3
$\begingroup$

I heard from string theorists thinking of the so-called "(1/2) K3 surface" or "half-K3 surface" as follows:

Let $T^2 \times S^1$ be a 3-torus with spin structure periodic in all directions. $T^2 \times S^1$, with this spin structure, is the boundary of a “1/2-K3 surface,” that is, a four-manifold $M^4$ that maps to a disc $D$ with generic fiber an elliptic curve. In particular, the map $$ M^4 \to D$$ has a section $$s : D \to M^4.$$ (possibly missing contexts) ...

Are these mathematically clear? Or do we require more clarification?

What are some other Mathematical way"s" to think about or to define "(1/2) K3 surface"?

Improve this question
$\endgroup$
1
  • 4
    $\begingroup$ Some K3 surfaces are elliptic, meaning that they are fibered over $\mathbb P^1$ with generic fiber an elliptic curve. Since $\mathbb P^1$ is glued from two disks it seems natural to consider the inverse image of (say) the unit disk to be "1/2" of the K3. This would somehow motivate their definition of a 1/2-K3 surface to be just any old elliptic fibration over a disk, except for the fact that an elliptic surface fibered over $\mathbb P^1$ is not in general a K3; the part of the quoted section after "that is" seems like it could just as well deserve to be called a "1/2-Enriques surface"... $\endgroup$
    – Dan Petersen
    Commented Feb 14, 2020 at 5:43

1 Answer 1

Reset to default
6
$\begingroup$

Of course what you've written is too vague to be a definition, but I can guess what they're talking about. In low-dimensional topology there's a 4-manifold called $E(1)$; this is a rational complex surface obtained from $CP^2$ by blowing up the nine basepoints of a cubic pencil. This manifold is fibred by elliptic curves (the proper transforms of the cubics). If you cut out one of these fibres, you're left with a 4-manifold whose boundary is the 3-torus (because the elliptic fibres have trivial normal bundle in the blown up surface). If you glue two of these 4-manifolds together along their common torus boundary, you get a 4-manifold called $E(2)$, also known as the K3 surface. As Dan Petersen mentions in the comments, this is the same as decomposing an elliptically-fibred K3 into two pieces by taking the preimages of the two hemispheres under the elliptic fibration, but additionally making sure that the two halves are diffeomorphic (each containing 12 of the 24 nodal fibres (with vanishing cycles a,b,a,b,a,b,a,b,a,b,a,b) if your fibration is Lefschetz).

Improve this answer
$\endgroup$
1
  • $\begingroup$ thanks +1 for the answer! $\endgroup$
    – wonderich
    Commented Feb 14, 2020 at 18:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .