20
$\begingroup$

Write $X_N$ for this blow up. Place the N points in 'general position' as needed. Then $X_6$ embeds in $CP^2$ as a smooth cubic surface. (See, eg, Griffiths and Harris.) But there is no other $N$ (except $N=0$) for which $X_N$ embeds in $CP^3$. (Proof: The topology of the blow-up disagrees with that of a smooth surface of degree $d$ in $CP^3$. (Gompf-Stipsisz p. 21.) On the other hand, $X_N$ embeds in $CP^5$ simply because any smooth algebraic surface $X$ so embeds. (Harris, `Algebraic Geometry, a first course', p. 193.)

Embarrassingly, I don't even know the answer for $N=1$ where $X_1$ is the 1st Hirzebruch surface! (I'm betting it does embed.)

Motivation: This question began in an attempt to better understand the 27 lines on the cubic and my initial surprise at how the construction described in GH of $X_6$ yielded a smooth surface in $CP^3$, and how all such surfaces arise through that construction by varying the 6 points. I am hoping answers might help me understand the moduli of blow-ups as I move the N points about the plane, and orient me as a novice to algebraic surfaces.

$\endgroup$

2 Answers 2

26
$\begingroup$

For $N=1$ the answer is yes: the embedding into ${\mathbb P}^4$ is given by the linear system of conics through the blown up point (the image has degree $d=3$). For $N=5$, the system of cubics through the 5 points gives an embedding ($d=4$).

ADDED: here are 2 slightly less obvious examples: For $N=8$ one can take quartics with an assigned double point and 7 simple base points ($d=5$). For $N=10$ take the quintics with 3 assigned double points and 7 simple base points ($d=6$; I did not check all the details here, because it's very boring, but I'm sure that it works).

In general, giving a satisfactory answer to your question seems very hard. There is a numerical equality, the so-called "double point formula" (Hartshorne, "Algebraic geometry", p.434), which is satisfied by all smooth surfaces of ${\mathbb P}^4$: $$d^2-10d-5HK+12\chi-2K^2=0,$$ where $H$ is the hyperplane section, $d=H^2$ is the degree, $K$ is the canonical divisor and $\chi$ the Euler characteristic of ${\mathcal O}_{X_N}$. In our case the formula becomes: $$d^2-10d-5HK+2N-6=0.$$ In addition there is result by G. Ellingsrud and C. Peskine [Invent. Math. 95 (1989), no. 1, 1--11] saying that only finitely many components of the Hilbert scheme of smooth surfaces in ${\mathbb P}^4$ contain smooth rational surfaces. So in principle it should be possible to classify all the smooth rational surfaces in ${\mathbb P}^4$. In practice, it is known that the degree is $\le 76$, [Cook, An improved bound for the degree of smooth surfaces in P4 not of general type. Compositio Math. 102 (1996)] and it is conjectured that $d\le 15$ (examples with $d=15$ do exist). There are also papers by several authors (Ranestad, Schreyer, Popescu, and others) that classify the smooth rational surfaces of ${\mathbb P}^4$ of degree $\le 11$. In these papers you can find examples of the kind you are looking for. For instance there are examples with $d=10$ and $N=18$.

$\endgroup$
11
$\begingroup$

Suppose we have a non-degenerate embedding $f:X\to \mathbb{P}^4$. Let $L = f^\ast \mathcal{O}_{\mathbb{P}^4}(1)$, and let $V \subset H^0(L)$ be the (base-point free) $5$-dimensional linear series giving $f$. There are two cases to consider.

Case 1: $V = H^0(L)$ is complete. I'm afraid here the problem seems quite difficult. There are two essential steps.

First, find all line bundles $L$ with $h^0(L) = 5$. Given general points $p_1,\ldots,p_N$ and multiplicities $m_1,\ldots,m_N$, this amounts to determining the dimension of the series of curves of a given degree $d$ having singularities of multiplicity $m_i$ at $p_i$ for each $i$. The Segre-Gimigliano-Harbourne-Hirschowitz conjecture provides an expected answer, but it is very much open.

Next, for each line bundle with $h^0(L) = 5$, we need to determine when the complete series gives an embedding. This question has also received a lot of attention (at least in more general formulations), but at least these more general versions are still active areas of research. A google search for ampleness of line bundles on blowups turns up many results.

Potentially an argument that tries to find some obstruction to embedding in $\mathbb{P}^4$ could sidestep this program, but being a codimension $2$ subvariety imposes much less structure than being a hypersurface.

Case 2: $V\subset H^0(L)$ is a proper subseries. In this case, choose a $6$-dimensional series $W$ with $V\subset W\subset H^0(L)$. Then $W$ gives a non-degenerate embedding of $X$ in $\mathbb{P}^5$, and the embedding of $X$ in $\mathbb{P}^4$ is the composition of the embedding in $\mathbb{P}^5$ with projection from a point. Since $V$ is base-point free, this projection is from a point lying off of the embedded surface in $\mathbb{P}^5$. Since the projection must be an isomorphism between the two images of $X$, the secant variety of the surface in $\mathbb{P}^5$ must be a proper subvariety of $\mathbb{P}^5$. But Severi showed the only smooth nondegenerate surface in $\mathbb{P}^5$ with deficient secant variety is the Veronese, isomorphic to $\mathbb{P}^2$ embedded by the complete series of conics. Thus this case never actually arises.

$\endgroup$

Your Answer

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

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