Three Conics Determine a Cubic

Abstract

Given a cubic K in the real projective plane. Then for each point P there is a conic $C_{{P}}$ associated to P. The conic $C_{{P}}$ is called the polar conic of K with respect to the pole P. We investigate the situation when three conics $C_1$, $C_2$, and $C_3$ are polar conics of K with respect to the poles $P_1$, $P_2$, and $P_3$, respectively. In particular, we give an elementary proof—without using any results from algebraic geometry—that any three conics $C_1$, $C_2$, $C_3$ in general position, satisfying only a non-degeneracy condition, determine a unique cubic K and three points $P_1$, $P_2$, $P_3$, such that $C_1$, $C_2$, $C_3$ are polar conics of K with respect to the three poles $P_1$, $P_2$, $P_3$. This can be seen as a higher degree variant of von Staudt’s Theorem.

1 Introduction

This work proceeds the article [3], in which it is shown that two given conics $C_0$ and $C_1$ can always be considered as polar conics of a cubic K curve with respect to corresponding poles $P_0$ and $P_1$. However, even though $P_1$ is determined by $P_0$, neither the cubic nor the point $P_0$ is determined by the two conics $C_0$ and $C_1$. This changes if we start with three conics $C_1$, $C_2$, $C_3$ in general position. In this situation, there is a unique cubic K and uniquely determined points $P_1$, $P_2$, $P_3$ such that $C_1$, $C_2$, $C_3$ are the polar conics of K with respect to the three poles $P_1$, $P_2$, $P_3$. Instead of formulating the result in the abstract language of algebraic geometry, we propose an elementary and explicit approach that shows a concrete method to calculate the resulting cubic curve K and the poles $P_1$, $P_2$, $P_3$, starting from the three given conic sections $C_1$, $C_2$, $C_3$. In particular, the condition for uniqueness and existence becomes visible in this way.

Our result can be seen as a higher degree variant of von Staudt’s Theorem which says that given three lines $\ell _1$, $\ell _2$, $\ell _3$ and three points $P_1$, $P_2$, $P_3$ in perspective position determine a unique conic C such that the points $P_i$ are the poles of the lines $\ell _i$ with respect to C (see [7, p. 135, Sect. 241]).

The setting in which we work is the same as in [3], but for the sake of completeness we recall the notation and terminology. We will work in the real projective plane $\mathbb {R}\mathbb {P}^2= \mathbb {R}^3 {\setminus } \{0\}/\sim$, where $X\sim Y\in \mathbb {R}^3 {\setminus } \{0\}$ are equivalent, if $X=\lambda Y$ for some $\lambda \in \mathbb {R}$. Points $X=(x_1,x_2,x_3)^T\in \mathbb {R}^3 \setminus \{0\}$ will be denoted by capital letters, the components with the corresponding small letter, and the equivalence class by [X]. However, since we mostly work with representatives, we often omit the square brackets in the notation. A non-degenerate conic in this setting is then given by an equation of the form $\langle X,AX\rangle =0$, where A is a regular, real, symmetric $3\times 3$-matrix with mixed signature, i.e., A has eigenvalues of both signs, and $\langle \cdot ,\cdot \rangle$ denotes the standard inner product of $\mathbb {R}^3$.

Let f be a non-constant homogeneous polynomial in the variables $x_1,x_2,x_3$ of degree n. Then f defines a projective algebraic curve

$$\begin{aligned} C_{f}:= \{[X] \in \mathbb {R}\mathbb {P}^2:f(X)=0\}, \end{aligned}$$

of degree n. For a point $P\in \mathbb {R}\mathbb {P}^2$,

$$\begin{aligned} Pf(X):=\langle P,\nabla f(X)\rangle , \end{aligned}$$

is also a homogeneous polynomial in the variables $x_1,x_2,x_3$. If the homogeneous polynomial f is of degree n, then $C_{Pf}$ is an algebraic curve of degree $n-1$. The curve $C_{Pf}$ is called the polar curve of $C_{{f}}$ with respect to the pole P; sometimes we call it the polar curve of P with respect to $C_{{f}}$. In particular, when $C_{{f}}$ is a cubic curve (i.e., f is a homogeneous polynomial of degree 3), then $C_{Pf}$ is a conic, which we call the polar conic of $C_{{f}}$ with respect to the pole P, and when $C_{{f}}$ is a conic, then $C_{Pf}$ is a line, which we call the polar line of $C_{{f}}$ with respect to the pole P (see, for example, the classical book of Wieleitner [8] or Dolgachev [2, Chap. 3] for a modern view). Note that $C_{Pf}$ is defined and can be a regular curve even if $C_{{f}}$ is singular or reducible. For some historical background, for the geometric interpretation of poles and polar lines, for the iterated construction of polar curves, as well as for the analytical method used today, see Monge [5, Sect. 3], Bobillier [1], and Joachimsthal [4, p. 373], or [3].

