0:0:1]$, and the tangent line at $P$ is given by $T_0 = 0$. The cubic then has equation $$T_2^2T_0 + T_2L_2(T_0,T_1) + L_3(T_0,T_1) = 0,$$ where $L_2$ is a quadratic form and $L_3$ is a cubic form. Since $T_0 = 0$ intersects $C$ at one point, we have that the $T_1^2$ coefficient in $L_2$ is zero, so in affine coordinates $x = T_1/T_0$ and $y = T_2/T_0$, the equation becomes $$y^2 + axy + by + dx^3 + ex^2 + fx + g = 0.$$ Since $d \ne 0$ by assumption, we can assume that $d=1$. Now replacing $y$ with $y + ax/2 + b/2$, we can assume $a = b = 0$. Moreover, by changing variables $x \mapsto x + e/3$, we can assume that $e=0$. We therefore have that the equation is of the form $$y^2 + x^3 + ax + b = 0.$$

On the other hand, if $C$ is singular, and we choose $[0:0:1]$ to be the singular point, the cubic has equation $$T_2L_2(T_0,T_1) + L_3(T_0,T_1) = 0.$$ By linear transformations, we get that either $L_2 = T_0^2$ or $L_2 = T_0T_1$. A similar argument as to the nonsingular case gives that these two cases can be reduced to cuspidal cubics of the form $$y^2 + x^3 = 0,$$ and nodal cubics of the form $$y^2 + x^2(x+1) = 0,$$ respectively. See Dolgachev for details. These give the three cases you stated.

Now, with respect to Newton's classification, I first give a reference: in Brieskorn/Knörrer's book Plane algebraic curves, in §2.5, starting at p. 87, you can find a leisurely explanation of how Newton went about classifying plane cubics. Apparently Newton found 72 different kinds of plane cubics, but overlooked six of them! You can even see some pages from Newton's work on pp. 93–98. The main difference between his classification and ours is that 1. Newton works over the reals, and 2. he does not work projectively. A good reason why we work projectively and over the complex numbers is so that classification problems like these are more manageable!

Takumi Murayama
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