Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For a smooth curve $C$ on a smooth, projective surface $S$ over $\mathbb{C}$, we have the genus formula:

$g(C) = 1 + \frac12(C^2 + C \cdot K_S)$

where $K_S$ is the canonical divisor. Is this formula still true for singular (e.g. reducible) curves on $S$ if one uses the arithmetic genus in the left hand side instead of the geometric genus?

share|improve this question
    
Sorry if I am ignorant, but what do you mean by the square of a curve?Thanks for clarifying. –  awllower Jun 3 '12 at 16:18
    
The self-intersection $C \cdot C$. –  Evariste Jun 3 '12 at 16:40
    
So the intersection is a number? It appears that I have to check out some definitions. Thanks again. –  awllower Jun 3 '12 at 16:41
2  
Dear @awllower: this is a long story! Given a smooth compact complex algebraic surface $S$, there is a bilinear form $Pic(S)\times Pic(S)\to \mathbb Z$, where $Pic$ denotes classes of divisors. A curve $C$ is a divisor so you can compute the value of that form on the pair $(class(C), class(C))$, and the result is (dangerously!) written $C.C$ or even worse $C^2$. It is rather technical but means intuitively that you somehow "deform" one copy of $C$ to $C'$ within $S$ and then $C.C$ is the ordinary cardinality of the set-theoretic intersection $C\cap C'$. –  Georges Elencwajg Jun 3 '12 at 17:25

1 Answer 1

up vote 3 down vote accepted

Yes, the formula is still true if $C\subset S$ is reduced, irreducible but not smooth.

The arithmetic genus is to be defined as $p_a(C)=dim_{\mathbb C}H^1(C,\mathcal O_C)$ and we then have $$p_a(C)= 1+\frac {deg[\mathcal K_S\otimes \mathcal O_S(C))\mid C]}{2} $$

You can find a proof in chapter II of this book.

share|improve this answer
    
Thank you Georges. –  Evariste Jun 3 '12 at 20:59

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.