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I'm reading about genuses from Liu's book "Algebraic Geometry and Arithmetic Curves". There he defines arithmetic genus of a projective curve and geometric genus of a smooth projective variety. As far as I understand, a projective curve is a projective variety.

So, if $m,n\in \mathbb{Z}$ and $m\geq 0$ are arbitrary, is there always a projective curve $C$ over a field $k$ such that $p_a(C)=n$ and $p_g(C)=m$? If not, can we somehow classify those integer pairs which satisfies $p_a(C)=n$ and $p_g(C)=m$ (probably with respect to $k$)?

selfstudying
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1 Answers1

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I assume a base field $k$ is fixed.

  1. There are (smooth) projective curves of geometric genus $g$ for every $g\geq 0$. They can be found as divisors of type $(g+1,2)$ on a smooth quadric surface (see Hartshorne, Remark 1.1.1 in chapter IV).
  2. Fix a smooth projective curve $C$ of genus $g$. You can get a projective curve $C_1$ with arithmetic genus $g+1$ and geometric genus $g$ by identifying two distinct points of $C$ (this creates one node, but you can get $n$ nodes similarly: identify a couple of smooth points on $C_{n-1}$, so to end up with a curve $C_n$ having $p_a=g+n$). Why does the geometric genus stay the same? By definition, $p_g(C_1):=p_g(\tilde C_1)$ where $\tilde C_1$ is the normalization. But $\tilde C_1$ is just "$C_1$ with the node unglued", i.e. $\tilde C_1=C$.

Unrelated: There are projective curves of any arithmetic genus.

  1. If you want $n=p_a\geq 1$ (ignoring item 1 above), just attach an elliptic curve transversally to a curve of geometric genus $n-1$. Indeed, such a glueing produces a nodal curve of compact type, for which the arithmetic genus is computed easily as the sum of the (geometric $=$ arithmetic) genera of the components.

  2. For $-n=p_a\leq 0$, take a rational curve $\mathbb P^1\subset \mathbb P^d$ and consider its disjoint union with $n$ distinct points in $\mathbb P^d\setminus\mathbb P^1$. Alternatively, consider the disjoint union of $n+1$ copies of $\mathbb P^1$. Indeed, in both cases, the resulting scheme $X_n$ has $h^0(X_n)=n+1$ (in the first case, there is one $h^0(\mathscr O_{\mathbb P^1})=1$ plus $n$ times $h^0(\mathscr O_{\textrm{point}})=1$; in the second case, there are $n+1$ copies of $h^0(\mathscr O_{\mathbb P^1})=1$). Hence, $$p_a(X_n)=1-\chi(X_n,\mathscr O_{X_n})=h^0(X_n,\mathscr O_{X_n})-h^1(X_n,\mathscr O_{X_n})=1-(n+1)=-n.$$

Brenin
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