Proof.
For the square lattice define
\[ x(z)=z^{-2}+\sum _{\substack {\lambda \in \mathbb Z+\mathrm i\mathbb Z\\ \lambda \ne 0}} \bigl ((z-\lambda )^{-2}-\lambda ^{-2}\bigr ), \qquad y=x'. \]
The summands are $O(|\lambda |^{-3})$ on compact sets away from the lattice, so the series converges normally. Pairing opposite lattice points shows that $x$ is even. The differentiated series is absolutely and normally convergent, and reindexing shows that $y$ is periodic. Thus $x(z+\omega )-x(z)$ is constant for $\omega =1,\mathrm i$; evaluating at $-\omega /2$ and using evenness shows that this constant is zero. Both functions are meromorphic on $E$, with their only poles at zero and expansions
\[ x=z^{-2}+a_2z^2+a_4z^4+\cdots , \qquad y=-2z^{-3}+2a_2z+4a_4z^3+\cdots . \]
Cancelling the Laurent expansions leaves an elliptic function with no poles and zero constant term, which vanishes by the maximum principle. This gives
E7FF7E63C71E
9.11.58
\begin{equation} y^2=4x^3-g_2x-g_3, \qquad g_2=20a_2,\quad g_3=28a_4. \end{equation}
The argument principle on a translated fundamental parallelogram shows that $x-c$ has two zeros, counted with multiplicity, for every finite $c$. Its fibers are therefore of the form $\{ z,-z\} $. If $y(z)=0$, the local multiplicity is at least two. Distinct points $z,-z$ would then give at least four zeros, so $z=-z$ on $E$, the multiplicity is exactly two, and $x''(z)\ne 0$. Since $y$ is odd, the map $(x,y)$ is injective and immersive away from zero: use $dx$ where $y\ne 0$, and $dy=x''dz$ otherwise.
Define
\[ \pi (z)=[1:x:y:x^2:x^3],\qquad \pi (0)=[0:0:0:0:1]. \]
In the last affine chart its holomorphic coordinates are
\[ [s^3:s^2:t:s:1],\qquad s=1/x=z^2+O(z^6),\qquad t=y/x^3=-2z^3+O(z^7). \]
Equation (9.11.58) becomes
\[ t^2=s^3(4-g_2s^2-g_3s^3). \]
Dividing $t$ by an analytic square root of the nonvanishing factor gives the ordinary cusp $t'^2=s^3$. Choose the local parameter $\zeta $ so that $s=\zeta ^2$ and $t'=\zeta ^3$. The local ring of the cusp is $\mathbb C\{ \zeta ^2,\zeta ^3\} $, with normalization $\mathbb C\{ \zeta \} $. To see this directly, note that division by $t'^2-s^3$ writes every germ as $A(s)+t'B(s)$, and the even and odd parts identify the image ring. Its fraction field contains $\zeta $, which is integral since $\zeta ^2=s$, and the larger ring is a finite module generated by $1$ and $\zeta $. An integral meromorphic germ cannot have negative order: in a monic equation, its highest power would have strictly smallest order. Hence $\mathbb C\{ \zeta \} $ is the full integral closure.
The image is the reduced projective locus
\[ \begin{aligned} Z_1^2& =Z_0Z_3,& Z_1Z_3& =Z_0Z_4,\\ Z_3^2& =Z_1Z_4,& Z_2^2& =4Z_0Z_4-g_2Z_0Z_1-g_3Z_0^2. \end{aligned} \]
On $Z_0\ne 0$ these equations describe the graph of the affine cubic, and since the fibers of $x$ have degree two, both square-root values of $y$ occur. On $Z_0=0$ they force $Z_1=Z_3=Z_2=0$, leaving only $\pi (0)$. In the last chart they give exactly the cusp displayed above. The finite local module at the cusp and the biholomorphic charts elsewhere show that $\pi $ is finite. Hence $\pi $ is the bijective normalization of a connected curve that is smooth away from this one cusp.
For the restricted Fubini–Study form, $\beta =b\beta _0$ is positive away from zero and
\[ C_1|z|^2\leq b(z)\leq C_2|z|^2 \]
near zero. Indeed, the derivatives of the coordinates in the last chart vanish at zero, and $ds=2z\, dz+O(z^5)dz$ gives the leading term. On the complement of zero, the Fubini–Study potential is
\[ L=\log (1+|x|^2+|y|^2+|x|^4+|x|^6) =-6\log |z|^2+\text{a smooth function near }0. \]
Globally $\mathrm{dd}^{\mathrm{c}}L=\beta -6\delta _0$, so integrating gives $H=6$. Since the normalization has generic degree one, the degree two of $x$ contributes no factor to the measure.
For the pole of weight one, the reference and the continuous correction are
\[ \Psi =G*((6-b)\beta _0)+G, \qquad v=-\frac5{2\pi }\cos (2\pi \operatorname {Re}z). \]
Direct differentiation gives $\beta +\mathrm{dd}^{\mathrm{c}}\Psi =5\beta _0+\delta _0$ and $\mathrm{dd}^{\mathrm{c}}v=(5f-5)\beta _0$. Since $F=1$, theorem 9.11.54 gives $c=5$, the mass, and uniqueness in the whole type. The residual is the descent of $v-\sup _ E(\Psi +v)$. Finally, $f/b$ is comparable to $|z|^{-2}$ near zero, and the integral of its $r$th power against $\beta =b\beta _0$ is comparable to
\[ \int _0^\epsilon \rho ^{3-2r}\, d\rho . \]
This is finite exactly when $r<2$, with logarithmic divergence at $r=2$.
$\square$
Pome
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.