Determine the dimension of the set of surfaces of $\mathbb{P}^{3}$ that contain certain conic.

Question

Let $C\subseteq\mathbb{P}^{3}$ be the conic of equations $$ C=V(X_{3}, X_{0}X_{2}-X_{1}^{2})=\{(t_{0}^{2}:t_{0}t_{1}:t_{1}^{2}:0)\in\mathbb{P}^{3}:(t_{0}:t_{1})\in\mathbb{P}^{1}\}. $$ I have to determine the dimension of the set of surfaces $S\subseteq\mathbb{P}^{3}$ of degree $d$ (that we identify with $\mathbb{P}^{\binom{d+3}{3}-1}$) such that $C\subseteq S$.

If $S$ is a surface in $\mathbb{P}^{3}$, we can write $$ S=V\left(\sum_{i_{0}+\cdots +i_{3}=d}a_{i_{0},\ldots,i_{3}}X_{0}^{i_{0}}\cdots X_{3}^{i_{3}}\right) $$ Then, $C\subseteq S$ if and only if $$ \sum_{i_{0}+i_{1}+i_{2}=d}a_{i_{0},i_{1},i_{2},0}t_{0}^{2i_{0}}t_{0}^{i_{1}}t_{1}^{i_{1}} t_{1}^{2i_{2}}=0 $$ for each $(t_{0}:t_{1})\in\mathbb{P}^{1}$. This is equivalent to $$ \sum_{i_{0}+i_{1}+i_{2}=d}a_{i_{0},i_{1},i_{2},0}Y_{0}^{2i_{0}+i_{1}} Y_{1}^{2i_{2}+i_{1}}=0 $$ as a polynomial in $K[Y_{0},Y_{1}]$. According to this, we have certain linear restrictions on the coefficients $a_{i_{0},\ldots,i_{3}}$. How many of them are linearly independent? If we had $r$, then the dimension of the set of surfaces that contain $C$ would be $$ \binom{d+3}{3}-1-r, $$ and we would have finished.

$\textbf{Remark.}$ I have read that the sought dimension is $\binom{d+3}{3}-1-(2d+1)$. Has this something to do with the Hilbert polynomial of $C$ (which is $P_{C}(l)=2l+1$)?

Answer 1

A surface $S\subset \mathbb P^3$ of degree $d$ is given as the vanishing of a homogeneous polynomial $f\in H^0(\mathbb P^3,\mathscr O_{\mathbb P^3}(d))$. For a fixed conic $C\subset \mathbb P^3$, you can see that $C\subset S$ if and only if $f|_C=0$, where $f|_C$ is the image of $f$ under the restriction map $$\rho_C:H^0(\mathbb P^3,\mathscr O_{\mathbb P^3}(d))\to H^0(C,\mathscr O_{C}(d)).$$ This is basically what you were correctly computing in your post. The target of $\rho_C$ can be identified, as a vector space, with $H^0(\mathbb P^1,\mathscr O_{\mathbb P^1}(2d))$, which has dimension $2d+1$. The map $\rho_C$ is onto, because $C\subset \mathbb P^3$ is a complete intersection (see Hartshorne Ex. II.8.4 or Ex. III.5.5), so one can easily compute the dimension of its kernel, and the dimension you are after is $$\dim\mathbb P(\ker\rho_C)=\binom{d+3}{3}-(2d+1)-1.$$ So, yes, the relation with the Hilbert polynomial $P_C$ is simply that $P_C(d)=h^0(C,\mathscr O_C(d))$, the dimension of the target of $\rho_C$ for fixed $d$.

Answer 2

Brenin's answer is probably the most geometric one. if you want a more algebraic one, you can proceed like this: let $I\subseteq S$ be the ideal of $C$, where $S=k[X_0,\dots,X_n]$. Then this is a homogeneous ideal and the graded piece $I_d$ is, by definition, the space of surfaces of degree $d$ which contain $C$.

Now, Hilbert's Nullstellensatz tells us $I = \sqrt{(X_3,X_0X_2-X_1^2)}$, but it is easy to show that the ideal $(X_3,X_0X_2-X_1^2)$ is actually prime, so that it is in particular radical. To compute the dimensions of the spaces $I_d$ we can now use the minimal free resolution of $I$: since the polynomials $f_1=X_3$ and $f_2=X_0X_2-X_1^2$ generate $I$, we have a surjective map of graded $S$-modules $$ S(-1)\oplus S(-2) \overset{(f_1,f_2)}{\longrightarrow} I $$ whose kernel is given by the set of $(k_1,k_2)$ such that $$k_1\cdot f_1 = -k_2 \cdot f_2$$ We observe that $f_1$ and $f_2$ are coprime in $S$, so that the above relation implies $k_2=f_1\cdot g$ and then we get $k_1=-f_2\cdot g$. This means that we have an exact sequence $$ S(-3) \overset{\begin{pmatrix} -f_2 \\ f_1 \end{pmatrix}}{\longrightarrow} S(-1)\oplus S(-2) \overset{(f_1,f_2)}{\longrightarrow} I \longrightarrow 0 $$ It is also clear that the map on the left is injective, so that we actually have a short exact sequence $$ 0 \longrightarrow S(-3) \overset{\begin{pmatrix} -f_2 \\ f_1 \end{pmatrix}}{\longrightarrow} S(-1)\oplus S(-2) \overset{(f_1,f_2)}{\longrightarrow} I \longrightarrow 0 $$ which is called the minimal free resolution of $I$. Since this is a short exact sequence of graded modules, we get $$ \dim I_d = \dim [ S(-1)\oplus S(-2) ]_d - \dim S(-3)_d = \dim S_{d-1} + \dim S_{d-2} - \dim S_{d-3} $$ and this we can compute easily.