Showing posts with label hypersurface. Show all posts
Showing posts with label hypersurface. Show all posts

Thursday, March 31, 2016

The Hodge decomposition, diamond, and Euler characteristics

 Seminar topic

Recall the sheaf of $r$-differential forms $\Omega^r_X$ on $X$ (with $\Omega^r_X(U) = \{fdx_{i_1}\wedge \cdots \wedge dx_{i_r}\ :\ f\ $is well-defined on $U\}$ and such sums) and the structure sheaf $\mathcal O_X$ on $X$ (with $\mathcal O_X(U) = \{f/g\ :\ f,g\in k[U],\ g\neq 0\ $on$\ U\}$). Then we may consider the sheaf cohomology of $X$, with values in $\Omega^r_X$ or $\mathcal O_X$.

Definition: Let $X$ be a smooth manifold of dimension $n$. The $(p,q)$th Hodge number is $h^{p,q}=\dim(H^{p,q})$, where $H^{p,q} = H^q(X,\Omega^p_X)$. These numbers are arranged in a Hodge diamond as below.



The Hodge diamond has a lot of repetition - by complex conjugation, we get that $h^{p,q}=h^{q,p}$, so it is symmetric about its vertical axis. By the Hard Lefschetz theorem (or the Hodge star operator, or Poincare duality), we get that $h^{p,q}=h^{n-q,n-p}$, so it is symmetric about its horizontal axis.

Proposition: Let $X$ be a Kähler manifold (note that all smooth projective varieties are Kähler) of dimension $n$. Then the cohomology groups of $X$ decompose as
\[
H^k(X,\C) = \bigoplus_{p+q=k}H^{p,q}(X),
\]
for all $0\leqslant k\leqslant 2n$. This is called the Hodge decomposition of $X$.

This decomposition immediately gives all the Hodge numbers for $\P^n$, knowing its cohomology. For a manifold of complex dimension $n$, there are several numbers and polynomials that may be defined. These are:
\begin{align*}
\chi_{top}(X) & = \sum_{i=1}^{2n}(-1)^i \dim(H^i(X,\C)) & \text{the (topological) Euler characteristic} \\
\chi^p(X) & = \sum_{q=0}^{n-1}(-1)^qh^{p,q} & \text{the chi-$p$ characteristic} \\
\chi_y(X) & = \sum_{p=0}^{n-1}\chi^py^p & \text{the chi-$y$ characteristic}
\end{align*}
Note the Euler characteristic is the alternating sum of the rows of the Hodge diamond, and the chi-$p$ characteristic is the alternating sum of the left-right diagonals of the diamond.

Example: In the case $X$ is a hypersurface in projective $n$-space $\P^n$ defined by a degree $d$ polynomial,
\[
\chi_y = [z^n]\frac{1}{(1+zy)(1-z)^2}\cdot\frac{(1+zy)^d-(1-z)^d}{(1+zy)^d+y(1-z)^d}.
\]
Since every row except the middle row of the Hodge diamond of a hypersurface is known (as it comes from the Hodge diamond of $\P^n$ by the Lefschetz hyperplane theorem), this expression gives all the unknown numbers. This particular formula is a simplification  of Theorem 22.1.1 in Hirzebruch, which itself comes from the Riemann--Roch theorem.

References: Huybrechts (Complex Geometry: An Introduction, Chapters 3.2, 3.3), Hirzebruch (Topological Methods in Algebraic Geometry, Appendix 1, Section 22)

Tuesday, March 1, 2016

The canonical bundle of projective space and hypersurfaces

Let $\P^n$ be projective $n$-space with coordinates $[x_0:\cdots:x_n]$. Cover $\P^n$ with affine pieces $U_i = \{x_i\neq 0\}$, each of which are $\A^n$, in coordinates $(y_1,\dots,y_n)$, where $y_j = x_j/x_i$. Recall that the canonical bundle of $\P^n$ is the $n$-fold wedge of the cotangent bundle of $\P^n$, or $\omega_{\P^n} = \bigwedge^nT^*_{\P^n}$. The canonical bundle for an arbitrary variety is defined analogously.

