44

Yes, such M exists. The boundary connected sum of 28 copies of the Milnor plumbing has boundary diffeomorphic to $S^7$ so it can be closed off with $D^8$ and you can let $M$ be the connected sum of the resulting closed manifold and 8 copies 7 copies of $S^4 \times S^4$.
Why does that work? Let $f: M \to \mathbb{O}P^1 = S^8$ have degree 240 and let $L$ ...

26

See Densities of Short Uniform Random Walks (with an appendix by Don Zagier) by
Jonathan M. Borwein, Armin Straub, James Wan, and Wadim Zudilin,
Canad. J. Math. Vol. 64 (5), 2012 pp. 961–990. http://cms.math.ca/10.4153/CJM-2011-079-2
Your integral is $p_4(1)$ in Equation (2.1). They write it in terms of hypergeometric functions (doing this for $p_4$ is one ...

20

More generally, the moduli space of algebraic K3 surfaces is a countable union on 19 dimensional subvarieties of the 20 dimensional moduli space of complex K3 surfaces, and exactly one of those components corresponds to K3 surfaces that embed as smooth quartic surfaces in $\mathbb{P}^3$.

answered Apr 1 '15 at 14:09

Joe Silverman

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18

The answer to your first question: "What is the relationship between $G_2$ and $\Delta$?" is
$$q\frac{d}{dq} \log \Delta = -24G_2 $$
where
$$\Delta = q\prod_{m=1}^\infty (1-q^m)^{24}$$ and
$$G_2 = -\frac{1}{24} +\sum_{d=1}^\infty \sum_{k|d}k q^d$$
(note that I've included the constant -1/24 which is in usual definition of $G_2$ which you admitted). ...

15

No. Consider a K3 surface with a polarization of degree 2 and with Picard rank 1. Since the tautological line bundle on $\mathbb{P}^3$ pulls back to a degree 4 line bundle, it follows that such a K3 surface cannot be embedded in $\mathbb{P}^3$.

15

Let me briefly expand on Jason's comment.
Actually, "formal scheme" is the right word here.
For any K3 surface $X$ over an algebraically closed field $k$ of any characteristic one has $$h^0(X, T_X)=0, \quad h^1(X, T_X)=20, \quad \, h^2(X, T_X)=0,$$
so the functor of Artin rings $$F \colon (\textbf{Art}) \to (\textbf{Sets})$$ describing the first-order ...

answered Jun 15 '15 at 13:41

Francesco Polizzi

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13

The group of symplectomorphisms $Aut(X)$ of a K3 is the group $O(\Lambda)$ of
automorphisms of its period lattice $\Lambda=H^{1,1}(M,{\Bbb Z})$. For each
(-2)-cohomology class $\eta\in H^{1,1}(M,{\Bbb Z})$, either $\eta$ or $-\eta$ is
represented by a curve (this follows from the Riemann-Roch formula). This curve
is unique, because its self-intersection is ...

13

I am posting my comments as an answer. I am concerned that I misunderstand the OP, so let me state first the result. There exists a finite field extension $K/\mathbb{Q}$ such that for every closed immersion $\mathbb{P}^1_\mathbb{C} \hookrightarrow X\otimes_{\mathbb{Q}}\mathbb{C}$, there exists a closed immersion $\mathbb{P}^1_K \hookrightarrow X \otimes_{\...

13

This is a standard argument and there probably exists a reference but it's not hard once you rephrase it in terms of the period domain.
The moduli space of K3 surfaces is locally isomorphic to its period domain.
The period domain is an open subset of the vanishing locus of a quadratic polynomial in $\mathbb P^{21}(\mathbb C)$. See, for instance, Huybrechts' ...

12

A preprint by Stefan Schröer came out today with the answer to this question: arXiv:2004.07025.
No such Enriques surface exists. In fact, there is no classical Enriques surface over $\mathbb F_2$ with 25 $\mathbb F_2$-points (and with the extension of $\mathbb Z^{10}$ by $\mathbb Z/2$ split in the Picard group, which you can also deduce).

11

I can answer your first question. In arXiv:1208.4074 by Dabholkar, Murthy and Zagier you can find a formula that implies
$H^{(2)}(\tau)= \frac{48 F_2^{(2)}(\tau)- 2 E_2(\tau)}{\eta(\tau)^3}$
where $E_2(\tau)$ is the quasi modular Eisenstein series and
$F_2^{(2)}(\tau)= \sum_{r>s>0,r-s\ \mathrm{odd}} (-1)^r s\, q^{rs/2}$ which you can use to compute $H^{...

10

To develop what Jason says: if your curve deforms in a family of rational curves, it means that you can find a dominant rational map from a ruled surface onto your K3. This is forbidden (over $\mathbb{C}$): e.g. because the nonzero 2-form of the K3 would lift to a nonzero 2-form on the ruled surface.

9

The answer for $K3$ surfaces is no. A counterexample, where the group is not even commensurable with an arithmetic group, was given by Totaro in Example 6.3 of this paper.

