Let $\mathbb{R} P^3$ be the real projective space of $\mathbb{R}^4$ and Q be a quadric defined as
$$ Q = \left\{\,\left[\,\left(\begin{smallmatrix} x_1 \\ x_2 \\ x_3 \\ x_4\end{smallmatrix}\right)\,\right] \in \mathbb{R}P^3 \mid x_1^2 + x_2^2 -x_3^2 -x_4^2 = 0 \,\right\} $$ The associated symetric bilinear form of the the quadratic form mentioned in the quadric above is non-degenerate. By using the polarisations theorem I arrive at the result $$B(x,y) = x_1y_1 + x_2y_2 - x_3y_3 - x_4y_4 \qquad \forall x,y \in \mathbb{R}^4$$ The polar hyperplane $E \subset \mathbb{R} P^3$ of a point $[\, p\,] \in \mathbb{R} P^3$ is defined as $$ E = \{\,\left[\, x \,\right] \in \mathbb{R}P^3 \mid B(p,x) = 0 \,\} $$ Show that

  • $\mathcal{C} = Q \cap E \neq \emptyset$
  • $\mathcal{C}$ is a non-degenerate conic (conic section: circle, ellipse, parabola, hyperbola)

I have tried a couple of things - here is a GeoGebra plot of mine I'm not allowed to post a picture (<>_<>)

With an image, it is somehow clear that the intersection is surely not empty. Here are the problems

  1. The quadric (even from the projective point of view) is a manifold of codimension 2 and the plane is a 2-dimensional subspace but this is a not enough to conclude the first question.
  2. I have tried to construct a point that lies in the intersection. I could not find anything
  3. To prove that the intersection is a conic, I can equivalently show that $|C| \geq 5$. Since $\mathcal{C}$ is not empty, there is at least one point $[\, r\,] \in \mathcal{C}$ so do $-r$. I failed to find another 3 points.
  4. My idea of how to show, that the conic is non-degenerate is to show, that it does not contain a line. This can be achieved by showing, that no 3 points in $\mathcal{C}$ lie on the same line.

Since I failed at showing 1, 3, I am just out of ideas. Here is the honor mention
For any $[\,x\,] \in \mathcal{C}$ - the line through $[\,p\,], [\,x\,]$ is a tangent of the quadric $Q$ at $[\, x \,]$ and only $[\, x\,]$. What worries me the most is, that none of these ideas involves the bilinear form (except for the last one). What am I missing?

  • $\begingroup$ Are you allowed to use matrix methods? E.g., $Q$ has matrix equation $\mathbf X^TA\mathbf X=0$ with $A=\operatorname{diag}(1,1,-1,-1)$ and you can parameterize $E$ as $\mathbf X = M\mathbf x$ via a $4\times3$ matrix $M$ (effectively, impose a coordinate system on $E$). The intersection is then given by the equation $(M\mathbf x)^TA(M\mathbf x)=\mathbf x^T(M^TAM)\mathbf x=0$, which is the equation of a quadric on $E$. You can then reason about $M^TAM$ to determine degeneracy. $\endgroup$ – amd Apr 25 '19 at 20:54
  • $\begingroup$ Thank you for your answer. Yeah, I can use the matrix method. It is just the diagonalization of the bilinear form, where $X$ is the orthogonal transformation made of eigenvectors but what is this $M$, that I can use to parameterize $E$ in such a way. To begin with $E$ lies in the projective space and $X$ is a real matrix and $x$ is a projective point - isn't it ? In which kind of way does this $M$ related to the bilinear form ? $\endgroup$ – Hirosanza Apr 25 '19 at 21:48
  • $\begingroup$ Choose any three linearly-independent points $\mathbf M_j$ on $E$. Every point on $E$ can then be written as a linear combination $\sum_j x_jM_j$, $M\mathbf x$ in matrix form. (This lifts $\mathbf x\in\mathbb{RP}^2$ to $\mathbb{RP}^3$.) It’s easy to work out that $(M^TAM)_{ij}=B(M_i,M_j)$. $\endgroup$ – amd Apr 25 '19 at 23:26
  • $\begingroup$ BTW, $Q\cap E\ne\emptyset$ doesn’t hold for all nondegenerate quadrics, so you’ll need to use the specific properties of this one to prove it (it’s a hyperboloid of one sheet). $\endgroup$ – amd Apr 25 '19 at 23:30
  • $\begingroup$ It's kinda clear that $Q \cap E$ does not hold for just any quadric - one of the easier cases to imagine is that of the ellipsoid. Thank you, I'll try to work it out. $\endgroup$ – Hirosanza Apr 26 '19 at 8:31

Thank you @amd ! $B$ can be expressed as \begin{align*} B\left(\left[\begin{matrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ \end{matrix}\right], \left[\begin{matrix} y_1 \\ y_2 \\ y_3 \\ y_4 \\ \end{matrix}\right]\right) &= \left[\begin{matrix} y_1 & y_2 & y_3 & y_4 \\ \end{matrix}\right] \left[\begin{matrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & -1& 0\\ 0 & 0 & 0 & -1\\ \end{matrix}\right] \left[\begin{matrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ \end{matrix}\right] = y^T A x \end{align*} Let $[\,p\,] \in \mathbb{R} P^3$ and denote the set $$E = \{\, [\,x\,] \in \mathbb{R} P^3 \mid B(x,p) = 0 \,\} $$ as the polar hyper plane w.r.t $P$. Let $[\,v^{(1)}\,], [\,v^{(2)}\,], [\,v^{(3)}\,] \in E$ be any arbitrary linear independent points. For some $r_1, r_2, r_3 \in \mathbb{R}\setminus\{\,0\,\}$ the representative $x$ of any point $[\,x\,] \in E$ can be characterized as $$ x = r_1 v^{(1)} + r_2 v^{(2)} + r_3v^{(3)} = \left[\begin{matrix} v^{(1)} & v^{(2)} & v^{(3)} \end{matrix}\right] \left[\begin{matrix} r_1 \\ r_2 \\ r_3\end{matrix}\right] = Mr $$ Where $M$ denote the matrix on the left and $r$ the vector on the right. We can write $$x^T A \,x = (Mr)^T A\, (Mr) = r^T (M^T A \,M) \,r$$ where \begin{align*} B^\ast = M^T A \,M = \left[\begin{matrix} B(v^{(1)}, v^{(1)}) & B(v^{(1)}, v^{(2)}) & B(v^{(1)}, v^{(3)}) \\ B(v^{(2)}, v^{(1)}) & B(v^{(2)}, v^{(2)}) & B(v^{(2)}, v^{(3)}) \\ B(v^{(3)}, v^{(1)}) & B(v^{(3)}, v^{(2)}) & B(v^{(3)}, v^{(3)}) \\ \end{matrix}\right] \end{align*} Since $B$ is non-degenerate and $v^{(1,2,3)}$ build a linear independent system, the matrix $B^\ast$ is symmetric with $$\textrm{det}(B^\ast) = \textrm{det}(M^T) \textrm{det}(A) \textrm{det}(M) \neq 0$$ representing a non-degenerate bilinear form in $\mathbb{R}^3$. $r$ can be seen as a representative of a point $[\,r\,] \in \mathbb{R} P^2$, therefore the set $$\mathcal{C} = \{\,[\,r\,] \in \mathbb{R} P^2 \mid r^T B^\ast r = 0 \,\}$$ defines a non-degenerate quadric in $\mathbb{R} P^2$ - a conic, especially $\mathcal{C} \neq \emptyset $


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