Definition: Let $X$ be a projective $n$-dimensional variety. The sheaf of regular functions on $X$ is $\mathcal O_X$, with $\mathcal O_X(U)=\{f/g\ :\ f,g\in k[x_1,\dots,x_n]/I(X), g\neq 0\}$, and the restriction maps are function restriction.

There is a natural grading on $\mathcal O_X$, given by $\deg(f)-\deg(g)$. A shift in the grading may be applied, called a {\it Serre twist}, to get a differently graded (but isomorphic) module: for $\varphi\in \mathcal O_X$ with $\deg(\varphi)=k$, set $\varphi\in\mathcal O_X(\ell)$ to have $\deg(\varphi) = k-\ell$.

Let $\alpha = dy_1\wedge\cdots\wedge dy_n\in \omega_{\P^n}$, which is well-defined on all of $U_i$. We claim this is well-defined on all of $\P^n$. We check this on the overlap $U_0\cap U_n$ (for nicer notation), but the approach is analogous for $U_i\cap U_j$.
\begin{align*}
U_0 & = \{(y_1,\dots,y_n)\ :\ y_i = x_i/x_0\} & y_i & = \frac{z_{i+1}}{z_i} & dy_i & = \frac{z_1dz_{i+1}-z_{i+1}dz_1}{z_1^2} \\
U_n & = \{(z_1,\dots,z_n)\ :\ z_i = x_{i-1}/x_n\} & y_n & = \frac1{z_1} & dy_n & = \frac{-dz_1}{z_1^2}
\end{align*}
Therefore
\begin{align*}
\alpha & = dy_1\wedge\cdots\wedge dy_n \\
& = \frac{z_1dz_2-z_2dz_1}{z_1^2}\wedge\cdots\wedge \frac{z_1dz_n-z_ndz_1}{z_1^2}\wedge \frac{-dz_1}{z_1^2} \\
& = \frac{dz_2}{z_1}\wedge\cdots\wedge \frac{dz_n}{z_1}\wedge \frac{-dz_1}{z_1^2} \\
& = \frac{(-1)^n}{z_1^{n+1}}dz_1\wedge\cdots \wedge dz_n.
\end{align*}
Since the transition function has a pole of order $n+1$ when $z_1 = 0$, which happens when $x_0=0$, we have that $\alpha$ has a pole of order $n+1$ at $\infty$. Therefore $\omega_{\P^n} \cong \mathcal O_{\P^n}(-n-1)$.

