12.30 Cotangent torsors on split elliptic ruled surfacesE4B698D2AEE3

Proof. Consider the vector bundle obtained from $\mathbb C_ z\times \mathbb C_ a\times \mathbb C_ b$ by the action

The coordinate $b$ gives a map to $\mathcal O_ E$, which defines an extension of $\mathcal O_ E$ by $\Omega _ E^1$, the kernel coordinate $a$ corresponding to $a\vartheta _ E$. The local lifts $s_ i=(z_ i,0,1)$ satisfy $s_ j-s_ i=m_{ij}\vartheta _ E$, as one sees by expressing $s_ i$ in the coordinates over $z_ j$. Thus the extension has class $\eta $.

If it split, the pullback of a holomorphic lift of $1$ to $\mathbb C$ would have the form $(z,g(z),1)$ with $g$ entire and

The derivative $g'$ would be lattice periodic, hence constant by the maximum principle on the compact torus. The first identity then forces this constant to be zero, contradicting the second. Hence $\eta \ne 0$. Serre duality on the compact connected curve gives $H^1(E,\Omega _ E^1)\simeq H^0(E,\mathcal O_ E)^*\simeq \mathbb C$, which proves the spanning assertion. $\square$

Proof. Let $A_ d$ be the $d$th tensor power of the degree-one bundle of example 12.29.2. Its factors are the stated exponential factors, and its degree is $d$. The bundle $F=A\otimes A_ d^*$ has degree zero. Start with a smooth Hermitian metric on $F$; its Chern curvature has integral zero. Solving the scalar Poisson equation on the flat compact torus gives a smooth real function whose addition to the weight of the metric makes the curvature vanish. The resulting Chern connection is flat and unitary.

On the simply connected cover $\mathbb C$, parallel transport gives a global parallel frame of unit length of the pullback of $F$. It is holomorphic, because the $(0,1)$-part of the Chern connection is the $\bar\partial $-operator of $F$. Each lattice element changes this frame by a constant unitary factor, and these factors form a character. Tensoring with the given frame of $A_ d$ gives factors of the stated form for $A$, and differentiation gives the last formula. $\square$

Proof. The relative cotangent sequence is

The two affine charts of a projective line give the direct images

For the last identity, a Laurent cocycle $g(z,t)\, \mathrm{d}t$ can be reduced, by subtracting forms holomorphic on the two charts, to its coefficient of $\mathrm{d}t/t$. Its residue is unchanged by $t_ i=b_{ij}t_ j$, with differentials taken relative to the base, so the local generators glue to the trivial line bundle. For functions, the two halves of a Laurent series exhaust the cocycle, which gives $R^1f_*\mathcal O_ M=0$. The direct images in degree zero follow because $\mathbb P^1$ has no nonconstant holomorphic functions and no nonzero holomorphic one-forms.

The Leray spectral sequences and the fact that the base is a curve now give

together with $H^0(M,\Omega ^1_{M/E})=0$. The cohomology sequence of the relative cotangent sequence is therefore the exact sequence

On a fiber, the class $\xi $ is represented on the two affine charts by $\mathrm{d}\log v=-\mathrm{d}t/t$, whose residue is nonzero, so $\xi $ splits the sequence. The pulled-back class $f^*\eta $ restricts to zero on every fiber. This proves both assertions. $\square$

Proof. Step 1: a model torsor. Over $\mathbb C_ z\times \mathbb P^1$, glue two affine cotangent charts by

Here $(u,y)$ are the coefficients of the covector $u\, \mathrm{d}z+y\, \mathrm{d}t$, and the linear part of the gluing is the change of these coefficients under $v=1/t$. If $s_0$ and $s_\infty $ are the zero sections of the two charts, the affine term says that

For $\lambda =m\tau +n$, give the first chart the lattice action

Its linear part is the transport of covectors under $(z,t)\mapsto (z+\lambda ,b_\lambda (z)t)$, and the cocycle identity for the factors and the invariance of $ty$ give the group law. In the other chart the action becomes

which is holomorphic also at $v=0$, and the two formulas agree on the overlap. Translation in $z$ makes the action free and properly discontinuous.

Step 2: its class. We compute the class of the quotient torsor using the local lifts $z_ i$. On the two types of overlaps over the base, the zero sections satisfy

For a mixed overlap, these formulas and $s_{j,\infty }-s_{j,0}=\rho \, \mathrm{d}v_ j/v_ j$ give

These are exactly the cocycles defining $Kf^*\eta +\rho \xi $. Two extensions with the same class have splitting cocycles that differ by a coboundary, and changing their local lifts by that coboundary identifies their inverse images of $1$. Thus the quotient torsor is $T_ W$.

Step 3: identification with the quotient. Since $\rho \ne 0$, set $Y=y/\rho $ and $Y_\infty =y_\infty /\rho $. Their transition is $Y_\infty =-t^2Y-t$. The maps

agree on overlaps and identify the glued charts with $Q$. In the first chart $\sigma =tY=ty/\rho $. Substituting $b_\lambda '/b_\lambda =-2\pi \mathrm{i}dm$ turns the lattice action into the action of proposition 12.29.3 with $\kappa =2\pi \mathrm{i}d\rho $. That proposition gives the assertions on the Stein property.

Step 4: Kähler classes. The restriction of a Kähler class to a ruling fiber is nonzero, because its integral over the fiber is positive. 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 the negative of that class, which still has nonzero restriction to the fibers; by lemma 12.30.3 its ruling component $\rho $ is nonzero. Finally, we use the quotient convention for projectivization: $\mathbb P(V)$ parametrizes the one-dimensional quotients of the fibers of $V$, that is, the lines in $V^*$. The complement $\mathbb P(V)\setminus \mathbb P(T_ M)$ consists of the lines in $V^*$ not contained in $\Omega ^1_ M$, and it identifies with $p_ W^{-1}(1)\subset V^*$ by normalizing a nonzero functional to take the value one on the distinguished copy of $\mathcal O_ M$. This proves the final assertion. $\square$