12.31 Symmetric tensors on split elliptic ruled surfaces7CF6C47CAA92

Proof. On the cover $\mathbb C\times \mathbb P^1$, the cotangent coordinates of the two ruling charts are related by

and the lattice acts by

both formulas follow by preserving the one-form $u\, \mathrm{d}z+y\, \mathrm{d}t$. In the other ruling chart the action is

which is holomorphic at $v=0$ and agrees with the formula in the finite chart. Both formulas preserve $h=ty=-v y_\infty $.

A function that is homogeneous of degree $r$ on the fibers is a sum of terms $g_ b(z,t)y^ bu^{r-b}$, $0\leq b\leq r$, with $g_ b$ holomorphic on $\mathbb C^2$. Since $y=-v^2y_\infty $, holomorphy at $v=0$ says that $v^{2b}g_ b(z,1/v)$ is holomorphic there; there is no cancellation between different $b$, because the monomials in $u_\infty ,y_\infty $ are linearly independent. Expanding in $t$ therefore gives

with entire coefficients.

Give $t$ weight $1$, $y$ weight $-1$ and $u$ weight $0$, so that $t^ ay^ bu^{r-b}$ has weight $\ell =a-b$. The chart changes and the lattice action preserve this grading; equivalently, it is the grading by the cotangent lift of the global action $t\mapsto ct$, $c\in \mathbb C^*$. The bounds $0\leq a\leq 2b$ and $b\leq r$ give $-r\leq \ell \leq r$.

For $\ell =-k<0$, a homogeneous function of weight $\ell $ can be written uniquely as $y^ kF(z,h,u)$ with $F$ homogeneous of degree $r-k$. Its invariance is equivalent to

which is the transformation rule for sections of $A^ k\otimes \operatorname {Sym}^{r-k}U$. Conversely, each such polynomial defines a function that is holomorphic in both ruling charts, because its exponents satisfy $0\leq a\leq 2b$. For $\ell =k>0$, the condition $a\leq 2b$ shows that a function of weight $\ell $ can be written uniquely as $t^ kh^ kF(z,h,u)$, with $F$ of the same remaining degree $r-k$. Its invariance involves the factor $b_\lambda ^{-k}$, which gives $A^{-k}\otimes \operatorname {Sym}^{r-k}U$. Weight zero consists exactly of the homogeneous polynomials $F(z,h,u)$ with the transformation rule of $\operatorname {Sym}^ rU$. Products of polynomial functions add degrees and weights, which proves the last assertion. $\square$

Proof. A splitting of $U$ would give an invariant section $u+a(z)h$ lifting $1$, with $a$ entire and

The derivative $a'$ would then be lattice periodic, hence constant on the compact torus. The first identity forces this constant to vanish, contradicting the second. Thus the extension does not split.

More generally, write an invariant polynomial of degree $s$ as

If its largest nonzero $u$-degree is $j>0$, then $a_ j$ is lattice periodic, hence a nonzero constant. Comparing the next coefficients gives

and the same derivative argument gives a contradiction. Hence $F=a_0h^ s$ with $a_0$ constant; this also covers $s=0$.

The powers of the invariant line $\mathcal O_ Eh$ give a filtration of $\operatorname {Sym}^ sU$ whose $s+1$ successive quotients are trivial line bundles. A line bundle of negative degree on a compact curve has no nonzero section, since the zero divisor of a nonzero section has nonnegative degree. Hence the filtration twisted by $A^{-k}$ has no global sections. Serre duality and $\Omega _ E^1\simeq \mathcal O_ E$ give $H^1(E,A^ k)=0$, and the cohomology sequences of the filtration twisted by $A^ k$ then give $H^1(E,A^ k\otimes \operatorname {Sym}^ sU)=0$. Riemann–Roch on the curve of genus one, for a bundle of rank $s+1$ and degree $kd(s+1)$, gives the dimension formula.

Finally, the leading coefficient gives the exact sequence of bundles

Indeed, translating $u$ by a multiple of $h$ does not change the leading coefficient, and a polynomial with zero leading coefficient is divisible by $h$. The vanishing of $H^1(E,A\otimes \operatorname {Sym}^{s-1}U)$ just proved gives the asserted surjectivity and kernel on global sections. $\square$

Proof. By lemma 12.31.1, lemma 12.31.2, the positive weights contribute nothing, the weight zero contributes one dimension, and the weight $-k$ contributes $kd(r-k+1)$ dimensions. Thus

The elliptic curve and its holomorphic line bundle $A$ are algebraic, so $M$ and $\mathbb P(T_ M)$ are projective. The identity $\pi _*\mathcal O_{\mathbb P(T_ M)}(r)=\operatorname {Sym}^ rT_ M$ identifies the section spaces. Since $\mathbb P(T_ M)$ has dimension three,

so the tautological line bundle is big. $\square$

Proof. Write $R_{r,k}$ for the component of tensor degree $r$ and weight $-k$. By the preceding results only $k\geq 0$ occurs, multiplication adds both indices, and the subalgebra of weight zero is $\mathbb C[h]$. Let $I$ be the ideal formed by all components with $k>0$, and put

Its component of weight zero is $\mathbb C$, and its ideal $J$ of positive weight index satisfies $J^2=0$, so $S=\mathbb C\oplus J$ as a vector space.

For every $r\geq 1$, products in $I^2$ have weight index at least two. Consequently the component of $S$ of degree $r$ and weight index one is

For $r=1$ this holds because $R_{0,1}=0$, and for $r\geq 2$ it is the last assertion of lemma 12.31.2 with $s=r-1$. Each of these components has dimension $d>0$, and they lie in distinct degrees, so $J$ is infinite dimensional.

If $R$ were finitely generated, its quotient $S$ would be finitely generated too. Write a finite set of algebra generators of $S$ as $c_ i+j_ i$, $1\leq i\leq N$, with $c_ i\in \mathbb C$ and $j_ i\in J$. Because $J^2=0$, every polynomial in these generators lies in the finite-dimensional span of $1,j_1,\ldots ,j_ N$. This contradicts the infinite dimension of $J$. $\square$