In geometry, a Hessian curve is a cubic plane curve similar to the folium of Descartes that belongs to the Hesse pencil. It is named after the German mathematician Otto Hesse. In projective coordinates, the projective equation of a Hessian curve is
X 3 + Y 3 + Z 3 − k X Y Z = 0. {\displaystyle X^{3}+Y^{3}+Z^{3}-kXYZ=0.}
For every nonsingular cubic plane curve there is a change of projective coordinates that transforms the equation of the curve into the above Hessian form. Theses curves are used in elliptic curve cryptography, because arithmetic in this curve representation is faster and needs less memory than arithmetic in standard Weierstrass form.
Definition
Let K {\displaystyle K} be a field and consider an elliptic curve E {\displaystyle E} in the following special case of Weierstrass form over K {\displaystyle K} :
Y 2 + a 1 X Y + a 3 Y = X 3 {\displaystyle Y^{2}+a_{1}XY+a_{3}Y=X^{3}}
where the curve has discriminant Δ = ( a 3 3 ( a 1 3 − 27 a 3 ) ) = a 3 3 δ . {\displaystyle \Delta =\left(a_{3}^{3}\left(a_{1}^{3}-27a_{3}\right)\right)=a_{3}^{3}\delta .}
Then the point P = ( 0 , 0 ) {\displaystyle P=(0,0)} has order 3. To prove that P = ( 0 , 0 ) {\displaystyle P=(0,0)} has order 3, note that the tangent to E {\displaystyle E} at P {\displaystyle P} is the line Y = 0 {\displaystyle Y=0} which intersects E {\displaystyle E} with multiplicity 3 at P {\displaystyle P} . Conversely, given a point P {\displaystyle P} of order 3 on an elliptic curve E {\displaystyle E} both defined over a field K {\displaystyle K} one can put the curve into Weierstrass form with P = ( 0 , 0 ) {\displaystyle P=(0,0)} so that the tangent at P {\displaystyle P} is the line Y = 0 {\displaystyle Y=0} . Then the equation of the curve is Y 2 + a 1 X Y + a 3 Y = X 3 {\displaystyle Y^{2}+a_{1}XY+a_{3}Y=X^{3}} with a 1 , a 3 ∈ K {\displaystyle a_{1},a_{3}\in K} . To obtain the Hessian curve, it is necessary to do the following transformation: First let μ {\displaystyle \mu } denote a root of the polynomial
T 3 − δ T 2 + δ 2 3 T + a 3 δ 2 = 0. {\displaystyle T^{3}-\delta T^{2}+{\delta ^{2} \over 3}T+a_{3}\delta ^{2}=0.}
Then
μ = δ − a 1 δ 2 / 3 3 . {\displaystyle \mu ={\delta -a_{1}\delta ^{2/3} \over 3}.}
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