2 Algebraic Curves and Multilinear Forms

Let $C_{{f}}$ be a conic given by the non-constant homogeneous polynomial

$$\begin{aligned} f(x_1,x_2,x_3):=\sum \limits _{1\le i\le j\le 3} c_{ij}\hspace{1.5pt}x_i\hspace{1pt}x_j. \end{aligned}$$

Then, the symmetric matrix

has the property that a point X belongs to $C_{{f}}$ (i.e., $f(X)=0$), if and only if $\langle X,T(X)\rangle =0$. Thus, the conic $C_{{f}}$ is represented by the matrix T. Since the expression $\langle X,T(Y)\rangle$ defines a bilinear form $\mathbb {R}^3\times \mathbb {R}^3\rightarrow \mathbb {R}$, $(X,Y)\mapsto \langle X,T(Y)\rangle$, we can consider the matrix T also as a purely covariant tensor of rank 2 (i.e., a tensor whose rank of covariance is 2 and whose rank of contravariance is 0). More precisely, if we consider the matrix T as a (0, 2)-tensor, where for $X=(x_1,x_2,x_3)$ and $Y=(y_1,y_2,y_3)$ we define

$$\begin{aligned} T(X,Y):= & {} \sum \limits _{1\le i,j\le 3} a_{ij}\hspace{1.5pt}x_i\hspace{1pt}y_j, \end{aligned}$$

then the expression $\langle X,T(X)\rangle =0$ is equivalent to $T(X,X)=0$. In order to obtain the coefficients of the (0, 2)-tensor $T=(a_{ij})_{1\le i,j\le 3}$ from a conic $C_{{f}}$ defined by a non-constant homogeneous polynomial f, we just set

$$\begin{aligned} a_{ij}:=\frac{1}{2!}\cdot \frac{\partial ^2 f}{\partial x_i\partial x_j}\quad \text {for all }1\le i,j\le 3. \end{aligned}$$

The next result shows that this relation between a conic $C_{{f}}$ and the corresponding (0, 2)-tensor $T_{{f}}=(a_{ij})_{1\le i,j\le 3}$ can be generalised to algebraic curves of arbitrary degree.

Lemma 2.1

Let $\Gamma _{{f}}$ be an algebraic curve of degree d given by the non-constant homogeneous polynomial

$$\begin{aligned} f(x_1,x_2,x_3):=\sum \limits _{1\le i_1\le \cdots \le i_d\le 3} c_{i_1\ldots i_d}\cdot \hspace{1.5pt}x_{i_1}\cdot \ldots \cdot x_{i_d}, \end{aligned}$$

and let $T_{{f}}=(a_{i_1\ldots i_d})_{1\le i_1,\ldots , i_d\le 3}$, where

$$\begin{aligned} a_{i_1\ldots i_d}:=\frac{1}{d!}\cdot \frac{\partial ^d f}{\partial x_{i_1}\ldots \partial x_{i_d}}\quad \text {for all }1\le i_1,\ldots , i_d\le 3. \end{aligned}$$

Then $T_{{f}}$ is a symmetric (0, d)-tensor and a point X is on the curve $\Gamma _{{f}}$ if and only if

$$\begin{aligned} T_{{f}}(\underset{d\mathrm {-times}}{\underbrace{X,\ldots ,X}})=0. \end{aligned}$$

