Timeline for K3 surfaces with no −2 curves

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Feb 2 at 1:20 comment added Cranium Clamp @YCor yes, take any K3 surface. The second cohomology with the intersection product is isomorphic as a lattice to E8^2 + U(-2)^3 for any K3 surface.
Feb 1 at 9:21 comment added YCor What is meant by $H^2(K3,\mathbb{Z})$? do you mean $H^2(S,\mathbb{Z})$ for some K3 surface $S$?
Feb 1 at 9:08 history edited Misha Verbitsky CC BY-SA 4.0
added 17 characters in body
Feb 1 at 9:06 comment added Misha Verbitsky Marco: any smooth rational curve on K3 has square -2. Conversely, using Riemann-Roch, you can prove that any line bundle $L$ with $c_1^2=-2$ satisfies $H^2(L)\neq 0$ or $H^2(L^*)'\neq 0$, and the zero divisor of its section contains a smooth rational curve as a component
Feb 1 at 9:04 comment added Misha Verbitsky Samir: many thanks, this is what I need I think.
Feb 1 at 9:04 comment added Misha Verbitsky Jason: for McMullen's nonprojective K3 surfaces, the Picard lattice is either degenerate or negative definite, and such lattices are easy to construct in any rank.
Feb 1 at 7:09 comment added Samir Canning For projective K3s, there are -2 curves whenever the Picard rank is at least 12. See, for example, Huybrechts's Lectures on K3 surfaces, Chapter 14, Corollary 3.8.
Feb 1 at 1:05 comment added Jason Starr . . . The rank is zero.
Feb 1 at 1:01 comment added Jason Starr Did you look at McMullen’s nonprojective K3 surfaces associated to Salem numbers? I will look now to see their ranks . . .
Jan 31 at 23:56 comment added Marco Golla @MishaVerbitsky: pardon my ignorance, but why does the second question imply the first?
Jan 31 at 22:56 comment added Misha Verbitsky the question is, for which $d$ we can be sure there are no -2 vectors? What about $d=12$? What about $d=20$?
Jan 31 at 22:50 comment added LSpice What is $d$ in "which has rank $\ge d$"?
Jan 31 at 22:49 history edited LSpice CC BY-SA 4.0
-2 -> $-2$, and consistent hyphenation
Jan 31 at 22:01 history asked Misha Verbitsky CC BY-SA 4.0