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Jan 23, 2011 at 17:14 comment added rita @Michael: that's right.
Jan 22, 2011 at 18:24 comment added Tong @rita: right. But I think using the technique of the proof of Noether inequality, maybe $M^2 \ge 2p_g-4$.
Jan 22, 2011 at 17:01 comment added rita If the canonical image of $S$ is a surface, then $M^2\ge p_g-1$. This improves the bound suggested by JC.
Jan 22, 2011 at 1:36 comment added Tong Yes, right. But as $p_g$ goes large, this bound seems not so beautiful. I do not know if there is a linear bound.
Jan 22, 2011 at 1:18 comment added J.C. Ottem Well, if $M^2>0$, the Hodge index theorem gives $$ K^2\le \frac{(K.M)^2}{M^2}\le (K.M)^2 $$
Jan 21, 2011 at 23:37 history edited Tong CC BY-SA 2.5
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Jan 21, 2011 at 23:26 history asked Tong CC BY-SA 2.5