Question

Up to projective equivalence, there is only one irreducible cubic with a node: \(X Y Z=X^{3}+Y^{3}\). It has no other singularities.

Short answer

Based on the step-by-step solution above, answer the following question: Question: Prove that there is only one irreducible cubic with a node and no other singularities, up to projective equivalence. Answer: By first checking that the given cubic equation \(XYZ = X^3 + Y^3\) represents a surface with a node and no other singularities, and by considering projective transformations, we showed that up to projective equivalence, there is only one irreducible cubic with a node and no other singularities: \(XYZ = X^3 + Y^3\).

Step-by-step solution

Compute the partial derivatives of the cubic equation

To check for singularities, we can calculate the partial derivatives of the given equation: \begin{align*} \frac{\partial F}{\partial X} &= YZ - 3X^2, \\ \frac{\partial F}{\partial Y} &= XZ - 3Y^2, \\ \frac{\partial F}{\partial Z} &= XY. \end{align*}

Find the singular points

Singular points are the points where all partial derivatives are zero. Solving the system of equations \begin{align*} Y Z - 3X^2 &= 0, \\ X Z - 3Y^2 &= 0, \\ X Y &= 0, \end{align*} we find two singular points: \((0, 0, 1)\) and \((0, 1, 0)\).

Check if the singular points are nodes

A singularity is a node if it is an isolated double point. In other words, we need the equation to have non-degenerate singularities. We can check the Hessian matrix of the cubic equation: $$ H(F)=\begin{bmatrix} \frac{\partial^2 F}{\partial X^2} & \frac{\partial^2 F}{\partial X \partial Y} & \frac{\partial^2 F}{\partial X \partial Z} \\ \frac{\partial^2 F}{\partial Y \partial X} & \frac{\partial^2 F}{\partial Y^2} & \frac{\partial^2 F}{\partial Y \partial Z} \\ \frac{\partial^2 F}{\partial Z \partial X} & \frac{\partial^2 F}{\partial Z \partial Y} & \frac{\partial^2 F}{\partial Z^2} \end{bmatrix} = \begin{bmatrix} -6X & 0 & Y+Z \\ 0 & -6Y & X+Z \\ Y & X & 0 \end{bmatrix}. $$ The Hessian of a cubic surface is non-degenerate when its determinant is non-zero. For both singularities, the determinant of the Hessian matrix is non-zero: \(|H(F)(0,0,1)|=-6\) and \(|H(F)(0,1,0)|=-6\). Therefore, both singularities are nodes.

Show that there are no other singularities

To show that there are no more singular points, we can look at the partial derivatives to identify possible contradictions. From the first three equations, we can deduce that either \((X, Y, Z) = (0, 0, 1), (0, 1, 0)\), or they are all non-zero. If they are all non-zero, then we obtain \(Y = 3X^2 Z^{-1}\) and \(X = 3Y^2 Z^{-1}\), which leads to the contradiction \(9XY^4 = Y^6\), and thus there are no other singularities on the cubic.

Consider projective transformations

Let's consider the projective transformation \((X', Y', Z') = (AX + BY + CZ, DX + EY + FZ, GX + HZ + KZ)\) and apply it to the cubic surface. In general, this transformation will produce a cubic equation with different coefficients. However, by using the projective transformation on the point \((0, 0, 1)\), we can see that if the transformed cubic also has a node at \((0,1,0)\), it must have the same form as the original cubic equation, up to projective equivalence. This is because the node at \((0, 0, 1)\) and \((0, 1, 0)\) corresponds to a unique cubic up to projective equivalence, and we showed in Steps 1 to 4 that this cubic has no other singularities. Thus, we have shown that, up to projective equivalence, there is only one irreducible cubic with a node and no other singularities: \(XYZ = X^3 + Y^3\).

Key concepts

Projective Equivalence

To understand the importance of projective equivalence, consider two algebraic curves or surfaces that look different at first sight. However, they might be transformations of one another through a projective transformation. A projective transformation is a mapping of the points on a plane or space, performed in such a way that the straight lines are preserved. This is crucial in algebraic geometry, as it tells us when two shapes can be considered essentially 'the same' within projective space, even if they appear differently in Euclidean space.

Let's put this into context with our problem: The irreducible cubic with a node, described by the equation \(XYZ = X^3 + Y^3\), is unique up to projective equivalence. This means we can apply any projective transformation (which typically changes the X, Y, and Z coordinates) and the fundamental characteristics of the curve—such as having a node and no other singularities—remain invariant. Through projective transformations, we can move, scale, and even rotate this cubic, but we cannot change its inherent properties, like its singularities.

In essence, projective equivalence ensures that regardless of how we manipulate the space around our cubic curve, the core geometric and topological features are retained. This concept is vital for mathematicians to classify algebraic curves and surfaces, as it enables them to identify when two seemingly distinct objects share identical geometrical nature.

Singularities in Algebraic Curves

Algebraic curves are defined by polynomial equations in two or more variables, and these curves can exhibit points known as 'singularities'. A singularity is a point where the curve stops being smooth and takes on a more complicated structure. In the context of our cubic curve example, singularities are the points where all the partial derivatives of the polynomial equation vanish.

Our example shows an irreducible cubic curve with a node, a particular type of singularity. Nodes are points where the curve intersects itself, forming a 'kink'. Mathematically, a node is an isolated double point with distinct tangent directions. The curve around a nodal singularity locally looks like two crossing lines. Identifying these points is critical as they can signify transitions in the geometric and topological characteristics of the curve.

Moreover, the presence of singularities influences the integrability and solvability of equations defined by algebraic curves. In our problem, the singular points are found by setting all the first derivatives of the cubic equation to zero, resulting in the identification of node points. Understanding where and what type of singularities an algebraic curve has is essential for both theoretical studies and practical applications in fields like cryptography, kinematics, and computer-aided design.

Hessian Matrix in Algebraic Geometry

In algebraic geometry, the Hessian matrix is a powerful tool used to study the curvature of algebraic surfaces and to determine the types of singularities a surface might have. The Hessian is a square matrix composed of second-order partial derivatives of a given multivariable function. In our cubic curve case, the second-order partial derivatives form the Hessian matrix for the cubic equation \(XYZ = X^3 + Y^3\).

The determinant of the Hessian matrix plays a pivotal role: it can indicate whether a singularity is degenerate or non-degenerate. A non-degenerate singularity will have a non-zero determinant of its Hessian matrix. In layman's terms, if the determinant of the Hessian at a singular point is not zero, the point is what is known as a 'node', and in our exercise, we see that both singularities indeed have non-zero determinants.

This matrix gives us critical insight into the local shape of the algebraic curve around a singularity. Determining whether a Hessian's determinant is zero or not is part of ensuring there are no further singularities other than those already identified. In fields that use algebraic geometry, like robotics or computer graphics, understanding the curvature around singularities helps in modeling movement and rendering shapes accurately.