# Invariants of the quadrics and Euclidean classification

## 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 $$q(x,y,z)$$.

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 $$Q$$.

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.