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.

Let $Y\subseteq \mathbb{P}^n$ be the zero locus of $f_1,...,f_k$ of degree $d_1,...,d_k$, and put $d=\sum d_i$. If $Y$ is nonsingular and a complete intersection, then $dim(H^0(\omega_Y))=dim(H^0(\mathcal{O}(d-n-1)))$, and hence we can compute its geometric genus.

I'm wondering about the case where $Y$ is not necessarily a complete intersection, but is still nonsingular. For instance if $Y$ is not codimension $k$ (ie the twisted cubic). Assuming we don't have any redundant $f_i$, is there anything we can say about the geometric genus? Even rough information would be helpful, ie a lower/upper bound, perhaps involving the difference between $codim(Y)$ and $k$.

Thanks

share|improve this question
add comment

1 Answer 1

up vote 1 down vote accepted

I think all you can really say in general is the following. Let $N$ denote the normal bundle of $Y$ in $\mathbb P^n$. Then by the adjunction formula you have that $\omega_Y \cong \bigwedge^c N (-n-1)$ where $c$ is the codimension of $Y$ in $\mathbb P^n$. Hence, you must have $\rho_g(Y)=h^0(Y,\omega_Y)=h^0(Y,\bigwedge^c N (-n-1))$ which could be pretty much any nonnegative number.

In order to say something substantive you need to make some assumptions on how $Y$ is embedded into $\mathbb P^n$ that is reflected in the positivity of the normal bundle $N$, specifically a bound on its first Chern class. You may also do this by making assumptions on the tangent bundle $T_Y$ of $Y$ and using the exact sequence $0 \rightarrow T_Y \rightarrow T_{\mathbb P^n} \rightarrow N \rightarrow 0$. But unless you make some more assumptions on $Y$ you can't say anything.

share|improve this answer
    
Okay I suspected I might need some more information. I'll see if I can figure some of that out in my situation. –  user16544 May 25 '12 at 16:19
add comment

Your Answer

 
discard

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