Proof

Since for every rearrangement $\pi$ of the sequence $\langle i_1,\ldots ,i_d\rangle$ we have

$$\begin{aligned} \frac{\partial ^d f}{\partial x_{i_1} \ldots \partial x_{i_d}}\;=\; \frac{\partial ^d f}{\partial x_{\pi (i_1)} \ldots \partial x_{\pi (i_d)}}\quad \text {and therefore}\quad a_{i_1\ldots i_d}=a_{\pi (i_1)\ldots \pi (i_d)}, \end{aligned}$$

we get that the tensor $T_{{f}}$ is symmetric. Furthermore, assume that the monomial $c_{n_1\hspace{.5pt}n_2\hspace{.5pt}n_3} \cdot x_1^{n_1}\cdot x_2^{n_2}\cdot x_3^{n_3}$ appears in f. Then $n_1+n_2+n_3=d$ and

$$\begin{aligned} \frac{1}{d!}\cdot \frac{\partial ^d ( c_{n_1\hspace{.5pt}n_2\hspace{1pt}n_3} \cdot x_1^{n_1}\cdot x_2^{n_2}\cdot x_3^{n_3})}{\partial x_1^{n_1} \partial x_2^{n_2}\partial x_3^{n_3}}\;=\; \frac{n_1!\cdot n_2!\cdot n_3!}{d!}\cdot c_{n_1\hspace{.5pt}n_2\hspace{1pt}n_3}. \end{aligned}$$

Now, it is easy to see that the number of coefficients $a_{i_1\ldots i_d}$ such that for $1\le i\le 3$ the number i appears $n_i$-times in the sequence $\langle i_1,\ldots ,i_d\rangle$ is given by the trinomial coefficient

$$\begin{aligned} \left( {\begin{array}{c}d\\ n_1,n_2,n_3\end{array}}\right) =\frac{d!}{n_1!\cdot n_2!\cdot n_3!}. \end{aligned}$$

This shows that for any point X we have $T_{{f}}(X,\ldots ,X)=0$ if and only if $f(X)=0$, or in other words, X is on the curve $\Gamma$. $\square$

Let us turn our attention now to polar curves. For this, we consider first polar curves of conics $C_{{f}}$ with corresponding (0, 2)-tensor $T_{{f}}=(a_{ij})_{1\le i,j\le 3}$. Above we have seen that for a given point $P\in \mathbb {R}\mathbb {P}^2$, a point X is on the polar curve $C_{Pf(X)}$ of $C_{{f}}$ with respect to the pole P if and only if

$$\begin{aligned} Pf(X):=\langle P,\nabla f(X)\rangle =0. \end{aligned}$$

Now, for $P,X\in \mathbb {R}\mathbb {P}^2$, a short calculation shows that $Pf(X)=2\cdot T_{{f}}(P,X)$, and hence, we get

$$\begin{aligned} Pf(X)=0 \iff T_{{f}}(P,X)=0. \end{aligned}$$

Since $T_{{f}}$ is symmetric, we have $T_{{f}}(P,X)=T_{{f}}(X,P)$, which shows that if X is a point on the polar curve of $C_{{f}}$ with respect to the pole P, then P is a point on the polar curve of $C_{{f}}$ with respect to the pole X. The next result shows that also this result can be generalised to algebraic curves of arbitrary degree.

Lemma 2.2

Let $\Gamma _{{f}}$ be an algebraic curve of degree d given by the non-constant homogeneous polynomial f, let $T_{{f}}$ be the corresponding symmetric (0, d)-tensor, and let $P\in \mathbb {R}\mathbb {P}^2$ be a point. Then

$$\begin{aligned} Pf(X)=0\iff T_{{f}}(P, \underset{(d-1)\mathrm {-times}}{\underbrace{X,\ldots ,X}})=0. \end{aligned}$$

In particular, a point $X\in \mathbb {R}\mathbb {P}^2$ is on the polar curve of $\Gamma _{{f}}$ with respect to the pole P if and only if $T_{{f}}(P,X,\ldots ,X)=0$.

