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Hi everyone

I was reading a little bit on the Horrocks Mumford surface (i.e. the abelian degree 10 surface, which is the zero section of an indecomposable rank 2 vector bundle on $\mathbb P^4$) and I want to ask a trivial question about it; Any references will be wellcomed. As is known, this vector bundle is the only one that is known to be indecomposable rank 2 vector bundle on $\mathbb P^4$: all the other are decomposable and hence their zero sections give surfaces in $\mathbb P^4$ which are complete intersections. However, there are known surfaces in $\mathbb P^4$ which are not complete intersections: taking two hypersurfaces which have a common component (e.g. a plane $\mathbb P^2$) their residual intersection is another surface. So what is so special about the Horrocks Mumford surface? I'm asking as it is also a residual intersection of a few hypersurfaces. Maybe the reason is that we cannot get these residual intersection surfaces (besides HM surface) as a zero section of a rank 2 vector bundle?

Thanks, Nick.

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  • $\begingroup$ The HM bundle $E$ has $c_1=5H$ and $c_2=10H^2$ where $H$ is the class of a hyperplane, and is the only known indecomposable rank two bundle with these Chern classes. But then if $f:\mathbb P^4\to\mathbb P^4$ is not constant and $deg(f)>1$ the pullback $f^*E$ will also be indecomposable and have larger Chern classes. (And two even more trivial remarks: "The" in line 1 is questionable: there is a $3$-dimensional moduli space of these surfaces. They do not arise as the zero-section of a vector bundle on $\mathbb P^4$, but rather the zero-locus of a non-zero section of such a bundle.) $\endgroup$
    – inkspot
    Commented Sep 21, 2011 at 9:29

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If you want to use the Serre construction to produce a rank 2 bundle from a codimension 2 subvariety, the necessary condition is that the determinant of the normal bundle should be isomorphic to a restriction of a line bundle from the ambient variety. Equivalently, the canonical class of the subvariety should be restricted from the ambient variety.

For the abelian surface this is automatically fulfilled --- the canonical class is trivial hence restricted. But for generic surface in $P^4$ this is not true. For example, we can consider $P^2$ blown up in a point and its map given by the linear system $|2h-e|$. The image will be a cubic surface in $P^4$. But its canonical class $e-3h$ is not restricted since it is not a multiple of $2h-e$. By the way, this cubic surface is the simplest example of a residual intersection you are asking about (take just two quadrics $zx - vy =0$ and $ux - zy = 0$ in $P^4$ with coordinates $(x,y,z,u,v)$ and remove the palne $x = y = 0$).

EDIT. Let me explain that the canonical class of the residual surface is almost never restricted. Indeed, let $X$ be the blowup of $P^4$ in $P^2$, let $H$ be the hyperplane class on $P^4$ (and its pullback to $X$ as well) and $E$ the exceptional divisor. Assume that the hypersurfaces we consider have degrees $a$ and $b$ respectively. Then they define divisors in linear systems $aH - E$ and $bH - E$ respectively, and the residual surface $S$ is the projection of their intersection. Since $K_X = -5H + E$, by adjunction $$ K_S = (-5H + E) + (aH - E) + (bH - E) = (a + b - 5)H - E. $$ If $a,b \ge 2$ and $S$ is smooth then this is not proportional to $H$ (so is not restricted from $P^4$). Indeed, for $a,b \ge 2$ the linear systems $aH - E$ and $bH - E$ are very ample, so by Lefschetz theorem the restriction map $Pic X \to Pic S$ is injective, hence $H$ and $E$ are linearly independent on $S$.

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  • $\begingroup$ Thank you. So if I understand correctly, besides the HM surface, for all the other surfaces in $\mathbb P^4$, which are residual intersection, the canonical class of the subvariety is not a restriction of the ambient variety? $\endgroup$
    – Nick
    Commented Sep 21, 2011 at 8:00
  • $\begingroup$ [correction: I meant "for all the other known surfaces in $P^4$"] $\endgroup$
    – Nick
    Commented Sep 21, 2011 at 8:03
  • $\begingroup$ Yes, I added the argument showing that this is almost always the case. $\endgroup$
    – Sasha
    Commented Sep 21, 2011 at 9:34

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