Let \(Y\) be a variety of dimension \(r\) in \(\PP^n\), with Hilbert polynomial \(P_Y\). Define the arithmetic genus of \(Y\) to be \[ p_a(Y) = (-1)^r \left( P_Y(0) - 1 \right). \] This is an important invariant which is independent of the projective embedding of \(Y\).
Show that \(p_a(\PP^n) = 0\).
If \(Y\) is a plane curve of degree \(d\), show that \(p_a(Y) = \frac{1}{2}(d-1)(d-2)\).
More generally, if \(H\) is a hypersurface of degree \(d\) in \(\PP^n\), show that \(p_a(H) = \binom{d-1}{n}\).
If \(Y\) is a complete intersection of surfaces of degrees \(a, b\) in \(\PP^3\), show that \(p_a(Y) = \frac{1}{2}ab(a + b - 4) + 1\).
Let \(Y^r \subseteq \PP^n\) and \(Z^s \subseteq \PP^m\) be projective varieties, and embed \(Y \times Z \subseteq \PP^n \times \PP^m \to \PP^N\) by the Segre embedding. Show that \[ p_a(Y \times Z) = p_a(Y) p_a(Z) + (-1)^s p_a(Y) + (-1)^r p_a(Z). \]
Solution
For an integer-valued polynomial we use \[ \binom{t}{m}=\frac{t(t-1)\cdots(t-m+1)}{m!} \] as a polynomial in \(t\); in particular \[ \binom{-q}{m}=(-1)^m\binom{q+m-1}{m}. \]
1. The arithmetic genus of projective space is zero.
The Hilbert polynomial of \(\PP^n\) is \[ P_{\PP^n}(t)=\binom{t+n}{n}. \] Hence \(P_{\PP^n}(0)=1\), and therefore \[ \boxed{p_a(\PP^n)=(-1)^n(1-1)=0}. \] This proves part (a).
2. If \(H\subseteq\PP^n\) is a hypersurface of degree \(d\), then\[ P_H(t)=\binom{t+n}{n}-\binom{t-d+n}{n}. \]
Let \(S=k[x_0,\ldots,x_n]\) and let \(F\in S_d\) be an irreducible homogeneous equation for \(H\). Multiplication by \(F\) gives an exact sequence of graded \(S\)-modules \[ 0\longrightarrow S(-d)\xrightarrow{\cdot F}S\longrightarrow S/(F)\longrightarrow0. \] Taking degree-\(t\) dimensions for \(t\gg0\) gives \[ P_H(t)=P_S(t)-P_S(t-d) =\binom{t+n}{n}-\binom{t-d+n}{n}. \]
3. A degree-\(d\) hypersurface in \(\PP^n\) has\[ \boxed{p_a(H)=\binom{d-1}{n}}. \]
The hypersurface has dimension \(n-1\). By step 2, \[ P_H(0)-1=-\binom{n-d}{n}. \] Using \[ \binom{n-d}{n}=(-1)^n\binom{d-1}{n}, \] we obtain \[ p_a(H) =(-1)^{n-1}(P_H(0)-1) =(-1)^n\binom{n-d}{n} =\binom{d-1}{n}. \] This proves part (c).
4. A plane curve of degree \(d\) has\[ \boxed{p_a(Y)=\frac{(d-1)(d-2)}2}. \]
A plane curve is a degree-\(d\) hypersurface in \(\PP^2\). Apply step 3 with \(n=2\): \[ p_a(Y)=\binom{d-1}{2}=\frac{(d-1)(d-2)}2. \] This proves part (b).
5. If \(Y\subseteq\PP^3\) is the complete intersection of surfaces \(F=0\) and \(G=0\) of degrees \(a\) and \(b\), then\[ P_Y(t) =\binom{t+3}{3} -\binom{t-a+3}{3} -\binom{t-b+3}{3} +\binom{t-a-b+3}{3}. \]
Because \(F,G\) define a complete intersection, they form a homogeneous regular sequence in \(S=k[x_0,x_1,x_2,x_3]\). The Koszul resolution of \(S/(F,G)\) is \[ 0\longrightarrow S(-a-b) \longrightarrow S(-a)\oplus S(-b) \longrightarrow S \longrightarrow S/(F,G) \longrightarrow0. \] Taking Hilbert polynomials and using \(P_S(t)=\binom{t+3}{3}\) gives the displayed formula.
6. The complete-intersection curve of step 5 has\[ \boxed{p_a(Y)=1+\frac12ab(a+b-4)}. \]
Since \(Y\) has dimension one, \[ p_a(Y)=1-P_Y(0). \] Substitute \(t=0\) into step 5: \[ P_Y(0) =1-\binom{3-a}{3}-\binom{3-b}{3}+\binom{3-a-b}{3}. \] For every integer \(c\), \[ \binom{3-c}{3}=-\frac{(c-1)(c-2)(c-3)}6. \] Hence \[ \begin{aligned} p_a(Y) &=\binom{3-a}{3}+\binom{3-b}{3}-\binom{3-a-b}{3}\\ &=1+\frac12ab(a+b-4), \end{aligned} \] where the last equality follows by expanding the three cubic polynomials and cancelling the pure \(a\)- and \(b\)-terms. This proves part (d).
7. For the Segre embedding of \(Y\times Z\), the Hilbert polynomial is\[ P_{Y\times Z}(t)=P_Y(t)P_Z(t). \]
Let \(R_Y\) and \(R_Z\) be the homogeneous coordinate rings of the given embeddings. The Segre coordinate ring has degree-\(q\) piece \[ (R_Y)_q\otimes_k(R_Z)_q. \] Therefore, for all sufficiently large \(q\), \[ H_{Y\times Z}(q) =H_Y(q)H_Z(q) =P_Y(q)P_Z(q). \] Two polynomials agreeing for all sufficiently large integers are equal, so the Hilbert polynomial is the product as claimed.
8. The arithmetic genus of the product satisfies\[ \boxed{ p_a(Y\times Z) =p_a(Y)p_a(Z)+(-1)^s p_a(Y)+(-1)^r p_a(Z)}. \]
Put \(A=p_a(Y)\) and \(B=p_a(Z)\). By definition, \[ P_Y(0)=1+(-1)^rA, \qquad P_Z(0)=1+(-1)^sB. \] Step 7 and \(\dim(Y\times Z)=r+s\) give \[ \begin{aligned} p_a(Y\times Z) &=(-1)^{r+s}\bigl(P_Y(0)P_Z(0)-1\bigr)\\ &=(-1)^{r+s}\bigl((-1)^rA+(-1)^sB+(-1)^{r+s}AB\bigr)\\ &=AB+(-1)^sA+(-1)^rB. \end{aligned} \] This proves part (e).
9. Q.E.D.
Step 1 proves part (a), step 4 proves part (b), step 3 proves part (c), steps <1>5–<1>6 prove part (d), and steps <1>7–<1>8 prove part (e).