12.32 Section algebras of cotangent extensions72E247F3A7CA

Proof. Set $s_1=y$ and $s_2=ty+\rho w$ in the finite chart. Since $\rho \ne 0$, these are linear coordinates that can replace $y,w$, with $w=(s_2-ts_1)/\rho $. In the infinite chart put

The transition is $(s_{1,\infty },s_{2,\infty })=t(s_1,s_2)$, so $s_1,s_2$ are the fiber coordinates of $\mathcal O_{\mathbb P^1}(-1)^{\oplus 2}$, while $u$ is the coordinate of a trivial line bundle.

By holomorphy in the infinite chart, a polynomial that is homogeneous of degree $j$ in $s_1,s_2$ has coefficients of degree at most $j$ in $t$. Distributing the exponent of $t$ among the $j$ fiber factors, every admissible monomial is a product of $s_1,s_2,ts_1,ts_2$. These four generators satisfy the determinant relation

and there are no further relations. Indeed, counting the $j+1$ monomials in $s_1,s_2$ and the $j+1$ admissible powers of $t$, the space of functions of degree $j$ has dimension $(j+1)^2$. The polynomial algebra in four variables modulo this nonzero quadratic relation has the same dimension

in degree $j$, where the second term is zero for $j<2$. The natural surjective graded map is therefore an isomorphism in every degree. Including $u$ adjoins a polynomial variable. The argument is uniform in $z$, and all coefficient functions are entire.

Identify the four generators as $c_-=s_1$, $h=ts_1$, $h+\rho w=s_2$ and $c_+=ts_2$; the determinant relation becomes the stated quadratic relation. These functions are polynomial also in the infinite chart, where

The lattice formulas follow by substitution in the finite chart. $\square$

Proof. Step 1: weights. Give $c_-$ weight $-1$, $c_+$ weight $1$, and $h,w,u$ weight zero. These weights come from the ruling action and are compatible with the quadratic relation and the lattice action. Using the relation to eliminate products $c_-c_+$, a polynomial of weight $-k$, $k\geq 1$, can be written as $c_-^ kF(z,h,w,u)$, and a polynomial of weight $k$ as $c_+^ kF(z,h,w,u)$, where $F$ has degree $r-k$ in tensor degree $r$; weight zero consists of the polynomials $F(z,h,w,u)$. These expressions are unique. Indeed, the substitution $c_+\mapsto h(h+\rho w)/c_-$ embeds the algebra of lemma 12.32.1 into

the reduced expressions involve distinct powers of $c_-$, and their coefficients vanish only if the original coefficients vanish, because $h(h+\rho w)$ is a nonzero polynomial. This proves that the weight decomposition is a direct sum.

Step 2: coefficient bundles. Let $U_3$ be the rank-three bundle on $E$ of linear polynomials in $h,w,u$ with the transformation rule

By the transformation of $c_-$ and $c_+$ in lemma 12.32.1, the block of weight $-k$ is $H^0(E,A^ k\otimes \operatorname {Sym}^{r-k}U_3)$, and the block of weight $k$ is $H^0(E,A^{-k}\otimes \operatorname {Sym}^{r-k}U_3)$; the inverse translation of $u$ appears when invariance is written out. The signs of the twists agree with those in lemma 12.31.1.

Step 3: reduction to the tangent bundle. Since $q_0\ne 0$, put $H=h+(K/q_0)w$. Then the translation of $u$ is $u\mapsto u-mq_0H$, and consequently

where the variable $h$ of the bundle $U$ defined above is replaced by $H$. The symmetric powers of $U_3$ are the direct sums of the bundles $w^ j\operatorname {Sym}^{s-j}U$, $0\leq j\leq s$. The vanishing for negative twists in lemma 12.31.2 eliminates every positive weight, and the same lemma identifies the algebra of sections of weight zero with $\mathbb C[H,w]$.

In each negative weight, expanding in powers of $w$ identifies the remaining spaces with the corresponding blocks for the tangent bundle in lemma 12.31.1, with a polynomial variable $w$ adjoined. Sending the Euler variable $h$ of those blocks to $H$ and their leading factor $y^ k$ to $c_-^ k$ gives bijections in every tensor degree and weight. Products of surviving blocks multiply their coefficients and add their powers of $c_-$, and the eliminated positive weights contribute no products. The bijections therefore preserve multiplication and give the asserted isomorphism of graded algebras. $\square$

Proof. The isomorphism of theorem 12.32.2 and theorem 12.31.3 give

where $\binom {2}{3}=0$. The surface $M$ is projective, and by coherent GAGA the holomorphic bundle $V$ is algebraic. Thus $\mathbb P(V)$ is a projective fourfold, and the volume of its tautological line bundle is the limit of $24h^0(M,\operatorname {Sym}^ rV)/r^4$, which equals $d>0$. Hence this line bundle is big. If the section algebra of $V$ were finitely generated, so would be its quotient by $w$, contradicting theorem 12.31.4.

For the canonical extension $0\to \mathcal O_ M\to V\to T_ M\to 0$ of a Kähler class, the dual extension of $\mathcal O_ M$ by $\Omega ^1_ M$ has nonzero ruling component, as shown in the proof of theorem 12.30.4. Its class therefore has $\rho \ne 0$ in the conventions used here, and the same argument applies. $\square$