Proof

Notice first that for $P=(p_1,p_2,p_3)$ and $X=(x_1,x_2,x_3)$ we have:

$$\begin{aligned} T_{{f}}(P,X,\ldots ,X)= & {} \sum _{j=1}^3 p_j\cdot \Biggl (\sum _{1\le i_2,\ldots ,i_d\le 3} a_{j\hspace{1.5pt}i_2\ldots i_d}\cdot x_{i_2} \cdot \ldots \cdot x_{i_d}\Biggr )\\= & {} \sum _{j=1}^3 \sum _{1\le i_2,\ldots ,i_d\le 3} a_{j\hspace{1.5pt}i_2\ldots i_d}\cdot p_j \cdot x_{i_2}\cdot \ldots \cdot x_{i_d}. \end{aligned}$$

Now, assume again that the monomial $c_{n_1\hspace{.5pt}n_2\hspace{.5pt}n_3} \cdot x_1^{n_1}\cdot x_2^{n_2}\cdot x_3^{n_3}$ appears in f. Then, for each $1\le j\le 3$ we have

$$\begin{aligned} \frac{\partial ( c_{n_1\hspace{.5pt}n_2\hspace{1pt}n_3} \cdot x_1^{n_1}\cdot x_2^{n_2}\cdot x_3^{n_3})}{\partial x_j}\;=\; n_j\cdot c_{n_1\hspace{.5pt}n_2\hspace{1pt}n_3} \cdot x_1^{n_1'}\cdot x_2^{n_2'}\cdot x_3^{n_3'}, \end{aligned}$$

where $n_j'=n_j-1$ and $n_i'=n_i$ for $i\ne j$. Without loss of generality we assume that $j=1$ and $n_1\ge 1$. Now, it is easy to see that the number of coefficients $a_{1\hspace{1pt}i_2\ldots i_d}$ such that for $1\le i\le 3$, the number i appears $n_i$-times in the sequence $\langle 1,\ldots ,i_d\rangle$ is given by the trinomial coefficient

$$\begin{aligned} \left( {\begin{array}{c}d-1\\ n_1-1,n_2,n_3\end{array}}\right) =\frac{(d-1)!}{(n_1-1)!\cdot n_2!\cdot n_3!}\;=\; \frac{n_1}{d}\cdot \frac{d!}{n_1!\cdot n_2!\cdot n_3!}. \end{aligned}$$

This shows that for any points $P,X\in \mathbb {R}\mathbb {P}^2$ we have

$$\begin{aligned} d\cdot T_{{f}}(P,X,\ldots ,X)=\bigl \langle P,\nabla f(X)\bigr \rangle , \end{aligned}$$

in particular, we get

$$\begin{aligned} Pf(X)=0 \iff T_{{f}}(P,X,\ldots ,X)=0. \end{aligned}$$

$\square$

It is obvious how the iterated construction of polar curves is carried out: If, for example, $P,Q,R\in \mathbb {R}\mathbb {P}^2$ are given and $\Gamma _{{f}}$ is an algebraic curve of degree $d\ge 3$, then the polar curve of the polar curve of the polar curve of $\Gamma _{{f}}$ with respect to the points P, Q, R, respectively, is given by the zeros of the $(0,d-3)$-tensor $T_f(P,Q,R,X,\ldots ,X)$. Notice that since $T_f$ is symmetric, the order of P, Q, R is irrelevant. As a consequence, we obtain the following

Fact 2.3

Let K be a cubic curve, let $P_1,P_2,P_3\in \mathbb {R}\mathbb {P}^2$, and for $1\le j\le 3$ let $T_j$ be the (0, 2)-tensor of the polar conic of K with respect to the point $P_j$. Then for $1\le j_1,j_2\le 3$ we have

$$\begin{aligned} T_{j_1}(P_{j_2},X)=0\iff T_{j_2}(P_{j_1},X)=0, \end{aligned}$$

in particular, if we consider the tensors $T_j$ as $3\times 3$-matrices, we obtain that

$$\begin{aligned}{}[P_{j_1}]=[(T_{j_2}^{-1}\cdot T_{j_1})\,P_{j_2}]. \end{aligned}$$

