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I cannot seem to find stated the following fact, which is surely well known to experts.

Let (S,L) be a polarized K3 surface. Then $M = L^{\otimes 3}$ is very ample and we can consider the embedding in the corresponding projective space $S \to P^N$.

Question: is the image of $S$ in $P^N$ a complete intersection? I suspect the answer is no but I haven't seen it stated explicitly. (by the way, I am certainly allowing $S = H_1 \cap ... \cap H_k \cap D_1 \cap D_2$, where the $H_i$ are hyperplanes and $D_1, D_2$ or more interesting divisors.

(information about other embeddings of S which are complete intersections, or statements about when one can know if S is a complete interesction would be highly appreciated as well)

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Suppose $X$ is a smooth surface complete intersection in $\mathbb{P}^n$, given by equations of degrees $d_1,\ldots ,d_{n-2}\ $ . I assume $d_i\geq 2$, because the image of a variety by a complete linear system cannot be contained in a hyperplane. By the adjunction formula the canonical divisor of $X$ is $(\sum d_i-n-1)$ times the hyperplane section. This is $\,0\,$ if and only if $(d_1,\ldots ,d_{n-2})= (4)$, $(2,3)$ or $(2,2,2)$. Thus the only polarizations for which the image can be a complete intersection are those of degree $4$, $6$ and $8$.

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  • $\begingroup$ thanks, I had done the computation but wasn't aware of the first result you mention. this will really my ignorance about projective geometry: how do you know a priori that the image of K3 cannot be contained in a hyperplane? $\endgroup$ – bananastack Sep 9 '14 at 18:39
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    $\begingroup$ This is completely general fact, nothing to do with K3 surfaces : you are considering a variety $X$, a line bundle $L$ on $X$ and the map $X--> \mathbb{P}(H^0(X,L)^*)$ which associates to $x\in X$ the hyperplane $H_x\subset H^0(X,L)$ of sections of $L$ vanishing at $x$. A hyperplane in this projective space corresponds to a line $\mathbb{C}s$ in $H^0(X,L)$, and saying that it contains $X$ would mean that $s(x)=0$ for all $x$, hence $s=0$, a contradiction. $\endgroup$ – abx Sep 9 '14 at 18:49
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That is not typically true. There are certainly more direct arguments, but you can always see this from Gritsenko-Hulek-Sankaran. If K3 surfaces were complete intersections, the moduli spaces would be unirational. In fact the moduli spaces are (typically) of general type.

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  • $\begingroup$ I see, thanks. Do you know if it's ever reasonable to expect some K3 surface to be a complete intersection in its polarization? Perhaps some criteria involving topological data? Maybe it's just unreasonable. $\endgroup$ – bananastack Sep 9 '14 at 15:06
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You get more complete intersections if you relax your conditions a little. It is certainly more natural to ask for complete intersections in weighted projective space (WPS), not just straight (ordinary) projective space. A weighted projective space ${\mathbb P}^n[a_0,\ldots, a_n]$ with positive integer weights $a_0, \ldots, a_n$ is defined, as ordinary projective space, to be the quotient of ${\mathbb C}^{n+1}\setminus \{0\}$ by ${\mathbb C}^*$, except that the ${\mathbb C}^*$-action has weights $(a_0, \ldots, a_n)$ instead of the usual $(1, \ldots, 1)$. In the language of an ample divisor $L$ on a projective variety $X$, this corresponds to considering the whole graded ring $R=\oplus_k H^0(X, L^{\otimes k})$; it is finitely generated, by elements $x_0, \ldots x_n$, where $x_i\in H^0(X, L^{\otimes a_i})$. Then $R$ is a quotient of the free polynomial ring $S=k[X_0, \ldots X_n]$, dual to an embedding of $X$ into ${\mathbb P}^n[a_0,\ldots, a_n]$.

This language immediately gets you a few more cases. For example, suppose $X$ is a K3 surface of degree $2$. Then there is a two-to-one covering $\pi: X\to {\mathbb P}^2$. In the language above, we picked three degree one variables $x_0, x_1, x_2$ but the map you get is not quite an embedding yet, corresponding to the fact that the polarisation is ample but not very ample. It turns out that we are missing another variable $x_3$ of degree $3$, and $X$ embeds into the WPS ${\mathbb P}^3[1,1,1,3]$ as a sextic hypersurface, given (wlog) by an equation $x_3^2 + f_6(x_0, x_1, x_2)=0$. Here the sextic $f_6$ gives you the branch locus of the original map $\pi$.

More generally, you might be able to get your K3 surface as a complete intersection in some key variety, following terminology by Miles Reid. This says that there is some fixed variety $G\subset {\mathbb P}^n$ (often a Grassmannian) so that every K3 surface with a polarisation of degree $d$ is a complete intersection inside $G$. In this sense, K3 surfaces up to degree about 18 are complete intersections. This is very much related to Mukai's work reinterpreting the classification of Fano threefolds. One reference for all this is Brown-Altinok-Reid's http://xxx.lanl.gov/pdf/math/0202092.pdf, see in particular Theorem 2.5 for the statement I just made.

All of this is more related to algebra than to topology, in the sense that the structure of the graded ring $R$ determines whether or not it can be anywhere near a complete intersection in any key variety. One obstruction to a variety possibly being a complete intersection can already be read off the Hilbert or Poincare series $P_X(t)=\sum_{k\geq 0} h^0(L^{\otimes k})t^k$. By Hilbert's theorem, $P_X(t)$ is a rational function in $t$, with denominator (in the above language of generators of $R$) being equal to $\prod_{i=0}^n (1-t^{a_i})$. If $X$ is a complete intersection of hypersurfaces of degrees $d_1, \ldots, d_l$ in ${\mathbb P}^n[a_0,\ldots, a_n]$, then the numerator of $P_X(t)$ is $\prod_{j=1}^l (1-t^{d_i})$; this is a standard fact about regular sequences. Now the point is that for $(X,L)$ a polarised K3 surface of degree $d$, Riemann-Roch and Kodaira vanishing completely specify $P_X(t)$ in terms of $d$ (exercise!) and so you have your obstruction. More generally, the key varieties $G$ can be recognised (or at least guessed) from the numerator of $P_X(t)$; all this is explained in the above paper of Brown et al.

As a further set of examples, I wrote out the story of polarised elliptic curves before in this language here: Algebraic geometry examples

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