3
$\begingroup$

Is there an explicit way of classifying (with regard to their compatibiliy with $\Omega_+$ or $\Omega_-,$ see below) the various families of equivariant holomorphic embeddings from $\mathbb{CP}^1$ to the quintic $Q$, such that $f_*[\mathbb{CP}^1]=2 \in H_2(Q,L;\mathbb{Z})?$

I'd like to know a phenomenology of what one can find, e.g., that only two or three kind of maps would have to be considered, for example the ones giving rise to an open surface, or to an unoriented surface with a crosscap.

Take a general quintic $Q$ (for which I don't know a concrete example to write down its equations), the antiholomorphic involution $$\sigma: \mathbb{CP}^4 \ni (x_1:x_2:x_3:x_4:x_5) \mapsto (\overline{x}_2:\overline{x}_1:\overline{x}_4:\overline{x}_3:\overline{x}_5),$$ $L$ the fixed locus of $Q$ under $\sigma,$ and $\Omega_\pm: \mathbb{CP}^1 \to \mathbb{CP}^1 \ (u:v) \mapsto (\overline{v}:\pm\overline{u});$ equivariance means $f\circ\Omega=\sigma\circ f.$

This should be possible in accordance to Clemens conjecture with $d=2,$ which is proved for small $d$ by Katz (see also Katz2): for a general quintic, we expect from Klemm $n^{g=0}_{d=2}=609250;$ it is known that Fermat quintic is not general in the sense of Clemens conjecture, since we have continuous families of maps.

Here $d$ can be thought more or less equivalently either as the homogeneous degree of the map or as the class in $H_2(Q;\mathbb{Z}):$ can this be proved more rigorously, perhaps using the fact that, in the exact sequence $$ H_2(Q;\mathbb{Z})=\mathbb{Z} \to H_2(Q,L;\mathbb{Z})=\mathbb{Z} \to H_1(L;\mathbb{Z})=\mathbb{Z}_2,$$ the first map is multiplication by 2?

Improve this question
$\endgroup$
11
  • $\begingroup$ You have to be careful with the equation of the quintic. The Fermat quintic that you wrote will have more embeddings than a general one. $\endgroup$
    – Lev Borisov
    Commented Feb 4, 2014 at 13:24
  • $\begingroup$ A holomorphic map $f$ into ${\mathbb P}_{\mathbb C}^4$ is given by $5$ homogeneous polynomials $p_i$ in $y_0,y_1$ of some degree $d$. I suspect that your $f_*[S^2]=2\in H_2(Q,L;\mathbb{Z})$ means that $d=2$. The equivariance means that the $p_i$'s have real coefficients. The fact that the map goes to the quintic means that $\sum_ip_i^5=0$. The requirement of $f$ being an embedding means that the matrices $\Big[\frac{d^2p_i}{dy_jdy_k}\Big]_{ij}$ have rank $2$ for $k=0,1$. The moduli space is given by explicit equations in an explicit Zariski open part of ${\mathbb P}_{\mathbb R}^{14}$. $\endgroup$
    – Sasha Anan'in
    Commented Feb 4, 2014 at 18:37
  • $\begingroup$ Well, one should also quotient the above set by the action of $\text{GL}_2{\mathbb R}$ (the real change of variables $y_0,y_1$). $\endgroup$
    – Sasha Anan'in
    Commented Feb 4, 2014 at 18:43
  • $\begingroup$ and also about the quotient; the dimension count should give zero, namely a finite number, as of Clemens conjecture. Can you provide some explicit examples for start, maybe disregarding equivariance for a second? regarding reality, I agree, even if my involution is not exactly complex conjugation $\endgroup$
    – jj_p
    Commented Feb 5, 2014 at 9:54
  • $\begingroup$ Your involutions $\Omega_{\pm}$ are complex conjugations on ${\mathbb P}_{\mathbb C}^1$ :all are equivalent modulo the action of holomorphic authomorphisms. (If you would like that your comments reach me, write please @nickname; I do not do so because at this place my comments reach you automatically.) $\endgroup$
    – Sasha Anan'in
    Commented Feb 5, 2014 at 12:14

0

Reset to default

You must log in to answer this question.