The question that we want to treat below, is embedded in a more general problem, namely the study of the relation of a hypersurface and its Hessian variety. In a recent work Sendra-Arranz [6] investigated the Hessian correspondence for the cases of hypersurfaces of degree 3 and 4 in an n-dimensional projective space. In particular, he showed that for degree 3 and dimension $n = 1$, the Hessian correspondence is two to one, and that for degree 3 and $n \ge 2$, and for degree 4, it is birational (see [6, Sects. 2.3, 2.4]). In particular, by introducing the variety of k-gradients as the variety of k-planes containing all the first order derivatives of a polynomial, he obtains algorithms which allow to reconstruct a hypersurface of degree 3 from its Hessian variety in the cases $n\ge 1$, and for degree 4 if n is even. More specifically, Sendra-Arranz proves in his Proposition 2.18 that for $n \ge 2$ a cubic can be recovered by the pencil spanned by its polars. Our main result in Theorem 2.4 is less general, but provides more specific information about the special case of degree 3 in 2 dimensions. Namely, what we show is that three conics in general position (i.e., three points of the Hessian variety) determine a unique cubic. More precisely, given three different conics $C_1,C_2,C_3$ which satisfy a non-degeneracy condition, we show how to construct the unique cubic K such that for three points $P_1,P_2,P_3\in \mathbb {R}\mathbb {P}^2$ determined by the three conics, the conic $C_j$ (for $1\le j\le 3$) is the polar conic of K with respect to the pole $P_j$. The construction we provide in the next section proves our main result, Theorem 2.4.

Theorem 2.4

Let $C_1,C_2,C_3$ be three non-degenerate conics and let $T_1,T_2,T_3$ be the corresponding (0, 2)-tensors given by $3\times 3$-matrices. Assume that the matrices $T_1,T_2,T_3$ satisfy the following condition:

(C) For all $P\in \ker \bigl (T_3\,T_1^{-1}\,T_2- T_2\,T_1^{-1}\,T_3\bigr )$, we have $\det \bigl (T_1 P,\;T_2 P,\;T_3 P\bigr )\ne 0$.

Then there are exactly three points $P_1,P_2,P_3$, determined by the conics $C_1,C_2,C_3$, and a unique cubic curve K, such that for $1\le j\le 3$, $C_j$ is the polar conic of K with respect to the pole $P_j$. The cubic K only depends on the two-dimensional pencil

$$\begin{aligned} {\mathcal {P}}=\{\lambda _1C_1 +\lambda _2 C_2+\lambda _3 C_3:\, (\lambda _1,\lambda _2,\lambda _3)\in \mathbb {R}^3\setminus (0,0,0)\}, \end{aligned}$$

generated by $C_1,C_2,C_3$: If $C_1,C_2,C_3$ are replaced by any other conics ${\tilde{C}}_1,{\tilde{C}}_2,{\tilde{C}}_3$ in ${\mathcal {P}}$ satisfying condition (C), then the same cubic K results.

Remark 1

With respect to condition (C), we would like to mention a few facts.

• (a) First, condition (C) is symmetric in the three indices: To see this, notice that $P\in \ker \bigl (T_3\,T_1^{-1}\,T_2- T_2\,T_1^{-1}\,T_3\bigr )$ is equivalent to

  $$\begin{aligned} Q=T_1^{-1}\,T_2P\in \ker \bigl (T_1\,T_2^{-1}\,T_3- T_3\,T_2^{-1}\,T_1\bigr ). \end{aligned}$$

  Replacing P in the determinant by the expression $T_2^{-1}\,T_1Q$ yields

  $$\begin{aligned} 0\ne \det (T_1P, \,T_2P, \, T_3P)= & {} \det (T_1\,T_2^{-1}T_1 Q, \, T_1 Q, \,T_3\,T_2^{-1}\,T_1 Q) \\ {}= & {} \det (T_1\,T_2^{-1}T_1 Q, \,T_1 Q, \,T_1\,T_2^{-1}\,T_3 Q)\\ {}= & {} \det (T_1\, T_2^{-1})\det (T_1Q, \,T_2Q, \,T_3Q). \end{aligned}$$

• (b) Observe also that (C) implies that $T_3T_1^{-1}T_2\ne T_2T_1^{-1}T_3$: Indeed, assume that $T_3T_1^{-1}T_2-T_2T_1^{-1}T_3=0$. Then the kernel of $T_3T_1^{-1}T_2-T_2T_1^{-1}T_3$ is $\mathbb {R}^3$. However, for $P=(x_1,x_2,x_3)$, $\det (T_1P, \,T_2P, \, T_3P)=0$ is a homogeneous cubic polynomial in the three variables $x_1,x_2,x_3$, which always has non-trivial solutions.

