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Let $Y$ be the smooth manifold underlying a K3 surface. As a manifold, $Y$ is diffeomorphic to $\{[x_0:x_1:x_2:x_3]\in\mathbb{C}P^3\colon X_0^4+x_1^4+X_2^4+X_3^4=1\}$. It is well known that $H^2(Y,\mathbb{Z})\cong U^{\oplus 3}\oplus(-E_8)^{\oplus 2}\cong \mathbb{Z}^{\oplus 22}$.

From the classification results of principal $S^1$-bundles $p:M\to Y$, we have know that they are classified by $H^2(Y,\mathbb{Z})$. Namely, by picking a connection 1-form $\theta$ on $M$, which satisfies $d\theta=p^*\omega$ for some closed $\omega\in \Omega^2(Y)$, we get a map $$\Phi:\{\text{Isomorphism classes of principal $S^1$-bundles $p:M\to Y$}\}\to H^2(Y,\mathbb{Z}),$$ by sending $[p:M\to Y]$ to $[\omega]$.

In the case that $Y$ is a K3 surface, I am wondering if there are concrete presentations of these principal $S^1$-bundles around. More concretely, suppose we have have a basis $c_1,\cdots,c_{22}$ of $H^2(Y,\mathbb{Z})$, can we construct 22 principal $S^1$-bundles with classes $c_1,\cdots,c_{22}$.

Since $S^1\to S^7\to \mathbb{C}P^3$ is an $S^1$-bundle, I tried to realize $Y$ as an $S^1$-bundle $S^1\to M\to Y$, where $M\subset S^7$ is some smooth submanifold. But this is probably not going to give me all bundles over $Y$, so any help or any link to literature would be helpful!

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  • $\begingroup$ One approach to giving explicit constuctions of $S^1$-bundles is to give, by algebraic geometry means, explicit constructions of algebraic line bundles. For the surface with equations you live, the space of algebraic line bundles has rank $20$, so one should be able to explicitly construct twenty independent classes this way. One can deform the K3 to make different classes appear algebraically but then one is face with choosing an explicit diffeomorphism to the deformation. $\endgroup$
    – Will Sawin
    May 17, 2022 at 14:41

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You can say a fair amount about the topology of the total spaces of the different bundles, although I suspect none of them is a particularly well-known manifold that has a `name'. (Except of course for the trivial bundle; I guess you could say $S^1 \times$ a K3 surface is a well-known manifold.)

The main observation is that you can read off the fundamental group and the basic homological invariants of $M$ (using your notation) in terms of the Euler class $e \in H^2(Y)$. It seems that the fundamental group is cyclic of order the divisibility of $e$. For the cohomology (and hence homology by duality) you can use the Gysin sequence for the cohomology

$\cdots \to H^j(M) \to H^{j-1}(Y) \stackrel{\cup e}{\rightarrow} H^{j+1}(Y) \to H^{j+1}(M) \to \cdots $

To go further, you'd need to know more about the automorphism group of the intersection form of $Y$. I think (but I'm not certain) that the automorphism group is transitive on element of a given square and divisibility. Assuming this to be the case, I believe that implies that any two Euler classes with the same square and divisibility are related by a self-diffeomorphism of $Y$. That certainly lessens the complexity of the problem.

For the automorphism group, I would look at Wall's On the orthogonal groups of unimodular quadratic forms. II. J. Reine Angew. Math. 213 (1963/64), 122–136. For realizing automorphisms by diffeomorphisms, see T. Matumoto, On diffeomorphisms of a K3 surface, Algebraic and topological theories (Kinosaki, 1984), Kinokuniya, Tokyo, 1986, pp. 616–621.

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  • $\begingroup$ Thank you for the extensive reply and resources! $\endgroup$
    – James
    May 18, 2022 at 7:23
  • $\begingroup$ The only bundles I have found so far are the Prequantization bundles, described in this paper by Crainic. projecteuclid.org/journals/journal-of-symplectic-geometry/… I hope to see if there are other "standard" bundles around on K3 surfaces $\endgroup$
    – James
    May 18, 2022 at 9:31

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