9

The minimal $s$ is $3$.
It is attained by several elliptic K3's,
including $y^2 = x^3 + (t^2-t)^4$ which has IV* fibers at
$t = 0, 1, \infty$ and no other singular fibers.
The comment by Ariyan Javanpeykar gives one argument that
$s$ can be no smaller. (See postscript.
This uses characteristic zero; in small positive characteristic
$s$ can be as small as $...

answered Oct 8 '18 at 14:24

Noam D. Elkies

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8

Yes, this is true: the smooth minimal model is a $K3$ surface.
In fact, let $\bar{X}$ be the resolution of the singularities of $X$. Then the following holds.
(1) Rational double points impose no adjunction conditions to canonical forms, hence $\omega_{\bar{X}}$ is trivial.
(2) Rational double points have simultaneous resolution, hence $\bar{X}$ is ...

answered Mar 16 '15 at 12:52

Francesco Polizzi

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7

Here is an alternative way to construct elliptic fibrations of quartic surfaces containing a line.
Let $X\subset \mathbb P^3$ be a smooth quartic surface and assume that it contains a line $L\subset X$. Let $\mathfrak d$ denote the $1$-dimensional linear system of hyperplanes in $\mathbb P^3$ containing $L$ restricted to $X$. Further let $H\in \mathfrak d$ ...

7

The standard reference is : H. Sterk, Finiteness results for algebraic K3 surfaces.
Math. Z. 189 (1985), no. 4, 507--513. Of course it uses the global Torelli theorem for K3's, plus some reduction theory for arithmetic groups. The paper is quite short and shouldn't be too hard to read.

7

You can start with
Hans Sterk
Finiteness results for algebraic K3 surfaces
Mathematische Zeitschrift, 1985, Volume 189, Issue 4, pp 507-513
The aim of this paper is to prove the following theorem:
Theorem. Let $X$ be an algebraic K3 surface over the complex numbers.
Then
a) the group $\mathrm{Aut}(X)$ of (biholomorphic) automorphisms is finitely ...

7

Likely this should only be a comment, but I don't have enough reputation for that...
J.C. Ottem has provided a wonderful reference about the basics of K3 surfaces in his comment. It's my personal experience though that when I'm working through notes such as Huybrecht's, it's instructive and motivating to have short and interesting papers to read once I've ...

7

This is probably not as explicit as what you requested, but there are recipes for how to build a Morse function on any smooth hypersurface in $CP^3$ with no index 1 or 3 critical points. There will be exactly 24 critical points, and any gradient vector field for this Morse function will be of the sort you asked about.
Such Morse functions are due to Rudolph ...

7

I think Verbisky proves a refined form of the Mirror Conjecture for hyperkaehler manifolds, not the conjecture in the strict form. This is explained at page 3 of the paper that you link.
In fact, the assumptions for the conjecture are known to hold only for a hyperkaehler manifold which is generic in its deformation class. More precisely, Verbitsky proves ...

answered Jun 1 '17 at 9:15

Francesco Polizzi

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7

A reference for the density of singular K3 surfaces in the period domain (see Will Sawin's answer) is
Piatetski-Shapiro, Ilya I.; Shafarevich, I. R., Arithmetic of K3 surfaces, Trudy Mezhdunarod. Konf. Teor. Chisel, Moskva 1971, Tr. Mat. Inst. Steklova 132, 44-54 (1973) ZBL0293.14010.
In particular, this is a screenshot of page 54:

answered Jul 6 '20 at 7:59

Francesco Polizzi

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6

[In comments guest2014 amended the question to ask not for a
$D_{14}$ fiber but for $I^*_{14}$, a.k.a. $\tilde D_{18}$]
The elliptic surface
$$
X : y^2 = x^3 + (t^3+2t) x^2 - 2(t^2+1)x + t
$$
over ${\bf C}(t)$ has a $I^*_{14}$ fiber at $t=\infty$.
(Note that the right-hand side is a cubic in $x$ whose discriminant
$4t^4 + 13 t^2 + 32$ has degree only $4$ ...

answered Feb 19 '14 at 3:29

Noam D. Elkies

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6

It is true that if $X\subseteq Y$ is a normal crossings divisor, then $Y$ has a log structure whose sheaf of monoids is the sheaf of regular functions invertible outside of $X$. It is also true that this log structure can then be restricted to $X$, giving a log structure on $X$.
On the other hand, if we start with $X$ a normal crossings variety, we do not ...

6

There are some silly, probably useless obstructions, at least in the "honest Fano" case: the Lefschetz hyperplane theorem equate the second Betti and third Betti numbers (resp. Hodge numbers of total weight $2$ and $3$) to the corresponding numbers of the Fano manifold. Fano manifolds of a given dimension are bounded, and explicit examination of the proof ...

6

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)....

6

That is the Noether-Lefschetz theorem. Searching online should find plenty of results in web pages and lecture notes. If you want a published source, how about: Mark Green, A new proof of the explicit Noether-Lefschetz theorem, J. Differential Geom. 27 (1988), no. 1, 155–159.

5

Interesting question. I think the answer is yes, let me try to prove it.
As you noticed, the ideal sheaf sequence shows that $h^1(D)=0$ is equivalent to the fact that $H^0({\mathcal O}_D)$ is 1-dimensional generated by the constant function $1$.
Considering, for every effective decomposition $D=A+B$, the exact sequence
$$
0 \rightarrow {\mathcal O}_B(-A) \...

5

It's always true with $\mathbb{Q}$ coefficients. It follows from a general result of Zucker, who proved Leray degenerates whenever you have a projective map to curve. But in this case, it's simpler to check it by hand. There is only one (two !) differentials to worry about
$$d_2:H^0(\mathbb{P}^1, R^1\phi_*\mathbb{Q})\to H^2(\mathbb{P}^1,\mathbb{Q})$$
But ...

5

Yes, this approach has been tried, and we're about to submit a paper [edit: the paper has now been submitted, see arXiv:1601.04238]. Alas, very little is known about hyperbolic (as we call them; those with a single positive eigenvalue) graphs, and currently the proof is heavily computer aided (too many cases) and still using some algebraic geometry (each ...

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