• (c) Consider the following example:

  $$\begin{aligned} T_1=\begin{pmatrix} 1&{}0&{}3\\ 0&{}2&{}0\\ 3&{}0&{}- 1 \end{pmatrix}\qquad T_2=\begin{pmatrix} 1&{}0&{}0\\ 0&{}1&{}0\\ 0&{}0&{}- 1 \end{pmatrix}\qquad T_3=\begin{pmatrix} 1&{}0&{}1\\ 0&{}1&{}0\\ 1&{}0&{}- 1 \end{pmatrix}. \end{aligned}$$

  Notice that $T_1$ does not belong to the pencil spanned by $T_2$ and $T_3$. Here, we have that $T_3T_1^{-1}T_2-T_2T_1^{-1}T_3=0$ and hence the kernel of $T_3T_1^{-1}T_2-T_2T_1^{-1}T_3$ is $\mathbb {R}^3$. But $\det (T_1P, \,T_2P, \, T_3P)=0$ whenever the second coordinate of P is 0. So, the example shows that condition (C) can be violated even in the case when the pencil of $T_1,T_2,T_3$ is two-dimensional. On the other hand, it is easy to see that condition (C) implies that the pencil of $T_1,T_2,T_3$ is two-dimensional.

3 Constructing a Cubic from Three Conics

Let $C_1,C_2,C_3$ be three non-degenerate conics and let $T_1,T_2,T_3$ be the corresponding (0, 2)-tensors given by $3 \times 3$-matrices matrices $T_1,T_2,T_3$ which satisfy condition (C) of Theorem 2.4.

Example

Let $C_1,C_2,C_3$ be given by the following three non-constant homogeneous polynomials $f_1,f_2,f_3$, respectively:

$$\begin{aligned} f_1(X)= & {} x_1^2 + x_2^2 + 4 x_1 x_3, \\[.6ex] f_2(X)= & {} 2 x_1^2 + 2 x_1 x_2 + 2 x_2^2 + 6 x_1 x_3 + 6 x_2 x_3,\\ f_3(X)= & {} x_1^2 + 6 x_1 x_2 + x_2^2 + 2 x_1 x_3 - 6 x_2 x_3. \end{aligned}$$

Figure 1 shows these three conics. Notice that all three conics meet in the origin, which is not excluded by the condition (C), as we will see below. Notice also that one of the conics is a circle, which is not a restriction since we can transform any conic by a projective transformation into a circle.

Fig. 1

The three conics $C_1$, $C_2$, and $C_3$ of the example

Then the corresponding matrices are:

$$\begin{aligned} T_1=\begin{pmatrix} 1&{}0&{}2\\ 0&{}1&{}0\\ 2&{}0&{}0 \end{pmatrix}\qquad T_2=\begin{pmatrix} 2&{}1&{}3\\ 1&{}2&{}3\\ 3&{}3&{}0 \end{pmatrix}\qquad T_3=\begin{pmatrix} 1&{}3&{}1\\ 3&{}1&{}-3\\ 1&{}-3&{}0 \end{pmatrix}. \end{aligned}$$

It is easy to verify that the matrices $T_1,T_2,T_3$ satisfy condition (C): Observe that $\ker \bigl (T_3\,T_1^{-1}\,T_2- T_2\,T_1^{-1}\,T_3\bigr )=[P]$ for $P=\big (\tfrac{6}{5}, -\tfrac{24}{5}, 1\big )$.

Let us turn back to our general construction and construct the three points $P_1,P_2,P_3$: By Fact 2.3, the points $P_1,P_2,P_3$ satisfy the following three necessary conditions

$$\begin{aligned} T_2\,P_1=T_1\,P_2,\qquad T_3\,P_2=T_2\,P_3,\qquad T_1\,P_3=T_3\,P_1, \end{aligned}$$

which is equivalent to

$$\begin{aligned} (T_1^{-1}\,T_2)P_1=P_2,\qquad (T_2^{-1}\,T_3)P_2=P_3,\qquad (T_3^{-1}\,T_1)P_3=P_1, \end{aligned}$$

and implies that $P_1$ satisfies

$$\begin{aligned} (T_3^{-1}\,T_1)(T_2^{-1}\,T_3)(T_1^{-1}\,T_2)P_1=P_1. \end{aligned}$$

(1)

