1
$\begingroup$

I was computing some GIT quotients and came up with the following question: to compute $\mathrm{Proj}(\mathbb C [f_1,f_2,f_3,f_4,f_5,f_6]/I)$ where $f_i$'s are homogeneous polynomials of same degree and $I$ is the ideal generated by $$\{f_3f_6-f_4f_5, f_1f_5-f_3^2-f_2f_3, f_1f_6-f_3f_4-f_2f_4, f_2f_4-f_3^2-f_1f_3, f_2f_6-f_3f_5-f_1f_5\}.$$

I thought it is isomorphic to $\mathbb P^2$, but here we have only 5 relations and don't know how to proceed from here. Is this isomorphic to a known projective variety ?

$\endgroup$
6
  • $\begingroup$ It is very difficult to identify a variety from equations alone. Tell us where they come from. $\endgroup$ Jun 15, 2016 at 6:11
  • $\begingroup$ $f_i's$ are polynomials in Plucker coordinates. They are from $G_{2,5}$ under the Torus action. $\endgroup$
    – Mathew
    Jun 15, 2016 at 6:15
  • 1
    $\begingroup$ A $4$-plane hyperplane section of the Grassmannian $G_{2,5}$ is typically an elliptic normal curve with degree $5$ and Hilbert polynomial $5t$. You could plug your equations into a computer algebra program and see if this is the correct Hilbert polynomial. $\endgroup$ Jun 15, 2016 at 11:08
  • 1
    $\begingroup$ I just realized that this is a 5-plane hyperplane section, not a 4-plane hyperplane section. Typically that is a quintic del Pezzo surface with Hilbert polynomial $(5/2)t^2 + (5/2)t + 1$. So you can see if Macaulay2 returns that Hilbert polynomial for your ideal. $\endgroup$ Jun 15, 2016 at 12:08
  • $\begingroup$ A good test that helps understanding the structure of the variety is the following. Let $Q_1,\dots,Q_5$ be your five quadratic polynomials in $f_i$. Let $x_1,\dots,x_5$ be five formal variables. Consider the sextic hypersurface $D := \{ \det(\sum x_i Q_i) = 0 \} \subset P^4$ and its singular locus. If you can describe these, it may help understanding the original intersection as well. $\endgroup$
    – Sasha
    Jun 15, 2016 at 20:58

1 Answer 1

4
$\begingroup$

The OP clarified that the surface $S\subset \mathbb{P}^n$ satisfies all of the following properties: (a) $S$ is smooth, (b) $S$ is rational so that $h^1(S,\omega_S)$ is zero, and (c) the Hilbert polynomial of $S$ equals $$ p(t) = 1+ d\frac{(t+1)t}{2},$$ for some integer $d$. Up to replacing $\mathbb{P}^n$ by the span of $S$, also assume that $S$ spans $\mathbb{P}^n$. (In the original question, $d$ equals $5$ and $n\leq 5$.)

Claim. Every surface $S$ in $\mathbb{P}^n$ that satisfies (a), (b) and (c) is abstractly a del Pezzo surface embedded in projective space by a sublinear system of the anticanonical linear system.

Proof of the Claim. By Bertini's theorems, a general hyperplane section $C$ of $S$ is a smooth, connected curve with Hilbert polynomial $p(t)-p(t-1) = dt$. Since the Hilbert polynomial of a smooth curve of degree $d$ and arithmetic genus $g$ equals $dt+1-g$, $C$ is a smooth, connected, genus $1$ curve of degree $d$. By adjunction, we have a short exact sequence of coherent sheaves on $S$, $$ 0 \to \omega_S \to \omega_S(\underline{C}) \to \omega_C \to 0.$$ Since $h^1(S,\omega_S)$ equals $0$, the associated map $H^0(S,\omega_S(\underline{C})) \to H^0(C,\omega_C)$ is surjective. Since $C$ is a smooth, connected, genus $1$ curve, $\omega_C$ has an everywhere nonzero global section. This is the image of a global section of $\omega_S(\underline{C})$ whose zero locus is disjoint from $C$. Since the zero locus of a nonzero global section of an invertible sheaf is an effective Cartier divisor, and since this effective Cartier divisor is disjoint from the ample divisor $\underline{C}$, this effective Cartier divisor is empty. Therefore $\omega_S^\vee$ is isomorphic to the invertible sheaf $\mathcal{O}_S(\underline{C}) = \mathcal{O}_{\mathbb{P}^n}(1)|_S$. Thus $S$ is embedded in projective space by a sublinear system of the anticanonical linear system. Since the anticanonical divisor class is ample, $S$ is abstractly a del Pezzo surface. QED Claim.

In the case of interest to the OP, $d$ equals $5$ and $n\leq 5$. If $n$ equals $5$, then $S$ is embedded by the complete linear system of the anticanonical divisor class. Otherwise $S$ is a linear projection of the anticanonically embedded quintic del Pezzo surface. However, it is straightforward to compute that for an anticanonical quintic del Pezzo surface $S\subset \mathbb{P}^5$, every point of $\mathbb{P}^5$ is contained in a secant line of $S$, cf. Exercise III.3.13, p. 177 of Kollár's book, "Rational Curves on Algebraic Varieties". Thus, a linear projection of an anticanonical quintic del Pezzo is singular. Since the surface $S$ is smooth, $S$ is a quintic del Pezzo surface embedded in $\mathbb{P}^5$ by the complete linear system of the anticanonical divisor class.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.