43

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 $L$ ...

39

[edited again mainly to add the Euler link, see last paragraph]
Yes, and indeed there are infinitely many rational points:
the birationally equivalent Diophantine equation
given by J.Ramsden in his partial answer to his own question,
$$
X + Y = Z + T,
\phantom{and}
XYZT = c,
$$
was already studied by Euler (in the equivalent form $xyz(x+y+z)=a$),
who in ...

answered May 19 '12 at 21:51

Noam D. Elkies

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24

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

Yes. By straightforward search the smallest example is
$\lbrace x,y,z \rbrace = \lbrace 45,64,180 \rbrace$, with
$$
(t+45) (t+64) (t+180) = t^3 + 17^2 t^2 + 150^2 t + 720^2.
$$
Given any solution $(x,y,z)$ we may produce infinitely many others
(other than the trivial scaling $(c^2 x, c^2 y, c^2 z)$) by using
the theory of elliptic curves to find rational $z'...

answered Apr 18 '12 at 2:16

Noam D. Elkies

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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). ...

16

I think the easiest place to see the $22$ is in a Kummer surface. Let $A$ be an abelian surface, so topologically $(S^1)^4$. This clearly has $h_2 = \binom{4}{2} = 6$, and there are obvious topological repreentatives for the $2$-cycles, given by $(S^1)^2$ in $6$ different ways.
Let $X$ be the quotient of $A$ by negation. This has $16$ singular points; the ...

answered Jun 29 '12 at 16:43

David E Speyer

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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

Thanks, YangMills, for the references to my papers. I want to elaborate, because I disagree with the statement that mirror symmetry is given by hyperkahler rotation. It may be the case for certain choices of K3, but I think this happens by accident and that it's not a useful principle. Here is how I view mirror symmetry for K3 surfaces. Choose a rank 2 ...

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_{\...

11

There are some papers of van Luijk, where he computes the ranks of some K3s over number fields. The trick is to note that $NS(X) \hookrightarrow NS(X_p)$, where $X_p$ is the reduction of $X$ modulo a prime ideal $p$. One can determine the rank of $NS(X_p)$ by counting eigenvalues of Frobenius which differ from $q$ (the size of the residue field) by a root ...

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^{...

11

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).

10

These $2$-spheres are called 'twistor lines'. They indeed cover the moduli space (in the non-polarized case) : more precisely, any two points of the moduli space may be linked by a chain of twistor lines.
A reference where this is nicely explained (and used !) is Huybrecht's Bourbaki talk about Verbitsky's Torelli theorem : http://arxiv.org/abs/1106.5573. ...

10

If you take the projective closure of your surface in $\mathbb{P}^3$ you find a singular quartic. This quartic has six singular points. Namely the two points found by Daniel and four points of the form $(\pm 1,\pm 1,0,1)$. Each of these points is an $A_1$ singularity. This implies that the resolution of the surface is a K3 surface.
This surfaces contains ...

answered May 19 '12 at 17:31

Remke Kloosterman

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10

The $\mathbb{P}^{2}$ contained in your cubic fourfold $X$ is cut out by linear forms (say) $L_{1},L_{2},$ and $L_{3}.$ Since the homogeneous ideal of $X$ is contained in the homogeneous ideal generated by $L_{1},L_{2}$ and $L_{3},$ there exist quadrics $Q_{1},Q_{2},$ and $Q_{3}$ such that $X=\{L_{1}Q_{1}+L_{2}Q_{2}+L_{3}Q_{3}=0\}.$ The octic K3 cut out by $...

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

A sketch of the proof goes as follows.
Let $G$ be a finite group of non-symplectic automorphisms on a $K3$ surface $X$. Since $G$ is non-symplectic, there exists $g \in G$ such that $g \omega \neq \omega$, where $\omega$ is the holomorphic $2$-form on $X$.
Then, setting $Y:=X/G$, one has $q(Y)=p_g(Y)=0$, since $q(X)=0$ and by the previous remark the ...

answered Oct 2 '12 at 10:41

Francesco Polizzi

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8

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.

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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8

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.

7

An algebraic (possibly singular) $K3$ surface is a normal algebraic surface whose minimal resolution is a smooth $K3$ surface. You can obtain these for example by blowing down $(-2)$-curves on a smooth algebraic $K3$. As long as you only do that you can even restrict the kind of singularities you allow. For example blowing down a $(-2)$-curve (and certain ...

7

Since the canonical class of $X$ is trivial, by adjunction one has $$2p_a(L)-2 = L^2,$$ where $p_a$ denotes the arithmetic genus.
Now assume that $L=\mathcal{O}_X(C)$, where $C$ is an effective curve. If $C$ is connected and reduced then $p_a(C) \geq 0$, so $L^2 < -2$ implies that $C$ is either non-reduced or non-connected.
As an example of the first ...

answered Mar 26 '13 at 17:35

Francesco Polizzi

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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

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

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 ...

6

The nodes do not modify the birational invariants of a surface. So if we blow-up the $k$ nodes of $X$ we obtain a smooth K3 surface $S$, containing $k$ $(-2)$-curves, whose topological Euler number is $24$. Coming back to $X$, we substitute each $(-2)$-curve (which is topologically a sphere, so has Euler number $2$) with a point. So the Euler number of $X$ ...

answered Dec 20 '12 at 22:00

Francesco Polizzi

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