Since the matrices $T_j$ are symmetric, for $M:=T_3\,T_1^{-1}\,T_2$ we have $M^T=T_2\,T_1^{-1}\,T_3$. Therefore, Eq. (1) is equivalent to $MP_1=M^TP_1$, which is equivalent to $(M-M^T)P_1=0$. Now, condition (C) ensures that $M\ne M^T$ (see Remark 1(b)). Since $(M-M^T)$ is a non-zero, real, anti-symmetric $3\times 3$-matrix, it has exactly one eigenvalue equal to zero. In fact, if

$$\begin{aligned} A=\begin{pmatrix} 0&{}a&{}b\\ -a&{}0&{}c\\ -b&{}-c&{}0 \end{pmatrix}, \end{aligned}$$

is an anti-symmetric matrix, then the eigenvalues of A are 0 and $\pm i\sqrt{a^2+b^2+c^2}$ and an eigenvector to the eigenvalue 0 is $(c,-b,a)^T$.

Hence, the pole $P_1$ is uniquely determined by Eq. (1), and we obtain $P_2=(T_1^{-1}\,T_2)P_1$ and $P_3=(T_1^{-1}\,T_3)P_1$. Before we proceed, let us compute the points $P_1,P_2,P_3$ in our example.

Example

With respect to $T_1,T_2,T_3$ we get $P_1=\bigg (\tfrac{6}{5}, -\tfrac{24}{5}, 1\bigg )$, $P_2=\bigg (-\tfrac{27}{5}, -\tfrac{27}{5}, 3\bigg )$, and $P_3=(\tfrac{39}{5}, -\tfrac{21}{5}, -10)$, which correspond to the affine points ${\bar{P}}_1=\bigg (\tfrac{6}{5}, -\tfrac{24}{5}\bigg )$, $\bar{P}_2=\bigg (-\tfrac{27}{15}, -\tfrac{27}{15}\bigg )$, and $\bar{P}_3=\bigg (-\tfrac{39}{50}, \tfrac{21}{50}\bigg )$, respectively. Figure 2 shows the conics with their poles.

Fig. 2

The three conics $C_1,C_2,C_3$ of the example with the three poles $P_1,P_2,P_3$

The goal of our construction is to find a (0, 3)-tensor $T_K$ of a cubic K, such that we have

$$\begin{aligned} T_K(P_j,X,X)=T_j(X,X)\quad \text {for } 1\le j\le 3. \end{aligned}$$

Since by condition (C), the points $P_1,P_2,P_3$ are not incident with a projective line, we may choose $\{P_1,P_2,P_3\}$ as a new basis. In other words, for $\tilde{P}_1=(1,0,0)$, $\tilde{P}_2=(0,1,0)$, and $\tilde{P}_3=(0,0,1)$, we map $P_i\mapsto \tilde{P}_i$ (for $1\le i\le 3$). For $1\le i\le 3$, let $T_i=(a_{jk}^i)_{1\le j,k\le 3}$ and let $\tilde{T}_i$ be the (0, 2)-tensors (i.e., the conics $\tilde{C}_i$) in this new basis. Since for any $1\le i,j,k\le 3$ we have $T_i(P_j,P_k)=T_i(P_k,P_j)=T_j(P_k,P_i)$, we also have

$$\begin{aligned} \tilde{T}_i(\tilde{P}_j,\tilde{P}_k)= \tilde{T}_i(\tilde{P}_k,\tilde{P}_j)= \tilde{T}_j(\tilde{P}_k,\tilde{P}_i). \end{aligned}$$

(2)

Now, let $T_{{\tilde{K}}}=(\tilde{a}_{ijk})_{1\le i,j,k\le 3}$ be a (0, 3)-tensor defined by stipulating

$$\begin{aligned} \tilde{a}_{ijk}:= \tilde{T}_i(\tilde{P}_j,\tilde{P}_k)\quad \text {for }1\le i,j,k\le 3. \end{aligned}$$

Then, by Eq. (2), the tensor $T_{{\tilde{K}}}$ is symmetric and has the property that for $1\le i\le 3$,

$$\begin{aligned} T_{{\tilde{K}}}(\tilde{P}_i,X,X)=\tilde{T}_i(X,X). \end{aligned}$$

For the corresponding cubic $\tilde{K}$ we therefore have that $\tilde{C}_i$ is the polar conic of $\tilde{K}$ with respect to the pole $\tilde{P}_i$.

