Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be a $K3$ surface and $\sigma$ an antisymplectic involution on $X$ and so $X$ is algebraic. 1.Why the signature of $Pic(X)$ is $(1,\rho -1)$? (This is well known but I cant find any direct proof) 2. Lets assume the set of fixed point set of $\sigma$ which we denote by $Fix(\sigma)$ contains a smooth curve say $D_g$ of genus $g\geq 2$, let's assume we have other curve in $Fix(\sigma)$ which is elliptic say $D$, i.e., we should have $D^2=0$ and $D.D_g=0$ then why this is a contrary to the fact that the signature of $Pic(X)$ is $(1,\rho -1)$. 3. How do we see curves inside $Pic(X)$? I mean curves are inside $X$ and $Pic(X)$ is a lattice!

share|improve this question
Dear Ru, the assertion of your first question is called the "Hodge Index Theorem". It is true for any smooth projective surface; it has nothing to do with $X$ being a $K3$. There must be a proof in any book on algebraic surfaces. About 3: since $X$ is a surface, curves in $X$ are (Cartier) divisors: their linear equivalance classes are by definition elements of Pic. –  Artie Prendergast-Smith Sep 9 '11 at 18:12
Thanks Artie, Ok, if I have a curve then I can associate an element of the picard group. I understand this now. Suppose I give a you an element of the picard group of $X$. How can I say if it is a curve of genus $g$ or it is an elliptic curve inside $X$? –  user13559 Sep 9 '11 at 18:45
Dear Ru, first one should say that given an arbitrary element x of Pic(X), there's no reason that it should be the class of a curve. (Maybe you already know that.) If you know that x is the class of a curve, and you know x^2, you can work out the genus of the curve using the adjunction formula. (See Hartshorne, V.1.5 for example.) –  Artie Prendergast-Smith Sep 9 '11 at 19:52
Let me add: in my last response, now I am using your assumption that X is a K3 surface. (Otherwise adjunction alone doesn't give enough information to answer the question.) –  Artie Prendergast-Smith Sep 9 '11 at 20:51
Dear Artie: I think your comments can be assembled into a good answer. –  S. Carnahan Sep 10 '11 at 3:49
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.