Lemma 12.19.1040DB573E7C1E

Proof. On the parameter supports, $|a-bu|\geq 3/4-2/16=5/8$. Hence the matrix $\left(\begin{smallmatrix}1& u\\ b& a\end{smallmatrix}\right)$ is uniformly invertible, and the projective line (12.19.9) lies in $\Omega $. Distinct deck translates of this line are disjoint: equality at a point fixes $[z_0:z_1]$, and invertibility of the matrix then forces the deck scalar to be one. Thus each map is an embedding onto a compact rational curve in $X$.

The evaluation map from $\mathbb P^1$ times the open parameter domain to $X$ is a holomorphic submersion. Indeed, after fixing a projective gauge, variations of $u$ and $a$ span the two $w$-coordinates at points with $z_1\ne 0$, variations of $\tau $ and $b$ span them at points with $z_1=0$, and the coordinate on $\mathbb P^1$ supplies the remaining direction. For $\varepsilon >0$, the average is the integral along the fibers of this submersion of a smooth compactly supported parameter density, which proves smoothness. As an average of closed positive integration currents, $T_{j,\varepsilon }$ is closed and semipositive.

All curve parameters lie in one compact subset of the domain where the matrix is invertible, so the $\omega $-areas of the curves are uniformly bounded. The weights in $\tau $ and $u$ have mass one, and the weight in $(a,b)$ has a fixed finite mass. This proves the degree bound.

When $u=0$, the curve is given in the product chart by $\zeta =\tau $ and $y=ax+b$. Integrating in $b$ and $\tau $ yields (12.19.11); the $(1,1)$-factor there is the integration current of the graph of $y=ax+b$. On $\{ |x|,|y|\leq 1/4\} $, the density $f(a,y-ax)$ is bounded below by a positive constant for $a$ in a fixed small disk about $1$. The covectors $\mathrm{d}y-a\, \mathrm{d}x$, for $a$ in that disk, have no common nonzero annihilator, so integrating their squared absolute values and using compactness of the unit sphere in the $(x,y)$-directions gives the asserted lower bound.

For general $u$, the curve is given by

On a fixed compact subset of the chart, $1+ux$ stays away from zero for all sufficiently small $u$. Solving for the parameters gives $\tau =\zeta -\log (1+ux)$ and $b=(1+ux)y-ax$. Regarding $\rho _ j$ as a smooth function on $E$, this change of variables also applies near the boundary of its support. The Jacobians, the density coefficients and any fixed number of their derivatives depend smoothly on $u$. Averaging against the approximate identity in $u$ therefore gives local $C^\infty $ convergence. This convergence is needed only for each fixed $j$; it is not uniform in $j$. $\square$