Since every point $\tilde{Q}=(q_1,q_2,q_3)\in \mathbb {R}\mathbb {P}^2$ can be written as ${\tilde{Q}}=q_1P_1+q_2P_2+q_3P_3$, we have

$$\begin{aligned} T_{{\tilde{K}}}(\tilde{Q},X,X)= & {} q_1\,T_{{\tilde{K}}}(\tilde{P}_1,X,X)+ q_2\,T_{{\tilde{K}}}(\tilde{P}_2,X,X)+ q_3\,T_{\tilde{K}}(\tilde{P}_3,X,X)\\= & {} q_1\,\tilde{T}_1(X,X)+q_2\,\tilde{T}_2(X,X)+q_3\,\tilde{T}_3(X,X), \end{aligned}$$

which shows that the polar conic of ${\tilde{K}}$ with respect to the point ${\tilde{Q}}$ belongs to the pencil spanned by the conics $\tilde{T}_1,\tilde{T}_2$ and $\tilde{T}_3$.

Now, the re-transformed cubic K has the property that the conics $C_1,C_2,C_3$ are the polar conics of K with respect to the poles $P_1,P_2,P_3$, respectively. Furthermore, by the observation above, if, for example, the conic $C_3$ is replaced by a conic ${\tilde{C}}_3$ in the two-dimensional pencil of $C_1,C_2,C_3$ such that $C_1,C_2,{\tilde{C}}_3$ satisfy condition (C), then the conics $C_1,C_2$ and ${\tilde{C}}_3$ are the polar conics of K with respect to the poles $P_1,P_2$ and some point Q, where the three points $P_1,P_2,Q$ are not collinear.

Example

In our example, $\tilde{K}$ in the affine plane is given by

$$\begin{aligned}{} & {} -2192-2919 x+264 x^2+122 x^3-1557 y+3384 x y+198 x^2 y+3726 y^2\\{} & {} \quad -81 x y^2-81 y^3=0, \end{aligned}$$

and finally, the sought cubic K is

$$\begin{aligned} -13 x^3 -66 x^2 y -27 x^2 -216 x y -39 x y^2 -27 y^2-22 y^3 = 0. \end{aligned}$$

Figure 3 shows the cubic K together with the three polar conics $C_i$ with respect to their three poles $P_i$. Recall that the lines connecting $P_i$ and the points of intersection of K with the polar curve $C_i$ are tangent to K.

Fig. 3

The cubic K together with the three poles $P_1,P_2,P_3$ and the three polar conics $C_1,C_2,C_3$ of the example. The tangents from $P_1$ to K are also displayed

Remark 2

We close this discussion by considering the situation when condition (C) is violated for three given conics $C_1$, $C_2$, $C_3$. Suppose that K is a cubic such that $C_j$ is the polar conic with respect to some pole $P_j$ for $j=1,2,3$. Then, $\det (T_1 P_1,\, T_2P_1,\, T_3P_1) = 0$ in condition (C) for $P_1\in \ker (T_3\,T_1^{-1}\,T_2- T_2\,T_1^{-1}\,T_3)$ means that the polar lines $g_1=T_1P_1$, $g_2=T_2P_1=T_1P_2$, $g_3=T_3P_1=T_1P_3$ of the conics $C_1$, $C_2$, $C_3$ with respect to the poles $P_1$, $P_2$, $P_3$ are concurrent, which in turn means that $P_1$, $P_2$, $P_3$ are collinear. Hence, $C_1$, $C_2$, $C_3$ are identical or span only a one-dimensional pencil. This shows that for the three conics in Remark 1(c), there is no cubic K with the property that $C_1$, $C_2$, $C_3$ are conic sections with respect to three poles. This means that (C) is a necessary condition in Theorem 2.4. On the other hand, if condition (C) is violated and $C_1$, $C_2$, $C_3$ span only a one-dimensional pencil, then a cubic K with the required properties exists, but this cubic is no longer unique: Just take an arbitrary conic ${\tilde{C}}_3$ such that $C_1$, $C_2$, ${\tilde{C}}_3$ satisfy condition (C) and apply Theorem 2.4 in order to obtain a cubic ${\tilde{K}}$ with respect to $C_1$, $C_2$ and $\tilde{C}_3$. Then there is a point $P_3$ on the line through $P_1,P_2$ and such that the polar conic of ${\tilde{K}}$ with respect to $P_3$ is $C_3$.