Let $X\subset \P^n$ be a smooth hypersurface defined by a degree $d$ equation $F(x_0,\dots,x_n)=0$. On the affine piece $U_0$ this becomes $f(y_1,\dots,y_n)=F(1,\frac{x_1}{x_0},\dots,\frac{x_n}{x_0})$ with $y_i = x_i/x_0$. The total derivative is
\[
\frac{\dy f}{\dy y_1} dy_1 + \cdots + \frac{\dy f}{\dy y_n} dy_n = \sum_{i=1}^n\frac{\dy f}{\dy y_i}dy_i = 0,
\]
and since $X$ is smooth, the terms never all vanish at the same time. Let $V_i=\{\frac{\dy f}{\dy y_i} \neq 0\}$, and set
\[
\beta_i = \frac{(-1)^{i-1}}{\dy f/\dy y_i} dy_1\wedge\cdots \wedge \widehat{d y_i}\wedge \cdots \wedge d y_n \in \omega_X,
\]
which is well-defined on all of $V_i\subset U_0$. We claim that the choice of $V_i$ does not matter, and indeed, assuming $i<j$,
\begin{align*}
\beta_j & = \frac{(-1)^{j-1}}{\dy f/\dy y_j} dy_1\wedge\cdots \wedge \widehat{d y_j}\wedge \cdots \wedge d y_n \\
& = \frac{(-1)^{j-1+i-1}dy_i}{\dy f/\dy y_j} \wedge dy_1\wedge\cdots \wedge \widehat{d y_i}\wedge \cdots \wedge \widehat{d y_j}\wedge \cdots \wedge d y_n \\
& = \frac{(-1)^{j-1+i-1}\frac{-1}{\dy f/\dy y_i}\left(\frac{\dy f}{\dy y_1}dy_1+\cdots + \widehat{\frac{\dy f}{\dy y_i}dy_i} + \cdots + \frac{\dy f}{\dy y_n}dy_n\right)}{\dy f/\dy y_j} \wedge dy_1\wedge\cdots \wedge \widehat{d y_i}\wedge \cdots \wedge \widehat{d y_j}\wedge \cdots \wedge d y_n \\
& = \frac{(-1)^{j-1+i-1+1}\frac{1}{\dy f/\dy y_i}\cdot \frac{\dy f}{\dy y_j}dy_j}{\dy f/\dy y_j} \wedge dy_1\wedge\cdots \wedge \widehat{d y_i}\wedge \cdots \wedge \widehat{d y_j}\wedge \cdots \wedge d y_n \\
& = \frac{(-1)^{j-1+i-1+1+j-2}}{\dy f/\dy y_i} dy_1\wedge\cdots \wedge \widehat{d y_i}\wedge \cdots \wedge d y_n \\
& = \frac{(-1)^{i-1}}{\dy f/\dy y_i} dy_1\wedge\cdots \wedge \widehat{d y_i}\wedge \cdots \wedge d y_n \\
& = \beta_i.
\end{align*}
Hence $\beta_i$ is well-defined on all of $U_0$, and we call it simply $\beta$. Next we claim it is well-defined on all of $X$. Again we only check on the overlap of $U_0\cap U_n$. On the affine piece $U_n$ this becomes $g(z_1,\dots,z_n)=F(\frac{x_0}{x_n},\dots,\frac{x_{n-1}}{x_n},1)=f(\frac{z_2}{z_1},\dots,\frac{z_n}{z_1},\frac1{z_1})$ with $z_i = x_{i-1}/x_n$. We employ the chain rule $\frac{\dy f}{\dy y_i}=\frac{\dy f}{\dy z_j}\frac{\dy z_j}{\dy y_i}$ and the results above to find that
\begin{align*}
\beta & = \frac{(-1)^{i-1}}{\dy f/\dy y_i} dy_1\wedge\cdots \wedge \widehat{d y_i}\wedge \cdots \wedge d y_n \\
& = \frac{(-1)^{i-1}}{\dy f/\dy z_j \cdot \dy z_j/\dy y_i} \frac{z_1dz_2-z_2dz_1}{z_1^2}\wedge \cdots \wedge \widehat{dy_i}\wedge \cdots \wedge \frac{z_1dz_n-z_ndz_1}{z_1^2}\wedge \frac{-dz_1}{z_1^2} \\
& = \frac{(-1)^{i-1}}{\dy f/\dy z_j \cdot \dy z_j/\dy y_i} \frac{(-1)^{n-1}}{z_1^n}dz_1\wedge\cdots \wedge \widehat{dz_i}\wedge \cdots \wedge dz_n \\
& = \frac{(-1)^{i+n}}{\left(\frac{1}{z_1}\right)^{d-1}\left(c+\cdots\right) z_1^n}dz_1\wedge\cdots \wedge \widehat{dz_i}\wedge \cdots \wedge dz_n \\
& = \frac{(-1)^{i+n}}{z_1^{n-d+1} \left(c+\cdots \right)}dz_1\wedge\cdots \wedge \widehat{dz_i}\wedge \cdots \wedge dz_n,
\end{align*}
for some constant $c$. This comes from expressing $f$ in terms of the $z_i$s and factoring. Since the transition function has a pole of order $n-d+1$ when $z_1 = 0$, which happens when $x_0=0$, we have that $\beta$ has a pole of order $n-d+1$ at $\infty$. Therefore $\omega_{X} \cong \mathcal O_{X}(-n+d-1)$.

References: Griffiths and Harris (Principles of Algebraic Geometry, Chapter 1.2)