Invariants of the quadrics and Euclidean classification


Given a quadratic polynomial $$q(x,y,z)$$, let $$\overline{A}$$ be its matrix and $$A$$ its main matrix. We define the real numbers $$D_i=D_i(\overline{A}), 1 \leq i \leq 4, d_i=d_i(A), 1\leq i \leq 3 $$ given by: $$$det(\lambda \cdot I_4-\overline{A})=\lambda^4 -D_1\lambda^3+D_2\lambda^2-D_3\lambda +D_4$$$ $$$det(\lambda \cdot I_3-A)=\lambda^3-d_1\lambda^2+d_2\lambda - d_3$$$ The expression $$D_4=det(\overline{A})$$ is called discriminant of $$q(x,y,z)$$. Similarly, the expression $$d_3=det(A)$$ is called discriminant of the main part of $$q(x,y,z)$$. We notice that $$$d_1=a+b+c \ d_2=ab+ac+bc-(f^2+g^2+h^2)$$$

In the classification of the quadrics there is still another factor, called an index, denoted by $$j$$, or index of the main part. The main $$q(x,y,z)$$ has index $$1$$ if $$d_1d_3 < 0$$ or $$d_2 < 0$$, and $$0$$ in another case.

Euclidean classification of the quadrics

$$$ \left\{\begin{array}{l} D_4=0 \left\{\begin{array}{l} d_3=0 \left\{\begin{array}{l} d_2\neq0 \left\{\begin{array}{l} d_2 > 0 \left\{\begin{array}{l} D_3\neq0 \left\{\begin{array}{l} d_1D_3 < 0 \text{ real elliptic cylinder } \\ d_1D_3 > 0 \text{ imaginary elliptic cylinder } \end{array}\right. \\ D_3=0 \text{ pair of conjugate imaginary planes } \end{array}\right. \\ d_2 < 0 \left\{\begin{array}{l} D_3\neq0 \text{ hyperbolic cylinder } \\ D_3=0 \text{ pair of parallel planes } \end{array}\right. \end{array}\right. \\ d_2=0 \left\{\begin{array}{l} D_3\neq0 \text{ parabolic cylinder } \\ D_3=0 \left\{\begin{array}{l} D_2 < 0 \text{ pair of real parallel planes } \\ D_2 > 0 \text{ pair of conjugate imaginary planes } \\ D_2=0 \text{ double plane } \end{array}\right. \end{array}\right. \end{array}\right. \\ d_3\neq0 \left\{\begin{array}{l} j=0 \text{ imaginary cone } \\ j=1 \text{ real cone } \end{array}\right. \end{array}\right. \\ D_4\neq0 \left\{\begin{array}{l} d_3\neq0 \left\{\begin{array}{l} j=0 \text{ ellipsoid } \left\{\begin{array}{l} D_4 < 0 \text{ real } \\ D_4 > 0 \text{ imaginary } \end{array}\right. \\ j=1 \text{ hyperboloid } \left\{\begin{array}{l} D_4 < 0 \text{ two sheets } \\ D_4 > 0 \text{ one sheet } \end{array}\right. \end{array}\right. \\ d_3=0 \text{ paraboloid } \left\{\begin{array}{l}D_4<0 \text{ elliptic } \\ D_4>0 \text{ hyperbolic }\end{array}\right. \end{array}\right. \end{array}\right.$$$

Obtaining the reduced equations from the unvariants

Let's suppose that we have a quadric given by the equation $$q(x,y,z)=0$$ in a system of rectangular coordinates$$(x,y,z)$$. Let's also suppose that we have determined the kind of quadric by means of the previous scheme and we have studied its eigenvalues $$\lambda_1, \lambda_2$$ and $$\lambda_3$$ of the main matrix of $$q(x,y,z)$$.

Quadrics of the centered type

If the quadric is of the centered type, the limited form $$$\displaystyle \lambda_1 x^2+\lambda_2y^2+\lambda_3z^2+\frac{D_4}{d_3}=0$$$with reference to a suitable system of rectangular coordinates, defines a quadric that coincides with $$Q$$.

Quadrics of the parabolic type

There are two non-zero eigenvalues $$\lambda_1>0$$ and $$\lambda_2$$. The quadric defined by the limited form$$$\displaystyle \lambda_1x^2+\lambda_2y^2-2z\sqrt{\frac{-D_4}{d_2}}=0$$$with reference to a suitable coordinates system, coincides with $$Q$$.

Degenerate Quadrics

The cones have already been considered. As far as other degenerate curves are concerned, obtaining a reduced equation from the invariants is equivalent to obtaining the limited equations of the conical ones. For the quadrics of the centered cylindrical type, for example, the limited form is $$$\displaystyle \lambda_1x^2+\lambda_2y^2+\frac{D_3}{d_2}=0$$$ and for those of the parabolic cylindrical type, the limited form is $$$\displaystyle \lambda_1x^2-2y\sqrt{\frac{-D_3}{d_1}}=0$$$

Finally, the limited form of a pair of parallel straight lines is $$$\displaystyle \lambda_1x^2+\frac{D_2}{d_1}=0$$$

For completeness, it is also possible to calculate the volume of a real ellipsoid Euclidean by means of the unvariants. Its formula is $$$\displaystyle A=\frac{4}{3}\pi \sqrt{\frac{D_4^3}{d_3^3}}$$$

Considering $$$x^2+y^2+2xz+6z-2=0$$$ classify the quadric by means of the euclidean invariants.

The matrix associated with the quadric is $$$\overline{A} = \begin{bmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 3 \\ 0 & 0 & 3 & -2 \end{bmatrix}$$$ As soon as the matrix is obtained, we are going to calculate the euclidean invariants. To do so, we are going to consider the following determinants: $$$det(x \cdot I-\overline{A})=x^4-13x^2+19x-7 \\ det (x \cdot I - A)=x^3-2x^2+1$$$ In view of two determinants, the euclidean invariants are the following ones: $$$\left \{ \begin{array}{l} D_1=0 \\ D_2=-13 \\ D_3=-19 \\ D_4=-7\end{array} \right.$$$ and $$$\left\{ \begin{array}{l} d_1=2 \\ d_2=0 \\ d_3=-1 \end{array} \right.$$$ Once obtained, we just need to calculate its index. As $$d_1d_3 < 0$$, the index $$1$$. Therefore, for the classification scheme, we have:

$$D_4 < 0, d_3\neq 0$$ and $$J=1$$.

Therefore, this is an elliptical hiperboloide.

Considering the quadric $$$q(x,y,z)=x^2+4xy+2xz+4y^2+4yz+z^2+2x=0$$$ we are going to classify it by means of the euclidean invariants.

We calculate the matrix associated with the quadric and then the characteristical polynomials associated with the matrix and the main matrix: $$$\overline{A} = \begin{bmatrix} 1 & 2 & 1 & 1 \\ 2 & 4 & 2 & 0 \\ 1 & 2 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix}$$$ $$$det(\overline{A}-x \cdot I-\overline{A})=x^4-6x^3-x^2+5x \\ det (A-x \cdot I)=-x^3+6x^2$$$ Therefore, the euclidean invariants are: $$$\left\{\begin{array}{l} D_1 = 6\\D_2=-1 \\ D_3=-5 \\ D_4 =0 \end{array} \right.$$$ and $$$\left\{ \begin{array}{l} d_1=6 \\ d_2=0 \\ d_3=0 \end{array}\right.$$$ Therefore, following the previous classification of the euclidean invariants, the quadric is a parabolic cylinder.