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According to the book of Harris, Algebraic geometry, A first course, page $9$, the twisted cubic is defined to be the image $C$ of the map $ v : \mathbb{P}^1 \to \mathbb{P}^3 $ given by $$ v : [X_0 , X_1 ] \to [ X_{0}^{3} , X_{0}^{2} X_{1} , X_{0} X_{1}^{2} , X_{1}^{3} ].$$

The author of this book says then, that $C$ lies on the $3$ quadric surfaces $ Q_0 $ , $ Q_1 $ and $ Q_2 $ given as the zero locus of the polynomials $ F_0 (Z) = Z_0 Z_2 - Z_{1}^2 $ , $ F_1 (Z) = Z_0 Z_3 - Z_{1} Z_2 $ and $ F_2 (Z) = Z_0 Z_3 - Z_{2}^2 $.

Could you explain to me why do we have this? How do we obtain those polynomials in that form?

Thanks in advance for your help.

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One can note that the twisted cubic $C$ is a determinantal variety; that is $C$ is the zero locus defined as: \begin{equation} C=\left\{[z_0:z_1:z_2:z_3]\in\mathbb{P}^3_{\mathbb{C}}\mid rank\begin{pmatrix} z_0 & z_1 & z_2\\ z_1 & z_2 & z_3 \end{pmatrix}=1\right\}=Q_0\cap Q_1\cap Q_2 \end{equation} as you denoted.

Then if: \begin{equation} z_0\neq0,\,\exists t\in\mathbb{C}\mid\begin{cases} z_1=tz_0\\ z_2=tz_1\\ z_3=tz_2 \end{cases}\iff\begin{cases} z_1=tz_0\\ z_2=t^2z_0\\ z_3=t^3z_0 \end{cases} \end{equation} that is: \begin{equation} C\cap U_0=\{[1:t:t^2:t^3]\in\mathbb{P}^3_{\mathbb{C}}\mid t\in\mathbb{C}\} \end{equation} where: \begin{equation} U_0=\left\{[z_0:z_1:z_2:z_3]\in\mathbb{P}^3_{\mathbb{C}}\mid z_0\in\mathbb{C}^{\times}\right\}; \end{equation} in other words, one can state that: \begin{equation} C\cap U_0=\left\{\left[1:\frac{t}{s}:\frac{t^2}{s^2}:\frac{t^3}{s^3}\right]\in\mathbb{P}^3_{\mathbb{C}}\mid t\in\mathbb{C},s\in\mathbb{C}^{\times}\right\}=\left\{\left[s^3:s^2t:st^2:t^3\right]\in\mathbb{P}^3_{\mathbb{C}}\mid t\in\mathbb{C},s\in\mathbb{C}^{\times}\right\} \end{equation} and therefore: \begin{equation} C=\left\{\left[s^3:s^2t:st^2:t^3\right]\in\mathbb{P}^3_{\mathbb{C}}\mid[s:t]\in\mathbb{P}^1_{\mathbb{C}}\right\}=v_{1,3}\left(\mathbb{P}^1_{\mathbb{C}}\right). \end{equation} An analogous reasoning holds for the general Veronese map: \begin{equation} v_{n,d}:[x_0:x_1:\dots:x_n]\in\mathbb{P}^n_{\mathbb{C}}\to\left[\mathbf{x}^d\right]\in\mathbb{P}^N_{\mathbb{C}} \end{equation} where $\displaystyle N=\binom{n+d}{d}-1$ and: \begin{equation} \mathbf{x}^d=\left(x_0^{e_0}\cdot\dots\cdot x_n^{e_n}\right)\mid\sum_{k=0}^ne_k=d,\,e_k\geq0. \end{equation}

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  • $\begingroup$ Thank you very much Sir. :-) $\endgroup$ – Lina45 Mar 10 '16 at 18:25
  • $\begingroup$ Not at all, lady! ;) (In Italy, Lina is a woman name.) And thank you for accepted my answer. :* $\endgroup$ – Armando j18eos Mar 10 '16 at 18:28
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    $\begingroup$ The feminine of Sir is Madam, not lady. Actually, calling a woman "lady" is as rude (or at least as extremely informal) as calling a man "mister" . $\endgroup$ – Georges Elencwajg Mar 11 '16 at 22:20

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